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Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permission Department, John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.
Chapter 1
11 (a) 98 Btu/(hrftF) x1.7307 = 170 W/(mK)
(b) 0.24 Btu/(IbmF) x4186.8=1.0 kJ/kgK
(c) 0.04 Ibm/(fthr)3600 sec/hr
x1.488 = 16.5 2Ns
mµ
(d) 1050 BtuIbm
x 41
9.48x10− JBtu
x 2.20462 Ibmkg
= 2.44 MJkg
(e) 12,000 BtuIbm
x 13.412
= 3.52 kW
(f) 14.7 2Ibfin
x 6894.76 = 101 kPa
12 (a) 120 kPa x 2lbf / in
6.89476kPa = 17.4 lbf/in2
(b) 100 Wm K−
x 0.5778 = 57.8 Btu/hrftF
(c) 0.8 2W
m K− x 0.1761 = 0.14 Btu/hrft2F
(d) 106 Ns/m2 x 11.488
= 6.7 x 107 lbmft sec−
(e) 1200 kW x 3412 = 4.1 x 106 Btu/hr
2
(f) 1000 kJkg
x 1 Btu1.055 kJ
x 1 kg2.2046 lbm
= 430 Btulbm
13 Hp = 50 (ft) x 0.3048 ( mft
) = 15.2 m
∆P = 15.2 m1000 Pa/kPa
x 9.8071
( Nkg
) x 1000 (kg/m3) = 149 kPa
14 P = ∆4
12 (ft) x 0.3048 ( m
ft) x 9.807
1 ( N
kg) x 1000 ( 3
kgm
)
∆P = 996 Pa 1.0 kPa ≈ 15 TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE
+ METER CHARGE
( ) ( ) ( ) ( )96,000 kw  hrs 0.045 $ /kw hr + 624 kw 11 50 $ /kw− − + $68 = $4,320 + $7,176 + $68 = $11,564
16 7 AM to 6 PM 11 hrs/day, 5 days/wk
hrs days(11) (22) 242 hrs /monthday months
=
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
3
( )
( )( )
624 kwratio = 1.57
96,000 kw hr242 hr
=⎛ ⎞−⎜ ⎟⎝ ⎠
17 This is a trial and error solution since eq. 11 cannot be solved
explicitly for i.
Answer converges at just over 4.2% using eq. 11
18 Determine present worth of savings using eq. 11
( )( )( )12 120.012$1000 1 1+
12P =
0.01212
P $134,000
−⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦⎛ ⎞⎜ ⎟⎝ ⎠
=
19 (a) Q VA= = 2 x 3.08 x 103 = 6.16 x 103m3/s m 6.16 x 10Q ρ= = 3 x 998 = 6.15 kg/s
(b) A= 4π (0.3)2 = 7.07 x 102 m2
Q 7.07x10= 2 x 4 = 0.283 m3 / s; ρ = 1.255 kq/m3
m = 1.225 x 0.283 = 0.347 kg/s
110 V = 3x10x20 = 600m3
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
4
= 600 x iQ 14
x 13600
= 4.17 x 102 m3/s
111
p p
3
q = mc T c = 4.183 kJ/(kgK)
= 983.2 kg/m ρ
∆
111 (cont’d)
( ) ( ) ( ) ( )
3 c3
m kg kJq = 1 983.2 4.183 5 20,564s kg Km
q = 20,564 kw
=−
kJs
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
112 = watq airq−
11,200(1)(10) =
= 25000x60x14.7x144x0.24(t 50)(53.35x510)
−
11,200 = 5601.5 (t250); t2 = (11,200/5601.5) + 50 = 70 F 113 Diagram as in 112 above. q wat = q air
1.5 (4186)(90t2) = 2.4 (1.225)(1.0)(3020)(1000) 6279(90t2) = 29,400
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Excerpts from this work may be reproduced by instructors for distribution on a not l purposes only to students enrolled in courses for which the textbook has been adopted. this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permissi
5
t2 = 90  29,4006279
= 85.3 C
114 q hA(t= s ) t∞ A= π (1/12) x 10 = 2.618 ft2
st = tsur 212 F ≈ = 10x2.618x(21250) = 4241 Btu/hr q
115 A= π x 0.25x4 = 3.14 16 m2
hA(tq = s ) t∞
h= s
qA(t t )∞
= 12503.1416(100 10)−
; h = 4.42 W/(m2 – C)
116 (tpq mc= 2t1) ; m Q x ρ= ρ = P/RT = 14.7x144/53.35(76+460) ρ = 0.074 lbm/ft3
m = 5000x0.074x60 = 22,208 lbm/hr = 0.24 Btu/lbmF pc q= 22,208x0.24(5876) = 95,939 Btu/hr Negative sign indicates cooling
117 (t1 pm c 3t1) +
forprofit basis for testing or instructionaAny other reproduction or translation of
on of the copyright owner is unlawful.
6
(t2 p2m c 3t2) = 0 = p1c p2c
t3 = 1 1 2 2
1 2
(m t m t )(m m )
++
1 2m Q 1ρ= = 1000x 14.7x14453.35(460 50)+
= 73.5 lbm/min
117 (cont’d)
2 2m Q 2ρ= = 600x 14.7x14453.35(460 50)+
= 46.7 lbm/min
3(73.5x80) (46.7 x 50)t 68.3 F
(73.5 46.7)+
= =+
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
7
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Chapter 2
21 through 220 Solutions are not furnished since many acceptable responses exist
for each problem. It is not expected that the beginning student can handle
these questions easily. The objective is to make the student think about
the complete design problem and the various functions of the system.
These problems are also intended for use in class discussions to enlarge
the text material.
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permission Department, John Wiley & Sons, Inc, 111 River Street, Hoboken, NJ 07030.
Chapter 3 31 (a) Pv = =srPφ 0.45(3.17)kPa = 1.43 kPa or 0.45(0.435) = 0.196 psia Pa = 101 – 1.43 = 99.57 kPa or 14.6960.196 = 14.5 psia
(b) =v
v
Pρ RvT or = = = 3v
v vv
P 1430ρ ; ρ 0.0104 kg/mR T 462.5(297)
or =0.196(144) 0.0006285.78(535)
lbv/ft3
(c) W = 0.6219 (1.43)
(99.57) = 0.00893 kgv/kga
or 0.6219(0.196) 0.00854 lbv/lba
14.5=
32 (a) English Units – t = 80F; P = 14.696 psia; Pv = 0.507 psia Table A1a
W = 0.6219 a
vPP
= 0.6219 (0.507)(14.696 0.507)−
= 0.0222 lbv/lba
i = 0.24t + W(1062.2 + 0.444t) i = 0.24 (80) + 0.0222[1061.2 + 0.444(80)] = 43.55 Btu/lbm
8
v = a
a
R T 53.35(460 80)P (14.696 0.507)144
+=
− = 13.61 ft3/lbm
(b) English Units – 32F, 14.696 psia Pv = 0.089 psia (Table A1) 32 (cont’d)
W = 0.6219(0.089) lbmv 0.00379
(14.696 0.089) lbma=
−
i = 0.24(32) + 0.00379 [1061.2 + 0.444(32)] = 11.76 Btu/lbma
v = 53.35(492)
(14.696 0.089)144− = 12.48 ft3/lbma
32 (a) SI Units – 27C; 101.325 kPa Pv = 3.60 kPa, Table A1b
W = 0.6219 v
a
P 0.6219(3.6) kgv0.0229P (101.325 3.6) kga
= =−
i = 1.0t + W(2501.3 + 1.86t) kJ/kga i = 27 + 0.0229(2501.3 + 1.86 x 27) = 85.43 kJ / kga
v = 3a
a
R T 0.287(300) = =0.88 m /kgaP (101.325  3.6)
(b) SI Units 0.0C; 101.325 kPa Pv = 0.61 kPa, Table A1b
W = 0.6219(0.61) =0.00377 kgv/kga
(101.325  0.61)
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9
i = 0.0 + 0.00377 (2501.3 – 1.86 x 0.0) = 9.42 kJ/kga
v = 30.287(273) 0.778 m /kga(101.325  0.61)
=
33 (a) English Units – 5000 ft elevation, P = 12.24 psia = 24.92 in.Hg t = 80 F; Pv = 0.507 psia (Table A1a)
W = 0.6219 v
a
P 0.6219(0.507) = P (12.24  0.507) = 0.0269 lbv/lba
i = 0.24(80) + 0.0269 [1061.2 + 0.444(80)] = 48.7 Btu/lbma
v = a
a
R T 53.35(540) = P (12.24  0.507) 144
= 17.05 ft3 / lbma
(b) English Units – t = 32 F, Pv = 0.089 psia ( Table A1a)
W = 0.6219(0.089)
(12.24 0.089) − = 0.00456 lbmv/lbma
i = 0.24(32) + 0.00456 [1061.2 + 0.444(32)] =12.58 Btu/lbma
v = 53.35(492)
(12.24 0.089)144− = 15.00 ft3/lbma
33 (a) SI Units 27 C, 1500 m elevation P = 99.436 + 1500(0.01) = 84.436 kPa Pv = 3.60 kPa, Table A1b
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
10
W = 0.6219x3.60 0.0277 kgv/kga(84.436 3.60)
=−
i = 27 + 0.0277 (2501.3 + 1.86 x 27) = 97.68 kJ/kga 33 (cont’d)
v = 30.287x300 1.065 m / kga(84.436  3.60)
=
(b) SI Units – 0.0C; 1500m or 84.436 kPa
Pv = 0.61 kPa; Table A1b
W = 0.6219 x 0.61 0.00453 kgv / kga
(84.436  0.61)=
i = 0.0 + 0.00453 (2501.3 – 0.0 x 1.86) = 11.33 kJ / kga
v = 0.287 x 273
(84.436  0.61) = 0.935 m3 / kga
34 (a) English Units – 70F, Pv = 0.363 psia Pv = φ Pg = 0.75(0.363) = 0.272 psia
W = 0.6219 (0.272) 0.0117 lbmv / lbma
(14.696  0.272) =
i = 0.24 (70) + 0.0117 [1061.2 + 0.444 (70)] 29.58 Btu / lbma= (b) Pv = 0.75 (0.363) = 0.272 psia; P = 12.24 psia
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
11
W = 0.6219 (0.272)(12.24  0.272)
= 0.0141 lbmv / lbma
i = 0.24(70) + 0.0141 [1061.2 + 0.444 (70)] 32.20 Btu/ lbma= 34 SI Units – (a) 20C, 75% RH, Sea Level 34 (cont’d) Ps = 2.34 kPa; Pv = 0.75 x 2.34 = 1.755 kPa
0.6219 x 1.755W = =(101.325  1.755)
0.0110 kgv / kga
i = 1.0 t + W(2501.3 + 1.86t) i = 20 + 0.0110(2501.3 + 1.86 x 20) = 47.92 kJ / kga (b) 20C, 75% RH, 1525m P = 99.436 – 0.01 x 1525 = 84.186 kPa Ps = 2.34 KPa; Pv = 0.75 x 2.34 = 1.755 kPa
W = 0.6219 x 1.755(84.186  1.755)
= 0.0132 kgv / kga
i = 20 + 0.0132(2501.3 + 1.86 x 20) = 53.51 kJ / kga 35 English Units – t = 72 Fdb; psia 14.696 P %; 50 ==φ
svs
v P P or PP φφ == ; Pv = 0.5(0.3918) = 0. 196 psia
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
12
Air dewpoint = saturated temp. at 0.196 psia = 52.6 F Moisture will condense because the glass temp. 40 F is below the dew point temp. 35 SI Units – t = 22C ; 50% ; P = 100 kPa Pv = φ Ps ; Pv = 0.5(2.34) = 1.17 kPa 35 (cont’d)
Air dewpoint = sat.temp. at 1.17 kPa = 9.17 C Glass temp. of 4 C is below the dewpoint of 9.17 C, therefore, moisture will ccondense on the glass
36 English Units  (a) At 55F, 80% RH, va = 13.12 ft3 / lba and ρ a = 0.0752 lbma / ft3
= 22,860 lbma / hr am 5000 (0.0762) 381 lbma / min= = (b) Using PSYCH ρ a = 0.0610 lbma / ft3 or va = 16.4 ft3 / lba = 5000 (0.061) = 305 lbma / min am 18,300 lbma / hr= 36 SI Units – (a) t = 13 C and relative humidity 80% then va 0.820 m≈ 3 / kga; am 2.36 / 0.82 2.88 kga / s= = (b) Assuming same conditions ; 3
av 0.985 m / kga= am 2.36 / 0.985 2.40 kga / s= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
13
37 English Units – t = 80F, 60% RH (a) v sP P 0.6 (0.507) 0.304 psiaφ= = = = 64.5 F dp sat vt (t @ P= ) (b) Same as (a) above 37 SI Units – (a) 27 C, 60% RH, Sea Level Ps = 3.57 kPa; Pv = 0.6 x 3.57 = 2.14 kPa dp sat vt =(t at P ) 18.4 C≈
(b) Same as (a) above 38 dpt 9C (48F)≤
42%φ ≤ ; W 0.0071 kgv / kga (lbv / lba)≤
Chart 1a & 1b
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
14
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50ENTHALPY 
BTU PER POUND O
F DRY A
IR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
dp Room
Problem 38
W=0.0071
72 (22)48 (9)
42 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04.08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82000
1000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
39 (a,b,d) Using the Properties option of PSYCH: Relative Humidity = 0.59 or 59%
Enthalpy = 30.4 Btu/lbma
Humidity Ratio = 0.0114 lbu/lba
(c) Again using the Properties option
At W=0.0114 lbv/lba; RH = 1.00 or 100%
The dew point = tdb or twb = 59.9 F Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
15
39 (cont’d)
(e) Using the Density of Dry Air option:
Mass Density = 0.070 lba/ft3 310 Using program PSYCH (a) tdb = 102.6; twb = 81.1F 75 Fdb; 65 fwb; 14.2 psia (b) m 58.7ν = lbm/hr
Q 2 = 1027 cfm 311 t1 = 80 / 67 F; t2 = 55 F and sat.; assume std. barometer (a) W1 – W2 = 0.0112 – 0.0092 = 0.002 lbv / lba (b) lq 31.5  29.3 2.2 Btu / lba= = (c) qs = 29.3 – 23.2 = 6.1 Btu / lba (d) q = l sq q 8.3 Btu / lba+ =
312 (a) *2
0.6219 (0.3095)W 0.0134 kgv / kga(14.696 0.3095)
= =−
10.24 (65  80) ( 0.0134 x 1056.5)W 0.00993 lbv / lba
(1096  33)+
= =
also W1 = 0.6219 Pv1 / (P – Pv1)
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
16
Pv1 = (0.00993 x 14.696) / ( 0.6219 + 0.00993) = 0.231 psia 312 (cont’d)
10.231 0.46 or 46%0.507
φ = =
(b) P = 29.42 – (0.0009 x 5000) = 24.92 in.Hg. or P = 12.24 psia
*2
0.6219 x (0.3095)W 0.01613(12.24  0.3095)
= = lbv/lba
W1 = 0.24(65 80) (0.01613 x 1056.5) 0.01265 lbv / lba
( 1096  33)− +
=
or kgv / kga Pv1 = 0.01265 x 12.24 / ( 0.6219 + 0.01265) = 0.244 psia
10.244 0.48 or 48%0.507
φ = =
313 (a) Sea Level
Dry Bulb, F
Wet Bulb, F
Dew point
F
Humid. Ratio, lba/lbv
EnthalpyBtu/lba
Rel. Humid., %
Mass Density lba/ft3
85 60 40.6 0.0053 26.6 21 0.072 75 59.6 49.2 0.0074 26.1 40 0.073
74.6 65.1 60.1 0.0111 30 60 0.073 88.6 70 60.9 0.01143 33.8 40 0.071 100 85.8 81.7 0.0235 50 56 0.068
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
17
(a) 5000 ft.
Dry Bulb, F
Wet Bulb, F
Dew point
F
Humid. Ratio, lba/lbv
EnthalpyBtu/lba
Rel. Humid., %
Mass Density lba/ft3
85 60 45.1 0.0076 28.7 25 0.060 75 58.6 49.2 0.0089 27.7 40 0.061
71.2 61.6 56.7 0.0118 30 60 0.061 102.7 70 55.8 0.01143 37.3 22 0.058 100 81.3 76.1 0.0235 50 47 0.057
(c) Note effect of barometric pressure. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
18
314
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
dp Room
Problem 314
72 (22)52 (11)
Max RH=49.6 %W=0.0083
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0.5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
1 00 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
315 Use Chart 1b, SI (a) td = 10 C; SHF = 0.62
(b) 1 22.4q m (i i ) (57.1  34)
0.867= − = = 63.95 kJ / s = 63.95 k W
sq 63.95 (0.62) 39.65 kW= = 315 Use Chart 1a, IP (a) td = 52 F; SHF = 0.63 Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
19
315 (cont’d)
(b) 5000(60)q = (32  22.6)= 203,317. Btu/hr
13.87
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
ADP2
Problem 315
80 (27)55 (13)52 (10)
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0.5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .8200 0
1000
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
sq 203,317 (0.63) 128,089. Btu/hr= =
316 (a) i1 = 30 Btu / lba; v1 = 13.78 ft3 / lba; W = 0.0103 lbalbv ; 50% 1 =φ
(b) i1 = 51.6 kJ / kga v1 = 0.86 m3 / kga
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
20
316 (cont’d)
W1 = kgakgv 0103.0
50% 1 =φ 317 Use the Heat Transfer option of program PSYCH:
q = 148,239 Btu/hr sq 102,235 Btu/hr=
SHF = 0.69
318 Use the Heat Transfer option of program PSYCH for sensible heat transfer only:
sq 178,911 Btu/hr= −
Negative sign indicates heating. 319 Use the program PSYC to compute the various properties at 85/68 F; sea level and 6000 ft elevation.
Elevation ft
Enthalpy Btu/lbm
Rel. Humpercent
Hum. Ratio lbv/lba
Density lba/ft3
0 32.2 42 0.0107 0.072 6000 36.3 45 0.0144 0.058
At sea level: am 5000 x 0.072 x 60 21,600 lba/hr= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
21
319 (cont’d) At 6000 feet:: am 5000 x 0.057 x 60 17,100 lba/hr= = Percent Decrease at 6000 ft:
(21,600 17,100)100PD 20.8%21,600−
= =
320 Use the program PSYC to compute the heat transfer rates at 1000 and 6000 feet elevation: (a) at 1000 ft, q 200,534 Btu/hr= (b) at 6000 ft, q 190,224 Btu/hr=
(c) PD = (200,534 190,224)100 5.1 %200,543−
=
321 (a) English Units – ; = 0 in.Hg. 29.92 PB = q
wi i 180.2 0.8 (970.2)
W∆
= = +∆
iw = 956.4 Btu / lbv From chart 1a; t2 = 91.5 F
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
22
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
2
Problem 321
98 (38)91.5 (32)
60 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82000
1 00 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
321 (a) SI Units – PB = 101.325 kPa
wi i 419.04 (0.8 x 2257)
W∆
= = +∆
iW = 2224.6 kJ / kg From chart 1b; t2 = 32 C (b) Use Humidification (adiabatic) option to obtain 91.5 F db. 322 PB = 29.92 in.Hg.; 0 q = (a) Using chart 1a Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
23
322 (cont’d)
wi i 1090 Btu / lbm
W∆
= =∆
From table A1
f
fg
ii 1090  196.1x = i 960.
=1
x = 0.931 or about 93 %
(b) x will be the same
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
a
bProblem 322
80
60
1090
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
100 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
24
323 Assume PB = 101.325 kPa; 0 q =
w∆i 272.1 i kJ / kg∆W 1000
= =
iw = 0.272 (on scale) t2 = 22.6 C
10 20 30 40 50
60
70
80
90
100
110
110
120
120
ENTHALPY  KJ PER KILOGRAM OF DRY AIR
10
20
30
40
50
60
70
80
90
100
ENTHALPY  KJ P
ER KILOGRAM O
F DRY AIR
SATURATIO
N TEMPERATURE  °
C
5 10 15 20 25
30 35 40 45 50
DR
Y B
ULB
TE
MP
ER
ATU
RE
 °C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
10% RELATIVE HUMID ITY
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 WET BULB TEMPERATURE  °C
30
0.78
0.80
0.82
0.84
0.86 VO
LUM
E  CU
BIC
ME
TER
PER
kg DR
Y A
IR
0.88
0.90
0.92
0.94
HU
MID
ITY
RA
TIO
 G
RAM
S M
OIS
TUR
E P
ER K
ILO
GR
AM D
RY
AIR
1
2
Problem 323
3822.6
20
80 %
0.272
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 101.325 kPaCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
1 .52.0
4 .0
4.02 .01.0
0 .5
0.2
0 .1
0.2
0.3
0.4
0.5
0.60.7
0 .85.0
2.0
0.0
1 .0
2.02.5
3.0
4.0
5.0
10.0
 SENSIBLE HEAT QsTOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
324 For adia. humidification
(a) w∆i = i 1131 Btu / lbw∆W
=
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
25
324 (cont’d) c a 2q = m (i  i )1 am 2000 x 60 / 13.14= am 9132 lba / hr= 1 2i 18.1 Btu / lba ; i 29.7 Btu / hr= = cq 9132 (29.7  18.1) 105,931 Btu / hr= = w a 3 2 3 2m m (W  W ) ; W = 0.0167; W 0.0032 lbv/lba= = wm 9132 (0.01 67  0.0032) 123.3 lbw / hr= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
26
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
2
3
Problem 324
1131
30 %
110 (43)60 (16)
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .8200 0
1 000
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
(b) Solution similar to (a) 325 English Units – See diagram for construction on chart 1a.
1
3
Q32 2000 2=3000 3Q12
= =
Layout 2L/3 on the chart and read: W3 = 0.007 lbv/lba Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
27
I3 = 22.2 Btu/lba
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50ENTHALPY 
BTU PER POUND O
F DRY A
IR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
2
3
Problem 325
40 (4) 100 (38)58.4 (15)
35
77
52
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .8200 0
1 000
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
325 SI Units – Same procedure as above, read: 3i 34 kJ / kga= 3W 0.007 kgv / kga= 326 English Units – Layout the given data on Chart 1a as shown for problem 325. a1m 2000(60) 12.66 9,479lba hr= = Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
28
326 (cont’d) a2m 1000(60) 14.44 4,155lba hr= =
a1
a1 a2
m32 9479= 0.695m +m 9479 415512
= =+
Layout distance 32 on line from 1 to 2 to locate point 3 for the mixture. Read: i3 = 21.5 Btu/lbm W3 = 0.0067 lbu/lba
For W, % Error = (0.007 0.0067)100 4.5
0.0067−
=
For I, % Error = (22.2 21.5)100 3.3
21.5−
=
327 250,000SHF 0.8200,000
= =
or SHF = 59 .8173
=
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
29
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
2
Problem 327
75 (24)
50 %
53 (12)
0.8
21.5
28.2
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
1000
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
328 Refer to diagram for 327 (a) a 1 2 1 2q = m (i  i ); i 28.2; i 21.5= = am 250,000 / (28.2  21.5) 37,313 lba / hr= = 3
a 2Q = m v 37,313 x 13.09 / 60 8,140 ft / min= = (b) similar procedure; 3Q 3.85 m / s= Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
30
329 (a) Use the AirQuantity option of program PSYCH, iterating on the relative humidity and setting the minimum outdoor Air Quantity to 0.01, NOT ZERO.
Use the properties option to find the entering wet bulb temperature of 62.6F. Then
φ = 0.852 (iterated) ts = 56F
= 9,360 cfm sQ (b) Proceed as above
φ = 0.882 ts = 56F
= 10,014 cfm sQ 330 Proceed as in 329 above.
φ = 0.92 ts = 56.1 56 F ≈
= 11,303 cfm sQ ≤
331 (a) 91.0000,550000,500SHF ==
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
31
331 (cont’d) (b) a 2 1q = m (i i ) or a 2m = q/(i i )1
a550,000m
(34.3 22.8)=
−
am =47,826lba hr
a 22
m v 47,826Q = = x 14.62=11,654 cfm60 60
or 5.5 m3/s
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
32
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
2
Problem 331
0.91
115 (46)72 (22)
30 %
22.8
34.3
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1.0
2.04.08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82000
1000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
332 a 2 1q = m (i i )
2a
qi = +im 1
a1400 x 60m 5,915.5
14.2=
2
5 x 12,000i = +38.55,915.5
Btu/lba 2i 2 8 .3= 6 Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
33
Then from Chart 1a, t2= 67F
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
2ADP
Problem 332
90
75
67
28.4
55
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30.20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82000
1 00 0
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
333 Use Adiabatic Mixing option of PSYCH with the Properties option to enter requested data. Assume volume flow rates of 3 to 1 to obtain. Tmix,db = 84.2 F Tmix,wb = 71.3 F 334 Use Program PSYCH at Sea Level elevation
Iteration on the supply volume flow rate is required. This is the same as the
leaving air quantity for the coil. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
34
334 (cont’d)
(a) Supply air quantity is 9,384 cfm.
(b) The outdoor air quantity is 938 cfm.
(c) Air enters the coil at 74.6 F db, 60.5 F wb at a rate of 9,740 cfm
(d) The coil capacity is 248,256 Btu/hr.
The amount of air returned is: (9,740 – 939) = 8,802 cfm.
335 Use Program PSYCH at 5,000 ft elevation
Iteration on the supply volume flow rate is required. This is the same as the
leaving air quantity for the coil.
(a) Supply air quantity is 11,267 cfm.
(b) The outdoor air quantity is 1,127 cfm.
(c) Air enters the coil at 74.6 F db, 62.1 F wb at a rate of 11,697 cfm
(d) The coil capacity is 334,143 Btu/hr.
The amount of air returned is: (11,697 – 1,127) = 10,570 cfm. 336 cfm 1000Q0 = (a) From Chart 1a st =120 / 74 F
ss r
q 200,000m =(i i ) (37.2 22.8)
=−
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
35
= 13,889 lb/hr = 1m 3
s s s sQ = m v = m (14.78)/60 = 3,421 ft /min (b) o o om = Q /v 1000 x 60 / 12.61 4758 lb/hr= =
r
1
m 13,889 4758 0.66;m 13,889
−= = 1From Chart 1a t 61/ 47 F=
3 1t  t (119 61)= − (c) w s s 2m = m (W W ) 13,889 (0.0075  0.0036) = = 54.2 lbm/hr (d) f 1 3 1q = m (i i ) =13,889 (32.8 18.6) 197,224 Btu/hr− =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
36
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
r
s
0
1
31 3
Problem 336
12072
30 %
40 61
47
0.8
1150
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82000
1 00 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
337 (a) st 120 / 71.4 F Use Chart 1Ha= s 1m 200,000 /(38.7 24.0) 13,605 lba/hr m= − = = sQ 13,605 x 17.85 / 60 4048 cfm= = (b) lba/hr 3947 60 x )2.15/1000(m0 ==
r1
1
m 13,605 3947 0.71; t 62.8 / 47 Fm 13,605
−= = =
3 1t t (119.5 62.8)= − (c) w s s 1m =m (w W ) 13,605 (0.0088  0.0046) 57.14 lbw/hr= =Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
37
(d) fq 13,605 (33.8  20.2) 185,028 Btu/hr= = 338 Assume fan power and heat gain are load on the space
s9384m x 60 = 42,915 lbm/hr; Prob 33413.12
=
fan duct s s cW q m (i i+ = − ) = (4 x 2545) + 1000 = 11,180 Btu / hr
c11,180i 20.8 20.54 Btu/lbm42,915
= − =
State c is required condition leaving coil Part a, b, and c are same as prob. 334; (d) coil 1 1 cq =m (i i ) 42,915 (26.8  20.54) 268,648 Btu/hr= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
38
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
s
r
0
1
c
1
Problem 338
1007255
20.54
50 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0.5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
1 00 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
339 )ii(mW );ii(mq cssfansrsr −=−=
(a) c ri 28 Btu/lbm; i 33.7 Btu/lbm= = Using Chart 1Ha rq 1,320,000 Btu/hr= fanW 30 x 2545 76350 Btu/hr= = fan a s cW 30 x 2545 76,350 = m (i i )= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
39
s aq = 1,320,000 = m r s(i i )
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50ENTH
ALPY  B
TU P
ER P
OUND OF
DRY AIR
SATU
RATION TE
MPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
35 40
40 45
45 50
5055
5560
60
65
65
70
70
75
75
80 WET BULB TEMPERATURE  °F
80
85
15.5
16.0
16.5 VO
LUM
E  CU
.FT. P
ER
LB. DR
Y A
IR
17.0
17.5
18.0
HU
MID
ITY
RA
T IO
 P
OU
ND
S M
OI S
TUR
E P
ER P
OU
ND
DR
Y AI
Rr
0
css
Problem 339
90 (32)80 (27)
50 %59 (15)
62.5 (17)
0.8
R R
ASHRAE PSYCHROMETRIC CHART NO.4NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
5000 FEET
0
1.0 1 .0
2.04 .08 .0
8.04.02.0
1 .0
0 .50 .4 0.30.20 .1
0 .1
0.2
0.3
0.4
0 .5
0 .6
0 .82000
1 000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
Two unknowns & two equations Solve simultaneous:
fa n s a r c
a
a
W + q = m (i i )
1 ,3 2 0 ,0 0 + 7 6 ,3 5 0m =(3 3 .7 2 8 )
m = 2 4 4 ,9 7 4 lb a /h r
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
40
s r s ai = i  ( q m )
s1,320,000i = 33.7  =28.3244,974
Btu/lba
Locate points on the condition line on Chart 1 Ha and point c is on cooler process line horz. to left of points. Read ts = 62.5 F, tc = 61.6F.
(a) s244,974Q = x16.2 = 66,143cfm
60
(b) 3
sQ 31.2 m= s
H
340 English Units –Tucson, Arizona, Elevation 2,556 ft. ; min 0i =i =31.1 Btu/lba and sat. air mint =64.5 F; PSYCH Shreveport, Louisiana, Elevation 259 ft. ; min 0i =i = 42.5 Btu/lba and sat. air mint 76.8 F; PSYC= SI Units – Tucson, Arizona ; min 0i =i 51.5 kJ/kga= mint =18.1 C; Chart 1b Shreveport, Louisiana ; min 0i =i =75.5 kJ/kga mint =24.8 C; Chart 1b
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
41
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  B
TU PER P
OUND OF D
RY AIR
SATURATION T
EMPERATURE 
°F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
13.0
13.5
14.0 VOLU
ME  C
U.FT. P
ER LB
. DR
Y AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
SL
TLO
Problem 340Shreveport, LA
9576.8
R R
ASHRAE PSYCHROMETRIC CHART NO.4NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.642 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
259 FEET
0
1.0 1 .0
2.04 .08 .0
8.04.02.0
1 .0
0 .50 .4 0.30 .20 .1
0 .1
0.2
0.3
0.4
0 .5
0 .6
0 .82 000
1000
0
5 00
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
42
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  B
TU PER P
OUND OF D
RY AIR
SATURATIO
N TEMPERATU
RE  °F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.00 2
.00 4
.00 6
.00 8
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
35 40
40 45
45 50
50 55
5560
6065
65
70
70
75
75
80 WET BULB TEMPERATURE  °F
80
85
85
14.0
14.5
15.0 VOLU
ME  C
U.FT. P
ER LB
. DR
Y AIR
15.5
16.0
16.5
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
TA
TLO
Problem 340Tucson, Arizona
10264.6
R R
ASHRAE PSYCHROMETRIC CHART NO.4NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 27.259 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
2556 FEET
0
1.0 1 .0
2.04 .08 .0
8.04.02.0
1 .0
0 .50 .4 0.30 .20 .1
0 .1
0.2
0.3
0.4
0 .5
0 .6
0 .82 000
1 000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
341 )ii(mq srs −= (a) = 1,319 lba/hr ton sm 12,000 /(28.2 19.1)= −
s1319 x 15.6Q 343 cfm/ton
60= =
o
s
m r1 13= 0.55 or 55%m 23.5r0
= =
(b) 3
sQ 0.046 m / s  kW≈ 0 sm /m 55%≈
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
43
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  B
TU P
ER P
OUND OF
DRY AIR
SATU
RATION TEMPER
ATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
35 40
40 45
45 50
5055
5560
60
65
65
70
70
75
75
80 WET BULB TEMPERATURE  °F
80
85
15.5
16.0
16.5 VO
LUM
E  CU
.FT. P
ER
LB. DR
Y A
IR
17.0
17.5
18.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y AI
R
r
s
0
1
Problem 341
100 (38)75 (24)50 (10)
10 %
40 %
0.7
R R
ASHRAE PSYCHROMETRIC CHART NO.4NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
5000 FEET
0
1.0 1 .0
2.04 .08 .0
8.04.02.0
1 .0
0 .50 .40.30 .20 .1
0.1
0.2
0.3
0.4
0 .5
0 .6
0 .82 000
1 000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
342 2 2 3 2
500,000q m (i i ); m(41.1 21.9)
= − =−
lba/hr 042,26m2 = = 6315 cfm 14.55/60 x 26042Q2 = lba/hr 6511 26,042 x 25.0m0 == F 5.49/5.67t ;25.0m/m mix30 == Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
44
342 (cont’d) Preheat Coil: ph 0 p 4 0q = m c (t t ) 6511 x 0.24 (606) 84,383 Btu/hr= = Heat Coil: h 2 5 1q = m (i i ) 26,042 (28.4  20) 218,753 Btu/hr= = Humidifier: w 2 2 5m = m (W W ) 26,042 (0.0144  0.0035) = 283.9 lbw/hr= (b) 3
2 ph hQ 2.98 m / s; q 24.7 kW; q 64.1 kW;= = =
kg/s 036.0mw =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
45
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
3
2
4
1
5
2Problem 342
1153
70 (21)
30 %
105 (40)60 (16)
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 000
100 0
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
343 Use Chart 1a; d a rq m (i is )= − or )ii/(qm srda −= (a) m a = 150 x 12,000 / (28.422) = 28,125 lbm/hr dQ 28,125 x 13.25/60 61,211 cfm= = m dQ 0.20 Q 1,242 cf= = m m mm = 1,242 x 60/13.5 5,521 lbm/hr [v assumed]= m r mi =i 1.8 x 12,000/5,521 24.5 Btu/lbm; t 62 / 57 F− = =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
46
(b) 3 3d m mQ = 2.93 m /s; Q = .59 m /s; t = 17/14 C
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
rm
s
Problem 343
0.8
0.6
75 (24)60 (16)
62 (17)
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
1000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
344 (a) a15.0 x 12,000m 29,508 lba/hr(31.2  25.1)
= =
d mQ 29,508 x 16.0/60 7,869 cfm; Q = 0.2 x Q= = s 1,574 cfm= m mm =1,574 x 60/16.2 5,829 lba/hr (v assumed)= mi 35.7 1.8 x 12,000/5,829 27.5 Btu/lba;= − =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
47
mt 62.5 /58 F= (b) 3 3
s m mQ =3.7 m / s; Q 0.74 m /s; t 17 /14.4 C= =
345 Use Chart 1a; r
1
m 100.8m 0r
= =
[Both design and min. load condition] is = ir  / mq sm
sr
ds ii
Qm
−= =
22.3)(29.3512,000 x 50
conditions all for constant is m lba/hr; 106,85m ss = Btu/lba 25.83 10612,000/85, x 2535.29i 's =−= (a) From Chart 1a; F 64t 's = (b) s'bc'1bss i )mm(imim +=+
271.07.318.258.252.24
)ii()ii(
mm
'1's
'ss
c
b =−−
=−−
=
(b) From chart 1a; cases both for F 49td =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
48
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
r
s
0
1
s
0'
1'
s
s'
Problem 345
0.9
95 (35)85 (29)77 (25)64 (18)55 (13)
50 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
100 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
346 Refer to problem 345. Results are similar. 347 (a) It is probably impossible to cool the air from 1 to 2 in one process. The extension of line 12 does not intersect the saturation curve. (b) Cool the air to state 1' and then heat to state 2.
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
49
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
1
2
1'
2
Problem 347
80 (27)60 (16)52 (11)
67
54
90 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
1000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
348 (a) c
s
m sh= =.83m ch
7
h
s
m cs= 0.16m ch
= 3
c
h
m 0.837 5.14m 0.163
= =
s r sq m (i i= − )
s50 x 12,000m 93,750 lba/hr(28.221.8)
= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
50
sQ 93,750 x 13.2/60 20,625 cfm= = (b) 3
sQ 9.7 m /= s
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
r
chs
Problem 348
90 (32)75 (24)52 (11)
90 %
0.65
20 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82000
1 00 0
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
349 See diagram of problem 348
(a) c h
s s
m m36 10.10.9; 0.10m 46.3 m 46.3
= = = = ; c
h
m 0.9 9.0m 0.10
= =
s50 x 12,000m 83,333 lba/hr(30.1  22.9)
= =
sQ =83,333 x 15.67/60 21,763 cfm=Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
51
(b) 3
sQ =10.3 m /s 350 (a) See diagram for problem 348 c
c c r c c ss
m = 0.837; q = m (i i ); m 0.714 x m 0.837 x 93,750m
= =
; cm 78,469 lba/hr= cQ 78,469 x 13.04/60 17,054 cfm= = cq 78,469 (28.220.6) 596,364 Btu/hr= = (b) 3
c cQ =8.1 m /s; q 175 kW= 351 SI Units
(a) On the basis of volume flow rate using Chart 1b:
2 313Q = Q 0.69 x 1.18 0.81512
= = m3/s
and m1 3 2Q = Q  Q = 1.18 0.815 0.365− = 3/s (b)
334 a3 4 3 4 3
3
34
Qq = m (i i ) = (i i )v
1.18q = (47.841.0) = 9.6 kW0.835
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
52
10 20 30 40 50
60
70
80
90
100
110
110
120
120
ENTHALPY  KJ PER KILOGRAM OF DRY AIR
10
20
30
40
50
60
70
80
90
100
ENTHALPY  KJ P
ER KILOGRAM O
F DRY AIR
SATURATION T
EMPERATURE  °C
5 10 15 20 25
30 35 40 45 50
DR
Y B
ULB
TE
MP
ER
ATU
RE
 °C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
10% RELATIVE HUMID ITY
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 WET BULB TEMPERATURE  °C
30
0.78
0.80
0.82
0.84
0.86 VO
LUM
E  CU
BIC
ME
TER
PER
kg DR
Y A
IR
0.88
0.90
0.92
0.94
HU
MID
ITY
RA
TIO
 G
RAM
S M
OIS
TUR
E P
ER K
ILO
GR
AM D
RY
AIR
1
2
3 4
Problem 351
Problem 351
292417.212
50 %
11
14.7
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 101.325 kPaCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
1 .52.0
4 .0
4.02 .01.0
0 .5
0.2
0.1
0.2
0.3
0.4
0.5
0.60.7
0 .85.0
2.0
0.0
1 .0
2.02.5
3.0
4.0
5.0
10.0
 SENSIB LE HEAT QsTOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
English Units (a) 1 34Q = 640 cfm; q = 33,684 Btu/hr
352 (a),(b)
From Chart 1b, states 1.4 and ADP are known. Based on approx. 11.8 C db, 11.2 C wb, and 90% RH locate state 2. Then for full load design condition air is cooled from 1 to 2 and the room process proceeds from 2 to 4.
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
53
For the high latent load condition, the air at 2 is reheated to state 3 where it enters the space and the process proceeds to state 4.
(c) 224 a 4 2 4 2
2
Qq = m (i i ) = (i i )v
2Q =35 x 0.817 (47.732) ; m2Q 1.8= 2 3/s
12 a 1 2
12
1.82q = m (i i ) = (60.632)0.817
q = 63.7 kW
34 a 4 3
34
23 24 34
1.82q = m (i i )= (47.739.4)0.817
q = 18.5 kWq = q  q = 3518.5=16.5 kW
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
54
10 20 30 40 50
60
70
80
90
100
110
110
120
120
ENTHALPY  KJ PER KILOGRAM OF DRY AIR
10
20
30
40
50
60
70
80
90
100
ENTHALPY  KJ P
ER KILOGRAM O
F DRY AIR
SATURATION T
EMPERATURE  °C
5 10 15 20 25
30 35 40 45 50
DR
Y B
ULB
TE
MP
ER
ATU
RE
 °C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
10% RELATIVE HUMID ITY
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 WET BULB TEMPERATURE  °C
30
0.78
0.80
0.82
0.84
0.86 VO
LUM
E  CU
BIC
ME
TER
PER
kg DR
Y A
IR
0.88
0.90
0.92
0.94
HU
MID
ITY
RA
TIO
 G
RAM
S M
OIS
TUR
E P
ER K
ILO
GR
AM D
RY
AIR 1
ADP 2 3
4
3
Problem 352
Problem 352 21
2723
17
19
14
11.8
11
9
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 101.325 kPaCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
1 .52.0
4 .0
4.02 .01.0
0 .5
0.2
0 .1
0.2
0.3
0.4
0.5
0.60.7
0 .85.0
2.0
0.0
1 .0
2.02.5
3.0
4.0
5.0
10.0
 SENSIBLE HEAT QsTOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
352 English Units (a),(b) See above (c) Btu/hr 2Q = 4103cfm ; 12q =221,243 Btu/hr; 34q 67,49= 8 223q 52,50= Btu/hr 353 English Units (a) s r s sq=m (i i ); m 5000 x 60/13.2 22,727 lba/hr= = (specific volume value of 13.2 ft3/lbm is assumed.) Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
55
s r si = i  q /m =28.2 10 x 12,000 / 22,727 22.9 Btu/lba− = s o s ot = t = 57.5 F; W =W 0.0083 lbv/lba=
(b) r r
m s
m m0m= = 0.462m m0r
rm =0.462 x 22,727 10,500 lba/hr= om 22,727 10,500 12,227 lba/hr= − = rQ 10,500 x 13.68/60 2,394 cfm= = oQ 12,227 x 12.11/60 2,468 cfm= =
(c) r
m'
m 0'm'= =0.578m 0'r
r om =0.578 x 22,727 13,131 lba/hr; m 9,596 lba/hr= =' r o'Q =13,131 x 13.68/60 2,994 cfm; Q 9,596 x 13.48/60= = = 2,156 cfm (d) c s m' sq = m (i i ) 22,727 (28.4  22.8) 127,271 Btu/hr= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
56
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BTU PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
r
0
m
s
0'
r
m'
ADP
Problem 353
0.8
1150
75 (24)
70 (21)
65 (18)57.5 (14)40 (4)
43 (6)
90 %
50 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
1000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
353 SI Units (a) ts = 14.2C; Ws = 0.0083 kgv/kga (b) 3
rQ =1.13m s; 3oQ =1.17m s
(c) 3rQ =1.41m s; 3
o'Q =1.02m s (d) cq = 37.3 kW 354 (a) Any combination that will yield an enthalpy less than 57.0 kJ/kga or 33 Btu/lba Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
57
(b) rs mkga/s 95.584.0/5m ===
o
r
m mr= =0.36m 0r
om 0.36 x 5.95 2.14 kga/s= = 3
oQ = 2.14 x 0.852 = 1.82 m /s = 3,857cfm (c) adt 15.4 C or 60F= (d) (Essentially, no difference) o n m s r sq /q = (i i )/(i i ) = 1.0
10 20 30 40 50
60
70
80
90
100
110
110
120
120
ENTHALPY  KJ PER KILOGRAM OF DRY AIR
10
20
30
40
50
60
70
80
90
100
ENTHALPY  KJ P
ER KILOGRAM O
F DRY AIR
SATURATION T
EMPERATURE  °C
5 10 15 20 25
30 35 40 45 50
DR
Y B
ULB
TE
MP
ER
ATU
RE
 °C
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
10% RELATIVE HUMID ITY
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 WET BU LB TEMPERATURE  °C
30
0.78
0.80
0.82
0.84
0.86 VO
LUM
E  CU
BIC
ME
TER
PER
kg DR
Y A
IR
0.88
0.90
0.92
0.94
HU
MID
ITY
RA
TIO
 G
RAM
S M
OIS
TUR
E P
ER K
ILO
GR
AM D
RY
AIR
r
s
0m
s
Problem 354
25 (77)18 (64)
20 (68)
0.6
57
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 101.325 kPaCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
1 .52.0
4 .0
4.02 .01.0
0 .5
0.2
0 .1
0.2
0.3
0.4
0.5
0.60.7
0 .85.0
2.0
0.0
1 .0
2.02.5
3.0
4.0
5.0
10.0
 SENSIBLE HEAT QsTOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
58
355 424,000SHF530,000 424,000
−=
− = 4
Construct condition line on Chart 1a with preheat and mixing processes. (a) sen s p r sq = 424,000 = m c (t t )
s424,000m 88,333 lba/hr
0.24 (75 95)−
= =−
3
sQ =88,333 x 14.07/60 20,714 cfm or 9.8 m /s=
(b) rr
m
m hm= =0.33; m 0.33 x 88,333 lba/hrm hr
=
r rm =29,150 lba/hr; Q 29,150 x 13.68/60= 36,646 cfm or 3.14 m /s=
hh
m
m =1 0.33 0.67; m 0.67 x 88,333m
− = =
h hm 59,183 lba/hr; Q 59,183 x 13.1/60= = 3
hQ 12,922 cfm or 6.1 m /s (at heated condition)= (c) ph h p h oq =m c (t t ) = 59,183 x 0.24 (6035) q= 355,098 Btu/hr or 104 kW (d) mq =88,333 x 0.24 (95  65) 635,998 Btu/hr or 186 kW= Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
59
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR
r
0 h
m
ss
Problem 355
4
95 (35)75 (24)
50 %
60 (16)35 (2)
20 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20.1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 000
100 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHUMIDITY RATIO
hW
356 Refer to chart 1a. (a)
34 a3 4 3 3 4 33
34 33
4 3
33
60q = m (i i ) = Q x (i i )v
q v (1750 x 13.23)Q x =60(i i ) 60(28.123)
Q = 75.7 or 76 cfm = 0.040 m /s
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60
(b) t3db = 58.5 F and 80% RH or 15 C
(c) 32 3
31Q = ; Q = 0.754 x 75.7 = 57 cfm or 0.028 m /s12
3
1Q = 76  57 = 19 cfm or 0.012 m /s
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR1
2
3
4
Problem 356
84
70
75
62
58.550
90 %
50 %
0.8
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82 00 0
100 0
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
357 (a) Refer to Chart 1
A reheat system is required. Process 12 is for the coil. Process 34 is defined by the SHF = 0.5
Process 23 represents the required heat. Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
61
State 3 is defined by the intersection of the reheat and space condition lines.
(b)
334 a3 4 3 4 3
3
34 33
4 3
33
Q x 60q = m (i i ) = (i i )v
q v 100,000 x 13.4Q = =60(i i ) 60(28.223.9)
Q = 5,194 cfm or 2.5 m /s
(c)
12 a 1 2
12
23
23
5,194 x 60q = m (i i ) = (34.220.2)13.4
q = 325,594 Btu/hr or 95.4 kW
5,194 x 60q = (23.920.2)13.4
q =86,050 Btu/hr or 25.2 kW
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
62
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALPY  BTU PER P
OUND OF D
RY AIR
SATURATION TEMPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE H UMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
40 45
45 50
50 55
55 60
6065
65
70
70
75
75
80
80
85 WET BULB TEMPERATURE  °F
85
90
12.5
13.0
13.5
14.0 VO
LUM
E  CU
.FT. P
ER LB. D
RY
AIR
14.5
15.0
HU
MID
ITY
RA
TIO
 P
OU
ND
S M
OIS
TUR
E P
ER P
OU
ND
DR
Y A
IR 1
2 3
4
ADP
Problem 357
85
70
75
62
66
56
5145
50 %
R R
ASHRAE PSYCHROMETRIC CHART NO.1NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 29.921 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
SEA LEVEL
0
1.0 1 .0
2.04 .08 .0
8.04 .02.0
1.0
0 .5 0.40.30 .20 .1
0 .10.2
0.3
0.4
0 .5
0.6
0 .82000
1 000
0
500
1000
1500
2000
3000
5000
SENSIB LE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
358 Assume room temperature humidity of 50% and layout the state & processes on required from point c to s. Supply Air: = senq 120,000 x 0.5 60,000 Btu/hr= = s p s rm c (t t )
s60,000m 53,192 lba/hr
0.24 (7570.3)= =
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
63
3sQ =53,192 x 16.33/60 14,477 cfm or 6.8 m /s=
Mixed Air: om 53,192 x 0.333 17,703 lba/hr= = 3
oQ 17,713 x 17.2/60 5,078 cfm or 2.4 m /s= = rm 53,192 17,713 35,479 lba/hr= − = 3
rQ =35,479 x 16.5/60 9,757 cfm or 4.6 m /s= Reheat: rh c p s cq = m c (t t ) 53,192 x 0.24 (70.355.2)= 192,768 Btu/hr or 56.5 kW= Coil: c m m cq =m (i i ) 53,192 (34.4  24.2)= 542,558 Btu/hr or 159 kW=
= % 1.5412,200
100)109,190412,200(=
−
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
64
10 15 20 25
30
35
40
45
50
55
55
60
60
ENTHALPY  BT U PER POUND OF DRY AIR
15
20
25
30
35
40
45
50
ENTHALP
Y  BTU
PER
POUND O
F DRY A
IR
SATU
RATION TE
MPERATURE  °
F
35 40 45 50 55 60
65 70 75
80
85
90
95 100
105
110
115
120
DR
Y B
UL
B T
EM
PE
RA
TU
RE
 °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10% RELATIVE HUMIDITY
20%
30%
40%
50%
60%
70%
80%
90%
35
35 40
40 45
45 50
5055
5560
60
65
65
70
70
75
75
80 WET BULB TEMPERATURE  °F
80
85
15.5
16.0
16.5 VO
LUM
E  CU
.FT. P
ER
LB. DR
Y A
IR
17.0
17.5
18.0
HU
MID
ITY
RA
T IO
 P
OU
ND
S M
OI S
TUR
E P
ER P
OU
ND
DR
Y AI
R
r
0
m
c ss
Problem 358
90 (32)
75
75 (24)
50 %
70 (21)
55 (13)
90 %
0.60.5
R R
ASHRAE PSYCHROMETRIC CHART NO.4NORMAL TEMPERATURE
BAROMETRIC PRESSURE: 24.896 INCHES OF MERCURYCopyright 1992
AMERICAN SOCIETY OF HEATING, REFRIGERATING AND AIRCONDITIONING ENGINEERS, INC.
5000 FEET
0
1.0 1 .0
2.04 .08 .0
8.04.02.0
1 .0
0 .50 .4 0.30.20 .1
0 .1
0.2
0.3
0.4
0 .5
0 .6
0 .82000
1 000
0
500
1000
1500
2000
3000
5000
SENSIBLE HEAT Qs
TOTAL HEAT Qt
ENTHALPYHU MIDITY RATIO
hW
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
Exοerpts from this wοrk may be reproduced by instructors for distribution on a notforproΓrt basis fortesting or instructional purposes only tο students enrolled in courses for which the textboοk has beenadopted. Αny other reproduction or trαnsΙαtiοn ofthis work beyond thαt permitted by Sections ]07 or ]08of the 1976 United Stαtes Copyright Αct withοut the permissiοn of the copyright owner is unΙαwful.Requests for permission or further informαtion should be αddressed to the Permission Depαrtment, JohnWiley & Sons, Ιnc, Ι ] ] Riνer Street' Hoboken, NJ 07030'
Chapter 4
41 (a) comfortable
(b) too warm
(c) comfortabΙe
(d) too dry
42 (a) comfortable
(b) too warm
(c) comfortable
(d) too dry
43 (a) Assume sedentary dry bulb of 78 F , clo = o.5, met. = 1 .8,
using equation 44a, to,act = 75  5.4(1 + 0.5)(1.8  1 .2) = 71 F
Relative humidity should be less than 50%
(b) Should wear a S\Ι/eater or light jacket and slacks.
(clo = 0.8)
44 Use fig 41
(a) Summer, to = 76 F or 24 C; Winter, to = 72 F or 22 C
(b) Use equation 44a aS a guide, \Λ/ith clo = 0.2,
met = 3.0, tdb :76 F
t
o/
to =76 5.4 (1+0.2)(31.2) = 64 F [winterorSummer]
45 From fig 43 temperature can rise about3.2 F.(j.g C)
t=68 +3.2=71.2 Fort=20+ 1.8= 21.8C
46 From fig 43 @200 fpm, temp rise ρ 5.3 F (2.9 C)
with t,,"t _ 9 F (5 c), temp rise ε 6.5 F (3.6 c)
47 to = (t, +t^r)2, then using Eq. 41
T,fn ='6* Cν\l2 σg _Tr) = (53s)4 +(O.103 x 109) (4o)1Ι2(78_74)
tmft:82For27'8C
to=(74+82)Ι2 = 78F or25.6C
48 Compute the operative temperature, to
Τ,xn = φ4q4 + (O.103 x 1o911eo11/'(εo _76)= 83.5 F or 28.6 C
to = (84 +76)12= 79.8 F or 26.5 C
From Fig 41, to = 79'8 F and 50 % R.Η. is out of the comfort
zone. Recommend lowering to to about 77 F or 25 C.
tu x72 F
49 Use Eq. 44 to estimate a value of the operative temperature
to, active, assuming to for sedentary activities is 78 F (25.6 C)
with met = 2.0. to, active = 78  5.4 (1 + 0.5) (2  1.2) = 71.5 F, (22C)
Exοerpts fiom this work may be reprοduοed by instruοtors for distribution on a not1brproΓit basis for testing or instruοtionaΙ puφoses only tostudents enrolled in οοurses for whiοh the textbook has been adopted' Αny οther reproduction οr trαnsιαtion of this wοrk beyοnd ιhaι permiιtedby Secιions 107 οr ]08 ofthe Ι976 Uniιed Stqιes CopνriPhι Αcιwithouι ιhe oermission οfthe cοpyrighι οwner is unlωυful.
Αs an approximation
Tmrt = 2To _Τ, and Tflx = Tno * ci1Ι21Tg _ Τ, ) Eq ' (41)
eliminating Tmrt between the 2 equations
2(Τo_T3)4 = Tno *CV1/21τn _Tr)
where all temperatures are absolute
Solve by trial and error with T, =72+ 460 = 532 R
and Te =(71.5+460)=531.5 R, C=0.103 x 1Oe, V=30
ta=85F(30C)
Cold surroundings require high ambient air temperature
for comfort, even with high activity level.
410 (a) Most occupants will be uncomfortable because the relative
humidity is more than 60%, even with trx = t,
(b) The lightest weight possible. Short sleeves, shorts,
open neck, etc.
(c) Lower relative humidity if possible by adjusting the cooling
system to remove more moisture. CouΙd also increase the
relative air motion to highest values, perhaps use fans.
