Solution Manual - Heat Transmission - 3rd Edition - William McAdams

SOLUTIONS MANUAL

HEAT TRANSMISSION (Third Edition)

Williams H. McAdams

By: Agerico U. Llovido

2013

CONTENTS

Chapter 2 Steady Conduction 1

Chapter 3 Transient Conduction 25

Chapter 4 Radiant Heat Transmission 48

Chapter 5 Dimensional Analysis 64

Chapter 6 Flow of Fluids 77

Chapter 7 Natural Convection 87

Chapter 8 Introduction to Forced

Convection 92

Chapter 9 Heating and Cooling Inside Tubes 107

Chapter 10 Heating and Cooling Outside

Tubes 133

Chapter 11 Compact Exchangers, Packed and

Fluidized Systems 145

Chapter 12 High-Velocity Flow: Rarefied

Gases 152

Chapter 13 Condensing Vapors 159

Chapter 14 Boiling Liquids 186

Chapter 15 Applications to Designs 192

1

CHAPTER 2

STEADY CONDUCTION

Chapter 2 Steady Conduction

2

2_001 A large flat furnace wall consists of three layers: 4 inches of kaolin insulating firebrick on the inside, 5 inches of kaolin insulating brick in the middle, and 5 inches of magnesite brick on the outside. If the inner surface is at 1375 F and the outer surface is at 215 F, what is the rate of heat loss per unit area through the wall? Use linear interpolation of the thermal-conductivity in the Appendix.

Solution:

Let x

a = 4 in Kaolin Insulating Firebrick

xb = 5 in Kaolin Insulating Brick

xc = 5 in Magnesite Brick

to = 1375 F

t3 = 215 F

Then,

Σ∆t = 1375 F - 215 F = 1160 F Assume average temperature from the following;

∆ta = ∆t

b = ∆t

c = Σ∆t / 3 = 386.67 F

t1 = t

o - ∆t

a = 1375 F - 386.67 F = 988.33 F

t2 = t

3 + ∆t

c = 215 F + 386.67 F = 601.67 F

Average temperatures t

a = (t

o + t

1) / 2 = (1375 + 988.33)/2 = 1,181.67 F use 1180 F

tb = (t

1 + t

2) / 2 = (988.33 + 601.67)/2 = 795 F

tc = (t

2 + t

3) / 2 = (601.67 + 215)/2 = 410.84 F use 410 F

Thermal Conductivities from Appendix, Table A-5. Linear Interpolation. Kaolin insulating firebrick @ t

a = 1180 F,

ka = .09925 Btu/(hr)(sq ft)(deg F per ft)

Kaolin insulating brick @ tb = 795 F

kb = 0.13712 Btu/(hr)(sq ft)(deg F per ft)

Magnesite brick @ tc = 410 F

kc = 2.19178 Btu/(hr)(sq ft)(deg F per ft)


3

Using Equation 2-13a:

+

+

Σ∆=

cc

c

bb

b

aa

aAk

xAk

xAk

x

tq

A

a = A

b = A

c = A

+

+

Σ∆=

c

c

b

b

a

ak

xk

xk

x

t

A

q

×+

×+

×

=

2.1917812

5

0.1371212

5

0.0992512

4

1160

A

q

q / A = 176.1 Btu/(hr)(sq ft) ----- (Ans) 2_002 A standard 2-inch steel pipe carrying steam is insulated with a 3-inch layer of glass wool. The outside surface of the

pipe is at 400 F, and the outside surface of the covering is at 150 F. Estimate the rate of heat loss from the pipe, expressed as Btu per hour per 100 ft of pipe.

Solution:

Outside diameter of 2 in steel pipe is 2x

1 = 2.375 in

Then: x

1 = 1.1875 in

x2 = x

1 + 3 in = 4.1875 in

Average Temperature = (400 F + 150 F)/2 = 275 F From Appendix, Table A-8, Thermal confuctivity of glass wool at 275 F k

m = 0.033 Btu/(hr)(sq ft)(deg F per ft)

Using Eq. 2-8.

)/xln(x

tk2

L

q

12

m∆π=

1.1875)ln(4.1875/

150)0(0.033)(402

L

q −π=

q/L = 41.132 Btu/(hr)(ft) q per 100 ft = 41,132 Btu/hr per 100 ft of pipe .... Ans


4

2_003. A large flat furnace wall is presently constructed of several layers of conventional insulating materials. The inner surface is maintained at 2400 F. Under normal operating conditions, the outer surface attains a temperature 350 F, and the heat loss through the wall is 700 Btu/(hr)(sq ft).

a. In order to make an economic evaluation of a proposal for further insulation on the outside of the furnace,

determine the percentage reduction in the heat loss as a function of the thickness and the mean thermal conductivity of the added insulating materials. Assume that the rate of heat loss per unit area is directly proportional to the difference between the temperature of the outer surface of the wall and the temperature of the surroundings (60 F).

b. What additional assumptions are necessary for such a calculations? c. In view of the general behavior of most insulating materials, are the assumptions of part b most like to make

the calculated percentage reduction greater or less than percentage reduction determined if these assumptions were not made?

Solution:

+

+

=

∑

cc

c

bb

b

aa

a

Ak

x

Ak

x

Ak

x

∆tq

cba AAA ==

∑∑

∆=

k

x

t

A

q

1

Σ∆t = 2400 F - 350 F = 2050 F [q/A]

1 = 700

Σ(x/k) = 2050 / 700 = 2.92857 a. With additional insulation. h = proportionality constant. k’ = thermal conductivity of additional insulation. x’ = thickness of additonal insulation.

Σ∆t2 = 2400 F - 60 F = 2340 F

∑

∑+

′

′+

=

h

1

k

x

k

x

∆t

A

q 2

2

Solving for h as constant; [q/A]

1 = h(350 - 60) = 700

h = 2.413793 Then:

2.413793

1

k

x2.928571

2340

A

q

2 +′

′+

=

k

x3.342857

2340

A

q

2

′

′+

=


5

Percentage reduction:

100

A

q

A

q

A

q

1

21 ×

−

=

100700

k

x3.342857

2340700

×′

′+

−

=

k

x3.342857

k

x100

′

′+

′

′

=

..... Ans

b. Use constant thermal conductivity as an additional assumption. .... Ans c. In view of thegeneral behavior of most insulating material which is k’=k

o’(1+ a

ot). Assumptions of part

b most likely increase percentage reduction if k’ at actual temperature is less than in b. Assumptions of part b most likely decrease percentage reduction if k’ at actual temperature is greater than in b. ......Ans

2_004 A long horizontal steam pipe 12 inches o.d. carries saturated steam at a temperature of 400 F. The outside of the pipe

may be assumed to be essentially at the saturation temperature of the steam. The pipe is centered in a 2-ft-square sheet metal duct. The space between the outside of the pipe and the duct is filled with powdered magnesia insulation, k = 0.35 Btu/(hr)(sq ft)(deg F per ft). it may be assumed that the sheet-metal duct is at a uniform temperature of 100 F. Calculate the rate of heat loss from the steam pipe per foot of pipe length by:

a. Mapping. b. Relaxation. Solution:

(a) By Mapping:


6

For the whole duct. N

L = 4 x 9 or 9 lanes for one-quarter

( )

( )

m

m

NomL

2700kL

q

100400k4

94

L

q

ttkN

N

L

q

=Σ

−

×=

Σ

−=Σ

Σq / L = 2700 (0.35) = 945 Btu / hr - ft . . . Ans. (b) By Relaxation:

Using Equation 2-17. (Reference)

2B2C3B2A1B2B

2B t4ttttqkz

q−+++==

∗

Using similarly to this problem, the following are subscript for temperatures. A6 to A8, B6 to B8, C6 to C8, D6 to D8, E5 to E8, F1 to F8, G1 to G8, and H1 to H8. (See Tabulation)


7

Table 1: A6, Step 1 to B8, Step 15

POINT A6 A7 A8 B6 B7 B8

t q* t q* t q* t q* t q* t q* Step 1 325 -12 250 -8 175 -4 319 -10 246 -6 173 -3

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7

Step 8

Step 9

Step 10

Step 11

Step 12

Step 13 325 -12 250 -8 319 -10

3 12 3 6

Step 14 328 0 -5 -4

Step 15


8

Table 2: A6, Step 16 to B8, Result.

POINT A6 A7 A8 B6 B7 B8

t q* t q* t q* t q* t q* t q*

Step 16

Step 17

Step 18 173 -3

-2

Step 19 246 -6 -5 -2

Step 20 250 -5 319 -4 246 -8 173 -5

-2 -2 -2 8 -2

Step 21 -7 -6 244 0 -7

Step 22

Step 23 328 0 250 -7 175 -4 244 0 173 -7

-2 -2 8 -4 -4 -2 8

Step 24 -2 248 1 175 -8 -4 171 1 -2 -2 8 -2

Step 25 -1 173 0 -1

Step 26

Step 27 328 -2 319 -6 244 -4

-2 -2 8 -2

Step 28 -4 248 -1 317 2 244 -6 171 -1

-2 -2 -2 8 -2 Step 29 -3 0 242 2 -3

-2

Step 30 0

Result 328 -4 248 -3 173 0 317 0 242 0 171 -3


9

Table 3: C6, Step 1 to D8, Step 15

POINT C6 C7 C8 D6 D7 D8

t q* t q* t q* t q* t q* t q* Step 1 304 -2 236 -1 168 0 288 -60 225 -38 163 -21

Step 2 288 -60

19

Step 3 288 -41 225 -38

13

Step 4 304 -2 288 -41 225 -25

-10 -10 40 -10

Step 5 -12 278 -1 -35

Step 6

Step 7 236 -1 278 -1 225 -35 163 -21

-9 -9 -9 36 -9

Step 8 -10 168 0 -10 216 1 163 -30

-8 -8 -8 32

Step 9 -8 -7 155 2

Step 10 278 -10 6

Step 11 -4

Step 12 155 2

5

Step 13 304 -12 236 -10 278 -4

3 12 3 3

Step 14 307 0 -7 -1

Step 15


10

Table 4: C6, Step 16, to D8, Result.

POINT C6 C7 C8 D6 D7 D8 t q* t q* t q* t q* t q* t q*

Step 16

Step 17 216 -7

2

Step 18 236 -7 168 -8 -5 155 7

-2 -2 8 -2

Step 19 307 0 236 -9 166 0 216 -5 5

-2 -2 8 -2 Step 20 -2 234 -1 -2

-2

Step 21 -3

Step 22

Step 23 234 -3 166 -2 278 -1 216 -7 155 -5

-2 -2 -2 -2 8 -2

Step 24 -5 -4 -3 214 1 3

Step 25

Step 26

Step 27 307 -2

-2

Step 28 -4 234 -5

-2

Step 29 307 -4 234 -7 166 -4 214 1 -2 -2 8 -2 -2

Step 30 -6 232 1 -6 -1

Result 307 -6 232 1 166 -6 278 -3 214 -1 155 3


11

Table 5: E5, Step 1, to F2, Step 15.

POINT E5 E6 E7 E8 F1 F2


Step 1 250 126 213 45 175 32 138 14 325 -12 319 -10

32 -128 32

Step 2 282 -2 213 77 175 32

38 19 -76 19

Step 3 282 36 232 1 175 51 138 14 13 13 -52 13

Step 4 232 14 188 -1 27

-10

Step 5 232 4

10

Step 6 282 36 232 14

9 -36 9

Step 7 291 0 23 188 -1

-9 Step 8 -10 138 27

-8

Step 9 188 -10 19

6

Step 10 291 0 232 23 -4

12 6 -24 6

Step 11 12 238 -1 2

6

Step 12 5 188 2 138 19

5 5 -20 Step 13 7 143 -1 325 -12 319 -10

3 12 6

Step 14 291 12 238 5 143 -1 328 0 -4

3 -12 -12 3

Step 15 294 0 -7 188 7 2

2


12

Table 6: E5, Step 16, to F2, Result

POINT E5 E6 E7 E8 F1 F2


Step 16 9

Step 17 238 -7 188 9 143 2

2 2 -8 2

Step 18 -5 190 1 4

Step 19

Step 20 319 -4

-2

Step 21 190 1 -6

2

Step 22 238 -5 3

2 Step 23 -3 190 3 328 0

-2 -2

Step 24 1 -2

Step 25

Step 26

Step 27 328 -2 319 -6

-2 -2 8 Step 28 -4 317 2

-2

Step 29 0

Step 30 143 4

2

Result 294 0 238 -3 190 1 143 6 328 -4 317 0


13

Table 7: F3, Step 1, to F8, Step 15

POINT F3 F4 F5 F6 F7 F8


Step 1 304 -2 288 -60 213 45 184 2 156 1 128 1

32

Step 2 288 -60 213 77 184 2 19 19 -76 38

Step 3 288 -41 232 1 40 156 1

13 13

Step 4 304 -2 288 -41 232 14 14

-10 -10 40 -10

Step 5 -12 278 -1 232 4 184 40 156 14

10 10 -40 10

Step 6 232 14 194 0 24

9

Step 7 278 -1 23 -9

Step 8 -10

Step 9 194 0 156 24 128 1

12 6 -24 6

Step 10 278 -10 232 23 194 12 162 0 7

6 6 -24 12

Step 11 -4 238 -1 194 24 162 0

6 6 -24 6 Step 12 5 200 0 6 128 7

5

Step 13 304 -12 278 -4 12

3 12 3

Step 14 307 0 -1 238 5 162 6 128 12

-12 3 3 -12

Step 15 -7 200 0 162 9 131 0

4 2 -8 2


14

Table 8: F3, Step 16, to F8, Result.

POINT F3 F4 F5 F6 F7 F8


Step 16 4 164 1 2

4

Step 17 238 -7 5

2 2

Step 18 -5 7

Step 19 307 0

-2

Step 20 -2

Step 21 200 4 164 7 131 2

4 2 -8 2

Step 22 238 -5 200 8 166 -1 4

2 2 -8 2 Step 23 278 -1 -3 202 0 1

-2

Step 24 -3

Step 25 131 4

2

Step 26 166 1 6

2

Step 27 307 -2 3

-2 Step 28 -4

Step 29 307 -4

-2

Step 30 -6 166 3 131 6

2 2 -8

Result 307 -6 278 -3 238 -3 202 0 166 5 133 -2


15

Table 9: G1, Step 1, to G6, Step 15.

POINT G1 G2 G3 G4 G5 G6


Step 1 250 -8 246 -6 236 -1 225 -38 175 32 156 1

Step 2 175 32

19

Step 3 225 -38 175 51 156 1

13 13 -52 13

Step 4 225 -25 188 -1 14

-10

Step 5 -35 156 14

10

Step 6 24

Step 7 236 -1 225 -35 188 -1

-9 -9 36 -9

Step 8 -10 216 1 -10

-8

Step 9 -7 188 -10 156 24

6 6 -24

Step 10 188 -4 162 0

6 Step 11 2 162 0

6

Step 12 188 2 6

5

Step 13 250 -8 236 -10 7

3 3

Step 14 -5 -7 162 6

3

Step 15 188 7 162 9

2 2 -8


16

Table 10: G1, Step 16, to G6, Result.

POINT G1 G2 G3 G4 G5 G6 t q* t q* t q* t q* t q* t q*

Step 16 9 164 1

4

Step 17 216 -7 188 9 5

2 2 -8 2

Step 18 236 -7 -5 190 1 7

-2

Step 19 246 -6 236 -9 216 -5

-2 -2 8 -2

Step 20 250 -5 246 -8 234 -1 -7 -2 -2 8 -2

Step 21 -7 244 0 -3 190 1 164 7

2 2 -8

Step 22 3 166 -1

2

Step 23 250 -7 244 0 234 -3 216 -7 190 3 1

-2 8 -4 -2 -2 8 -2

Step 24 248 1 -4 -5 214 1 1

-2 Step 25 -1

Step 26 166 1

2

Step 27 244 -4 3

-2

Step 28 248 -1 244 -6 234 -5

-2 -2 8 -2

Step 29 -3 242 2 234 -7 214 1

-2 -2 8 -2 Step 30 0 232 1 -1 166 3

2

Result 248 -3 242 0 232 1 214 -1 190 1 166 5


17

Table 11: G7, Step 1, to H4, Step 15.

POINT G7 G8 H1 H2 H3 H4


Step 1 132 -2 119 -1 175 -4 173 -3 168 0 163 -21

Step 2

Step 3

Step 4

Step 5

Step 6

Step 7 163 -21

-9

Step 8 163 -30

-8 32

Step 9 138 -2 155 2

12

Step 10 10

Step 11

Step 12 155 2

5

Step 13 7

Step 14 119 -1

3

Step 15 138 10 2

4 :


18

Table 12: G7, Step 16, to H4, Result

POINT G7 G8 H1 H2 H3 H4


Step 16 138 14 119 2

4 -12 4

Step 17 142 2 6

Step 18 173 -3 168 -8 155 7 -2 -2 8 -2

Step 19 -5 166 0 5

-2

Step 20 173 -5 -2

-2

Step 21 142 2 -7

4

Step 22 6

Step 23 175 -4 173 -7 166 -2 155 5

-4 -2 8 -2 -2

Step 24 175 -8 171 1 -4 3

-2 8 -2

Step 25 142 6 119 6 173 0 -1

4 2 -8

Step 26 142 10 121 -2

2 -8 2

Step 27 144 2 0

Step 28 171 -1

-2

Step 29 -3 166 -4

-2

Step 30 121 0 -6

4

Result 144 2 121 4 173 0 171 -3 166 -6 155 3


19

Table 13: H5, Step 1, to H8, Step 15.

POINT H5 H6 H7 H8 t q* t q* t q* t q*

Step 1 138 14 128 1 119 -1 109 2

Step 2

Step 3 138 14

13

Step 4 27

Step 5

Step 6

Step 7

Step 8 138 27

-8

Step 9 19 128 1

6 Step 10 7

Step 11

Step 12 138 19 128 7

5 -20 5

Step 13 143 -1 12

Step 14 143 -1 128 12 119 -1

3 3 -12 3 Step 15 2 131 0 2

2


20

Table 14: H5, Step 16, to H8, Result.