411 (a) Even if the suit was heavy weight, many executives would be
ΕXceφts from this work may be reproduοed by instτuctors for distribution on a notforpro1'it basis for testing or instructional purpοses onΙy tostudents enrollοd in οourses for which thΘ teΧtbook has been adοpted. Αny other reproducιion or trαnsιαtion οf ιhis νοrk beyond ιhαt permittedby Secιions ] 07 or 108 o{ the ] 97 6 United Stαtes Copyrighι Αcι ινithout the permissiοn of the coρyright oινner is unΙωνfuΙ.
cool if sedentary.
(b) Would definitely be cold, especiaΙly hands and feet.
(c) Probably would be comfortable in typical work cΙothes
(d) Probably would be comfortable since they would keep
their coats on and would be walking around.
(e) Cold to very cold
412 Determine relative temperatures difference between inside and outside.
68  45 237 4 _ 45 = 29 Costs are79o/o of that for increased setting, or
74  45 2968  45 23
Costs are increased by 26o/o if thermostat is raised.
413 Too much air motion in the cold winter months tends to cause drafts and
make people uncomfortabΙe. Air velocity just sufficient to prevent large
temperature gradients from floor to ceiling is best for winter. Τhe opposite
is true for hot summer months. Higher air velocity tends to compensate
for high temperature and humidity.
414 (a) Raising the chiΙled water temperature will cause the cooling coil to
operate with a higher surface temperature and the relative humidity in
the space will tend to rise if the latent heat gain is signifΙcant such as
would be the case with many occupants, this could lead to
u ncomfortable cond itions.
(b) Yes, during the unoccupied hours the space load may be almost totalΙy
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SensibΙe heat gain and the load is much less than the design value. ln
this case the chiΙled water temperature may be increased.
415 Τhese fans may bring air down in the Summer, increasing the
velocity of air in the occupied zone and providing improved comfort.
ln the winter, air may be drawn upward, pushing the warm air at the
ceiling downward where it can increase the temperature in the
occupied zone without increasing significantly the air motion below
the fan.
4'16 (a) Τable 42 gives a minimum required amount of ventilation air
of 15 ft3 /min per occupant. this is the minimum amount of
outdoor air that should be used under any circumstances.
Therefore, (Qo)rin = 15(30) = 450 ft3/min
(b) on the basis of floor area, the occupancy wouΙd be 25 and the
minimum ventilation requirement would be
Q, = 15 (25) = 375 ft3 /min. lt would be better to design for
floor area if lowest air flow is desired. With 30 actuaΙ student air
flow is such a case wouΙd be insufficient.
417 Use Eq. 45, Solving for C,
Cs = (QtC" + N)/Qt = C" + (N/at)
= (2001196 + (O.25l9oο)
Ξ 478 x 1o6 = 478 ppm
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or using Sl Units
c, = (2oo / 106)+ (0. 118 t 0.472x 9OO)
= (2OO / 106) + (278 t 106):478 ppm
418 n = number of people to occupy a room
N=n(5.Oml/s)
Solving Eq. 45 for Ν
N = Qt (C,  C") = n (5.0) ml/sPerson
n : Qt (C,  C") / (5.0)
: 2.8 (1000280) / 5.ο
n = 403 persons or 0.0069 m3 /s  person
For English Units:
n = 6000 (1oOO  28Ox 106) / O.O107
= 404 persons or 14.8 cfm/person
419 Use the M100 media of fig. 48. From table 43, select a
12x24 x 8 unit; 650 cfm, ΔP = 0.4 in. wg
At ΔP = 0.25 in. wg. each unit will handle
Q = Ql o'25 Ι o.40]1l2 = 650 [O.25 t o'4oJ1l2
:514 cfm/unit. Then the number of units
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;g μ=(2000 l 514) = 3.89 or 4. This is a satisfactory number.
420 Use the M100 media from TabΙe 43 select a O.3 x O.6 x O.2 unit.This is rated at 0.3 m'/s with 1oo pa pressure drop.
Αt ΔP = 60 pa the alΙowabte flow rate for each unit would be
Q = (0.3) (60/1 OOf tz = 0.23 m3/s
1.OO m3/s wouΙd require 1'Oolo'23 = 4'34 units. This requires atΙeast 5 filter units, but since this is an odd number, recommendusing six units.
Trying the 0.6 x 0.6 x 0.2 filter the allowable flow per unit would be
Q = (0.62) (60/1}q1t2 = 0.48 This would require more than two
units of this size. Εconomies would determine the best choice.
421 Solving Εq' 41O for Q
Q = Qr [ΔP / ΔP,]1'2 = 9OO [o'1 l0'15]12 =735 cfm/module
N = ss00/235 = T.4g [must be integer] Use g modules
VeΙ = Q/Α = ΨΨ = 344fpm = 5.7 fps(2)(8) '
422 Solving Εq' 41O for Q
Q = Qr [ΔP / ΔP,J1l2 = (o'42) 24 l 37 '4]1t2 = 0.336
m=(2.8)/0.336=8.3
Use 9 modules, a 3 x 3 arrangement.
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/3
VelocitY =
423 M200; 0.6x0.6
Use Eq. 410
ΔP = Δη ta / Qι.]' :1OO
0.4Velocity =
a (2.8)m3 /s=1.73m1s
FAcΕ AREA (0.3)(0.6)(e)m2
xO.2; O.4Om3 /s/module
a
lO,4OtO.42l2 =90.7 Pa
=2.22m1s
424
425
A (0,3)(o 6)
No solution exists due to the fixed air quantity for the unit. Thispart of the problem is intended to show the student that typicaldirect expansion equipment cannot be used in this \May. lt alsoshows that the load due to outdoor air is very large.
Γho : o'25 rh"; Locate point 1 on psychrometric Chart at82'4 F db and
66.8 F wb
it = 31.4 Btu / lbm and v1 = 13.9 ft3 /lbm
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by Sectiοns 107 οr ]08 ofthe t976 Llnited Stαtes Copyrilht Αctv,ithοut ιhe peιmissiοn ofιhe copyright oνner is unlανful'
exhaust
sHF= 0.7
74
Q1, = rhi (ii  is) = at /V (60) (i1  is)
Φl = 35ο Ι 12,ooo Qls , = (350 l 12'000) (6ο / v1) (i1 _i.)
is = 31 .  ''''J8?rε;
'' = 23'46 Btu / lbm
Locate on psychrometric chart' ts = 65'6 F db' 55'5 F wb
Q.r = lil, (ir.  is) = 36'000; ir =27 '6 Btu / lbm
tr, = Γil1 = ΨΨ ^ :8695'7 lb / hr(27 .6  23.46)
O, = rh, (vr, = Ψ(13.4) = 1940 cfm
Qt" = 8695.7 (31.4  23.46) = 69,000 Btu / hr = 5.75 tons
Qr = 5.75 (350) = 2014 cfm
(ο) Design filters for 2014 cfm, use M200 media of fig 48.
Try the 24x24x8 units of table 43. 920 cfm @0.4 in. wg.
For max. ΔP of 0'125 in.wat.
Q = 920 [0. 125 tO.4O] 1t2 =514 cfm / module;
n = 2014 I 514 = 3'92, use 4 modules
426 Use the M15 media, η = 93 % from fig' 43'
From table 42,60 cfm / person is required, outdoor air.
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A fresh air balance on the filter gives QrEt + Qo = Qs
where Q,. is recirculated air, Qs is outdoor air and
Q. is supply air.
8, = (60  20) 10.93 = 43.0; Q, = 43.0 +20 = 63.0 cfm / person
or the total amount of air supplied is
Qτ = 63.0 x 55 = 3465 cfm; Try the 12x24x8 unit of table 43
Q/unit=9oO[O'1 /O.35]1Ι2=481 cfm; n =4755 l481
= 7 .2 modules
Use 8 modules [Note: The M24 media could also be used]
427 Q, = (25  15) / 0.S = 12.5 cfm / person
Φ. = 15 + 12.5 = 27 .5 cfm / person
428 Filter location is B, figure 49
Use Eq. 412, solve for RQ.. since
RQΓ = { QoEv[C, _(1_Et)Co] + N}/ (EvEfcS)
RQr={2OOxO.85[180(10.8)0.0]+(10x150x35'32)]l
(0.85 x 0.8 x '180) where Co = 0.0
RQr = 185 ft3 / min or cfm
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students enroΙΙed in οourses fοr which the textbook has been adopted. Αny other reprodiction or ιrαnslαιiοn of this νork beyond ιhαι permitted
by Sectiοns Ι 07 or 1 08 οf the Ι 976 tJnited Sιαtes Copyrιght Αcι ιiithout the permissiοn of ιhe copyrighι oνner is unlιτνν.ful'
75
76
Qo = 2OO cfm, Qs : (1S5 + 2OO) = 385 cfm
42g Solve Εq' 411 for RQ,
RQr = (_Qo)(Eu)(cr)+ N / ΕrEiC,
RQr = [ (20) (0.65) (220) + (125) (35.32 ft3/m3)]
(0.65X0.7)(220)
RQ. = Ψ9*_!1!5^, = 15.53 cfm/person' (0.65)(0.7)(220)
43o For filter location A, use Εq. 41 1, solving for RQ,
RQr = (QoEvCs + N) / (EvEfCs)
RQr = t (2OO (0.85) 180) + (10 x 120 x 35.32 ft3/m3 )l t
(0.85 x 0.8 x '180 )
RQr = 183 cfm, Qo = 2OO cfm; d, = 383 cfm
431 (a) This type of space will require a high ventilation (supply air)
rate to handle the load, air cleanliness is not the main criterion.
Therefore, a low efficiency filter with low pressure drop is
acceptable. From table 42, assume occupancy will be about 30
persons / l OOO ft2. So the total design occupancy is 90
persons. Τhe design will be based on this occupancy although
the cooling requirements may dictate a larger supply air rate.
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students enrolΙed in courses Γor which the textboοk has been adopted. Αny οther reprοducιion or trαnsΙαtion of ιhis νork beyond ιhαt permitted
by Secιiοns 1 07 or ] 08 of the Ι 97 6 [Jniιed Stαtes Cοpyright Αcι νιthout the permissiοn οf ιhe copyright oνner is unlανful.
77
A ''fresh air'' balance on the filter gives Φ, = (Q" _ Qo) / Ef
d," = (20 15) / 0.5 = 10 cfm / person recirculation rate
431 (continued)
Φ, : 1O + 1 5 =25 cfm / person supply rate
Qτ = 25 x 90 = 2250 cfm total supply rate
Net face area, Αf = 2250 / 35o = 6'43 ft2
(b) A higher efficiency would reduce the total amount of air and
reduce the required face area. However this is not desirable in
this case. First the filter system would have to be enlarged to
handle the greater amount of air. A lower filter efficiency could
be used and still maintain the required air quality.
For example, suppose the load dictates 4000 cfm instead of
2250 cfm, then for 90 PeoPle
Φ, = 4ooo / 90 = 44'4 cfm / person
Using a minimum of 15 cfm / person of outdoor air.
Qr. = 44.4  15 = 29 '4 cfm / Person
8. = 29.4: (20  1s) / Er
Et = 5 Ι29'4 : 0'17 or 17oλ required
432 (a) Q = (Q, / v) 60 (i. ir)l5x
I
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studeπs enrolled in courses for which the textbook has been adopted. Αny oιher reprοduction ο
by Sections ] 07 or ] 08 of ιhe 1976 Uniιed SιαteS Copyrighι Αcι ν)ithouι ιhe permission οfιhe cop;
225 people
75FRH=5ο%
125,ο0ο
78
Υ x13 ft3 / Ιba
Φ. = (125,oOo x13) Ι [ 60 x (28  1e.4 )]
Qs = 3,149 cfm
Φ, = Φo = 15 x225
Φ, = 3,375 cfm
Q. must be 3,375 cfm, find ne\Λ/ Supply air condition
125,000 = (3,375 I 13) 60 (28  i.)
i' = 28  (125,000 x 13 ) / ( 3,375x 60) = 20 Btu i lba
Locate new condition on chart aS Sho\Λ/n' Coil must cool oDA
down to this new condition.
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(b)
(c)
D 50 52
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Chapter 5
k = CΔx =0.2 Χ4 =O.8 (Btu _ in) / ( hr  ft2 _ F)
k = 1 .14 x 0.1 = 0.114 W / (mC)
C=k l Δx=o.3o / 5.5 = o.o55 Btu / 1ft2nr_ 11
C = O.O43 / ο. 14 = .307 W/ (m2 _ C)
(a) R = 1lC= 1/0.055 = 18.3 (ft' hr F)/ Btu
R' = R l A= 1 l CA= 18.3/ 1ο0 = 0.183 ( hrF)/Btu(b) R ='1 I .307 = 3.26(m'C)/W
R'=3.26l9.3=0.35C/W
5'1 (a)
(b)
52 (a)
(b)
53
54 R = ΣRi , Rgyp =1ΙC=1l3'1=o'32Rbtd = !0.33 = 3.03; Rair = 0.68
R = 0.68 + 0.32+ 3.03 + 0.32+ 0.68
R = 5.03 (hr ft2 F) / Btu
R0.68 R=0.68
R=0.32
55
tnb ιn2R'= Γ2 + η
2πk 2πkoL
AssumeL=1ft
kι:0'2 Btu  in' t(ft2 _hr_F); kp =314 Btuin tσe _hr_F)
81
lnside Surface (7 m/s )
overalI Τhermal Resis.
0120
= 0.652 m2clW
510
511
Outside Surface
4 in. Face Brick
Sheathing
lnsulation
2x4 stud
Gypsum board
lnside surface
Τotal
Between Frame
0.17
0.65
1.32
1 1.0
0.32
ο.68
14.14
Αt Framinq
0.17
0.65
1.32
4.27
0.32
0.68
7.41
UA:U1Α; +U1Α1, U = UiAi /Α + U1Α1/A
Αr =
14'5
^nd
A' =lΞanα υ = LA 16 A 16 R
, =ΓΨ " : ) ΓΨ " +1= 0 o77Btu / (r',..tt' r)110 "14.14) 116 7.41J
An ordinary walt with ε = O.9 has a unit resistance of 0.68. A
highIy reflective wall, ε = O.05, has a unit resistance of 1'70.
Assume radiation heat transfer is zero for reflective wall. Τhen
the resistance due to convection alone is approximately
Rc=1.7; hc=1/Rc=0.59; h.*r:1/0'68 = 1'47
Frac. Conv. = ha lh. *, = 0.59 I 1'47 = 0'4
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*nιοh the texibook has been adopted' Αny οther reprοclucliοn o^:.:::Ei::::":!,::iχ:,i''Ψ'y''o"!:*! ,   ,
82
512 Αssume 15 mph wind. Rr (2 x 6) RzQ x 4\
1. Outside surface, 15 mPh 0'17 0'17
2. Siding 0'79 0'79
3. Sheathing 1'32 1'32
4. lnsulation, 19'0 11'0
2x4 4'27
2x6 6.7
5. GYPsum wall board 0'32 0'32
6. lnside surface 0.68 0'68
Total 28.98 18'55
Ut = 0.035 Btu / (hr  ft2  F)
υ2 __O.o54 Btu / 1ιlr  ft2  F1
% DiffereΠCΘ = [o'O54r_0035) ι'' ool = 35'2[ 0.0sη ) '
513 Air space will be near the indoor temperature with small
Δt across the air Space.
Use t."rn = 50 F and Δt = 10 F and read
R = 1.oz(rrr f( r) I Atu [Tabte S3a] or 0.18 m2clW
514 Assume tr"rn = 50 F; Δt = 10 F
R = 3.55 (hr  ft2 _F) / Btu or 0.62 (.2 _c/W) [Τabte 53a]
515 qc/Α = U"Δt
Find U for highly reflective surfaces because radiation will be
minimal. This will give a good approximation for the convection
component. From Table 52a,l1orz', heat flow down
83
Uc=1/R = 1l(2x4.55) = 0'11
q./Α
or
U^ _ l' = o.625; q. /Α = 0.625(63  43) = 12'5 W/m2" (2x0
Γ, _ '4 / a '.4lQ/A.. =σ'n l( l l _[ l' 'l l'
'Lι1oO] _ι1οo]
]'
for ε1= t2=O'9, E : 0.82, σ'
(q/A),. = O.1 713 x9.s2 t635 o _s.τol
m
Radiation heat transfer is about 10 times greater'
516 U* = O.O7 Btu / (hr  tt2  11
Ud = O.4O Btu / (hr  ft2  F)
Uwin = O'81 Btu / (hr  ft2  F)
Ad = 17 '78 ft2; Awin =25'0 ft2;
Α* = 117 '2 f(Parallel heat flow Paths
UΑ = U*A* + U6Α6 + U*;nΑ*in
ι ι (o.O7 x117'2) + (O.4 x 17.78) + (0'81 x25'0)I I_
117.2
= O.3O Btu /(hr  f( _F) or about 1 '72\Ν t(m' _ c)
517 q/Α = U(ti _to)
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students enrolled in courses tbr wbich the texibook has been uJot.α. Αny οιher reprολucιion or ιrαnsιαtion of ιhis work beyond ιhαι peπnitted: : ' ι.41^ ^^^'',.:/Ιa! ^''''aΔv;" "'Ι6!$ιΙ
20'
g'
84
From Table 54b, construction 2, R = 8.90 (hr  ft2  F) / Btu
Assume insulation does not fitl the airspace'
Remove R for metal bath and plaster of 0.47 (f''ι.  tt'  f)/Btu and\"' )
add R for acoustical tile and insulation'
Ceiling, R" = 1 / 0'8; insulation' R='1 1'00;
R1e61=20.68; U = 1/R = 0'048 Btu / (hr  ft2 Fiq/Α = o'o48 (72  5) = 3'22 Bτυ / (hr _ ft2)
518 From Table 54a, Construction 1
Uw= 9!+=oe71wr(m2c)0.1761 \ /
Ud = 2.27 Wl(m2 a)' Table 58
Uwin = 4'62w1(m2  ")'
Table 55b
Αw = 35 m2;Αwin =8m2;Αd = 2m2
UΑ = U*A* + U6 + Αα + UwinΑwin
u _ Q.e7 1x35) + (2.27x2) + (4'62x8) _ 2.16 w I (m2  c)
35
519
U = O.14 Btu / (hr _ ft2 _ F1τable5 _ 4a, Construction No. 2
R=1ΙO.14 = 7'14, Rn =7'14_(1 to'44) + (1/0.55) =6.69
Un = 0.15 Btu/(h r  ft2  F) or about ο.85 W l 1m2c1
520 Αssume Ηardwood, k = 1.25 (Btuin) / (hr  ft2  F)
Summer Winter
Rι = 0.68 Ri = 0.68
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85
Rα = '1 '375 l 1.25 Rα = 1 '375 l 1'25
Ro = 0.25 Ro = 0.17
R, = 2.03 R* = '1.95
U, =0.49 Btu / (hrft2 F) U* =0.51 Btu /(hrft2F)
Both values are greater than the value given in
Τable 58 of O.39 Btu / (hr _ ft2  F), but acceptable.
521 Computed: Ri = 0.68, Rs = 0.03 (estimate); Ro = 0.25
Ri +RgaRo=0.96=R
U = 1.04 Btu / (hr  tt2 r); or 5.92 wl(m2 c) computed
Utub : 1.O4 Btu / (hr  tι2 _r); Table 55a
or 5.91 \Ν l(m2 _c); Table 5_5b; Same result\ /'
522 (a) From Table 55 U=1.08 Btu / 1nr  ft2  F)
(b) Αssume tr"rn : 50 F; Δt = 10 F
Ras = 1.ol (nr f( r) r atu
' ., = ++ 1 .O1 = 1'94, Un = O.52 Btu/1ιlrft2F)Rn =UR', 1.og
523 (a) Uw = o.o89 Btu /(hr  tt2 _ 11 or O.51 W/(m2 _ c)τaοle 59
l l _ ο.o29 Btu / (hr  ft2 _F) or O'16 W/ ('_c) τante 51o,.Jfl ν'νLζ) lJιur\ιll ιL _ l ιJlv'l\J vYl1ιιl )
(b) Q=UΑ(ti _tg); tr= t"urA
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.._''r] ^_':'_^' ! ^'.'.ι ^'.4'ι
86
tavg =35'8 F (2'44 C) Table 511}Chicaαo.lllinois
A :22 F (12 C) Figure 57 ) *
523 (continued)
tg__35'822 = 13.8; tι:72re2c)q* = 0.ο89 (4 x 20 x7) (72  1 3'8) = 2,900 Btu / hr or 0.85 kW
qn = 0.029 (20 x20) (72  13.8) = 675 Btu /hr or 0.2 kW
524 R" = R1, Un = 0.029 Table 510
Rrin = 1l 0.48 Table 51a (Fibrous Pad)
Re = #"+ 2.08 = 36.6
U" = 0.027 Btu / (hr  ft2  F) or 0.16 Wl(m'z  C)
525 (a) R*=: +11+ (1 t g.1) = 22.6"0089 \
U* = 0.044 Btu/(hrftuF) or 0.25 w1m2c)
Rfι = : + (1 tO.4s): 36.60029 \
Un = 0.027 Btu/(hr  tt2 f) or 0.155 Wl(m'zc)
(b) Refer to problem solution 523
Q* = 0.044 (4 x20 x7) (72  13.8) = 1434 Btu / hr or 0'42k\^Ι
Qn = 0 027 (20 x20) (72  13.8) = 629 Btu / hr or 0.18 kW
526 Rins = # = 4'1z (rrr _f( _r)l εtu
C = O.24Btu/(hr 'f( F) or 1.36 Wl(m'z  c)
Excerpts fτom this work may be reproduced by instruοtors for distribution on a notforprοfit basis for testing οr instructional purposes only tostudentsenrolledincoursesforwhichthetextbοokhasbeenadopted. ΑnyοιherreproductionοrιrαnsΙαtionofιhisνοrkbeyondιhαιpermiιιedbνSectiοη.s1[]7ny ιnρn{t]ιo 1076'ΙΙb;l.)ζll"/^4'')ι"} 'l''':''' :"'''.Ι^''Ω'Ι l λ / .*'
87
Then from Fig. 58, U' = 0.85 Btu/(hr f( F) or 1.47 W(mC)
Q = U'P (ti  to) = 0.85 x 300172 101 = 15,8'10 Btu/hr or 4.63 kW
527 tι = 72 F (22 C) Assumed
R" = R5 + R1η, Ub = o.ο52 Btu / 1nr tt2 _ 11 Table 59
Rfi,.,, =R1 +Ru6*Rqyp=(5.0) + 0.0 +(1 12.22) = 7.22
R^ 1 +7.22=26.5 0.052
U" = o.o38 Btu / (hr ft2 _F) or o'22ννl(m'z_c)
528 Ub : 1.14 from Table 59
Rn = ++ O.7 + (1 t 12'6)= 1.66 (m2c)Λ//, Un = 0.60 W(m21.14
c)
or Un = O'1OO Btu / 1nr ft2 F)This does not account for the walls above grade.
529 U = 0'16 Τable 5'10 (no finish)
Rn:++(t69)+ (1 t4.6) =6.611m2c1 lw0.16 \ '
Un = 0.15 W/(m2C) or 0.027 Btu / (hr ft2  r)
530 q/A= Un(ti ts)= (ti tt)/Rt=(ttt)lR2Rl=Rgyp+R1nr+R1, R1 = (112.6)+ 0.7 +0.12= 0.90
tl = ti _UnR1(ti _tg) =20  [1.05 x 0.9 (2ο  10)]
EΧcerpts frοm this work may be reproduced by instructors for distribution on a notfor_pΙofit basis fοr testing or instruοtional purposes only to
students enτolled in courses for which the textbook has been adopted . Αny οther reproduction or trαnsltιιion of ιhis νοrk beyοnd ιhαt permiιιedhιi 9λ"};." ιηa ^'' ]Λ9 ^.}L ιoaA ι L;.^) c'''^' '.'':ι'ι ^'"^:' "'ι5υ{ιi ' . . .
88
531
t1= 14.6 C or 58F
R2 = Rrr, +Ri = ( I 12.6) +0'12='20
tz = 20_ [O.60 x O'2 (2o1ο)] = 18.8 c or 65.8 F
q/A=Un (ti _tg)=(tι _t'')/R1; Rl=Ri +Rc
= 0.12 + (1 I 4.6)  0.34
tι=20_ (O.15) (O.34) (2010) = 19.5 c or 67 F
532 C = 0.2 Btui(hrft2F); Figure 58
8 = U'P (t1 _to); Ui, = 0.81 Btu/ (hrftF) or 1.a W(mC)
U'ni = 1.37 Btu / (hr  ft F)
(a) q/P = 0.81 (70  5) = 52'7
(b) q/P = 1.37 (70  5) = 89'1
533 Q = Δt / R' ; Eq. 525; L>>Ζ'
R'=
Btu / (hr  ft) or 50.7 W/m
Btu / (hr  ft) or 85.6 W/m
L = 100ft
. Γzoo x121Γ^ tn(12x1OO/2x3o)l
_'nLo@J_3"100Ι12)
R' = 8. 12 x 1O3 thr  F) / Btu
of pipe wall.
Which neglects the resistance
Excerpts from this work may be reproduοed by instructors for distribution on a not_forprofit basis for testing or instruοtional purposes only to
Students enΙollod in courses fbr ινh'iοh the textbook hu, υ"., uJoi"a. Αny oιher reprολucιiοn οr ιrαnslαιioi οf this νork beyond ιhαt permitιed
L Q.nl;n" ιnη ^.' 'Λο
: '' ^ ' '1"4'ι "'
2πkL
6 7042  =3,4488tu / hr;' 8.12 x 1O'
orq=1.01 kW; 9=33 1W/mL
q/L = 34.488tu/(hrft)
534 Q=Δt /R'
535
R'g = 2π (1.4) 100=4.98x103 C/W
Neglect resistance of the inside film and the tube wall'
. 605 =11.04kwO = c' 4.98 x 10
Moisturewillmovetowardtheinside.Locatethevapor
retardent on the outer side of the insulation'
The insulation will beοome wet if the retardent is placed on the
inside or left out entirely and the plywood would probably \Λ/arp
and rot.
536 (a) Q/A = Uo(ti _to)=(tι _t1)/R1 =(ti _t)lR2
Ro = O'68 + O'45 +1 1 + 1'O +O'8 + O'17 = 14'1
Uo = O'o71 Btu /( ιrr  ft2 _F1
Rr = 0'68 + O'45 = 1'13( hr  f( F) / Btu
R2 = O'68 + O'45 +11 12'13(hr  ft2 F) / Btu
tt=tι_R1Uo(tι_to)=7O_(1.13xO'O71)(7O1O)=65.2F
Excerpιs Γrom ιhis \νοrk lnay be reproduced by insιructors for distribution οn a noιforproΓlι basis for ιesting or instructiona' purposes only to
students enrolled in courses foΙ ^γηι:ι lh. j:*tbook has *9,i'"Ιl'Ji."o"'i,:';;;;;;;;;;;"Ζii"111i117; γ,:y:γ:'::'"*o
ιyt.pern:ted
90
tz=70(12'13 x 0.071) (7010) = 18'3 F
(b) At 70 Fοο, 3O%o R.Η. and possible leakage of air to surfaces 1
or 2
tdp=37F<65FΠocondensationexpected
(c)Sincez=lS.3Fismuchlessthanthedewpoint,condensationwould oοcur'
Place vapor retardent at the location of interface 1'
537 Assume infiltration is negligible
ufAf (ti  t") = UwA*(t.  to)+ U'P(t.  to) = rilcp(tc  to)
* _ UrAtt, +(U*Α* +U'P+rhcp)to',ti=72F; to = 1o Fιc _
(UtAt + U*Α* + U'P + rhc, )
Αssume 1.5 in. of wood floor, Pine; Rwooο = '1'5l0'8 =1'88
U, =1; Rt =O'92+1'88+O'92 =3'72;Ut=0'27'Rf
UΛr= 0'27 x 30 x 60 = 484 Btu/(hrF)
rr a'R* =0.68+(6/15)+ O'17 =1'25',U* = O'80"*_Rw νv
U*A* = O.8O x 2(30 + 60)2  288 Btu/(hrF)
U'P=1.8x(30+60)2=324,h.p = 20 x 0.075 x 60 x o '24 = 21 '6 Btu/(hrF)
+ 484 x72 + (288 + 324 + 21'6)10
= 36.85 Fιc_ 484+288+324+21'6
538 (a) Q/A = U(ti to) = (trto)/R1;
Uz=0.112', Rr=O.17 + 0.33 + 4'17 + 2'22 = 6'89 (construction 2)
ExcΘrpts from this work may be rοproduced by instructors for distribution on a notforpτofrt basis foτ 1e^sting or lnstructional purposes only to '
students enτolled in courses fοr whiοh the textbo"t ι'", υ"* "J"p
t.d. ,ιny otlr", ,rproEu.ιiοn οr ιrαnslαιioλ of ιhis νork beyond ιhαt permitιed
91
t1 = (6.89 x 0.112) (720) = 55.6 F
(b) U : 0.211 (construction 1)
Rr = 0.17 + 0.33 +2'22 = 2'72
ti =to+R1U(ti to)=O + (2.72x0.211)(720) = 41.3F
(c) lf room air leaks into the air space for the case of no roof deck
insulation (b) there could be some condensation since t6p = 50
F at72 F and 45% RΗ. With the insulation, no condensation
would be exPected.
539 (ti to) / Rr = (ti ti) / R1
R, = 4'5 or O.79 1m2cyw ; Τable 54a (Const. No.1)
Rτ = 0.68 + O.45 + O.94 =2.O7 1nrtt2_F1/Btu or 0.365 1m2cμru
Between Furring and block
tl = ti _
ft,' _ to) = 22_ ffiιr'+17) = 3.98 C or 39 F
tdp = 9.5 C, Assuming room air can diffuse into the air space,
condensation likely will form on the concrete block surface.
Therefore, place vapor retardant on inside surface of gypsum
board. Use foil backed retardent. Retardent must not touch
concrete blocks!
Exοerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instruοtional purposΘS onιy to _
;;#;;;Ι;ln .o,,'",'to. *ni: lhΘ j:;boοΙ has been adopted. o'l
"!!1: ,.,r.':'!!i',?! ":::::i:::::":!,*r::!'Ρ'n! 'y,'o*Ψ*o ,
92
540 U1A1(titn) + UaΑ+(titn) = 2U3Α3(tnto) + lJzAzftnto)
,^ _ (UlAl + UzΑz )ti + 2UgΑgto + UzΑzto'' 2υsΑg + UzAz * UlΑl * UqAι
UrAr = 0.09 x 8 x 20 = 14.4',
UzAz = 0.09 x 8.54 x20 = 15.4',
U3Α3 =0.09x3x8
UιAι=0.09x3x20=5'4
r _(4.4+15.4)70+(2 x 0.8 x o) + (15.a x o)L6= S5.8F" (2 x 1.08) + 15.4 + 14.4 + 5.4
Place water pipes in this space with some caution.
Uf Αf (titb ) = (U*Α* + UυtΑt )(tυ tg ) ; ti = 72 F
tg = turg Α  37.6 _23 = 14'6 F or B C
1Ut =
& ; Rf = (2 x 0.92) + (1 5/0.8) + 2.1 = 5.82
Carpet and Fibrous pad assumed, Ur = 0.172 Btu/ (hr  ft2  F)
U* =.164 Table 5 _ 9; Uot = 0.029 Τable 5 _ 1o
ltΑrti + ( U*Α* + Ubf Αf )tg
= 1.08
541
tb=U1Α1 + U*Α* *UυrΑr
r, _(0.172x400 x72) + (.1G4 x 80 x 7 + 0.029 x 4OO)1a.6l'b
to = 30.3 F or 0.95 C
Exοerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instruοtional purposes only tostudents enrolled in courses fοr whiοh the textbook has bοen adopted. Αny oιher reproλucιion or ιrαnslαιiοi qf ιhis νork beyond ιhαι nermiιιedby Seαiοns Ι 07 οr ! 08 n! n." t o7A t Ιr;lDs <}t^ .^^',]Ll λ οt 1η: ιΙ^^...'Ι 
Excerpts frοm this work may be reproduοed by instructοr9 jor distribution on a notforprοΓrt basis fοr
testing or instructiοnal purposes on1y to studeπs enrοlled in οourses for whiοh the textbook has been
adopted. Αny other reproduction or trαns,lαtion of this work beyond thαt permitted by Sections ]07 or
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Requests fοr permissioln or furthir"infοrmαtion should be'αddressed ti the Peimission Depαrtment' John
iiria sonλ, nr, ] 1 ] Riνer Street, Hoboken' NJ 07030'
61
CHAPTER 6
Refer to Table B1 . The computer program PSYC may be used to find the
humidity ratio from t66 and assumed 100% RΗ'
Design relative humidity is determined by possible condensation on inside
of glass. Find glass surface temperature (which is the maximum dew
poi;t temperature of the inside air allowed)'
q/A = U(t'  to) = Cr(tr  b)
tι= 72"F; t., = glass Surface temperature
U = 0.65 Btu/(hrft2F)' Table 55a
111, hi = 1.46 Btu/inrft2f)
c1 uhiCι = 1'172 Btυl (rrrft2F1
62
WindDirection, deg.
CCW from N0.0
0.0
0.0
0.003
ο.0
0.003
140
290
10
340
360
20
6
13
4
12
I7
11
2
9
24
18
28
(a) Pendleton, OR
(b) Milwaukee, Wl
(c) Anchorage, ΑL
(d) Norfolk, VA
(e) Αlbuquerque, NM
(f) Charleston, SC
94
tr=Uti + to (Cr  U)
cl
63
* RH = 60 o/o would probably be uncomfortableRΗ = 40 to 50% would be more realistic
Assume that the weather strip does not change the conveοtive heat loss.
From Figure 62, Cp = 0.3. Using Eq. (67b) with the air density of 0 'F,the pressure difference due to wind is
ο: [o
.086ψ!\'( ls*ot * .467 fi l'\ft')\ mph)
ΔP. =
z.(y.rrΙbm_ ft)ι lbf_s')
(o.rnr.o t"'tt )ι lbf l ft')
ΔP* = 0'037in'wg
Αssuming slight stack effect, ΔP ^y
0.04 in. water
Using Table 61 and Fig. 61,
Exceφts from this work may be reprοduced by instruοtors for distτibution οn a notforprofit basis Γor testing or instructional puφoses only to
studentsοnrolledincοursestbrwhichthetextbookhasbeenadopted. ΑnyοtherreproducιionorιrαnslαιionofιhisνοrkbeyondthαtPermiιιedby Sections Ι07 or Ι08 ofthe ]976 United Stαtes Copyright Αctνιthout the permιssiοn ofιhe copyrighι oνner is unlaννful
CitylndoorTου, oF
OutdoorΤου, oF
tr=top,
oF
Design orMax.
RH%(a) Caribou, ME
(b) Birmingham, ΑL
(c) Cleveland, oΗ(d) Denver, CO
(e) San Francisco, CΑ
(g) Boise, lD
Rapid City
72
72
72
72
72
72
72
10
23
b
3
39
'16
I
35.5
50.2
42.6
41.3
57.3
32.8
44.0
26.2
46.1
34.6
32.9
59.9*
23.6
36.5
95
Loose fit with non\Λ/eatherstripped, K  6; a lL = O.75 cfm/ft
Loose fit with weatherstripped, K= 2; Q/L = O'24 cfmtft
Total length of crack, ι = [(3 x 3) + (2 x 5)] x 9 = 171 ftUsing Ll2for calculation, then
Q,, = 0.75 x 17112= 64.1 cfm, Q, = 0'24x17112= 2O'5 cfm
INow Q, = rh cr(t1  to) = v cp(t;  tr)
Q,r8,, At4, Q,, Qt
ΔP., =
64.t20.564.r
= 0.68
64
or a reduction of 68% in sensible heat loss.
Also, (Kl  K)lΚ1= β2)16 = 0.67 or 670/o Reduction.
From Fig. 62, Cp = 0'52tor windward wind'
Assuming standard sea level air density, the pressure difference due to
the wind speed of 13 m/s is
_ 53.6Pα
z.( ι.okg _ *
t i/s'
(a) From Table 62, K = 1 for tightfitting.
Then, from Fig. 61, Q/L = 0.60 L/ms
Q = 0.60 x (0.9 + 2'0) x2 = 3l8_LΔ
Αssuming that the wind speed and wind direction are the Same as the
given conditions for the bank at Rapid City, SD, the heating load (at 20'6
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students enτol1ed in οourses fbr which the textbook has been adopted. Αny oιher reprοiucιion or ιrαnslαιion οf ιhis ννοrk beyond ιhαt permitιed
by Sections 107 or ]08 ofthe Ι976 rJnited States Copyright ΑcιΙνilhouι ιhe permission οfιhe copyrighι owner is unlαwful'
Βry]s)
a, = (: + ε * ο. o ο 1
Τ ) (' .#l
ι r r,
* }r)rr, _ 1_zo.o1)" c = 393.tW
σ6.C outdoor temp. and 72'C indoor temp.) due to the door infiltration can
be calculating using Eq. (62b) as:
(b) From Table 62, Κ = 2 for averagefitting'
Then, from Fig. 61, Q/L = 1'25 L/ms
Q = '1 .25x (0.9 + 2.0) x 2=725Lls
α, _(', .zs* 0 OO'+) ('.rrfil ι rrrn?)o, _ e20.6))" C = 819 0W
(c) From Table 62, K = 6 for averagefitting'
Then, from Fig. 61, Q/L = 3.40 L/ms
e = 3.40 x (0.9 + 2.0) x2 = 19f2Lls
α, =(lν τ2* '001c) ('.#l ιrr, *}a)o'_e20.6))"C
_2227'6W
From Figure 62, Cρ = 0.52' Using Eq. (67b) with the standard air density,
the pressure difference due to wind is
r2
Λ p _ o'sz (o oτ οs#) (zzen ι κl #)' ( o.rnro :n:γ
r, ^\ι\ΓΙυ ( b^_f) ι lbflft')
z. ιz.17':::_! 
ι lbf _s'z)
ΔP' = 0't35in'wg
Neglecting stack effect and pressurization, ΔP  0.135 in' water
From Table 61, K = 2 for averagefitting with nonweatherstripped.
From Fig. 61, Q/L = 0.60 cfm/ft.
L" = [(3 x2.5) + (2 x 4)]x3 = 46.5 ftExcerpts from this work may be reproduced by instruοtors for distribution on a notfor_profit basis for testing or instruοtronal puφoses only tο
studentsenrοlledinοourses1brwhiοhthetextbookhasbeenadopted. Αnyotherreproducι'ιonorιrαnsιαιionοfthisτνοrkbeyondιhαtpermiιted
by Sectiοns 107 or ]08 oftnn isri {Jniιed SιαιeS Copyrighι Αcιτνιιhοuι ιie permisiiοn ofιhe cοpyrighι oνner is unlαwful'
65
bb
Q = 0.60 x 46.5 = 27.9 cfm.
(a) The wind effect is assumed to be independent of height and pressure
differences due to wind are the same as those given in Ex. 61.
3'd Floor: ΔP"/Cο = 0.037; ΔP, = 0.037 x 0.8 = 0.03 in. water
orientation ΔP, ΔP* ΔPτWindward 0.03 0.066 ο.ο96SidesLeeward
0.03 0.066 0.0360.03 0.033 0.003
gth Fιoor: ΔP./Co _ 0.100; ΔP,  0.100 x 0.8 = 0.08 in. water
orientation ΔP. ΔP, ΔPτWindward 0.08 0.066 0.ο14
SidesLeeward
0 08 0.066 0.'146
0.08 0.033 0.1 13
(b) For Bitlings, MT, design conditionS are to = 7oF, tι= 72"F, Φι= 28o/o.
From Table 63, K = 0.66 for conventional οurtain wall.
Αir will infiltrate on windward side only on 3'd floor.
Windward  3E floorQiA = 0.15 cfm/ft2; Q = 0.15(120 x 10) = 180 cfmThen 9 "
= (1 80 x 60/1 2.4)(0.24)(72  (7)) = 16,514 Btu/hr
Q,. = (180 x 60/12.4)(0.005  0.000)1060 = 4,616 Btu/hr
Qt = Q, + 8r. = Ζ*13oBtu1hΙ [3'd Floor]
gth Floor  All exfiltration on this floor.
Qt = oοΞtuΔr 19th FlooηEXοerptS from this wοrk may be reproduced by instructors for distributioIr on a notforpro1it basis for testing or instruοtional puφoses only tο
students enrolled in courses tbr which the textboοk has been adopted. Αny οιher reprotluctiοn οr trαnslαtion of this νork beyond ιhαι permitted
by Secιions Ι07 or ] 08 of ιhe Ι 976 t'Jnited Sιαtes Copyrighι Αcι 1υiιhouι ιhe permission οf the cοpyrighι οlυner is unΙιrννful.
98
67
(a) Windward Doors: Double vestibule type
ΔPτ = o.146 in' water, assume 1/8 in' οracks
Q/L = 16 cfm/ft tFιg. 67] , L  32 ft [Ex' 62]
Q = 16 x 32 x0'7 = 358 cfm
(Assume30o/oreductionforvestibuledoors)
67 (Cont.)
Side Doors: Double vestibule tYPe
ΔPτ = o'o52 in' water' 1/8 in' cracks
Q = O.O (negative pres' dιff')
(b) Windward: ΔPτ = O146 in' water' K = 0'66 lΤable 63]
Q/A = o.2o cfm/ft2 [Fig. 66]
A=120x10=1200ft2Q = 0.29(12ο0) = 240 cfm
Sides: ΔPτ = O'O52 in' water' K = O'66 [Table 62]
8 = O.O (negative pres' dιff')
UULeeward: ΔPτ = O'047 in' water, K = O'66 [Fig' 66]
Q/Α = O.1Oo cfm/ft2, A = 1200 ft2
Q = O.1οO(1200) = 120 cfm;
Total infiltration for the walls is
8* = 240 + O.O + 120 = 360 cfm
(c) Totat infiltration is sum for doors and walls'
From Εx.62'for leeward door' Q = 179 cfm'
Then the total door filtration is
Qo=358+179=537cfm(neglectinfil.duetotraffic).Andthetotalwallinfiltrationis360cfm,thenQτ=897cfm
Εxcerpts tiom this wοτk may be reproduced by instτuctors foτ dlstribution on a notforprofit basis for testing oΙ instructional puφoses only to
students enιolled in οourses tυ. *ι'λ}, the textbook has been uaoρt,i 'ιny ornr, ,rprorturιιon'o''trαnsΙαtιonξthιs νork beyond ιhcιt permiιιed
by Sectiοns 1 07 or ] 08 o7 ιn' to:ii'inιird Sιαιes Cοpyrιgh' 'ι'i"rii'i"'' 'i" perission of ιhe cοpyright oινner is unΙανfuΙ'
68
9V
For Charleston, WV: to = 1 1oF' ti = 70oF
q, = (897 x οοll \ 'τz)(o'z4)(7011) = 65'ο25 Btu/hr
q. = (897 x 60/1 1'72)(1060)(O'OO5  O OOO) = 24'338 Btu/hr
q = q" + 9. = 89,363 Btu/hr
(a) Assuming standard sea level air density, the pressure difference due
to the wind sPeed of 20 m/s is
( o.o,urtψ\'(zo*pt *1.461ιL:\ /
,,_ υ.Ι9/+ tbfl fi,J =o 197in,water
UUWindward: ΔP*=O197xO25=ooo:
χ } E:lι, n*l:Leeward: ΔP* = O"197 x (0'5) = 0'099 in
AssumPtions:1) temperature differencΘ, tι _ to' = 40oF
zi tιle neutral pressure level is at floor 9'
3) the floor height is 12 ft', and
4) Cο = 0'80'
Then, from Fig.65,
Floor 1: h = 108 ft., ΔPr/Co = O.13, and ΔP" = O'13 x O'8O = 0'104 in' water
Floor 5: h = 60 ft., ΔP, = 0'065 x O'8O = 0'052 in' water
Floor 15: h = 72ft., ΔP, = O'O85 x 0'8O = 0'068 in' water
Floor 20: h = 132ft., ΔPS = Q'160 x O'80 = 0j28 in' water
Windward Leeward
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students enrolled ιn couΙSes tor whiοh the textbook *' γγ119iJ]1'^"'Ιi";:i:j!i::;:;:;:':""":jfi,':i:,i{:x:λx#,.beyond ιhαι permιtted
;}i::j;μιfii,Z'i'"fλ!"in:ii'ini,a s**, copy,ign,'a;,\iιiio"i''Ι, i,.ι'in of ιhe coρyrighι owner is unlαwful
& r *,,
Flnοr ΔP* ΔP" ΔPτ ΔP* ΔP" ΔPτ
1
5
1520
0.0490.0490.0490.049
0.1 ο40.0520.0680.128
0.1 530.1010.0190.ο79
0.099ο.0990.ο990.099
0.1040.0520.0680.128
0.0050.0470.1670.227
100
(b)
68 (Cont.)
o.2 ο.1 0.0 ο.1 Ε.2
ΔR. ilt. ιlrateι
*E*lΛfiΠ$ffard ".,t  LEθΛrard
lnfiltration  Windward Sides, from 1't to 13th Floor
Leeward Sides, 1't Floor onlY
Exfiltration _ Wind\Λ/ard Sides, from 14th to 2oth Fιoor
Leeward Sides, from 2nd to 2Oth Floor
(c) 1rt floor, lnfiltration on all sides  through doors, walls and fixed
windows
Windward Walls: from Table 63, K = 0'22 for tight fitting.
f rorn f ig. 66, Q/A = O.OB cfm/ft2'
A=(1οο+60)1 2=1920f(Q = O.O8 x 1920 = 154 cfm
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.* \λd& tl
#
1υ1
Leeward walls: from Table 63, K = 0'22 for tight fitting'
From Fig. 66, Q/A = O'OO5 cfmlft2'
A = (1OO + 60)12 = 1920 f(Q =0.005x1920= 10cfm
windward Doors: from Fig. 67, Q/L = 17 cfm/ft for 1/8 in' crack'
68 (Cont.)
For vestibule doors, assume a 35% reduction'
Q/L = 17 x 0.65 = 1 1'05 cfm/ft
L = (3 x 6.75) + (2 x 6) = 32'25 ft
Q = 11.05 x 32.25 = 356 cfm
Leeward Doors: from Fig. 67, Q/L = 1.5 cfm/ft for 1/8 in' crack'
Γor vestibule doors, assum e a 35o/o reduction'
Q/L = 1.5 x 0.65 = 0'975 cfm/ft
Φ = O.975x32'25 = 31 cfm
Then,totalinfiltration(neglectingtrafficeffect)isQtot = 154 + 1O + 356 + 31 = 551 cfm'
(d) and (e) lnfittration rate is zero due to negative pressure differentials for
the 1Sth and 2οth floors'
69
For Minneapolis, MN: to = 1 1oF' t1 = 70"F '
[Note:Δt=70(11)=81oFisinconsistentwithProblem68whereΔt = 40"F was used; however, an error is assumed to be minor]
(a) From Prob. 68, Qτ = 551 cfm for 1't floor'
q, = (5s1 , dotl 2'15)(0'24)(70  (11)) = 52'896 Btu/hr
o, = lbSt x 60/1 2.15)('1060)(0.O04  o.ooo) = 11,537 Btu/hr
Qt= Qr* Qr. =64,433 Btuihr
(b) and (c) qt = O'O due to zero infiltration
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by Sectiοns 107 οr 1 08 οf the Ι 97 6 t]nited SιαιeS δ"iyriii ii 'i,thout
tie permιsiιon of the copyrighι owner is unlα:wful'
102
610
For Des Moines, lA: to = 4oF, ti = 70oF'
Transmission heat loss (negtecting infiltration) through windows, doors,
walls, and roofs can be determined by Eq. 519 as:q = UA(tι to)
Windows: A = (3 xa)1 2= 144ft2;
From Table 55a, U = 0.55 Btu/(hr  ft2 'F1;q = 0.55(144)(70  (4)) = 5,861 Btuihr
Doors: A = (3 x 6.75)1 2=243ft2',From Table 58, U = O'28 Btu/(hr  ft2  'F);
(assume panel with metal storm door)q = 0 28 (243)(70  (4)) = 5,035 Btu/hr
wails: A = 8[(36 + 64)21 144 60.75 = 1395.25ft2',
From Table 54a, U = 0.14 Btu/(h r ftz  'F);
Q = O. 14(1 395. 25)(70  (4)) = 14,455 Btu/hr
Roof/Ceilinq: A = 36 x 64 = 2304 ft2',
Ξrorn Example 53, U = O.83 Btu/(hr _ ft2 _ 'F1;
q = 0.083(2304)(70  (4)) = 14,151 Btu/hr
Transmission heat loss through the slatongrade floor can be determined
bY Eq 523 as:Q = U'P(ti to)
Floor: p = (36 +64)2=200ft;u' = o.8o Btu/(hr  ft  F), from Fig' 58 (assume insulation Rvalue
of 5.4 (hr  ft2 'F)/ Btu and d = 2ft)'q = 0.8(2OOX7O  (4)) = 11'840 Btu/hr
Finally, total transmission heat loss is the sum of all heat losses;
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61 1
612
LP, 
ΔP* =
qt = 5l.ΞΖBtuΔt
From Figure 62, Cp= 0'52. Using Eq'
the pressure difference due to the wind
103
(67b) with the standard air density,
of 15 mph is
0.0765ψy\fr' )
(0 5η[
9s =
Qr=
Qt=
z.(nlιlbry_ ι_ ι lbf _s2
0.058 in. water
For a lowrise building, neglect stack effect and pressurization' thus
ΔP1 = 0.058 in. water
From TabIes 6'1 and 62, Κ = 1 for tightfitting windows and doors'
From Fig. 61, Q/L = O.'13 cfm/ft'
L" = [(3 x 3) + (2 x a)]x3 + (3+6.75)x2x3 = 109'5 ft
e = 0.13 x 109.5 = 142 dm,
(14.2 x 60/1 2.15)(0'24)(70  (4)) = 1,245 Btu/hr
(ιι'zx60/12'15)(1ο6OXο.Oο5_0.00ο)=372BtulhrQ, + Qr. = 1,617 Btu/hr
For Ηalifax, Nova Scotia: to = 2oF, ti = 70oF'
Refer to Problem 610 for other data'
Windows: q = 0.55(1 44)(70  2) = 5'386 Btu/hr
Doors:q=0.28(243)(702)=4'627Btulhrwails: Q = 0.14(1 395. 25)(70  2) = 13,283 Btu/hr
κootrcuιιιno: q = O.083(23O4)(70 2)= 13'004 Btu/hr
Γ*r, q = O.8(2OοX70 _2) = 10'880 Btu/hr
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/ii i"i.d Sror", copyrιsnt ii 'Ιiiiouι ιi" per*isiion of ιhe copyrighι οwner is unΙαwful'
104
613
614
615
616
Total: qt = 4Ζl€oBtu/hr
Memphis, TN; to = 21 oF; ti = 70"F
R*= 0.92+ 1.55 + 0.99 + 1.77 +0.17 = 5.4 (Tables 51 a'52a)
U* = 115.4 = O.'185 Btu/(hr  ft2  'F)Us = 0.81 Btu/(hr  f(  "F) (Table 55a)
Αs = 6xax3 = 54 ft2
n* = (40xl O)54 = 346 ft2
q,= 0.185 x 346 x (70  21) = 3,136 Btu/hr
qs = 0.81 x 54 x (70  211= 2,143 Btu/hr
Qtotrl = 5zβ auk!
Concord, NH; to = 2F, ti = 70oF
R*= 5.40.99+ 3.0=7.41U* = ο.135 Btu/(hr _ ft2  "F)
Us = 0.81 Btu/(hr f(  'F) (From problem 613)
q* = 0.135 x 346 x [70  (2)]= 3,363 Btu/hr
qn = 0.81 x 54 x (72) = 3,149 Btu/hr
Qtotrι = οβ1ΖBuer
lnstructor suPPlies solution.
(i'  iu)(a) q=9svs
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105
. qν" (280'00Ο)(14.6)Ο^= :=:_ =^251 cfm19(S _
i._iu β2.7 _21.8)60 v!
(b) q = rh cp(t,  t..) = 9ε
cp(ts  tr)vs
. qν, (250,00Ο)(14.6)Q' = ;r=δ= (O24X1 15_7O)ω = 5'633 cfm
617
SHF = Q./(9. + α") = 100'999 ===,
= 3.O34st (1 33, ooo  1 oo, ooo)
Locate states, and οondition line and heating pΓocess on psychometric
chart.Q. = rh cp(t, _ tr) or Γh . = q
'/cr(t, _ t')
Γh. = 100,000/(0'24 x20) = 20,833 lbm/hr
Q. = Γh. X vr/6O = 20,833 x 14.05/60
Q" = 4,878 cfm or about 4,900 cfm
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106
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tffini
p
71
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Chapter 7
First, find longitude from Table B1a
Then, convert Daylight Saving Time to Local Standard Time using Eq. 75
Next, determine the equation of time from Table 72
Finally, determine Local Solar Τime using Eq. 76
Τhe following table Summarizes the solutions of the problem.
Norfolk, VA 76.2Lincoln, NE 96.75
Casper, WY 106.47
Pendleton, OR 118.85London, UK 0.45
75 9:00:00 ΑM90 1:00:00 PM105 10:00:00 ΑM120 3:00:00 PM0 7:00:00 PM
Local^."":' Eouation LocalSolarδtanοarα : 'oτ llme llmeilme
8:00:00 ΑM 2'41 min 7:52:47 A'Λ
12:00:00 PM 2.41min 11:30:35ΑM9:00:00 AM 2'41 min 8:51:43 ΑM2:00:00 PM 2'41min 2:02:11PΜ6:00:00 PM 2.41min 5:55:47 PM
Standard DaylightLocation Longltuαe' Meridian, Savingsoνν Τime
72Ηour angle (negative for morning and positive for afternoon) can bedetermined by
h:Ι5* (LST _l2)
(a) h = 15*(8.19 '12:00) = 15*(3.683) = 55.25 deg.
(b) h = 15*(10:03  12'.00) = '15*(1 .950) = 29.25 deg.
(c) h = 15*(15:46 12:00)= 15*(3.767)= 56.50deg.
107
(d) h = 15"(12.01  12:00) ='15*(0.017)= 0.25 deg'
73 Αt sunset and sunrise, β =0"; sin(B) = ρ
From Eq. 78; οos(/) ο οos(h). οos(δ)  _sin(/). sin(δ)
οos(h,,) = οos(ft,,) = _ tan(/)' tan(δ)
The following table summarizes the solutions of the problem.
Location Latitude,'N 'i"J;::',3" Cos(h) ^ffi::.