POINT H5 H6 H7 H8

t q* t q* t q* t q*

Step 16 2 119 2

4

Step 17 143 2 6

2

Step 18 4

Step 19

Step 20

Step 21 131 2

2

Step 22 4

Step 23

Step 24

Step 25 131 4 119 6 109 2

2 121 -8 4

Step 26 6 240 -2 6

2

Step 27 0

Step 28

Step 29

Step 30 143 4 131 6 121 0 109 6

2 -8 4 2 -8

Result 143 6 131 -2 121 4 111 -2


21

Duct section with temperatures.

Consider the outer surface of the quarter section.

m

m

m

662kL

q

2

7371665543332111213343556671

2

73k

L

q

∆tkL

q

=

++++++++++++++=

= ∑

For the wholw duct: q/L = 4 (662)K

m = 2,648k

m

q/L = (2648)(0.35) = 926.8 Btu/hr per ft. . . Ans.


22

2_005 Consider a rod having an o.d. of 1.0 inch, in which heat is generated internally according to the equation

−=

2

1R

rww

1

where w is the local generation per unit volume at the local radius r, R is the total radius , and w

1 is the local generation

per unit volume at the center line. The total amount of heat leaving the surface is uniform along the longitudinal axis of 500,000 Btu/(hr)(sq ft). Calculate the temperature drop from the center line of this rod to its surface. The thermal conductivity of the rod is 19 Btu/(hr)(sq ft)(deg F per ft).

Solution:

q

1/A = 500,000 Btu/(hr)(sq ft)

R = 0.5 inch k = 19 Btu/(hr)(sq ft)(deg F per ft)

−=

2

1R

r1ww

dr

dtkA

R

r1qq

2

1 −=

−=

−=−

2

1R

r1q

dr

dtkA

drR

r1qkAdt

2

1

−=−

R

0

2

3

13R

rrqtkA

−=∆−

=∆− R

3

2qtkA 1

=∆− R

3

2

A

qtk 1

-(19)(∆t) = (500,000)(2/3)(0.5/12)

∆∆∆∆t = - 731 F .... Ans.


23

2_006 A long steam pipe of o.d. D1 is covered with thermal insulation having an o.d. D

2. The outer surface of the pipe remains

at a constant temperature t1 as more insulation is added (D

2 is increased). Is it ever possible that adding insulation will

cause an increase in heat loss? If possible, under what conditions does this situation arise? The insulating material has a thermal conductivity of k.

The heat loss per unit area of outside surface of the insulation is assumed to be directly proportional to the difference between the temperature of the outer surface of the insulation and the temperature of the surroundings t

s

(assumed to be constant). The proportionality constant is called the surface coefficient of heat transfer and is designated by the symbol h.

Solution:

Equation 2-8.

∆=

∆π=

1

22

om

1

2

m

D

DlnD

t2Ak

D

Dln

tL2kq

Equation 1-4.

t)(tAhq wom −=

Or combining gives.

h

1

2k

D

DlnD

tt

A

q

1

22

s1

o

+

−=

Where: t

1 - t

s = constant

h = constant Getting maximum heat loss:

( )

h

1

2k

D

DlnD

)(L)D(ttq

1

22

2s1

+

π−=

( )

2

1

2

s1

hD

1

2k

D

Dln

L)(ttq

+

π−=


24

Let us use denominator only since numerator is constant.

2

1

2

hD

1

2k

D

Dln

+

=y

Determine D

2 that gives y as minimum.

First derivative equate to zero:

2k/hD

0hD

1

D

1

2k

1

dD

dy

2

2222

=

=−

=

Check if D

2 = 2k/h is at minimum.

Second derivative:

3

22

22

2

2

hD

2

D

1

2k

1

dD

yd+

−=

Substituting D

2 = 2k/h,

( ) 32

2

22

2

h(2k/h)

2

2k

h

2k

1

dD

yd+

−=

( ) ( )

0>=+−=3

2

3

2

3

2

22

2

2k

h

(2k)

2h

2k

h

dD

yd

Therefore D

2=2k/h gives minimum value of y as above or a maximum value of q. Then it is possible that adding

insulation will cause an increase in heat loss if D2 > 2k/h. ..... Ans.

- end -

25

CHAPTER 3

TRANSIENT CONDUCTION

Chapter 3 Transient Conduction

26

3_001 A very long rod of homogeneous solid has a constant cross section, is hexagonal in shape, and is perfectly insulated from the surroundings on one end and on all faces parallel to its axis. The other end is not insulated. Originally the rod is at a uniform temperature of 70 F. Suddenly the temperature of the surface of the uninsulated end is increased to and maintained at 170 F. After 2.5 hr, the temperature 1 ft from this surface is 160 F. How much additional time will be required for the temperature 2 ft from the end to reach 160 F?

Solution: Solve equation 3-2a:

θ∂

∂=

∂

∂⋅

t

x

t

ρc

k2

2

p t = t

a = 170 F @ x = 0, t = 0 F @ x = ∞

t = tb = 70 F @ θ = 0, t = t

a = 170 F @ q = ∞

t = 160 F, θ = 2.5 hrs, x = 1 ft Find q when t = 165 F , x = 2 ft Using Fig. 3-6 or

dZeπ

2Y

Z

0

Z2

∫−=

ba

a

tt

ttY

−

−=

αθ2

xZ =

Then @ t = 160 F, θ = 2.5 hrs, x = 1 ft

0.1

70170

160170Y =

−

−=

0.0889Z =

( )

0.0889α2.52

1Z ==

12.653=α Then when t = 160 F, x = 2 ft

0.170170

160170

tt

ttY

ba

a =−

−=

−

−=

0.0889Z =

αθ2

xZ =

( )

0.0889θ12.6532

2Ζ ==

q = 10 hrs . . . Ans.


27

3_002 It is proposed to heat dry sand by passing it steadily downward through a vertical pipe which is heated by a vapor condensing on the outside. The sand is assumed to flow through the pipe with a uniform velocity profile. At the exit of the pipe the sand flows into a mixer, and its temperature becomes uniform. The inside wall of the pipe is maintained at 220 F, and the thermal resistance between the pipe wall and the dry sand is assumed to be negligible. The sand initially at a temperature of 120 F is fed to the pipe at a rate of 1.18 cu ft/hr. The pipe is 18 ft long and has an i.d. of 1.2 inches.

Estimate the temperature of the mixed sand leaving the heater. Sand properties: Density = 100 lb/cu ft Thermal conductivity = 0.2 Specific heat = 0.24 Solution:

Sand Properties: ρ = 100, k = 0.20, cp = 0.24

ta = 220 F, t

b = 120 F

Volume = (π/4)(1.2 / 12)2(18) = 0.141372 cu ft

Time , θ = (0.141372 cu ft) / (1.18 cu ft / hr) = 0.119807 hr r

m = (1/2)(1.2 / 12) = 0.05 ft

From fig. 3-9, Cylinder L = ∞ . (Actual L = 180)

2

mprc

kX

ρ

θ=

2)(0.05)(100)(0.24

19807)(0.20)(0.1X =

X = 0.40

0.161560tt

ttY

ba

mam =

−

−=

0.161560

120220

t220Y m

m =−

−=

tm = 203.844 F .... Ans.


28

3_003 A large flat glass slab 6 in. thick (k = 0.63, c = 0.3, and ρ = 139) has been cooled very slowly in an annealing oven so that its temperature is substantially uniform at 900 F. It is proposed that the slab be further cooled from 900 F by passing air parallel to the flat faces normal to the 6-in. dimension. To minimize thermal strains, the maximum temperature gradient allowable in the slab is 50 F/in. The air would be blown past the slab at a rate such that its temperature rise would be negligible and the coefficient from hot glass to air would be 5.0. Radiation may be neglected. a. When the slab is at 900 F, what is the lowest temperature of air that may be used for cooling? b. If air at the temperature in (a) is used for 3 hr, what is the lowest air temperature that could then be used? c. If the air temperature were so regulated that the slab was always being cooled at the maximum allowable

rate, what would be the air temperature at the end of 3 hr? d. Sketch curves of temperature vs. thickness of the slab at the end of 3 hr for (b) and (c). Solution:

k = 0.63, c = 0.3, ρ = 139 a. t

b = 900 F

( )bsm

sa ttr

k)th(t −=−

ts = 900 F - (50 F / in)(3 in) = 750 F

rm = 3 in

h = 5.0 b. n = 1

( )( )

0.505

123

0.63

hr

km

m

===

θ = 3 hrs

( )( )0.30139

0.63

c

k;

rX

p2

m

=ρ

=ααθ

=

α = 0.015108

( )( )

( )212

3

30.015108X =

From Figure 3-3 Y

m = 0.516

From Figure 3-4 Y

s = 0.24

Mid-Plane Temperature:

ba

mam

tt

ttY

−

−=


29

0.516900674.4

t674.4Y m

m =−

−=

tm = 790.81 F

Surface Temperature:

ba

sa

tt

ttYs

−

−=

0.24900674.4

t674.4Y s

s =−

−=

Y

s = 728.54 F

( ) ( )mt−=− sm

sa' tr

ktth

( ) ( )( )790.81728.54

123

0.63728.54t5 a' −=−

ta’=697.16 F . . . Ans.

c. θ = 3 hrs, 50 F / in maintained. Equation 1.

0.516tt

ttY

ba

mam =

−

−=

Equation 2.

0.24tt

ttYs

ba

sa =−

−=

Equation 3. t

m - t

s = 3 x 50 = 150 F

Then,

0.465120.516

0.24

tt

tt

Y

Y

ma

sa

m

s ==−

−=

t

a - t

s = 0.46512 ( t

a - t

m )

0.53488 ta = t

s - 0.46512 t

m

0.53488 ta = t

s - 0.46512 (150 + t

s)

0.53488 ta = 0.53488 t

s - 69.768 - - - Eq. 4.

From Eq. 2.

0.24tt

ttYs

ba

sa =−

−=

t

a - t

s = 0.24 ( t

a - 900 )

0.76 t

a = t

s - 216

Substitute in Eq. 4 0.53488 t

a = 0.53488 (0.76 t

a + 216) - 69.768

0.128371 ta = 45.766

ta = 356.5 F .... Ans.

ts = 0.76 (356.5) + 216 = 486.94 F

tm = 150 + 486.94 F = 636.94 F


30

d. m = 0.5, X = 0.725 For (b)

900674.4

t674.4Y

−

−=

For (c)

900356.5

t356.5Y

−

−=

Tabulated Value: n = r / r

m

n Y Temp. in (b) Temp. in (c)

0 0.516 790.8 637

0.2 0.474 781.3 614.1

0.4 0.445 774.8 598.4

0.6 0.4 764.6 573.9

0.8 0.326 748 533.7

1 0.24 728.6 487


31

3_004 A composite slab consists of 9 inches of magnesite brick (A) and 8.70 inches of alumina brick (B), in good thermal

contact. The initial distribution of temperature is specified below. Suddenly the exposed side of A is brought to, and maintained at, 100 F, while the exposed side of B is suddenly brought to, and maintained at, 0 F. Find the temperature distribution at the end of 3.05 hr, and when the steady state is reached.

Data and Notes: The initial distribution of temperature in A is given by the equation t = 22.2x - 2.46x2, where x is

distance (inches) from the bare face of A, toward B, and t is temperature in degrees Fahrenheit. The initial distribution of temperature in B is given by t = 10x - 90.

The following table shows t at various values of x:

x in A .. 0 1 2 3 4 4.5 5 6 7 8 9

t .......... 0 19.78 34.6 44.5 49.4 50 49.4 44.5 34.6 19.78 0

x in B .. 9 10 12 14 16 17.7

t .......... 0 10 30 50 70 87 It is desired to use at least three slices in A and B.

A B

k ,Btu/(hr)(sq ft)(deg F pe r ft) .... 2.15 0.58

ρ , lb/cu ft ...................................,. 159 90

cp ,Btu/(lb)(deg F) ........................ 0.22 0.2

α , sq ft / hr .................................. 0.0615 0.0322 Solution:


32

( )0.0615α;

Mα

∆x∆θ A

AA

A ==

( )0.0322α;

Mα

∆x∆θ B

BB

B ==

use ∆x

A = 1 in, M

A = M

B = 2

( )( )( )

( )( )( )

0.7236∆x

20.0322

∆x

20.0615

1∆θ

B

2B

2

=

==

0.725

12

917.7∆xB =

−=

( )( )0.05646

20.0615

12

1

∆θ

2

=

=

Number of time steps = 3.05 hrs / 0.05646 hrs = 54 steps. Temperature and Points

x Point Temperature, F

0.000 0 0

1.000 1 19.78

2.000 2 34.60

3.000 3 44.50

4.000 4 49.40

5.000 5 49.40

6.000 6 44.50

7.000 7 34.60

8.000 8 19.78

9.000 9 0.00

9.725 10 7.25

10.450 11 14.50

11.175 12 21.75

11.900 13 29.00

12.625 14 36.25

13.350 15 43.50

14.075 16 50.75

14.800 17 58.00

15.525 18 65.25

16.250 19 72.50

16.975 20 79.75

17.700 21 87.00


33

At Point 9.

( )( )( )( )2.150.7236

0.581

k∆x

k∆xR

AB

BAr ==

Rr = 0.3728, M

A = M

B = 2

Equation 3-13a

108

'9 t

1.3728

0.3728t

1.3728

1t +=

1.3728

t0.3728tt 108'9

⋅+=

Table 1: Temperature t

0 to t

9, Step 0 to step 28

Time Steps 0 1 2 3 4 5 6 7 8 9

∆θ

0 100 19.78 34.60 44.50 49.40 49.40 44.50 34.60 19.78 0.00

1 100 67.30 32.14 42.00 46.95 46.95 42.00 32.14 17.30 16.38

2 100 66.07 54.65 39.55 44.48 44.48 39.55 29.65 24.26 14.57

3 100 77.33 52.81 49.56 42.01 42.01 37.06 31.90 22.11 21.86

4 100 76.40 63.44 47.41 45.79 39.54 36.96 29.59 26.88 20.05

5 100 81.72 61.91 54.62 43.47 41.37 34.56 31.92 24.82 25.08

6 100 80.95 68.17 52.69 47.99 39.02 36.65 29.69 28.50 23.27

7 100 84.08 66.82 58.08 45.85 42.32 34.35 32.57 26.48 27.15

8 100 83.41 71.08 56.34 50.20 40.10 37.45 30.42 29.86 25.34

9 100 85.54 69.87 60.64 48.22 43.82 35.26 33.65 27.88 28.80

10 100 84.94 73.09 59.05 52.23 41.74 38.74 31.57 31.23 27.00

11 100 86.55 71.99 62.66 50.39 45.48 36.65 34.98 29.29 30.33

12 100 86.00 74.60 61.19 54.07 43.52 40.23 32.97 32.66 28.53

13 100 87.30 73.59 64.34 52.36 47.15 38.25 36.45 30.75 31.82

14 100 86.80 75.82 62.98 55.75 45.30 41.80 34.50 34.13 30.03

15 100 87.91 74.89 65.78 54.14 48.77 39.90 37.97 32.26 33.30

16 100 87.44 76.85 64.51 57.28 47.02 43.37 36.08 35.63 31.51

17 100 88.42 75.98 67.06 55.77 50.32 41.55 39.50 33.79 34.76

18 100 87.99 77.74 65.87 58.69 48.66 44.91 37.67 37.13 32.96

19 100 88.87 76.93 68.22 57.27 51.80 43.17 41.02 35.32 36.18

20 100 88.47 78.54 67.10 60.01 50.22 46.41 39.24 38.60 34.39

21 100 89.27 77.78 69.28 58.66 53.21 44.73 42.51 36.82 37.56

22 100 88.89 79.27 68.22 61.24 51.69 47.86 40.77 40.03 35.79

23 100 89.64 78.56 70.26 59.96 54.55 46.23 43.95 38.28 38.90

24 100 89.28 79.95 69.26 62.41 53.09 49.25 42.26 41.42 37.14

25 100 89.97 79.27 71.18 61.18 55.83 47.68 45.34 39.70 40.19

26 100 89.63 80.58 70.22 63.50 54.43 50.58 43.69 42.76 38.44

27 100 90.29 79.93 72.04 62.32 57.04 49.06 46.67 41.06 41.42

28 100 89.96 81.16 71.13 64.54 55.69 51.86 45.06 44.05 39.70


34

Table 2: Temperature t

0 to t


29 100 90.58 80.54 72.85 63.41 58.20 50.37 47.95 42.38 42.60

30 100 90.27 81.72 71.98 65.53 56.89 53.07 46.38 45.28 40.90

31 100 90.86 81.12 73.62 64.43 59.30 51.63 49.18 43.64 43.74

32 100 90.56 82.24 72.78 66.46 58.03 54.24 47.64 46.46 42.07

33 100 91.12 81.67 74.35 65.41 60.35 52.83 50.35 44.85 44.83

34 100 90.84 82.73 73.54 67.35 59.12 55.35 48.84 47.59 43.18

35 100 91.37 82.19 75.04 66.33 61.35 53.98 51.47 46.01 45.87

36 100 91.09 83.20 74.26 68.20 60.16 56.41 50.00 48.67 44.26

37 100 91.60 82.68 75.70 67.21 62.30 55.08 52.54 47.13 46.87

38 100 91.34 83.65 74.94 69.00 61.14 57.42 51.10 49.71 45.29

39 100 91.83 83.14 76.33 68.04 63.21 56.12 53.56 48.19 47.83

40 100 91.57 84.08 75.59 69.77 62.08 58.39 52.16 50.70 46.28

41 100 92.04 83.58 76.92 68.84 64.08 57.12 54.54 49.22 48.75

42 100 91.79 84.48 76.21 70.50 62.98 59.31 53.17 51.65 47.23

43 100 92.24 84.00 77.49 69.59 64.90 58.07 55.48 50.20 49.64

44 100 92.00 84.87 76.80 71.20 63.83 60.19 54.14 52.56 48.15

45 100 92.43 84.40 78.03 70.31 65.69 58.99 56.37 51.14 50.49

46 100 92.20 85.23 77.36 71.86 64.65 61.03 55.06 53.43 49.03

47 100 92.62 84.78 78.55 71.00 66.45 59.86 57.23 52.05 51.31

48 100 92.39 85.58 77.89 72.50 65.43 61.84 55.95 54.27 49.89

49 100 92.79 85.14 79.04 71.66 67.17 60.69 58.06 52.92 52.09

50 100 92.57 85.92 78.40 73.10 66.18 62.61 56.81 55.07 50.71

51 100 92.96 85.48 79.51 72.29 67.86 61.49 58.84 53.76 52.85

52 100 92.74 86.23 78.89 73.68 66.89 63.35 57.62 55.85 51.50

53 100 93.12 85.81 79.96 72.89 68.52 62.26 59.60 54.56 53.59

54 100 92.91 86.54 79.35 74.24 67.57 64.06 58.41 56.59 52.26 Table 3: Temperature t

10 to t


10 11 12 13 14 15 16 17 18 19 20 21

7.25 14.50 21.75 29.00 36.25 43.50 50.75 58.00 65.25 72.50 79.75 0

7.25 14.50 21.75 29.00 36.25 43.50 50.75 58.00 65.25 72.50 36.25 0

15.44 14.50 21.75 29.00 36.25 43.50 50.75 58.00 65.25 50.75 36.25 0

14.54 18.59 21.75 29.00 36.25 43.50 50.75 58.00 54.38 50.75 25.38 0

20.23 18.14 23.80 29.00 36.25 43.50 50.75 52.56 54.38 39.88 25.38 0

19.10 22.01 23.57 30.02 36.25 43.50 48.03 52.56 46.22 39.88 19.94 0

23.54 21.33 26.02 29.91 36.76 42.14 48.03 47.13 46.22 33.08 19.94 0

22.30 24.78 25.62 31.39 36.03 42.40 44.63 47.13 40.10 33.08 16.54 0

25.97 23.96 28.09 30.82 36.89 40.33 44.76 42.37 40.10 28.32 16.54 0

24.65 27.03 27.39 32.49 35.58 40.83 41.35 42.43 35.34 28.32 14.16 0

27.92 26.02 29.76 31.48 36.66 38.46 41.63 38.35 35.38 24.75 14.16 0

26.51 28.84 28.75 33.21 34.97 39.14 38.40 38.50 31.55 24.77 12.38 0

29.58 27.63 31.02 31.86 36.18 36.69 38.82 34.98 31.64 21.96 12.38 0

28.08 30.30 29.75 33.60 34.28 37.50 35.83 35.23 28.47 22.01 10.98 0

31.06 28.92 31.95 32.01 35.55 35.05 36.36 32.15 28.62 19.73 11.00 0

29.47 31.51 30.46 33.75 33.53 35.96 33.60 32.49 25.94 19.81 9.86 0

32.40 29.97 32.63 32.00 34.85 33.57 34.22 29.77 26.15 17.90 9.91 0

30.74 32.52 30.98 33.74 32.78 34.54 31.67 30.19 23.84 18.03 8.95 0

33.64 30.86 33.13 31.88 34.14 32.23 32.36 27.75 24.11 16.39 9.01 0

31.91 33.38 31.37 33.63 32.05 33.25 29.99 28.24 22.07 16.56 8.20 0

34.78 31.64 33.51 31.71 33.44 31.02 30.74 26.03 22.40 15.13 8.28 0

33.02 34.14 31.68 33.48 31.37 32.09 28.53 26.57 20.58 15.34 7.57 0

35.85 32.35 33.81 31.52 32.78 29.95 29.33 24.55 20.96 14.08 7.67 0

34.07 34.83 31.94 33.30 30.73 31.06 27.25 25.14 19.31 14.31 7.04 0

36.87 33.00 34.06 31.34 32.18 28.99 28.10 23.28 19.73 13.18 7.16 0

35.07 35.47 32.17 33.12 30.16 30.14 26.14 23.91 18.23 13.44 6.59 0

37.83 33.62 34.29 31.17 31.63 28.15 27.03 22.18 18.68 12.41 6.72 0

36.03 36.06 32.39 32.96 29.66 29.33 25.17 22.85 17.30 12.70 6.20 0

38.74 34.21 34.51 31.03 31.15 27.41 26.09 21.23 17.78 11.75 6.35 0


35

Table 4: Temperature t10

to t21

, Step 29 to step 54

36.95 36.63 32.62 32.83 29.22 28.62 24.32 21.93 16.49 12.06 5.88 0

39.61 34.79 34.73 30.92 30.72 26.77 25.28 20.41 17.00 11.18 6.03 0

37.84 37.17 32.85 32.72 28.84 28.00 23.59 21.14 15.79 11.51 5.59 0

40.46 35.35 34.95 30.85 30.36 26.22 24.57 19.69 16.33 10.69 5.76 0

38.71 37.70 33.10 32.65 28.53 27.47 22.95 20.45 15.19 11.04 5.35 0

41.26 35.90 35.18 30.82 30.06 25.74 23.96 19.07 15.74 10.27 5.52 0

39.54 38.22 33.36 32.62 28.28 27.01 22.41 19.85 14.67 10.63 5.13 0

42.05 36.45 35.42 30.82 29.81 25.34 23.43 18.54 15.24 9.90 5.32 0

40.35 38.73 33.64 32.62 28.08 26.62 21.94 19.34 14.22 10.28 4.95 0

42.80 36.99 35.68 30.86 29.62 25.01 22.98 18.08 14.81 9.59 5.14 0

41.14 39.24 33.93 32.65 27.94 26.30 21.55 18.89 13.83 9.97 4.79 0

43.53 37.53 35.94 30.93 29.47 24.74 22.60 17.69 14.43 9.31 4.99 0

41.91 39.74 34.23 32.71 27.84 26.03 21.22 18.51 13.50 9.71 4.66 0

44.25 38.07 36.22 31.03 29.37 24.53 22.27 17.36 14.11 9.08 4.85 0

42.65 40.23 34.55 32.80 27.78 25.82 20.94 18.19 13.22 9.48 4.54 0

44.94 38.60 36.52 31.17 29.31 24.36 22.01 17.08 13.84 8.88 4.74 0

43.38 40.73 34.88 32.91 27.76 25.66 20.72 17.92 12.98 9.29 4.44 0

45.61 39.13 36.82 31.32 29.29 24.24 21.79 16.85 13.61 8.71 4.65 0

44.08 41.21 35.23 33.05 27.78 25.54 20.55 17.70 12.78 9.13 4.35 0

46.26 39.65 37.13 31.50 29.30 24.16 21.62 16.66 13.41 8.57 4.56 0

44.77 41.70 35.58 33.21 27.83 25.46 20.41 17.52 12.62 8.99 4.28 0

46.90 40.17 37.46 31.71 29.34 24.12 21.49 16.51 13.25 8.45 4.49 0

45.44 42.18 35.94 33.40 27.92 25.41 20.32 17.37 12.48 8.87 4.22 0

47.51 40.69 37.79 31.93 29.40 24.12 21.39 16.40 13.12 8.35 4.44 0

46.09 42.65 36.31 33.59 28.02 25.40 20.26 17.26 12.38 8.78 4.18 0

48.12 41.20 38.12 32.17 29.50 24.14 21.33 16.32 13.02 8.28 4.39 0 Temperature distribution after 3.05 hrs


36

3_005 A large slab of dry wood, 0.75 inch thick, initially at 88 F, is suspended horizontally in a tunnel. Suddenly the upper

face is subjected to a uniform radiant flux of 393 Btu/(hr)(sq ft), from a number of infrared lamps suitably arranged above the upper face. At this same zero time, dry air at a mean temperature of 114 F is blown past the upper and lower faces, giving surface coefficients of heat transfer h equal to 4.48 Btu/(hr)(sq ft)(deg F) on each face. It is desired to predict the temperature distribution within the slab as a function of time. a. Derive the necessary equations for an approximate numerical solution of this problem.

b. Taking ∆x = 1/8 inch, proceed with the calculation of the temperature profile for three time increments, using an M of 3.

c. Compute the temperature profile after an infinite time interval. Wood properties: k = 0.101

a = k/ρcp = 0.00318 sq ft/hr

Solution: a. Upper half slice temperature (Eq. 3-10).

( )[ ]M

P

M

2tto22NMt2Nt iao +

+⋅+−+⋅='

Eq. 3-10a

k

∆x

dA

dq2P r≡

where

k

∆xhN

⋅=

sq.ft.Btu/hr393

dA

dqr ⋅=

Interior temperature (Eq. 3-6a)

( )M

tt2Mt't 21o

1

+⋅−+=

Lower half slice surface temperature (Eq. 3-7a)

( )[ ]M

2tto22NMt2Nt iao

+⋅+−+⋅='

where t

a = 114 F, t

b = 88 F


37

b. ∆x = 1/8 in = 0.125 in No. of Points = (0.75 in / 0.125 in) + 1 = 7 Points: t

0 up to t

6.

Upper half slice surface temperature: Point t

0.

( )[ ]M

P

M

2tto22NMt2Nt iao +

+⋅+−+⋅='

( )( )81.0644

0.101

0.125/123932

k

∆x

dA

dq2P r ==≡

( )3

α∆θ

∆xM

2

==

( )( )0.462

0.101

0.125/124.48

k

∆xhN ==

⋅=

( ) ( )( )[ ]3

81.0644

3

2tto20.46223t0.4622t ia'o +

+⋅+−+⋅=

( )81.06442t0.076t0.924tt 10a3

1'o +++=

Points t

1 to t

5:

( )M

tt2Mt't 21o

1

+⋅−+=

( )3

tt23t't 21o

1

+⋅−+=

( )2103

1 ttt ++='t1 Point t

6:

( )[ ]M

2tt22NMt2Nt 56a6

+⋅+−+⋅='

( )56a3

1'6 2t0.076t0.924tt ++=

t

a = 114 F

For three time increment:

∆θ

Step t0 t1 t2 t3 t4 t5 t6

1 88 88 88 88 88 88 88

2 123.03 88.00 88.00 88.00 88.00 88.00 96.01

3 123.92 99.68 88.00 88.00 88.00 90.67 96.21 c. Temperature profile for infinite time interval. Continue b until there is no significant change in temperatures.


38

Tabulated value from step 4 to 40.

∆θ


4 131.72 103.86 91.89 88.00 88.89 91.63 98.00

5 134.71 109.16 94.59 89.59 89.51 92.84 98.68

6 138.32 112.82 97.78 91.23 90.65 93.67 99.50

7 140.85 116.31 100.61 93.22 91.85 94.61 100.08

8 143.24 119.26 103.38 95.23 93.22 95.51 100.72

9 145.27 121.96 105.95 97.28 94.65 96.49 101.34

10 147.12 124.39 108.40 99.29 96.14 97.49 102.00

11 148.79 126.64 110.69 101.28 97.64 98.54 102.69

12 150.33 128.71 112.87 103.20 99.15 99.63 103.41

13 151.75 130.63 114.93 105.08 100.66 100.73 104.15

14 153.07 132.44 116.88 106.89 102.16 101.85 104.90

15 154.30 134.13 118.73 108.64 103.63 102.97 105.67

16 155.46 135.72 120.50 110.33 105.08 104.09 106.43

17 156.55 137.23 122.19 111.97 106.50 105.20 107.20

18 157.58 138.65 123.79 113.55 107.89 106.30 107.96

19 158.56 140.01 125.33 115.08 109.25 107.38 108.71

20 159.49 141.30 126.81 116.55 110.57 108.45 109.46

21 160.38 142.53 128.22 117.98 111.86 109.49 110.18

22 161.22 143.71 129.58 119.35 113.11 110.51 110.90

23 162.02 144.84 130.88 120.68 114.32 111.51 111.60

24 162.80 145.91 132.13 121.96 115.50 112.48 112.28

25 163.53 146.95 133.34 123.20 116.65 113.42 112.94

26 164.24 147.94 134.49 124.39 117.75 114.34 113.59

27 164.92 148.89 135.61 125.55 118.83 115.23 114.21

28 165.57 149.81 136.68 126.66 119.87 116.09 114.82

29 166.20 150.69 137.72 127.74 120.87 116.93 115.41

30 166.80 151.53 138.71 128.78 121.85 117.74 115.99

31 167.38 152.35 139.67 129.78 122.79 118.52 116.54

32 167.94 153.14 140.60 130.75 123.70 119.28 117.08

33 168.48 153.89 141.49 131.68 124.58 120.02 117.60

34 169.00 154.62 142.36 132.58 125.43 120.73 118.10

35 169.50 155.32 143.19 133.45 126.25 121.42 118.59

36 169.98 156.00 143.99 134.30 127.04 122.09 119.06

37 170.44 156.66 144.76 135.11 127.81 122.73 119.52

38 170.89 157.29 145.51 135.89 128.55 123.35 119.96

39 171.32 157.89 146.23 136.65 129.26 123.95 120.39

40 171.74 158.48 146.92 137.38 129.96 124.53 120.80


39

Tabulated value from step 320 to 360.

∆θ


320 183.35 174.85 166.36 157.86 149.37 140.88 132.38

321 183.35 174.85 166.36 157.86 149.37 140.88 132.38

322 183.35 174.85 166.36 157.86 149.37 140.88 132.38

323 183.35 174.85 166.36 157.87 149.37 140.88 132.38

324 183.35 174.85 166.36 157.87 149.37 140.88 132.38

325 183.35 174.85 166.36 157.87 149.37 140.88 132.38

326 183.35 174.85 166.36 157.87 149.37 140.88 132.38

327 183.35 174.85 166.36 157.87 149.37 140.88 132.38

328 183.35 174.85 166.36 157.87 149.37 140.88 132.38

329 183.35 174.85 166.36 157.87 149.37 140.88 132.38

330 183.35 174.85 166.36 157.87 149.37 140.88 132.38

331 183.35 174.85 166.36 157.87 149.37 140.88 132.38

332 183.35 174.85 166.36 157.87 149.37 140.88 132.38

333 183.35 174.85 166.36 157.87 149.37 140.88 132.38

334 183.35 174.85 166.36 157.87 149.37 140.88 132.38

335 183.35 174.85 166.36 157.87 149.37 140.88 132.38

336 183.35 174.85 166.36 157.87 149.37 140.88 132.38

337 183.35 174.85 166.36 157.87 149.37 140.88 132.38

338 183.35 174.85 166.36 157.87 149.37 140.88 132.38

339 183.35 174.85 166.36 157.87 149.37 140.88 132.38

340 183.35 174.85 166.36 157.87 149.37 140.88 132.38

341 183.35 174.85 166.36 157.87 149.37 140.88 132.38

342 183.35 174.85 166.36 157.87 149.37 140.88 132.38

343 183.35 174.85 166.36 157.87 149.37 140.88 132.38

344 183.35 174.85 166.36 157.87 149.37 140.88 132.38

345 183.35 174.85 166.36 157.87 149.37 140.88 132.38

346 183.35 174.85 166.36 157.87 149.37 140.88 132.38

347 183.35 174.85 166.36 157.87 149.37 140.88 132.38

348 183.35 174.85 166.36 157.87 149.37 140.88 132.38

349 183.35 174.85 166.36 157.87 149.37 140.88 132.38

350 183.35 174.85 166.36 157.87 149.37 140.88 132.38

351 183.35 174.85 166.36 157.87 149.37 140.88 132.38

352 183.35 174.85 166.36 157.87 149.37 140.88 132.38

353 183.35 174.85 166.36 157.87 149.37 140.88 132.38

354 183.35 174.85 166.36 157.87 149.37 140.88 132.38

355 183.35 174.85 166.36 157.87 149.37 140.88 132.38

356 183.35 174.85 166.36 157.87 149.37 140.88 132.38

357 183.35 174.85 166.36 157.87 149.37 140.88 132.38

358 183.35 174.85 166.36 157.87 149.37 140.88 132.38

359 183.35 174.85 166.36 157.87 149.37 140.88 132.38

360 183.35 174.85 166.36 157.87 149.37 140.88 132.38 No. of time steps = 323 up to infinity, there is sign of change in temperature.

( )3

α∆θ

∆xM

2

==

( )3

∆θ0.101

12

0.125

M

2

=

=

∆θ = 0.000358 hrs or Minimum time interval to reach no change in temperature = 323 x 0.000358 hrs = 0.115634 hrs = 6.938 minutes


40

Equivalent Equation for temperature profile: t = 183.3486 - 67.9543x . . . Ans. 3_006 A long hollow flue of square cross section is initially at a uniform temperature of 60 F. Suddenly hot gases are blown

through the flue at a high velocity, maintaining the inner surface of the walls at 500 F. The inside cross section is a 2-ft square, and the total cross section is a 5-ft square, the walls being 1-1/2 ft thick. The wall has the following three

properties: k of 2.2, ρ of 115, and cp of 0.18. It is assumed that the outside surface coefficient of heat transfer is given

by h = 0.27(∆t)1/4, where h is the local surface coefficient in Btu/(hr)(sq ft)(deg F) and ∆t is the difference between the local surface temperature and the ambient temperature of the surrounding, 60 F.

Derive the necessary equations for a numerical solution to this problem. Outline in detail the procedure for calculating the temperature distribution within the walls of the flue at the end of any time interval and the total heat transferred from the hot gases to the wall since time zero.

Solution:


41

k = 2.2, ρ = 115, cp = 0.18, h = 0.27(∆t)1/4

a. Derivation of necessary equation for a numerical solution. Consider one quarter of the flue.

( )k

xhN;

α∆θ

∆xM

2∆

==

Assume M = 5 t

A2 = t

B2 = t

C0 = t

C1 = t

C2 = 500 F

Using equation like Eq. 3-14a for inside the thickness temperatures.

( )( )A3A4B3A25

1A3 tt2tt't +++=

( )( )B3C3B4A3B25

1B3 ttttt't ++++=

( )( )C3D3C4B3C25

1C3 ttttt't ++++=

( )( )D3E3D4C3D25

1D3 ttttt't ++++=

( )( )D2E3D3C2D15

1D2 ttttt't ++++=

( )( )D1E1D2C1D05

1D1 ttttt't ++++=

( )( )D0E0C0D15

1D0 ttt2t't +++=


42

( )( )A4A5B4A35

1A4 tttt't +++= 2

( )( )B4C4B5A4B35

1B4 ttttt't ++++=

( )( )C4D4C5B4C35

1C4 ttttt't ++++=

( )( )D4E4D5C4D35

1D4 ttttt't ++++=

( )( )E4F4E5D4E35

1E4 ttttt't ++++=

( )( )E3F3E4D3E25

1E3 ttttt't ++++=

( )( )E2F2E3D2E15

1E2 ttttt't ++++=

( )( )E1F1E2D1E05

1E1 ttttt't ++++=

( )( )E0F0D0E15

1E0 ttt2t't +++=

Using Equation like Eq. 3-15 for outer surface temperature with

h = 0.27(∆t)1/4

( )k

xt0.27N

1/4∆∆

=

∆x = 6 in = 0.5 ft.