',1:ff ',tι''iBillings, MT 45.8 20.6 0.3865 112.7 4:29 AM 7:30 AM
orlando, FL 28.43 20.6 0.2035 101.7 5:13 AM 6:46 AM
Anchorage, AL 61.17 20.6 0.6829 133.1 3:07 ΑM 8:52 AM
Honolulu, Ηl 21.35 20.6 0.1469 98.4 5:26 AM 6:33 AM
Note earlier sunrise at greater latitudes
74/ = 33.0 deg. N
h = 15.(912) = 45.0 deg.
on Sep 21, δ = 0.0 deg.
From Eq. 78; sin(B) : οos(/). οos(ft). οos(δ) + sin(/). sin(δ) = 0.593
β = 36'37 deg'
sin δοos/ _ cos δsin / οos hFrom Εq' 711 ; υυs ψ =
"o,
βcos{ = 0.478
Φ = 118.57 deg. (clockwise from north)
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108
75At sunris e, β _ 0"; sin(B) = g
From Eq. 78; οos(/) ' οos(h) ' οos(δ) _ _sin(/) ' sin(δ)
οos(h,. )  _tan(/)' tan(δ)
(a) June 21'' δ 23'45 deg; / = 58 deg'
cos h = O'6942; h = 133'96 deg' or 8'93 hours
Sunrise is at 3:04 AM (Solar Time)
76
= O.751
Φ = 41.33 deg. (clockwise from north)
(b) Dec21'' δ= 23'45 deg; / = 58 deg'
cos h = 0.6942; h = 46'04 deg' or 3'07 hours
Sunrise is at 8:55 AM (Solar Time)
. sinδcos/ _οosδsin lcoshFrom Εq'711; οosΦ=" =0'751
Φ= 138.67 deg. (ctockwise from north)
Maximumsolaraltitudeangle,βwilloccuratsolarnoon,h=0
^^^ λ_ sin δ cos /  οos δ sin / οos h
From Eq. 711; ν"oo _ cos β
77
]09
From Eq. 71O, β^u*:90 _ Min\, _ u,)
From Table 72, δ'u"l = 23.45
(a) Denver, CO: I = 39.75 deg. N.
For north latitude, / is positive and greater than δrrr so we need largestpositive value of δ.
From Table 72, δ'u" = 23'45 deg. and hence β'", = 73.70 deg.
Therefore, maximum solar altitude angle occurs at solar noon on June 21.
(b) Lansing, Ml: l= 42.77 deg. N.
For north latitude, / is positive and greater than δrrr so we need largestpositive value of δ.
From Table 72, δ'u"= 23.45 deg. and hence β'u"= 70.68 deg.
Therefore, maximum solar altitude angle occurs at solar noon on June 21.
(c) Sydney, Αustralia: / = 33.95 deg. S.
For south latitude, / is negative and / is greater than lδ'u"l So \Λ/e need
largest negative rralue of δ.
From TabΙe 72, δ'u, = 23'45 deg. and hence β'", = 79.50 deg.
Therefore, maximum solar altitude angle occurs at solar noon on Dec 21.
Longitude'. Lt= 100 deg. W
Local Standard Time. LCT = 3:30 pm
on Nov 21, Eoτ = '13.8 min
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110
78
79
Using Eq. 76, LSf = 15.50  (10090).4 /60+ '13.8/60 = 15.063 Hr or 3:04pm.
Latitude: l = 37 '5 deg. Ν
Ηour angΙe: h = 15('15.06312) = 45.95 deg.
on Nov 21, δ = 19.8 deg.
Using Eq' 78 to calculate solar altitude, β= 21.36 deg.
Then using Εq.711 to calculate solar azimuth; Φ = 226'56 deg. (clockwisefrom north)
Surface azimuth; ψ = 12+180 = 192 deg. (clockwise from north)
Finally, using Εq.712 to calculate wallsolar azimuth
y= 1226.561921= 34.56 deg.
Using Εq' 713b to calculate angle of incidence for a vertical surface
θ = 39.92 deg.
Using Εq' 713a to caIculate angle of incidence for an inclined surface
For surface tilt = 70", θ= 32.30 deg'
For Ottawa, Ontario on July 21,
Longitude. Lt= 75.67 deg. W
Latitude: Ι = 45.32 deg. N
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Equation of Τime: ΕoT = 6.2 min
Decination: δ= 20.6 deg.
(a) Eastern Daylight Savings Time: EDSI = 4:00 pm
Using Eq. 76, LS7 = 14.852 Hr or 2'.51 pm.
Hour angΙe: h = 15*(14'85212) = 42.78 deg'
Using Eq. 78 to calculate solar altitude, β= 47"16 deg.
Using Εq'713c to calculate angle of incidence for a horizontal surface,
θ= cos1(sin(47.16)) = 42'84 deg.
(b) At sunset, β = 0 and sinp = Q
cos(fr ): _tan(/). tan(δ)
Hour angle: h = 112.34 deg.
Solar time at sunset: LSf = 12 + hl15 = 19.49 hr or 7:29 pm.
Εastern Daylight Savings Τime can be calculated by
ΕDST = LST + (L' _ ΕSη(4min/degt4l)  ΕoT +L
EDSΓ = 19.49 +(75.6775)(4/60)(6.2/60)+1 = 20.638 hr or 8:38 pm.
710For Philadelphia, PA on July 21,
Longitude'' Lι= 75.25 deg. WLatitude: / = 39.88 deg. NEquation of Time: ΕoT = 6.2 minDeclination: δ= 20.6 deg.Eastern Daylight Savings Time: EDSI = 10:30 am
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lνΛf.l^ιi^,^ ^{lL]d 'ιl^yL λo1'^') tι1nt ^trmιtted
Using Eq. 76, LS7Γ = 15.852 Hr or 3:51 pm.
Hour angle: h = 15(15.85212) = 57.78 deg.
Using Eq. 78 to calculate solar altitude, β = 49'42 deg.
Using Εq' 711 to find solar azimuth; Φ = 114.30 deg. (clockwise from north)
(a) Using Εq' 713c to calculate angle of incidence for a horizontal surface,
θ = cos1(sin(49.42)) = 40.58 deg.
(b) For vertical surface facing southeast, Surface Tilt; α = 90 deg., and
Surface azimuth; ψ= 135 deg. (clockwise from north)'
Using Εq' 712 to calculate wallsolar azimuth , y= 114'3135ι = 20.7 deg'
Using Εq' 713b to calculate angle of incidence for a vertical surface,
θ = cosl(cos(49.42)cos(20 7ο)) = 52'52 deg'
(c) For inclined surface faοing south, Surface Tilt; α = (9040) = 50 deg.,
and Surface azimuth', ψ= 180 deg. (clockwise from north)'
Using Εq' 712 to calculate wallsolar azimuth, y= l114.3180 = 65.7 deg.
Using Εq.713a to calculate angle of incidence for an inclined surface,
d = cos1 (cos(49.42)cos(65.70)sin(50)+sιn(+g.42)cos(50)) = 46.'t 1 deg.
711
112
712
713For Calibou, MΑ on July 21,
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113
Longitude' Lt= 46.87 deg. WLatitude: l= 68.02 deg. NEquation of Time: EOT = 6.2 minDeclination: δ= 20.6 deg.Solar Parameters; Α = 346'4 Btu/hrft2 or 1093 Wm2, B = 0.186,
and C = 0.138Eastern Daylight Savings Time: ΕDSr = 2:00 pm
Surface Tilt; α = 60 deg.Surface azimuth, SW; ι/ = 225 deg' (clockwise from north)
Using Eq. 76, LSI = 14'72 Ηr
Hour angle: h = 15.(14.7212) = 41 .58 deg.
Using Eq. 78 to calculate solar altitude, β = 36.04 deg.
Using Εq'711 to find solar azimuth; Φ__ 230.2 de9. (clockwise from north)
Using Εq.712 to calculate wallsolar azimuth, y= 5'2 deg'
Using Εq' 713a to calculate angle of incidence, θ = 7.45 deg.
Using Εq' 715 and clearness number of 1, Gryρ = 252'51 Btu/hr_ft2 or796.75 Wm2
Using Εq' 716a, Gρ = 250.28 Btu/hrft2 or 790'03 Wm2
Using Eqs, 718 and 72O, Gα= 26'13 Btu/hrft2 or 82'46 Wm2
Therefore, total clear sky irradiation is276.51 Btu/hrft2 or 872.49 Wm2
714Given lnformation:
Date: June 21Longitude' Lt= 96'0 deg' WLatitude: / = 36.0 deg. NΕquation of Τime: ΕoT = 1 .4 minDeclination: δ= 23.45 deg.
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114
Solar Parameters; Α = 346.1 Btu/hrft2 or 1092 Wm" ' B = 0' 1 85,
and C = 0.137Central Daylight Savings Time: CDSI = 8:00 pm
Surfaοe Τilt; α = 90 deg.
Surface azimuth, SW; ι/ = 225 deg' (clockwise from north)
Reflectance from water; Ps = 0'25
Using Eq. 76, LSf = 18.58 Hr
Hour angle: h = 15(18.5812) = 98'65 deg'
Using Eq. 78 to calculate solar altitude, β = 7 '02 deg'
Using Εq' 715, G,νρ = 76'24 Btuihrft2 or 240'5 Wm2
lrradiation reflected from the ground can be determined by
Gn: PrF.r(sinβ +C)G'o
where F'ncan be determined from Εq'724'
ThereforΘ, Gκ = 2'47 Btu/hrft2 or 7'8 Wm'
715Given lnformation:
Date: lιΛar 21
Latitude: / = 56.0 deg. NEquation of Time: ΕoΤ = 7'5 min
Declination: δ= 0.0 deg.Solar Parameters; Α = aoε.g Btu/hrft2 or 1164\Nlm', B = 0'149'
and C = 0.109Local Solar Time: LSf = 12:00 PmSurface Tilt; α = 90 deg'Surface Azimuth, S; ψ= 180 deg' (clockwise from north)
Clearness number; CN = 0'95Diffuse Reflectance from Sno\Λ/; ρn = 0'7
Hour angle: h = 0.0 deg.
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115
Using Eq. 78 to calcuΙate solar altitude, β= 34'0 deg'
Using Εq' 711 to find solar azimuth; Φ = 180.0 deg. (clockwise from north)
Using Εq' 712 to calculate wallsolar azimuth, 7r= 0.0 deg.
Using Εq' 713b to calculate angle of incidence, θ = 34.0 deg.
Using Εq.715, Gtvρ = 268.5 Btu/hrft2 or 847 '1 \Νlm2
Using Εq' 7 16a, Gρ = 222'6 Btu/hrft2 or 702'3 Wm'
Using Εqs' 721 and 722, Gα = 33.O Btu/hrft' or 104. 1 \Νlm2
lrradiation reflected from the ground can be determined by
G^: PrF'r(slnβ+C)G'o
where F*n can be determined from Εq.724'
Therefore, GR = 62'8 Btu/hrft2 or 198.1 \Νlm'
716Given lnformation:
Date: Aug 2'1
Latitude: I = 32.0 deg. NEquation of Time: ΕoT = 2'4 minDeclination: δ= 12.3 deg.Solar Parameters; Α = 350.9 Btu/hrft2 or 1107 \Λllm2, B = 0'182,
and C = 0.134Local Solar Time: LSf = 10:00 amSurface Tilt; α = 45 deg.Surface azimuth, SW; ι/ = 225 deg. (clockwise from north)Diffuse Reflectance from ground; ρn = 0.3
Hour angle: h = 30.0 deg.
Using Eq. 78 to calculate solar aΙtitude, β = 56.1 deg.
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116
Using Εq.711 to find solar azimuth; Φ= 118.7 deg. (clockwise from north)
Using Εq.712 to calculate wallsolar azimuth, r= 106.3 deg.
Using Εq.713b to calculate angle of incidence, θ = 61.5 deg.
Using Εq' 715, Gruo = 281'8 Btu/hrft2 or 889'1 Wm2
Using Εq'716a, Gρ = 134'4 Btu/hrft2 or 424'0 Wm2
Using Eqs. 718 and 72o, Gα= 32'2 Btu/hrft2 or 10'1 '7 \ΝΙm"
Using Eqs. 723 and724, GR = 11'9 Btu/hrft2 or 37 '7 \'ΙΥlm2
Using Eqs. 725 , Gt= (34.4 + 32.2 + ',11.9) = 178.6 Btu/hrft2 or
= (424.0 + 101 .7 + 37.7) = 889.1 Wm'z
717The following results are determined from a computer program employing
equations in the book from Eqs. 76 to 726'
Following tables summarize input and output data calculated for southwest
facing vertical window at32 deg. N latitude, 90 deg. W longitude, for all
daylιght hours of a clear day on July 21with ground reflectance of 0.2 and
clearness number of 1.
lnput DataLongitude 90 deg
Standard Meridian 90 degEOT 6.2 min
Latitude 32 deg
Declination 20.6 deg
Surf Azimuth 225 degSurf Tilt 90 deg
A 346.4 Btu/hrft2
B 0,186c 0,138cN1
RΗoG 0'2
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117
Output Data
cDsr LsI h, o β,
. Φ,. Ψ, o θ, " Gruo* Go* Ga* Gρ* G,*
7.00 5.90 91.55 9.50 71.57 153.43 151.90 112.19 O.OO 6.97 3.40 10.378.00 6.90 76.55 21.78 78.63 146.37 140.64 209.84 o.OO 13.03 10.68 23.719.00 7.90 61.55 34.38 85.69 139.31 128.74 249.18 o.OO 15.47 17.51 32.9810.00 8.90 46.55 47.09 93.60 131.40 116.76 268.71 o.OO 16.69 23.39 40.071 1.00 9.90 31.55 59.65 104.24 120.76 104.98 279.23 o.OO 17.34 27.g5 45.2s12.00 10.90 16.55 71.33 123.59 101.41 93.63 284.65 o.OO 20.57 30.90 51.4613.00 1 1 .90 1.55 78.52 172.69 52.31 83.01 286.52 34.88 24.03 32.03 90.9414.00 12.90 13.45 73.44 229.79 4.79 73.4s 285.30 81.05 27.54 91.28 139.8715.00 13.90 28.45 62.18 252.83 27.83 65.62 280.70 115.85 30.36 28.70 174.9116.00 14.90 43.45 49.71 264.52 39.52 60.08 271.44 135.41 31.69 24.45 1g1.5417.00 15.90 58.45 37.00 272.79 47.79 57.55 254.30 136.46 30.69 18.81 185.9718.00 16.90 73.45 24.37 279.93 54.93 58.44 220.69 115.51 26.33 12.15 153.9919.00 17.90 88.45 12.00 286.94 61.94 62.60 141.60 65.16 1s.97 4.90 86.03*Unit of lrradiation is Btu/hrft,
718Using the developed program, following tables summarize input and outputdata caΙculated for southfacing Surface tilted at 45 deg. on Apr 21 inLouisville, KY.
lnput DataLongitude
Standard MeridianEoΤ
LatitudeDeclination
Surf ΑzimuthSurf TiΙt
ABc
CNRHOG
tSI h, o β'.
Output Data
Φ," ψ," θ," Gruo*
18.8 161.2 164.4 0.035.7 144.3 150.0 0.049.8 130.2 135.4 0.0
85.73 deg90 deg1.1 min
38.18 deg1 1.6 deg180 deg45 deg
358.2 Btu/hrft20.1 640.12
1
0.2
1.02.03.0
165.0 38.3150.0 32.9'135.0 24.8
ιJD
0.00.00.ο
Gα*
0.00.00.0
^ * ^*ιJR ιra
0.0 0.00.0 0.00.0 0.0
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118
4.0 120.0 _15.1 61 .55.0 105.0 _4.3 71.66.0 90.0 7.1 80.87 .0 75.0 18.9 s9.98.0 60.0 30.6 99.79.0 45.0 42.0 111.310.0 30.0 52.3 126.811 .0 15.0 60.2 149.312.0 0.0 63.4 180.013.0 15.0 60.2 210.714.0 30.0 52.3 239.215.0 45.0 42.0 248.7'16.0 60.0 30.6 260.317 .0 75.0 18.9 270.118.0 90.0 7 .1 279.219.0 105.0 4.3 288.420.0 120.0 15.1 298.521 .0 135.0 _24.8 310.222.0 150.0 _32.9 324.323.0 165.0 38.3 341.224.0 180.0 _40.2 360.0
*Unit of lrradiation is Btu/hrft,
118.5 120.7 o.o o.o o.o1ο8.4 106.0 o.o o.o o.o99.2 91.4 95.7 o.o 9.890.1 76.8 215.8 49.2 22.180.3 62.5 259.6 120.0 26.668.7 48.4 280.3 186..1 28.753.2 35.1 291.1 238.3 2g.830.7 23J 296.5 271.5 30.40'ο 18'4 298'2 282.9 3o.530.7 23.7 296.5 271.5 30.453.2 35.1 291.1 238.3 2g.868.7 48.4 2S0.3 186.1 28.780.3 62.5 259.6 120.0 26.690.1 76.8 215.8 49.2 22.199.2 91.4 95.7 o.o 9.8
108.4 106.0 o.o o.o o.o1 18.5 120.7 o.o o.o o.o130.2 135.4 o.o o.o o.o144.3 150.0 o.o o.o o.o161.2 164.4 o.o o.o o.o180.0 175.2 o.o o.o o.o
0.0 0.00.0 0.00.7 10.52.8 74.14.8 151.46.5 221.37.8 275.98.6 3,10.58.9 322.38.6 310.57.8 275.96.5 221.34.8 1s1.42.8 74.10.7 10.50.0 0.00.0 0.00.0 0.00.0 0.00.0 0.ο0.0 0.0
719Using the developed program,.following tables summarize input and outputdata calcuΙated for an eastfacing windδw, 3 ft. wide by 5 ft. high, with noset baοk on a clea r Jul21 day inBoise, lD.
Ιnput DataLongitude
Standard MeridianΕoT
LatitudeDeclination
Surf AzimuthSurf Titt
ABc
CNRHOG
LSr h, o β,.
5.0 ,105.0 3.86.0 90.0 14.0
116.22 deg120 deg6.2 min43.57 deg20.6 deg90 deg90 deg
346.4 Btu/hrft2ο.'1860.1 38
1
0.2
Φ,'65.074.8
ψ,"25.015.2
Output Data
θ, o Guo* Go" Gα*
25.3 21.5 19.5 3.620.6 160.9 .150.6
27.4
GR* Gt* e0.4 23.5 352.46.'1 184.1 2761 .3
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7.0 75.0 24.7 84.4 5.68.0 60.0 35.6 94.8 4.89.0 45.0 46.2 106.9 16,910.0 30.0 56.1 123.0 33.01 1.0 15.0 63.8 146.7 56.712.0 0.0 67.0 18ο.0 90.013.0 15.0 63.8 213.3 123.314.0 30.0 56.1 237.0 147.015.0 45.0 46.2 253.1 163.116.0 60.0 35.6 265.2 175.217 .0 75.0 24.7 275.6 .185.6
18.0 90.0 14.0 285.2 195.219.0 105.0 3.8 295.0 205.0
*Unit of lrradiation is Btu/hrft2
25.3 222.0 200.7 36.835.8 251.6 203.9 38.548.6 267.7 177.2 36.162.1 276.8 129.6 31.476.0 281.6 68.2 26.290.0 283.0 0.0 21.5104.0 281.6 0.0 17 .5117 '9 276.8 ο.0 17 '2131.4 267.7 0.0 16.6144.2 251.6 0.0 .15.6
154.7 222.0 0.0 13.8159.4 160.9 0.0 10.0154.7 21.5 0.0 1.3
119
12.3 249.9 3747.918.1 260.6 3908.923.0 236.3 3544.726.8 187.8 2817.229.2 123.6 18s3.630.0 51.4 771.729.2 46.6 699.726.8 44.0 659.823.0 39.7 594.818.1 33.7 505.912.3 26.1 391.96.1 16.1 241.60.4 1.8 26.7
oE is the rate at which solar energy strike the window in Btu/hr
7 20Given lnformation:
Latitude: l= 32.47 deg.Surface azimuth, S; ζz=Windowwidth;W=4ft.Windowheight; H=6ft.Setbackdistance; b=1
N180 deg. (clockwise from north)
ft.
(a) On April2lDeclination: δ= 12.3 deg.Local Solar Τime: LSr= 9:00 am
Hour angle: h = 15.(912) = 45.0 deg.
Using Eq. 78 to calcuate solar altitude, β = 43'82 deg'
Using Εq' 71'1 to find solar azimuth; Φ= 106'27 deg. (clockwise from north)
Using Εq' 712 to calculate wallsoΙar azimuth, y= 73'73 deg.
Using Εqs. 728 to 730 to calcuΙate shaded dimensions,
Χ = (1 ft.)tan(73.73) = 3.43 ft.y = (1 ft ).tan(43.82)/cos(73.73) = 3.42 ft.
v
6'Shaded area can be calculated by
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'7
X
!tr
4'
120
Α,h =W * H _ (W  x) * (H _ y) = 22'52 ft2
720 (Cont.)
Therefore, the percentage of the window that is shaded is 93"8%.
(b) On July 21Declination: δ= 20.6 deg.Local Solar Time. LSl. = 12:00 pm
Hour angle: h = 15*(1212) = 0.0 deg.
Using Eq. 78 to calculate solar altitude, β = 78.13 deg.
Using Εq.711 to find solar azimuth; Φ= 180.0 deg. (clockwise from north)
Using Εq' 712 to calculate wallsolar azimuth, r= 0.0 deg.
Using Eqs. 728 to 730 to calculate shaded dimensions,
x = (1 ft.)*tan(0.0) = 0.0 ft.
y = (1 ft.)*tan(78.13)/cos(0.0) = 4.76 ft.
Shaded area can be calculated by
Α,h=W*H_(W _x)*(ΙΙ_y) = 19.03ft2
Τherefore, the percentage of the window that is shaded is 79.3%.
(c) On Sep 21Declination: δ= 0.0 deg.Local Solar Time: LSl. = 5:00 9m
Hour angle: h = 15*(1712) = 75.0 deg.
Using Eq. 78 to οalculate solar altitude, β= 12.61 deg.
Using Εq'711 tofind solarazimuth; Φ=261.81 deg. (clockwisefrom north)
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studentsenrolledincoursesforwhiοhthetextbοokhasbeenadopted. ΑnyotherreproducιionortrαnsΙαιionοfthisνοrkbeyondthαιpermιιtedby Secιions Ι07 or ]08 ofιhe Ι976 [JnitedιSιαιesCοnνriqhl Α' \υithn1ιt thο hovh'i'CiΛ' n{tho '^:nιot '1ιlv' ;a ''"Ι^''t\'1
121
Using Εq' 712 to calculate wallsolar azimuth , y= 81.81 deg.
7 20 (Cont.)
Using Εq.728 to calculate the horizontally shaded dimension, X,
x = (1 ft.).tan(81.81) = 6.95 ft.
Since x is greater than W, the window is completely shaded.
Therefore, the percentage of the window that is shaded is 100%'
721Given: Problem 720 with a tong 2 ft overhang located 2ft above the top of
the window.
For this problem, bo for overhang is the sum of the overhang depth and the
setback; henοe, bo = /+] = 3 ft.
(a) Τhe vertically shaded dimension on the window due to the overhang
can be calculated bY:
lo=botanβlcosy_!o.
where 1rr, is the distance of the overhang above the window. Therefore,
Υo = (3 ft.)*tan(43.82)/cos (73'73)  2 _ 8'27 ft'
Sinοe η is greater than H, the window is completely shaded.
Therefore, the percentage of the window that is shaded is 100%.
(b) Similafly, yo= (3 ft.) tan(78.13)/cos(0.0)  2= 12'27 ft'
Since y, is greater than H, the window is completely shaded'
Therefore, the percentage of the window that is shaded is '100%'
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students enrolled in courses fbr whiοh the texibook has been aJopted. Αny οιher reprολucJioι or ιrαnslαιiοn ο this νork beyond ιhαt permiιιed
b,ySectiοns Ι07 οr Ι08of the 1q76τΙito)qf'''"?""";*13 l^'!'1 ^ ' '1
122
(c) Since the window is completely shaded due to the setback' there is no
need to calculate Yo'
7 22Given. Problem 72owith 6 in. setback instead of 1 ft' setback'
(a) Using Eqs. 728 to 730 to calculate shaded dimensions,
x = (0.5 ft.)*tan(73.73) = 1'71 fL
, = (o.s ft.)tan(43.82)/οos(73'73) = 1'71 ft'
Shaded area can be calculated bY
Α,n =W * H _ (W _x)* (H _ y) = 14.19 fi2
Therefore, the percentage of the window that is shaded is 59'1%'
(b) Using Eqs. 728 to 730 to calculate shaded dimensions,
γ = (O.5 ft.)tan(0'0) = ο'0 ft'
, = (ο.s ft.)*tan(78.13)/cos(0'0) _ 2'38 ft'
Shaded area can be calculated bY
Α,l, =W * H _ (W _'T )
* (H _ y) = 9'52 ft2
Therefore, the percentage of the window that is shaded is 39'7%'
(c)UsingEqs.T28to73Otocalculateshadeddimensions,
x = (0.5 ft.).tan(81'81) = 3'48 ft'
, = (O.S ft.)tan(12.61)/cos(81 'S1) = 0'79 ft'
Shaded area can be calculated bY
Α,h =W * H _(W _.χ)* (H _ y) = 21'27 ft2
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studentsenrolledinοoursestoιwhichtnetexδookhasbeen ^l"i'r'i λiy'nrrrrproλur^ιιoλo,ιrαnslαιiοiofιhisνorkbeyondthαtρermitted
hιι'Sρrtinnrl07nrl0Rn{tι1o Ιo7AΙΙbi+^)c'!"^^f''";..1'! ''',.'"''^""i'''""'''''"'^^{i"^^^'^'';ι^1 '''''"iο"ι^'{"l
123
Therefore, the percentage of the window that is shaded is 88'6%'
7 23on December 21, Declination: δ= 23'45 deg. Using the same procedure
as described in Problem 720, the following table summarizes the
calculated data.
:"^::j Hour Solar Solar Surface '"Jlff#:"' ι"#::γ Shaded %Shaded:,ol1' Angle, Αltitude, Azimuth, Solar Dimension Dimension Area, ft2 ΑreaTime, ""; o o Azimuth,' "hr f\ZιlΙluιl '' 1x1, ft (y)' ft
8.OO 6O.O0 9.98 126.22 53 78 1'37 o'30 8'98 37 '4
9.OO 45.00 1g.4g ',136.52 43 48 o'95 o'49 7 '18 29'9
1o,oo 3o.oo 27.17 148.96 31'04 0'60 0'60 5'65 23',5
'1 1.OO 15.00 32.27 163.69 16'31 o'29 0'66 4'20 17 '5
12.00 O.OO 34.08 18o.OO O'OO o oo 0'68 2'71 1 ',1 ',3
13.00 15.00 32.27 196.3',1 16'31 0 29 0'66 4'20 17 '5
14.00 3O.OO 27.17 211'04 31'04 o 60 0'60 5 65 23'5
15.00 45.00 19.49 22g 48 43'48 0 95 0 49 7 18 29',9
16.00 60.0O 9.98 23g.78 53'78 1'37 0 30 8 98 37 '4
7 24This problem is similar to ProblemT21 but the overhang depth is 3 ft
instead of 2ft. Since the window in ProblemT21 is completely shaded in
all cases. Τhe window in this problem is also completely shaded in all
cases since the overhang depth is greater in this problem'
7 25
7 26
7 27First, we need to know angle of incidence and solar irradiation. Using Eqs'
78 to 726 (or a computeiprogrφ developed for previous problem),
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124
incidence angle and solar irradiation on a southwestfacing window for
Boise, lD on a clear July 21 day at 3:00 pm solar time are
Angle of Incidence θ= 52.4 deg',Direct Solar lrradiation: Gp = 163'4 Btu/hrft2'
Diffuse Solar lrradiation: Ga + Gκ = 34'5 + 23'0 = 57 3 Btu/hrft', and
Total Solar lrradiation: G1 = 163'4 + 57 '3 = 220'9 Btu/hrft2'
Then, the area of the glazing and of the frame is calculated to be 12'44 ft2
and 2.56 ft2, resPectivelY.
From Table 73, solar heat gain coefficients for the glazing system lD 21c
are
SHGGgo(52.4") = 0.548 and SHGGgα= 0'52'
From Table 52, the outside surface conductance may be estimated to be
4.0 Btu/hrft2'F.
From Τable 56, the Uvalue for the fixed, double glazed window having
aluminum frame with thermal break utilizing metal spacers is 1.13 Btu/hr
ft2'F.
From Table 71, solar absorptance of the aluminum frame (assuming the
window is not a nev/ one) is 0.8.
Αssuming the window with no setback (Ar,u'" = Aruπ), the SHGC for the
frame can be calculated using Eq' 731 as:
SHGGr = 0.8*(1 '1314'0) = 0'226'
Then, using Εq.732, the total solar heat gain is
Qsnc = (0.548.1 2.44 + 0.226*2.56)*163'4+ (0.52*1 2.44 + 0.226*2.56)57.5 = 1613'68 Btuihr'
7 28From Table 73, the glazing transmittance and absorptances for the glazing
system lD 21c are
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125
7 28 (Cont.)
Tρ6(52'4") = O.+1 56, Αtwοβ2Α) = O'140, Αfzυοφ2'4) = 0'1524'Ta= O'40, Af ια= o.'13, and Αfza= O.15.
Using Eq. 735, total transmitted solar heat gain is
Qrru"', = (0'4156163.4 + 0.4ο*57.5)12'44 = 1130.9'1 Btu/hr.
Using Eq. 736, total solar heat gain absorbed by the glazing is
8 oroo,, = [1 63'4*(0"1 4+0' 1 524) + 57'5*(0"1 3+0' 1 5)]1 2'44
= 794.64 Btu/hr.
From Table 55a, the Uvalue for the center of glass is 0.42 Btu/hrft2"F.
Similar to the previous problem, the outside surface conductance may beestimated to be 4.0 Btu/hrft2'F.
Then, the inward flowing fraction for glazing layer 1 can be calculated by:
Nt=0.42 14.0=0.105
From Table 52a, the inside surface conductance may be estimated to be1.46 Btu/hrft'"F.
The conductance from the inner pane to the outdoor air can be calculatedby:
,11flo'2= 1 1=1 1 =o'59Btu/hrft2'F
U hi 0.42 ι.46
Then, the inward flowing fraction for glazing layer 2 can be calculated by:
Nz= 0.42 / 0.59 = 0.71
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students enrolled in οourses fbr which the textboοk has been adoρted' Αny oιher reproduction οr lrαnslαtiοn of ιhis wοrk beyond thαι permitted
126
728 (Cont.)
Using Εq. 738, the inward flowing fraction of the gΙazing system is
N = [1 63.4*(0.'1 0S*0. 14+0.7 1*0.1 524) +
57.5*(ο.'1 05*0. 13+0.710. 1 5)] l 220 '9= 0.122
Using Eq. 739 and the SHGGr calculated from the previous problem, thesolar irradiation absorbed by the frame is
Qoroo,f = (163.4 + 57.5)*2.56*0.226 = 127 .80 Btu/hr.
Using Εq' 740, the total absorbed solar heat gain of the fenestrationsystem is
Quruo,ur = 794.64*0.122 + 127 .80 = 224.75 Btu/hr.
The total solar heat gain is then
Qsrc = 1130.91 + 224.75 = 1355.66 Btu/hr.
7 29From Table 74, lAC for a lightedcolor Venetian blind installed on aresidential doublepane window is 0.66.
Using Εq' 741, the total solar heat gain is
Q suc = φ'2262'56220'9)+ [0.548*12.44*163.4 + 0.52* 12.44*57 .5]*0.66
= 1108.48 Btu/hr.
7 30From Table 76, for a Ιightedcolor Venetian blind, shade transmittance,reflectance, and absorptance are 0.05, 0.55, and 0.40, respectiveΙy.
Using Εq' 742, the transmitted solar heat gain is
127
730 (Cont.)
Qrroo = 0.05*1130.91 = 56.55 Btu/hr.
Using Εq' 743, the absorbed solar heat gaΙn is
4or"" = 224.75 + 0.40*1 130.91+ 0.55*1 130.91 *0.122*(0.'13+0.I S)
= 698.36 Btu/hr.
731From Table 73, solar heat gain coefficients for the glazing system lD 5bare
SHGGgo(52'4") = 0.6256 and SHG Ggα = 0.60.
Similar to Problem 727 , SHGGr = 0.226.
Then, using Εq'732, the total solar heat gain is
Qsμc = (0.625612.44 + 0'226*2.56)*1 63.4+ (0.60.12.44 + 0.226*2.56)*57.5 = 1828.64 Btu/hr.
7 32From Table 73, the glazing transmittance and absorptances for the glazingsystem lD 5b are
TDθ(52'4") = 0.5332, 'Af ,οoβ2'4) = 0.1924, 1froθβ2'4) = O'12,Ta= 0.51, Atro= 0.19, and Arzd= 0.11.
Using Εq. 735, total transmitted solar heat gain is
Qrsac,g = (0.5332*'163.4 + 0.51*57.5)*1 2.44 = 1448.64 Btu/hr.
Using Eq. 736, total solar heat gain absorbed by the glazing is
λQ πllc'g = [1 63.4*(0.1924+0'12) + 57.5(0. 1 9+0. 1 1)112.44
===Ξ7 32 (Cont.)
= 849.60 Btu/hr.
From Table 55a, the Uvalue for the center of gtass is 0.55 Btu/hrft2"F.
similar to the previous problem, the outside surface conductance may beestimated to be 4.0 Btu/hr_ft2_.F.
Then' the inward flowing fraction for glazing layer 1 can be caΙculated by:
Nz=0.SS /4.0=0.1375
SimiΙar to the prevΙous probtem, the inside surface conductance may beestΙmated to be 1.46 Btυ/hrft2"F.
The conductance from the inner pane to the outdoor air can be calcuΙatedby.
ho,z=t\ =T] 1 = 0.88 Btu/hrft2"F(]_τ O55  146
Τhen' the Ιnward flowing fraction for gtazing layer 2 canbe calcutated by:
/vz=0.55/0.gg=0.625
Using Εq. 738, the inward flowing fraction of the glazing system is
N = []63 4(O.1375ο. 1924+o.625*0'12) +
= 'Jr';Δ'' 1375*0' 19+0'625*0' 1 1 )] l 22o'g
Τhe solar irradiation absorbed by the frame is the Same as the previousprobΙem, and is equaΙ to 127 .80 Btu/hr'
Using Εq' 74O, the totat absorbed solar heat gain of the fenestrationsystem is
7 32 (Cont.)
Qo'"o'gt = 849'60*0'1OO + 127 '80 = 212j6 Btu/hr'
The total solar heat gain is then
Qwc = 1448'64 + 212J6 = 1661'4 Btu/hr'
B1
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis fortesting or instruοtionaΙ purposes only to students enrolled in courses for which the textbook has beenadοpted. Αny other reprοduction οr trαnslαtion of this wοrk beyond thαt permitted by Sections ] 07 or ] 0Bof the 1976 United Stαtes Copyright Αct without the permission of the cοpyright owner is unlαwful.Requests for permission or further infοrmαtion should be αddressed to the Permission Depαrtment, JοhnWiley & Sons, Ιnc, ] Ι ] Riνer Street, ΙΙoboken, NJ 07030.
Chapter 8
The heat gain is generally gΓeater than the cooling load during themorning hours M/hen sunlight first strikes a building and the internalloads first begin. Heat is being stored in the building structure,furnishings, etc.Late at night when occupants are not present, lights and equipment areoff and solar radiation is zero, the building gives up stored heat to theair, which the equipment removes as cooling load. The heat gain maybe quite small, zero, or negative.At some time during the day, probably early evening, as heat gain isdecreasing, and equilibrium condition can be established when heatgain and cooling load are equal. Or, some interior zones, where thecooling load is driven only by internal heat gains may reach equilibrium ifthe heat gain remains constant for a number of hours.
MultipΙe design conditions should be checked, including peak dry bulbalong with mean coincident \Μet buΙb, and peak wet bulb along with meancoincident dry bulb temperature.
ΑSHRΑE 90.1 specified the2'5% design conditions, which roughlycorresponds to the 1% design conditions in the current Handbook ofFundamentals and the textbook.
a)
b)
c)
82
83
Location OutdoorDB.'F
OutdoorWB,'F
lndoorDB.'F
lndoorRH. %
Elevation, ft Latitude,ON
Norfolk, VA 91 76 75 50 30 36.90
84
select materials; some may need to be entered into the tayer library' The
resulting wall construction ,pp"rrt as shorrun here' (Note that not
everything is specified exacity, .o that a student using a density of 120
ιυit1yt"i υiicκ wiιι get a different set of CTF coefficients')
program in executefor this wall.
131
for room mode, we
B5
After running the HvacloadExplorerobtain the following CTF coefficients
nxn,
Btu/hft2"F
Yn,Btu/hft2"F
zn,Btuihft2"F
Φn
0 4.276507 0 ο0ο445 0.642344
1 5.36497 0.ο1 1581 0.98287 0.638772
2 1.141149 0.01 1845 0.376555 0.02179
3 0.027 59 0.001 134 0.01 101
4 7 .7Ε05 0.0ο0017 5E06
This problem is solved in the Same \Λ/ay aS Problem 84,
13 insulation is changed to 5.5" thick R19 insulation.except that the R
Εxοerpts fιom this work may be reproduοed by instΙuοtοrs for dtstribution on a notforprofit basis for testing or instructional puφoses only to
students enrolled in courses tor whiοh the teΧtbook t,u, υ"., uaipt.α' iny oιh* upiurrιλno, oαnsΙαtionΖf ιhis work beyοnd thαι permιιιed
by Secιions 107 or ] 08 o7 ιne ii;i' initid Sror^ copyrιgrt 'aι rr"ιiiλuι ιn, prr^ι''ιλn of the cοpyrighι owner is unlανful'
".."...τns1] 130
0 5α]  "183 5nΟ 3.
86
132
The folΙowing CTF coefficients are obtained:
Αgain, this problem follows the procedure of the last two problems. Thethickness of the roll roofing must be estimated, and the conductivity chosento match the overall conductance. (k=thickness*conductance)
The following CΤF coefficients are obtained:
Εxceφts from this work may be reproduced by instruοtors for distτibution on a notforprofit basis for testing or instruοtional puφoses only tostudents eι'ιrolled in courses for whiοh the textbook has been adopted. Αny oιher reproduction or ιrαnsιαιion of this ινork beyond ιhαι permiιtedby Secιiοns ] 07 οr 1 08 of ιhe Ι97 6 Uniιed Sιαtes Cοpyrighι Αcι w ithοuι ιhe permissiοn οf ιhe copyrighι oνner is unlανιful'
nxn,
Btu/hft2"FYn,
Βtu/hft2'FΖn,
Btu/hft2"FΦn
0 4.277384 0.000071 0.644513
1 5.95084 0.004622 1.08666 0.779066
2 1.847897 0.008936 0.510931 0.10021
3 0.16027 0.001835 0.05401 0.001435
4 0.001331 0.00004 0.000734
133
87
nxn,
Btu/hft2'FYn,
Btu/hft2'FΖn,
Btu/hft2'F Φn
0 1.014657 0.006092 0.6445131 1.09939 0.029838 0.6816 0.1505942 0.126521 0.006044 0.0791043 0.000256 0.000071 0.00003
The following CΤF coefficients are obtained:
nxn,
Btu/hft2'FYn,
Btu/hft2'Fzn,
Btu/hft2'F Φn
0 1.014651 0.00468 0.6544711 1.12785 0.027234 0.71129 0.1 781 592 0.1 51 609 0.00674 0.0955263 0.000351 ο.0001ο6 0.000053
ln this case, a reasonable value for the resistance of the airspace must beselected. For the air^space, an Rvalue of 1 is chosen; thus conductivity isset to '12 Btuin lhr ft2 F, and the thickness Ιs set to 12 in Density anα bpare set to zero and 0.24, respectively.
Νntε: Ιayers listed fιnm tοp tο bοfiom :epr*s*ni {rα:n thg nutsiιJ* tο inside ot lhe sur{εce
Excerpts from this work may be reproduced by instruοtors for.distribution on a notfor_profit basis for testing or instruοtional puφoses only tostudents enrolled in courses Γor which the textbook has been adopted' Αny οther reproλcιion or trαnsl(]ιiοn of ιhis νοrk beyontl thαι permiιιedby Sectiοns ]07 οr ]08 ofιhe 1976 Uniιed StαιeS Cοpyrighι Αcιw'ithouι ιhi permissιλn οfιhe cοpyrighι ονner ιi unlαινful'
88
134
Again, reasonable values must be assumed for the density of theacoustical tile and the specific heat of the limestone concrete.
The foΙlowing CTF coefficients are obtained:
nxn,
Btu/hft2'FYn,
Btu/hft2'FZn'
Btu/hft2'FΦn
CI 3.162792 0.002232 0.2851 '16
1 3.76069 0.01895 0.38995 0.71ο366
2 0.633425 0.007779 0.137 459 0.01912
3 0.00642 0.000149 0.0ο352
First, appy the exterior οonvective heat transfer correlation, Equation 818a, to determin Θ h"' Assume the 15 mph wind is windward on the surface,which results in h" = 2.3 Btu/(hft2F).
Estimate the sky temperature as 10.8 R below the outdoor ambienttemperature = 546.87 R. Then estimate the effective sky temperature for avertical surface from Εquation 825
tsky,o= cos (90/2)λry +(1cos(90/2))f, = 550.0 R
EXceψtS from this work may be reproduοed by instructors for distτibution on a notfor_profit basis for testing or instructional puφoses only to
students enrolled in οοurses for which the textbook has been adopted. Αny oιher reproduction or trαnsιαιion of this wοrk beyοnd ιhαι permitιed
by Sectiοns Ι 07 οr Ι 08 οf the 1976 tJnited Stαtes Copyrι1hι Αcι \νitlιout ιhe permission of the copyrighι oιιner is unlαινfuΙ.
89
136
DeclinationSurf Αzimuth
Surf TiltΑBc
CNRΗoG
20.6 deg270 deg90 deg
346.4 Btu/hrft20.1 860.1 38
1
0.2
MDST LSI h, o β, "
1 .00 23.79 176.83 34.272.00 0.79 168.17 33.233.00 1.79 153.17 28.804.00 2.79 138.17 21.655.00 3.79 123.17 12.546.00 4.79 108.17 2.11
7.00 5.79 93.17 9.198.00 6.79 78.17 21.059.00 7.79 63.17 33.2210.00 8.79 48.17 45.4911.00 9.79 33.17 57.5112.00 10.79 18.17 68.4613.00 11.79 3.17 75.2814.00 12.79 11.83 72.1915.00 13.79 26.83 62.3616.00 14.79 41 .83 50.6317.00 15.79 56.83 38.4118.00 16.79 71 .83 26.1719.00 17 .79 86.83 14.1520.00 18.79 101 .83 2.5821.00 19.79 1 16.83 8.27
22.00 20.79 '131.83 18.0023.00 21.79 146.83 26.0624.00 22.79 161.83 31.74*Unit of lrradiation is Βtu/hrft2
Output Data
Φ,' Ψ,o θ,o
356.41 86.41 87.0313.26 256.74 101.0628.82 241.18 114.9942.20 227.80 128.6353.39 216.61 141.5862.87 207.13 152.8071 .22 198.78 159.1779.02 190.98 156.3786.89 183.11 146.6595.77 174.23 134.23107.55 162.45 120.81127.34 142.66 106.97168.24 101.76 92.97218.87 51.13 78.94245.60 24.40 65.01259.78 10.22 51.37269.56 0.44 38.42277.71 7.71 27.20285.45 15.45 20.83293.49 23.49 23.63302.43 32.43 33.35312.83 42.83 45.77325.24 55.24 59.19339.93 69.93 73.03
* ^ * ^ * ^ * ^*ιr/vD ιJD ιrd ιJR ιrt
0.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.00
108.07 0.00 6.71 3.22 9.93206.38 0.00 12.82 10.26 23.08246.69 0.00 15.32 16.92 32.24266.87 0.00 16.57 22.71 39.29277.85 0.00 17.25 27.27 44.53283.62 0.00 17.61 30.29 47.91
285.80 0.00 20.83 31.59 52.42284.93 54.68 25.38 31.06 111.11
280.80 1 18.63 30.63 28.75 178.01272.32 170.01 35.51 24.81 230.32256.78 201.20 38.43 19.50 259.13227.20 202.07 37.19 13.16 252.42161 .85 151.27 27 .51 6.19 184.975.53 5.07 0.93 0.10 6.100.00 0.00 ο.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.000.00 0.00 0.00 0.00 0.00
The hourly drybulb temperature is calculated using Equation 82. Here,the hour nearest to the local solar time has been used to determine thetemperature. A spreadsheet is used to obtain the solution. lteration isaccomplished by simpΙy pasting the calculated values of Io" back into the" Io", estimated" column.
lnput Data
UValue 0.1 Btui(hft2F)
Solar absorotivitv 0.8
Ihermal emissivitν 0.9
Exceφts from this work may be reproduced by instruοtors for distrrbution on a notforproΓit basis for testing or instructional puφoses only tοstudents enrolled in courses for whiοh the textbook has bοen adopted. Αny other reproducιion or trαnsιαtion οf ιhis wοrk beyond thαι permittedby Secιions 107 or ]08 ofthe Ι976 United Stαtes Cοpyriqht Αctιliιhοuι ιhe permissiοn οfthe cοpyright oνner is unlανful.
Τis 72.O F
Peak temperature 96.0 FDailv Ranqe 25.4 F
Mean Wind Soeed 10.0 mph
811
137
This problem uses the same solution procedure as Problem 810.
lnput DataLongitude 116.22 deg
Standard Meridian 105 degEOT 6.2 min
Latitude 43.57 degDeclination 20.6 deg
Exceφts from this woτk may be reproduced by instructors fbr distrrbution on a notfor_profit basis for testing or instructional puφoses only to
students enrolled in οourses for whiοh the textbook has been adopted. Αny other reproduction οr trαnsιαtion of this ινοrk beyond ιhαι permittedby Secιions ] 07 or ]08 of ιhe Ι 976 United Sιates Cοpyright Αcι ν,ithout ιhe permission of the cοpyrighι oνner is unlαwfuΙ.
Output Data
ClockTime
LocalSolarTime
OutdoorDrybulbTemp.
(F)
skvΤemp.
(F)
Effectiveskv
Temp.(F)
To",estimated
(F)
hc(Btu/(hft'F))
hrsky
(Btu/(hfrF))
hrg..o
(Btu/(hft'F))
To",calculatedfrom 824
(F)
9conduction(Btu/(h
tt'))
1.00 23.79 75.2 64.4 67.5 74.60 1.58 0.46 0.47 73.14 0.1 1
2.00 0.79 73.9 63.1 66.3 73.15 1.58 0.46 0.47 71.93 0.01
3.00 1.79 72.6 61.8 65.0 71.94 '1.58 0.45 0.46 70.72 0.13
4.00 2.79 71.6 60.8 64.0 70.72 1.58 0.45 0.46 69.75 0.23
5.00 3.79 70.9 60.1 63.2 69.75 '1.58 ο.45 0.46 69.02 0.306.00 4.79 70.6 59.8 63.0 69.03 1.58 0.45 0.46 68.78 0.32
7.00 5.79 71.1 60.3 63.5 69.'18 1.58 0.45 0.46 72.34 0.038.00 6.79 72.4 61.6 64.7 74.55 1.58 0.46 0.47 77.55 0.569.00 7.79 74.7 63.9 67.0 78.99 1.58 0.47 0.48 82.47 1.0510.ο0 8.79 78.0 67.2 70.3 83.53 1.58 0.48 0.49 87.66 1.5711.00 9.79 8'1.8 71 .0 74.1 88.42 1.59 0.49 0.50 92.76 2.0812.00 10.79 86.1 75.3 78.5 93.24 '1.59 0.5ο ο.51 97.77 2.58
13.00 11.79 90.2 79.4 82.5 98.33 1.59 0.51 0.52 102.84 3.0814.00 12.79 93.2 82.4 85.6 109.05 1.59 0.53 0.54 122.43 5.0415.00 13.79 95.2 84.4 87.6 131.23 1.61 0.57 0.58 142.11 7.01
1 6.00 14.79 96.Ο 85.2 88.4 148.77 1.62 0.60 0.61 155.98 8.4017 0ο 15.79 95.2 84.4 87.6 159.92 1.62 0.62 ο.63 162.15 9.0218.00 16.79 o?ξ 82.7 85.8 162.49 1.62 0.62 ο.63 158.51 8.6519.00 17.79 90.7 79.9 83.0 153.00 1.62 0.60 0.6'1 138.39 6.6420.00 18.79 87.4 76.6 79.7 118.72 1.60 0.54 0.55 86.48 1.45
21.0ο 19.79 84.1 73.3 76.4 84.81 1.58 0.49 0.50 81.63 0.9622.00 20.79 81.3 70.5 73.6 81.65 1.58 0.48 0.49 78.96 0.7023.0ο 21.79 78.7 67,9 71 .1 78.97 1.58 0.47 0.48 76.54 0.45
24.00 22.79 76.7 65.9 69.1 76.55 1.58 0.47 0.48 74.60 0.26
138
Surf ΑzimuthSurf Tilt
ABc
CNRHOG
180 deg90 deg
346.4 Btu/hrft20.1 860.1 38
1
0.2
MDST LSr h, " β, "
1.00 23.15 167.23 24.772.00 0.15 177 .77 25.803.00 1.15 162.77 23.914.00 2.15 147.77 19.345.00 3.15 132.77 12.596.00 4.15 117.77 4.217.00 5.'15 102.77 5.318.00 6.15 87 .77 15.609.00 7.15 72.77 26.3210.00 8.15 57 .77 37 .171 1 .0ο 9.15 42'77 47 '7612.00 '1 0.15 27 .77 57 .4213.00 11.15 12.77 64.6814.00 12.15 2.23 66.9515.00 13.15 17.23 62.9116.00 14.15 32.23 54.7117.00 15.15 47 .23 44.6718.00 16.1 5 62.23 33.9519.00 17 .15 77 .23 23.1020.00 18.'15 92.23 12.4821.00 19.15 107 .23 2.3822.00 2015 122.23 6.8523.00 21.15 137.23 14.7924,00 22.15 152.23 20.95*Unit of lrradiation is Btu/hrft'z
Output Data
Φ, " Ψ, o θ, " Grρ* Go* Gα*
346.83 166.83 152.14 0.00 0.00 0.002.32 177 '68 154.11 ο.00 0.00 0.0017.66 162.34 150.59 0.00 0.00 0.0031.94 148.06 143.20 0.00 0.00 0.0044.76 135.24 133.87 0.00 0.00 0.0056.15 123.85 123.75 0.00 0.00 0.0066.47 113.53 113.42 46.47 0.00 2.8976.20 103.80 103.29 173.45 0.00 10.7785.92 94.08 93.66 227.72 0.00 16.4596.42 83.58 84.88 254.61 22.70 20.78108.99 71.01 77.37 269.45 58.92 24.56125.92 54.08 71.58 277.79 87.76 27.57151 .07 28.93 68.02 281.98 105.54 29.47185.34 5.34 67.06 283.00 1 10.30 29.99217 .50 37.50 68.82 281.09 101.56 29.04239.78 59.78 73.09 275.81 80.22 26.78255.07 75.07 79.44 265.88 48.70 23.50266.84 86.84 87.38 248.28 11.34 19.55277.01 97.01 96.45 215.64 0.00 15.02286.67 106.67 106.26 146.49 0.00 9.10296,51 116.51 116.49 3.97 0.00 0.25307 .11 127.11 126.80 0.00 0.00 0.00318.90 138.90 136.76 0.00 0.00 0.00332.16 152.16 145.67 0.00 0.00 0.00
^ * ^*ιJR Lz1
0.00 0.000.00 0.000.00 0.000.00 0.000.00 0.000.00 0.001.07 3.967.06 17,8313.24 29.6918.90 62.3823.67 107.1527.24 142.5829.38 164.3929.95 170.2428.90 159.5026.32'133.3122.36 94.5717 .29 48.1811,44 26.465.19 14.280.07 0.320.00 0.000.00 0.000.00 0.00
Output Data
Exceφts from this work may be reproduοed by instructors fbr distribution on a notforprofit basis for testing or instructional puφosΘs only tostudents enrolled in courses for which the textboοk has been adopted. Αny other reprοducιion or ιrαnslαtiοn of ιhis work beyond thαι permiιιedby Sectiοns ]07 οr Ι08 ofιhe 1976 Uniιed Stαtes Cοpyright Αctτιiιhout the permission ofιhe cοpyright oνner is unΙαwful.
lnput Data
UValue 0.1 Btu/(hft2F)
Solar absorptivitv 0.9Γhermal emissivih ο.9
Tis 72.0 FPeak temoerature 96.0 F
Dailv Ranqe 30.3 FVlean Wind Soeec 1 1.0 mph
139
ClockTime
LocalSolarTime
OutdoorDrybulbTemp.
(F)
skvTemp.
(F)
EffectiveSkv
Temp.(F)
To"'estimated
(F)
h"(Btu/(hft'F))
hrsky
(Btu/(hft'?F))
hrgrα
(Btu/(htt'ε))
To.,calculatedfromS24
(F)
Qconduction(Btu/(h
ft2))
1.00 23.15 73.0 62.2 65.3 71.15 1.72 0.45 0.46 71.15 0.092.O0 0.15 71.2 60.4 63.5 69.41 1.72 0.45 0.46 69.41 0.263.00 1.15 69.6 58.8 62.0 67.96 1.72 ο.45 0.46 67.96 0.404.00 2.15 68.'1 57.3 60.5 66.51 1.72 0.44 0.45 66.51 0.555,00 3.15 66.9 56.'1 ξoa 65.35 1.72 0.44 0.45 65.35 0.676.00 4.15 66.0 55.2 58.4 64.48 1.72 0.44 0.45 64.48 0.757.00 5.15 65.7 54.9 58.1 64.1 I 1.72 0.44 0.45 65.5'1 0.658.00 6.15 66.3 55.5 58.7 68.60 1.72 0.44 0.45 70.67 0.139.00 7.15 67.8 57.0 60.2 74.15 1.72 0.45 0.46 75.96 0.4010.0ο 8.15 70.5 59.7 62.9 81.31 1.73 0.46 0.47 89.08 1.71
1 1.0ο 9.15 74.5 63.7 66.9 99.67 1,74 0.49 0.50 106.51 3.4512.00 10.15 79.0 68.2 71.4 116.02 1.74 0.52 0.53 121.05 4.9013.00 11.15 84.2 73.4 76.5 129.01 1.75 0.55 0.56 131.75 5.9714.ο0 12.15 89.ο 78.2 81.4 137.21 1.75 0.57 0.58 137.47 6.5515.00 I 3.15 92.7 81.9 85.0 '139.67 1.75 0.58 0.59 137.47 6.551 6.00 14.15 95.1 84.3 87.5 136.52 1.75 0.58 0.59 132.07 6.0117.00 1 5.15 96.0 85.2 88.4 127.91 1,74 0.57 0.58 121.67 4.9718.00 16.15 95.1 84.3 87.5 114.37 1.73 0.54 0.55 107.09 3.5119.00 17.15 93.0 82.2 85.3 100.24 1.73 0.52 0.53 98.57 2.6620.ο0 18.15 89.6 78.8 82.0 93.64 1.72 0.51 0.52 91.61 1.9621.00 '19.15 85.7 74.9 78.1 85.63 1.72 0.49 0.50 83.43 11422.00 20.15 81.8 71 .0 74.1 79.56 1.72 0.48 0.49 79.56 0.7623.00 21 15 78.4 67.6 70.8 76.37 1.72 0.47 0.48 76.37 0.4424.00 22.15 75.4 64.6 67.8 73.47 1.72 0.46 0.47 73.47 0.15
B12
This problem is solved in the same manner as Example 82. The results
Day 1
0.3120.4630.5080.4940.4540.4100.3870.4020.473
Day 2 Day 3
1.126 11260.954 0.9540.804 0.8040.673 0.6730.56't 0.5610.475 0.4750.426 0.4260.426 0.4260.487 0.487
(conduction heat fluxes for each hour in Btu/(hrft2)) may be summarized intabular form as:
Hour
1
2
345
67
B
9
Exceφts from this work may be reproduοed by instructors for distrrbution οn a nοtforprofit basis for testing or instructional puφoses only tοstudents enrolled in courses for whiοh the textbook has been adοpted. Αny οther reproduction or trαnslαtion οf ιhis νork beyond ιhαι permiιιedby Secιiοns Ι07 or Ι08 οfιhe Ι976 Uniιed Stαιes Cοpyright Αct wiιhouι the permission ofthe copyrighι oνner is unlανful.

140
1011
12131415161718192021
222324
0.6120.8201.0891.3991.7151.9982.2232.3622.4052.3522.2152.0161.7861.5511.327
0.6200.8251.0921.4011 .7161.9992.2232.3622.4052.3532.2152.0161.7861.5511.327
0.6200.8251.0921.4011 .7161.9992.2232.3622.4052.3532.2152.0161.786'1.551
1.327
813
Because the wall is Ιightweight, the results converge rapidly.
This problem is soΙved in the Same V/ay aS the previous problem. Note thatthe additional insulation substantially reduces the conduction heat flux, asexpected. The resuΙts (conduοtion heat fΙuxes for each hour in Btu/(hrft2))may be summarized in tabular form as:
Hour1
2
34567o
I1011
12
131415161718192021
22
Day 1
0.2030.3290.3790.3800.3560.3230.2980.2960.3280.407ο.5350.7110.9251.1541.3731.5591.6911.7571.7541.6851.5621.406
Day 20.9150.7780.6590.5540.4640.3900.3400.3220.3440.4170.5420.7150.927'1 .1561.3741.5591.6921.7581.754'1.685
1.562'1.406
Day 30.9150.7780.6590.5540.4640.3900.3400.3220.3440.4170.5420,7150.9271 .1561.3741.5591.6921.7581.7541.6851.5621.406
Excerpts from this work may be reproduced by instructors for.distribution οn a notforprofit basis for testing or instruοtional purposes only tostudents enrolled in οourses fοr which the textbook has been adopted. Αny other reprοdλCιion or ιrαnslαtion of ιhis lιork beyond thαι permiιιedby Sections ] 07 οr ] 08 οf ιhe 1 976 (Jniιed Sιαιes Cοpyright Αcι ιυithouι thi permissιλn ο7 ιhe copyrighι ονner ιi unlαwfut'
2324
141
1.2371.070
1.237 1.2371.070 1.070
814
B'15
The solution to this problem is similar to that of Problem 89, except that toestimate the maximum possible surface temperature, the surface may beassumed to be adiabatic, and U is then zero. Also, the surfacetogroundradiation coefficient is zero, and no correction is necessary for the skytemperature, as the surface is assumed to be horizontal. Assume the windis windward, h"= 1.3 Btu/(hft2F). Then, the final converged answer for thesurface temperature is:
hr,sky= 1.361 Btu/(hft2F) fr"= 201.0 F
From Table 82, heat gain for occupants that are "Seated, very light work"have 245 Btulhr (72 W) sensible heat gain, and '155 Btu/hr (45 W) latentheat gain. The sensible portion is assumed to be 70% radiative/ 30%convective.
The sensible heat gain from people is72 Wperson x 30 people = 2160 W.
The radiative portion is 0.7 x 2160 = 1512νν '
Τhe convective portion is 0'3 x2160 = 648 W.
The latent heat gain from people is 45 Wperson x 30 peoPΙe = 1350 W.
The sensibΙe heat gain from lighting is '1 .5 ννft2 x 4OOO sq. ft. = 6000 W;20o/o is assumed to enter the plenum space directly, leaving 4800 W whichis assumed to be 59% radiative I 41% convective.
The radiative portion is 0.59 x 4800 = 2832\'tΥ.
The convective portion is 0.41 x 4800 = '1968 W.
The sensible heat gain from equipment is 1 ννfi( x 4OOO sq. ft. = 4O0O W,which is assumed to be 20o/o radiative I 80% convective. (Note this
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816
142
assumption is based on the assumption that most of the equipment is fancooled. Students are likely to make varying assumptions.)
Τhe radiative portion is 0'2x 4000 = 800 W.
The convective portΙon is 0.8 x 4000 = 3200 W.
The total sensible heat gain is 2160 + 4800 + 4000 = 10960 W.
The radiative portion is 1512 + 2832 + 800 = 5144 W.
The convective portion is 648 + 1968 + 3200 = 5816 W.
The total latent heat gain is 1350 W.
From Table 82, heat gain for occupants that are involved in "Sedentary\ι/ork" is275 Btu/hr (81 W) sensible heat gain, and 275 Btulhr (81 W) latentheat gain. The sensible portion is assumed to be 70o/o radiative/ 30%convective.
The sensible heat gain from people is 81 Wperson x 35 people = 2835 W.
Τhe radiative portion is 0'7 x 2835 = 1984.5 W.
The convective portion is 0.3 x 2835 = 850.5 W.
The latent heat gain from peopΙe is 81 Wperson x 35 peoPle = 2835 W.
Τhe sensible heat gain from lighting is '15 \,ΙΥlm2 x 75O m' = 11250 W; 50%is assumed to enter the plenum Space directly, Ιeaving 5625 W that isassumed to be 59% radiative I 41% convective.
The radiative portion is 0.59 x 5625 = 3319 W.
The convective portion is 0.4'l x 5625 = 2306 W.
The sensible heat gain from office equipment is 7000 W, which is assumedto be 20o/o radiative I 80% convective. (Note this assumption is based on
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143
the assumption that most of the equipment is fancooled. Students arelikely to make varying assumptions.)
The radiative portion is 0.2x 7000 = 1400 W.The convective portion is 0.8 x 7000 = 5600 W.
The total sensible heat gain is 2835 + 5625 + 7000 = 15460 W.
The radiative portion is 1984.5 + 3319 + 1400 = 6703.5 W.
The convective portion is 850.5 + 2306 + 5600 = 8756.5 W.
The total latent heat gain is 2835 W.
817
Heat gain to the space = 0.8 x 6000 W = 4800 W
Problem 818
At 4.00 p.m., 70 people are present. Assuming "seated, light offiοe \Mork",the sensible heat gain per person is245 Btu/hr (72νν) and the latent heatgain per person is 200 Btu/hr (59 W).
Sensible heat gain = 245 Btu/hr/person x 70 people = 17150 Btu/hr.
Latent heat gain = 200 Btu/hr/person x 70 people = '14000 Btu/hr.
Αt 6:00 p.m., no one is present; sensible and latent heat gains are O Btu/hr.
819
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144
First, compute the properties of the corresponding fictitious surfaces, usingEqns 835, 836, 837. ResuΙts are shown in the shaded table entries,beΙow.
Surface Area (ft') Aε T(F) AεT λ εl]f τ;lF)1 Νorth roof 639.7 0.9 575.8 122 70241.8 1983.7 0',9 , 1,10.62 South roof 639.7 0.9 575.8 143 82332.6 1983.7 0.9 1ο3.93 West wall 84.0 0.9 75.6 102 7711.2 2539.4 0.9 113,84 East wall 84.0 0.9 75.6 92 6955.2 2539;4, 0.9 114.15 Αttic floor 1176.0 0.9 1058.4 95 100548.0 1'4;47.',4' 0:9 128,4
Τhen, compute the radiant interchange factor and radiation heat transfercoefficient using Eqns. 838 and 839. Using Eqn. 840, estimate theradiative heat flux from each Surface (Q,.uα), then determine the radiativeheat transfer from each Surface (Q,"rα). Then, compute the total radiativeheat transfer from all surfaοes = 69,769.5 Btu/hr. Divide by the totalSurface area, 2623.4 ft2, to get the baΙancing factor, 26'6 Btu/(hrft'1, whichmust be subtracted from the previously caΙculated heat flux from eachsurface to determine the "balanced" radiation heat flux from each surface(q,rοlbal). Multiply by the area to determine the radiation heat transfer fromeach Surface (Q,"α/bal)' Cheοk to see that they noνv Sum to zero.
Suι"face Fit Trus (R) hriQraο
(Btu/(hrft2))
Q,uo(Btu/hQ
qr"6/bal(BtΨ
(hrft'))
Q,u6lbal(Btu/hr)
I North roof 0.872 576.0 11.4 129.7 82950.9 156.3 99964.12 South roof 0.872 583.1 11.9 463.7 296624.7 490.3 313637.93 West wall ο.897 567.6 11.2 132.6 11140.0 '106.0 8906.14 East wall 0.897 562.7 1 1.0 242.5 20368.1 215.9 18134.25 Αttic floor 0.832 571.4 10.6 355.3 417837.0 328.7 386561.8
820
First, compute the properties of the corresponding fictitious surfaces, usingEqns 835, 836, 837. Results are sho\Mn in the shaded table entries,below.
Surface Area(m2) Aε T(C) ΑεΤ A, t1 Ti(e)
1 North roof 120.7 ο.9 108.7 43 4672.9 ' 372:7, 0.:,9, 38.3
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821
145
2 South roof 120.7 0.9 108.7 50 5433.6 372.7 0.9 36.02 West wall 18.0 0.9 16.2 36 583.2 475t.:5 ,0.9 39;64 East wall 18.0 0.9 16.2 38 615.6 4Ι5.5 0.;9 39.55 Attic floor 216.0 0.9 194.4 32 6220.8 277.5 ο.,9 45,3
Τhen, compute the radiant interchange factor and radiation heat transfercoefficient using Eqns. 838 and 839. Using Eqn. 840, estimate theradiative heat flux from each Surface (9,"rα), then determine the radiativeheat transfer from each Surface (Q,"α). Then, compute the total radiativeheat transfer from all surfaces = 3027.9 W. Divide by the total surfacearea, 493.5 m2, to get the balancing factor, 6.1 Wmz, which must besubtracted from the previously calculated heat flux from each surface todetermine the "balanced" radiation heat flux from each surface (q,u6lbal).Multiply by the area to determine the radiation heat transfer from eachsurface (Q,"α/bal). Check to see that they no\M Sum to zero.
Surface Fir Tuus (K) hrt Qraα ^(Wm'\ Q,"o (il4 qru6/baΙ
(Wm')Q,,ο/bal
(w)1 North roof 0.872 313.8 6.'1 28.6 3459.0 34.8 4199.92 South roof 0.872 316.2 b_J 87.2 10534.5 93.4 11275.4J West wall 0.897 310.9 6.1 22.0 395.4 15.8 285.04 East wall 0.897 31 1.9 6.2 9.4 168.4 3.2 58.05 Attic floor 0.835 31 1.8 57 76.2 16457.6 70.1 15132.3
The solution procedure is identicaΙ to that of Problem 819, except theemissivities for surfaces 1 and 2 are 0.1. Fictitious surface properties areshown in the first table.
Surface Area (ft') ε Aε T(F) AεT At ε:1 T^,(F)1 North roof 639.7 0.1 64.0 122 7804.6 1983.7 0.6 97.62 South rool 639.7 0.1 64.0 143 9148.1 1983:7, ο.6 96.6J West wall 84.0 0.9 75.6 102 7711.2 2539;4 0'5 98.64 East wall 84.0 0.9 75.6 92 6955.2 2539.4 ο.5 99.25 Attiο floor 1176.0 0.9 1058.4 oξ 100548.0 14:47.'.4 t 0,2 11'3.3
Τhe total radiative heat transfer from al surfaces = 3476.1 Btu/hr. Thebalancing factor is 1. 3 Btu/(hrft2).
Surface Fir Τ*s (R) hri 9rao(Btu/
Q,rα(Btu/hr)
Qrrα/bal(Btu/
Q,,a/bal(Btu/hr)
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146
(hrft')) (hrft'))
1 North roof 0.098 569.5 1.2 30.3 19378.4 31.6 20226.02 South rooi 0.098 579.5 1.3 60.8 38903.3 62.1 39751.03 West wall 0.874 560'ο 10.5 35.5 2984.3 36.9 3095.64 Εast wall 0.874 555.3 10.3 74.1 6220.9 72.7 6109.65 Attic floor 0.222 563.8 2.7 49.8 58521.2 48.4 56963.0
Note that the radiative heat fluxes from surfaces 1 and 2, and to surface 5
are significantly lower. (The heat flux incident on surface 5 has beenreduced by s5%.) Τhe catch is that "in real life", everything else does notremain the same. ln particular, the temperatures would changesignificantly.
The solution procedure is identical to that of Problem 820, except theemissivities for surfaces 1 and 2 are 0.1. Fictitious surface properties areshown in the first table.
SurfaceArea(m2) Aε r(c) AεT ,Λ ε',Ι Tτ(c)
1 North roof 120.7 0.1 12.1 43 519.2 372'7' ο.6 33;62 South roof 120.7 0.1 12.1 50 603.7 372,7 0'6 33ι'2
J West walΙ 18.0 0.9 16.2 Jb 583.2 .475.5 0,5 ,,33.9
4 East wall 18.0 0.9 16.2 38 615.6 ,475,,5' 0.5 33,85 Attic floor 216.0 0.9 194.4 32 6220.8 277,,5 o'.2 41.1
The total radiative heat transfer from all surfaces = 341.5 W. Thebalancing factor is 0.7 Wm2.
Surface Fy Trrs (K) hrl 9raα ^/Wm') Q,"ο (h4
qru6/bal(Wm')
Q.,a/bal(w)
1 North roof 0.098 311.4 0.7 6.3 765.2 7.0 848.7
2 South roof 0.098 314.8 0.7 117 1407.1 12.3 1490.7
J West wall 0.870 308.1 5.8 12.1 217.6 12.8 230.04 East wall 0.870 309.0 5.8 24.6 443.7 25,3 456.1
5 Attic floor 0.241 309.7 1.6 14.7 3175.0 14.0 3025.6
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822
823
147
Convective heat transfer coefficients are determined from Table 88.Coefficients for the pitched roof surfaces are based on the "Sloping  45degrees" surface position. Α more sophisticated approach would involveinterpolation. The resuΙts are summarized below.
Surface Area (ft2) τ(F) SurfacePosition
Direction ofHeat Flow
h"Btu/(hrft2F)
9""onu""tion(Btu/(hrft"))
1 North roof 639.7 122Sloping 
45 deqreesDownward o.42 15.54
2 South roof 639.7 143 Sloping 
45 deοreesDownward 0.42 24.36
3 West wall 84.0 102 Vertical Horizontal 0.56 9.52
4 East wall 84.0 92 Veftical Horizontal 0.56 3.92
5 Attic floor 1 176.ο 95 Horizontal Downward 0.18 1.80
Convective heat transfer coefficients are determined from Table 88.Coefficients for the pitched roof surfaces are based on the "Sloping  45degrees" surface position. A more sophisticated approach \Mould involveinterpolation. The results are Summarιzed below.
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824
B25
Surface Area (m2) r(c) SurfacePosition
Direction ofHeat Flow
hc(wm2K)
Qt'"onu""1onrιΛ//m')
1 Νorth roof 120.7 43Sloping 
45 deqreesDownward 2.39 33.46
2 South roof 120.7 50Sloping 
45 deqreesDownward 2.39 50.1 9
J West wall 18.0 36 Vertical Horizontal 3.18 22.26
4 Εast wall 18.0 38 Vertical Horizontal 3.18 28.62
5 Attic floor 216.0 32 Horizontal Downward 1.02 3.06
r_
826
148
First, the solar irradiation on the window is obtained in the same manner asthe solution for Problem 717. The following tables show results for thewestfacing window.
lnput DataLongitude 101.7 deg
Standard Meridian 90 degEoτ 6.2 min
Latitude 35.23 degDeclination 20.6 deg
Surf Αzimuth 270 degSurf Tilt 90 deg
A 346.4 Btu/hrft2B 0.186c 0.138cN1
RHOG 0.2
Output Data
cDSr Lsr h, " β, " Φ, " Ψ, o θ, " G,νρ" Go* Ga* Gr* G,*
15.00 13.12 16.75 69.25 229.59 40.41 74.35 283.92 76.59 27.06 30.47 134.12
The layer absorptances of the doublepane \Mindo\Λ/ v/ith '1l8 in. sheet glass(lD5a) can be found from Τable 73 as:
GDirect,outer: 7f ,774 dοg) : O.13 (χdiffuse,outer: .fiyaφrn_ 0.11
σDirect,inner :'dr(7 4deg) : 0.06 acΙiffuse'inne, : .fz'aixur, : 0.07
Then, the solar radiation absorbed by each pane of the doublepanewindow may be determined by (neglecting incident solar radiation from theinside):
Q"itob,o,b,d,outer, j,ρ: 0.13(76.5g) + O.11(57.53) : 16.29 εtu/(hr_ft^2)
Q"itob,o,b,d,n*',," i, θ :0.06(76.5g) + 0.07(57.53) : 8.62Βtνl(hrft2)
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149
First, the solar irradiation on the window is obtained in the same manner asthe solution for Problem 717 ' Τhe following tables show results for thewestfacing window.
lnput DataLongitude 108.53 deg
Standard Meridian 105 degEOT 6.2 min
Latitude 45.8 degDeclination 20.6 deg
Surf Αzimuth 270 degSurf Τilt 90 deg
A 346.4 Btu/hrft2B 0.186c 0.138cN1
RHOG 0.2
Output Data
MDST LSr h, " β,' Φ,' Ψ, o θ, o Grvo* Gρ" Gd* Gπ* Gt*
15.00 13.66 24.92 57.57 227.35 42.65 66.77 277.89 109.60 29.57 27.29 166.46
The layer absorptances of the doublepane \Λ/indo\M \Λ/ith 1/8 in. sheet glass(lD5a) can be found from Table 73 as:
&Direct,outer: .ir(67 deg) : 0.L27 ddffise'outer: fι,aι1urr: 0.11
aDirect,inner: 7t1167 deg) : 0.073 σc]iffuse,inner: .lz,aιρr, : 0.07
Τhen, the Solar radiation absorbed by each pane of the doublepanewindow may be determined by (neglecting incident solar radiation from theinside):
Q" it ob,o,bnd,outer' j, ο : 0.Ι27 (0g.6) + 0. 1 1 (56.s6) : 20.1 7 Btu/(hrft2)
Q"ιt ob,o,bud,inner, j, θ : 0.073(109.6) + 0.07(56.86) : 1 1.98 Btu/(hrft2)
827
RεsULTs BY ΤΗΕ Ηts ΜΕΤΗΟD UsιΝG T}*Γ ΗVΑCεXpLORΕRpRΟBRΜΑ ARΕ l{lcllΕR Τι.ΑΝ RΕ$ULΤ$ βY ΤΗf; RΤS ΜETl*iΟDUsl ΝG ΤltΕ spRΕΑD$ι* HΕΤ.
828
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150
The RTS method is used to obtain the cooling load results for this problem.
The following table shows total cooling loads and cooling loads due to
window heaigains for both lowe and regular doublepane windo\Λ/S' Αιl
cooling loads due to other heat gains are the same as those shown in
Example 816.
Hour
LowE Windows(from ExamPle 816)
Regular Windows
WindowConduοtion
(Btu,hr)
WindowSΗG
(Btu,&r)
Total(Btu/hr)
WindοwConduction
(Btu,&ιr)
WindowSΗG
(Btu,4lr)
Total(BtuAr)
1 186 364 4418 228 420 4516
2 146 299 3843 179 345 3921
2 110 246 3352 135 284 3414
4 79 203 2940 97 234 2989
5 58 167 2623 72 193 2662
6 52 tJo 2419 63 159 2452
7 61 254 2465 75 293 2518
8 92 465 2737 112 537 2829
9 145 710 8'190 178 820 8333
10 215 978 9562 toc 1129 9761
11 300 1247 1 0883 JOO 1437 11141
12 389 1492 12143 477 1720 12458
13 469 1694 13275 574 1 951 I 3637
14 533 1 833 14250 654 2111 14648
15 577 1897 1 5007 707 2185 15425
16 593 188'1 1 5486 726 2167 1 5905
17 585 1787 15701 717 2060 16105
1B 553 1624 1 0635 677 1 873 11008
19 503 139'1 9550 616 '1604 9877
20 444 1 089 8460 544 1256 8727
21 386 839 7477 472 968 7692
22 327 674 6588 401 777 6765
23 274 546 5777 336 629 5922
24 228 445 5057 279 513 5176
As shown in the above table, usιng the regular \Μindo\Μ Would resuιt in
slighily higher cooling loads than using the lowe window. The following
figure illustrates the 'hcrease in cooling loads due to changing the type of
w]ndow from the lowe window to the reguιar window.
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151
Cooling Load Comparison
αldΦocr)
:Ξooζ)
18000
'15000
12000
9000
6000
3000
0't0 '13 '16
Tirne, Hour
829
830
SolutΙon to be provided by an instructor.
First, the solar irradiation must be determined and is the same as thatshown for Problem 810. Τhen, the hourly dry bulb temperature iscalculated using Εquation 82. Here, the hour nearest to the local solartime has been used to determine the temperature. Finally, the solairtemperature is calculated using Εquation 863 with the thermaΙ radiationcorrection term being zero for a vertical surface.
Local lnsolation Outdoor SolairSolar (Btu/h Drybutb Temp
Clock Time Time ft2) Τemp (F) (F)1.00 23.79 0.00 75.2 75.22.00 0.79 0.00 73.9 73.93.00 1.79 0.00 72.6 72.64.00 2.79 0.00 71.6 71.65.00 3.79 0.00 70.9 70.96.00 4.79 0.00 70.6 70.6
Exceφts from this work may be reproduοed by instruοtors for distribution on a notforprofit basis fοr testing or instruοtional puφoses only tostudents enrolled in courses flοr whiοh the tΘΧtbook has been adopted. Αny oιher reprοduction or trα.nslαtion οf this νork beyond ihαι permiιιedby Sectiοns ] 07 or ] 08 of the 1 976 tJnited Stαtes Copyright Αct y'iιhouι the peιmission of ιhe copyright ονner ii untαwful.
152
7.008.009.0010.001 1.0012.0013.0014.0015.0016.00'17.00
'18.00
19.0020.0021.0022.0023.0024.00
ClockTime1.002.003.004.005.006.007.008.009.0010.0011.0012.0013.0014.0015.00'!6.00
17.0018.0019.0ο20.0021.0022.00
5.796.797.798.799.7910.7911.7912.7913.7914.7915.7916.7917.7918.7919.7920.7921.7922.79
9.9323.0832.2439.2944.5347.9152.42111 .11
178.01230.32259.13252.42184.976.'100.000.000.0ο0.00
73.177.081 .1
85.890.795.7100.6115.4130.8142.1147.1143.9127.788.684.1
81.378.776.7
SolairTemp (F)
73.071.269.668.166.966.066.369.072.379.990.6100.4108.8114.61 16.6115.1110.2102.396.991.885.781.8
71.172.474.778.08'1.8
86.'1
90.293.295.296.095.293.590.787.484.181.378.776.7
831
This problem uses the same solution procedure as Problem 830. Notethat the solar irradiation is the same as that shown for Problem 81 1.
LocalSolarTime23.150.1 5
1.152.153.154.155.156.157.158.1 5
9.1510.1511.1512.151 3.1514.1515.'1 5
16.1517.15'18.15
19.1520.15
OutdoorΙnsolation Drybulb(Btu/hft2) Temp (F)
0.00 73.00.00 71.20.00 69.60.00 68.10.00 66.90.00 66.03.96 65.717.83 66.329.69 67.862.38 70.5107.15 74.5142.58 79.0164.39 84.2170.24 89.0159.50 92.7133.31 95.194.57 96.048.18 95.126.46 93.014.28 89.60.32 85.70.00 81.8
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153
23.0024.00
21.1522.15
78.475.4
78.475.4
LongitudeStandard Meridian
EOTLatitude
DeclinationSurf Azimuth
Surf ΤiltAparBparCparCN
RHOG
0.000.00
106.62 deg105 deg6.2 min35.05 deg20.6 deg0 deg0 deg
346.4 Btu/hrft20.1 860.1 38
1
0.2
832
This problem uses the simiΙar Solution procedure as Problem 83O. First,the solar irradiation is determined for the flat roof using the proceduredescribed in Chapter 7. The resuΙts are shown below.
Ιnput Data
MDST Lsr h, o β,.1.00 23.79 176.83 34.272.00 0.79 168.17 33.233.00 1 .79 153.17 28.804.00 2.79 138.17 21.655.00 3.79 123.17 12.546.00 4.79 108.17 2.117.00 5.79 93.17 9.'198.00 6.79 78.17 21.059.00 7.79 63.17 33.2210.00 8.79 48.17 45.491 1.00 9.79 33.17 57 51
12.00 10.79 18.17 68.46'13.00 11.79 3.17 75.2814.00 12.79 11.83 72.1915.00 13.79 26.83 62.3616.00 14.79 41.83 50.6317.00 15.79 56.83 38.4118.00 16.79 71.83 26.1719.00 17 .79 86.83 14.1520.00 18.79 101 .83 2.5821 .00 19.79 1 16.B3 8.2722.00 20.79 131.83 '18.0023.00 21.79 146.83 26.06
Output Data
Φ, " Ψ, o θ, o Gruo* Go* Gα* Gπ* Gt*
356.41 356.41 124.27 0.00 0.00 0.00 0.00 0.0013.26 13.26 123.23 0.00 0.00 0.00 0.00 0.0028.82 28.82 1 18.80 0.00 0.00 0.00 0.00 0.0042.20 42.20 111.65 0.00 0.00 0.00 0.00 0.0053.39 53.39 102.54 0.00 0.00 0.00 0.00 0.0062'87 62'87 92'11 0.00 0.00 0.00 0.ο0 0.0071 .22 71.22 80.81 108.07 17 .26 14.91 0.00 32.1779.02 79.02 68.95 206.38 74.12 28.48 0.00 102.6086.89 86.89 56.78 246.69 135.17 34.04 0.00 169.2195.77 95.77 44.51 266.87 190.31 36.83 0.00 227.14107.55 107 .55 32.49 277 .85 234.37 38.34 0.00 272.71127.34 127.34 21.54 283.62 263.81 39.14 0.00 302.95168.24 168.24 14.72 285.80 276.42 39.44 0.00 315.86218.87 218.87 17.81 284.93 271.28 39.32 0.00 310.60245.60 245.60 27.64 280.80 248.75 38.75 0.00 287.50259.78 259.78 39.37 272.32 210'52 37.58 0.ο0 248'10269.56 269.56 51.59 256.78 159.55 35.44 0.00 194.98277 '71 277 '71 63.83 227 .20 10ο.20 31 .35 0.00 131 .56285.45 285.45 75.85 '161.85 39.s6 22.33 0.00 61.90293.49 293.49 87.42 5.53 0.25 0.76 0.00 1.01302.43 302.43 98.27 0.00 0.00 0.00 0.00 0.00312.83 312.83 108.00 0.00 0.00 0.00 0.00 0.00325.24 325.24 116.06 0.00 0.00 0.00 0.00 0.00
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154
24.00 22.79 161.83 31.74 339.93 339.93 121.74 0.00*Unit of lrradiation is Βtu/hrft2
Then, the solair temperature is determined using Equation 863 with thethermal radiation correοtion term being 7 "F for a horizontal surface.
0.000.000.000.00
Clock Time1.002.003.004.005.006.007.008.0ο9.0010.0011.0012.0013.ο014.0015.ο016.0017.0018.0019.0020.0021.0022.0023.0024.00
Local lnsolationSolar (Btu/hTime tt2)23.79 0.000.79 0.ο01.79 0.002.79 0.003.79 0.004.79 0.005.79 32.176.79 102.607.79 169.218.79 227.149.79 272.7110.79 302.9511.79 315.8612.79 310.6ο13.79 287.5014.79 248.1015.79 194.9816.79 131.5617.79 61.9018.79 1 .01
19,79 0.0020.79 0.0021.79 0.0022.79 0.00
Outdoor SolairDrybulb Temp
Τemp (F) (F)
75.2 68.273.9 66.972.6 65.671.6 64.670.9 63.970.6 63.671 .1 70.572.4 85.974.7 101.578.0 116.481.8 129.386.1 139.790.2 146.393.2 148.395.2 145.796.0 138.695.2 127.293.5 112.890.7 96.087.4 80.684.1 77.181.3 74.378.7 71.776.7 69.7
833
For hour 15, Equation 864 is used to find the conduction heat flux.
Q"conduction = .0052 x (151 '274) + .001 44 x (138'174) +
.00645 x (120.374)...
= 2.897 Btu/(hrft2)
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834
Equation 864 is used to find the conduction heat flux for each hour.
Hourq"
(Btu/(hrft2))
Hourq"
(Btu/(hrft2))
1 1.835 13 0.828
2 1.824 14 0.798
J 1.772 15 0.791
4 1.693 16 0.810
5 1.595 17 0.861
6 1.486 '18 0.948
7 1.372 19 1 .071
8 1.259 20 1.225ο 1.149 21 1.396
10 1.047 22 1.563
11 0.956 23 1.704
12 0.882 24 1.797
835
For hour 12, Εqυation 864 is used to find the conduction heat flux.
Q"conduction = 0.0061 92 x (143.972) + 0.044510 x (1 34.372) +
0.047321 x (1 21 .472)...
= 7 .028 Btu/(hrft2)
836
Equation 864 is used to find the conduction heat fΙux for each hour.
Ηour
q"(Btu/(hr
ft2)) Hour
q"(Btu/(hr
ft2))
1 0.674 13 1.050
2 0.401 14 1.544
3 0.1 99 15 2.0124 0.051 16 2.409
155
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156
5 0.058 17 2.6946 0.138 18 2.841
7 0.197 19 2.8348 0.232 20 2.671
9 0.209 21 2.361
10 0.075 22 1.936
11 0.194 23 1.466
12 0.583 24 1.031
Equation 864 is used to find the conduction heat flux for each hour withsoΙair temperatures calculated in Problem 831.
Hour
q"(Btu/(hr
ft2)) Hour
q"(Btu/(hr
ft2))
1 3.492 13 0.055
2 3.147 14 0.1 65
3 2.758 15 0.563
4 2.348 16 1.112
5 1.937 17 1.754
6 1.536 18 2.417
7 1.154 19 3.026
I 0.796 20 3.510
9 0.472 21 3.823
10 0.1 99 22 3.958
11 0.00ο 23 3.931
12 0.098 24 3.765
Using the simplified approach, the solution procedure is the same as that ofProbΙem 727. First, we need to know angΙe of incidence and Solarirradiation. Αssuming a westfacing window, the incidence angle and solarΙrradiation for Albuquerque, NM on a cear July 21 day at 3:OO pm solartime are (see solution in Problem 810 for reference)
Angle of lncidence θ = 65.0 deg.,Direct Solar lrradiation. Gρ = 1 18.6 Btu/hrft2,
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837
838
B39
157
Diffuse Solar lrradiation: Ga a Gκ = 30.6 + 28.9 = 59'4 Btu/hrft2
Then, the area of the glazing and of the frame is calculated to be 27 .2 ft2and 4.8 ftz, respectively.
From Table 73, solar heat gain coefficients for the glazing system lD 5bare
SHGG9ο(65") = 0.515 and SHGGsα = 0.60.
From Τable 52, the outside surface conductance may be estimated to be4.0 Btu/hrft2'F.
From Table 56, the Uvalue for the fixed, double glazed window havingaluminumclad wood/vinyl frame with insulated spacers is 0.48 Btu/hrft2"F.
From Τable 71, solar absorptance of the vinyl frame painted white is 0.26.
Αssuming the window with no setback (Ar,r'" = Asuπ), the SHGC for theframe can be calculated using Eq. 731 as:
SHGG1= 0.26*(0.4814.0) = 0.031.
For an unshaded window, the total solar heat gain is calculated using Eq.7 32 as
Qsμc = (0.51 5*27 '2 + 0.0314.8)'1 '18.6
+ (0.6027 '2 + 0.031*4.8)*59.4 = 2657 '2 Btυlhr'
This problem uses the same solution procedure as the previous problem.Assuming a southfacing window, the incidence angle and solar irradiationfor Boise, ]D on a clear Jυly 21 day at 3:00 pm solar time are (see solutionin Problem 81 '1 for reference)
Angle of lncidence' θ = 68.8 deg.,Direct Solar lrradiation: Gρ = 101.6 Btu/hrft2,Diffuse Solar lrradiation: Ga + Gκ = 29'0 + 28.9 = 57 '9 Btu/hrft2, and
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t '158
The area of the glazing and of the frame is the same as that calculated inProblem 838.
From Table 73, solar heat gain coefficients for the glazing system lD 29aare
SHGG,ο(68.8") = 0'408 and SHGGsα= 0.57.
From Table 56, the Uvalue for the fixed, triple glazed window havingaluminumclad wood/vinyl frame with insulated spacers is 0.44 Btu/hrft2"F.
The outside surface conductance and solar absorptance of the frame areassumed to be the same as those in Problem 838.
Assuming the window with no setback (Ar,u'" = Asuπ), the SHGC for theframe can be calculated using Eq. 73'1 as:
SHGGr = 0.26*(0 .4414.0) = 0.029.
For an unshaded window, the total solar heat gain is calculated using Eq.732 as.
Qsuc = φ'40827 '2 + 0.0294.8)*101.6+ (0.57*27.2 + 0.029*4.8)57.9 = 2047.4 Btu/hr.
840
First, determine conduction heat gain by multiplying fluxes from Problem 833 by the surface area, 8OO ft2. Then, from Table 820, select theradiative/convective split to be 63%137o/o Apply the split to determine theconvective and radiative heat gains. Then, apply Equation 867 to theradiative heat gains to determine the radiative cooling load. Sum theradiative cooling load and the convective heat gain to get the cooling load.
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HourConductionHeat Gain(Btu/h0
ConvectiveHG
RadiativeHG
RadiativeCoolingLoad
CoolingLoad
1 5462.3 2021.0 3441.2 2903.7 4924.72 4899.6 1812.8 3ο86.7 2813.1 4625.9
3 4334.5 1603.8 2730.7 2702.9 4306.7
4 3796.4 1404.7 2391.7 2583.0 3987.7
5 3300.5 1221 2 2079.3 2460.2 3681.4b 2854.1 1056.0 1798.1 2339.0 3395.07 2460.6 910.4 1550.2 2222.4 3132.9
8 2123.8 785.8 1338.0 2113.0 2898.99 1854.2 686.1 1168.2 2014.0 2700.1
10 1673.6 619.2 1054.4 '1930.6 2549.9
11 1598.0 591.3 1006.7 1868.0 2459.212 1629.0 602.7 1026.3 1829.3 2432.1
13 1759.4 651.0 1 108.4 '1815.8 2466.8
14 1983.0 733.7 1249.3 1827.8 2561.5
15 2318.0 857.6 1460.3 1868.9 2726.616 2803.7 1037.4 1766.3 1947.4 2984.817 3450.5 1276.7 2173.8 2068.9 3345.518 4215.9 1559.9 2656.0 2230.3 3790.2
19 5016.0 1855.9 3160.1 2419.2 4275.220 5741.3 2124.3 3617.0 2614.7 4739.0
21 6266.5 23'18.6 3947.9 2789.4 5108.022 6473.8 2395.3 4078.5 2913.3 5308.6ZC 6345.1 2347.7 3997.4 2969.4 5317.1
24 5971.2 2209.4 376'1.9 2961.2 5170.6
'159
Cooling Loads and Heat Gains
7ο00.0
60ο0.0
5000.0
4000.0
3000.0
2000.0
1 000.0
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Ιt
ΦΘG6)Ιoδ
ωoJ
0.0
16ο
841
First, determine conduction heat gain by multiplying fluxes from Problem 8
35 by the surface area, 1OOO ft2. Then, from Τable 820, select the
radiative/convective split to be 84oλl16o/o' Apply the split to determine the
convective and radiative heat gains. Then, apply Εquation 867 to the
radiative heat gains to determine the radiative cooling load. Sum the
radiative cooling load and the convective heat gain to get the cooling load.
HourConductionΗeat Gain(Btuihr)
ConvectiveHG
RadiativeHG
RadiativeCooling
Load
CoolingLoad
1 2227.2 356.4 1870.8 4864.4 5220.7
2 1338.2 214.1 1124.1 4530.5 4744.6
J 627.7 100.4 527.3 4222.8 4323.2
4 57.3 9.2 48.2 3938.6 3947.7
5 397.4 63.6 333.8 3676.2 3612.7
6 746.0 119.4 626.7 3436.0 3316.7
7 935.8 149.7 786.0 3228.9 3079.2
o 610.8 97.7 513.1 3120.5 3022.8
9 507.2 81.2 426.1 3179.7 3260.8
10 2313.5 370.2 1943.3 3417.1 3787.2
11 4567.1 730.7 3836.3 3808.3 4539.0
12 7028.4 1124.5 5903.8 4316.0 5440.6
13 9455.0 1512.8 7942.2 4893.4 6406.2
14 116ο9.7 1857.6 9752.2 5486.8 7344.4
15 13293.3 2126.9 I 1 166.3 6042.8 8169.7
16 1 4350.1 2296.0 12054.1 651 1.5 8807.6
17 14672.7 2347.6 12325.1 6849.5 9197.1'18 14222.5 2275.6 I 1946.9 7024.8 9300.4
19 13ο18.3 2082.9 1Ο935.4 7018.0 9101.0
20 11142.9 1782.9 9360.0 6824.5 8607.3
21 8809.6 1 409.5 7400.1 6467.5 7877.0
22 6593.1 1ο54.9 s538.2 6042.8 7097.7
z3 4782.8 765.3 4017.6 5622.2 6387.5
24 3353.'1 536.5 2816.6 5228.0 5764.5
Exc€φts frοm this work may be reprοduοed by instructors for distribution οn a notforprofit basis for testing or instructional puφoses only to
students enτo]led in courses for which the textbook has been adopted. Αny oιher reprοducιion or trαnsιCιιion of thιS work beyond ιhαt permitted
η s;"iι'n' ]07 οr ]08 οfthe Ι976 Uniιed Stαtes Copyrighι Αctwiιhout ιhe permission οfιhe cοpyright oνner is unlαwful.
161
cooling Loads and Ηeat Gains
L
flαl
(!(,(!Φ!οδ
ΦoJ
1 6000.0
1 4000.0
1 2000.0
'10000.0
8000.0
6000.0
r Conduction Heat Gain(Btu/h0
*x* Cooling Load
842
First, determine conduction heat gain by multiplying fΙuxes from Problem 836 by the surface area, 1200 ft2 ' Ther , from Τable 82o, seΙect theradiative/convective Split to be 84γoμe% Αpply the split to determine theconvective and radiative heat gains. Then, apply Equation 867 to theradiative heat gains to determine the radiative cooling load. Sum theradiative cooling load and the convective heat gain to get the cooling load.
Hour
ConductionHeat Gain(Btιl/hr)
ConvectiveHG
RadiativeΗG
RadiativeCoolingLoad
CoolingLoad
1 809.2 129.5 679.7 1203.0 1332.42 481.0 77.0 404.1 1 105.3 1182.33 238.5 38.2 200.4 1018.3 1056.54 61.2 9.8 51.4 941.3 951.15 69.3 11 .1 58.2 872.8 861.76 166.0 26.6 139.5 811.4 784.97 237.0 37.9 199.0 756.2 718.3o 278.7 44.6 234.1 707.9 663.3I 250.9 40.2 210.8 674.0 633.810 90.4 14.5 75.9 667.9 653.411 233.3 37.3 196,0 699.7 737.112 700.2 112.0 588.2 771.3 883.3
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911131517192123
13 1260.3 201.6 1058.6 877.1 1078.8
14 1852.6 296.4 1556.2 1007.4 1303.8
15 2414.9 386.4 2028.5 1149.7 1 536.1
16 2890.5 462.5 2428.0 1290.4 1752.9
17 3233.3 517.3 2716.0 1416.6 1933.9
18 3409.7 545.6 2864.2 1516.5 2062.019 3401.3 544.2 2857.1 1580.7 2124.920 3205.0 512.8 2692.2 1602.7 2115.5
21 2833.6 453.4 2380.2 1579.7 2033.0
22 2323.3 371.7 1951 .6 1513.7 1885.5
23 1759.4 281.5 1477.9 1417.5 1699.0
24 1237.6 198.0 1039.6 1309.3 1507.3
162
Gooling Loads and Heat Gains
4000.0
3500.0
3000.0
2500.0
2000.0_ο_ Conduοtion Heat Gain
(Btu/hr)
*_α_* Cooling Load1500.0
'1000.0
500.0
0.0
500.0
First, the hourly Soιar heat gains are determined using the same solutionprocedure Shoν1/n in Problem 838. The results are Sho\Λ/n below. Notethat the SolaΓ irradiation on the window is the Same aS that shown in
Problem 810. Also, note that the calculated Soιar gain at 3:00 p.m. isslightly different from that shown in Problem 838 due to rounding errors.
αt
ΦΘ(τΦΙcδ
ιl1oJ
843
lnput DataGlass AreaFrame Αrea
27.2 ft',
4.8 ft2
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students enrolled in courses for whiοh thο textboοk has been adopted. Αny οιher reproducιiοn or ιrαfiSlαιiοn of this work beyond thαι permiιιed
by Secιions ]07 οr ]08 ofthe Ι97 i United Stαtes Copyrιghι Αct wiιhout the permission οfιhe copyright oνner is unΙαwfuΙ.
163
Diffuse SHGCAngular SHGC  0'Angular SHGC  40'Angular SHGC  50'Angular SHGC  60"Angular SHGC  70"Αngular SHGc  80"
Frame sΗGc
0.60.70.670.640.580.450.23
0.03'1
Output Datalnc. Dir lrradiation, Diff lrradiation, Solar Heat Gain,
Αngle, ' Btu/hrft2 Btu/hrft2 Btu/hr87.03 0.00 0.00 0.00101 .06 0.0ο 0.00 0.00114.99 0.00 0.00 0.00128.63 0.0ο 0.00 0'00141.58 0.00 0.00 0.00152.80 0.00 0.00 0.00159.17 0.00 9.93 163.51156.37 0.00 23.08 380.04146.65 0.00 32.24 530.96134.23 0.00 39.29 647.01120.81 0.00 44.53 733.28106.97 0.00 47 .91 788.9892.97 0.00 52.42 863.2878.94 54.68 56.44 1314.4465.01 1 18.63 59.38 2657 .0551.37 170.01 60.32 3940.0638.42 201.20 57.93 4657.1227.20 202.07 50.35 4594.5520.83 151.27 33.70 3393.4023.63 5.07 1.03 111.7433.35 0.0ο 0.00 0.0045.77 0.00 0.00 0.0059.19 0.00 0.00 0.0073.03 0.00 0.00 0.00
ClockTime1.002.003.004.005.006.007.008.009.0010.0011.0012.0013.0014,0015.0016.0017.0018.0ο19.0020.0021.0022.0023.0024.00
ln the original RΤS methodology, two types of radiant time faοtors v/ereutilized to convert Soar heat gains into cooling Ιoads. The Solar.RTS wasused to convert the beam transmitted solar gain whiΙe the NonsoΙarRΤSV/aS used to convert all other Solar gains' However, to simpΙify thecalculations, only one RTS (NonsolarRTs) is used in this edition. Sincethe calculated Solar heat gains include both transmitleΞ aιd absorbed Solargains, the recommended radiative and convectivΞsplits shown in Table 820 would not be applicable. For this problem, it is assumed that theradiative fraction of the combined solar heat gain is about 0.9. Therefore,
Exοeφts from this wοrk may be reproduced by instructors fοr distribution on a notforprofit basis for tΘSting or instructional purposes only tοstudentsenrolledincoursesforwhichthe textbookhasbeenadopted. Αnyοtherreproductionοrtrαnslαtionοfιhisνοrkbeyondιhαιpermiιιedby Sectiοns 107 or ]0B ofthe Ι976 Uniιed Stαtes Copyrighι Αcιwιιhοuι ιhe permissiοn ofιhe cοpyright oνner is unlατνfuΙ.
164
the radiative/convective split is 90%110%. Then, apply the split todetermine the convective and radiative heat gains and apply Equation 867to the radiative heat gains to determine the radiative cooling load. Αnd,finally, sum the radiative cooling load and the convective heat gain to getthe cooling load.
Hour
SolarHeat Gain
(Btu/hr)Convective
HGRadiative
HG
RadiativeCooling
LoadCooling
Load
1 0.0 0.0 0.0 141 0 141.0
2 0.0 0.0 ο.0 91.6 91.6
3 0.0 0.0 0.0 59.8 59.8
4 0.0 0.0 0.0 39.2 39.2
5 0.0 0.0 0.0 25.9 25.9
6 0.0 0.0 0.0 17.2 17,2
7 163.5 16.4 147.2 87.6 103.9
I 380.0 38.0 342.0 215.2 253.2
9 531.0 53.1 477.9 339.6 392.7
10 647.0 64.7 582.3 450.5 515.2
11 733.3 73.3 660.0 543.8 617.1
12 789.0 78.9 710.1 615.8 694.7
13 863.3 86.3 777.0 687.3 773.6
14 1314.4 131.4 1 183.0 932.5 1064.0
15 2657.1 265.7 2391.3 1661.8 1927.5
16 3940.1 394.0 3546.1 2566.6 2960.617 4657.1 465.7 4191.4 3304.7 3770.4
18 4594.6 459.5 4135.1 3630.8 4090.3
19 3393.4 339.3 3054.1 3261.0 3600.4
20 111.7 11.2 100.6 1624.6 1635.8
21 0.0 0.0 0.0 912.1 912.1
22 0.0 0.ο 0.0 549.4 549.4
23 0.0 0.0 0.0 343.2 343.2
24 0.0 0.0 ο.0 218.7 218.7
Exοerpts from this work may be reproduοed by instructors for distribution on a notforprofit basis for testing or instructional puφoses onιy to
students enrolled in courses }or whiοh the textbook has been adopted. Αny other reproducιiοn or ιrαnsιαιion of ιhis νοrk beyond thαι permiιιed
by Secιions 1 07 or ] 08 of the 1 97 6 Uniιecl Sιαιes Copyrιghι Αcι uithοuι the permissiοn οf ιhe copyright owner is unΙαινful.
165
IΦ
(!ΘGΦJ.Φ(!oJ
5000.0
4500.0
4000.0
3500.0
3000.0
2500.0
2000.0
1500.0
1 000.0
500.0
Cooling Loads and Heat Gains
OutPut DataDir lrrad, Btu/hr Diff lrrad, Btu/hr
ft2 f1'
0.00 0.000.00 0.000.00 0.00
_ο Soiar Heat Gain (Btu/hr)
*x Cooling Load
Solar Heat Gain,Btu/hr0.000.000.00
0.0
B44
This problem uses the same solution procedures as Problem 843. Note
that the solar irradiation on the window is the same as that shown in
Problem 811. Also, note that the calculated Soιar gain at 3:00 p.m. is
slightly different from that shown in Problem 839 due to rounding errors.
ThΞ radiative/convective split of 90%l10% is also used for this problem.
lnPut DataGlass Area 27.2 ft'Frame Area 4.8 ft2
Diffuse SHGC 0.57Angular SΗGc  0" 0.68Angular SHGC  40' 0.65Αngular SHGO  50' 0.62
Angular SHGC  60' 0.54
Angular SHGC  70' 0.39Angular SHGC  80" 0.18
Frame SHGC 0.029
ClockTime1.002.003.00
lnc.Angle,'152.14154.11'150.59
EΧοeφtS from this work may be reprοduοed by instructors fbr distribution on a notforprofit basis for testing or instructional puφoses only to
students enrolled in courses fbr which the textbook has been adopted. Αny other reproiucιιon or trαnslαιiοn οf ιhis νork beyond thαι Permiιιed
by Sectiοns Ι 07 or Ι 08 οf the ] 97 6 (]niιed Stαtes Copyright Αct wiιhouι the permission οf the copyrighι oινner is unlωυful.
166
4.005.006.007.008.009.0010.001 1.0012.00'13.00
14.0015.0016.0017.0018.0019.0020.0021.0022.0023.0024.00
143.20133.87123.75113.42103.2993.6684.8877.3771.5868.0267.0668.8273.0979.4487.3896.45106.26116.49126.80136.76145.67
0.000.000.000.000.000.0022.7058.9287.76105.541 10.30101.5680.2248.7011.340.000.000.000.000.000.00
0.000.000.003.9617.8329.6939.6848.2354.8158.8559.9457.9553.1 0
45.8636.8426.4614.280.320.0ο0.000.0ο
0.00ο.00ο.00
6'1.91278.89464.39680.731139.731721.362140.212255,352046.811551.04978.1 5592.46413.94223.454.970.000.000.00
Ηour
SolarHeat Gain(Btιl/hr\
ConvectiveHG
RadiativeΗG
RadiativeCooling
LoadCooling
Load1 0.0 ο.ο 0.0 356.3 356.32 0.0 0.0 0.0 341.8 341.83 0.0 0.0 0.0 328.6 328.64 0.0 0.0 0.0 316.4 316.45 0.0 0.0 0.0 305.0 305.0o 0.0 0.0 0.0 294.1 294.17 61.9 6.2 55.7 297.0 303.1I 278.9 27.9 251.0 338.9 366.8I 464.4 46.4 418.0 391.2 437.610 680.7 68.1 612.7 457.1 525.211 1139.7 114.0 1025.8 583.2 697.212 1721.4 172.1 1549.2 763.0 935.113 2140.2 214.0 1926.2 935.3 1149.314 2255.4 225.5 2029.8 1046.5 1272.015 2046.8 204.7 1842 1 1067.8 1272 4to 1551.0 1 55.1 1395.9 991 0 1146.117 978.1 97.8 880.3 853.6 951.418 592.5 59.2 533.2 723.9 783.119 413.9 41.4 372.6 633.6 675.020 223.4 22.3 201.1 551.8 574.121 5.0 0.5 4.5 465.4 465.922 0.0 0.0 0.0 421.3 421.323 0.0 ΑΔ 0.0 393.5 393.524 0.0 0.0
..o.0373.0 373.0
Exceφts from this wοrk may be reproduοed by instructors Γor distribution on a notforprofit basis for testing or instruοtional puφoses only tostudents enrolled in courses for whiοh the textbοok has been adoρted' Αny oιher reprοducιion or ιrαnslαιιon of ιhis νork beyond thαι permιιιedby Secιions ]07 or ]08 ofιhe Ι976 Uniιed Stαtes Cοpyrighι Αcιlνithout lhe permissiοn ofιhe copyright oνner is unΙcrννful'
Cooling Loads and Heat Gains
L
Φ
ιEΘ(EΦΙΦΦoJ
2500.0
2000.0
I 500.0
1 000.0
500.0
0.0
l_Solar Ηeat Gain (Btu/hr)
,x Cooling Load
845
167
Hourlnternal
Heat Gain(w)
ConvectiveHG
RadiativeHG
RadiativeCoolingLoad
CoolingLoad(Vψ
I 200.0 100.0 100.0 125.3 225.32 200.0 100.0 100.0 116.6 216.63 200.0 100.0 100.0 111.O 211.04 200.0 100.0 100.0 107.3 207.35 20ο.0 100.0 100.0 105.0 205.06 200.0 100.0 100.0 103.4 203.47 200.0 100.0 100.0 102.4 202.4δ 2000.0 1ο00.0 1000.0 566.7 1566.7I 2000.0 1000.0 1000.0 753.8 1753.810 2000.0 1000.0 1000.0 85'1.1 1 851 .1
11 2000.ο 10οο.ο '10οο.ο 906.9 1906.912 2000.0 1000.0 1000.0 940.8 1940.813 2ο00.0 1000.0 1000.0 962.1 1962.1
14 2000.0 1ο00.0 1000.0 975.6 1975.615 2000.0 1000.0 1000.0 984.3 1984.316 2000.0 1000.0 1000.0 990.0 1990.017 2000.0 1000.0 1000.0 993.6 1993.618 2ο00.ο 1000.0 1000.0 996.0 1996.019 200.0 100.0 100.0 532.6 632.6
EXceφtS from this wοrk may be reprοduced by instructors fοr distribution on a notfοrprofit basis for testing or instructional puφοses only tostudents enτolled in οourses for which thθ teΧtbook has been adopted. Αny οιher reproduction or trαnsΙαtion of this τνοrk beyond ιhαι permittedbySectiοns Ι07 or Ι08of ιhe ]976UniιedStαtesCοpyrightΑctwithοutthepermissiοnοf thecοpyrighιownerisunlcrννful.
20 200.0 100.0 100.0 346.2 446.221 200.0 100.0 100.ο 249.2 349.222 200.0 100.ο 1ο0.ο 193.6 293.623 200.0 100.0 100.0 159.9 259.924 200.0 100.0 100.0 138.7 238.7
'168
Cooling Loads and Heat Gains
2500.0
2000.0
1500.0+ lnternal l1eat Gain (W)
*x Cooling Load (\Λ/)
Hour
lnternalHeat Gain
rw)Convective
HGRadiative
HG
RadiativeCooling
Load
CoolingLoad(w)
1 200.0 100.0 100.0 405.'1 505.12 200.ο 1οο.0 '100.0 392.9 492.93 2ο0.ο 100.0 100.0 381.7 481.74 200.0 100.0 100.0 371.2 471.25 200.0 100.0 100.0 361.3 461.36 200.0 100.0 100.0 352.1 452.17 200.0 100.0 100.0 343.3 443.3I 2000.0 1000.0 1000.0 518.9 15'18.9o 2000.0 1000.0 1000.0 562.8 1562.810 2000.0 1000.0 1ο00.0 59ο.6 1590.611 2000.0 1000.0 1000.0 612.3 1612.312 2000.0 1ο00.ο 10ο0.ο 630.8 1 630.8'13 2000.0 1000.0 1000.0 647.1 1647.1
14 2000.0 10οΟ.0 1000.0 661.9 1661.9
Εxοeφts from this work may be reproduced by instΙuctors for distribution on a notforprofit basis for testing or instructional puφoses only tostudents enrolΙed in οourses for which the textbook has been adopted' Αny oιher reprοductιοn or ιrαnsl(tιion of ιhis work beyond thαt permiιιedby Sectiοns Ι 07 or ] 08 of the 1 976 Uniιed Stαtes Copyrιghι Αcι'ννiιhouι ιhe permissionbf the cοpyrighι οwner is unΙαlνful.