( ) ( ) ( )1/41/4

∆t0.0613642.2

0.5∆t0.27N ==

M = 5, t

a = 60 F

( )M

t2N4M2t2t2Nt't A5A4B5a

A5

−−+++=

( ) ( ) ( )

−−+++−= A5A5A4B5A55

1A5 t60t0.12272812t2t60t7.36368't 4

141

Then:

( ) ( ) ( )

−−++++−= B5B5B4C5A5B55

1B5 t60t0.12272812ttt60t7.36368't 4

141

( ) ( ) ( )

−−++++−= C5C5C4D5B5C55

1C5 t60t0.12272812ttt60t7.36368't 4

141

( ) ( ) ( )

−−++++−= D5D5D4E5C5D55

1D5 t60t0.12272812ttt60t7.36368't 4

141

( ) ( ) ( )

−−++++−= E5E5E4F5D5E55

1E5 t60t0.12272812ttt60t7.36368't 4

141


43

( ) ( ) ( )

−−+++−= F5F5F4E5F55

1F5 t60t0.12272812tt60t7.36368't 4

141

2

( ) ( ) ( )

−−++++−= F4F4E4F3F5F45

1F4 t60t0.12272812ttt60t7.36368't 4

141

( ) ( ) ( )

−−++++−= F3F3E3F2F4F35

1F3 t60t0.12272812ttt60t7.36368't 4

141

( ) ( ) ( )

−−++++−= F2F2E2F1F3F25

1F2 t60t0.12272812ttt60t7.36368't 4

141

( ) ( ) ( )

−−++++−= F1F1E1F0F2F15

1F1 t60t0.12272812ttt60t7.36368't 4

141

( ) ( ) ( )

−−+++−= F0F0E0F1F05

1F0 t60t0.12272812tt60t7.36368't 4

141

2

Rate of heat loss equation for one-quarter section:

( ) ( ) ( ) ( ) ( )5E4E5D4D5C4C5B4B5A4A21

m tttttttttt[kL

q−+−+−+−+−=

( ) ( ) ( ) ( ) ( )]tttttttttt 0F0E2

11F1E2F2E3F3E4F4E −+−+−+−+−+

and

( ) ( ) ( ) ( ) ( )'5E'

4E'

5D'

4D'

5C'

4C'

5B'

4B'

5A'

4A21

m tttttttttt[kL

q−+−+−+−+−=

( ) ( ) ( ) ( ) ( )]tttttttttt'

0F'

0E21'

1F'

1E'

2F'

2E'

3F'

3E'

4F'

4E −+−+−+−+−+

b. Procedure for calculating the temperature distributionwith with M = 5, ∆x = 6 in = 0.5 ft.

( )θ∆α

∆=

2x

M

( )( )( )

0.106280.18115

2.2

ρc

kα

p

===

( )( )( )

5∆θ0.10628

0.5M

2

==

∆θ = 0.470455 hr Assume 20 time interval or 9.4091 hrs Step 1: Make tabulation of temperature profile stepwise starting from zero, including heat transferred as derive in part

(a). For 20 time interval (9.4091 hrs).


44

Table 1: tA2 to tB3, Step 0 to 20.

∆θsteps tA2 tA3 tA4 tA5 tB2 tB3tB3

0 500.00 60.00 60.00 60.00 500.00 60.00

1 500.00 148.00 60.00 60.00 500.00 148.00

2 500.00 200.80 77.60 60.00 500.00 200.80

3 500.00 236.00 98.72 67.04 500.00 232.48

4 500.00 259.94 119.84 79.43 500.00 255.71

5 500.00 278.24 138.93 94.59 500.00 272.75

6 500.00 292.53 156.80 109.59 500.00 286.62

7 500.00 304.51 172.82 124.33 500.00 298.03

8 500.00 314.68 187.50 137.88 500.00 307.94

9 500.00 323.61 200.72 150.45 500.00 316.61

10 500.00 331.51 212.76 161.84 500.00 324.37

11 500.00 338.60 223.64 172.23 500.00 331.35

12 500.00 344.99 233.55 181.64 500.00 337.69

13 500.00 350.78 242.55 190.21 500.00 343.47

14 500.00 356.05 250.76 197.99 500.00 348.77

15 500.00 360.87 258.26 205.08 500.00 353.63

16 500.00 365.28 265.12 211.56 500.00 358.10

17 500.00 369.32 271.41 217.48 500.00 362.23

18 500.00 373.04 277.19 222.91 500.00 366.04

19 500.00 376.46 282.50 227.90 500.00 369.56

20 500.00 379.62 287.40 232.49 500.00 372.82

Table 2: tB4 to tC4, Step 0 to 20.

tB4 tB5 tC0 tC1 tC2 tC3 tC4tC4

60.00 60.00 500.00 500.00 500.00 60.00 60.00

60.00 60.00 500.00 500.00 500.00 148.00 60.00

77.60 60.00 500.00 500.00 500.00 183.20 77.60

98.72 67.04 500.00 500.00 500.00 211.36 91.68

117.73 79.43 500.00 500.00 500.00 230.37 107.87

136.12 92.90 500.00 500.00 500.00 246.00 122.18

152.58 107.30 500.00 500.00 500.00 258.43 136.29

167.92 120.89 500.00 500.00 500.00 269.25 149.17

181.77 133.88 500.00 500.00 500.00 278.64 161.40

194.50 145.76 500.00 500.00 500.00 287.12 172.68

206.05 156.76 500.00 500.00 500.00 294.80 183.26

216.64 166.76 500.00 500.00 500.00 301.87 193.06

226.29 175.94 500.00 500.00 500.00 308.38 202.23

235.14 184.31 500.00 500.00 500.00 314.44 210.76

243.24 191.97 500.00 500.00 500.00 320.06 218.72

250.69 199.00 500.00 500.00 500.00 325.31 226.15

257.54 205.44 500.00 500.00 500.00 330.20 233.08

263.86 211.37 500.00 500.00 500.00 334.77 239.55

269.68 216.83 500.00 500.00 500.00 339.04 245.60

275.07 221.86 500.00 500.00 500.00 343.03 251.24

280.05 226.51 500.00 500.00 500.00 346.76 256.52


45

Table 3: tC5 to tD5, Step 0 to 20.

tC5 tD0 tD1 tD2 tD3 tD4 tD5tD5

60.00 60.00 60.00 60.00 60.00 60.00 60.00

60.00 148.00 148.00 148.00 60.00 60.00 60.00

60.00 200.80 200.80 183.20 95.20 60.00 60.00

67.04 236.00 232.48 211.36 116.32 70.56 60.00

75.21 259.94 255.71 230.37 136.03 79.71 65.63

86.47 278.24 272.75 246.00 151.24 90.69 71.84

97.64 292.53 286.62 258.43 164.92 100.94 79.96

109.21 304.51 298.03 269.25 176.73 111.45 88.13

120.12 314.68 307.94 278.64 187.63 121.48 96.76

130.60 323.61 316.61 287.12 197.57 131.37 105.20

140.36 331.51 324.37 294.80 206.91 140.85 113.59

149.54 338.60 331.35 301.87 215.64 150.03 121.68

158.06 344.99 337.69 308.38 223.89 158.82 129.53

166.02 350.78 343.47 314.44 231.66 167.25 137.03

173.42 356.05 348.77 320.06 239.01 175.28 144.21

180.32 360.87 353.63 325.31 245.94 182.93 151.03

186.75 365.28 358.10 330.20 252.48 190.19 157.52

192.74 369.32 362.23 334.77 258.66 197.08 163.65

198.31 373.04 366.04 339.04 264.47 203.59 169.45

203.52 376.46 369.56 343.03 269.95 209.73 174.92

208.37 379.62 372.82 346.76 275.09 215.53 180.07

Table 4: tE0 to tF0, Step 0 to 20.

tE0 tE1 tE2 tE3 tE4 tE5 tF0tF0

60.00 60.00 60.00 60.00 60.00 60.00 60.00

60.00 60.00 60.00 60.00 60.00 60.00 60.00

77.60 77.60 77.60 60.00 60.00 60.00 60.00

98.72 98.72 91.68 70.56 60.00 60.00 67.04

119.84 117.73 107.87 79.71 64.22 60.00 79.43

138.93 136.12 122.18 90.69 68.73 62.82 94.59

156.80 152.58 136.29 100.94 75.15 66.33 109.59

172.82 167.92 149.17 111.45 81.94 71.52 124.33

187.50 181.77 161.40 121.48 89.58 77.27 137.88

200.72 194.50 172.68 131.37 97.42 83.79 150.45

212.76 206.05 183.26 140.85 105.55 90.56 161.84

223.64 216.64 193.06 150.03 113.67 97.56 172.23

233.55 226.29 202.23 158.82 121.77 104.55 181.64

242.55 235.14 210.76 167.25 129.70 111.50 190.21

250.76 243.24 218.72 175.28 137.44 118.29 197.99

258.26 250.69 226.15 182.93 144.91 124.88 205.08

265.12 257.54 233.08 190.19 152.11 131.22 211.56

271.41 263.86 239.55 197.08 158.99 137.31 217.48

277.19 269.68 245.60 203.59 165.55 143.11 222.91

282.50 275.07 251.24 209.73 171.79 148.63 227.90

287.40 280.05 256.52 215.53 177.70 153.86 232.49


46

Table 5: tF1 to tF5, Step 0 to 20.

tF1 tF2 tF3 tF4 tF5 q/kL

60.00 60.00 60.00 60.00 60.00 0.00

60.00 60.00 60.00 60.00 60.00 0.00

60.00 60.00 60.00 60.00 60.00 123.20

67.04 67.04 60.00 60.00 60.00 200.64

79.43 75.21 65.63 60.00 60.00 261.18

92.90 86.47 71.84 62.82 60.00 293.42

107.30 97.64 79.96 66.33 62.25 317.64

120.89 109.21 88.13 71.52 65.45 331.90

133.88 120.12 96.76 77.27 70.10 343.44

145.76 130.60 105.20 83.79 75.40 351.51

156.76 140.36 113.59 90.56 81.37 358.61

166.76 149.54 121.68 97.56 87.59 364.39

175.94 158.06 129.53 104.55 94.01 369.74

184.31 166.02 137.03 111.50 100.43 374.51

191.97 173.42 144.21 118.29 106.78 379.01

199.00 180.32 151.03 124.88 112.98 383.19

205.44 186.75 157.52 131.22 118.99 387.16

211.37 192.74 163.65 137.31 124.77 390.90

216.83 198.31 169.45 143.11 130.29 394.44

221.86 203.52 174.92 148.63 135.55 397.80

226.51 208.37 180.07 153.86 140.55 400.99

Step 2: Make new temperature distribution layout after specified time interval, say 20 time intervals.

Step 3: Make graph for heat transfer rate against time interval. Remember that ∆θ = 0.470455 hr.

Use q/kL as ordinate and time interval steps (one step = ∆x) as the abscissa. Use Simpson rule to get the total heat transferred after the specified time interval.

Using 20 time interval (9.4091 hrs).


47

Simpson’s Rule:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]n1n4321o

b

axgx4g...x2gx4gx2gx4gxg

3

∆xdxxg +++++++≈ −∫

Where:

( )

kL

qxg =

hr470455.0x =θ∆=∆

n = 20, n-1 = 19

( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )

+++++

+++++

++++

++++++

=∫

400.99397.804394.442390.904387.162

383.194379.012374.514369.742364.394

358.612351.514343.442331.904317.642

293.424261.1824(200.64)123.32040

3

0.470455dxxg

b

a

( ) 2924dxxgb

a=∫

For one quarter of flue.

Btu/ft6426k2924L

Σq=⋅=

For one whole flue:

Btu/ft25705k11684

L

Σq=⋅=

Ans

- end -

48

CHAPTER 4

RADIANT HEAT TRANSMISSION

Chapter 4 Radiant Heat Transmission

49

4_001 A large plane, perfectly insulated on one face and maintained at a fixed temperature T1 on the bare face,

which has an emissivity of 0.90, loses 200 Btu/(hr)(sq ft) when exposed to surrounding at absolute zero. A second surface having the same size as the first is also perfectly insulated on one face, but its bare face has an emissitivity of 0.45. When the bare face of the second plane is maintained at a fixed temperature T

2 and

exposed to surroundings at absolute zero, it loses 100 Btu/(hr)(sq ft). Let these two planes be brought close together, so that the parallel bare faces are only 1 in. apart, and let the

heat supply to each be so adjusted that their respective temperatures remain unchanged. What will be the net heat flux between the planes, expressed in Btu/(hr)(sq ft)?

Solution:

4

11

411

100

T0.171εσTε

A

q

==

ε

1 = 0.90

( )4

1

100

T0.900.171200

=

T

1 = 600.41 R

4

22

422

100

T0.171εσTε

A

q

==

ε

2 = 0.45

( )4

2

100

T0.450.171100

=

T

2 = 600.41 R

Equation 4-5.

( )

++

−=⇔

1ε

1

ε

1

1TTσAq

21

42

41121

Since T

1 = T

2 = 600.41 R

sqft perhrperBtu 0q 21 =⇔ Ans.

4_002 A furnace having walls 28.25 in. thick contains a peephole 7 by 7 in. in cross section. If the temperature of

the inner walls of the furnace is 2200 F, what would be the heat loss through the peephole to surroundings at 70 F?

Solution:


50

A

1 = A

2 = 7 in x 7 in = 49 sq in = 0.340278 sq ft.

Assume there is wall reradiation: Use Figure 4-11. Ratio = (Smaller Side) / (Distance b/w planes) Ratio = 7 in / 28.25 in = 0.2478 Line No. 6, Square planes connected by non-conducting plae but reradiating walls.

0.2367F =

( )4

24

112121 TTFσAq −=⇔ T

1 = 2200 F + 460 F = 2660 F

T2 = 70 F + 460 F = 530 F

−

=⇔

4

2

4

112121

100

T

100

TF0.171Aq

( )( )

−

=⇔

4

421

100

530

100

26600.23670.3402780.171q

ftsqBtu/hr6885q 21 −=⇔ Ans.

4_003 An electric furnace of rectangular cross section is to be designed for batch heating of a stock from 70 to 1400

F. The hearth, covered with stock, is 6 by 12 ft in area. The refractory side walls are well insulated. parallel to the plane of the roof, in a plane several inches below it, is a system of round-rod resistors, each 10 ft long and 0.5 in. in diameters, spaced on 2-in. centers. The plane of resistors is 4 ft above the top of the stock.

What is the heating time for a 6-ton batch of stock having a mean specific heat of 0.16, when the resistor temperature is maintained at 2000 F?

Notes: Assume that the emissivities of the resistors and stock are 0.6 and 0.9, respectively, and neglect heat losses and heat storage in the walls of the furnace.

Solution:


51

A

1 = Area of resistors

A2 = Imaginary Plane

A3 = Area of stock

Other data:

ε1 = 0.6

ε3 = 0.9

T

1 = T

2 = 2000 F + 460 F = 2460 F

T3 = (1/2)(70 F + 1400 F) + 460 F = 1195 F

First part: Solving for emissivity of equivalent plane A

2.

212 ℑ=ε

Equation 4-33:

2121122212 FA

11

1

A

11

1

A

1

A

1+

−

ε+

−

ε=

ℑ

2111

2

221 F

11

ε

1

A

A1

ε

11+

−+

−=

ℑ

( )( )

( )( ) 5π

288

1012

0.5π

122

A

A

1

2 ==

e1 = 0.6 e2 = 1.0 From Fig. 4-12, Line No. 5 Ratio = 2/0 in / 0.5 in = 4.0

0.6F21 =

0.6

11

0.6

1

5π

2881

1

11

21

+

−+

−=

ℑ

0.621 =ℑ

0.6ε 212 =ℑ=


52

Second Part: New Area: A

2 = (6)(12)= 72 sq ft

ε2 = 0.072

A3 = (6)(12) = 72 sq ft

ε3 = 0.9

Equation 4-33: A

2 = A

3

233223 F

11

ε

11

ε

11+

−+

−=

ℑ

Figure 4-11, Line No. 7 6 ft x 12 ft, 4 ft apart Ratio = 6 ft / 4 ft = 1.5

0.69F23 =

0.69

11

0.9

11

0.072

11

23

+

−+

−=

ℑ

0.6923 =ℑ

T

2 = T

1

( )4

34

223232 TTσAq −ℑ=⇔

( )( )( )

−

=

44

net100

1195

100

24600.0692720.171q

q

net = 294.640 Btu/hr

let θ is the heating time. Weight = 6 tons = 12,000 lb

qnet

= (12,000)(0.16)(1400 - 70) / θ

qnet

= 2,553,600 / θ

Then:

2,553,600 / θ = 294,640

θθθθ = 8.667 hrs.... Ans. 4_004 An annealing furnace 10 ft long has a cross section normal to length, as shown in Fig. 4-28.The firebox a is

at a uniform temperature of 2200 F, and the 10- by 6-ft hearth b is covered with stock at a temperature of 1400 F. So far as radiant-heat transfer is concerned, assume that the firebox acts like a uniformly black plane c, 2 ft high over the bridge wall. Neglect the contribution of the combustion products to the radiant-heat interchange in the system, and neglect convection. Assume no external losses from the furnace, and assume that the refractory surfaces have emissivities of 0.65. a. Calculate the direct interchange of heat by radiation between the firebox and the stock, if the latter

is a black surface.


53

b. Calculate the total net interchange between the two if the stock is a black surface. c. Repeat b, assuming that the stock has an emissivity of 0.75

d. Calculate the average temperature of the working chamber walls and roof for condition c.

Solution: T

1= 2200 F + 460 F = 2660 F (Firebox)

T2 = 1400 F + 460 F = 1860 F (Stock)

A

1 = 2 ft x 10 ft = 20 sq ft

A2 = 6 ft x 10 ft = 60 sq ft

ε1 = 1.0

(a) ε2 = 1.0

Direct Interchange of Heat:

( )4

24

112121 TTFσAq −=⇔ Solving for F

12


54

ft7.211164BD

ft8.485366AD

ft4BC

ft6AC

22

22

=+=

=+=

=

=

( )

( ) ( )L

2

BCADBDACFAFA 212121

+−+=≡

( )

( ) ( )( ) ftsq3.62910

2

48.48537.21116FAFA 212121 =

+−+=≡

Then: F

12 = 3.629 sq ft / 20 sq ft = 0.18145

F21

= 3.629 sq ft / 60 sq ft = 0.06048

( )( )

−

=⇔

44

21100

1860

100

26603.6290.171q

Ans.Btu/hr236,404q 21 −=⇔

(b) Total net interchange if stock is a black surface.

( )4

24

112121 TTFσAq −=⇔ Equation 4-21a.

+

+=

2R21R1

121121

FA

1

FA

1

1FAFA

−=

=

12

12121

12

1R1211R1

F

F1FA

F

FFAFA

( )

−=

0.18145

0.1814513.629FA 1R1

ftsq16.371FA 1R1 =

−=

=

21

21212

21

2R2122R2

F

F1FA

F

FFAFA

( )

−=

0.06048

0.0604813.629FA 2R2

ftsq56.3743FA 2R2 =

+

+=

2R21R1

121121

FA

1

FA

1

1FAFA

+

+=

56.3743

1

16.371

1

13.629FA 121

ftsq16.316FA 121 =


55

0.8158ftsq20/ftsq16.316F12 ==

( )( )

−

=⇔

44

21100

1860

100

266016.3160.171q

Ans.Btu/hr1,062,872q 21 −=⇔

(c) Repeat (b) if stock emissivity, ε2 = 0.75.

ε1 = 1.0

( )4

24

112121 TTσAq −ℑ=⇔ Equation 4-33.

1212211121 FA

11

1

A

11

1

A

1

A

1+

−

ε+

−

ε=

ℑ

1222

1

112 F

11

ε

1

A

A1

ε

11+

−+

−=

ℑ

0.8158

11

0.75

1

60

201

1

11

12

+

−+

−=

ℑ

0.690612 =ℑ

( )( )( )

−

=⇔

44

21100

1860

100

26600.6906200.171q

Ans.Btu/hr899,754q 21 −=⇔

(d) Average temperature of the working chamber wall and roof for condition (c). Chamber emissivity =

0.65. Equation 4-92.

ba

bTaTT

42

414

R+

+=

( )

⋅+++

−−= 2R1R2R1R121R

ε1

εA

ε1

εAa

2

22

1

11

( )

⋅+++

−−= 2R1R2R1R122R

ε1

εA

ε1

εAb

1

11

2

22

0.75ε

1ε

2

1

=

=

ftsq56.3743FA2R

ftsq3.629FA12

ftsq16.371FA1R

2R2

121

1R1

==

==

==

A

1 = 20 sq ft, A

2 = 60 sq ft

( )( )( ) ( )( ) ( )( )

+++

−−= 56.374316.37156.374316.3713.62916.371

0.751

0.7560

ε1

20εa

1

1


56

1

1

ε1

82674εa

−=

( )( )( ) ( )( ) ( )( )

+++

−−= 56.374316.37156.374316.3713.62956.3743

ε1

20ε

0.751

0.7560b

1

1

+

−= 1187

ε1

1127.5ε180b

1

1

Note: It is mentioned that the solutions for problems involving gray gas in a gray enclosure have never involved the

emissivity of the no-flux surfaced so chamber emissivity is useless here.

ba

bTaTT

42

414

R+

+=

++

++

=

1187ε-1

1127.5ε180

ε-1

82764ε

T1187ε-1

1127.5ε180T

ε-1

82764ε

T

1

1

1

1

42

1

141

1

1

4R

( )( )( )( )111

4211

4114

Rε-111871127.5ε18082764ε

Tε-111871127.5ε180T82764εT

++

++=

ε

1 = 1.0

285624

202950T82674TT

42

414

R

+=

( ) ( )285624

1860202950266082674T

444

R

+=

T

R = 2190 R or 1730 F . . . . Ans.

4_005 On a clear night, when the effective black-body temperature of space is minus 100 F, the air is at 60 F and

contains water vapor at a partial pressure equal to that of ice or water at 32 F. A very thin film of water, initially at 60 F, is placed in a very shallow well-insulated pan, placed in a spot sheltered from the wind (h = 0.46), with a full view of the sky.

State whether ice will form, and support this with suitable calculations. Solution: Using Equation 4-57.

( )4

1G14

GG Tα-Tεq/A σ=

Equation 4-58 for water vapor only.

wLP,Tw,G CεεwG

⋅=

wG1 αα =

w

0.45

1

G/TLTP,Tw,w C

T

Tεα

G1w1

=


57

TG = 60 F + 460 F = 520 F

T1 = -100 F + 460 F = 360 F

Water vapor at 32 F, P

w = 0.006 atm.

Using Fig. 4-15 to Determine P

wL when will ice will form.

(1) P

wL = 0.01 ft.atm

TG = 520 F,

0.034ε LP,Tw, wG

=

T1 = 360 F

PwLT

1/T

G = 0.007 ft.atm

0.022ε

G1w1 /TLTP,Tw, =

Fig. 4-16, Cw ~ 1.0

εG = (0.034)(1) = 0.034

αG1

= αw = (0.022)(520 / 360)

0.45 (1.0) = 0.026

( ) ( )

−

=

44

100

3600.026

100

5200.0340.171

A

q

q/A = 3.5042 Btu per hr per sq ft Let t

p = temperature of water in pan.

q/A = h( tG - t

p )

3.5042 = 0.46 ( 60 F - tp )

t

p = 52.4 F > 32 F

(2) P

wL = 0.1 ft.atm

TG = 520 F,

0.13ε LP,Tw, wG

=

T1 = 360 F

PwLT

1/T

G = 0.07 ft.atm

0.126ε

G1w1 /TLTP,Tw, =

Fig. 4-16, Cw ~ 1.0

εG = (0.13)(1) = 0.13

αG1

= αw = (0.126)(520 / 360)

0.45 (1.0) = 0.149

( ) ( )

−

=

44

100

3600.149

100

5200.130.171

A

q



q/A = h( tG - t

p )

11.9743 = 0.46 ( 60 F - tp )

t

p = 34 F > 32 F


58

(3) P

wL = 0.12 ft.atm

TG = 520 F,

0.14ε LP,Tw, wG

=

T1 = 360 F

PwLT

1/T

G = 0.10 ft.atm

0.149ε

G1w1 /TLTP,Tw, =

Fig. 4-16, Cw ~ 1.0

εG = (0.14)(1) = 0.14

αG1

= αw = (0.149)(520 / 360)

0.45 (1.0) = 0.176

( ) ( )

−

=

44

100

3600.176

100

5200.140.171

A

q



q/A = h( tG - t

p )

12.449 = 0.46 ( 60 F - tp )

t

p = 33 F > 32 F

Then, P

wL = 0.1 ft.atm , t

p = 34 F

PwL = 0.12 ft.atm, t

p = 33 F

Extrapolation at 32 F, P

wL = 0.24 ft.atm

Mean beam length = L = (0.14 ft.atm) / (0.006 atm) = 23.4 ft. Table 4-3, P

wL = 0.14

Ratio for H

2O = L / L

o = 0.9141

Table 4-2, L

o = 2X

Distance b/w parallel planes, X.

( ) ( )ft12.8

0.91412

23.4

0.91412

L

2

LX o ====

The ice will form if the distance between the blackbody space of -100 F is 12.8 ft or more above the shallow

pan. 4_006 The convection section of an oil pipe still on which performance data are available is composed of a bank of

tubes 24 ft long, 3.5 in. i.d., and 4.0 in. o.d. There are six tubes in each horizontal row; center-to-center spacing of the tubes, arranged on equilateral triangular centers, is 8 in. The minimum free area for gas flow is


59

53.8 sq ft. Oil enters the bottom row of tubes at 420 F and flows upward through each row in series, leaving the convection section at 730 F. Flue gas at atmospheric pressure from the combustible chamber, flowing transverse to the tubes, enters the top of the bank at 1550 F, with a mass velocity of 810 lb / (hr)(sq ft of minimum free area) and leaves at the bottom of the section at 590 F.

The flue gas contains 7.1 per cent CO2 (dry basis); the ratio of H

2O:CO

2 in the flue gas is 1.38; the molecular

weight of the flue gas is 27.5; the mean molal heat capacity between 1550 and 590 F is 7.77. Calculate the number of rows of tubes required. Additional Data and Assumptions. 1. External heat losses in the section are negligible.

2. The emissivities of all surfaces are 0.9. Because of the high emissivities, radiation received on a surface by reflection may be ignored.

3. Preliminary calculations indicate that the temperature drop through the oil film and tube wall is 30 F.

4. Average length of radiant beams through the gas will differ for the tube surface and for the refractory surface, but the latter will be assumed equal to the former. Likewise, the convection coefficient of heat transfer from gas to refractory will be assumed the same as from gas to tubes.

5. The effect of the refractory end walls, through which the tubes pass, will be neglected. 6. For convection heat transfer from flue gases, simplified Eq. (10-11b) is satisfactory.

Solution:

Equation 4-109 Modified to petroleum heaters:

( ) ( )1G11

41

4G1G1G TT'AhTTσAq −+−ℑ=

Equation 4-110 Modified:

( ) ( )G1G2pG0GpGLG TTcwTTcwqiq −=−−−=

T

1 = (1/2)(420 F + 730 F) + 30 F + 460 F = 1065 R

T

G1 = 1550 F + 460 F = 2010 R

T

G2 = 590 F + 460 F = 1050 R

c

p = 7.77 / 27.5 = 0.2825 Btu/lb-F

T

G = (1/2)(T

G1 + T

G2) = 1530 R


60

Equation 4-103.

⋅

−

+

+

+

−

=ℑ

R1G

G

R1

G

11

1G1

F

1

x

ε1

x

ε

1

AA

x

ε

11

ε

1

A

1

xA

ε1 = 0.9

Flow area = 53.8 sq ft 53.8 sq ft = ( s - 6 (4/12))( 24 ) s = 4.242 ft Let N be the number of rows of tubes. Refractory end wall neglected. A

R = 2N(6.93 / 12)(24) = 27.72 N

A1 = A

1’ = 6π(4/12)(24)(N) = 150.80 N

Figure 4-12, Ratio = 8 in / 4 in = 2, Line 2. F

R1 = 0.30

Equation 4-58.

GwGcG TwLP,Tw,cLP,Tc,G ∆εCεCεε −⋅+⋅=

Table 4-2 X = 2.121 ft, Lo = 2X = 2 (2.121 ft) = 4.242 ft Table 4-3 P

G = 1 atm, P

c = 0.071 atm

Pw = 1.38 (0.071 atm) = 0.099 atm

Solving PcL for CO

2 by Table 4-3.

assume ratio L/L0 = 0.7954

L = 0.7954 (4.242) L = 3.374 ft PcL = (0.071 atm)(3.374 ft) = 0.24 ft.atm Then Table 4-3, Ratio L/L

0 = 0.7954

Solving PwL for H

20 by Table 4-3.

assume ratio L/L0 = 0.9052 L = 0.9052 (4.242)

L = 3.840 ft PcL = (0.099 atm)(3.840 ft) = 0.38 ft.atm Then Table 4-3, Ratio L/L

0 = 0.9052

Then: PcL = 0.24 ft.atm, PwL = 0.38 ft.atm


61

TG = 1530 R

Figure 4-13

0.10ε LP,Tc, cG

=

Figure 4-15

0.155ε LP,Tw, wG

=

Figure 4-14: P

T = 1 atm:

Cc = 1/0

Figure 4-16 (P

w + P

T)/2 = (0.099 + 1)/2 = 0.55

PwL = 0.38

Cw = 1.05

Figure 4-17: P

w / (P

c + P

w) = (0.099) / (0.071 + 0.099) = 0.5824

PcL + P

wL = 0.24 + 0.38 = 0.62 ft/atm

TG = 1530 R (1070 F)

0.01∆εTG =

Substitute in Equation 4-58.

εG = (0.1)(1.0) + (0.155)(1.05) - 0.01 = 0.2528

Solving for e

2.G (Double the path)

2P

cL = 2(0.24) = 0.48

2PwL = 2(0.38) = 0.76

TG = 1530 R

0.12ε LP,Tc, cG

=

0.22ε LP,Tw, wG

=

Figure 4-14: P

T = 1 atm:

Cc = 1.0

Figure 4-16 (P

w + P

T)/2 = (0.099 + 1)/2 = 0.55

2PwL = 0.76

Cw = 1.02

Figure 4-17: P

w / (P

c + P

w) = (0.099) / (0.071 + 0.099) = 0.5824

2PcL + 2P

wL = 0.48 + 0.76 = 1.24 ft/atm

TG = 1530 R (1070 F)

0.02∆εTG =


62

Substitute in Equation 4-58.

εG = (0.12)(1.0) + (0.22)(1.02) - 0.02 = 0.3244

Equation 4-102.

G2G

2G

ε2ε

εx

⋅−=

Then:

⋅

−

+

+

+

−

=ℑ

R1G

G

R1

G

11

1G1

F

1

x

ε1

x

ε

1

AA

x

ε

11

ε

1

A

1

xA

⋅

−

+

+=

⋅

−

+

+

30

1

0.3527

0.25281

0.3527

0.2528

1

27.72N150.80N

F

1

x

ε1

x

ε

1

AA

R1G

G

R1

153.74N

F

1

x

ε1

x

ε

1

AA

R1G

G

R1 =

⋅

−

+

+

( )

35.947N

153.74N0.3527

0.2528

11

0.9

1

150.8N

1

0.3527A 1G1 =

+

−

=ℑ

Equation 10-11b.

0.4

o0.6

maxp1 /DG0.133ch =

G

max = 810 lb/hr-sq ft

cp = 0.2825 Btu/lb-F

Do = 4 in = 1/3 ft

( )( ) ( )0.40.6

1 31/8100.28250.133h =

h

1 = 3.242 Btu / hr-sq ft- F

q

G = w

Gc

p( T

G2 - T

G1 )


63

wG = (810 lb/hr-sq ft)(53.8 sq ft) = 43,578 lb/hr

q

G = (43,578)(0.2825)(2010 - 1050) = 11,818,354 Btu/hr

( ) ( )1G11

41

4G1G1G TT'AhTTσAq −+−ℑ=

( )( )

( )( )( ) 485,098N10651530150.8N3.242

100

1065

100

15300.17135.947Nq

44

G

=−+

−

=

Then: 485,098N = 11,818,354 N = 24.4 say 25 rows of tubes - Ans.

- end -

64

CHAPTER 5

DIMENSIONAL ANALYSIS

Chapter 5 Dimensional Analysis

65

5_001 A gas having a molecular weight of 86 is flowing at a rate of 1.38 kg/min through a tube having an actual i.d. of 1.20 in. At a cross section where the absolute pressure is 330 atm and the temperature is 340 F, the true density is 1.2 times

that predicted by the perfect-gas law, and the viscosity is 2.0 x 10-6 force-pound x seconds per square foot. Calculate

the numerical value of the dimensionless Reynolds number. Solution: Reynolds number:

NRe

= DG/µ

µF = 2.0 x 10-6 Force-pound x seconds per square foot

µ = µFg

c

µ = (2.0 x 10-6)(32.2)

µ = 6.44 x 10-5 lb matter / sec-ft D = 1.2 in = 0.1 ft

G = 4w/πD2

w = 1.38 kg/min = 0.050715 lb/sec

G = 4(0.050715)/[π(0.1)2] = 6.457234

NRe

= (0.1)(6.457234)/6.44 x 10-5

NRe

= 10,027 . . . Ans.

5_002 It is desired to investigate the drag force on geometrically similar solid objects placed in a free stream of flowing fluid.

In the general case, the force F exerted on the body is thought to depend on the characteristic length of the object x,

the approach velocity V, the acceleration due to gravity g, and the following properties of the fluid: the density ρ, the

viscosity µ, the surface tension σ, and the velocity Va at which sound waves are propagated in the liquid.

a. What is the number nπ of algebraically independent dimensionless groups which may be used to correlate

the variables? b. Determine a set of these n

π dimensionless groups, using the force F in only one group.

c. Calculate the numerical value of each of these groups except the one in which the force F appers, for the condition in which x = 7 inches, g = 32.2 ft/(sec)(sec), V = 11 ft/sec, and the fluid is water at 68 F (assume V

a = 4900

ft/sec). Solution: Variables -

aVσ,µ,ρ,g,V,x,F,

Then:

( ) 0Vσ,µ,ρ,g,V,x,F, a =φ

Dimension of Factors: MLθF

Dimensions Factors

F x V g ρ µ σ Va

M 1 1

L 1 1 1 -3 1 -1 1

θ -1 -2 -1 -1

F 1 1 a. n

f = no. of physical factors of importance

nf = 8

ni = nd = number of dimension

ni = 4


66

nπ = no. of independent dimensionless group

nπ = n

f - n

i = 8 - 4 = 4

b. Incompatible factors

( ) ( ) ( )Lθ

M&,T

1,L

M,Lx 3 =µ=β

=ρ=

FF

µ

ρx

xµ

ρx

Lµ

Mθ

ρxρLM

xL

23

33

=

===

==

=

Remaining factors:

aVσ,g,V,

µ

xV;

ρx

µ

µρx

x

θ

LV

F

ρx;

x

F

L

Fσ

µ

gxρ;

xρ

µ

µρx

x

θ

Lg

µ

xV;

ρx

µ

µρx

x

θ

LV

a42a

3

2

32

232

2

22

2

12

ρ=Π===

=Π==

=Π=

==

ρ=Π===

c. Values: x = 7 inches = 0.583333 ft g = 32.2 ft/(sec)(sec) V = 11 ft/sec V

a = 4900 ft/sec

at 68 F

ρ = 62.282 lb/cu ft

µ = 6.75 x 10-4 lb/(sec)(ft)

σ = 72.7 Dynes/cm2 = 0.67534 N/ft

2

( )( )( )

( ) ( ) ( )

( )( )( )263,737

106.75

62.28249000.583333

µ

ρxV

36,730106.75

32.20.58333362.282

µ

gxρ

592,063106.75

62.282110.583333

µ

xV

4

a4

4

32

2

32

2

41

=×

==Π

=×

==Π

=×

=ρ

=Π

−

−

−

5_003 Below are some data from an early article on the performance of a packed regenerator. In heating runs, the packed

spheres were initially at an elevated uniform temperature, and air at room temperature entered at constant mass rate. In cooling runs, the spheres were initially at room temperature, and air at approximately 200 F entered at constant mass rate. In the data given below the relation between the temperature of entering air (t

1), the temperature of exit air


67

(t2), and initial uniform temperature (t

0) of the bed of spheres is given by the dimensionless term y = (t

2 - t

0)/(t

1 - t

0). The

heat capacity of the vertical cylindrical column was small compared with that of the spheres used for packing, and the column was well insulated.