Ξ'6o(!(Σ)Ιoδ
(υoJ
846
Γ
'169
15 2000'ο 100ο.0 1000.0 675.5 1675.516 2000.0 1000.0 1000.0 688.2 '1688.2
17 2000.0 1000.0 1000.0 700.0 1700.018 2000.0 1000.0 1000.0 711 .1 1711 .1
19 200.0 100.0 100.0 537.6 637.620 200.0 100.0 100.ο 495.7 595.721 20ο.ο 100.0 1ο0.ο 469.9 569.922 200.0 100.0 100.0 449.9 549.923 200.0 100.0 100.0 433.2 533.224 200.0 100.0 100.ο 418.4 518.4
Gooling Loads and Heat Gains
2500.0
2ο00.0
1500.0
1000.0
500.0
0.0
ο lnternal Fleat Gain (W1
*s* οooling Load (W)
&x** *
; :. .::l L.*
ΕΧcerpts tiom this ινork may be reproduοed by instructοrs for distributiοn on a not_forprofit basis fοr testing or instruοtional puφoses only tostudents enrolΙed in οouτses 1br whiοh the textbook has been adopted. Αny oιher reproduction or trαnsιαιιon of this work beyond ιhαι permiιιedby Secιiοns ]07 or ]08 οfιhe Ι976 LΙnited StαιeS Cοpyright Αcι νiιhouι ιhe permiSSiοn οfιhe copyrighι oνner is unlαwful'
ΞoΞ(!o
σ,:=oooτ,(τ,
aι'
'6Θ(τ,α)t
170
847
Gomparison of LW and Mνι/ 1 Zone Responses
2000.0
1500.0
1000.0
50ο.ο
Ι
ι*.\Η\ 5**+
ι ξ'. _l)*ι ':'*Ι^__x
r lnternal Heat Gain β)_α.* MW1 Zone Clg. Ld αv)_.* ΗW Zone Clg. Ld. (W)
232119171511 '13
Hour
As shown in the figure, there is a signifrcant difference in the response ofthe two Zones, with the ΗW zone having substantiatly more damping andtime delay.
Exceφts f?om this work may be reproduοed by instructors for distribution on a notforprofit basis for testing or instructional puφoses only tostudents enrolled in courses fοr whiοh thΘ textbook has been adopted. Αny oιher reproclucιiοn or ιrαnsιαιionbf ιhιs work beyoλd ihαι permiιιedby Sections ]07 οr ]08 ofιhe }976 United Stαιes Cοpyright Αctwithοut the permissiοn οfιhe cοpyrighι οwner ιi unlανful'
171
848
Assumptions applied to each heat gain are discussed in the solution toProbΙem 815. The equipment heat gain is assumed to be continuous. Thetotal convective and radiative heat gains are determined in the folΙowingtable. Τhe Ιatent cooΙing Ιoads are equivalent to the latent heat gainsshown in the last column.
Name: ξ99p]9_ Liοhtino Eοuipment ΤotalTotal
RadiativeTotal
Convective
Latentfrom
PeopleRadiativeFraction: 0.7 0.59 0.2
lloυr
ΗeatGain(\Λ/)
ΗeatGain/w)
Heat Gain(w)
HeatGain (W)
HeatGain (W)
Heat GainrιΛΛ
HeatGain(w)
1 0 0 4000 4000 800 320C) 02 U 0 4000 4000 800 320ο 03 0 0 400ο 4000 800 3200 04 ο 0 4000 4ο00 800 3200 05 0 0 4000 4000 800 3200 06 0 0 4000 4000 800 3200 07 ο 0 4000 4000 800 3200 n
o 2160 4800 4000 1 0960 5144 581 6 1 350I 2160 4800 400ο 1 0960 5144 581 6 1 35010 2160 4800 4000 1 0960 5144 581 6 1 35011 2160 4800 4000 1 0960 5144 5816 1 35012 2160 4800 4000 1 0960 5144 581 6 1 35013 2160 4800 4000 1 0960 5144 581 6 1 35014 2160 4800 4000 1 0960 5144 581 6 1 35ο15 2160 48CI0 4000 1 ο960 5144 5816 1 35016 2160 4800 4000 1 0960 5144 581 6 1 35017 2160 4800 4000 1 0960 5144 58l 6 1 35018 0 4800 4000 8800 3632 5168 019 0 0 4000 4000 800 3200 020 U n 4000 40ο0 80ο 32ο0 021 ο 0 4000 400ο 800 3200 022 0 4οο0 400ο 800 3200 023 0 0 4000 4000 800 3200 024 0 0 4000 4000 800 32ο0 0
The sensible loads are then determined from the radiative and convectiveheat gains using Equation 867 and the radiant time factors from Τable 821, as shown in the next table.
ΕΧceφts liom this work may be reproduced by instructors 1br dlsιribιltlon οn a noιtbrpτofit basis for testing or instructiona] purposes onΙy tosιuderrts en:olled itι οοιlrses fοr ιhich the teΧιbook has bοen adοptοd. lπ;l οιΙιer reprοdιιction or trαnslcιιioιι cf ιhis ιι'ork beyond ιhαι ρerfrιiιtedby Sectiοlιs ] 07 οr Ι08 οf ιhe Ι 97 6 Uniιed StαιeS Copyrighι Αcι ινiιhouι ιlιe ρernιission οf the cοpyrighι ο'!'ner iS unlαιν.ful'
172
Hour
lnternalHeat Gain
(w)Convective
ΗGRadiative
HG
RadiativeCooling
Load
CoolingLoad(w)
1 4000.0 32ο0.0 800.0 2218.4 5418.4
2 4000.0 3200.0 800.0 2162.5 5362.5
3 4000.0 3200.0 800.0 2110.6 5310.6
4 40ο0.0 3200.0 800.0 2062.1 5262.1
5 4000.0 3200.0 800.0 2016.7 5216.7
6 4000.0 3200.0 800.0 1973.8 5173.8
7 4000.0 3200.0 800.0 1933.3 5133.3
8 10960.0 5816.0 5144.0 2782.4 8598.4
9 10960.0 5816.0 5144.0 2996.'1 8812.1
10 10960.0 5816.0 5144.0 3131.2 8947.2
11 10960.0 5816.0 5144.0 3237.5 9053.5
12 10960.0 5816.0 5144.0 3328.ο 9144.0'13 10960.0 5816.0 5144.0 3408.1 9224.1
14 1096ο.0 5816.0 5144.0 3480.7 9296.7
15 10960.0 5816.0 5144.0 3547.6 9363.6
16 '10960.0 58 1 6.0 5144.0 3609.6 9425.6
17 10960.0 5816.0 5144.0 JOO /. / 9483.7
18 88ο0'0 5168.0 3632.0 3413.3 8581.3'19 4000.0 3200.0 800.0 2799.1 5999.1
20 4000.0 320ο.0 8ο0.0 2625.5 5825.5
21 4000.0 3200.0 800.0 2512.4 5712.4
22 4000.0 3200.0 800.0 2422.9 5622.9
23 4000.0 3200.0 800.0 2346.8 5546.8
24 4000.0 3200.0 800.0 2279.4 5479.4
Cooling Loads and Heat Gains
Ξ'6ΘΦΦΞoδ
π,oJ
120ο0.0
10000.0
8000.0
6000.0
4000.0
2000.0
0.0
ι
Εxοeφts fiom this woτk may be reproduced by instructofs for distrlbution on a notforprofit basis for testing or instructional purposΘs only to
students enrolled in οouτses foτ which thΘ textbοok has been adopted' Αny other reprοducιion or trαnsιαιion of this νοrk beyond ιhαt permiιted
by Secιions Ι07 or Ι08 οfthe ]976 ιJnited Stαtel Cοpyrighι Αct\'ιithout the permissiοn ofιhe cοpyrighι owner is unΙαlυfuΙ.
849
173
Assumptions applied to each heat gain are discussed in the solution toProblem 8'16. The equipment heat gain is assumed to be continuous; thelighting heat gain is assumed to occur from 8 a.m.6 p.m. The totalconvective and radiative heat gains are determined in the following table.The latent cooling Ιoads are equivalent to the latent heat gains shown in thelast column.
Name: People Liohtinο Eουioment TotalΤotal
RadiativeTotal
Convective
Latentfrom
Peοοte
RadiativeFraction: 0.7 ο.59 0.2
Hour
HeatGain(w)
HeatGain(w)
Heat Gainrw)
HeatGain (W)
HeatGain (W)
Ηeat Gain/\ΛΛ
HeatGain(w)
1 0 0 70ο0 70οο 1400 5600 0
2 0 ο 7000 7000 1400 5600 0
3 0 0 7000 7ο00 1400 5600 0
4 0 0 7000 7000 1400 5600 0
5 U 0 7000 70ο0 1400 5600 0
6 0 0 7000 7000 1400 560ο 0
7 0 0 7000 7ο00 140ο 5600 0
I 2835 5625 7000 1 5460 6703.25 8756.75 28359 2835 5625 7000 1 5460 6703.25 8756.75 2835't0 2835 5625 7000 1 5460 6703.25 8756.75 283511 2835 5625 700ο 1 5460 6703.25 8756.75 283512 2835 5625 7000 1 5460 6703.25 8756.75 283513 2835 5625 7000 1 5460 6703.25 8756.75 283514 2835 5625 7000 1 5460 6703.25 8756.75 2835
15 2835 5625 7000 1 5460 6703.25 8756.75 2835'16 2835 5625 7000 1 5460 6703,25 8756.75 283517 2835 5625 7000 '15460 6703.25 8756.75 2835
18 0 5625 70ο0 12625 4718.75 7906.25 0
19 0 0 7000 7000 1 400 5600 0
20 0 0 7000 7000 1400 5600 0
21 0 0 7000 7000 1400 5600 0
22 0 ο 7000 7000 1 400 5600 0
23 0 ο 7000 700ο 1400 5600 ο
24 0 0 7000 7000 1400 5600 0
The sensible loads are then determined from the radiative and convectiveheat gains using Εquation 867 and the radiant time factors from Τable 821 , as shown in the next table.
EXceφtS fτοm this work may be reproduοed by instruοtors for distribution on a not_for_prοfit basis Γor testing or instructional puφoses only to
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by Secιions ]07 or ]08 ofthe Ι976 LΙniιed Stαtes Cοpyright Αctlυithοut the permissiοn οfιhe cοpyright oνner is unlανful.
174
Hour
lnternalHeat Gain
(w)Convective
ΗGRadiative
HG
RadiativeCooling
Load
CoolingLoad(w)
1 7000.0 5600.0 1400.0 2888.2 8488.2
2 7000.ο 5600.0 1400.0 2830.6 8430.6
3 7000.0 5600.0 1 400.0 2777.5 8377,5
4 7000.0 5600.0 14ο0.0 2727.8 8327.8
5 7000.ο 5600.0 14ο0.0 2680.7 8280.7
6 7000.0 5600.0 1 400.0 2635.8 8235.8
7 7000.ο 5600.0 140ο.0 2592.7 8192.7
8 15460.0 8756.8 6703.3 3814.6 12571.3
9 15460.0 8756.8 6703.3 4292.6 13049.3
10 15460.0 8756.8 6703.3 4539.8 13296.5
11 15460.0 8756.8 6703.3 4694.4 13451.2
12 15460.0 8756.8 6703.3 4806.9 13563.6
'13 1546ο.0 8756.8 6703.3 4897.4 13654.2
14 15460.0 8756.8 6703.3 4975.3 13732.1
15 15460.0 8756.8 6703.3 5045.3 13802.0
ιo 15460.0 8756.8 6703.3 5109.7 13866.5
17 15460.0 8756.8 6703.3 5170.2 '13926.9
'18 12625.0 7906.3 4718.8 4754.8 12661.0
19 7000.0 5600.0 1400.0 3825.2 9425.2
20 7000.0 5600.0 1400.0 3446.7 9046.7
21 7000.0 5600.0 1400.0 3246.7 8846.7
22 7000.0 5600.0 1400.0 31 19.9 8719.9
23 7000.0 5600.0 1400.0 3027.2 8627.2
24 7000.0 5600.0 1400.0 2952.4 8552.4
Cooling Loads and Heat Gains
Ξ'6Θ(!q)ΞoE
IEoJ
18000.0
16000.0
14000.0
'12000.0
10000.0
8000.0
6ο00.0
4000.0
2000.0
0.0
G lnternal Fteat Gain (W;
*x* Cooling Load (νγ)
Εxcerpts tioιτ this work may be reproduced by instΙuctors fοτ distribution οn a notforprofit basis for tΘsting or instruοtional puφoses only to
students enrolled in οourses tbr which the textbook has been adopted. Αny other reprodλc'tiοn or trαnsιαιιon of thιs νork beyond ιhαt permitted
iysr,itιon, 107 οr ]0Bοf the 1976ΙJnitedStαtesCopyrighιΑctlνithoutthipermissiοnof thecopyrightoνnerisunΙαινful'
850
175
Heat gain to the space = 0.8 x 6000 W = 4800 W, assumed 59% radiative,41o/o convective, from Τable 820' The sensible loads are then determinedfrom the radiative and convective heat gains using Equation 867 and theradiant time factors from Table 821, as shown below. Τhere are no latentcooling loads.
Ηour
lnternalHeat Gain
(w)Convective
ΗGRadiative
HG
RadiativeCooling
Load
CoolingLoad(w)
1 0.0 0.0 0.0 79.8 79.8
2 0.0 0.0 0.0 52.4 52.4
3 0.0 0.0 0.0 34.7 34.7
4 0.0 0.0 0.0 23.1 23.1
5 0.0 0.0 0.0 '15.6 15.6
6 4800.0 1968.0 2832.0 1473.9 3441.9
7 4800.0 1968.0 2832.4 2060.5 4028.5
8 4800.0 1968.0 2832.0 2365.5 4333.5
9 4800.0 1968.0 2832.0 2540.5 4508.5
10 4800.0 1968,0 2832.0 2646.7 4614.7
11 4800.0 1968.0 2832.0 2713.2 4681.2
12 4800.0 '1968.0 2832.0 2755.6 4723.6
13 4800.0 1968.0 2832.0 2782.9 4750.9
14 4800.0 1968.0 2832.0 2800.5 4768.5
15 4800.0 1968.0 2832.0 2812.0 4780.0
16 4800.0 1968.0 2832.0 2819.5 4787.5
17 4800.0 1968.0 2832.0 2824.3 4792.3
18 48ο0.0 1968.0 2832.0 2827.5 4795.5't9 0.0 0.0 0.0 1366.5 1366.5
20 0.0 0.0 0.0 777.9 777.9
21 0.0 0.0 0.0 471.8 471.8
22 0.0 0.0 0.0 296.0 296.0
23 0.0 0.0 0.0 189.3 189.3
24 0.0 0.0 0.0 122.5 122.5
A plot showing the lighting heat gain and resuιting cooling loads follows.
Excerpts frοm this work may be reproduced by instructors for distribution οn a notforpro1it basis 1br testing οr instruοtional puφoses onΙy to
students enτolΙed in courses tbr ινhiοh the tsxtbook has been adopted. Αny οther reproclucιion or trαnsιαtιon οf this ινork beyοnd ιhαt permitιed
by Secιionι 107 οr ]08 ofthe Ι976 United Sιaιes Copyri?hι Αcι1,ithouι the permission ofιhe copyrighι olνner is unlανful.
176
t6000.0
5000.0
4000.0
3000.0
Ξ,6
Θ(υΦΙoδ
ιl,oJ
{ nternal Fleat Gain (\Λl)
___x* Cooling Load (\Λ/)
57 9 11 13 15 17 19 21 23
Hour
ι
85'1
The schedule described in problem 818 is reduced to the number of
peopιe present per hour in the table below. Assuming "Seated, light office
*ork'', the sensible heat gain per person is 245 Btu/hr (72νν) and the latent
heat gain per person is 2bO Btu/hr (59 W). lnternal heat gains from
occupants are assumed to be 70o/o radiative. The latent cooling loads are
equivalent to the latent heat gains shown in the table.
The sensible loads are then determined from the radiative and convective
heat gains using Equation 867 and the radiant time factors from Table 8
21 , as shown below.
As is readily evident from the plot, the heavyweight zone significantly
damps the response to the heat gains'
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by Secιiοns ]07 or ]08 ofιhe Ι976 United SιαιeS Copyri?ht ΑcιΙyithouι ιhi permissiοn οfthe copyrighι owner is unlανful'
Cooling Loads and Heat Gains
177
HourPeopleοresent
lnternalHeat Gain
(w)Convective
HGRadiative
HG
RadiativeCooling
Load
CoolingLoad(w)
Latentheatgain(\Λ/)
1 0 0.0 0.0 0.0 20.6 20.6 0
2 0 0.0 0.0 0.0 12.5 12.5 ο
3 0 0.0 0.0 0.0 7.8 7.8 0
4 0 0.0 0.0 ο.0 5.0 5.0 0
5 0 0.0 0.0 0.0 3.4 3.4 0
6 0 0.0 0.0 0.0 2.5 2.5 0
7 0 0.0 0.0 0.0 2.0 2.0 0
I 0 0.0 0.0 0.0 1.6 1.6 0
9 40 2880.0 864.0 2016.0 1021.8 1885.8 2360
10 40 2880.0 864.0 2016.0 1484.5 2348.5 2360
11 60 4320.0 1296.0 3024.0 2233.7 3529.7 3540
12 60 4320.0 1296.0 3024.0 2593.8 3889.8 3540
13 60 4320.0 '1296.0 3024.0 2784.5 4080.5 3540
14 70 5040.0 1512.0 3528.0 3143.9 4655.9 4't30'15 70 5040.0 1512.0 3528.0 33'17.9 4829.9 4130
16 70 5040.0 1512.0 3528.0 3410.8 4922.8 4130
17 10 720.0 216.0 504.0 1931 .3 2147.3 590
18 0 0.0 ο.0 0.0 1010.7 1010.7 0
19 0 0.0 0.0 0.0 552.7 552.7 0
20 0 0.0 ο.0 0.0 309.2 309.2 0
21 0 0.0 0.0 0.0 175.8 175.8 0
22 0 0.0 0.0 0.0 101.1 101.1 0
23 0 0.0 0.0 0.0 58.8 58.8 0
24 0 ο.0 0.0 0.0 34.6 34.6 0
Cooling Loads and Heat Gains
Ξ'ΞΘπ,ΦΞοδ
π,oJ
6000.0
5000.0
4000.0
3000.0
2ο00.0
1000.0
* lnternal Heat Gain (W1
*s* Cooling Load (W)
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students enrolled in course s for which the textbook has been adopted. Αny oιher reproducιiοn or ιrαnsιαιion of ιhιs νork beyond ιhαt permitιed
by Secιiοns ] 07 or ] 08 οf the Ι 97 6 (Jnitecl Sιαιes Copyrιghι Αcι Ιiιhοuι the permissiοn οf ιhe copyright οwner is unlανfuΙ.
178
Solution to be provided by an instructor.
Excerpts from this wοrk may be reprοduced by instru:lor:.fo, distribution on a notforprofit basis for
testing or instruοtional purposes on1y ω stuαeλs enrο1led i, "or.r"s for which the textboοk has been
adopted. Αny other repiodλction "y γ""ititi"λ
o}thιs ινork beyo'nd thαt permitted by Sections ]07 or 10B
of the 1976 (Jnitecl itotus Copyright Αct 'ιiinλ, the permissiono/'λu copyrιgit owner is unlαwful'
h'uouurt, for permission or further ιnfοrmαtionilrouιa i" oaarrrruiΙo ιhe Piimiision Depαrtment, John
'';i,;;;'';:"{,i", il l Riνir Street' Hoboken' NJ 07030'
CHAPTER 9
From Table 91, the number of average degree day is 6283'
From Fig. 91, Co = 0'60Using Εq' 92,
f
Or F = 438.7 mcf of natural gas
= 438,727 std ft3
= 102,867 kwhr
(24hr l dαy)(6283' F _ dαy)\2^2^1000Β tu l hr)(}'60)
(0 η(7ο_ 12"tr)(10ο \Βtu l stdff)
f (24hrldαy)(6283'F_dαy)(Ζ^?5,000Βtulhr)(0'60)(1 J) ql o _Γz' r)βaL2Βtu l kW _ hr)
$Elec = 102,867(0.1 O) = $1 0'287
$Gas = 438.7(4.5) = $1,974
SΕΙec_$Gαs 0287_ι974= 4.2
$Gαs ι914
or the electric cost is about 5'2 times as much'
(1 02,s 67 kt4l _ hDβ a lΨ !u_
l kY' hr)
@at)(000stdff lmΦSource energy using eιec' =
= 1063'6 mcf
Source energy using gas = 438'7 mcf
>_Ξ
93
94
ΕSΕ,_ΕSG 1063.6 _438'7
"o,"ff=tiU  = 1.42
That is eleο. heat uSeS 242% more Source energy'
The following are information for Washington' DU'
From Table gl,1he number of average degree day is 4224'
From Fig. 91 , CD= O'62'
From Table B1a, the outdoor temperature is 20 "F'
Forenergyefficientfurnace,assume85o/oeffiοiencyfactor,Using Εq' 92,
04hr l d αν)(4224" F _ dαy)(20'000Βt u l hr)(0'62\,,  , 'o'=177,468stdft3" =
10.85X7O_
20"trX1O00Βtu l stdft3)
Or F = 177 .5 mcf of natural gas
Qro" = ato + 5
120,OO0=a(20)+bg=2(gQ)+b
12ο,ooQ =(20 _60)aa = 3OOO, b = 180,000
Q uo" = 180,000  3'000 to
8o. = Qrn  Qιnt = 18O'oOO  3'Oo0 to  20'0ο0
ξo" = 160,000 _ 3,000 to
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Αny oιher *pa""riλλ"]r'i,o^ιoιι*7 *is work beyond ιhαt permiιted
by Secιions ]07 or 108 o7 m, l'oii'iλιi'rλi,'λ,r, copyrιgn, 'a}iΙ:i';;u' iλ'! irrλ'*iλ" of ιhe copyright oνner is unlανful'
Load Profiles
1 80000
 160000
ξ tηooooαt_ l z000ο€ lοοοoο6j sooooE' 6oooo
3 4ooooΞ 2OoOo
030 4o
Outdoor TemP, "F
95
186
96
Group
Sunday
Monday
Tuesday
WednesdayThursday
Friday
Saturday
lll lιl lVVVl
shift 2 shift 1
1481012162024Hour
Αssumes Sunday and Saturday in shift 2
See Table 92
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students enrolled in courses for whrch the textbook has been adopted Αny other reprodiction or ιrcιnsιαιion οf this wοrk beyond ιhαt permitιed
bySecιiοns ]07 οr ]08of the Ι976(JnitedStαιeSCopyrightΑcινιthoutthepermissionof ιhecopyrighιowneriSunlανfuΙ.
Groupshift 1
hrs inea. qp
shift 1
Days in
ea. qp.
Totalshift 1
hrs ea. qp
Totalhrs in
ea. οp.
Frac. ofshift 1
hrs ea. gp
Frac. ofhrs in
ea. qp.
I
il
Πt
IV
V
VI
0
0
2
4
4
4
0
0
5
5
5
5
0
0
1ο
20
20
20
28
28
28
28
28
28
0.0
0.0
0.36
0.71
0.71
0.71
'1.0
1.0
0.64
0.29
0.29
0.29
lo/
96 (Cont.)
99
97
98
The procedure is the same as Problem 96. Use appropriate bin data fromΑpp. B in last step as per Table 93.
Refer to Εxample 92, insert shift hours of Problem 96 in column 2 and 3of Table 95 and recalculate.
Reconstruct Τable93 for the appropriate city to obtain Shift A and Shift Bhours. lnsert the hours in columns 2 and 3 of Table 95 and recalculate.
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Bin.Temp
Shift t hrs in each GrouplII III IVVVI
shift I
hrsshift 2
hrs6257524742373227221712
0
0
0
0
ο
0
00
0
0
0
0
0
0
0
0
0
00
00
0
3438493635322710
I61
9677674748382817121
0
68829462655436171640
69588687997566282211
1
2672552962322471991577258222
3753463883374204223471571137316
TotaΙ: 1807 2994
188
910
Reconstruct Table 93 using the shift hour fractions from Problem 96 andbin hours and temperatures for the appropriate city. lnsert the shift hoursin column 2 and 3 of Table 95 and bin temperatures in column 1 andrecalculate.
91 1
Solution furnished by an instructor.
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by Secιions ] 07 or 108 ofιhe Ι976 [Jniιed Sιαιes Copyrighι Αcι ινithout the permission ofthe copyrighι oνner is unlανful.
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Requests for permission or further informαtion should be αddressed to the Permission Depαrtment,
John Wiley & Sons, Ιnc, ] ] ] Riνer Street, Hoboken, NJ 07030'
CHAPTER 1O
1O1. (a) ',* o+ + ρΖι_Pz+'Ψ + ρZz+ ρWp + ρ tt, Vl = VziZι=Ζz
neglect (! iιz; Wp = Η, g/g/" = 80 (ft  lbf)/lbm
Pz = Pl + ρWp = 20 + 62.4(80)1144 = 54.7 psig
Ps = Pz  PΖs_ U:)rg= 54.7 _ (62'4 X 50)/1 44 _ (20 x 62'41144)
Pg= 54.7 _ 3ο.3 = 24'4 psig
Pι=Pz PZι((izg_ (1't)g+ =54.7 ffiX25) (#)(20+ '15) =
28.7 psig
(b) Neglecting the pump, the pressure or head required for this pipe is:
ΔP = 28'7  20 psi or ΔP = 8.7 psi
Δ Η 20.1 ft.
Note:20 1
This is thecharacteristicfor only partof the totalsystem.
150
(8 x 2.31)  250 + 30 + 300 =
I87IΓlotlοlσlο)ΞΞ
0L_ oa
102. Ηl = Hz * Ηο + (. trZz=
190
'103.
104.
105. (a)Ηl=alΦι+ZιHz=ΔzQz+Ζz
Series ConnectionQl=Qz;SumΗ
Ηl_Ηz=Hp+ !,t+Ζz=0
H^ =  ('2, = 25 300 = 325ft: H^ = 325ft of head
= 970 kPa
100
Ηl = 98.5 ft of waterPι = 42'6 psig ε 294 kPa Jt
H2
z2Hq
Ξ(σοΞ5o
75
ιLI
U(σο.)c.
zaa H=Ηlr
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(Ι +ΙΙ )+ 12 + 1b
191
H=Hl *Hz
Η=Qz(aι+Az)+Gι+zz)
Parallel ConnectionHl = Ηz, Ζι=ZzorZ=0Sum Q
a
4.026106.
12
Ql= ^E,Φz= ^EYar YazQ = Q l + Q r= JH_Ζ(Jil q * ^[ν η) .
6ra*M)'+Ζ
t t=f!Y1 ; V = 6. 3ft/sec, L = 3OO ft, D =' D2g
Referring to Figures 102a and 1 02b
Re = ρνD _ 62'4(1 .04)6'30(4'026 l12) = 75,696
μ (2'7 11490)
ε = 0.00015 ft; Table 101
ε/d=o.ooo1 5" 12
4.026
ε/D = 0.00045; f = 0'022 Fig. 101
/3OO) (6.30)2( , = 0.022 \' = 12.1 ft of 30o/o E.G.sol.(4.026 t 12) 2(32.17)
= 12.6 ft of water : 38 kPa
'.ΞExcerpts frοm this work may be reproduοed by instructors for djstribution on a nοtforprofit basis for testing or instruοtional puφoses only to
students enrolled in courses Γor which the textbook has been adopΙed' Αny οther reproduction or ιrαnsιαtion of ιhis νork beyond ιhαιpermitted by Secιions ] 07 or 1 08 of the ] 976 United Sιαtes Cοpyright Αct νithouι ιhe permission of ιhe copyright οwner is unlανful.
192
107. (a) so
Ξ(σ
925J
Qn = 48 gpm;
28ft
Qn = 60 gpmQs = 32 gpmQc = 41 gpm
100
gpm
Qs = 24 gpm; Qc = 32 gpm;
125 gpm
a2
125 150
(b)
(c)
(d)
108.
} ο_
Refer to Problem 107; Ζι = Ζz _ 0
(a) Hn= uoQi, θn= 9= + =O.9O8Oai 50'
Hs = auQ3, Θs = 9= =
= o'o278a6 30'
Hc = a.Q3, θc = +=+ =o.O'148aa 45'
H 02
(^n/o!o8 + JΤt ozτa + J.roιua)2 645.06= 0.001 55Q2
(b) Q = 100 gpm; H = O.OO155 x (1OO),  1S.5 ftExcerpts from this work may be reproduced by instruοtors for distribution on a nοtforprοfit basis for testing οr instructional purposes only tostudents enro]Ιed in courses for which the textbook has been adopted' Αny other reproduction or ιrαnslαtion of this work beyoncl ιhαιpermiιιedbySectiοns Ι07 or l08 of ιhe ]976LlniιedSιalesCοpyrightΑctιι]iιhouιιhepermissionοf ιhe copyrighιονnerisunlανfuΙ.
193Hn = Hs = Ηc = 'l5.5 ft
Qn= JΠ/rΑ =.,/l55/O.OO8 =44gpm
Q g = .'m lo'o2?8 = 23'6 οpm
Qc = Jss/oJl4g = 32.4 gom
(c) From (a) above. H = 0.001 55(12q2 = 24.2 tt
Qn= =55qpm
Qg = Jπ2loβ278 = 29.5 gpm
Qc = J%2toβ148 = 4O.4 qpm
e = 125 gpm
1o9. Q = CαA r2g"(P, _Pr)l'''' D2
=84.8 = O.55'L ρ ] 'D1 154.1 v'vv
aSSume Cα = 0.638 using Fig. 109;
Ar= ζ (O.os4s)2 = O.00565 m24Pι_Pz
= O.O98(13.55)9.s = 13.O13 J/kgρ
Q = 0.63S(0.0056 5)t2(13,013)11/2 = 0.0184 m3/s x 292 gpm
V z = 3.26 m/s; Re  999(3'26)(0:085) = 1.98 x'105
1.4x103
C6 ^:
0.638 From Fig. 109.
Τherefore the original assumption is satisfactory.
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ι
194
1o1o. (a)V.. = Γ2g" (Po, _ρ,ll"' _z x s?'ιτ x o.os x
L" [ ρ )] L ι0 +91x 144)144
1t2
=3.97 ft/sec
Table A1a; ρ = 59'83 lbm/ft3
(b)rir = ρVA(O. 821= 59.83 x 3.97 r+][qΨ''l2 * o.εzι4rι 12 )rh = 39 lbm/sec or 140,674 lbm/hr
1011 (a)
Read from Fig. 1011a at 35 ft and '125 gpm, W. = '1.6 HP
Q = 180 gpm, H, = 20 ft; 1.8 ΗP
Τhis is actually out of the operating range of the pump and the
efficiency is very low. ln situations like this there is a danger of
overloading the pump motor; however, that does not appear to be
a problem in this case since the motor is probably a 2HP model.
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40
#35Ξ
(σΦT2s
(b)
(c)
(d)
195
1012. NPSHR =
NPSHA =
to20='uΧ
12
NPSHΑ = 20 ft (Figure 10_1 1b)
[e"e") 2,r, f&9"]\ ρg / ι ρg)
( l'soτxιt_\ az.z13.55x62.4 Ζ"_2_ 44
62.2
Ζ"=32'85_20_2_1'17 = 9.68 ft; (2.)rr, = 9.68 ft
'1013. (a) 231 gpm, ξp = 73'4%W, = 12* tlP
(b) 225 gpm ηo = 73 '3o/o
W, = 12 ΗP
1014. η, = 73.3oλ; W, = 11. 5 HP
225 gpm; '185 ft of head
ηr=73.5%;Ws=14ΗP
(c) ηo= 73o/o, W. = 14 ΗP 225 231
gpm
1015 From Problem 1014b, the original system defined by 225 gpm and 149ft of head and would operate at 242 gpm, 173 ft of head and require14.4 HP with the 7in impellor. Τhen,
EΧcerpts 1iom this work may be reprοduced by Ιnstruοtors for distribution on a notforprofit basis for testing or instruοtiοnal puφoses οnly tostudents enrolled in οοurses for which the textbook has been adopled' Αny οιher reproducιion or ιrαnsιαιion οf ιhis wοrk beyοnd thαιpermiιted by Sections ] 07 or Ι 08 of ιhe Ι 97 6 United Sιαtes Copyrighι Αcι ινithouι ιhe permission of ιhe cοpyright οlυner is unlαwful.
_.ι _Ι
200
140
100
185
149
(a)
(b)
196
ΓPln= 35οo[+.l =3o37\242 )
H^ = Π3 (go37 )' = l3O ft" ι3500/
w^ 1 4.4(s037)' = e.4 HP" ι3500,The Εfficiency wouΙd not Chan ge,74.2oλ
1o16 Dn = "r(#)=
7(o e7) = 6 8 in
Hn = *, (Ψ)' = 173 (o'g44)= 163 ft
η 1. .[Ψ)u = 14'4(o'g17) = 13 2 ΗP
The Εfficiency would not Chan ge,74.2oλ
1017. Uses Fig. 1 020 or program PIPE
(a) 25 gρm; 1 % in., V < 4 ftlsec,2in dia. or less(b) 40 gpm; 2in. V < 4 fVsec,2in dia. or less(c) 15 gpm; 1Υιin., v < 4 fUsec, 2in dia. or less(d) 60 gρm;2% in', '(,'1< 4 pVsec; dia > 2 in.
(e) 2OO gpm; g %in., .(,'1< 4 ftllOO pUsec; dia > 2in.(η 2ooo gpm; 8 in., /1 slightly > 4 ftl1oo ft
Exοerpts 1iom this work may be reprοduced by instructors fbr distribution on a notfοrprofit basis for testing or instructional puφosos only tostudents enrolΙed in courses tbr which the textbook has been adopted' Αny οιher reproduction or trαnsιαιion of this νork beyond ιhαιpermiιιed by Secιions ] 07 or Ι 08 of ιhe ] 97 6 Unιted Sιαιes Copyrighι Αcι ν ithοuι the permission of ιhe cοpγ,ι'ighι ονner is unlαwful.
1971ο18. (a) K = 30 ft, ft = 0.019; K = 0.57 (Table 102; Figure 1022a)
V = 3.82 fVsec; ! r = 0'57(3'822lβ2.2x2)= O.13 ft
(b) K = 340 ft, ft= 0.017; K = 5.78V = 5.0 f/sec; ! r = 5.78 x 5.02t132.2 x2) = 2.24 ft
(c) K= 60ft, ft= 0.018; K = 1.08V = 6.5 ft/sec; ! r= 1.08 x 6.52t(2x32.2) = 0.71 ft
'101e. ! r = 2.31 (#)' = 10.8 ft of water or 4.7 psi.
1020. Assume com. stl. pipe
Q = O.O3 mt/s = 108 m3/hr, size pipe for about 4 mllOO m
From Fig. 1020, use 5 inch pipe, lD = 130 mm
nt  a,λ' f _ ..25 ml100 m; [1= (3'251100)200 = 6.5 m of water or 63.7 kPa
ΔPg = 35 kPaΓ o991'
= ι7 '27 kPaFor strainer. ΔP"   '" 1 0.00722 J
Then for the pump:
ΔPp = 63.7 + 35 + '17 .3 + 3(1000)(9 .807)11000 = 145'4 kPa
Ηp = 145.419'807 = 14.8 m
Q = O.O3 mu/s  30 L/s
1021. Size the pipe using Fig. 1020 or program PIPE. Fitting equivalentlengths found using Fig. 1022a; 1022b and Τable 102' ProgramPιPE could be used to solve the complete problem including fitting
losses. Data for hard calculations are summarized below:
Εxcerpts from this work may be reproduced by instruοtors for distribution on a notforpro1it basis for testing or instruοtional puφoses οnly tostudents enrolΙed in οourses Γor whiοh the textbook has been adopted' Αny other reproducιion or ιrαnsΙαtion of ιhis νork beyond ιhαtpermitted by Sections ]07 or Ι08 ofιhe Ι976 Uniιed SιζlιeS Cοpyright Αcιlνiιhοuι ιhe permission ofιhe cοpyrighι oνner is unΙανful.
L
ro(3) I o(3)
lO(3) Θ/o\
1ot3] Θ \:/
198
] 5ι5)
Sec.No.
qpm itft./'100 ft
Left.
ιtft.
Con.Valve ft.
Coilft.
Τotalft.stze tn.
1120
3.38 45 1.5 1.53
570
3.64 '15 0.6 0.62.5
640
3.'1 24 0.7 11.4 12.12
740
3.'1 13 0.4 12.0 12.42
490
5.84 27 1.6 1.62.5
10120
3.38 42 1.4 1.43
25ο
4.7 22 1 10.0 11.02
350
4.7 26 1.2 10.0 11.22
I 306.3 28 1.8 14.4 16.21.5
930
6.3 13 0.8 15.0 15.81.5ch 120 20
Exοerpts lrom this work may be reproduced by instruοtors for distribution on a not1brprofit basis for testing or instructional purposes only tostudents enrol]ed in courses Γor whiοh the tΘXtbook has been adopted' Αny οιher reproducιiοn or trαislαtiοn of ιhis νori beyoncl ιhαιpermitted by Secιions ] 07 or Ι 08 of ιhe ] 976 (Ιnιted SιαιeS Copyrighι Αct wiιhοut the peimission οf ιhe cοpyrighι owλer is unΙcnνful.
199
The head losses forsame.
the three parallel runs are approximately the
For run (156Z410), Hp = 49.6 ftFor run (15Sg10), Hp = 55.5 ftFor run (123410), Hp = 46J ft
Therefore, a pump should be selected to provide about 56 ft of head at120 gpm.
1022. 500 gpm, Use 5 inch pipe; !'f = 4.17 fil1OO ft
V = 8.0 ftlsec
Length of pipe = 160 + 3O + 12 = 202 ft
65 in elbows = ,lS ft (Figure 10_22)
35 in gate valve = 12 ft
15 in gΙobe valve = 130 ft; Total equivalent length = 419 ft
. 4.17(41e\/, = Ι_1}J "/ = 17 '5 ft of water
For strainer: ! "=
2.31 [#)' = e.24ft of water
For cond  ! "o
= 20 ft of water
Τhen Hp = 17 '5 + 9.24 + 20 + (3o _ 12) = 64.7 ft at 5OO gpm
1023'Use Εq. 1033
L
Exοerpts frοm this work may be reproduοed by instructors for distribution on a notΓorprofit basis fbr testing or instructional puφosοs only tostudents enrolled in οourses fοr which the textbook has been adopted Αny οιher Ιeprοdactιon or ιrq'nsιαιion of ιhis νork beyοnd thαιpermiιιed by Sections ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyright Αct νiιhλut ιhe peimission of the cοpyτ.ighι owner is unlανfuΙ.
,,_ _ 6oott*:;*g ,l)  3x6 sxlou (1,0 oull
".  = 19.4 9a1. = 74 L
I gz.οgο og.οgο ]
οoo[[9 Ψξg _,'l _ 3x6.5x1 o_u (1,, o _ ou1l1024' Use Eq. 1034 v, = L[ 0'ο1ο0zz ) '____J
ι '_ 69:ο%,Vr_8.Ζgal = 33 L
200
1025. Use Eq. 1033tl = 60oF, P2= 50 psig, P1 = 20 psig, v1 = O.O16ο53 ft3/lbm
vz = O.O1 6772 ft3/lbm, tz = 220"F
Vτ= = _11_5_gal.  435 L
1026(a) Use Eq. '1016
P,+ PτPzι= Pzgzz +ρνv+ρg'g"9cι
'' = ff (zzzι)+ ρw +
Exοerpts from this work may be reproduced by instruοtors for distribution on a nοt_forprofit basis for testing οr instructional purposes only tostudents enrolled in οourses for which the textbook has been adopted,' Αny other reproductιon or ffαλιαιion of ιhis wori beyondιhαιpermiιιed by Secιiοns ]07 οr ]08 οfιhe Ι976 L]niιed Sιαιes CopyrightΑctνΙιhouι ιhe peimission οfthe copyrighι owλer is untιιυful'
lL9c
Τ24o tΙ,
ΙP9n
LE
9cl

(b)
= # e4o) #(60) + # es1=Be psis or 61 2 kpa201
P,'+ρg!= P2+ρg29c  gc
Pz=Pι t (zιzz)=89.o effiPz = 15 psig = i03 kpa or about o absorute
(c) No, makeup water is not available to overcome a pressure of ggpsig. However, the domestic water system probabry has abooster pump.
1027 (a) Pι = Pz * o* (zzzι) + ρνν * ρ9ι,9c gc
= 5 + 62.4(240)144
=Pz+ !99c
Pl = 109 psig
62.4(60) . 62.4(25)144 144 'n0.8
Pr = 93.8 psig or 647 kpa
(b) η (zzzι) = 5 *62(,2ra0)
or 752 kPa
= 5 + 104 =.t09 psig
Excerpts from this work may be repτoduced by instructors for distributiοn οn a notforprofit basis for testing or instructiοnal puφoses only tostudents enrolled in courses for which the iextbook t,u. υ".n 1dγνωj Ατιl λrir, Ιrpr"a""rιon or ιrαnslαιion οf ιhis work beyond ιhαιpermiιιed by Sections ] 07 or Ι 08 of the ] 976 United Stαιes copyrighi Αcι withλuι ιhe p,e1miιsιon of ιhe copyright ονner is unlαινful.
202
1028
(c) This location is at Ιeast workable. Ηowever the pressure at thepump is still very high. The domestic service waterpressure would have to be boosted to a higher pressure at the 2othfloor.
(oo , sz) +
Qo *Q. =
x +o)= (ao x ιτ)100
(a,
Qα=
Solve Simultaneous
57Qb + (O x 1OO)  4OQb = 100 x47
= Ψ = 41.2 say 41gpm17
Q. = Q.. = 1OO  41 = 59 gpm
Size all pipe for 100 gpm
D = 3 in. from Fig. '1 020 or PIPE
1029. (a) Each chiller requires 600 gpm. Since chiller 2 is partially loaded it
must have the full flow of 600 gpm.Therefore, Q"p = 1200  750 = 450 gpm
(b) (150  60) + (450 x 42) = 600 ts, ts = 46.5 F
(c) LR = 150/600 = 0.25
(d) Main pipe to and from sec. Circuits: D = 8 in. com. stl.
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ab
(b)
(c)
203Dns = Dco = O in. com. sfl. (S in. a litile small)
D"p = 6 in. com. sfl.
Dsc = DRo = 6 Ιn. com. stl. ( could be 5 in. but easier to make all 6in.)
(e) Rpm, = ΓPIΠl 050l12oo) = 35OO(75o/12o0) = 2188
(η ιW = *#=1ff =,l_[ffiJ' =, (##) = O 756
or 75o/o
1030. (a) Q"n = 12OO 750 = 450 gpm
(b) Qrtr + Qztz = Qsts; ,. = (450x42)+ (750x60)
= 53.3 F1200
Both chillers receive the same temp. water
(c) Load ratios are the same:
LR= ##=0628or63%1031.
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204
10 (13) 2 2Ο(6) 3 *{s)T'r^;n ^ !ι }iJJυα'
**ntΓ*if Φr
u?
ι$
BaΙ*n*sve,v*
{typi*aii
Note: Piping is type L copper
1031.
Αll engths are total equivalent lengths
1*{S}
Ω*{s} s 2Ο(s) 8
**rnnr** pip*
CoilFΙow rateopm (L/s)
Lost head ft (m)
Coil Con. valveABc
40(2.5)40(2.5)50(3.2)
12(3.7)15(4.6)18(5.5)
10(3)12(3.7)15(4.6)
(continued)
SectionNo. gpm
Dia.in.
irftl1ο0'
L"ft
!.rft
12
2334^,^il C
1309050
3
2%2
3.74.85.0
602030
2.2
r; ) 355
' he reproduced by instruοtors for distribution on a not_Γorprofit basis fortiting or instructionai puφoses onΙy tohiοh the textbook has been adopted' Αny oιher λeprοduction οr ιrαislαιiοn of ιhis wori beyond ιhαι
''/ United Sιαιes Cοpyrighι Αcι ιιithouι ιhe permission of ιhe copyv.ight oνλer is unlωυful.
Con. C45Com. oi
1300
33
3.7 40151.50.0 39.2 ft (tote
26Coil ACon. A
40 2 3.4 30 1.012
10 23 ft (total)37Coil BCon. B78
40
80
2
2%
3.4
3.9
10
20
0.515120.8 28.3 ft (totat)
205
Circuit 123451 is the path of greatest lost head. From Fig. 1O1 1 choose at40 ftof head and 130 gpm the 7 in., 1750 rpm model which froduces about 43ft of head.
rfι*{}ιJ
Ξp^q 2ΟΟ 2ΟΟlbu] 2(rΟ)3(rΟ}
Ιs*ianc*vxlγ*
{typi*ai}
*οntr*i. valv*{ typi*xi}
1032.Notes: PΙpe is schedule 40, commercial steel.
Αll lengths are total equivalent lengths excuding control valves.
Circuit Flow rategpm (L/s)
Control valvehead loss ft (m)
Α 60(3.8) 40(12)B 70(4.4\ 5ο(15)c 70(4.4\ 50(15)
t p,ih*:y +ο* 4QPumP {12)l t1i
*hill*rp ιl Π]$} $
'' be reproduοed by instruοtors for distribution on a notfor_profit basis for testing or instructional puφoses on1y to"hiοh the textboοk has been adopted. Αny oιher )eproducιiοn or ιrαλhιiοn o7 tnπ 'orκ beyond thαι
'976 United Stαtes Cοpyrighι Αcι ννithouι the peλission of ιhe copyright ονner is unlαwful.
206
SectionNo. gpm
Dia.in.
ifft/'l00'
L"ft
(.t
ft
122334Con.41
20014070
200
432%
4
2.44.23.5
2.4
200200240
400
4.88.48.4509.6 81 .2 ft (total)
25Con. A54
60
140
2%
3
2.5
4.2
240
200
o408.4 54.4 tt (total)
36Con. B
70 2% 3.5 40 1.450 51.4 ft (total)
Circuit 12341 has the largest head loss of alΙ paths. Select pump for 2OOgpm at 81 ft of head. From Fig. 101 1, use: 5Υ' in., 3500 rpm model. Willoperate at 96 ft at 200 gpm.
1033 (a) qst = 20 x 12,000 x2= 480,000 Btu
/1 _ gst 480000= 512.8 ft3
ρc, (t,.t.) 62.4(1) (6045)
orQ=3,8369a1
(b) Vol = 513 ft3 ora Space 8 ftx 8 ftx 8 ft ora cyΙindrical tank 8 ftdia. x10.2 ft
1033. (continued) Solution  Sl:
(a) Qg1 = (352_280) (2)= 144k\^lhr= mc, (trt.) _ Qρcp(tt.)
q = Qst  144 x3600 = 14 m3
ρο, (t1ts ) 980(4. 184) (1 67)
' 1le reproduοed by instruοtors for distribution on a notforprofit basis for testing or instruοtional purposes only to'.'hiοh the textboοk has been adopted. Αny oιher reproduction οr trαnslαtion of ιhis wοrk beyond thαt
^76 (Ιnιιed Sιαιes Copyrighι Αcινιιhout ιhe permission οfthe cοpyrighι oιυner is unlανful'
207
(b) Vol. = 2Αmx2.4mx2.4m
1034. Solutions may/can Vary. Α typical solution is:
(a) Use 2 chillers of '15 tons total capacity in a reverse return systemsimilar to figure 1032. The piping would be routed overhead aroundthe complex with supply and return running parallel, starting andreturning to the equipment room.
(b) Total flow rate is
Qτ =16 x2'25 = 36 gpm Using PIPE or Fig 1021; Dia. = 2in'
(c) Estimated length =225x4x2= 1s00ft. Τotal Eq. Length=2x 18OO =3600 ft
Assuming an average Ιoss ofabout 2.5 ft1100 ft; The pump head required would be:
Hp = 2.5 x 3600/100 = g0 ft with flow rate of 36 gpm
'1035 Solutions may vary
(a)Figure 1034 is a schematic of what the system wouΙd be.
However, there would be 3 chillers and the secondary piping would
be routed in a square fashion around the outside of the parking
garage in reverse return.
(b)Τhe primary system would appear as in Figure 1034 with the
common pipe as shown because of the expected variable and light
load at night.
(c)The tertiary circuits would be as shown in Figure 1034 and piped in
a reverse return manner.
(d)For each building:Excerpts Γrom this work may be reproduοed by instructors Γor distribution on a notfοrprofit basis for testing or instruοtiοnal puφoses only tostudents enrolΙed in οourses for whiοh the textboοk has been adopted' Αny other reprοductiοn οr ιrαislαιion of ιhis νork' beyοnd thαtpeιmitted by Sections Ι 07 or Ι 08 of the ] 976 tJnited Sιαιes Cοpyrighι Αcι 1ψiιhouι the permission of ιhe cοpyrighι owner is unΙα:wful.
208
Qi =1500 x 't2000
= 600 gpm4x500(6045)QΤ=4x600 =2400 gpm
(e) Dia. = 10 in., Figure 1020 or plpE
1036.
1037 ' Αssume boiler pressure of 2'O psig with ΔP/L = 2'O oz or 0'125 psi/1OOft' (TabΙe 104a). Τhen, ΔP = o'125x 175l1OO = 3'5 ozor 0'22 psi ΔPis about Υzthe alΙowabΙe from Table 104a'
Assume boiler pressure of 1.o psig with ΔP/L Ξ 0'125 psi/1Oo ft.(TabΙe 104a)' Then ΔP = 0'125 x 175l1OO = 0'22 psi which is nearthe maximum in Τable 104a' Either boiler pressure could be used,but select2.0 psig to be conservative.
Εxcerpts fiom this work may be reproduced by instructors fοr distribution on a notforprofit basis for testing or instructional purposes only tostudents enrolled in cοurses for whiοh the textboοk has been adοpted. Αny oιher reproducιiοn or ιrαnsιαιion of ιhis wοrk beyoncl thαtpermiιted by Sectiοns Ι 07 or 108 of ιhe ] 97 6 (]nitect Stαιes Cοpyrigh} Αcι νith'οuι ιhe peimission of ιhe cοpyright ονλer is untωνfuΙ.
;ControΙ valve (Typical)IY φ a_Air Vent (Typical)
,+9*Π4 Heating Device (Typical)
T<TypicatTrapr€
IEι o r5erξι] Possible! ξ2 < Vacuum Breaker on each;   <'' Heatinq Device
1039. Refer to Table '105a.
The available head is = 2 x'100/110 = 1'82ftl1ο0 ft. Then at 850 lb/hr
of condensate flow, D = 1 in. nominal is adequate.
1040 (a)
q = rhcp(t,t,) = ga##(1)(6s  42)
^  1200x 1200x7'48  λ ^F^ ..ι'ι = 60 X 624(654η = 1'250 gpm
209
From Figure 1048a at 850 lb/hr; ΔP/L = 0.'125 ρsil100 ft, and boilerpressure of 2.0 psig: Pipe diameter = 4 in., with steam velocity of4,00ο ftlmin at zero psig. Correct velocity to 2'0 psig (Fig. 1049a)V = 3,800 ftlmin
1038. For each unit at full load:ft. = 283 lb/hr
Pipe size depends on slope of line, Τable 105a. For slope of 1/8 to Υιin./ft, D = 1 in. nominal specify slope of % in./ft (conservative).
(b) Αssuming no changes in the temperatures, the total flow rate wouldbe:
Φ., = Ξ9t 1l25O) = 937.5 or 938 gpmP 1200 \ '/
Τhe chillers could share the flow:
^938(J, = =469gpm'2
and be above their minimum flow of 70o/o.
jι  469 = 0.75 or or Ts%
Qmin 625EΧcerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instruοtional puφoses only tostudents enrolΙed in οourses for whiοh the textbook has been adopted. Αny οther reproducιiοn or ιrαnslαιion οf ιhis νork beyond thαιpermitted by Secιions ]07 οr Ι08 οfthe E976 LΙniιed Stαtes Copyright Αct\νithout ιhe pe'ιmission οfιhe copyrighι οwner is unΙαlvful.
210
Τhis is probably the best way to operate. Τhere would be no bypassflow and the parallel pumps couΙd operate at:
RPM, = *x 35OO = 2,626625
(c) At 6ο% full load, again assuming no change in temperature, the
totaΙ flow rate would be:
Qp = 0.6(1 250) = 750 gpm
This is too much flow for one chilΙer and not enough for two chillers
at minimum flow of 875 gpm or 438 gpm each. Τherefore, both
chilΙers will have to operate at least at 438 gpm each and some ftow
bypassed equal to:
Φυ, = 875  75O = 125 gpm
The pumps could both be slowed to:
RPMP = ,9'? ι35oο) = 2'450' 1250
(d) Αt 25% of fuΙΙ capacity, again assuming the temperatures do not
change, the flow rate for the load is:
d, = 0.25(1250) = 313 gpm
which is less than the minimum flow rate for even one chiller.
Τherefore, shut down one chiller and operate the other at least its
minimum flow of 438 gpm. The bypass flow would be:
Qop = 438  3]3 = 125 gpm
One pump would be shut down and the speed reduced for the other
pump to:
RPM, = *(3sOO) =2,453' 625Excerpts 1iom thrs work may be reproduced by instructors for distributiοn on a notforprofit basis for testing or instructional puφoses only tostudents enrolled in οοurses for whiοh the textbook has been adopted. Αny other reproclucιion οr ιrαislαιion of this work beyοnd'thαtpermitted by Sections ] 07 or Ι 08 of the ]976 Uniιed Sιαtes Cοpyrighι Αcι withοuι ιhe permission of ιhe copyright owner is unlανfut.
Excerpts from this work may be reproduοed by instructors for distribution on a notfοrprοfit basis fortesting or instructional pulposes only to students enrolled in courses for which the textbοok has beenadopted. Αny other reproduction or trαηslαtion of this work beyond thαt permitted by Sections ]07 orΙ0B of the Ι976 United Stαtes Copyright Αct without the permission of the copyright iwner is unΙαιυful.Requests for permission οr further informαtion should be αddressed to thi permission Depαrtm'ent,John Wiley & Sons, Ιnc, ] Ι ] Riνer Street, Hobοken' NJ 07030.