Correlate all the values of y given below, plotted as ordinates against some suitable abscissa:

Material Steel Steel Steel Steel Lead Glass

Bed depth, inches ..... 5.125 10.25 5.125 2.56 9.55 9.71

Bed diam, in .............. 4 8 4 2 8 8

Particle diam, in ......... 0.125 0.25 0.125 0.0625 0.233 0.237

Sp ht .......................... 0.111 0.111 0.111 0.111 0.0347 0.168

Sp gr ......................... 7.83 7.83 7.83 7.83 11.34 2.6

k, Btu/(hr)(ft)(deg F) .. 26 26 26 26 19.5 0.4

Air superficial mass

vel., lb/(sec)(sq ft) ..... 0.15 0.15 0.3 0.3 0.3 0.3

y θ, θ, θ, θ, θ, θ,

sec sec sec sec sec sec

0.1 300 600 140 . . . 113 127

0.2 360 680 160 . . . 133 149

0.3 380 740 190 . . . 147 166

0.4 410 810 205 100 162 182

0.5 450 890 220 . . . 176 199

0.6 480 970 240 . . . 191 215

0.7 510 1050 250 . . . 206 232

0.8 570 1140 300 140 222 250

0.9 650 1300 330 170 248 280

The quantity θ is the elapsed time at which the indicated y was measured. Solution: L - ft D

t - ft

Dp - ft

cp - Btu/lb-ft

ρ - lb/cu ft k - Btu/sec-fr-deg F G

o - lb/sec-sq ft

θ - time, sec

MLθTFH System:

( )Hoppt Kq,,Gk,ρ,,c,D,DL,y φ=

p

Hnm

oife

pc

pb

ta KθGkρcDDαLy =

( ) ( ) ( ) ( )p

2

2n

m

2

if

3

ecba

Hθ

MLθ

θL

M

θLT

H

L

M

MT

HLLL1

=

(1) ΣL: a + b + c - 3f - i - 2m + 2p = 0


68

(2) ΣL: e + i - p = 0

(3) ΣM: -e + f + m + p = 0

(4) ΣT: -e - i = 0

(5) Σθ: -i - m + n - 2p = 0 (4) i = -e (5) e + i - p = 0 e - e - p = 0 p = 0 (3) -e + f + m + p = 0 -e + m + n = 0 e = m - n Equating: e = f + m = m - n (1) a + b + c - 3f - i - 2m + 2p = 0 a + b + c - 3f + e -2m = 0 a + b + c - 3f + e - 2(e - f) = 0 a + b + c - 3f + e - 2e + 2f = 0 a + b + c -f -e = 0 a + b + c = f + e f = a + b + c -e Then: p = 0 a = a b = b c = c e = e f = a + b + c -e i = -e m = e -f = 2e - a - b -c n = -f = e - a - b -c Then:

cbaecba2e

oeecbae

pc

pb

ta θGkρcDDαLy −−−−−−−−++=

e2

op

c

o

p

b

o

t

a

o rk

θGc

θG

ρD

θG

ρD

θG

Lραy

=

Note: Strouhal Numbers:

θG

LρN

oLSl, =

θG

ρDN

o

tDSl, t

=

θG

ρDN

o

pDSl, p

=

Ratio of Peclet Number to Strouhal Number:

ρk

θGc

N

N2

op

Sl

Pe =


69

Steel # 1: L = 5.125 in = 0.4271 ft D

t = 4 in = 0.3333 ft

Dp = 0.125 in = 0.01042 ft

cp = 0.111 Btu/lb-F

ρ = 7.83 x 62.4 = 488.6 lb/cu ft

k = 26 Btu/hr-ft-deg F = 7.2222 x 10-3 Btu/sec-ft-deg F

Go = 0.15 lb/sec-sq ft

( )( )θ

1391.2

0.15θ

488.60.4271

θG

LρN

oLSl, ===

( )( )θ

1085.7

0.15θ

488.60.3333

θG

ρDN

o

tDSl, t

===

( )( )θ

33.9414

0.15θ

488.60.01042

θG

ρDN

o

pDSl, p

===

( )( )( )( ) θ107.07754

107.2222488.6

θ0.150.111

ρk

θGc

N

N 4

3

22op

Sl

Pe −

−×=

×==

Table No. 1

X1N LSl, =

X2N

tDSl, =

X3N

pDSl, =

X4N

N

Sl

Pe =

y θ X1 X2 X3 X4

0.1 300 4.637333 3.619000 0.113138 0.212326

0.2 360 3.864444 3.015833 0.094282 0.254791

0.3 380 3.661053 2.857105 0.089319 0.268947

0.4 410 3.393171 2.648049 0.082784 0.290179

0.5 450 3.091556 2.412667 0.075425 0.318489

0.6 480 2.898333 2.261875 0.070711 0.339722

0.7 510 2.727843 2.128824 0.066552 0.360955

0.8 570 2.440702 1.904737 0.059546 0.403420

0.9 650 2.140308 1.670308 0.052218 0.460040 Steel # 2: L = 10.25 in = 0.8542 ft D

t = 8 in = 0.6667 ft

Dp = 0.25 in = 0.02083 ft

cp = 0.111 Btu/lb-F

ρ = 7.83 x 62.4 = 488.6 lb/cu ft




70

( )( )θ

2782.4

0.15θ

488.60.8542

θG

LρN

oLSl, ===

( )( )θ

2171.7

0.15θ

488.60.6667

θG

ρDN

o

tDSl, t

===

( )( )θ

67.8503

0.15θ

488.60.02083

θG

ρDN

o

pDSl, p

===

( )( )( )( ) θ107.07754

107.2222488.6

θ0.150.111

ρk

θGc

N

N 4

3

22op

Sl

Pe −

−×=

×==

Table No. 1

X1N LSl, =

X2N

tDSl, =

X3N

pDSl, =

X4N

N

Sl

Pe =

y θ X1 X2 X3 X4

0.1 600 4.637333 3.619500 0.113084 0.424652

0.2 680 4.091765 3.193676 0.099780 0.481273

0.3 740 3.760000 2.934730 0.091690 0.523738

0.4 810 3.435062 2.681111 0.083766 0.573281

0.5 890 3.126292 2.440112 0.076236 0.629901

0.6 970 2.868454 2.238866 0.069949 0.686521

0.7 1050 2.649905 2.068286 0.064619 0.743142

0.8 1140 2.440702 1.905000 0.059518 0.806840

0.9 1300 2.140308 1.670538 0.052193 0.920080 Steel # 3: L = 5.125 in = 0.4271 ft D

t = 4 in = 0.3333 ft

Dp = 0.125 in = 0.01042 ft

cp = 0.111 Btu/lb-F

ρ = 7.83 x 62.4 = 488.6 lb/cu ft



( )( )θ

695.6

0.30θ

488.60.4217

θG

LρN

oLSl, ===

( )( )θ

542.8

0.30θ

488.60.3333

θG

ρDN

o

tDSl, t

===

( )( )θ

16.97

0.30θ

488.60.01042

θG

ρDN

o

pDSl, p

===

( )( )( )( ) θ102.83102

107.2222488.6

θ0.300.111

ρk

θGc

N

N3

22op

Sl

Pe 3−

−×=

×==

Table No. 1


71

X1N LSl, =

X2N

tDSl, =

X3N

pDSl, =

X4N

N

Sl

Pe =

y θ X1 X2 X3 X4

0.1 140 4.968571 3.877143 0.121214 0.396343

0.2 160 4.347500 3.392500 0.106063 0.452963

0.3 190 3.661053 2.856842 0.089316 0.537894

0.4 205 3.393171 2.647805 0.082780 0.580359

0.5 220 3.161818 2.467273 0.077136 0.622824

0.6 240 2.898333 2.261667 0.070708 0.679445

0.7 250 2.782400 2.171200 0.067880 0.707755

0.8 300 2.318667 1.809333 0.056567 0.849306

0.9 330 2.107879 1.644848 0.051424 0.934237 Steel # 4: L = 2.56 in = 0.2134 ft D

t = 2 in = 0.1667 ft

Dp = 0.0625 in = 0.00521 ft

cp = 0.111 Btu/lb-F

ρ = 7.83 x 62.4 = 488.6 lb/cu ft



( )( )θ

347.6

0.30θ

488.60.2134

θG

LρN

oLSl, ===

( )( )θ

271.5

0.30θ

488.60.1667

θG

ρDN

o

tDSl, t

===

( )( )θ

8.485

0.30θ

488.60.00521

θG

ρDN

o

pDSl, p

===

( )( )( )( ) θ102.83102

107.2222488.6

θ0.300.111

ρk

θGc

N

N3

22op

Sl

Pe 3−

−×=

×==

Table No. 1

X1N LSl, =

X2N

tDSl, =

X3N

pDSl, =

X4N

N

Sl

Pe =

y θ X1 X2 X3 X4

0.4 100 3.476000 2.715000 0.084850 0.283102

0.8 140 2.482857 1.939286 0.060607 0.396343

0.9 170 2.044706 1.597059 0.049912 0.481273 Lead:


72

L = 9.55 in = 0.7958 ft D

t = 8 in = 0.6667 ft

Dp = 0.233 in = 0.01942 ft

cp = 0.0347 Btu/lb-F

ρ = 11.34 x 62.4 = 707.6 lb/cu ft

k = 19.5 Btu/hr-ft-deg F = 5.4167 x 10-3 Btu/sec-ft-deg F


( )( )θ

1877

0.30θ

707.60.7958

θG

LρN

oLSl, ===

( )( )θ

1572.5

0.30θ

707.60.6667

θG

ρDN

o

tDSl, t

===

( )( )θ

45.81

0.30θ

707.60.01942

θG

ρDN

o

pDSl, p

===

( )( )( )( ) θ108.14797

105.4167707.6

θ0.300.0347

ρk

θGc

N

N3

22op

Sl

Pe 4−

−×=

×==

Table No. 1

X1N LSl, =

X2N

tDSl, =

X3N

pDSl, =

X4N

N

Sl

Pe =

y θ X1 X2 X3 X4

0.1 113 16.610619 13.915929 0.405398 0.092072

0.2 133 14.112782 11.823308 0.344436 0.108368

0.3 147 12.768707 10.697279 0.311633 0.119775

0.4 162 11.586420 9.706790 0.282778 0.131997

0.5 176 10.664773 8.934659 0.260284 0.143404

0.6 191 9.827225 8.232984 0.239843 0.155626

0.7 206 9.111650 7.633495 0.222379 0.167848

0.8 222 8.454955 7.083333 0.206351 0.180885

0.9 248 7.568548 6.340726 0.184718 0.202070 Glass: L = 9.71 in = 0.8092 ft D

t = 8 in = 0.6667 ft

Dp = 0.237 in = 0.01975 ft

cp = 0.168 Btu/lb-F

ρ = 2.60 x 62.4 = 162.24 lb/cu ft

k = 0.4 Btu/hr-ft-deg F = 1.1111 x 10-4 Btu/sec-ft-deg F


( )( )θ

437.6

0.30θ

162.240.8092

θG

LρN

oLSl, ===

( )( )θ

360.6

0.30θ

162.240.6667

θG

ρDN

o

tDSl, t

===


73

( )( )θ

10.68

0.30θ

162.240.01975

θG

ρDN

o

pDSl, p

===

( )( )( )( ) 0.838766θ

101.1111162.24

θ0.300.168

ρk

θGc

N

N4

22op

Sl

Pe =×

==−

Table No. 1

X1N LSl, =

X2N

tDSl, =

X3N

pDSl, =

X4N

N

Sl

Pe =

y θ X1 X2 X3 X4

0.1 127 3.445669 2.839370 0.084094 106.523282

0.2 149 2.936913 2.420134 0.071678 124.976134

0.3 166 2.636145 2.172289 0.064337 139.235156

0.4 182 2.404396 1.981319 0.058681 152.655412

0.5 199 2.198995 1.812060 0.053668 166.914434

0.6 215 2.035349 1.677209 0.049674 180.334690

0.7 232 1.886207 1.554310 0.046034 194.593712

0.8 250 1.750400 1.442400 0.042720 209.691500

0.9 280 1.562857 1.287857 0.038143 234.854480 By curve fitting:

e2

op

c

o

p

b

o

t

a

o rk

θGc

θG

ρD

θG

ρD

θG

Lραy

=

α = 0.003593 a = -29.130232 b = 30.947910 c = -3.944493 e = -0.394509 Therefore:

-0.3945092

op

-3.944493

o

p

30.947910

o

t

-29.130232

o rk

θGc

θG

ρD

θG

ρD

θG

Lρ0.003593y

=

5_004 Two metals, A and B, are being considered for use in constructing tuning forks which are to produce vibrations of a

given frequency. What would be the ratio of the costs of the metal required for the two tuning forks, expressed as a function of the frequency, the physical properties of the metals, and the cost per unit mass of each metal (assumed independent of the quantity purchased)? The tuning forks are to be geometrically similar.

Solution:


74

Let R = ratio of the cost of the metals. f = frequency, Hz U

1, U

2 = cost per unit mass.

Other properties:

ρ1, ρ

2 = densities of material

V1, V

2 = sound velocities

( )212121 U,U,V,V,ρ,ρf,R φ=

h

2g

1e

2d

1c

2b

1a UUVVρραfR =

hgedc

3

b

3

a

M

1

M

1

θ

L

θ

L

L

M

L

M

θ

11

=

(1) Σθ: -a - d - e = 0

(2) ΣM: b + c - g - h = 0

(3) ΣL: -3b - 3c + d + e = 0 (1) d + e = -a (2) -3b - 3c - a = b + c = -a / 3 (3) b + c - g - h = 0 g + h = -a / 3 Then:

ga

2g

1da

2d

1ba

2b

1a 3

131

UUVVρραfR−−−−−−=

g

2

1

d

2

1

b

2

1

a

222U

U

V

V

ρ

ρ

UVr

fαR

31

31

=

g

2

1

d

2

1

b

2

13

a

23

22

3

U

U

V

V

ρ

ρ

UVr

fαR

=

Therefore:

φ=

2

1

2

1

2

1

23

22

3

U

U

V

V

ρ

ρ

UVr

fR ,,,


75

5_005 It is agreed to assume that the thermal conductivities of gases and vapors depend on the following factors: cp, specific

heat at constant pressure; cv, specific heat at constant volume; µ, viscosity; T, absolute temperature; α = ðρ/ρ(ðp)

T,

coefficient of compressibility; b = ðρ/ρ(ðT)p, coefficient of thermal expansion.

List all conclusions which may be drawn from dimensional consideration. If one additional physical variable were to be included, what should it be?

Solution: units c

p - Btu/lb-R

cv - Btu/lb-R

α - sq ft / lb

β - 1/R additional variable is the viscosity m - Btu/sec-ft-R

MLθTF system

( )Hvp Kµ,β,α,,c,ck φ=

f

Hedcb

va

p1 Kµβαccαk =

f2edc

2ba

Hθ

ML

Lθ

M

T

1

M

L

MT

H

MT

H

θLT

H

=

(1) ΣH: a + b - f = 1

(2) Σθ: -e - f = -1 e + f = 1

(3) ΣL: 2c - e + 2f = -1

(4) ΣT: -a - b - d = -1 a + b + d = 1

(5) ΣM: -a - b - c + e + f = 0 (1) and (4) -f = d f = -d (5) -a - b - c + e + f = 0 a + b + c - e - f = 0 a + b + c -1 = 0 a + b + c = 1 (2) e = 1 - f e = 1 + d e = 1 + c but c = 1 - a - b Then: a = a b = b c = 1 - a - b d = 1 - a - b e = 1 + c = 2 - a - b f = a + b -1


76

-1ba

Hb-a-2b-a-1b-a-1b

va

p1 Kµβαccαk+=

=

H

b

Hv

a

Hp1

K

αβµ

αβ

Kc

αβ

Kcαk

KH = Jg

c = (778)(32.2) = 25,000 (lb matter)(ft)

2 / (Btu)(sec)

2

Final results.

φ

=

αβ

Kc

αβ

Kc

K

αβµk HvHp

H

,

- end -

77

CHAPTER 6

FLOW OF FLUIDS

Chapter 6 Flow of Fluids

78

6_001 Water at 68 F flows from a lake through 500 ft of 4-inch i.d. cast-iron pipe to a water turbine located 250 ft below the surface of the lake. After flowing through the turbine, the water is discharged into the atmosphere through a horizontal 50-ft section of the same pipe. The turbine power output is 10 hp when the water in the discharge pipe is flowing at 5 ft/sec. What is the turbine efficiency, defined as the actual power output of the turbine divided by the power output that would be withdrawn if there were no friction within the turbine?

Solution: D = 4 inch = 1/3 ft

@ 68 F, µ = 2.43 lb/hr-ft = 6.75 x 10-4 lb/sec-ft

νm = 0.016056 cu ft / lb

p1 at the lake = 0 gauge pressure

p2 before turbine

p3 after turbine

p4 at discharge = 0 gauge pressure

( )

m

12

chc

1m2

m21

zz

g

g

r2g

LGfpp

ν

−+

ν=−

where: z

2 - z

1 = -250 ft

L1 = 500 ft

g = 32.17 ft/(sec)(sec) g

c = 32.2 ft x pounds matter/(sec)(sec)(pounds force)

G = V / νm = 5 / 0.016056 = 311.41 lb/sec-sq ft

Solving for f

m:

( )( )153,783

106.75

311.41

µ

DGN

4

31

Re =×

==−

Equation 6-8e.

( ) 1.2fN3.2logf

1Re10 +=

( ) 1.2f153,7833.2logf

110 +=

f = 0.00502

( )( ) ( )( )( )( )

( )( )

( )( )0.016056

500

32.2

32.17

32.22

5000.016056311.410.00502p0

121

2

2 +=−

p

2 = 14,828 lb/sq ft

( ) ( )( ) ( )( )( )( ) 3

121

2

hc

m2

m43 p

32.22

500.016056311.410.00502

r2g

LGfpp ==

ν=− 2

p

3 = 73 lb/sq ft

p

2 - p

3 = 14,828 - 73 = 14,755 lb/sq ft


79

Volumetric rate of flow = q

q = AV = (π/4)(1/3)2(5) = 0.43633 cu ft/sec

( ) ( )( )550

0.436314,755

550

qpphp 32 =

−=

hp = 11.70 hp Turbine Efficiency = (10 hp / 11.70 hp) (100 %) Turbine Efficiency = 85.5 % . . . Ans. 6_002 If the turbine in Prob. 1 were by-passed, what would be the mass flow rate of water through the 550 ft of 4-inch pipe? Solution: If the turbine is by-passed:

( )0

zz

g

g

r2g

LLGfpp

m

12

chc

21m2

m41 =

ν

−+

+ν=−

( ) ( )( )( )( )

( )( )

( )( )

00.016056

250-

32.2

32.17

32.22

5500.016056G0.00502

121

2

=+

G = 1,372.3 lb/sec-ft w = GA

w = (1,372.3)(π/4)(1/3)2 w = 119.76 lb/sec . . Ans. 6_003 Air is to be delivered from a compressor to a distribution line by means of standard 1-1/2-in. steel pipe, 300 ft long. The

maximum rate of consumption by the equipment connected to this distribution system is 600 cu ft of free air (measured at 68 F) per minute. In order for the pressure in the distribution line to be always 70 lb/sq in. gauge or higher, what must be the pressure rating of the compressor? Assume that the air flow is isothermal.

Solution: Air at 600 cu ft per minute at 68 F g

c = 32.2 ft x pounds matter/(sec)(sec)(pounds force)

L = 300 ft D = 1 1/2 in = 1/8 ft

p

TRG=ν

T = 68 F + 460 F = 528 F R

G = 53.34

rh = D/4 = 1/32 ft

p

2 = 70 lb/sq in gauge = 84.696 lb/sq in abs.

p2 = 12,196 lb/sq ft abs.

A = (π/4)D2 = (π/4)(1/8)2 A = 0.012272 sq ft


80

( )( )12196

52853.34=ν2

ν

2 = 2.309 cu ft /lb

Air viscosity at 68 F

µ = 0.0440 lb/hr-ft = 1.2222 x 10-5 lb/sec-ft

( )( )5

81

Re101.2222

353

µ

DGN

−×

==

3,610,293NRe =

Therefore, α = 1 Steel pipe:

( ) 1.2fN3.2logf

1Re10 +=

( ) 1.2f3,610,2933.2logf

110 +=

f = 0.003036

( )( )

111

G

p

28,164

p

52853.34

p

TR===ν1

( )212

1m ν+ν=ν

1.1545p

14,0822.309

p

28,164

1121

m +=

+=ν

Use Equation (6-9d) assuming ν

2/ν

1 > 2.

hc

2m

c

2

mG

22

21

r2g

LGfln

αg

G

T2R

pp+

ν

ν=

−

1

2

12,196

p

p

28,164

2.309 1

1

=

=

ν

ν

1

2

( )( )( )

( )( )( )

( )( )( )( )( )32

1

21

2221

32.22

3533000.003036

12,196

pln

32.21

353

52853.342

12,196p+

=

−

( )56,395

12,196

p3,870ln

56,327

12,196p 122

1 +

=

−

By Trial and Error: p

1 = 60,621 lb/sq ft abs.

p1 = 406.31 lb/sq in gauge. . . Ans.

6_004 Glycerol at 68 F, specific gravity 1.26, is pumped at a rate of 28,000 cu ft/hr by a single pump through two horizontal

pipes connected in parallel. Both pipes have a length of 100 ft. One is a standard 4-in. steel pipe, the other a standard 9-in. steel pipe. It is agreed to neglect pressure drop due to fittings. What is the velocity in the smaller pipe? What is the pressure drop through the lines? If the lines discharge into an open tank, what head must be developed by the pump?

Solution:


81

Use Line a for 4 in pipe and line b for 9 in pipe

Glycerol at 68 F, µ = 7 cp = 4.704 x 10-3 lb/sec-ft

q = volumetric flow rate = 28,000 cu ft/hr = 7.777778 cu ft/sec q = q

a + q

b

ρ = (1.26)(62.4) = 78.624 lb/cu ft Line a:

ρr2g

LGf

r2g

LGfpp

hac

2aa

hc

m2

aa21 =

ν=−

r

ha = D

a/4

For 4 in dia. pipe, D = 4.026 in = 0.3355 ft r

ha = D

a/ 4 = 0.083875 ft

gc = 32.2 ft x pounds matter/(sec)(sec)(pounds force)

( )( )( )( )78.6240.08387532.22

100Gfpp

2aa

21 =−

2

aa21 G0.235465fpp =−

Line b:

ρr2g

LGf

r2g

LGfpp

hbc

2bb

hc

m2

bb21 =

ν=−

r

hb = D

b/4

For 9 in dia. pipe, D = 8.941 in = 0.7451 ft r

hb = D

b/ 4 = 0.186275 ft

gc = 32.2 ft x pounds matter/(sec)(sec)(pounds force)

( )( )( )( )78.6240.18627532.22

100Gfpp

2bb

21 =−

2

bb21 G0.106024fpp =−

Then:

2

bb2

aa G0.106024fG0.235465f =

2

aa2

bb G2.220865fGf =

2πD

4qρ

A

ρqG ==

( )

( )2

aa

0.3355π

q78.6244G =

G

a = 889.4q

a

( )

( )2

bb

0.7451π

q78.6244G =

G

b = 180.3q

b


82

Then:

( ) ( )2

aa2

bb 889.4q2.220865f180.3qf =

b

aab

f

f7.351272qq =

Equation 6-8e.

( ) 1.2fN3.2logf

1Re10 +=

Then:

( )( ) 1.2fN3.2log

1.2fN3.2log

f

f

aRea10

bReb10

b

a

+

+=

( )( )a3

aaaRea 63,434q

104.704

889.4q0.3355

µ

GDN =

×==

−

( )( )b3

abbReb q

104.704

180.3q0.7451

µ

GDN 559,28=

×==

−

( )( ) 1.2f63,474q3.2log

1.2f28,559q3.2log

f

f

aa10

bb10

b

a

+

+=

( )( ) 1.2f63,474q3.2log

1.2f209,945q3.2log

f

f

aa10

aa10

b

a

+

+=

By trial and error, try

13.1=b

a

f

f

q

b = (7.351272)(1.13)q

a

qb = 8.306937q

a

7.777778 = qa + 8.306937q

b

qa = 0.8357 cu ft/sec

( ) 53,0120.835763,434NRea ==

( ) 1.2f53,0123.2logf

1a10

a

+=

0.07826fa =

f

a = 0.006125

( )( )( )[ ]( )( )( )[ ]

1.131.20.078260.835763,4343.2log

1.20.078260.8357209,9453.2log

f

f

10

10

b

a =+

+=

Therefore trial value is okay.

13.1=b

a

f

f

13.1=bf

0.006125


83

fb = 0.004797

a. Velocity of the smaler pipe.

Ga = V

aρ

Ga = 889.4q

a

Ga = V

aρ = 889.4q

a

Va(78.624) =(889.4)(0.8357)

Va = 9.4535 ft/sec . . . Ans.

b. Pressure drop thru line a or b.

2

aa21 G0.235465fpp =−

( )2

aa21 889.4q0.235465fpp =−

( ) ( )( )[ ]2

21 0.8357889.40.0061250.235465pp =−

p1 - p

2 = 796.76 lb/sq ft . . . Ans.

c. Head developed by pump. Head = (796.76 lb/sq ft) / (78.624 lb/cu ft) Head = 10.134 ft . . . Ans. 6_005 It is desired to heat 27 lb/min of dry air from 40 to 1040 F (moving-stream temperatures) by passing it through a

heated horizontal section of smooth pipe, having an i.d. of 2 inches. The heat is to be supplied by electrical wires wrapped around the outside of the pipe, and these wires will supply 4.0 kw of electrical power/ft of heated pipe, distributed uniformly over the heated length. The air is to leave the heated section at a pressure of 1 atm. abs. What must be the length of the heated section and the pressure of the air entering the section?

Solution: w

a = 27 lb/min

= 1,620 lb/hr = 0.45 lb/sec c

p = 0.24

t1 = 40 F

t2 = 1040 F

a. Heated Length Q = w

ac

p(t

1 - t

2)

Q = (1,620)(0.24)(1040 - 40) Q = 388,000 Btu/hr = 113.95 kw L = heated length = 113.95 kw / 4 kw/ft L = 28.5 ft . . . Ans. b. Pressure entering the section, p

1.

p

2 = 1 atm abs. = 2116.8 lb/sq ft abs.

D = 2 inches = 0.166667 ft


84

rh = D/4 = 0.041667 ft

gc = 32.2 ft x pound matter / (sec)(sec)(pound force)

( )

( )22

a

0.041667π

0.454

πD

4wG ==

G = 330 lb/(sec)(sq ft)

( )( )

11

1G1

p

4604053.34

p

TR +==ν

11

p

26,670=ν

( )( )2116.8

460104053.34

p

TR

2

2G2

+==ν

37.82 =ν

( )2

21m

ν+ν=ν

2

37.8p

26,670

1

+

=νm

18.9p

13,335

1

+=νm

Viscosity of air at average temperature. t

ave = (1/2)(40 + 1040) = 540 F

µ = 0.06978 lb/hr-ft = 1.93833 x 10-5 lb/sec-ft

( )( )2,837,496

101.93833

3300.166667

µ

DGN

5Re =×

==−

α = 1 Smooth pipe: Equation 6-8.

0.32NRe

0.1250.00140f +=

( )

0.0024762,837,496

0.1250.00140f

0.32=+=

Use Equation 6-9d, T

m = 540 F + 460 F = 1000 R

hc

2m

1

2

c

2

mG

22

21

r2g

LGfln

αg

G

T2R

pp+

ν

ν=

−

( )( )( )

( )( )( )

( )( )( )( )( ).04166732.22

33028.50.002476

p26,670

37.8ln

32.21

330

100053.342

2116.8p2

1

2221 +

=−

( )2864

705.56

p3382ln

106,680

2116.8p 122

1 +

=

−

By trial and error: p

1 = 42,269 lb/sq ft abs.

p1 = 20 atm abs. . . . Ans


85

6_006 Water is pumped over a high pass in mountain country from a lake in the valley. The pump is located 10 ft above the lake surface. From the pump, the pipe line extends 2000 ft to the top of the pass, which has an altitude of 1010 ft above the lake surface. The pipe line then runs 1000 ft down the other side of the pass, losing 500 ft in altitude. The pipe line next travels another 1000 ft horizontally and empties into a reservoir.

Neglecting end losses and assuming that the lengths of pipe are reported as “equivalent lengths” (thus containing allowance for bends, etc.), compute the shaft horsepower which must be delivered to the pump to move water through this system at a rate of 120 gal/sec. The pipe line has an i.d. of 1.5 ft and is made of steel. the over-all pump efficiency is 75 per cent. The discharge end of the pipe is several feet above the surface of the reservoir.

Solution:

D = 1.5 ft steel q = 120 gal/sec Pimp-Efficiency = 0.75 Equation 6-9a.

m

12

chc

m2

m21

zz

g

g

r2g

LGfpp

ν

−+

ν=−

z

2 - z

1 = 1010 ft - 500 ft - 10 ft = 500 ft

L = 2000 ft + 1000 ft + 1000 ft = 4000 ft q = 120 gal/sec x 231 cu n/gal x 1 cu ft / 1728 cu in q = 16.042 cu ft/sec

w = ρq = (62.4)(16.042) = 1001 lb/sec

( )

( )221.5π

10014

πD

4w

A

wG ===

G = 566.45 lb/sec-ft r

h = D/4 = 0.375 ft

g = 32.17 g

c = 32.2

νm = 1/62.4 = 0.01603 cu ft/lb

viscosity at 68 F, µ = 2.43 lb/hr-ft = 6.75 x 10-4 lb/sec-ft

( )( )1,258,778

106.75

566.451.5N

4Re =×

=−

Equation 6-8

( ) 1.2fN3.2logf

1Re10 +=

( ) 1.2f1,258,7783.2logf

110 +=


86

f = 0.00354

( )( ) ( )( )( )( )

( )( )

( )( )0.01603

500

32.2

32.17

0.37532.22

40000.01603566.450.00354pp

2

21 +=−

p

1 - p

2 = 34,178.3 lb/sq ft

( ) ( )( )550

16.04234,178.3

550

qpphp 21 =

−=

hp = 996.9 hp Checking if height 1000 ft is reached. If L’ = 2000 ft z

2’ - z

1 = 1010 ft - 10 ft = 1000 ft

( )( ) ( )( )( )( )

( )( )

( )( )0.01603

1000

32.2

32.17

0.37532.22

20000.01603566.450.00354pp

2

2'1 +=−

p

1 - p

2’ = 63,832.9 lb/sq ft

( ) ( )( )550

16.04263,832.9

550

qpphp 2'1 =

−=

hp = 1,861.8 hp > 996.9 jp Therefore: Shaft hp = 1,861.8 hp / 0.75 = 2,482 hp Ans.

- end -

87

CHAPTER 7

NATURAL CONVECTION

Chapter 7 Natural Convection

88

7_001 A bare horizontal steam pipe with an o.d. of 12 inches carries saturated steam at 240 F. The temperature of the ambient air is 70 F. Calculate the rate of heat loss by natural convection, expressed as Btu per hour per foot of pipe length.

Solution:

Pr,fGr,f NNX ⋅=

2

f

f2

f3

oGr,f

µ

∆tgβρDN =

f

fpPr,f

k

µcN =

Table A-25. Film temperature = tf = (1/2)(t

s + t

a) = (1/2)(240 + 70) = 155 F

cp = 0.2407

µf = 0.0494

kf = 0.01702

0.6986k

µcN

f

fpPr,f ==

6p2

100.858µk

βgcρY ×==

ρ

f = 0.065 lb/cu ft

( ) ( )( )( )( )

682

100.8580.017020.0494

0.2407104.17β0.065Y ×=

×=

β = 0.001701 D

o = 12 in = 1.0 ft

∆t = 240 F - 70 F = 170 F

( ) ( ) ( )( )( )( )

3208,767,120.0494

1700.001701104.170.0651N

2

823

Gr,f =×

=

Pr,fGr,f NNX ⋅=

( )( ) 8101.50.69863208,767,12X ×==

Equation 7-7a: X from 103 to 10

9

0.25

oc D

∆t0.27h

=

( ) 0.975

11700.27h

0.25

c ==

q/L = hcπD∆T = (0.975)(π)(1)(170)

q/L = 521 Btu/hr-ft . . . Ans. 7_002 The parallel wooden outer and inner walls of a building are 15 ft long, 10 ft high, and 4 inches apart. The inner surface

of the inner wall is 70 F, and the inner surface of the outer wall is at 0 F. a. Calculate the total heat loss, expressed as Btu per hr, neglecting any leakage of air through the walls. b. Repeat the calculation for the case in which the air space is divided in half by a sheet of aluminum foil, 0.001

inch thick, stretched parallel to the walls.

Solution: A = 15 ft x 10 ft


89

L = 10 ft a. Total heat loss.

2

f

f23

xGr,µ

∆tgβρxN =

t

f = (1/2)(70 + 0) = 35 F

cp = 0.2396

µf = 0.0419

kf = 0.01408

NPr

= 0.713

6p2

102.166µk

βgcρY ×==

ρ = 0.080 lb/cu ft

( ) ( )( )( )( )

682

102.1660.014080.0419

0.2396104.17β0.080Y ×=

×=

β = 1.994 x 10-3

x = 4 in = 0.3333 ft

( ) ( ) ( )( )( )( )

6

2

-3823

xGr, 1087,856,2830.0419

0-70101.994104.170.0800.3333N ×==

××=

L = 10 ft high

Then, For NGr,x

from 2.1 x 105 to 1.1 x 10

7

C = 0.071, n = 1/3 Equation 7-9b.

( )

n

p

2f

f2

f3

f

'c

k

µc

µ

∆tgβρx

xL

C

k

xh

91

=

( )( )

( )

( )( )( )[ ] 3

1

0.7137,856,283

0.333310

0.071

0.01408

0.3333h

91

'c =

hc

’ = 0.365

Equation 7-9.

qc = h

c

’A(t

s1 - t

s2)

qc = (0.3650(150)(70 - 0)

qc = 3,832.5 Btu/hr . . . Ans.

b. With air space divided in half by a sheet of aluminum foil, 0.001 inch thick.

qc/A = h

c

”(t

s1 - t

sf) = h

c

”(t

sf - t

s2)

qc/A = h

c

’(t

s1 - t

s2)

Try t

sf= 35 F


90

For 70 F side: t

f = (1/2)(70 + 35) = 52.5 F

Properties of air: c

p = 0.2398

µ = 0.04300 k = 0.0145 N

Pr = 0.7111

6p2

10µk

βgcρY ×== 91.1

ρ = 0.0774 lb/cu ft

( ) ( )( )( )( )

682

100.01450.04300

0.2398104.17β0.0774Y ×=

×= 91.1

β = 1.989 x 10-3

x = (1/2)(4 in - 0.001 in) = 1.9995 in = 0.1666 ft

( ) ( ) ( )( )( )( )

5

2

-3823

xGr, 104.35434,9200.0430

35-70101.989104.170.07740.1666N ×==

××=

L = 10 ft high

Then, For NGr,x

from 2.1 x 105 to 1.1 x 10

7

C = 0.071, n = 1/3 Equation 7-9b.

( )

n

p

2f

f2

f3

f

'c

k

µc

µ

∆tgβρx

xL

C

k

xh

91

=

( )( )

( )

( )( )( )[ ] 3

1

91

0.7111434,920

0.166610

0.071

0.0145

0.1666h'

c =

hc

’ = 0.265 at 70 F side.