GHAPTΕR 1 1
111 . (a) Using Eq. 1 '11b
, = 1=L959ll,
Ao = Ψ= O.353 ft2; Αssumed K = 6ζJΑ" ' 'ν s5ο
X5o = ' 1: =ug^o)
= 68.5 ft; x1ρρ = 34.2ft; xlso = 22.8ft50.,/0.353
(b) Q, = CQov"/v,, C = 2; Εq. 112a
(Q,)so = 2(3OO)850/SO = 10,2OO ft3/min
(Q,)loo = 6Oο(85O)/1OO = 5,1O0 ft3/min
(Q,)rso = 6OO(850)/150 = 3,400 ft3/min
11_2' Using Εq. 113
t, _ t, = 0.8(to _ tι.) (V,/Vo), Δt, = 0.8(10075) V,ll'tοO
(Δt,)so = 0.8(28)50l1100 = 1'02 F
(Δt*)loo = 2'04 F
(Δt )lso = 3.06 F
113. 50 ftlmin throw = 24  6 = 18 ft
212
114. Q, =
Vo=
V,=
115.
Qo_ Xζ Q"λ;=1l3Kro'ffi = *v*  18 x 5o
= 132.71.13K '1 .13x6
From Eq. '1 1'1 and Qo = Vo x Αo; Assume K = 6
Αny combination shown would be
acceptable. The size would depend
on the available total volume flow
rate of air and the size of the space.
Q oC V "lV, , Eg. 11=2a
Q o/Ao = ιzstl L r9)'l= οsο fVminL4 \ιz1
1
V"(1.13)Kl lx ; K= 6, x = 12ft
vr=
Qr=
r636 x 6(1 .i q
^l+x(O.S)2 t12 = 159 ft/min' 'γ4
125x2x6361159 = 1000 cfm
A ceiling type diffuser system has the ability to handle large
quantities of air because the air is discharged radially and
diffuses the high velocity jet in a short distance.
116. (a) A perimeter type system would be necessary to achieve a
satisfactory heating performance. Αny other type of system
would lead to a cold and drafty floor.
ExοeΙpts Γrom this wοrk may be reproduced by instructors for distribution on a notfοr_profΙt basis for testing or instruοtional puφoses only tostudents enrolled in οοurses for which the textbοok has been adopted. Αny other reprοduction or trαnsι{ιιion οf this work beyοnd ιhαιpermiιted by Sections ]07 οr l08 ofthe Ι976 tJnited Stαtes CopyrighιΑctνiιhoutιhe permission ofιhe copyrighιονneris unlαιllfuΙ.
D^
δ ln. ft.
29 aJ 0.2539 4 0.3349 5 0.4t759 6 0.500118 t2 1.00
213(b) An overhead type system would be preferred because of thegreater need for cooling during the summer and less needfor heat durΙng the winter.
117 ' Α perimeter type system would be the best choice. This typesystem is required to do a good job of heating. Α spreading jetshould be used when heating and a nonspreading jet shouldbe used when cooling.
118' Some kind of overhead system wouΙd be preferable sincecooΙing would be the dominant mode of operation. However,ceiling diffusers with radial discharge woutd not be required dueto a low volume of circuΙated air' Α high side walt type of systemor ceiling diffusers with discharge in only one or two directionswith a Ιarge throw would be preferred. This would give themaximum air motion with a smaΙl amount of circulated air.
119. 10 in. round diffuser, TabΙe 114;650 cfm
lnterpolation between 600 & zoo cfm is required
NC=0.5x(21 17)+17=19
x5o = 0.5(1 110) + 10 = 10.5 ft
p=oo62(ffi)'=oo73in wq
1110.
For 150 cfm/ft, ΔPo = O.08 x (150116712 = O.ο65 in' wg.Excerpts from this work may be reproduced by instructors for distributiol on a notforprofit basis fοr testing or instructionaΙ puφoses only tostudents enrolled in courses for which the textbοοk has been adopted Αny oιh,er Ιλaurtιon or ιrαnslαιion οf this wοrk beyοnd thαιpermitted by Secιions ] 07 or Ι 08 of ιhe ] 976 ('nited Sιαιes Cοpyright Αct w iιhλuι ιhe p,e{mssιon of ιhe copyrighι oνner is untιnνfut.
214Throw values are for a 4 ft active length then
X5ο = 21 _ 0.6(4) = '18.6 ft;
Τhe uncorrected NC for a 10 ft length is NC = 23 _ 0.6(5) = 20.
For a length of 6 ft the correction is 2.
Corrected: NC = 20 2 = 18
1111. Model 28, 448 TBar; Table '1 16, 270 cfmlnterpolate:
NC = 0.7 (36  32) + 32 = 34.8 or 35
xso = 0.7(11 10) * l0 = 10'7 or 11 ft
, ^^'t2p=O.r[ +l =o.13in.wqι245 )
1112' From Table 111, L = 12 ft' Then from Τable 112 at
Q = 40 Btu/(hrft21, lxuo/L1rr*
=1.3 and the range is1.2  1.8, and X5s = 1.3 x12= 15.6ft
A good solution would be to use the 4 in. size with
'150 cfm/ft. with uncorrected throw of 18 ft and NC = 19.
The corrected throw is:
Xso ='18 x 0.85 = 15.3 ft and NC = 19  4= 15
P = 0.057rΨ)' = O.066 in. wqι139 ) 
1113. (a) Room char. Length = 14 ft, Table 1 1'1
(x5ρ/L)rr, = 0.8, Table 112
Range of xso/L = 0.5 to 1.5; xso = 0.8 x 14= 11.2ft
EΧοerpts fτom this work may be reproduced by instΙuctoΙS fοτ distribution on a notforprofιt basis for testing or instruοtional purposes only to
students enrolled in courses for which the textbook has been adopted. Αny other reprοducιιon or ιrαnsιαιion οf this ιυork beyond thαt
permiιted by Secιions ] 07 or Ι 08 οf the ] 976 ιJniιed Stαtes Copyrighι Αcι lυ iιhouι the permission οf the cοpyrighι οwner is unlαwful.
215The best choice would be a 12 in. size with 600 cfm
(b) xso = 13. :9 Q) = 14.3ft;x5e/L  14'3 = 1.02(in the range)80'ι ''νν'
14
ΔPo = o.o81 (ffi)' = 0.0g6 in. wg., NC = 22+ffιol =24.5
1114. Room char. Length is 26 ft,Table 112
(a) (x5ρ/L),,, = 1.6 (Table 112); range of (x5g/L) = 1'2  2'3
Xso = 1.6 x 26 = 41.6 ft; Q/diff = 60012 = 3OO οfm
From Τable 115, the 18 x 4, 14 x 5, or
12 x 6 sizes may be acceptable aΙthough the throw is
less than desired. Xso = 31 ft
Xso/L = 31.6126 = 1.2 (barely in the acceptable range)
(b) Xso = 31 ft (zero defΙection)
NC = 22*, ^po:
o.o6e (#)'= 0.065 in. ws.
1115 lt is good practice to keep the core veΙocity below 5OO ftlmin. Asolution is the 18 x '1 2; Table 117
ΔPo o.O45 [,Ψ)' = O.O57 in. wgι535/
NC= 21 + 65 fZl =24135' ι
Note that static pressure and ΔPo are negative.
1116. Guidelines:
1Place diffusers under or between double windows.
2Select throw using the ΑDPl procedure. Characteristic length
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216= floor to ceiling.
3Noise criteria (Nc) should usuaΙly be less than 30.4Be Sure that the totaΙ pressure required is compatiblewith the pressure characteristics of the system. For example,a smalΙ commercial system may have a fan that produces onlyabout 0.6 in. wg. total pressure while a large commerciat systemmay operated at25 in. wg. total pressure. The diffusertotalpressure Ιosses should be no more than abo ut 10% of thefan total pressure.
5 Use data from Table 1 13
1117 ' GuideΙines:
1center diffusers in square or nearly square spaces. Dividelarge or irregular spaces into imaginary square spaces andpΙace a diffuser in each Space. Select throw using ΑDPl procedure.
2Τry to obtain a balance between many small diffusers versusa few very rarge diffusers to be cost effective.
?_vι
4_ t See Problem 1116
5 Use data from Table 114
1118. Guidelines:
1Locate diffusers about 12 in. below ceiling on inside waΙls. Setectthrow using ADpl procedure.
2Τhe jet may be spread with this type diffuser. However, morethan one diffuser should be used where the room width is atleast two times the room depth.
3ζ.I See problem 1116
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v&
2174
5 Use data from Table 1 '1S
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ffi
218
1119. Guidelines:'1Locate grilles in ceiling near the inside wall.
2Noise criteria (NC) should be less than 30.
3The negative static pressure should be held to minimum,
especially for light commercial systems with small fans.core velocities of less than 5oo ftlmin will usually yield aquiet system with a reasonably lowpressure loss. Highervelocities and pressure ross may be tolerated with heavycommercial applications.
4 Use data from Tale 117.
1120. (a) H24, Mod 28at 89 cfm each
2Ψ'windows
(b) X1oo= 3ft; L =7 ft; X19ρ/L =3l7 = O.43; o.3 < (x19ρ/L) < 1.o89 cfm/diffuser; ΔPo  0.06 in. V/g.; ΝC = 20Νote: other acceptable soΙutions also exist.
1121. SimiΙar to 112O Diffusers shouΙd throw air towards the windows Arrange to obtain uniform air motion Might use diffusers with short throw around exposed walls with
larger units in the interΙor.
1122. L = 9 ft; Table'l i1Exοeιpts from this work may be reproduοed by instructors for distribution on a notforprofit basis for testing or instruοtional puφoses only tostιιdents enrolled in courses for whiοh the textbook has been adopled,' Αny othe, )eprocturtιon or trαnsιαtion of ιhis νork beyond ιhαιpeιmiΙιed by Secιions ] 07 or Ι 08 οf the ] 97 6 tJniιezι Sιαιes Copyrιgh'ι Αct νιthλut ιhe peλission οf the cοpyrighι oνner is unlαwful.
+l+ <l+ +l+
*l+ +te +lr
€l+ +l, €Ι_+
219X59/L = 0.9; Τable 11=2, straight Vanes ( Assume light load for a
secondary system)xso=0.9x9=8.'1 ft
Α solution: 94 ft length diffusers with 50 cfm/ft, 2 in. size,Τable 113, x = 8.5 ft (no correction required); NC = 15  4 = 11ΔPo = β0l4q2 x O.036 = O'047 in. wg.Place 3 diffusers on each exposed wall
1123. Use 4'12 in. size from Τable 114
650 cfm/diffuser; L = 20 ft 80
Room Load = 18 Btu =(hr _ ftΖ )
x5sil = 0.8, Table 112
x = 16 ft (desired)
(650  630)Xactual =
F5 _ 63οi (7 _ '15) + 15 = 15.5 ft
xact_15.5_.,fl : ^
= 0.78 (in acceptable range)
NC = 27; ΔPo=O 105 rΨ]'= O.112 in. wg.ι630/
1124' Use 14H48, Model 28 diffusers from Τable 116;
229 cfmldif. as shown. L = 20 ft, xlse/L
= 0.3 and acceptable range is 0.3 to 1.0.
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78
220Desired throw: xloo = 0.3 x 20 = 6 ft.
Actual throw: Xact = 6.5, TabΙe '1 '16
Xr"t/L = 6'5Ι20 = 0.33, o.K. 80
NC = 29, ΔPo = 0.095 in. wg. ltLr78
1125. Refer to Problem 1123, Q = 2600 cfm; refer to Table 117.
Αssume a Ιayin ceiling with 2 ftx 4 ft tiles. Τo assure a quiet
return, limit NC to about 20. Use 224 in. x24 in. grilles with
1300 cfm each. Nc < 25, ΔPo = 0.048 in. wg.
1126. Refer to Problem 11 24,3200 cfm.
Assume a2ftx4 ft layin ceiling.
Use 24 in. x 24 in. size from Τable 117 ' Using three units,
cfm/grille = 320013 = 1067; Nc < 2ο
ΔPo = 0.033 + 0.006 = 0'027 in. wg.
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adopted. Αny other reproduction or trαnslαtion of this work beyond thαt permitted by Sections ]07 or
Ι 0B of the t 976 United Stαtes Copyright Αct without the permission of the copyright owner is unlαwful.
Requests for permissiοη or further informαtion should be αddressed to the Permission Depαrtment,
John Wiley & Sons, Ιnc, ] ] ] Riνer Street, Hοboken, NJ 07030.
CHAPTER 1 2
(a) W. = rt (Pl _ Pz)lρ Αssume standard air
= Q(Pl P2) _ 2ooο(1.9)
= o.60 HP  O.45 kW
121.
6350 6350
(b) r,= ξffi =#ffi+ = O 54 or 54oλ
(c) V = 2Ooo/O '84=2,381ft/min, Pu = Q381l4ooq2 = ο.35 in wg
P. = 1.9  0.35 = 1.55 in wg
W" =2000x'1.55
= 0.496350
Ιs = Ws/Wrr, = 0'4911'1 = 0'44 or 44o/o
(d) From (c) abovet P, = 1.55 in wg
122 Qz = a, #ffi = 2ooo (ffi#) =24oocrm = 1,133 L/s
ΔP,z = ΔP.l [Hffi)'
= 1.55 (ffi#)' = 223in wg ρ 555 Pa
ΔPoz = ΔΓ (κpv")2 ι o( *Ψ9\ =z'τιin wg =682Pa".[ffi)= ι1οOo/
W, = \iν, [RPM, ]' = l '1 (12o0lu = l.9 HP = 1'42k\ΝιRPM1
' ι1000/
123. (a,b) Qz = Q (750i900) = 0.833 Q r
Poz= PO1(75ol9OO)2 = ο'694 Pο1
HPz ΗP1(75oi9oo)3 = 0.579 HPl
123
2.4
2.0
'1.6
1.2
0.8
0.4
221
(d) Po = 130 in. wg
Φ = 9625 cfm
ΗP = 2.34(LΞ
σJΞ.Ξ
I
ΦΞΦU)ΦαEΞot
6
5
4
3
2
1
00.002 468111214cfm x 103
124. Since pressure in in. wg. is plotted on the ordinate instead of head the
pressure must be adjusted to reflect the barometric pressure at 5280 ft
elevation.
Po = (Po)rtο(ρ/P.tα) = (Po).tα(Po/Po,'tα)
also, W = W.tα(ρ/Ps*o)= Wr16(P6/P5'r16)
Po,,tα = 14.696 psia; Po = O.491(29'42 _ 0.0009 x 5280); Eq. 34
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permiιιed by Sectiοns ] 07 οr Ι 08 of ιhe Ι976 LΙnitec] Sιαιes Cοpyrigh} Αcι ν)ithouι the permissiοn οf ιhe copyrighι oνner is unΙανful'
80ο RPM 7OO RPM
a Po HP a Po HP
6,000 2.3 2.75 5,250 1.76 1.84
10,000 1.87 3.5 8,750 1.43 2.34
14,000 1.15 3.45 12,250 0.88 2.31
222Po = 12.112 psia
Τhen in Denver, Co the ne\M characteristics may be obtained by
computing Po and W at various volume flow rates from Fig. 128.
Po = (Po),ro(2.112t14.696) = 0'824(P6)916 ?Πd W  0824 Wstο.
Q cfm
6,000 '10,000 14,000
Po W Po W Po W
Sea Level 2.3 2.75 1.87 3.5 1.15 3.45
Denver 1.9ο 2.27 1.54 2.88 0.95 2.84
. :. (w",0  w)1oo _ 3.5  2.88(b) ΔW=ff= 35
ΔW = 18% (decrease)
125. Refer to Problem 124 for explanation.
P6 = (99.436  0.10 x 1618) = 83.256 kPa
Po = (Po)stο = (83.256/101 '325) = 0.822(Po).16
\Λ/ = Wstο x0'822
(a)
Q m3/min
125 155 180
Po W Po W Po W
Sea Level 400 1 350 320 1 600 260 2000
Αlbuquerque 329 1110 263 1315 214 1644
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students enrolled in οourses fοr whiοh the textbook has been adopted. Αny oιher reproducιion or ιrαnsιαιiοn of ιhis wοrk beyond ιhαι
permiιted by Sections ]07 or Ι08 ofιhe Ι976 Uniιed Stαιes Copyrighι Αctwiιhout ιhe permission ofιhe copyright oνner is unlαινful.
223
(b) ΔW = (16001315)
10O = 17'ro/odecrease1 600
126. (a) This is at the limit of the good selection range. lt would be
better to choose a different fan.
(b) Α near perfect match with the fan capable of producing about
1.85 in. V/g. totaΙ pressure at 10,000 cfm.
(c) A bad application and out of the recommended range.
Would probably be unstable.
127 [From Fig. 129]
(a) No, fan is too small.
(b) No, not a good application, fan is too large.
(c) YeS, near perfect appΙication; moderate fan speed,
high efficiency.
128. 150 m3/min, 4OO Pa [From Fig 1 210J
The fan would be acceptable and is reasonable.
η1 = 55%; RPM = 85O; W, = 185o W
129. (a) At 1418 cfm = 1420 cfm, Ve = 2OOO ft/min
P" = [+Ψl' = o.zs in. wg., Po = P, + P, = O.88 in. wg.ι4005,
P. = 0.88  0.25 = 0,63 in. wg.  518 in. wg.EΧcerpts from this work may be reproduced by instructors 1br distribution οn a notforprofit basis for testing or instructional puφoses only tostudents enrolled in οourses for whiοh the textbook has been adopted. Αny οther reproducιion or ιrαnslαιion of this νοrk beyοnd thαιpermitted by Sections ] 07 οr Ι 08 of the ] 976 United Stαtes Copyright Αcι νithοuι ιhe permission of ιhe copyright ονner is unlαιιful.
224
(b) From Τable 121 in col. For 5/8 in. wg.
The rpm is 1092 and power is 0.39 ΗP
1240 1420cfm
1210' (a) ΔPo = 3.0 + 0.3 + 0'20 = 3.50 in. wg.
0.94O.BB
0.80
(b)
3.
3.CI
. 't3,500 15,00ο
Q cfm
1211.
(c) 13,500 to 14,000 cfm
/2istem, actual
\o10e2 rPm
desired fanr t.ι ,' svstem with svstem{z' P eifect factor
'osystem wlthout systemι effect faοtor
iffu" setection withoutI system effect factor
βε,,fοr whiιih tlοr ]08 οfthe ]9
225
System eff. Factor = 610  430 = .1gO pa
1212' D" = (4 x 12 x 16lπ)112 = 15.6 in
Αssume blast area ratio = 0.7, Table 123\/V
" = 4ο00(1 2 x 161144) = 5333 fVmin
One eff. Duct length = 5.3 diameters, table 12_2
or L" = 5.3 x '15.6 = 83 in.
% Εff ' Duct length = 100 x 30/83 = 36
Elbow in position C, Fig. 1213
EΙbow loss factor = 0'79, Τable 125
ΔPo = 0.79(5333l4o0q2 = 1.40 in. wg.
1213' V ι = 4OO 0l π x 142l14 x 1aa)] = 4,276 ftlmin
Duct length = 28 in.; R/D = 10.5114 = o.7s; L/D = 29114 = 2.0
Elbow and duct loss factor = 1.O, Τable 126
ΔPo  1'0(4276l4ooq2 = 1'14 in. wg.
1214' Blast area ratio = 0.7, TabΙe 123
D" = (4 x 20 x 2Olπ)1l2 = 22'6 in.
V = 1O,OOO x 1441(20 x20) = 3,600 fVmin
L" = 3.6 dia., Table 122
L/Le  (0122.6)13.6 = 0.12
Co = 0.4, Table 124
ΔPo = 0. 4(3600l4ooq2 = O.32in. wg.
1215' D" = (4 x 12 x 12lπ)1l2 = 13.5 in.
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226V" = 25001(12x 121144) = 2500 ft/min
One eff. Duct length = 2.5 diameters,Table 122
L = 2.5 x 13.5 = 33.9 or 34 in.
1216. From Problem 1215, Vr= Ve = 2500 ft/min (assumed)
ΔPo = C"(v /4005)2; Co = O.16/(250O t400q2 = O'41
From Τable 126, LlH = 4'3
Length = 4.3 x 12 = 51.5 in.
1217. (a) Τhe design condition and the observed condition are on
nearly the same system characteristic. Therefore, it is
probable that the fan is not running at the desired speed
of about 920 rpm but at a lower speed of about 6'10 rpm.
(b) The fan is operating near the 920 rpm characteristic
but something related to the duct system has changed.
Possibly a damper is closed, a duct has collapsed or some
other obstruction is present.
(c) Both the system and the fan characteristic have
changed. The duct system has probably
become fouled or slightly damaged is some way while
the fan speed has decreased slightly due to \Μear and tear.
1218' Wsn,l = 16 HP; Wsh,2 Ξ 1.5x5000= 1.62
6350x0.73
% Diff = rco(16_,!62) = +οO% [decrease from 1 to 2!ι 16 )
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students enrolled in courses for whiοh the textbook has been adopted. Αny other reprοductiοn or ιrαnslαιion of ιhis work beyond thαt
permiιιed by Sections ] 07 οr Ι 08 of ιhe 1 976 United Sιαιes Cοpyrighι Αcι Ιithοuι ιhe ρerfiιissiοn οf ιhe cop1'ν'ighι oνner is unΙανfuΙ.
227
1219. (a) Assume 15,000 cfm is an equivalent value for the day.
Forfull load point 1: Wr = 16 x0.746x24=286.5 kwh
For part load cond.: Wp = 6.7 x 0]46 x24 = 120.Okwh
ιW = (28q'9_ ]20) x '1OO = 58o/o (decrease)
286.5
(b) No, the fan would be forced to operate to the left of the maximum
pressure and would probably be unstable.
1220' W l = 28.5 ΗP; W z= 17 '5 HP (static po\Μer used)
ΔW = (28'?:17 '5) 1OO = 39o/o (decrease)
28.5
1221. (a) Wr = 28.5 x 0.746 x24 = 510 kwh
W, = 27 '0 xO'746 x24 = 483 kwh (vanes assumed "λ open)
ιW = (510_:183)
1OO = 5'3% (decrease)510
(b) W, = 27 x0.746x24 = 483 kwh
ΔW = (510_l_483)
1OO = 5'3% (decrease)51ο
(a) and (b) essentiaΙly the same.
1222' The actual inside dimensions are 10 x 8 in. or D" = 9.8 ιn., Table 127
For duct, unlined, ΔPo/L = 1 .8 in. wg./100 ft (Fig. 1221)
O  2ooox144 = 3600 fvmin
1 0x8
From Fig. 1223, roughness corr. Factor = 1.51, then for the
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228lined duct, ΔPo/L = 1 .8(1 .5'11 = 2'72 in' wg./100 ft. and
ΔPo  50 x2'7211ο0 = '1.36 in. wg. or about 338 Pa
1223. ΔPo = (ΔPo)rl X P/P.l, ρ _]!}Psι Pο,rz
Pυ = 0.491(29'42  0.0009 x 5000) = 12.236 psia
ΔPo = 1.36(12'236114.696) = 1'13 in. wg. or about 282 Pa
1224.
Q = 600 cfm
tl
=t'
()Θ
Dz = Ds = 10 in'', A2lA1= 0.6 = Α3/Aa
ΔPd/L = 0.185 in. wg./100 ft; Fig' 1221
ΔPzs = 0.1 85 x20l100 = ο.037 in. wg.
For contraction, AzlAι = 0.6, Coz= 0'21 ] Table 12gAFor expansion, A+/As = 1'67, Cρa = O.80 'Υz= Vs= 600  = 11OOfUmin; Υι=Vgx ff =660ft/min
ι(ιo1'4112 )
ΔPιz= 0'21(11}ol4oO5)2 = O.O16 in. wg.
ΔPοo = 0.80(660/4o05)2 = a'O22 in. wg.
ΔPo = 0.037 + 0.016 + 0'022 = 0.075 in. wq.
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2291225' (a) Bellmouth, Co = 0'2i Αbrupt, Co = 0.5; Table 12109 & 10Α
ΔPos = O'2(1ooo/4oo5)2 = O.o125 in' W9.  3.1 Pa
ΔPon = O.5(1oOo/4oO5)2 = O.0313 in. wg. x7'8Pa
o/o Diff .  (0'0313  0'0125) (1oo) = lsoo/o0 0125
(b) ΔPoa = O'2(40O0l40oq2 = O.2O in. wg. ε 50 Pa
ΔPon = O.5(4OOol4ooq2 = O.50 in. wg' x 124 Pa
o/o Diff.  (0.5  0.2) (1oo) = 15Oo/oo.2\/
1226. Table 128a, Co = 0.25
V o = 1200l[(πla)x(1 4ln)2! = 1122'5 ftlmin
ΔPo = O.25(1122.5l4oO5)' _ O.O2 in. wq.
also
Vo = 0'6l@lξ(0.35)2] = 6'24 mls
ΔPo  0.25(6.2411.2q2 = 5.8 Pa
1227. (a) Co = 0.15, Table 128b
Vo = 25OO x 1441(16 x 16) = '1406 ftlmin
ΔPo  o.15(1 40614005)2 = O.O'185 in. wq.
or Vo = 1.21(0.4 x 0.4) = 7.5 m/s
ΔPo = 0'15(7 '5l1'2q2 = 5.1 Pa
(b) Co = 1 '2 Τable 128C
ΔPo = 1.2(140614005)2 = O.148 in. wq.
or ΔPo = 1'2(7.5l1'29)'= 40.6 Pa
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230
1228. a/a c= 25O18OO = 0.3125
Aυ/Ac=(6112)2=O.25
Co = 0.345, Τable 1211A
Vυ = 25Ol(πl4)(6t2)2]= 1273 ft/min
Vυ = 0'12[(ila)(ο.15)2] = 6.8 m/s
ΔPoο = 0.345(1 273t4o05)2 = O.O35 in. wg.
ΔPoο = 0.345(6 '8t1'2q2 = 9.6 Pa
Qr/Qο=55O/80O=0.6875
ΑS/Ac = (0112)'= 0.694
C" = 0.135, Table 1211A
V. = 550l(πla)(0l12)'! = 1οo8 ftlmin
V, = o.26ll(ila)(0.25)21 = 5.3 m/s
ΔPo" = O.135(1oO8/4οο5)2 = O.OO9 in. wq.
ΔPo" = O.135(5.3/1 2972 = 2'3 Pa
1229. From Problem 1228
a/aο = O.31 25; A/A"= 0'25
Vo = 1273 fVmin or 6.8 m/s
Cο = 0.93, Τable 1211F
ΔPoo = 0.93(1 273l4oo5)2 = O.O94 in. wq.
or ΔPoο = O.93(6.811'2q2 = 25.8 Pa
Qr/Qc = 0.6875; Αr/A" = 0.694
C, = 0.135; Table 12118
V, = lOOB fVmin or 5.3 m/s
ΔPo, = O.135(1oOs/4oo5)2 = O.OO9 in. wq.
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or
or
or
or
or ΔPo, = o.135(5.3/1 29)2 = 2'3 Pa231
1230' (a) Ao/Al = 6.0, θ = 180 deg., Co = 37'4, Table 1298
vr = o?90"'^!o
= 2,ooo ft/min(18x18)
ΑoVo = ΑlVl, Vo = 1, x 2,O0O = 2000/6 = 333 fVminΑo
ΔPo = 37.4(333l4o0q2 = 0.260 in. wq.
(b) Co = 14.35 (Table 1298)
ΔPo = 14.35(333t4ooq2 = O.O99 in. wq.
1231. ao/Q" = 5OO/1OOO = O.5o
Αο/A" = (8Ι12)'= o'444
ΑS/Αc = (8/1 2)' = 0'444
Q,/Qc = 500/1OOO = O'5
(a) Cυ = 0.755, Table 1212A 1 lnterpolation required or use
C" = O.2'15, Table 1212A l ASHRΑE Duct Fitting Data
Vο = 50Otπt4)(8t2)2]= 1432 ft/min  v,or Vυ= O.24l(πl4χo.2)'] =7'64mls
ΔPoυ = 0.755(1 4g2l4o05)2 = O.O97 in. wο.
or ΔPoo = O'755(7 .64112972 = 26'5 Pa
ΔPo, = O'215(1432I4OO5)2 = O.O28 in. wq.
or ΔPo, = O'215(7.641129)2 = 7 '54 Pa
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1
2
3
55
72
20
0.50
0.16
2.0
1020
630
550
23335
14
13
0.'135
0.055
0.050
0.07 4
0'ο40
0.010
2(0.26)
2(0.26)
0.17
0.032
0.004
0.038
0.034
0.013
0.003
0.140
0.057
0.051
1233' L = D x Co/f; f = 0.019, ΤabΙe 1213
Bellmouth: L = 1 x0.210.019 = 10.5 ft
Αbrupt Int.: L = 1 x 0.5/0.019 = 26.3 ft
Q = 1OOO x πl4 = 785 cfm; ΔPo/L = O.12in'wg'l1oo ft, Fig' 1221
ΔPog = 0.12x 10'5l100 = 0.0126 in. wg. or about 3.'1 Pa
ΔPon = 0.'12x26'3l100 = 0.0316 in. wg. or about 7'9 Pa
1234' From ProbΙem 1226' Co = 0.25, D = 14 in'
Lu = DxC olfi f = 0.017 , Table 1213
γ'= lx 0'25 = 17 '2ft 12 0.017
Q = 12OO cfm; ΔPο/L = 0.'13 in. wg./1OO ft
ΔPo = 0.13 x 17 '2l100 = 0.022 in. wg. or about 5.6 Pa
Note: Most of following duct sizing problems can be solved with the computer
program, DUCT.
1235. From Figure 1236.
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234
Estimate Tota] Εquivalent Length of Run 123 to be approximately
132 ft, Τable 1214' Then ΔPo/Le = (0.13 x 1 00)1132
= 0.'10 in. wg./100 ft size ducts using Figure 1221 and record the
actual ΔPo/L from Figure 1221'
Section
No.
Le
ft
acfm
D
in.
ΔP/L
in. wq.
100 ft
ΔPn
in. wg.
1
2
3
4
5
45
16
71
55
55
300
220
100
80
120
II6
5
6
0.084
ο.090
0.083
0.14
0.125
0.ο38
0.014
0.059
0.077
0.069
Run 125 actually has the greatest lost pressure.
ΔPlη = 0.038 + 0'077 = 0.115 in. wg.;
ΔPιzs = 0.038 + 0.014 + 0.059 = 0.121 in. wg.
ΔPlzs = 0.038 + 0.014 + 0.059 = 0.1'1 '1 in. \Λ/g.
1236. The design pressure loss is (0.25  0.1 ) = 0.1 5 in. wg. (for supply ducts)
Assume the run with the largest equivalent length is:
12345, Le = 185 ft
Τhen for design: ΔPo/Le  (0'15 _ 0'03) x 1OO = 0.065 in. wg./'1oO ft185
Exοerpts fτom thls wοrk may be reproduοed by instructors for distribution on a notforprofit basis for testing or tnstruοtional puφoses only tostudents enrolled in courses fοr whiοh the textbook has been adοpted. Αny oιher reproduction or trαnsΙαιion of this νork beyond ιhαιpermitted by SecιΙοns ]07 οr ]08 οfιhe Ι976 United Sιαιes CopyrightΑcιιι,iιhοuι the permissiοn οfιhe copyι'ighι ονner is unlnνful.
235
Section Θ has a total flow of 845 cfm. Τherefore, the maximum
velocity in section Θ wjl! be about 800 fVmin if a 14 in. duct is used.
Exοerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional puφoses only tostudents enτolled in οourses for which the textbook has been adοpted. Αny other repsφluction or ιrαnsιαιion of ιhis νork beyond thαιpermitted by Sections ] 07 or Ι 0B of the 1976 Uniιed Snrcs Copyright Αcι wiιhout ιhe permis\ o/*e copyright oνner is unlαwful.
\
236
1236. (continued)
(a)
Section
No.
Le
ft
acfm
D
in.
ΔP/L
in. wο.
100 ft
ΔPn
in. wg.
1
2
3
4
5
6
7
II
88
18
16
17
46
51
43
49
49
845
595
395
275
125
250
200
120
150
14
12
12
9
7
I8
7
I
0.065
0.07ο
0.034
0.065
0.065
0.060
0.072
0.065
0.040
0.ο57
0.013
0.0054
0.0'1 1
0.030
0.031
0.031
0.032
0.020
With the equal friction method, every branch should have a damper for
balancing purposes.
Αctual total pressure loss:
ΔPo  ΔP1 + ΔP2 + ΔP3 + ΔPa + ΔPu + ΔPοs
ΔPo = 0'146 in' wg'
Note that run 12347 actuaΙly has the greatest loss in total pressure
but the difference is not significant. Use ΔPo = 0.15 in. wg.
(b) Sizing of the longest run, 1 2345, is the same as (a) above where
ΔPo/L = 0.065 in. wg./100 ft. Construct a new table as follows:
EΧcerpts fiοm this wοrk may be reproduοed by instructors tbr distribution on a notforpro1lt basis for testing or instruοtional puφoses only tostudents enrolled in οourses fοr which the textbook has been adopted. Αny oιher reproducιiοn or ιrαnslαtiοn of ιhis work beyond ιhαtpetmiιιed by Secιions Ι07 or 108 ofιhe Ι976 Uniιed Sιαιes Copyrι4hιΑcι'!ιιhouι ιhe permission οfthe copyrighι oνner is unΙιrwfuΙ.

237MAIN DUCT RUN BRΑNCH DUcτS
(1)
Sec
No.
(2)
Le
ft.
(3)
cfm
(4)
DJwxh
in.
(5)
ΔP
L
(6)
vfpm
(7)
ΔPo
(2)(5)
100
(8)
ΣΔPo
Σ(7)
(e)
Br.
Seο
No.
(10)
ΔPι
ΔPoot
(8)*
ΔPα
(1 1)
Le
ft.
(12)
ΔPi
L
('10)100
(11)
(1 3)
οfm
(14)
D"/wxh
in.
(1 5)
vfpm
I BB 845 14 0.065 800 0.057 0.057 f) 0.ο39 51 0.076 250 9 550
2 '18 595 12 0.070 760 0.013 0.070 7 0.036 43 0.084 200 I 570
J 16 ?oξ 12 0.034 500 0.005 ο.ο75 8 0.035 49 0.071 120 7 500
4 17 275 I 0.065 600 ο.01'1 ο.086 9 0.020 49 0.041 150 I 420
5 46 125 7 ο.065 500 0.ο3 0.116
ffuser G 0.030 0,146
The left 8 columns are the same as (a) above. The branches, 6789,
are sized to balance in the right hand 7 columns.
(c) Equal Friction Method
 Design Procedure Sysιem type: Supp}y
Duct Sizing Method: Equal FrictionRounding Method: Round Nearest
 Ean Selection Knoι\Ιn Fan Parameter: F'an TotaΙ Ρressure : 0 . 250 in. wg
Fan Αlrflow: B45.0 cfmFan or Externa1 Total Pressure: Ο.25Ο in. wg
Coi1 Lost PreSsure: Ο . ΟΟ0 in. wgΕi1ter Lost Pressure: 0 . 000 in. wgMisc. Lost PreSSure: 0.ΟΟΟ in. wα
ΑΗU External Total Pressure: Ο.25Ο in. wg
AΗU Pressure for Supply System: 0.150 in. wg  or 60.0 %
AHU Pressure for Return System: 0.1Ο0 in. wg  or 40.0 z
 Lost Pressure from Αir HandJ_ιng Unit to Diffuser Diffuser ΙD Q Tota1 Delta P
(cfm) (1n. wg)
71 125.0 A.L20Excerpts from this work may be reproduced by instruοtοrs for distribution οn a notforprοfit basis for testing or instruοtional puφoses only tostudents etrrolled in οourses for which the textbook has been adopted. Αny other reproducιion or ιrαnsιαιion οf this wοrk beyond thαtpermitted by Sections ] 07 οr 108 οf the ] 976 United Stαιes Copyrighι Αcι ''νiιhout ιhe permission οf ιhe cοpyrighι oνner is unlανful.
23822 150.0
120.a 0.200.0 0.250.A 0.
0.154130t25Ι21
t63034
Total B45. Ο
ΙD Fitt1ng Type
1 Air Ηandling Unit2 Straight Duct3 ConicaΙ Contraction4 Εlbow5 Elbow6 Tee / νΙΥe main
branchcommon
7 Straight DuctB Tee / wye main
branchcoτnmon
9 Straight Duct1Ο Tee / wye main
branchcoΙτιmon
11 Straight Duct12 Tee / ,νΙye il?}l.n
coΙnmon13 Straight Duct14 Εlbow15 Straight Duct16 Elbow17 Diffuser / GrilΙe18 Straight Duct19 Elbow20 Straight Duct21 EΙbow22 Diffυser / GrΙΙ23 Elbow24 Straight Duct25 ΕΙbow26 Diffuser,/ Gri11e21 Ε]boν'τ28 Straight Duct29 Elbow30 Diffuser / GritΙ31 Elbow32 Straight Duct33 EΙbow34 Dlffuser / GrilΙ
 Calculated Fitting VaΙues 
Dia.( in)
0.014. Ο
14. Ο
1Δ a
14. Ο
72 .4oΔ
14.072 .010.08.0
72 .010.09.01 .0
10.09.01.0?n9.07.0'7.01 .07.0
1.0'1 .01 .07.0
?n7.Ο7.0
8.08.0
Q Velocity(cfm) (ftlmin)
845.0 0.0845.0 '79Ο.4845.0 19A.4845.0 '790.4845.0 19A.4595.Ο '75'7.6250.0 565. 9B45.Ο 19a.4595.0 151 .6395.0 124.220Ο.0 573.0595.0 151 .6395.0 124.2215.0 622.5L20.0 449.0395.0 '724.2215.0 622.5L25 .0 461 .1150.0 561.3215.0 622.5125 .0 461 .1L25 .0 461 .1L25 .0 461 .'7L25 .0 461 .1] 2η n
15Ο.0 561.31s0.0 561.315Ο.0 561.315Ο.0 561.315Ο.0L24.0 449.0ι20.0 449.a120.0 449.0120 .02Ο0.0 573.02AA. 573.020Ο.0 573.02Ο0.0250.0 565.9250.0 565.9250.0 565.9250.0
Delta P(in. wg)
Ο.0000.0130. Ο130. Ο060.0060. Ο050.018
0. ΟΟ70.0040.01_7
0.0070.0030. Ο16
0.0070.0020.009
0.0080.0020.0020.0030.0300.0060.0050.0160.0050.0400.0020. Ο110. Ο03Ο.036Ο.0Ο30.0100.0050.0400.0030.0130.0040. Ο50
AP /L(in. wg)
0.06415
0.011_61
0.08259
0.07138
0.05817
Ο.05B17
Ο. Ο8082
Ο. ΟB0B2
0.05405
0. ο7106
0.060049.09.09.0
(c) Balanced Capacity Method
 Design Procedure 
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____
System type: SupplyDuct Sizing Method: Balanced Capacity
Rounding Method: Round Nearest
 Fan Selection 
239 
Known Fan Parameter: Fan Total Pressure : 0.250 in. wg
Fan Αirflow: B45.0 cfmFan or External Total Ρressure: Ο.250 in. wg
CoiΙ Lost Pressure: 0.Ο00 in. wgE'iΙter Lost Ρressure: 0.000 in. wgMisc. Lost PreSSure : 0 . 0Ο0 in. ι^rq
ΑHU Εxterna1 Total Pressure: Ο.25Ο in. wg
AΗU Pressure for SuppΙy System: 0'150 in. wg  or 60.0 %
ΑHU Pressure for Return System: 0.1Ο0 in. wg  or 40.0 %
 LoSt Pressure from Αir Ηand1ing Unit to Diffuser _
Diffuser ΙD Q Tοtaι Delta P(cfm) (in. wq)
L] 125.0 0.13822 150.0 Ο.15426 720.0 0.14530 200.0 0.14034 250.0 0.141
Total 845.0
 Cal cuLated Fitting Values 
ΙD Fitting Type Dia. Q Velocity DeΙta P ΔP/L(in) (cfm) (ftlmin) (in. wg) (in. wg)
1 Air Handling Unit 0. Ο 845.0 0.0 0.002 Straight Duct 14.0 845.0 '790.4 0.013 Ο.064153 Conicaf Contraction 14.0 845.0 190.4 0.0134 E}bow 14.0 845.0 190.4 Ο.0065 trlbow 14.0 845.0 '190.4 0.0066 Tee / Wye main 1'2 .0 595 . 0 '7 57 .6 0 . 005
branch 8.0 250.0 116.2 0.016common 14.Ο 845.Ο '790.4
7 Straight Duct 72.a 595.0 '751 .6 Ο.0Ο7 0.071678 Tee / \NΥe main 10.0 395.0 124.2 0.004
branch 1 .0 200.0 1 48.4 0.016common 72.0 595.0 15'1 .6
9 Straight Duct 10.0 395.0 124.2 0.007 0.0825910 Tee / Wye main 9.0 215.0 622'5 0.0Ο3
branch 6.0 720.0 6lL.2 Ο.013common 10.0 395.0 124.2
11 Straight Duct 9. Ο 215 '0 622.5 0.0Ο7 0.07138L2 Tee / Wye main 6.0 \25.0 636.6 0.003
branch 1 .0 150.0 561.3 0.009coΙnmon 9.0 21 5.0 622.5
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24013 Straight Duct
0.72304L4 Elbow15 Sιraight Duct76 Εlbοwl1 Diffuser / Grille
1 8 Sιraight Duct19 Elbow2a Straighι Duct27 Elbow22 Diffuser / Grille23 Elbow24 Straight Duct25 Elbow26 Diffuser / GrilΙe2"7 Elbow28 Straighι Duct29 Elbow30 Diffuser / Grille31 Εlbow32 Strarght Duct33 Elbow34 Diffuser,/ Grille
1237.
(a)
6.0
6.06.06.0
7.0/.u7.01.0
6.06.Ο6.0
1.41.07.0
8.08.08.0
125 .0
L25 .0L25 .01rη n
1,25 .0
150.0150.0
15Ο.0150.01"20 .0120.0t2a .0120.0200. Ο
200.0200.0200. Ο
250.025Ο.0250.0250.0
636.6
636.6636 .6636 .6
561.3561.3561.3561.3
6tt .26L7.26L1.2
148.4'1 ΔQ Δ
"7 48 .4
'716.21L6 .2'Ι 16 .2
0.016
0.0040.0050.007Ο.030
0.0060.00s0.0160.0050.0400.0040.0230.0060.036Ο.005n nT q
0.0090. Ο400.005^ ^410.0070.050
0.72344
0.08082
Ο.08082
0.ΙL427
0.13629
0.10661
ΔPos + ΔPon = ο.70 _ 0.35 = 0.35 in. wg.
ΔPos = 0.65(0.35) = 0.23 in. wg.
ΔPoκ ^y
0.35 _0'23 = 0.12 in. wg.
Τhe method of Solution is similar to Problem 1236' An
acceptable solution follows:
Longest run  1 234567891 1 13
The summation of equivalent lengths may vary v/ith designers.
ΔPo/Le =(0.23  0.03)
100 = 0.092 in. wg./'100 ft217
Size all Suppιy ducts for this pressure loss per unit length.
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241
150
100
100
75
100
200
16
14
14
12
12
12
10
I7
5
6
5
6
5
1200
1 050
850
750
650
550
475
225
175
50
125
50
75
50
1
2
3
4
5
b
7
8
9
10
11
12
13
14
Εxοerpts frοm this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instruοtional puφoses only ω
students enrolled in courses for wt'iοh the textbook has been adopted. Αny oιher reproducιion or ιrαnsl(tιion of ιhis work beyond thαι
permiιιed by Secιiοns ] 07 or l δε i7 in, Ι976 United Stαιes Cοpyrigh't Αct ιlithouι the peimissiοn ofιhe copyrighι oνner is unΙανful'
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1242. (continued)
ο

1237. (continued)
ΔPo for the longest run will be about 0'22 in. wg' for the above sizes"
Τherefore, size the return system for a pressure loss of (0.35 _0'22)
or 0.13 in. wg.
For the return system:
(L")r* x 230 ft, ('1 r  2r  3r)
then ΔPo lL"= r0'13__0'05'] ''oo = O.O35 in. wg./'1oo ftι230)
Using the equal friction method:
Section
No.
acfm
D
in.
L"
ft
ΔP"/L ΔPo
in. wg.
1r
2r
3r
4r
5r
1200
800
400
400
400
18
'16
12
12
12
115
70
44
14
28
0.038
0.033
0.ο36
0.036
0.036
0.044
0.023
0.0'16
0.005
0.010
Return system is the same for parts (a) and (b).
ΔPo for return = 0.133 in. \Mg.
(c) Equal Friction Method
 Desiqn Procedure 
System type: SupplyDuct Sizing Method: Equal Γriction
Rounding Method: Round Nearesι
 Ean SeΙection 
Known Εan Parameter: Εan Total Pressure : 0.700 in. wg
244
Εan Airflow: 1000.0 cfmFan or ExternaΙ Total Pressure: 0.700 in. wg
Coil Lost Pressure: 0.25Ο in. wgF'iΙter Lost Pressure: 0 . 10Ο in. wgΜisc. Lost ΡresSure: 0.0Ο0 in. wg
ΑHU Externa1 Total Pressure: 0.350 in. ι^rg
AΗU Pressure for SuppΙy System: 0.228 in. wg  or 65.0 %
AHU Pressure for Return System: 0.123 in. wg  or 35.0 %
 Lost Ρressure from Αir Ηand1ing Unlt to Diffuser Diffuser ΙD Q Total DeΙta Ρ
(cfm) (in. wg)
2'7 75.Ο 0.22130 75. Ο 0.20134 75.0 0.2Ι138 s0.0 0.27256 75.0 0.17160 100.0 0.15863 100.0 0.17161 100.0 0.L4212 200.0 0.2a215 150.0 0.131
Total 1000. Ο
_ Calculated Fittinα Values 
ΙD Εitting Type Dia. Q VeΙocity Delta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)
1 Αir Ηandling Unlt Ο.0 1000.0 0. Ο 0.0002 Conical Contraction 14.0 1000.0 935.4 0.0113 Straight Duct 14.0 1Ο00.0 935.4 0.007 0.087454 Tee / wye main l2.a 850. Ο 082.3 O. O1O
branch 1 .0 15Ο. Ο 561.3 0.061coΙτΙnon 14.0 1000.0 935.4
5 Straight Duct 12.0 850.0 L082.3 0.003 0.138196 Tee / Wye main 12 .0 650 . 0 82'7 .6 0 . 008*<10>
branch 1 '0 2a0.0 1 48.4 Ο.058coΙnmon L2 '0 B50.0 1082.3
7 Straiqht Duct l2.0 650.0 B2'7 .6 0.005 0.08429B Tee / vfrze main 10.0 550.0 1008.4 Ο.0Ο9
branch 6.O 1OO.O 5O9.3 Ο.Ο37common L2.0 650.0 821.6
9 Straight Duct 10.0 550.0 1008.4 0. ΟΟ6 0.1516410 Tee ,/ Wye main 10.0 450.0 825.! O. O1O*<10>
branch 6.0 100.0 509.3 0.064coΙπnon 10.0 550. Ο 1008.4
11 Straight Duct 10.0 450.0 825.7 0.004 0.10485L2 Tee ,/ Wye main 9. 0 350 . Ο 192 .2 O . ΟΟ5
branch 6.0 100.Ο 509.3 a.024conτnon 10.0 450.0 825.L
13 Straight Duct 9. 0 350 . 0 1 92.2 0 . 014 0. 11082
Excerpts from this work may be reproducοd by instruοtors for distribution on a not_forprofit basis for testing or instructional puφoses only tostudents enrolled in οοurses for whiοh the textbook has been adopted. Αny οther reproductiοn of trαnslαtion of ιhis νork beyοnd thαιpermiιιed by Secιions Ι07 or l08 οfthe Ι976 Uniιed Stqιes Copyright Αcι'ι)iιhout the permission ofthe cοpyrighι oνner is unlcrwful.
24514 Tee ,/ Wye main 8. Ο 215.a "781 .8 Ο. Ο05
branch 5.0 75.Ο 550.0 0.019common 9.0 35ο.Ο 192.2
15 Straight Duct B.0 215.0 '7B1 .8 0.0Ο6 0.7268'7L7 ΕΙbow B.0 215.0 1B1.B 0.0Ο818 Straight Duct 8.0 215.0 181 .B 0.013 0.L268'719 Tee ,i Wye main B.0 225.a 644.6 Ο.0Ο9*<10>
branch 4.0 5Ο. Ο 573.0 a .021common 8.0 215.4 l1i .B
20 Straight Duct B.0 225.a 644.6 0. Ο07 0.0880Ο27 Tee / Wye main 7.0 150.0 561.3 0.003
branch 5.0 75.0 550. Ο 0.011coΙnmon B.0 225.0 644.6
22 Straight Duct 1 .0 150.0 561.3 0.018 0.080822? Tρρ / Ιn1lzo main 5.0 75.0 550.0 0.002f νν / νγli ν
branch 5.Ο 75.0 550.0 0.007coΙnmon 1.a 150.0 561.3
24 Εlbow 5. Ο 75.0 550.0 0.00625 Straight Duct 5. Ο 75.0 550.0 0.072 Ο.1186926 Rectangular Transition 5. Ο 75.0 210.a 0.0Ο52'l Diffuser / GrilΙe 75.0 0.03028 Straight Duct 5.0 75.0 550.0 0.0Ο7 0.1186929 Rectangular Transition 5. Ο 75.0 210.a 0.0Ο53Ο Diffuser / Grille 75.0 0.02531 Elbow 5.0 75. Ο 55Ο.0 0.0Ο632 Straiqht Duct 5. Ο 75.0 550.0 Ο.018 0.1186933 RectanguΙar Transition 5.0 75.0 21a.0 0.00534 Dif fuser / Grille 75.0 A.02535 Elbow 4.0 50.0 573.0 0.00736 Strarght Duct 4.0 50.0 573.0 0.017 0.1691631 Rectangular Transition 4.0 5Ο.0 180.0 Ο.01038 Dif fuser ,/ cril1e 50. Ο 0.02053 Elboιv 5.0 75.0 550.0 0.00654 Straiqht Duct 5.0 75.0 550. Ο 0.01B 0.1186955 RectanguΙar Transition 5.0 75.0 210.0 0.00556 Diffuser / cril1e 75.0 0.03051 Ε1bow 6.0 100.0 509.3 0.00458 Straight Duct 6.Ο 100.0 509.3 0.008 a.0122l59 Rectangular Transition 6.0 1Ο0.0 360.0 0.0036Ο Diffuser / cri]ιe 100.0 0.0456L Straight Duct 6. Ο 100.0 5Ο9.3 0.072 a.0B221'62 Rectangular Transition 6. Ο 100.0 36Ο.0 Ο.00363 Diffuser / Gri1Ιe 1Ο0.0 0.03264 E1bow 6.0 100.0 509.3 0.0Ο465 Straight Duct 6.0 1Ο0.0 509.3 0.008 a.0822166 Rectangular Transition 6.0 100.0 360.0 0.00361 Diffuser / Gril]_e 1Ο0.0 0. Ο456B Straight Duct '7 .A 200.0 148.4 0.02'7 0.l.362969 Elbow 1.0 200.0 148.4 0.01370 Straight Duct 1 .0 20Ο. Ο "7 48 ' 4 0 .a21 0.1362911 Rectangu1ar Transition 1 .0 20Ο. ο 120.0 0.00112 Diffuser / Gritle 20Ο. Ο 0.04513 Straight Duct 1 .0 150.0 561.3 0.032 0.080821 4 Rectangular Transition 7.0 150.0 54Ο.0 0.00115 Dif fuser ,/ Grille 150.0 Ο.02016 Elbow 8.0 215.0 181 .B 0.005
*  De}ta P ιfas computed using the fitt1ng equivalent lenqth
Balanced Capacity Method
Exοeφts from this wοrk may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only tostudents enτοlled in οourses Γor whiοh the textbook has beοn adopted. Αny οther reproduction or trαnslαιion of ιhis νοrk beyond thαtpernιiιιed by Secιions ]07 or l 08 οf the ]976 Uniιed Stαtes Copyright Αcι νιιhout ιhe permissiοn οf ιhe cοpyrighι owner is unlcrνιful.
246
 Desiqn Procedure 
Note that almost alΙ branch ducts need a damper to increase the diameter andreduce ve1οcity.
System type: SupplyDuct Sizing Method: Balanced Capacity
Rounding Method: Round Nearest
 Fan Selection Known F'an Parameter: Ean Tοta1 Pressure : 0.700 in. wg
Εan Αirflow: 100ο.Ο cfmFan or Εxternal Total Pressure: Ο.7ΟΟ in. wg
CoiΙ Lost Pressure: 0.25Ο in. wgFilter LoSt Pressure: 0 . 1Ο0 in. wgMisc. Lost Pressure: 0.000 in. wq
ΑHU External Total Ρressure: 0.350 in. wg
AHU Pressure for Supply System: 0.228 in. wg  or 65.0 ?ΑHU Pressure for Return System: 0.L23 in. wg  or 35.0 ?
 Lost Pressure from Αir Handling Unit to Diffuser 
Diffuser lD Q TotaΙ Delta P(cfm) (in. wq)
2'7 75.0 0.22130 75.0 0.24334 75.0 0.2713B 50.0 0.2L256 75.0 0.23260 100.0 0.21963 100. Ο 0.28561 100.0 0.25'712 2Ο0.0 0.20215 150.Ο 0.17Ο
TotaΙ 10Ο0.0
 CaΙcuΙaιed Γiιting Vafues 
ΙD Εltting Type Dia. Q VeΙocrty DeΙta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)
1 Αir Ηand1ing Unit Ο.0 10Ο0.0 0.0 0.0002 Conical Contraction 14 .0 1000.0 935.4 0.0113 Stralght Duct 14. Ο 1000.0 935.4 0.007 0.08745/ .Γaa / τlτ.,o main 72.a 850.0 L082.3 0.010
branch 6. Ο 15O . O 1 63 .9 O. O57coττunon 14.0 1000.0 935.4
5 Straight Duct l2.0 850. Ο 1082.3 0.0Ο3 0.138196 Tee / wye main !2.a 650.0 821.6 Ο.OΟ8*<1O>
branch 1.a 20Ο.Ο '748.4 0.058
Excerpts from this work may be reproduοed by instruοtors for distribution on a nοt_forprofit basis fοr testing or instruοtional puφoses only tostudents enrοlΙed in οourses fοr whiοh the textbook has been adopted. Αny οther reProduction or ιrαnsΙαιiοn of ιhis work beyond ιhαιpermiιιed by Secιiοns Ι07 or ] 08 οf the Ι 97 6 United SιαιeS Copyrighl Αct νiιhouι ιhe peιmission of ιhe cοpyrighι olνner is unΙcrννful.
coΙnmon7 Straight DuctB Tee / wye main
branchcoΙnmon
9 Straight Duct10 Tee / Wye main
branchcoΙnmon
11 Strarght Duct'r ^^ / r^ir7^ malnΙ99 / νlyν
branchcoΙτιΙnon
13 Straight Duct74 Tee / Wye main
branch
ι5L1LBL9
202L
2223
coΙnmonStraight DuctElbowStraight DuctTee / Wye main
branchcoτnmon
Straiqht Duct/ l1lrΣ^ ηa1nιγ ] ν
branchcommon
Straight DuctTee / Wye main
branchcoτΙυnon
24 Elbow25 Straight Duct26 Rectangufar Transition21 Diffuser / crille28 Straight Ducι29 Rectangular Transition30 Diffuser / Gritle31 tr1bow32 Straight Ducc33 Rectangular Transition34 Diffuser / Grille35 Elbow36 Straight Ducι31 Rectangular Transition38 Diffuser / Grille53 Elbow54 Straighι Duct55 Recιangufar Trans1tion56 Diffuser / GriΙ1e51 ΕΙbow58 Sιraighι Duct59 Rectangular Transition60 Diffuser / GriΙ]e67 Straight Ducι62 Rectanqular Transition63 Diffuser / Grille64 ΕΙbow65 Scraight Duct66 Rectangular Transition61 Diffuser / crille68 Straight Ducι
72 .01_2 .010.0Δo
L2 .010.010.04.Ο
10.01Ο.09.04.0
10.09.08.04.0on8.08.08.08.04.ο8.08.07.05.Οo.u1.44.05.0?n5.Ο5.05.0
4.04.Ο
ηn5.ΟtrΛ
4.04.0ΔΓ\
4.44.04.0
85Ο.0650. Ο
10Ο.0650.0550.0450.0100.055Ο. Ο
450.0350.01Ο0.0450.035ο. ο275.075.0
350.0t1\ aο' tr Λ
215.0225.050.0
215 .0aa tr Δ
150.075.0
225.015Ο.075.Ο75.Ο
150.075.075.0?tr Λ
75.075.075.075.ο7η n
75.075.075.Ο5Ο. Ο
50. Ο
50.050.075.075.075.Ο'tr
Λ
1Ο0.01Ο0.010Ο. Ο
100.01ο0.010Ο.0100.0100.010Ο.0100.0100.02ο0. Ο
ι082.3821 .6
1008.4ι145.9821 .6
10ΟB.4825 .7
1145. 91ο0B.4825.L192.2
1145. 9825.17q2 2
181 .8859.41A) )181 .8181 .8181 .8644 .6573.0'7 81 .8644 .6561.3550.0644 .6561.3859.4550.0561.355Ο.0550.0210 .0
859.4214 .0
550. Ο
550.0214 .0
573.0573.0180.0
0.0050.0090.039
0.006
247
0.08429
0.15164
Ο.104B5
0.11082
0.010*<10>a .062
0.0040.0050.031
0.0140.0050.017
0.0060.0080.013
4.04.44.0
4.04.Ο
Δol4.Ο4.0
7.Ο
859.4859.4210 .0
1145.91145.9360.0
L1_45 .9360.0
1145. 97L45 .9360. Ο
148.4
0.L2681
0.!268'70.009*<10>0 .021
0.0Ο7 Ο. Ο88Ο00.0030.011
0.018 0.080820.0070.007
Ο.0060.at2 0.118690.00s0.0309.02! 0.352660 .0240.025Ο.0060.018 0.11869Ο.0Ο50. Ο250. ΟΟ70.017 Ο.169160.0100.0200.015Ο.053 0.352660 .4240.03Ο0 .0210.060 0.591440.0420.0450.090 0.591440 .0420.032Ο a2'10.06Ο 0.59'7440.0420.045a.021 0.!3629
Exοerpts from this work may be reprοduced by instructors for distribution on a notfοrprofit basis for testing or instruοtional puφoses only to
students enrolled in courses for which the textboοk has been adopted. Αny other reprοducιiοn or ιrαnsιαtion οf this νork beyοnd thαιpermitted by Sections ] 07 οr l 08 οf the 1 976 United StαιeS Copyrighι Αcι'withouι ιhe permission of the cοpyrighι owner is unlανυful.
24869 EΙbow 1.0 200.a ']48.4 0.013
70 Straight Duct 1 .0 200.0 148.4 0.02'1 0.13629lt RectanguΙar Transition 1 .0 2Ο0.0 120.0 0.00112 Diffuser / Grille 200.0 0.04513 Straight Duct 6. Ο 150.0 '7 63.9 0.069 0.171391 4 Rectangular Transition 6. Ο 150.0 540.0 0.00715 Diffuser / Griι1e 150.0 Ο.02016 EΙbow B.0 2'75.a '7B'7.B 0.0Ο5
*  Delta Ρ was computed using the fitting equivalent Ιength
Return Ducts, Equal Friction Method
 nηοj ^_ D:ocedure __vνυΙYlΙ L ]
System type: ReturnDuct Sizing Method: Εqual Friction
Rounding Method: Round Nearest
 Fan SeΙection 
Known Εan Parameter: Εan Tota1 Pressure : 0.700 in. wg
Fan ΑirfΙow: 1200,0 cfmFan or External Tοtal Ρressure: Ο.7ΟΟ in' wg
Coil Lost Pressure: Ο.25Ο in. wqΕiΙter LoSt Pressure: 0.100 in. wgMisc. Lost Pressure: 0.000 in. wg
AΗU Εxternal Tota1 Pressure: 0.350 in. wq
AΗU Pressure for SuppΙy System: 0.228 in. wg  or 65.0 %
ΑΗU Pressure for Return System: 0.123 in. wg  or 35.0 %
 Lost Ρressure from Αir Ηandling Unit to Diffuser 
Diffuser ΙD Q TotaΙ Delta P(cfm) (in. wg)
11 400.0 0.12514 400.0 0.113l"7 400.Ο 0.096
Tota1 120Ο.0
_ CaΙcu1ated F'itting VaΙues _
TD Fitting Type Dia. Q Velocity Delta P LP/L(in) (cfm) (ftlmin) (in. wg) (in. wg)
1 Αir Handllng Unit 0. Ο 1200.0 Ο.0 Ο. ΟΟ0
2 Rectangular Transition 18.0 72A0.0 679.1 0.0013 Straight Ducι 1B.0 L200.a 619.1 0.002 0.035124 Tee / \ηye main 1'2.a 400.0 509.3 0.026
branch 16.0 800.0 573.0 0.026coπlmon 18.0 1200.0 679.1
5 Straight Duct 16. Ο 800.0 573 ' 0 0.006 0.03a246 Tee / wye main 12.0 4ΟΟ.0 509.3 0.0]5
branch \2.0 400.0 509.3 0.013coΙnmon 16.0 800.0 573.0
Exceφts from this work may be reproduced by instruοtors for distribution on a notforprofit basis for testing or instructiοnal puφoses only to
students enrolled in οourses for whiοh the textbook has been adopted. Αny other reproducιion or ιrαnslαιion of ιhis νοrk beyond thαιpermitιed by Sectiοns ] 07 or Ι 08 of ιhe 1976 United Sιαtes Cοpyright Αcι νithouι the permission οf ιhe copyright oνner is unΙανfuΙ.
7 Stralght Duct0. Ο3476 B ElbowΟ. Ο039 Straight Duct10 Rectangular Transition11 Diffuser / criΙle72 Straight DucL13 Rectangular TransitionL4 Diffuser / GrilΙe15 Straight Dυct16 Rectangular Transition11 Diffuser / Griι1e
12 .0
L2 .0L2 .0
L2 .0L2 .0
L2.AL2 .0
400.012 .0
4Οο.04Ο0.04Ο0.0400.040Ο.0400.0400.0400.0400.0
0.007509.3
0.0010.0140.0500.0010.0140.0500.0030.0140.05Ο
249
0 .034'7 6
0.03476
0.03476
509.3400.0
s09.3l00.0
5Ο9.3100.0
509.3100. Ο
1238. The three branches from the plenum
possibιe for the Same preSSUre ]oss.
to be more extensive than Α or c.
For B: ΔPo/Le
For Α: ΔPo/Le
must be designed as close as
Start \Ι/ith B since it appears
=[
=(
0.18  0.025
104
'100 = 0.107 in. wg./100 ft
'100 = 0.109 in. wg./10ο ft
100 = 0.149 in. wg./100 ft
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instruοtional puφoses only tostudents enrolled in courses for whiοh the textbook has been adopted. Αny other reproductiοn or trαnsιαtion of ιhis wοrk beyond ιhαιpermiιted by SecιioιιS ] 07 or ] 08 οf the Ι 97 6 United SιαιeS Copyrighι Αct νithouι the peιmission of ιhe cop1',ι"ighι oνner is unlcrνful'
025
145
0.180.
025
142
0.18  0.For C: ιP"/L" = (
B
Α
c
2501238. (continued)
Note that the resulting total pressures losses turn out to be:
(ΔPo)a = 0.144 in.wg., (ΔPo)n = 0.157 in. wg.; (ΔPo)c = 0.161 in. \Μg.
Within the accuracy of the calculation these are appΓoximately equal'
It may be necessary to use a damper in branch B, sec. 8.
1238.
Branch A, Balanced Capacity Method__ Γ)Αq i α pηocedure ννJf Y11 ! l
System type: SupplyDuct Sizing Method: Balanced Capacity
Rounding Method: Round Nearesι
__ Εan Sefection _
Known Εan Parameter: Plenum Tota1 Ρressure : 0.1B0 in. wg
Fan Αirflow: 4Ο0.0 cfm
Excerpts from this work may be reproduced by instructors for distribution on a nοtforprofit basis fοr testing or instruοtional puφoses οnly to
students enrolled in οourses 1br whiοh the textbook has been adopted' Αny other reproduction or trαnsιαιion of this νork beyοnd ιhαt
permittecι by Secιions ]07 or Ι08 ofthe ]976 Uniιed Sιαtes Cοpyright Αcινiιhouι ιhe permission ofιhe cοpyrighι owner is unlαwful.
MAN DUcτ RUN BRΑNCΗ DUCTS
(1)
Seο.
No.
(2)
Le
ft.
(3)
cfm
(4)
D"
in.
(5)
ΔP
L
actual
(6)
vfpm
*r"
1
(7)
ΔPo
(2)(5)
100
(8)
ΣΔPo
Σ(7)
(e)
Br.
Sec
Νo.
(1 0)
ΔPi
ΔPo6+
(8)+
ΔPα
(11)
Le
ft.
(12)
ΔPi
L
(1 0)1 00
(1 1)
(13)
cfm
(14)
De
in.
(1 5)
vfpm
8 44 500 12 .057 650 .025 025 14 0.094 55 0.171 125 b 660
o 22 375 10 .085 700 .019 .044 10 0.ο75 52 0.144 200 7 760
11 25 175 7 0.1 1 630 o28 .072 12 0.047 38 0.124 75 5 550
IJ 54 100 o .ο87 520 .047 0.119
Tot 145 ΔPο .025 0 144
1 50 40ο 10 .095 760 ο.048 0.048 7 0.084 38 0.221 100 ξ 750
2 19 300 .ο92 700 0.018 .ubb b 0.ο66 48 0.1 38 100 b 510
25 200 ο8 59ο 0.02 .086 4 0.046 q7 0.ο8'1 '100 o 500
48 100 o .095 530 0.046 . t5z
Tot. 142 ΔPο 0.025 157
15 56 225 7 17 850 0.095 .095 to 0.041 38 0.1 ο8 125 ο 610
17 48 100^
.085 510 0.04'1 '136
Tot. 104 ΔPο 0.025 161
Fan or External TotalCoil Lost
E'iΙter LostMisc. Lost
AΗU External Total Pressure:
ΑHU Pressure for Supply SystemΑΗU Pressure for Return System
Diffuser ΙD
PressureΡreSSurePressurePressure
0 .269 1n.0.000 in.0.000 in.0.Ο00 in.
0.269 in.
0.180 in.0.089 in.
Dia.(in )
0.Ο9.ο9.09.Οqn8.05.0on8.01.4trΛ
8.0'7.0ξn5.01.05.05.05.0
5.0ζn5.0
5.0trΛ
5.0
5.05.Ο5.Ο
a(cfm)
4ΟΟ. Ο
40Ο. Ο
400.040Ο. Ο
400.0300.0100.0400.03Ο0.0200.0100.0300.0200.0100.0100.0200. Ο
100.0100.01ΟΟ.010Ο.010Ο. Ο
100.0100.0100.0100.01ο0. Ο
100.0100.0100.01Ο0. Ο
10Ο.Ο100.0
Velocity( ftlmin)
Ο.0905.4905.4905.4905.4Qξo /
733.4onξ /
859.4t48.47?? Ia ξo 1
148.4133.41)) Λ
148.4133.41aa Λ
1)) Λ
360. Ο
7?? Δ
a 1a Λ
36Ο.0
??? Δ
aa1 Λ
360.0
111 Λ
1" 1 Λ
360.0
0.0000.0110.0030.0110.0140.0060.039
251
0 . 14157
0.14157
0.1_3629
0.1_9911
0 .1997'7
0.Ι99'71
0.L9911
0.3991"7
wg1^rg
wg\^rg
1^rg
wgwg
or 6'7or 33
0%0%
Lost Ρre$sure from Αir Handling Unit to Diffuser 1
15192329
Total 400.0
ΙD Εitting Type
1 Αir Handllng Unit2 Conlcal Contraction3 Straight Duct4 Elbow5 Stralght Duct6 Tee / wye main
branchcoΙnmon
7 Straight Duct8 Tee / wye main
branchcoΙnmon
9 Straiqht Duct10 Tee ,/ Wye main
branchcoτnmon
11 Straighι Duct72 E1bοw13 Straight Duct1'4 RecιanguΙar Transition15 Diffuser / Grille16 EΙbowt'7 Sιraight Duct18 RectanquΙar Transition79 Diffuser / cr1lιe2A Efbow2L Straiqht Duct22 RectanguΙar Transition23 Diffuser / cri]_ιe24 Sιraight Duct2Ι ButterfΙy Damper28 RectanguJar Transition
^ r^^1 n^l ^ nν ΙOιdL ιreΙLd r
(cfm) (in. wg)
100. Ο 0.171100.0 0.155100. Ο 0.141100.0 0.187
 Calculated Fittlng Values 
Delta P ^P/L(in. wg) (in. wg)
0.018 0.148780.0050.020
0.0160.0040. Ο]_3
0.0200 .0!20.0160.0100.0250.0Ο70.0160.0100 .4250.0070.0160. Ο10a.0250.0160.0580. Ο10
Exοeφts from this work may be reproduοed by instruοtors for distributiοn on a not_for_profit basis for testing or instructional purposes only tostudents enrolled in οourses for which the textbook has been adopted. Αny οther reproducιion or trαnslaιiοn of ιhis work beyond ιhαιpermiιted by Secιions ] 07 or Ι 08 οf the ]976 Uniιed SιαιeS Copyrιghι Αct'wiιhouι ιhe permissiοn of ιhe copyrighι oνner is unlανfuΙ.
25229 Diffuser / Grille
Branch B, Balanced Capacity Method/t"
\. Desιgn Procedure _
System tyρe: SupplyDuct Sizing Method: Balanced Capacity
Rounding Method: Round Nearest
 Εan Sefection __
Κnown Ean Parameter: Plenum Total PressurΘ : 0.180 in. wg
Fan Αirflow: 500.0 cfmFan or Εxternal Total Pressure: Ο.18Ο in. wg
Coil Lost Ρressure: 0 . 000 in. wgΓi1ter Lost Pressure: 0.000 in. wgMisc. Lost Pressure: 0.Ο00 in. Wg
AHU External Total Pressure: 0.180 in. wg
AΗU Pressure for SuppJ_y System: Ο.18Ο in. wg  or 1OO.O %
AHU Pressure f or Return System: 0 . 000 in. \^Ιg  or . ο %
 Lost Ρressure from Αir Ηandling Unit to Diffuser Diffuser ΙD Q TotaΙ De]ta P
(crm1 (1n. wg)
13 10Ο.0 0.185L6 75.0 0.17919 200.0 0.\4223 125.a Ο.15B
Total 5Ο0.0
 Calculated Εittinα VaΙues 
ΙD Εitting Type Dia. Q Velocity Delta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)
1 A1r Ηandling Unrt 0. Ο 500.0 0.0 Ο. OOO2 ConicaΙ Contraction 10.0 5Ο0.0 916.1 Ο. O113 Straight Duct 10.0 500.0 976.1 0. Ο18 0.72'723Λ 'Γaa / τι;l',o main 9.0 375.0 848.8 O.Ο06
branch 6.O Ι25.O 636.6 O.o42coπιrrion 1Ο.0 500.0 916.7
5 Straight Duct 9.0 375.0 B4B.B 0.019 0.725"756 Tee / wye main 6.ο 2ΟΟ.0 1O18.6 Ο.OO9
branch 6.0 175.Ο 891.3 0.019coΙnmon 9.Ο 375.0 84B.B
7 Elbow 6.0 175.0 891.3 0.0088 Straight Duct 6.0 175.0 891.3 O.014 0.221759 Tee / wye main 4.0 75.Ο 859.4 0.OO6
branch 5.0 1ΟΟ.0 '733.4 0.020coΙnmon 6.0 175.Ο B91.3
Εxceφts from this work may be reproduοed by instruοtors fοr distribution on a not1brprofit basis for testing or instructional puφoses only tostudents enτolled in οourses for which the textbοοk has been adopted. Αny οther reproducιion or ιrαnslαιion of ιhis νork beyond thαιpermiιιed by Sections ]07 οr Ι08 οfιhe 1976 Uniιed Stαtes Copyright Αcιννιιhouι ιhe permission οfthe copyrighι olνner is unΙιrwfuΙ"
100.0 0.025
10 Elbow11 Straight Duct\2 RectanguJar Transition13 Diffuser,/ Gril1eL4 Straight Duct15 Rectangular Transition16 Diffuser / GriΙ1e71 Straight Duct18 RectanguΙar Transition19 Diffuser,/ Griιle20 Straight Duct27 Butterfly Damper22 Rectangular Transition23 Diffuser / critle
141B
5.0ξn5.0
4.04.0
6.06.0
b.u6.06.0
100.0100.0100.010Ο.075.075.075.0
200.0200.0200 .0125 .4125.01oξ Λ
L2s.A
'733.41?2' Δ
360.0
859.4210 .0
1018 . 6120 .0
636.6636.6450.0
0.0070.0280.0100 .0250 .0280 .0250 .0250.0410.0140.0250 .0L20.0440.005a .02s
2530.1_99'71
0.35266
0 .29022
0.72304
AP /L(1n. wg)
 Design Ρrocedure System type: Supply
Duct Sizing Method: Balanced CapacityRounding Method: Round Nearest
 Εan Se1ecιion 
Knoιtn Ean Parameter: Plenum Total Pressure : 0.180 in. wg
Branchffilanced Capacity Method
Εan ΑirfΙοw:F.an or External Total Pressure:
ColΙ Lost Pressure:Filter Lost Pressure:M1sc. Lost Ρressure:
AΗU ΕXterna1 Tota1 Pressure:
ΑHU Ρressure for Supply System:AHU Pressure fοr Return System:
225.0 cfm0. 180 in. wgΟ . 0Ο0 in. wg0 . 000 in. \^Ιg
0.000 in. wg
0 . 18Ο in. wq
0.1800. Ο0Ο
or 10Ο.0or .C)
ln. \.^/g
in. wgzz
 Lost Pressure from Αir Ηandlinq Unit to DiffuserDiffuser ΙD a Totat DeΙta
(cfm) (in. wg)
10Ο.0 Ο.191725.0 0.186
Total aatr Λ
 Ca1cu1ated Fittinα Values 
ΙD Fitting Type Dia.(in )
a(cfm)
Velocity( ftlmin)
Delta P(in. wq)
1 Air Handling Unit 0.0 225.A O. O O. OOO2 Conical Contraction 8 . Ο 225 .0 644 .6 O . OO5Excerpts from this work may be reproduced by instruοtors for distribution on a notforprofit basis for testing or instructional puφoses only tostudents enrolled in courses Γor which the textboοk has been adopted. Αny οther reproducιion οr ιrαnslαιion of this νori beyondιhαιpermiιιed by Secιions ] 07 or Ι 08 of ιhe 1 976 ιJnited Stαtes Copyrighι Αct ιιiιhout ιhe permissiοn of the copyrighι oινner is untαwfuΙ'
2543 Stralght Duct
0.088004 Butterfly Damper5 Straight Duct6 Elbow7 Straight Duct8 Elbow9 Straight Duct1Ο Tee / Wye main
branchcoΙτ'ιmon
11 Straight Duct72 Efbow13 RectanguΙar Transitjon74 Diffuser / Grille15 Sιraight Duct71 Rectangular Transition18 Diffuser / GriΙle
8.0
8.08.08.08.08.08.06.06.08.Ο6.06.Ο6.0
6.06.0
225 .0
22\ Λna tr n
225 .0225 .0C1ζ
^1_25 .01Ο0.0225.01ΟΟ. Ο
100.010Ο.0100.014tr
^
L25 .0L25 .0
644 .6
644 .6644 .6644 .6644 .6644 .6644 .6636.6509.3644 .6509.3509.3360.0
636.6450.0
0.003
0.!120.003 0.08800Ο.0060.004 0.088000.006Ο.005 0.088000.0030.011
o. oo7 o. 0822L0.0030.0030.02s0.010 0.123040.00s0 .025
1239. Solution follows Example 1214 closely.
1240 Solution follows Exampte 1214 closely
Exceφts from this work may be reproduοed by instructors for distribution on a notforprofit basis for testing or instruοtional puφoses only tostudents enrolled in courses for which the textbook has been adopted. Αny οιher )eproducιion or ιrαλhιion of ιhis wοrk beyond ιhαιpermiιιed by Secτiοns ] 07 or Ι 08 of ιhe 1976 (]niιed Stαtes Copyrighι Αcι ιιiιhouι ιhe peλission of the cοpyrighι oιιλer ii unΙαw'ful.

2551241.
1242.
1
Po=o
1
2
2
S
Supply fan: ΔPo = 4 in. wg.
Return fan: ΔPo = 1.75 in. wg.
1
Pυ=0
1
2
4
2
SF
S Space Pressure
A
le
S
Fan, ΔPo = 5.75 in. wg.
Exοorpts from this work may be reproduοed by instruοtors for distribution on a notforprofit basis for testing or instructional puφoses only tostudents enτolled in οourses for which the textbook has been adopted. Αny other reproduction or trαnsιαti()n οf this work beyond ιhαιpermitιed b) Sections ] 07 or Ι08 of the 1 976 Uniιed Stαtes Copyright Αct lνiιhout ιhe permission ofthe copyrighι owner is unΙawfuΙ.
SF
F
M c E
A
8,,
2571243.
1
Pυ=0
1
2
Supply fan: ΔPo = 4 in. wg.
Return fan: ΔPo = 1.75 in. wg.
1244.
1
Pυ=0
1
2
Excerpts from this work may be reproduced by instructors for distribution on a notforprοfit basis for testing or instructional purposes onΙy to
students enrolled in οοurses for whiοh thΘ tΘxtbook has been adopted. Αny other reprοduction or ιrαnslαιion οf ιhis νοrk beyond thαιpeιmiιted by Secιions ] 07 οr ] 08 of ιhe Ι 976 Uniιed Stαtes Copyright Αcι νithout ιhe permission of the cοpyrighι oινner is unΙαw/ul'
4
2
6
4
R F
M c Ε
S
SI
258Fan, ΔPo = 5'75 in. wg.
1245.
(a) Αssume a reasonable duct velocity of about '1200 fpm. ΔPo/L = 0.095
in. wg./100 ft. and D" = 18 in. (may be converted to 20x14 in. for
example)
For the duct: ΔPο  (0.095 x 40)1100 = 0.038 in. wg.
For elbows: Co = O.15; ΔP" = 2x o.15(118O/4OO5)2 = 0'026 in. wg.
For damper: Co = O.52, ΔPα = 0'52(118o/4OO5)2 = 0.045 in. wg.
For grille: ΔP, = 0'25 in. wg
For expansion: Vo = V.'(A.'/Αo) = 118Ol2 =59O fpm
ΔPu  1'2(59ol4oo5)2 = 0.026 in. wg.
overall: ΔPo = 0.038 + 0.026 + 0.045 + 0'25 + 0.026 =
ΔPo  ο.385 in. wg.
(b) For 18 in. duct with 1,ο00 cfm, ΔP/L = 0'027 in. wg./10ο ft
For duct: ΔP6 = 0'027 x 401100 = 0.01 1 in. wg.
For elbows: ΔP" = 2x0'15(59ol4og5)2 = 0.006 in. wg.
For griΙle: ΔP, = O'25(1Qoo/2oοο)2 = 0.063
For expansion: Vo = 59012 = 295 fpm
ΔP" = 1'2(295l4oO5)2 = O.OO7 in. wg.
Εxοerpts frοm this work may be reproduced by instruοtors Γor distribution on a nοtforprofit basis for testing or instruοtional puφoses only tostudents enrolled in courses for which the textbook has been adopted Αny oιher reproduction or trαnsιαιion of this νork beyond ιhαιpermitted by Sections Ι07 οr 108 οfthe ]976 Uniled Sιαtes Copyrighι Αcιwiιhοuι ιhe permissiοn οfthe copyrighι oνner is unlωνful.
259For damper: ΔP6 = 0.385 _ (0.01 '1 + 0.007 + ο.063 + 0.007) =
ΔP6 = 0'297 in' wg. = Co"(59O lAooqz
(c) Co" = 0.29710.022 = 13.7
1246. Equal Friction Method
Note that a damper has been inserted in duct 6 (Nο. 34 betow) tocause an increase in duct diameter from 8 to 9 1n. with a consequentdecrease in velocity to an acceptable 1evel.
 Deslgn Procedure System type: Supply
Duοt Sizing Method: trqua1 Εr1ctionRounding Method: Round Nearest
 Fan Selection Known Fan Parameter: E'an Tota1 Pressure : ο.900 in. wg
Fan Αirflow: 845.0 cfmΕan or Ext.ernaΙ Total Pressure: 0.9ΟΟ in. wg
Coil Lost Pressure: Ο.500 in. wqΕilter LoSt Pressure: 0.100 in. wgMisc. Lοst Pressure: 0.050 in. wα
AΗU External Total Pressure: 0.250 in. wg
AΗU Pressure for Supply System: 0.150 in. wg  or 60.0 %
AΗU Pressure for Return System: Ο.10Ο in. wg  or 40.0 %
 Lost Ρressure from Αir Ηand1ing Unit to Diffuser Diffuser ID Q TotaΙ Delta Ρ
(cfm) (in. wq)
19 150.0 Ο.14124 L25.0 0.12828 120.0 a.12332 200.0 0.11s38 25Ο.0 0.143
Total 845.0
 Ca1cu1ated Fittinα Values 
TD Εitting Type Dia. Q Velocity Delta P ^P/L(in) (cfm) (ftlmin) (in. \^/g) (in. wg)
EΧοerpts from this work may be reproduced by instruοtors for distributiοn on a not_fοr_profit basis for testing or instruοtional puφoses only tostudents enrolled in courses fοr which the textbοok has been adopted. Αny oιher reproducιiοn or ιrαnslαιion of this work beyond ιhαιpermitted by Secliοns ] 07 or 1 08 of the ] 976 Uniιed Sιαιes Copyrighι Αct wiιhοuι the permissiοn οf ιhe cοpyrighι ονner is unlωυful.
260
1 Αir ΗandΙing Unit2 Conical Contraction3 Straight Duct4 Elbow5 Straight Duct6 Elbow
7 Straight Duct8 Tee / vlye main
branchcoΙτιmon
9 Straight Duct10 Tee / Wye main
branchco]πnon
11 Straight Duct:L2 Tee ,/ Wye main
branchcoΙτιmon
13 Stralght Duct14 Tee / Wye main
branchcoΙττnon
15 Straight Duct16 ΕΙbow11 Strai ght Duct18 Rectangular TransitionL9 Diffuser,/ Grltle20 Sιraight Duct21 EΙbow22 Straight Ducι23 Rectanqular Transition24 Diffuser / Gri1Ιe25 EΙbow26 Straight Ducc21 Rectangular Transition28 Diffuser,/ crill_e29 ΕlbowJU SιraΙqhι ι]uct31 Rectaigular Transition32 Diffuser,/ Grille3 3 E]_bοw34 Stralght Ducι35 Butterfly Damper36 Straight Duct3'/ RectanguΙar Transition38 Diffuser / Gri1le
0.0L4 .014.014.014.014.0
14.0ι2 .0on
74 .012.A10. Ο
8.012 .010. Ο
9.07.0
10.09.Ο1.01.09.01.01.01.41.0
1.41.07.01.0
1.07.07.0
845.0845.0845.0845.0845.0845.0
845.0595.0250 .0845.0ξoζ n
395.0200.0595.0395.0215 .012A.A395.Ο215.0150.0L25 .0
15Ο.015Ο.0150.0150.0150. Ο
125 .0L25 .01atr ΛrZJ . υ12η nT rξ n
L20 .072A.A720 .0120.0200.020Ο. Ο
240.A200.0250.0250.0250.0
250 .0250. Ο
0.0190 .4190 .4190.4190 .4190 .4
190.415'7 . 656η q
190.4151 .61'Δ
'573.0151 .6'1
'Δ 'G22 \449 .0124 .2622 .5561.346'7 .'7
561.3561.3561.3450.0
46'7 .1461 .1Δ61 1
375.0
449.A449 .036Ο.0
573.0573.0600.0
565.9565.9ξ6η q
565.9750.0
0.0000.0080.0030.0Ο60.005Ο.006
0.0060. Ο050.018
0. Ο070.0040.017
0.0Ο70.0030.016
0.0Ο70. ΟΟ30.011
0.0060.0050.0160.0040.0400.013Ο.0Ο20.0020.002Ο.0300.0020.0110 .002Ο.0360.0030.0100.0000.04Ο0. ΟΟ3Ο. Ο130 .0240.0010.0010. Ο5Ο
0.06415
0. Ο6415
Ο. Ο6415
0 .01 L61
0.08259
0.07138
0.08082
0.0BΟ82
0.05817
0.05817
0.05405
0.07106
0.06Ο04
0.06004
8.08.0B.Ο
9.09.09.0oΛ9.Ο
Balanced Capacity Method
Note that dampers have been inserted in ducts 6 and 7 (No. 31 and36 below) to cause an increase in duct diameter and a consequentdecrease in velocity.
 Design Procedure System type: Supply
Duct Sizing Method: Balanced CapacityRounding Method: Round Nearest
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261
_ Εan Selection Known Γan Parameter: Fan Tota1 Pressure : 0.9OO in. wg
Εan Airflow: 845.0 cfmFan or Εxternal Total Pressure: 0.900 in. wg
Coil Lost Pressure: Ο . 500 in. wgΕi1ter Lost Pressure: 0.100 in. wgMisc. Lost Pressure: 0.Ο50 in. wα
ΑΗU Externa1 Total Ρressure: 0.250 in. wg
AΗU Pressure for Supply System: 0.150 in. wg _ or 60.O ?AΗU Pressure for Return System: Ο.100 in. wg  or 4O.O %
_ Lost Pressure from Αir Ηandling Unit to Diffuser Diffuser ΙD Q Total De]ta P
(cfm) (in. wg)
19 150.0 0.14124 725.0 0.72828 720.0 0.14033 200. Ο 0.14038 2s0.0 0.131
Total 845.0
 Calculated Fittlng Values 
ΙD Εitting Type Dia. Q VeΙocity De1ta P ^P/L(in) (cfm) (ftlmin) (in. wg) (in. wg)
1 Α1r ΗandΙing Unit 0. Ο B45. O O. O O. OOO2 Conical Contraction 14. Ο B45. O 19O.4 Ο. OΟ83 Straight Duct 14.0 845. O '79a.4 O. OO3 O. Ο64154 Elbow 14 . 0 845. 0 '7 90 .4 O . OO65 straight Duct 14.0 845.0 i90.4 o.oo5 0.064156 Ειbow 14.0 B45.O 190.4 O.OO67 Straight Duct 14.ο 845.0 19O.4 O.OO6 O.O64158 Tee / wye main 72.A 595. O j5j .6 O. OO5
branch 10.0 250.0 458.4 Ο.020coΙnmon 14.0 845.Ο 190.4
9 Stra1ght Duct 72.0 595.Ο 151 .6 O.ΟΟ7 0.0'716'710 Tee / Wye main 10.0 395. O j24.2 O. OO4branch 8.0 20Ο.0 573.0 O. O17coΙτunon ι2.0 595.0 151 .6
11 Straight Duct 10.0 395. Ο 124.2 O. OO7 O. O825912 'Tοο / τιι',o main 9.0 21 5.O 622.5 O.OO3branch 6. O L2a.O 6Ιl.2 O. O13coπ]Ιnon 10.0 395.0 '724.2
13 Stralqht Duct 9. Ο 2']5.O 622.5 O. OO7 O. O713B\4 τee / Wye main 1 .0 15O. Ο 561.3 O. OO3branch '7 .0 L25.0 46i .i 0.011co]nmοn 9.Ο 2'75.0 622.5
15 Straight Duct 1 .a 150.0 561.3 O. OO6 Ο. O8Οs2
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26216 Elbow j .0
11 Stralght Duct i .01B Rectangular Transjtion i.019 Diffuser / Grilte20 Straiqht Duct j.A2ι Ε1bow 1.O22 Straight Duct 1.023 Rectangular Transition i.024 Diffuser / eriΙle25 Elbow 6. 026 Straight Duct 6. 021 Rectangular Transition 6.028 Diffuser / Grill_e29 Elbow 8.030 Butterfly Damper 8.031 Straight Duct 8.032 Rectangu1ar Transitiοn 8. Ο
33 Diffuser / Gri11e34 E1bow 1Ο.035 Butterfly Damper 10. Ο
36 Straight Duct 10.03l Reclangular Transit1on 1Ο.038 Diffuser,/ critle
15ο.0150.0150.0150.012η n
125.01t\ cl
725 .0125.0120 .0120 .0120.0120 .0200 .0200. Ο
200 .0200.0200.0250.0250.0250.A250. Ο
250 .0
561.3561.3450.0
461 .'7461 .'7461 ."7
375. Ο
677 .2G1 1 2
36Ο.0
573.0573.0ξ?? n
600.0
458 .4458 .4458.4750. Ο
0.0ο50.016 0.080820.0040.0400.013 0.058170.0020.002 Ο.058170.002Ο.0300.0040.023 0.1\4270.0070.0360.0030 .0240.010 0.071060.00Ο0.0400 .0020.0L60.008 0.035990.0010.050
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CHAPTER 13
131. From Ξq' 132
h= tr* = lb* = ft'Α (C* _ c ) r.,,. _ t2 (lο* , n, ) ft2 _hr
Now C and W are related by Eq 1314
C = Wρ" ψ* % = lb*/ft3
lba fto
Τhe density of dry air must be used. Then from Εq. ,1 317
hο = hrP" = ft3 X9 = ]b,' ftz _ hr ft3 ft2 _hr
Consider Eq. 1313 which is dimensionless
h Btu ft3 lb,F ft2  hr ηzι ^Ξ^;  
P3Cp,hη ftz _ hr _ F b, Btu ft3
Clearly dimensionless when C* is used.
132. using Eq. 13'18,
h = Le2t3 = 1; hα = η = ,,'9. = 41'7lba/(hr  ftr)cpahα Cp, 0.24
also h, = hα/ρ" = 41 .7lO'O75 = 555.6 ft3/1hr _ ft')
h6 : 0.057 kgal(m' s)
133. hd = 0.61s p"oa7
k
2610,075x100x60x(1 t14_
= g52ΚΞo = oβ44
k = 0.0147 Btul(fthrF) (Τable A4a)
n = Ot.O,::!,
x 0.61 5(852)047 = 2.S9Btu/(hrft2F)(1t12) \ '
h6 = h/cpa =2.5910.24= iO.B lba/(ft2h11
h, = h6/ρ" = 1O.8/0 .O75 = 144 ff lσt'hr)
134. Nu = 0.023 Re0.8 pro 3 0r h = 0.023 (k/D) Reo.s pro 3
and h6 = h/cpa, assuming Le = '1
Re = ρ V olμ; V = 600/ (il$ = 471 ft/min or 28,260 ftthr
νL= 0.044lbmlfthr; ρ = 0.075 (TabΙe A4a)
Re = 0'075Ι?8'260x1
= 48,1700.044
Pr = 0.7; k = 0.01 47 btulhrftF (TabΙe A4a)
h = O.O23 ΨΨ (48,17o)os(O.7)o., = 1.7 Btu/(hrft2F)1
hα = 1'710'24 = 7'1 lba/(ft2hr)
h* = h6/ρ^=7'1l0'075 = 95 ft3/(ft'hr)
135. 43,560 ft2 = 1 acre; Γi'l * = hdΑ(Wr,  W)
Use j factor anaΙogy, h/crh6 = Le2l3
Αssume: Le = 0.85; Cρ + 0'24E
Thenho ' =23.221bal(ft2hr)o.24(o.8q2t3
Using chart 1: W = 0.013 Ιbu/lba
W*  0.0223lb"ilba (assume sat. air at 80 F)
Excerpts from this work may be reproduced by instruοtors fbr distribution on a nοt_forprofit basis for testing or instructiοnal purposes οnly tostudents enroΙled in οourses Γor rνhiοh the textbook has been adopted. Αny oιher reproductiοn o, ιroλιotion of ιhis νοrk beyond ιhαιpemitιed by Sections ] 07 or Ι 08 of ιhe ] 976 Uniιed Sιαιes Copyrighι Αct y]iιhout ιhe permission of the copyright owner is unlανful.
__.Ξ_262
Γh, = (23'22)1ooo(43,56OXo.o223  o.o1 3)fr,,, = 9,415,ooO lbr/hr Ξ 19 gpm/acre
136. Use analogy of Eq. 13_1g
Ql= hαΑ(W  Wr)irgi Q. = hΑ(t  t*)
q = qr + Q., W, = 0.00765; W = O.O'1 10; Chart 1
ho= h 144Ε\ 9(1.15)=, =49'23lba/(ft2hηcoLe2l 3 \ l'  ν'il
024(σ82τπ =
Q/Α = 49.23(0.o 110.00765)1o65 = 176 Btu/(hr _ ft')Q./A = 9(1 .1 5)(7550) = 25g Btu/(hrft2)
q/Α = 435 Btui(hr _ ft2) = 1'37 kWlm2
137. Qr= rh,(i*i,,)
Γh, = hdΑ(WV,  W)
h6 = h/(cr"Lezu); cp, = 0.24 Btul(lba _ F); Le2t3 = 1
hα = '1 '510'24 = 6'25lbal(hr  ft')W, = 0'0223lb,/lba W* = o.0096 lb,/Ιba
i = 28.4 Btu/lba; Chart 1
Γh, = 6.25(300 x 15O)(O. 0223  0.0096)Γh, = 3,572lbWhr
Q1 = 3,572(1,05o) = 3,75O,600 Btu/hr oΓ x 1,o99 kWΑny water on the deck and occupants neglected.
138' lt is assumed that the blanket is folded in half over the clothes line withone side exposed to air.
ho = h Le2t3 = #(0.g3)2l3 = 18.g7tbai(hr ft )cp
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Γh* =
Γhr, =
hdΑ(Wb  W,); Wn = 0.0312;W^= 0'0152
} 'ΔΘ=m*/[hdΑ(Wv_W")]
263
12. A. = rh r/G,
4000 x 0.071
6O/1OOO = 17 ft2
or instruοtional puφoses only toαιion οf ιhis wοrk beyond thαtghι οwner is unlαwful.
Δθ=(16  4)
= 0.71 hrI 8. 87(56)(0. 03 1 2  0.01 52)
Δθ = 42'6 min Say 45 min.
139. The procedure is the same as example 131 except that the
energy balance line AB will have a positive slope
and tl ι=75"F, tlz = 90"F
Ans: 68162"F; 17 .4 ft2:4.8 ft
13'10. The solutions to this problem closely follows example 132.
Ans: 77169"F; 17.4 ft2; 4.8 ft
Αns. 31126 C; '1.6 m2; 3 m
1311. The procedure is the Same aS example 132 exοept that the energy
balance line AB will have a negative slope and the inlet and outlet
water temps. are reversed.
Αns: 71t69"F; 17 '4 ft2;5.5 ft
13 50
45
40
35
30
25
20
Xhαam =5510.24=229.2
h1/h6 = 3.05 =#,
t
Ε_ο
J
οοι
oτΙ_ιlJ
'lI
JI
t2tι
YΕxοerpts fromstudents enrοl]permiιted by Sι
9060 70 80 100
264Υ : 2'7, Then
L = Gry/hdθm = 10ο0 x2'71229'2 = 11 .8 ft
1313. The solution to this problem closely follows example 133
1314. Ans: 1.4 to 1.5
1315. Solution of this problem follows example '134 closely.
'13'16. Αns: 5OO ft2; 12'2 ft
1317. Extrapolate the 72 F wb curve in Fig. 139. The largest cooling
tower modeI ''M'', iS not Ιarge enough to handle 2ο00 gpm.
Therefore use two towers of 1000 gpm each. Select the
model "L" which ls rated at about 1100 gpm.
13'18 See example 133; the cooΙing tower must be larger.
1319. See example '133; the cooling tower must be larger.
1320. (a) Model B or C using Fig. '139
(b) Cooling Range = t 1 t. 2= 10085 = 15 FExοerpts Γrom this work may be reproduοed by instructors for distribution on a notforprofit basis fοr testing or instruοtional purposes only tostudents enrolled in courses for which the textbook has been adopted' Αny οther reprοduction or trαnsιαtion of ιhis νοrk beyοnd thαιpeιmitted by Secιiοns ]07 or l 08 of ιhe ] 976 United Sιαιes Cοpyright Αcι
'yiιhouι the permission ofιhe copyrighι owner is unlαινful'
265Αpproach = t. ztwol = 85 _ 76 = 9 F
Tower capacity = Q
q = (200 x 60 x 8.33)(1)(15) = 1,499,400 Btu/hr
1321 q  5OO qpm x Δt; qpm = '=??o,o^o = 50500x1 0
qpm/ton= 50 ^^. =3.0
(250,000 / 1 5,000)
Note: !n this case, '1 ton = 15,000 Btu/hr
Cold water temperature; t" = 70  10 = 60 F
From Fig. 137; t*o = 42 F
1322. Αlbuquerque, NM; tψ6 = 64 F (Τable B1a)
(a) From Fig. 137; cold water temperature = 73 F;
gpm/ton = 2'5i \Λ/arm water temperature Ξ (73 + 10) 83 F
(b) Charleston, SC; tψ6 = 79 F (Table B1a)
From Fig. 137; cold water temperature = 84 F,
gpm = ΨΨ x2'5= 83 gpm (a & b)15,000
1324. Model G, nominal rating  600 gpm & 250 tons (Table 132).
Using Figure 139; assume gpm is constant.
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1323' (a) tons = ''?9?99o = 80, gpm/ton = 24OΙ8O= 3.O, maximum two = 72F15,000
(b) gpm/ton = Ψ = 4'o',max. t,5 = 65 F80
266With cooting range of (97  85) = 12
Max. two = 76 F (Figure 139)
With cooling range = 15, tu, = 100 F
Max. two = 74 F (Figure 139)
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_=\ 
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CHAPTER 14
141. (a) P  12060= 0.43
200  60
* = 200180
= 0.33120  60
F = 0.985 {Fig. 14191
LMTD _ (1so  6_q)_:(299_ 120)(180  60)
ln
LMTD = 98.7oF
(200 120)
Btu/hrF275(7.48)16'500
=2η5ft3/hr; Q  = 34 gpm
(b) C" = (fr "r)",,
= 50oo *uo (29'92x0'491x144) φ'24)
53.35(520)
= 5490 Btu/hrFcn = cc(1,2  tu1)/(t*z  t*r) = 5490(1 20 60y(200  180)
= '16,500 Btu/hrF
(rh cp)* = (Q pcp)* = 16,500(c) Cn =
Q=(6OlΧ1) 60
(d) q = UAF(LMTD)
UA = Cι., (t*zt*l) _
F(LMTD)
16,500(2ο0 _ 180)
0.985(e8.7)
UA
= 3390 Btu/hrF
UA UΑ(e) NTU =
NTU =
Cc Cair
= 0.62
Cmin
33905490
^267
(f) "=
12060 0.43\ / 20060
268142' (a) q = UΑF(LMTD) = (rh cp),i,(1 1050)
Γh, = 4000 x 14'7 x 144l(53.35 x 510) = 311'2 lbΙminor 18,672b/hrQ = 18,672(0.24)(11050) = 268,874 Btu/hrq = (rh cr,,)('l80  tr.,o) = Q5 x 8.33)(1)(180  tho)60
tι.o = 1 80  =?9y+ = 158.5 F or 159 F25x8.33x60
p= 11050 =0.46: *='180159 =0.35
180 50 1 10  50F = 0.98; Fig. 141
LMrD=S# =88
't[ ^JA = 9/(UF x LMΤD) ='!u!!!^o== = 312 ft2' 10x0.98x88
(b) Cair = 18,672(0'24) = 4481 = Cmin
Cwat = 25 x (60.817 .48) x 60 = 12,193 = Cr*, =
110  5o = 0.401'
Cmin  4481 = o.3T
180  50 crr, 12,193
NTU = 0.7, Fig. 1 418
UA/C,1. = O'7i μ= 0'7\!481 = 314 ft2, ,10
143. (a) Γh air = 32οo x 60 x ::'::'!! = 13,726lb/hr53.35x555
Cair=Cmin=3294
NTU  1ox3oo = 0.91 125
3294ε = 0.615, Fig. 1418 T
at C.inlCrr, = 0
l refriο. l
atr.
+
125
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students enrolled in courses for whiοh the textbook has been adopted. Αny oιher reprοduction or trαnslαtiοn of this νork beyond ιhαιpermitted bν Secιιons ] 07 or Ι 08 οf ιhe Ι 97 6 LΙniιed Sιαιes Cοp1τ'ighι Αct Ιψiιhouι the permissiοn of ιhe copyrighι oνner is unlcτwfuΙ.
r_.
(b) ,., = taηh(mrl)
, ο = [E ,][.'+ O.35rnBj = 1.243\mrφ) \r )\ r/mrΦ  (18.26)(0 '5112)1.243 = 0.9454
,_ tanh(0.9454) =0.7g' 0.9454
(c) Within readability of Fig. 144 the answers are the same
269143. (continued)
0.61s = ,':"='^? , tco = t"o = 0.6'1 s(125 gs) + g5125  95
t,o = 113'5'F
(b) Q = Cui,(tao  tai)  rh,.irg = 3294(11 3.5  95). 3294(1 13.5  95)
= 928 tb/hrlrl,. = .6s.s
144' (a) ,= ΓηlL κy ] L 90(0.00 8 l 12) J
lm = (1.0  0.5)
18.26= 0.7612
R/r = 1l0.5 = 2.0i η = ο.8, Fig' 144
145. ξs = 1 + (1  η); η = O.78 from proplem 144A
ξs = 10.9(10.78) = 0.80
1 1 Δx 1 1 (0.01 5112)'1 Λ^
Uo ho?ro k(Α, /Αo ) hi(Αi /Αo ) 10x0.8 (100x1)
1J = 0.17; Assumes Α; = Αo and k.opp",. = 100
200x(1/9)
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270The second term may be neglected
Uo = 5.9 Btu/(hrft2F)
147. '1 =! =0.133Uo 10x0.8 1 100(1/9)
Uo = 7.5 Btu/(hrft2F)
148. ,  tanh(m/)' m!.
, = Γeηl''' = z"sτ _1''' =64.18 m1L κy ] [ l za1o.'t 6x1 o_3 l]
v
mI = 64.18(6 x 1O') = 0.385; η = 0.953
14g' ns=1 *(1 η)_1o.s5(1 _O.95)
η. = 0.96
Ar = 2ΗLWP, mm2; Α, = LW mm2; A = Ar + (LW _ tLWPS)
Where P" = fin pitch in fins/in. and L = W = 1
Δ = 2HLWP' + LW _ tP, _2x6x0'47 +1_0'16x'47  '1 '1AΛ'
Αo  ,ΗL\Λ/P' 2x6x'47 l' lo+
+=\:γJ^ = O.O19; U = 52.3 W(m' _ c)u 1400 57(0.96)
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1411 .
tan h(m r/ )
mrΦ
R" = 1.2s ψ(β  0'2)'''; m =
r
Γ znl1l2
Lπ]*=},β=;;L>M
(a) Diml =;'=Ψ = 0.56 in
Dim, = rΓrg)' *or1'''' 2Lι2l l
'Φ= [}_,][,+O35.(})]
e0(0.01/ 12)
η=
271
= 16.33 ft1
= jιιo.uu)'* (1'35)'1'''= O.73
Then L = Dimz = 0.73in.; M = Dimr = 0.56 in.
ψ= 0'56 =1'75:3=0'73 =1.3
(0.64 t2) 0.56R"
= 1.27(1.Ts)(L30.3)1 t2=2.22r
φ = (2.22 1)t1 + O.35ln( 2.22)!= 1'56; . = ΓL
mrΦ  16.33(0.32112)1.56 = 0.631
n_ tanh(0.762) =0.869' 0.762
(b) Dim.a12.5mm'2Dim2 = ! lZZt + 12.5'ltt2 = 12.65 mm' 2'L = Dimz = 12.65 mmM = Dimr = 12.5 mm
t=12'5 =2.s:g= 12'65
=1.0125 12.5R"
= 1.27(2.s)(1.01 2  0.g)1t2 = 2.69r
f'''2x10
Excerpts fiom this work may be reproduοed by instructors for distribution on a notforprofit basis for testing or instruοtional puφoses on1y tostudents enτolled in οourses for whiοh the textbook has been adopted' Αny οther reproducιιon or ιrαnsιαtion ofιhis lνοrk beyond thαιpermitted by Sections ] 07 οr ] 08 οf ιhe Ι 97 6 United Sιαιes Copyrighι Αcι y,ithοuι the permission of ιhe copyright owner is unlωνful.
b = 1.35 in
b=22mm
272
Φ = 2.69  1)t1 + 0.35 ln(2.69)] = 2.26Γ zxoε l _^ _m=l E'\vv l=66.67m11170(0.00018)_j
mrQ = 66.67(0.005)2.26 = 0.753_ tanh(O.753)
11 =' 0.753
η = 0'85
1412. + = 1, ; neslecting tube wall resistanceUo hoΠo h; (Α, 7no ) ι  v_ _{
(a) 1o=1 *(1 η)1_o.9(1 _O.s4)=0.86
1 1 :^ = 0.120; Uo = 8.60 Btu/(hrft2F)%
= 1o10€6
* οoo(l o)
(b) ro = 1 0.9(1 0.81) = 0.831 1
^4^ = 17.8; Uo = 0.056 kW(m2  c)% =
0,.068ro s3 *
1a.+rl o1
1413. (a) Rct = 2.222 x 106ο.010
Rct = 4.15 x 10a 1hrft2r1/εtu
(b) '12 fins/in = 0.472 fin/mm
Rct=3.913x107
1414' Re= ρνD;ρ=60.6 lbm/ft3μ
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0.18
]""'
1o
uott=
1 .oe3 x 1o41m2  cyw
l
0 64( 1t12  1)'
ι 0.010 )
rc(1t0 72_ 1)'ι 0.18 )
273Dr = 0.545112 = 0.0454 ft; Table C2
V = e lA= 2'5 x+ =3.44fVsec' ι 7 '48^ 60(π t4)(o'o454)2
μ = o.93 lbm/(ft  hr) = 2'58 x 1Oa lbm/(ftsec) Τable A1a
Re = 60.6(3.44)0'0454
= 36,6g3i Re = 36,7002.58x10a
(L/D),in = 410'0454 = 88 ftPr = 2'43 (Pr = crμ/k)ιlo'',u = O.ο23 κ"Bu Pro'; k = 0.383 Btu/(hrftF)k
h = O.O23 ,Ψf''='J. (36,7oo)o'12.431o' = 1,136 Btu/(hrft2F1(0.0454)
1415' ρ = 1'o1(62.4) = 63.02 LBM/FT2 [Fig. 1 02a]
μ= 0.7l1490 = 4'7 x '1o4 lbm/ftsec [Fig. 1O2b]
Cp = 0.93 Btu/lbmF [Fig. 14ε];
K = 0.93 Btu/lbmF [Fig. 149]
Υ = 3.44 fVsec [ProbΙem M1aJ;
D = 0.0454 ft [Problem 1414]
63.02(3.4 4)0 '045ΔRe=ffi=20,940(L/D),1n = 88 ft [Problem 1414]
o _Cpβ _4'7x1o41sοoo)o.93 _ΕΕ.)lrν''νia k ο.285
h = O.O23 ,!o:',u=u], (2ο,94O)ou15'521o'= 690 Btu/(hrft2F)(0.0454)
1416' ρ = 1.O45 x 62.4 = 65'21 lbm/ft2 ;
μ = 1 '3l1490 = 8'725 x 1Oa lbm/(ftsec)
co = 0.81 ;k= 0.22; V = 3.44 ft/seC, D = 0.ο454 ft
Re=#=11,670
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274
,," _ 0.81x8.725x1 043600 =
.1 ..16
0.22
h = 0.023 !:?3=). (i 1 ,670)0.(.1 ..t6)0. = 209 Btu/(hrft2F10.0454 '
1417. Use hydraulic dia. for rectangular channel
Dη = 4rn = a(AJP) = 4(3/8) = 1.5 in. = 0.125 tt
ρ = 62.4lbm/ft3 [Table A1a];
μ = 3.45 lbm/(fthr) [Table A1a]
Re = u':o!!_),Ψ
'?? = 32,556(3.45l3600)
cp = 1.003 Btu/lbmF [Table a1a]
k = 0.338 Btu/(fthrF); Pr = 3.45 x 1 .003/0.338 = '10
(a) For cooΙin9, h = O.O23 ! R"o'Pro'D
h = O.O23 9Ψ (32,556)0r11o.21o. = 5O9 Btu/(hrft2F)0.125
(b) For heating;
h = O.O23 9Ψ (32,556)0s11O.21o. = 642Btu/(hrft2F)0.125
1418. Dr, = 0.125 ft [From problem 14171
ρ = 62.4(1.o45) = 65'2lbm/ft3 [Fig. 1 O2a]
μ = 3.5/149O = 2.35 x 1o_3 lbm/ftsec [Fig ' 102b]
Re=
c, = 0.89 btu/(lbmF) [Fig. 148]
k = 0.28 Btu/(fthrF) [Fig. 1ag];
Pr = 2.35 x 1 o3136oο)(o.s9) l0 '28 = 26.9
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275
(a) Cooling
h = 0.023 # ('13,900)0t126.910t = 28s Btu/(hrft2F)
(b) Heating: h = 285 (26'9):1
= 396 Btu/(hrft2F)'"" (26'910'e
141g' (a) Re = ρΥD _ 99ο'2(1 '5)(0'012) _ ^'''μ o'uffi =29'905
Assume L/D > 60 ,
Then Φ = o.o23 Reo.8Pro4, Pr = [o'sgοxlo3x+'lεzl = e.glk ι 6373 )
π = 9 9?: (O.637X31 ,157)o s(3.91)o 4
0.o12h = 8287 W(m2  C) = 8.29 kw(m', C)
Data from Figures 102a, 102b, 148 and 149.
(b) Re = (1 'ο28)999(1 '5)(0'012)
= 15,4OO1.2x103
1 .2x103x3.7x103,r=ffi=8.9: o.o23h = Ξ'Ξ]Ξ (ο.5o)(15,4oO)08(8.9)oo = 5140 W(m2 _ c)
0.012= 5.1 4 k\Νl(m2  c)
1420. (a) Re = 62'4(0'5)(0'3!112)
= 922< 2soo(3.45l3600)
hD =
.1.86tReP. D .,lls [ra)o 'o : Αssume ( n)o'o
= ,τ  l.vvιl.,, L, ιr',,l,,1oΦLΙl,, ιλJ
Εxοerpts from this work may be reproduοed by instructors for distribution on a nοtforprοfit basis for testing or instruοtional purposes only tostudents enτolled in courses Γor whiοh the textbook has been adopted. Αny oιher reproductiοn or trαnslαιion of this work beyond ιhαιpermitted by Secιiοns ] 07 or Ι 08 of the ] 976 Uniιed Sιαιes Copyrighι Αct νiιhout ιhe permission of ιhe cοpyright οlιner is unlcrlνfuΙ.
276
,,=ξ1# =1O'4
π = sff# p22(1o.ol(ffi),"' = 66 Btu/(hrft2F)
(b) Same procedure as part a using data for 30o/o ethyleneglycol from Figures 102a, 102b,148 and 149.
14_21. Re  992.2(0.1.0)x103 = 1519
653There is a question about the flow regime. lt is probably
in the transition region. Assume it is laminar and use
7 10.14
Eq. 1 424 and assu me  U I = 1.
[ρ'J
Pr  o.653X'1ο_3(4.182) = 4'34
0.63τ 1 89φΞ3)
1519(4'34)rΨ']]1/3  328 W(m' c)n=O.O1 ι'_\ 'ι 3 )'
1422. Use average values for Gu and G. and Eq. 1426.
G. =:o;(rh. )"us= 0.912;(Gr)",s = ln't,,u =237.81bm/(ft2hr)' Α.'' "9'v z(o.sε9)'4\12)
(rh,),us  (1 + 0.1)12 = 0.55 lbm/hr
(Gu),us = 0.55/4" = 290'6 lbm/(ft2hr)
DG, _ r0.589)r?ΞΞ) = 12 3lιt ι 12 ,ι 0.95 /
DGu _r0 589)rΞ9ψ) = 15
βι ι 12 ,ι 0.95 /
Εxcerpts from this work may be reprοduced by instruοtors for distribution on a notfοrprοfit basis for testing or instruοtional purposes οnΙy to
students enrotled in οourses for whiοh the textbook has been adopted. Αny other reprοduction or trαnsιαιιon οf ιhis work beyond thαι
permiιιed by Sections ] 07 or Ι 08 οf the ] 97 6 Uniιed Sιαtes Copyrighι Αct νilhout ιhe permission οf the copy'ighι oνner is unΙαινfuΙ.
H= = 138(pr.,,. [#j"'L? E)"102
277
pρ'( e")"' = .,u.o1, 61 o )"' = 1oo8
βι lρ") ι0.0135/
Pr _ 0.95(1.001) =2.48' 0.384
i,, = 1001 Btu/lbmΔt ρ 80 = 1'160 _ s0); twall Ξ 80'F (Using water outside the tubes)
6 = 13.8 0.384  '4n' Γ 1φ]l"u ,,,oo8]o
2
  ο.58 g t12
QΑ8) '' L,1 ,(uη] "
Uυδl'
h = 888 Btu/(hrft2F)
1423. Use average values of G. and
A"= ζ (O.o15)'= 1'767 x 1Oa"4(G. )"us =
(Gu)"ug =
(ο.s8)o. 126x1o_3 Ιz1.767x10a
0.126x103 (1 + O. 1 2) t 2
1.767x10a
Gu and Eq. 1426
mt
= 0.314 kg/(m2  s)
= 0.399 kg/(m2  s)
DGr _ ο 015(0.311) = 12'1
βι 0.390x10_"
DGu ( ρr)"' =
O.o15(0.39_9 [ gzο
l1
Ι2 = 1024
βι ο" ) O.39Ox1o_3 L0'219J
Pr.  O'39Ox1 9jΙ1' 19x1 03 = 2'46; ig = 2326kJ/kg' 0.665
Δt = 45C = (73 _ 28); liquid water assumed outside tubes
h=13gφj95)(2'46)1t3lffi].,u..o24)o2=5022W(m2c)
h = 5.02 kw(m2  c)
1424' Use Εq. 1428
R_22 1 At inlet x : 0.20; at outlet 10oF superheat
Excerpts from this wοrk may ,l"rΓj],' ξ,ο,Jhm/fu p,ειJ*FsnRιε 1"Q,psi?i k,':. Ω if,'$"t.nω puφoses only to
students enrolled in οourses'for which tbe textbook has been adopted. Αny oιher reproduction or trαnsιαιιon of this wοrk beyοnd ιhαιpermιtted by Sections ]07 or Ι08 ofιhe 1976 United Sιαιes CοpyrightΑcι\υithout ιhe permission ofιhe copyright ονner is unlαννful.
278
Since X" Σ 1.O; Cl = 8'2 x 1O3; n = O.4
Αssume tube wall thickness of 0.016 in.
Τhen D; = 0.375  2(0.0161 = 0.343 in.
Ai= πg
= 6'417 x 1o4 ft2'4rh 80
rJ = : =

= 124,700 lbm/(ft2  hr)A j 6.417x10a
μ. = 0'52lbm/(fthr) at 30'F (sat. temp. at 70 psia)GD
= 124,700(0.343 I 12)
= 6855βι 0.52
k = 0.056 Btu/(hrftF) at 30'F (sat. temp.); Table Α3ai1g = 88'5 Btu/lbm
h = 8.2x 1O3 (0.o56)
Γrοεssl, (ττaφ'qaa'sβz'ιτ))lo o
(O.343, r)L'ooccΓ ι 5(32l 7) J.]
h = 779 btu/(hrft2F)
1425. Use Equation '1 428
R22;G = 2OO kg/(m2  s); Dr = 8.5 mm; L = 2 m; P,= 210 kPa
ξ = 30%, Xe = 100%
Tsat = 24C a210 kPa abs. Pres. [Table A3b]
μ. = O.27O x 1O3 Ns/m2 [Table Α3b]; extrapolate
k. = 0.107 W(m  c) [Table A3b]
iβ= 223 kJ/kg [Table A3b]
GD=200(0.0085_) =6296βι 0 '270x10_3
Cι = 8.2 x '1O3; n = O.4
h = 4106 W(m2 _ C) = 4.11 kννlm',_ c
Excerpts from this work may be reproduced by instructors for distribution on a notforprofιt basis for testing or instruοtional puφoses only to
students enrolled in οoursοs for which the textbook has been adopted. Αny other reprοducιion or ιrαnsιαιion οf ιhis work beyond thαιpermitlecl by Sectiοns ] 07 or Ι08 οf ιhe ] 976 tJnited Stαtes Cορyright Αct \νiιhout ιhe permissiοn οf the copyrighι oνner is unΙπwfuΙ.
6 = 8'2x 1O3 (o'1O7)Γ
β296\20.0085 L'
. r0.
J]
0007
7(223)1
2x9.80Ie
279
1426. lr, = f !!1 , Assume isothermat" D2g
Re = 36,700; ProbΙem 1414; smooth tubes
f =0.022, Fig. 101;L=(6x6)+(5xl)=41 ft
Dr = 0.0454 ft; V = 3.44 fUsec, Problem 1414
lr, =o.o2z* !1 *lt!!l,t =3.65ft0.0454 2x32.17
1427' l ι. = f !ζ, Αssume isothermal" D2g
V =O.5ft/sec; f =64/Re =641922=O.OOg
L = (10 x 10) + (9 x 1.5) = 113.5 ft; Di = 0.34 in.
lι. = 0.06 n*11?^'?x12,  Ψ', = 1'07 ft0.34 2x32.17
1428. Refer to Fig. 1410
Load/circuit = 10 x 1 2,000110 = 1 2,000 Btu/hr
Length/οircuit = (6 X 5) + (5 x 0.75) = 33.75 ft
(a) ΔP. /L = 0.10 psi/ft; CF = 1.25 Fig. 1410
ΔP. = 0.10 x 33.75 x 1'25 = 4.22 psi
(b) ΔP. /L = 0.04 psi/ft
CF = 1.25
ΔP. = 0.04 x 33.75 x 1'25 = 1.7 psi
1429. Gt,^ = 18OO lbm/(hrft'); t, = 7O"F,tz= 120"F
(a) Figure 1412
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280
Re, = G'Xo
, Gc = G"
 ΨΨ = 3214 lbm/(ft2hr)μ σ 0.56 ''_'''\
ψ = 4.6 x 102 lbm/(fthr) at t = 95oF (Table A4a)
Re, 3214(1 .083 t12)= 6306
4.6x102
j = 0.0091;f = 0.021 Fig. 1412
I = jG" c, Pr2l3 = O.OO91 (3214)0.24(O.T)2t3
= 8.91 Btu/(hrft2F)
(b) Re6 = 6306 x 0.52511.083 = 3057; assumes expanded tubes
plus fin collars.
A 4xoxrα _ 4 1.25x1.083x0.56 = 1O.O8At πD6D π 0'0152x0 '525x12
JP = (3057)04(1O.OS)015 = 0.0285 (Eq. 1439)
j= 9. 1x1ο3 Fig. 1414; h = O.OO91(3214)(0.24)(0.7γzl!
= 8.9 Btu/(hrft2F)
D*=0.525x10 08(1.25  0.525) = 0.904
116.7
Using Eq. 1444.
1+
( 0.525,0 " I(3o57)0 2'[αoo+.1
t
1.25  0.525
]"
 ι 'zs _,,l_o u
_10.e04 I FP=
4( L o o06)\67 )
0.173
f = 4.2 x 1O2 or f = 0.042 (Note that f may be in error up to
!35o/o (Figure 1415)
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281
1430. (a) G" = 4.5 kg/(m'.); tp = 2OC; Re = Gc(0'0275)
'
μ
μ = 1 8'2 x 106 Νs/m2 1Table A4b]
Re = 4'5(0'027Ξ)
= 6800; Cp = 1.OO5 kJ/kgC18.2x10o
j = 0.009; f = 0.020 [Fig. 14121
h = jG" copr2t3 = o.oog0(4.5)1.005(0.7)2t3  0.051 kJ/(m2sc)
6 = 0.051 kw(m2c)
(b) See problem 1429b for prcedure
1431. Use Eq. 1442 or 1445
l  elΓr,r * o2{ eι_,l*'+4l,n  2gρρ1L\,*" )lo''_' )_, η λ,
ρ, = #!S#L = 0.O75 lbm/ft3; ρ2 = O.068 lbm/ft3' 'i 53.35(530)
P,, = (ρl + ρ2)l2= 0'072lbmlft3
A _ αΥ _147(1.083112)5 = 118'45Aο σAf, (1)0.56
Where V = Αt. xL; A1," = 1 ft2;L = 5 x'1.083/1 2 = 0'451
l h = β214)2 _Γ(, * ιo.u6), )r9Ψ _ l)*'' 2(32'17)(0'072)(0.075)(3600)Ζ L\ /\0.068 )
o 042(1 1s 5)(0 075)lo 072 l
h = 12.2ft of air
ΔPo = Ω, (1i::?)Α;'z = 0 18 in. wg
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2821432. Use Equation 1 442 or 14_45
ti = 10C; t2 = 30C, ffΑ" = 1 18.5 from problem 1431
1O1x103oι = ffi1 = 1.244 kg/m3; Pz= 1.161 kg/m3;
pm = 1.203 kg/m3
t' = ffi[(,'. (0 56)2 )(# 1).
o.O19(1 18.5) 1μ!1\ / 1.203 )
I r, = 1.67 m of air
ΔPo = 1 6, (#lι',rool = 2'O8mm of wg.
or ΔPo = 1'67(1'244)9'807 = 20.4 Pa
14g3. Re = G.Dr,
, Dι' = O.O101 ftμ
Αt 65.F, μ = 4.39 x 1O2 Ιbm/(fthr) [Table A4a]2700(0.0101)κe=lffi =621
From Fig. 1416;j = 0.013, f = 0.053
h = G" co j Pr2t3 = 2700(0.24)O.013(0.72)2t3 = 1 O. S Btu/(h rft2F)
Where co= 0'24 Btu/(lbmF); Pr = O.72 From Τable A4a
1434' Use Eq. 1433; tυ = 65 F
14.6x144o Ξ ff = 0'074lbmlft3; ρ2= 0.077 lbmlft3' 53.35(535)
Pm = (ρl + ρ")l2 = 0.076 lbmift3
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28s
A 4L ψI2 = 132π=π= oοlοl
Assume a contraction ratio of 0'5
Then Κi  0 '32', Κ" = 0 '27 Fig ' 1 417
ΔPo _ e7oq2 _ t(o.32 * l os2)
π _
2β2.17 )1 4'6(1 44)(0.07 4)(3600)'
+ 2 (# 1)  o 053(1 3r)ιffi#] _ r _ (o 5)2 _'''1#'\
ΔPo/Pg1 =4'126x10a
ΔPo = 4'126x 1o4114.6)(1 44)(12)t62'4 = 0'17 in' wg'
1435. (a) Coil DescriPtion'
Type of coil = Refrigerant condenser
Tube pattern = Staggered platefintube coil
Material = Αluminum fins with copper tubes
Refrigerant type = Refrigerant 134
Finned side fluid = air
Finned side air pressure = 29'92 inches of Hg
Face area  4'44 square feet
Height of heat exchanger = 20'0 inches
Width of heat exchanger = 32'0 inches
Numberofrowsoftubesintheairflowdirection=4
Numberoftubesperro\Λ/=16circuitsontubeside=4
Fin pitch = 8 fins/inch Fin thickness = '006 inches
Vertical tube spacing = 1'250 inches
ΕxcerDts from this work may be reproduced by instructors for distribution on a notfbrprofit basis for testing or instructional puφoses only to
sιudenιs enτolled in courses ι", *L1.λ ,ι'. ιexibook ιr, υ..nrjopi.α*' λny o,rnr, ,rρroλur'ioλirιron,Ιαιιoi of ιhis νork beyond ιhαι
permiιιed by Sections ] 07 * ,oi'iiiλ)jili'bnii i,o*, ciiilii, iri ''ιhout
ih" prr^ιr,rio' oj'n" copyrighι oνner is unlωυful'
284Horizontal tube spacing = 1'083 inches
Tube outside diameter = .500 inches
Tube wall thickness = .016 inches
lnside tube fouling factor = 'OOOO BTUHRSQFTF
'/6C
1495. (continued)
Total heat transfer rate = 48783'2 Btu/hr
Sensible heat transfer rate = 48783'2 Btu/hr
Entering air conditions:
Dry bulb temPerature = 95'0 F
Face velocitY = 650'00 FPM
Air volume flow rate = 2888'9 CFM
Leaving air conditions:
Dry bulb temPerature = 111'4 F
Tube side conditions:
Refrigerant saturation temperature = 125'0 F
Air pressure loss = '393 inches of water
Tube side Pressure loss = 1'07 PSI
Fin efficiency = '3gg Surface effectiveness = '824
Tubesideheattransfercoefficient=388.88tu/hrSQFTF
Finnedsideheattransfercoefficient=12'5Btu/hrSQFTF
Mean temperature difference = 20'7 F
(b) Yes
1436. Coil DescriPtion'
Type of coil = water or brine solution
Tube pattern = staggered platefintube coil
Material = aluminum fins with copper tubes
Tube side fluid = water
:i:.J:fJ:f;':T:J$E::',i:;i1:i1i:'*iJ$ιι'{ij:{..Ti ι:;i:i'b:;,ff:;;i!:'ii:i1r,*"Ιir;?j':ii,:!;;iiii:i!jΙ:'i#ii!*''
ρermiιιed by Secιiοns Ι 07 "''iοi')i 'i i ii6 t'Jniιed Sιαιes Copyrighι Αcι w ilhouι ιhe permι
286Finned side fluid = air
Finned side air pressure = 29.92 iches of Hg
Face area = 5.56 square feet
Height of heat exchanger = 20.0 inches
Width of heat exchanger = 40.0 inches
Number or ro\Λ/S of tubes in the air flow direction = 2
Number of tubes per ro\Μ = 16 Circuits on tube side = 4
Fin pitch = 7 fins/inch Fin thickness Ξ .008 inches
Vertical tube spacing = 1.250 inches
Ηorizontal tube spacing = 1.083 inches
Tube outside diameter = .500 inches
Τube wall thickness = .016 inches
lnside tube fouling factor = .0000 BtuhrSQFTF
Diameter of inlet pipe/header = 1.0 inch(s)
Total heat transfer rate = 95759.1 Btu/hr
Sensible heat transfer rate = 95759.1 Btu/hr
Entering air conditions:
Dry bulb temperature Ξ 7ο.ο F
Face velocity = 650.00 FPM
Air volume flow rate = 3611.1 CFM
Leaving air conditions:
Dry bulb temperature = 94.6 F
Tube side conditions:
Entering fluid temperature = 150.0 F
Leaving fluid temperature = 128.2 FExcerpts from this work may be reproduced by instructors for distribution on a notfor_prοfit basis for testing or instructional purposes only to
students enrolled in courses for whiοh the textbook has been adopted . Αny oιher reprοduction οr trαnsΙαtion of ιhis 'work beyond thαι
permitted by Sections l 07 οr Ι 08 of ιhe ] 976 t]niιed Stαιes Copyrighι Αct wιthouι ιhe permissiοn οf ιhe copyrighι oιυner is unΙαwful'
287Τube side fΙuid velocity = 4.00 FPSCooling or heating liquid flow rate = 9.0 GPM
Air pressure loss = .187 inches of water
Tube side head loss = 8.20 feet or water
Fin effiοiency  .831 Surface effectiveness = .846
Tube side heat transfer coefficient = '1368.8 Btu/hr,SQFTF
Finned side heat transfer coefficient = 14.4 Btu/hrSQFTF
Mean temperature difference = 56.8 F
1437. From problem 1429, Re = 6306 (based on xp)
and h5 = 8.9'1 btu/(hrft2F)
jn _.1: = r 1280 Nr(Re)1 2 (Eq. 1442)jι
i
then Ψ _ 1_ 1280 x 5 (Re)1'2 = 1_ 6400 Re_1 '2
iι
i
and Jξ_ 1_(8x12so)(6306)_1 1 = o 872, ^ _ ιJ.ιr, gjs 1 _ (5x1 2s0)(63o 6l_ι 'z
now hε/hs = jε/js = 0'872
hε = 0'872(8.91)
ha = 7.77 Btu/(hrft2F)
1438. From problem 1430, Re = 6800 and h5 = 51W(m' C)
jn  1 128oNr(Re)l 2
[From sotution to 14371js 1 64OORe1 2
jο _ 1_(6x128ο)(680)] 2 = 0.96js 1 64OO(6801r z
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288ιl ur rr s = jo/js = 0.96
hο = 0'96 x 51 = 49W(m2  C) or o.o49 kW(m2  c)
1439' Re, = ρ%xo = =o':1:= rΨ (1.os3/12) x 60 = 8225
μ 0.0445 0.54
j = 0.0095 (Fig. 1412)
ιl  jcco Pr2l3 = O.OO85 x
h = 10.5 Btu/(hrft2F)
ο.o73, ffi x60x o24(o'η2t3
c.. _ 0.0123  0.0092 =
.1 .033 x 1Oa,) Using Chart 1 orv ι
85 _ 55
.^^  0.01230.0063 = 1.5 x 1Oa
) PSYCνz  85_q5
Cavg = 1'27 x1Oa; Use Eq. 1 47o &1473',k = 1zε*s}η;
From Table 51a.
M2 = ##l, Ψ] = u, 3; M = 22'7 ft1
Ψ = 1'27 ψ(β o.3)"' = # (1  o.3)1t2 = 1'265r
* = [E'_ l)[l + O.35ιn&l'ιr /ι r)Φ = (1 '265 _ 1)[1 + 0.35 ln(1.265)] = 0'287
MrΦ = 22.7 xΨ xO'287 = 0'285
 tanh(mrl) = 0.g74 or g7.4o/oΠm_'' mrΦ
ξ,s = 1 * (1  η,) = f _O.94(1 _ o'g74)= O.98 or 98%
1440' For 80/67oF; tοp = 60oF
surface temperature must be equal to or less than 60"F'
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students enrolled in οourses for whiοh the texδook has bee, uJopt.α. Αny οιher reproλucιion οr trαnsιαιiοn of this wοrk beyond thαι
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289
Moisture would condense at the base of the fin on the
tube outer surface if it condenses at all'
Let t* be this temPerature.
9 = UiAi(t, tr) = h6η9A(i,  ir)
where 1 =1' *,1*,Αr=Ai"'_ UιAi hiAι kΑ,
,nα 1 =!*41= :_ '9o]:^
= O.OO1"" U, hi ' k 1000 12x190
Ui = l OOO Btu/(hrft2F); where k"opp", = 190 Btu/(fthrF)
t* = tr . ffi(i, i*) = 50 + ffi (31.7  i*)
Αssume a value for tr, read i* from chart 1 and compute t* to
check assumption' Assume t* = 55'8'F then i' = 23'7 Btu/lba
and the calculated t* checks O.K. Therefore moisture will
condense at the base of the fin and on some portion of the fin.
There will probably be no condensation near the outer edge of the
fin.
1441. For 27Ι19 g, tοο = 15 C
Solution is similar to problem 1440
UiAi(t, _ tr) = h6η6(i3 _ i*)
1 =
1 +0'0005 = O.O1888; Uι= 52.98ui 53 58
t*=ti  ffi (i"_i*) =14'3,g4,,* β4.2_i,) *H
For t* = 16, i* = 45. Checks O.K.
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students enrοlΙed in οourses 1br which the textiook has been adopted. Αny oιher reprοduction or ιrαnsιαιion οf ιhis νοrk beyond thαl
prr^ιt'rca υy srctiοns ] 07 or l οε ij rni l ozο LΙniιed Sιαtes cip)rιgnt 'ιci lνιthοuι ihe permissiοn ξ ιhe copyright ονner is unlαwful'
290There will be no condensation because the tube outside wall is
greaterthanthedewpointtemperatureoftheair.
1442" This problem is intended for computer solution because
considerable iteration is required'
Coil DescriPtion:
Type of coil = water or brine solution
Tube pattern = staggered platefintube coil
Material = aluminum fins with copper tubes
Tube side fluid = water
Finned side fluid = air
Finned side air pressure = 29'92 inches of Hg
Face area = 12.50 square feet
Height of heat exchanger = 30'0 inches
Width of heat exchanger = 60'0 inches; W = 2H
Number of rows of tubes in the air flow direction = 5
Number of tubes per ro\Λ/ = 24 Circuits on tube side = '12
Finpitch=12fins/inchFinthickness=.008inchesVertical tube spacing = 1'250 inches
Horizontal tube spacing = 1'083 inches
Tube outside diameter = '500 inches
Tube wall thickness = '016 inches
lnsidetubefoulingfactor=,OOO0BtuhrSQFTF
Diameter of inlet pipeihead er = 2'5 inch(s)
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students enrolled in courses fbιwhich the texibook has υ*,' "J"pi"α. 'q
iy oιη" "p'oλucιιon
or trαnsιαιion of ιhis work beyond thαι
permitιed by Sectiοns ] 07 * l οε'"iiλr''i ii'6inι,"d S,o'r, coiirιgn, 'a"i 'ιthout ihe permission οf the copyrighι owner is unlανful'
29114.42. (continued)
Total heat transfer rate = 232885.0 Btu/hr
Sensible heat transfer rate = 164919.4 Btu/hr
Entering air conditions:
Dry bulb temperature = 80.0 F
Wet bulb temperature Ξ 68.0 F
Enthalpy = 32'3 Btu/LBMΑ
Humidity ratio = 83.3 grains/LBMA
Face velocity = 550.00 FPM
Air volume flow rate = 6875.0 CFM
Comment: coil is 34.3 percent dry
Leaving air conditions:
Dry bulb temperature = 57.4 F
Wet bulb temPerature = 57.1 F
EnthalPY = 24'4 Btu/LBMA
Ηumidity ratio = 68.7 Grains/LBMA
Tube side conditions:
Entering fluid temperature = 45.0 F
Leaving fluid temperature = 62.4 F
Τube side fluid velocity = 4.00 FPS
Cooling or heating liquid flow rate = 26.9 GPM
Air pressure loss = .774 inches of water
Τube side head loss = 14.09 feet of water
Fin efficiency = .gg9 Surface effectiveness = .819
Tube side heat transfer coefficient = 822.3 Btu/hrSQFTF
Finned side heat transfer coefficient = 10.0 BtuihrSQFTFExοeΙpts from this work may be reproduced by instructors 1br drstribution on a notforpro1lt basis for testing or instructional purposes only to
students enrolled in courses fοr whiοh the textbook has bοen adopted. Αny oιher reproducιion or τrαnsιαtion of ιhis wοrk beyond thαι
permitled by Sections ] 07 οr Ι 08 of ιhe ] 976 tJniιed Sιαtes Copyright Αct without ιhe permissiοn οf the cοpyrighι ονner is unΙανfuΙ.
292
1443. Coil DescriPtion.
Τype of coil = Direct expansion
Tube pattern = Staggered circularfintube coil
Material = Aluminum fins with copper tubes
Refrigerant tYPe = refrigerant22
Finned side fluid = air
Finned side air pressure = 29.92 inches of Hg
Face area = 10.31 square feet
Height of heat exchanger = 24.8 inches
Width of heat exchanger = 60.0 inches
Number of rows of tubes in the air flow direction = 4
Number of tubes per ro\Μ = 16 Circuits on tube side = 16
Fin pitch = 12 Fins/inch Fin thickness Ξ .014 inches
Vertical tube spacing = 1.500 inches
Horizontal tube spacing = 1.300 inches
Tube outside diameter = .625 inches
Tube wall thickness = .022 inches
lnside tube fouling factor = .0000 BtuHRSQFTF
Total heat transfer rate = 241221.0 Btu/hr
Sensible heat transfer rate = 162201'8 Btu/hr
Entering air conditions:
Dry bulb temperatuΓe = 82.0 FEΧceφts fiοm this work may be rοprοduced by instructors for drstribution on a notforprofit basis for testing or instιuοtional purposΘS only to
students enrolled in courses for which the texibook has been adopted. Αny oιher reproducιion or trαnsιαtiοn of this νork beyοnd thαt
permitted by Secιiοns ] 07 or Ι 08 οf ιhe ] 976 (Ιnited Sιαιes Cοpyrιght Αct \υιιhouι the permissiοn of ιhe copyrighι oνner is unlανful'
293Wet bulb temPerature = 67.0 F
Enthalpy = 31.4 Btu/LBMA
Ηumidity ratio = 74'9 Grains/LBMΑ
Face velocity = 500'00 FPM
Air volume flow rate = 5156.3 CFM
Comment: Coil is .0 Percent drY
Leaving air conditions:
Dry bulb temPerature = 52'2 F
Wet bulb temperature Ξ 50.7 F
Enthalpy = 20.6 Btu/LBMA
Humidity ratio = 52.1 Grains/LBMA
Tube side conditions.
Refrigerant saturation temperature = 35.0 F
Air pressure loss = .623 inches of water
Tube side pressure loss = 1.36 PSI
Refrigerant quality entering/leaving evaporator = '29
Enthalpy change in evaporator = 62'75 Btu/LBM
Fin efficiency = .679 Surface effectiveness = '885
Tube side heat transfer coefficient = 375.5 Btu/hrsQFTF
Finned side heat transfer coefficient = 9.3 Btu/hrSQFΤF
1444. Coil DescriPtion:
TYPe of coil = Steam
Tube pattern = Triangular platefintube coil
Material = Aluminum fins with copper tubes
Τube side fluid = Steam
Finned side fluid = Air
Finned side air pressure = 29'92 inches of Hg
Exceφts frοm this work may be reproduced by instructors for distτibution on a notforprofit basis for testing or instructional purposes only to
Students enroιled in courses for whiοh the texibook has been adopted. Αny οther reprοducιion or trαnslαιiοfi ofthiswοrkbeyοnd ιhαt
permitted by Sectiοns Ι 07 οr l οε iiin, ] 9t7 6 (Jnitecl Stαtes Copjrιght ,ιci τιithout ihe permission of ιhe copyrighι owner is unlιrννful'
294
Face area = 12.00 square feet
Height of heat exchanger = 24'0 inches
Width of heat exchanger = 72'0 inches
Number of rows of tubes in the air flow direction = 2
Number of tubes per ro\M = 16 Circuits on tube side = 16
Finpitch=8Fins/inch Fin thickness = .006 inches
Vertical tube spacing = 1.500 inches
Ηorizontal tube spacing = 1'299 inches
Τube outside diameter = '625 inches
Tube wall thickness = .0'18 inches
lnside tube fouling factor = .0000 BtuhrSQFTF
Total heat transfer rate = 554503'6 Btu/hr
Sensible heat transfer rate = 5545ο3'6 Btu/hr
Entering air conditions:
Dry bulb temperature Ξ 60.0 F
Face velocitY = 750'00 FPM
Αir volume flow rate = 9000.0 cFM
Leaving air conditions:
Dry bulb temperature Ξ 116'1 F
Tube side conditions:
Steam temPerature = 227 '1 F
Steam saturation pressure = 5'000 PSIG
Air pressure loss = '269 inches of water
Fin efficiency = .738 surface effectiveness = '756
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students enrolled in οourses tbr w*ch the texibook has b..n uJot"d. Αny οιher reprολucιion or ιrαnslαtion οf ιhis τυork beyond thαι
permitted by Sections ] 07 οr l οε^"i 'n,
isizο LΙniteιl Sιαιes cοpyrιglιt ,ιci νιthout the permission of ιhe copyright ονner is unΙανful'
295
1444. (continued)
Τube side heat transfer coefficient = 2051.7 Btu/hrSQFTF
Finned side heat transfer coefficient = 14.6 Btu/hrSQFTF
1445. Coil DescriPtion:
Type of coil = Water or brine solution
Tube pattern = Staggered platefintube coil
Material = Aluminum fins with copper tubes
Τubesidefluid=3Oo/oethyleneglycolsolution
Finned side fluid = Αir
Finned side air pressure = 29'92 inches of Hg
Face area = 5.56 square feet
Height of heat exchanger = 20'0 inches
Width of heat exchanger = 40'0 inches
Number of rows of tubes in the air flow direction = 2
Number of tubes per ro\Λ/ = 16 Circuits on tube side = 4
Fin pitch = 7 Fins/inch Fin thickness = .008 inches
Vertical tube spacing = 1'250 inches
Horizontal tube spacing = 1'083 inches
Tube outside diameter = .500 inches
Tube wall thickness = .016 inches
Insidetubefoulingfactor=.OOO0BtuhrSQFTF
Diameter of inlet pipe/header = '1'0 inch(s)
Total heat transfer rate = 9ο610'1 Btu/hr
Sensible heat transfer rate = 90610'1 Btu/hr
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students enrolled in οourses fbr which the texibook has been aJopteα. Αny other reproλucιion or ιrcιnsιαιion of ιhis νork beyond ιhαt
permiιted by Secιions ] 07 or l οε^"i 'n,
i szο (Jniιed SιαιeS cip)rιgΙιt 'ιci νιιhοuι ihe permission of ιhe copyrighι oνner is unlωυfuΙ'
296Entering air conditions:
Dry bulb temperature = 70.0 F
Face velocity = 650.00 FPMΑir volume flow rate = 3611.1 cFMLeaving air conditions:
Dry bulb temperature = 93.3 F
Tube side conditions:
Entering fluΙd temperature = 150.O F
Leaving fluid temperature = 128.4 F
Tube side fluid velocity = 4.00 FPSCooling or heating Ιiquid flow rate = 9.O GPM
Αir pressure loss = '186 inches of water
Τube side head loss ='10.13 feet of water
FΙn efficiency  .83'1 Surface effectiveness = .846
Tube side heat transfer coefficient = 796.O Btu/hrSQFτFFinned side heat transfer coefficient = 14.4 Btu/hrseFTFMean temperature difference = 57.6 F
Τhere is a 5 percent reduction in capacity and increased pressure losson the tube side.
1446. Coil Description:
Type of coil = Water or brine solution
Τube pattern = Staggered platefintube coil
Material = Αluminum fins with copper tubes
Tube side fluid = 30oλ ethylene glycol solutionExc€φts from this work may be reproduced by instruοtors for distribution on a nοtforprofit basis Γor testing or instruοtional purposes only tostudentsenroΙIedincoursesfοrwhiοhthetextbookhasbeenadopted. Αnyοtherreprολucιionοrιrαnsιαtioiofιhislυorkbeyλnithαιpermitιed by Sectiοns ] 07 or Ι 08 of ιhe 1 976 United Sιαιes Cοpyrighι Αct ιυ ithοuι ihe permission of ιhe copyright olνner is unlανful.
297Finned side fluid = Air
Finned side air pressure = 29.92 inches of Hg
Face area = 12.50 square feet
Height of heat exchanger = 30.0 inches
Width of heat exchanger = 60.0 inches
Number of rows of tubes in the air flow direction = 5
Number of tubes per ro\Λ/ = 24 Circuits on tube side = 12
Fin pitch = 12 Fins/inch Fin thickness Ξ .008 inches
Vertical tube spacing = 1.250 inches
Horizontal tube spacing = 1.083 inches
Τube outside diameter = .500 inches
Τube wall thickness = .016 inches
lnside tube fouling factor = .0000 BtuhrSQFTF
Diameter of inΙet pipe/header = 2'5 inch(s)
Total heat transfer rate = 211374.2 Btu/hr
Sensible heat transfer rate = 155955.9 Btu/hr
Entering air conditions:
Dry bulb temperature Ξ 80.ο F
Wet bulb temperature Ξ 68.0 F
EnthalPY = 32'3 Btu/LBMA
Ηumidity ratio = 83.3 Grains/LBMA
Face velocity = 550.00 FPM
Αir volume flow rate = 6875.0 cFMComment: Coil is 43.6 percent dry
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2981446. (continued)
Leaving air conditions:
Dry bulb temperature Ξ 58.6 F
Wet bulb temperature Ξ 58.2 F
Enthalpy = 25'1 Btu/LBMA
Humidity ratio = 71.4 Grains/LBMΑ
Tube side conditions:
Entering fluid temperature = 45.0 F
Leaving fluid temPerature = 62.0 F
Τube side fluid velocity = 4'00 FPS
Cooling or heating liquid flow rate = 26'9 GPM
Αir pressure loss = .756 inches of water
Tube side head loss = 18.13 feet of water
Fin efficiency = .699 Surface effectiveness = '819
Tube side heat transfer coefficient = 476.4 Btu/hrsQFTF
Finned side heat transfer coefficient = 10.0 BtuhrSQFTF
The capacity is reduced by about 9 percent, the pressure loss on the
tube side is increased and the leaving air temperatures have increased
by about 1 degree.
1447. Check Examples 141 through 145
Coil Description:
Type of Coil = Water or Brine Solution
Τube Pattern = Staggered Plate_FinTube Coil
Material = Αluminum Fins With Copper Tubes
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students enroΙled in οourses for whiοh the texibook has been adopted. Αny oιher reproducιion or ιrαnsιαtion οf ιhis work beyond thαι
permitιed by Sectiοns ] 07 or l oε i7 'n,
]976 Uniιed Sιαιes Copyrighι ,ιci νithouι ihe permission of ιhe copyright oνner is unlιτwfuΙ'
299Tube Side Fluid = Water
Finned Side Fluid = Air
Finned Side Air Pressure = 0. FT. of Εlevation
Face Area = 2.17 Square Feet
Ηeight of Heat Exchanger = 12.5 ]nches
Width of Heat Exchanger = 25.0 linches
Number of Rows of Tubes in the Αir Flow Direction = 5
Number of Tubes Per Row = 10 Circuits on Τube Side = 5
Fin PΙtch = 8 Fins/lnch Fin Thickness = .006 lnches
Vertical Tube Spacing = 1.250lnches
Horizontal Tube Spacing = 1.083 lnches
Tube Outside Diameter = .525 lnches
Tube Wall Thickness = .015 lnches
lnside Tube Fouling Factor = .0000 ΗRFTΛ2_F/Btu
Diameter of Inlet Pipe/Ηeader = 1.3 lnch(s)
Total Heat Τransfer Rate = 133026.9 Btu/HR
Sensible Heat Transfer Rate = 133026.9 BtuiHR
Entering Αir Conditions:
Dry Bulb Τemperature = 50.0 F
Face Velocity = 950.00 FPM
Αir Volume Flow Rate= 2061.6 cFMLeaving Air Conditions:
Dry Bulb Temperature = 107.6 F
Tube Side Conditions:
Entering Fluid Temperature = 150.0 FEXcerpts fiοm this work may be reproduced by instruοtοrs Γor drstribution on a notforprofit basis for testing or instruοtional purpοses only tostudents enrolled in courses for whiοh the textboοk has been adopted. Αny οther reproducιιon or trαnsιαιion of this ιιork beyond thαιpermitιed by Sections ] 07 οr Ι 08 of ιhe Ι 976 United Sιaιes Cοpyrighι Αct withοuι ιhe permission of ιhe copyright oνner is unlα:ινful.
300Leaving Fluid Temperature = 128.3 F
Τube Side FΙuid Velocity = 4.00 FPS
Cooling or Heating Liquid Flow Rate = 12.5 GPM
Αir Pressure Loss = 1.13'1 lnches of Water
Tube Side Head Loss = 6.77 Feet of Water
Fin EfficieΠCy = '750 Surface Effectiveness = '771
Tube Side Ηeat Transfer Coefficient = '1 354.2 Btu/hrSQFΤF
Finned Side Ηeat Τransfer Coefficient = '19.5 Btu/hrSQFΤF
Mean Temperature Difference = 58.5 F
Τhe above results show that a 5 row coil would easily satisfy
the specified requirements. Τhe manual calculation of the
examples are very conservative.
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151
Exceφts from this wοrk may be reproduced by instruοtors for distribution on a notforproΓrt basis fοr
testing or instruοtional purpδsο, on1y to students enrolled in courses for whiοh the textbook has been
adoptΞd. Αny οther repiodiction οr trαnslαtion of this work beyond thαt permitted by Sections ]07 or ]08
o7 ilιe lιzο (]nited Smtes Copyright Αct without the permission of the copyright owner is unlαwful.
iequests for permission or furtieridormαtion should be αddressed to the Permission Depαrtment, Jοhn
Wiley & Sons, Ιnc, 11Ι Riνer Street, Hoboken, ]\ΙJ 07030.
Chapter 15
COP : qe/\rv; UsePidiagram
_νV = i+_ig =119.5_105.5
=  14.0 Btu / lbm p
Q"=il iι=42'5119'5
= 77 Btu / lbm
g"=Q"+\M=7714
= 63 Btu / lbm
COP=63 114=4.5
CoP"".n.1 = 50ο l (570 5οο) = 7'14
14 ooFtlooF
4 ooF
(a)
(b)
4.50r= = 0.63 or 63 %
7.14
(c)
ψ= Q" =coP
10 (12,000) _26,667 Btu lhr = 1ο.5 HP N 7.8 kW4.5
then HP =
10'5 = 1.O5 and Ψ =O.22ton 10 kW
(d) Qe = ΓhQe oΓ Γh =10 (12,000)
= 1905 lbm / hr = 0.24 kg /s63
300
153
(e) W = 10.5 HP from (c) above = 7.8 kW
(f) PD = rhv. = 1905_ι0'68)
= 21'6 ft3 / min x O.O1 m3 / s60
152 W = 2.5 kW; rh = 0.05 kg / s
(a) 8c=il_iη, _νv  i+_iο
i+ =ig _W / rir : 398.4 + 2'5 / 0.ο5 = 448'4 kJ / kg
Qc = i1 i4:260'3  448'4 = 188'1 kJ / kg
w = i3 i4 = 398.4  448'4: 50 kJ / kg
9e : 9c \M  188.150 : 138.1 kJ / kg
(b) COP = 138.1 I 50 = 2.76
( coP )carnot = #*:3.86iι = 448'4 kJ / kg from part (a)
0r = 2.76 / 3.86 = 0 '72 or 72 o/o
R 134a; Energy Balanοe
rh1i1 + mηiι =m2i2+ rh5i5 , rh1 = fi2= rh3 = rhη = rhs
is = i.l +iι _iz = 44'94 + 101 '54 _ 37.98 : 108'5 Btu / Ibm
Ps = P+ :16.6 Psia
w=i6is,So=Ss
! 100 F sat. liquidP1: 138.8 Pjsla
r0 Fsat. vapor 80 F
Excerptsfromthiswοrkmaybereproduοedbyinstructorsfordistributionona.notforprofitbasisfortestingorinstructionalpurposesonlyto _,l _...__^ll^r;^^^llfaAqfnru,lιiοhthetextbοοkhasbeenadonteιl. Αnνotherreυroducιionortrαnslαιiono{thiswοrkb?νοbdthηι^2v":1'^)
(c)
(d)
5
301
io :'127 Btu / lbm lChart 3]
\Ι/ : 127  108.5 = 18.5 Btu / lbm
Qe = i+ _i3 : iι _iz = 1o1 '54 _ 37'98 = 63'6
153 (continued)
154
(a)
(b)
HP^t TT8ton rhge
HP =
(18.5) 778 (12,ooo)=1.37
132
Pt9
r80P
2
0.75
0.90 = 0.15
ton (63.6) 60 (33,000)
Γ. ^ rr..1l"' v30ν = 1+ C _ c
IL \PoJl vb
n = k = 1.17; 1 ln= 0.855
P./P6:180120:9
v3 = ?
1I: o.9o [chart 4 tabv6 2.38
nu =ι+ O.O3  o.03 (9)o uuu]
,.. : ftua or ]1ι1 =
0u lb / ft3'v PD PD v3
rh 0.75PD 2.14
Πν = Γ,
* o.15_ o.15 (91o'εss
le Α3a]
0.90 =
Exοemtsfrοmthiqrνnrltra\,L__^j,'^^lL.:__}_'.^a^.^.f,j_+:!_,'+:^____^}ε^.nfi'+λ"iafΛf tροt;fσΛrinctrlotinnql nllrnnsεsnnlvt^
155
156
GπlΦ() a β00)1728
= 52.36 ft3 / min
rh / PD = 0.15 t 214 = O'07 lb / ft3
(e\ fr,  tro = 1 o'07
= o.8o or 80 %\v" rha 0.35
(d) Power is directly proportional to the mass flow rate
therefore, Power compares as in (c) above'
0ν =o'7o
4 cyl  3" bore, 4" stroke, 800 rPm
Pι = 49] psia (chart 3)
Pι = 138.8 psia (chart 3)
Πν = frΥ2, / PD; Υ2a = 1 'o4 ft3 / Ibm;
Table Α2a @ 55 F / 52 psia
Qιz = rh (i2 _ η)
PD=
iι __iι = 46 Btu / tbm
iz = 112 Btu / lbm
1ιz=ηPt" (iz_,.,l = Ξ#f4 (.ε'46) = 2g26 Btu / minΥ2a
or 912=139,560 Btu/hr = 11'6tons
R22, assume suPerheat = 20 F
2 qsoF
_. Ξ __^fi+ L.]. f^f tea1inσ nr instnlctiοnaι nurooses onlv to
303
Subcooling = 10 F
t. = 130 F, te = 45 F
(a) 3t
90
'156 (Continued)
tl:13010=120F
iι:iz:46 Btu / lb
(b) Q" = 144,000 Btu / hr (Fig. 157)
Wc = 14.8 kW = 50,498 Btu / hr
(c) te = 32.5 F (Fig. 1 5τ); W = 13.3 kW
157 Refer to Fig ' 157 cΑP = 133,000 Btu/hr
3\PΦ
P b*Ρ;.}
4\'d
ι1 35
te= 47 F
120 E
^ l1 .'jL''L:^.^_^^+f^rnrnfithasisfοrtestinsorinstructional purposesonΙyto
ΞΞι_ΦΞoo
t
Ξφα)o()o
o(dο(τ,
ζ)
304
158
Excerpts from this work maystudents enrolled in courses
orator Tι
ΞΥ13.7 ;
=oc(ι)
Ξoα
cΞΦooo
'δωo(σo
only toιhis work
Evapl a),( c)
( b) (a)( c)io
Design Pointo o(
Measured OperatingPoints
by Secιions ]07 οr ]08 οf ιhe 1976 United Sι*eΝwff#{&ψeff&Etrβwib,gΦeruεgffi'"'^''n''' ιhαι permiιιed
305
Suction valve, ΔP:2 psia
Discharge valve, ΔP = 4 psia
'10 F S.H. in intake man. and cyl.
Piston clearance=5oλ'
Γ ι r"']"nl γ"ηu= 1+C_.ι ]1 rPoJ.] uo
159
Pe = 69 psia; te = 30 F (Τable A4)
(a) Τhe condensing temperature is still about '1 '15 F, but the
evaporating temperature is low, about 30 to 31 F.
(b) (qα _q) / aα = 1_*=0.36 or 36 % low305
(c) lt appears that the evaporator is not loading the compressor.
Check for proper air fΙow over evaporator. Fan speed may be
low or an obstruction exists.
Glzτ s
sοsia,Ι
sat R22 νao450F '
P=90.73psia
ExcΘrpts from this work may be reproduοed by instructors for distribution on a notforprofit basis for testing or instructional purposes only tostudents enrolΙed in cοtlrses fnr whinh tLΔ +__}L^^l' L^^ L^^ ^j^_ι' l
V3 = Vg at 45 F; vg = 0.604 ft3 / lbm (Τable Α3a)
vo = 0.66 ft3 / lbm (Chart 4 at 55 F / 89 psia)
Pc=275+4= 279 psia; Pb = 90.73 2= 88.73 psia
n:1.16, C=0.05
(a) o" = 1+ O.O5 _ o.O5 (Ξ+\1/1 16l /o οo+\
L "_v'vι'ιs&rcJ ] ι .* .,l
+ 0'εeε
N 0.2 kg /s
res. loss
(b)ri'r 
(c)
159 (cont
'1510 (a)
0.604 : 27.75bmlmin
.n1 IP. I
n 11Pυ)
]l, ,oi6 I
.66) l ( 2'79 )l 'ο _.' I'(8εzs)
1
31 kJ/kg
1.0 HP x 8.2 kW
)'''o] ι*l>f k assumed and 2 psi p
ry, (PD) / v. : (ο.838) 20 /
ψ=!Ψ, W=*o",[[
* = 1J9(ss.73) (144) (o0.16'
inued)
= 1 0,466 ft  lbf /lbm x
w _27.75(10,4666) _ 1.0.80(33,ο00)
Γ
ou =l+ O.04 _ o.o4 (Ψ'ν L 'u'Note: Αn average value c
Excerpts from this work may be reproduοed by instruοtors for distribution on a not_forprofit basis for testing or instruοtional purposes only toctrr.lan+. ^^ir
i
307
assumed in suction header and valve.
Tv = 0.90, Γh = (PD) ηu lνz 19
= 9'4 (ο.90) l 0'74
rh ='l 1.44 lbm / min
β*p=zooirδ oort"II
\l60.oF
91( ft _ lbη / lbm
Btu / min
o6o'55
Γ  1'4_1
w = _1! (53) 1 44 (o'77) l [Ψ']
*(1'4_1) \ / \ /Lι53/
vιl = fr*
= 9491 (11.44) / (0.9 x778) =0m
ι6.I
=94
55 I
tz
1t
1510 (Continued)
QH=ψ+Qι=(155x60) + 3O,OOO = 39,3ooBtu/hr
or qH = 655 Btu / min
9491 =125.2Btu / ΙbmW23 =i2 i3, iZ =iZt W = 133 +
j
778
9Η = ia _i3i iη = 9μ +iu = _#+ +125'2= 68 Btu / lbm11.44
(b) lteration is required
P3 will decrease with the lighter load but Pz is also lower and
(P3 / P2) t'n will be about the same as part (a);
ν2 l ν6 will be about constant. Then Tv : Constant.
Excerpts from this work may be reproduced by instructors for distribution on a notforprofit basis flοr testinρ οr l'nstnlctiοnaΙ nιlrnοses onlv to+,,f,^+..rl
j
308
However , Υ2= O.85 and rh : (PD)ην l νe = (9.4) 0.87 / 0.85
= 9.62 lbm / min.
ΠoWW =9491 (48)(1 03) : 14.78 Btu / tbm
53 (0.77) 778
W = (1 4.78) (9.62) (601 = 8530 Btu/hr;
Qn = 8530 + 24,000 = 32,530 Btuihr
Which assumes Ps I Pz is constant and 2 psi pres. loss in the
valve.
ie = iz t w :83.5 + 14'78 94'3 Btu / lbm
iι =iz_ 9ιz = 111_24,000 / 9'62 (60) = 69'4 Btu / lbm
,, = # (50) = 188 psia 18
Ps
1511 Reduced air flow reduces the load on the evaporator. Without
suction pressure control the evaporator pressure will decrease
until condensate will freeze and completely block the
evaporator air flow. Liquid refrigerant will return to the
Εxοerpts from this work may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to
41.9 .83.5 9 4.3t
309
1512
15'13
1514 (a)
compressor and eventually cause the compressor suction
valve to faiΙ.
lnstall an evaporator pressure regulator set to maintain a
pressure such that the temperature of the evaporator surface
will not decrease below the freezing point for water.
lnstall a suction pressure regulator on the compressor inlet.
Τhe regulator shouΙd be set to limit the suction pressure to a
level compatible with the compressor capacity.
Using chart 2 with the construction shown,
1515
the final temperature is 9ο F
(b) 'v  L=!=0.18s orm (ν 20 i
18.5 o/o vapor (Use chart 2)
Using chart 2
X3 = 0.495; ts = 125 F
28ooFl8ooF
0 0.25 .495χ+
Excerpts from this rπork may be reproduced by instructors for distribution on a notfor_profit basis for testing or instruοtionaΙ purposes only tostudents enrolled in courses fbr whiοh the textbook has been adopted. Αny other reproduction or trαnslαιiοn of this work beνond ιhαΙ nemilto)by Sectiοns ]07 or ]08 οfιhe !Q76 [Ιnifa'] Slnlo" r'^'.'':'ι'' ι ''
tb ammonia/lb sol.
TzΞτ7 l6TTsia
310
1516 (coP)ma, = Ψ:fd: T^ = 180 + 460 : 64OR;\ _  /Ιlιaλ Tg(To _ ξ )' '9 
Te=75 + 460 = 535 R; To =]00 + 460 = 560 R
(COP)ma x = 2.675
1517 Refer to Chart 5 for saturated vapor at 10 mm hg.
Vapor must first be condensed to sat. liquid at '10 mm hg.
Q" = irs, Table Α1a;
i'n = 1ο64.8 Btu/lbmu for 1 lbm of vapor oΓ 9. = 1064.8 Btu
at 50 F, P = 0.178 psia or
1517 (continued)
P = 10 mm hg
Locate point I at x = 0; P = 10 mm hg
Locate point s atx = 0.6 ; P = 10 mm hg
ffru=1;m.=5
ms5mvmm6ms
5 Elv = ; tΠS = i x 45'5 Ξ 37.g (depends on scale used)ob
(a) x = 0.50
(b) Q, = i,  io = 50  (70) = 20 Btu / lbm of solution
Exοerpts Γrom this work may be reproduced by instruοtors for distribution on a notfοrprofit basis for testing or instruοtional purposes only tostudents enτolΙed in οourses for which the textbook has been adopted. Αny oιher reproducιion or ιrαnslαιion of ιhis νοrk beyond thαι pemιιιedby Sectiοns Ι07 or ]08 ofιhe ]976 tΙnited Sιοιρs Canιιriohι Δ?r1υ!a!"^''a n'^ ' '
311
Qtot : 1064.8 + (6 x 20) = 1 ,184.8 Btu
i
70δh9
0.5 0.6χ+
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ADDENDUMto
Solutions ManuaΙ fοr McQuiston, ΙΙVAC 6e
ProbΙem 610
For the floor, it is unοlear what2 in. vertiοal edge insulation means (whether 2 in. is the
thiοkness of insulation or the depth of the edge insulated).. The solution assumes that the insulation has Rvalue of 5.4 hrft2oF/Btu and the
depth of the edge is 2 ft.
For the door, Table 58 in the 6th edition does not have Uvalue for the wood storm doorand there are three types of the wood door with 1 % in. thickness.
ο The solution assumes that the doors are panel doors with metal storm doοr; henοe,
its Uvalue is 0.28 But/hrft''F.
ΡrobΙem 79
The standard time zone for ottawa, ontario is Εastern Standard Time instead of CentralStandard Time.
. The solution uses Eastern Standard Time.
Prοblem 714
For the specified loοatiοn, the sunset oοοurs before 9:00 p.m. CDST on June 21.
ο The solution uses 8:Ο0 p.m. CDST instοad of 9:00 p.m.
Prοblems 825 and 826
Both problems do not specifu the window orientation.. The solutions assume the westfacing window for both prοblems.
Tabte 820
Reοommended radiative and conveοtive fraοtions for solar heat gains should be revisedsinοe the 6th edition uses the SHGC values in the calοulation of the (οombined) solar heatgain for the RΤS methοd.
Example 816
Τhe example actually uses 90%/10% of radiative/οonveοtive split of the cοmbined solarheatgain. Ηowever,thetext (page270) says 100%/0Υoforthetransmittedsolarheatgainand 630Λ1370Λ for the absorbed sοlar heat gain.
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312
ProbΙems 8_25 and 826
The solutions for both problems use 90%11,0% for the combined solar heat gain.
ExampΙe 9_1
The οalοulation for this example should be
" =
('1Ι11?2Ψoi19.']!o:9Φ _ Ι22'606(0.ssx70 0)(1000)
(Changing 13 to24 and 122790 to 122606).