For 0 F side: t

f = (1/2)(35 + 0) = 52.5 F

Properties of air: c

p = 0.2396

µ = 0.04070 k = 0.01361 N

Pr = 0.7165

6p2

102.75µk

βgcρY ×==

ρ = 0.083 lb/cu ft

( ) ( )( )( )( )

682

100.013610.04070

0.2396104.17β0.083Y ×=

×= 75.2

β = 2.213 x 10-3

x = (1/2)(4 in - 0.001 in) = 1.9995 in = 0.1666 ft


91

( ) ( ) ( )( )( )( )

5

2

-3823

xGr, 106.21621,1230.0407

0-35102.213104.170.0830.1666N ×==

××=

L = 10 ft high

Then, For NGr,x

from 2.1 x 105 to 1.1 x 10

7

C = 0.071, n = 1/3 Equation 7-9b.

( )

n

p

2f

f2

f3

f

'c

k

µc

µ

∆tgβρx

xL

C

k

xh

91

=

( )( )

( )

( )( )( )[ ] 3

1

91

0.7165621,123

0.166610

0.071

0.01361

0.1666h'

c =

hc

’ = 0.281 at 0 F side.

Then:

qc/A = h

c

”(t

s1 - t

sf) = h

c

”(t

sf - t

s2)

(0.265)(70 - tsf) = (0.281)(t

sf - 0)

tsf = 34 F

q

c/A = (0.265)(150)(70 -34)

qc/A = 1432 Btu/hr . . . Ans.

- end -

92

CHAPTER 8

INTRODUCTION TO FORCED

CONVECTION

Chapter 8 Introduction to Forced Convection

93

8_001. Hot oil having a specific heat of 0.5 flows through a reliable meter indicating a steady rate of 50,000 pounds/hr, enters a well-insulated counterflow heat exchanger at a gauge pressure of 80 pounds/sq inch and at a temperature of 380 F, and leaves the exchanger at a gauge pressure of 70 pounds/sq inch and at a temperature of 150 F. Cold oil, having a specific heat of 0.5, flows through a reliable meter indicating a steady rate of 80,000 pounds/hr, enters the exchanger at a gauge pressure of 95 pounds/sq inch and at a temperature of 100 F, and leaves the exchanger at a gauge pressure of 85 pounds/sq inch and a temperature of 300 F. The total cooling surface is 8000 sq ft. It is agreed to overlook the fact that the over-all heat-transfer coefficient may vary with temperature.

Both oils are nonvolatile, even at atmospheric pressure, no chemical changes occur, and the steady state prevails. All thermometers are properly located to measure bulk-stream temperatures, and the thermometers have been recently calibrated.

What sound conclusions should be drawn from these facts/ Make conclusions quantitative if possible. Solution: Hot oil: c

ph = 0.5

wh = 50,000 lb/hr

ph1

= 80 psig

th1

= 380 F

ph2

= 70 psig

th2

= 150 F

Cold oil: c

pc = 0.5

wc = 80,000 lb/hr

pc1

= 95 psig

tc1

= 100 F

pc2

= 85 psig

tc2

= 300 F

A = 8000 sq ft

( ) ( )

−

−

−−−=

c1h2

c2h1

c1h2c2h1oL

tt

ttln

tttt∆t

( ) ( )F63.83

100150

300380ln

100150300380∆t oL =

−

−

−−−=

0.7143100380

100300

tt

ttη

c1h1

c1c2H =

−

−=

−

−=

1.15100300

150380

tt

ttZ

c1c2

h2h1 =−

−=

−

−=

Figure 8-6 (C). Three shell passes and 6,12,18, etc., tube passes. F

G = 0.75

Figure 8-6 (D). Four shell passes and 8,16,24, etc., tube passes. F

G = 0.85

Figure 8-6 (E).


94

Six shell passes and 12,24,36, etc., tube passes. FG = 0.95

Conclusions: 1. The heat transfer rate of hot oil and cold oil is not balance. Hot oil: q

h = w

hc

h(t

h1 - t

h2)

qh = (50,000)(0.5)(380 - 150)

qh = 5,750,000 Btu/hr

Cold oil: q

c = w

cc

c(t

c2 - t

c1)

qc = (80,000)(0.5)(300 - 100) q

c = 8,000,000 Btu/hr

hc qq ≠

2. Available number of shell passes are as follows: a. Three shell passes, and 6, 12, 18, etc., tube passes. F

G = 0.75.

b. Four shell passes, and 8, 16, 24, etc., tube passes. FG = 0.85.

c. Six shell passes, and 12, 24, 36, etc., tube passes. FG = 0.95.

3. Over-all heat-transfer coeffcient are as follows: A. Hot Oil, q

h = 5,750,000 Btu/hr

a. Three shell passes.

( )( )( )

1563.830.758000

5,750,000

∆tAF

qU

oLG

ho ===

b. Four shell passes.

( )( )( )

13.2563.830.858000

5,750,000

∆tAF

qU

oLG

ho ===

c. Six shell passes.

( )( )( )

11.8563.830.958000

5,750,000

∆tAF

qU

oLG

ho ===

B. Cold Oil, q

c = 8,000,000 Btu/hr

a. Three shell passes.

( )( )( )

20.963.830.758000

8,000,000

∆tAF

qU

oLG

co ===

b. Four shell passes.

( )( )( )

18.4363.830.858000

8,000,000

∆tAF

qU

oLG

ho ===

c. Six shell passes.


95

( )( )( )

16.563.830.958000

8,000,000

∆tAF

qU

oLG

ho ===

To make it balance, consider vapor condesation on the hot oil side: 80 psig, 380 F, sat. = 323.76 F 70 psig, 150 F, sat. = 315.86 F 95 psig, 100 F, sat. = 334.4 F 85 psig, 300 F, sat. = 327.49 F Therefore ar 380 F hot oil temperature there are some vapor condensing. Then: q = 50,000c

p(380 -150) = 80,000(0.5)(300 - 100)

cp = 0.69565 Btu/lb - specific heat of hot oil with vapor combined.

Then q = 8,000,000 Btu/hr See item B of Number 3 for U values. 8_002. An oil is being cooled by water in a double-pipe parallel-flow heat exchanger. The water enters the center pipe at a

temperature of 60 F and is heated to 120 F. The oil flows in the annulus and is cooled from 260 to 150 F. It is proposed to cool the oil to a lower final temperature by increasing the length of the exchanger. Neglecting heat loss from the exchanger to the room, and making any reasonable simplifying assumptions which are necessary, but stating all such assumptions clearly, determine:

a. The minimum temperature to which the oil may be cooled.

b. The exit-oil temperature as a function of the fractional increase in the exchanger length. c. The exit temperature of each stream if the existing exchanger were switched to counterflow operation. d. The lowest temperature to which the oil could be cooled with counterflow operation. e. The ratio of the required length for counterflow to that for parallel flow as a function of the exit-oil

temperature. Solution: Oil: t

h1 = 270 F, t

h2 = 150 F

Water: tc1

= 60 F, tc2

= 120 F

a. Minimum temperature of oil for parallel flow approaches 120 F. b. Assumptions: 1. U

o is constant.

2. ch and c

c are constants.

Parallel Flow:

( ) ( )

−

−

−−−==

c2h2

c1h1

c2h2c1h1oLom

tt

ttln

tttt∆t∆t

( )( ) ( )

−

−

−−−==

c2h2

c1h1

c2h2c1h1oh2h1hh

tt

ttln

ttttπDLUt-tcwq


96

( )

( ) ( )c2h2c1h1

c2h2

c1h1h2h1

hh

o

tttt

tt

ttlnt-t

cw

πDLU

−−−

−

−

=

( )

( ) ( )120t60270

120t

60270lnt-270

cw

πDLU

h2

h2h2

hh

o

−−−

−

−

=

( )

( )h2

h2h2

hh

o

t-330

120t

60270lnt-270

cw

πDLU

−

−

=

hh

o

cw

πDUK =

Let F be the fractional increase in length based on 150 F leaving oil temperature = (L-L

o)/L

o

th2 KL F

150 1.2972734 0

148 1.3506493 0.041175

146 1.4078118 0.0852386

144 1.469359 0.1326821

142 1.5360443 0.1840863

140 1.6088357 0.2401974

138 1.6890058 0.3019964

136 1.7782759 0.3708099

134 1.8790552 0.4484955

132 1.9948673 0.5377689

130 2.1311657 0.6428342

128 2.2970721 0.7707228

126 2.5096575 0.9345936

124 2.8071783 1.1639368

122 3.3114718 1.5526702

121 3.8120527 1.9385418

120.5 4.3103488 2.3226521

120.4 4.4704407 2.4460586

120.3 4.6766654 2.6050263

120.2 4.9670663 2.8288812

120.1 5.4630249 3.2111896

120.05 5.9586036 3.5932053

120.04 6.1180933 3.7161475

120.03 6.3236853 3.8746276

120.02 6.6134127 4.0979632

120.01 7.1086344 4.4797037

120.005 7.6038025 4.8614028

120.004 7.763204 4.984277

120.003 7.9687048 5.1426868

120.002 8.2583368 5.365949

120.001 8.7534568 5.7476109


97

By curve fitting, a tenth order poynomial gives

th2

= 197.1815368e-10F

- 1071.722386e-9F

+ 2509.935e-8F

- 3318.383799e-7F

+ 2729.319811e-6F

- 1456.425193e-5F

+ 517.9791641e-4F

- 136.5134068e-3F

+ 58.29677484e-2F

+0.3333119694e-F

+ 119.9992369 c. Exit temperatures if switched to counter flow. For counter flow:

( )

( ) ( )1.2973

tttt

tt

ttlnt-t

cw

πDLU

c1h2c2h1

c1h2

c2h1h2h1

hh

o =−−−

−

−

=

( )

( ) ( )1.2973

60tt270

60t

t270lnt-270

cw

πDLU

h2c2

h2

c2h2

hh

o =−−−

−

−

=

( )

( )1.2973

t-t-330

60t

t270lnt-270

h2c2

h2

c2h2

=

−

−

Assume Cp

oil = 0.5

( ) ( )

( )( ) ( )( )

( )( )

c2h2

h2c2

hc

hc

h2h1hhc1c2cc

2t390t

t2700.560t

ww

1502700.5w601201w

ttcwttcw

−=

−=−

=

−=−

−=−

Then:

( )

( )1.2973

60-t

2t330

t270ln120-2t

c2

c2

c2c2

=

−

−

By Trial and Error Method: t

c2 = 127.85 F

th2

= 390 - 2(127.75) = 134.3 F

d. Minimum temperature of oil for counterflow approaches 60 F.

e. Assumptions: 1. U

o is constant.

2. ch and c

c are constants.

Counter Flow:

( ) ( )

−

−

−−−==

c1h2

c2h1

c1h2c2h1oLom

tt

ttln

tttt∆t∆t


98

( )( ) ( )

−

−

−−−==

c1h2

c2h1

c1h2c2h1oh2h1hh

tt

ttln

ttttπDLUt-tcwq

( )

( ) ( )c1h2c2h1

c1h2

c2h1h2h1

hh

o

tttt

tt

ttlnt-t

cw

πDLU

−−−

−

−

=

( )

( ) ( )60t127.85270

60t

127.85270lnt-270

cw

πDLU

h2

h2h2

hh

o

−−−

−

−

=

( )

( )h2

h2h2

hh

o

t-202.15

120tlnt-270

cw

πDLU

−=

15.142

Let Lc = Counterflow Length and Lp = Parallel Flow Length ratio of lengths as a function of oil exit oil temp.

( )

( )

−

−=

120t

210lnt-202.15

120t

142.15lnt-330

Lp

Lc

h2h2

h2h2

8_003. Sulfuric acid in the amount of 10,000 lb/hr (specific heat of 0.36) is to be cooled in a two-stage countercurrent cooler of

the following type: Hot acid at 174 C is fed to a tank, where it is stirred in contact with cooling coils; the continuous discharge from this tank at 88C flows to a second stirred tank and leaves the second tank at 45 C. Cooling water at 20 C flows into the cooling coil of the second tank and from there to the cooling coil of the first tank. The water is at 80 C as it leaves the coil in the hot-acid tank.

Calculate the total area of cooling surface necessary, assuming U of 200 and 130 for the hot and cold tanks,

respectively, and neglecting heat losses. Solution:

U

A = 200

UB = 130

w

h = 10,000 lb/hr


99

ch = 0.36

th1

= 174 C = 345.2 F

th2 = 88 C = 190.4 F th3 = 45 C = 113 F t

c1 = 20 C = 68 F

tc3

= 80 C = 176 F

cc = 1.0

Solving for w

c and t

c2.

Heat Balance thru hot tank A. q

A = w

hc

h(t

h1 - t

h2) = w

cc

c(t

c3 - t

c2)

qA = (10,000)(0.36)(345.2 190.4) = w

c(1)(176 - tc2)

qA - 557,280 = w

c(176 - t

c2)

w

c = 557,280 / (176 - t

c2) Equation 1

q

B = w

hc

h(t

h2 - t

h3) = w

cc

c(t

c2 - 68)

qB = (10,000)(0.36)(190.4 - 113) = w

c(1)(tc2 - 68)

qB = 278,640 = wc(tc2 - 68)

w

c = 278,640 / (t

c2 - 68) Equation 2

Equating: w

c = 557,280 / (176 - t

c2) = 278,640 / (t

c2 - 68)

2tc2

- 136 = 176 - tc2

tc2

= 104 F

Hot Tank:

ALAAA ∆tAUq =

( ) ( )

−

−

−−−=

c2h2

c3h1

c2h2c3h1L

tt

ttln

)tttt∆t

A

( ) ( )F9.122=

−

−

−−−=

104190.4

176345.2ln

104)190.4176345.2∆t

AL

ALA

AA

∆TU

qA =

( )( )122.9200

557,280A A =

A

A = 22.7 sq ft

Cold Tank:

BLBBB ∆tAUq =


100

( ) ( )

−

−

−−−=

c1h3

c2h2

c1h3c2h2L

tt

ttln

)tttt∆t

B

( ) ( )F63.465

68113

104190.4ln

68)113104190.4∆t

BL =

−

−

−−−=

BLB

BB

∆TU

qA =

( )( )63.465130

278,640AB =

A

B = 33.8 sq ft

Total area of cooling surface: = A

A + A

B

= 22.7 + 33.8 = 56.5 sq ft . . . . Ans. 8_004 It is desired to heat a cold fluid stream from 20 to 180 F with hot water at a temperature of 200 F. A countercurrent

double-pipe heat exchanger is to be used, with the hot water in the inner pipe and the cold fluid in the annulus. The specific heat of the water may be assumed to be constant at 1.0. The true specific heat of the cold fluid is given by the equation

cp = 0.5 + 0.014 ( t - 20 )

a. What is the minimum quantity of water, per pound of cold fluid, that may be used to accomplish this heating?

What is the corresponding exit-water temperature? b. If the water rate is set at 2 pounds of water per pound of cold fluid and the overall coefficient of heat transfer

is 200, what is the required area of the exchanger, expressed as square feet of surface area per pound of cold fluid

flowing per hour? What is the mean ∆t? Solution: t

c1 = 20 F

tc2

= 180 F

th1

= 200 F

ch = 1.0

Countercurrent, Eq. 8-19

( ) ( )

−

−

−−−=

c1h2

c2h1

c1h2c2h1oL

tt

ttln

tttt∆t

( )20t0.0140.5c cc −+=

a.

minimumw

w

c

h =

q = w

cc

c∆tc


101

( )[ ]∫ −+=c2

c1

t

tccc dt20t0.0140.5wq

( )[ ]∫ −+=

180

20ccc dt20t0.0140.5wq

( ) ( )

+−=

2c 160

2

0.014201800.5wq

q = 259.2w

c

q = whc

h(t

h1 - t

h2)

q = wh(1)(200 - t

h2)

259.2w

c = w

h(200 - t

h2)

h2c

h

t200

259.2

w

w

−=

th2 = exit temperature = 20 F + 5 F = 25 F

1.4825200

259.2

w

w

c

h =−

=

b.

2w

w

c

h =

2t200

259.2

w

w

h2c

h =−

=

t

h2 = 70.4 F

( ) ( )

−

−

−−−=

c1h2

c2h1

c1h2c2h1oL

tt

ttln

tttt∆t

( ) ( )F32.89

2070.4

180200ln

2070.4180200∆t oL =

−

−

−−−=

Then:

( )( )32.89200

259.2

UDtw

q

w

A

oLcc

==

A / w

c = 0.0394 sq ft per pound of cold cluid per hour.

F32.89∆toL =

8_005 A vertical shell-and-tube heat exchanger is to be used to condense steam. The exchanger consists of 86 eight-foot

lengths of standard 1-inch 18 BWG copper condenser tubing. The steam is to be condensed on the outside of the tubes by water flowing through the inside of the tubes in one pass. The water is available at a rate of 600,000 lb/hr and a temperature of 60 F, and the steam will be condensed at a pressure of 1 atm.

Estimate the capacity of the condenser, expressed as pounds of steam condensed per hour. Data: The mean individual coefficients of heat transfer may be assumed to be: Water side = 1250


102

Steam side = 1750 Scale on water side = 1800 Scale on steam side = 1800 Solution: D = 1 inch, 18 BWG copper condenser tubes n = 86 pcs l = 8 ft w

c = 600,000 lb/hr

cc = 1.0

tc1

= 60 F

Steam pressure = 1 atm For condenser tubes: D

o = 1.0 in = 0.08333 ft

Di = 0.902 in

xw = 0.049 in

For steam at 1 atm h

fg = 970.3 Btu/lb

th1

= th2

= 212 F

For copper, say k = 221 Btu/hr-sq ft-deg F per foot Heat Transfer Area:

Ao = πD

oLn = π(0.08333)(8)(86)

Ao = 180.11 sq ft

( ) ( )

−

−

−−−=

c1h2

c2h1

c1h2c2h1oL

tt

ttln

tttt∆t

( ) ( )

−

−

−−−=

60212

t212ln

60212t212∆t

c2

c2oL

−

−=

152

t212ln

t60∆t

c2

c2oL

−=

c2

c2oL

t-212

152ln

60t∆t

Then:

ic

o

icd

o

avew

ow

hdho Dh

D

Dh

D

Dk

Dx

h

1

h

1

U

1++++=

where D

ave = 0.951 in

hh = 1750


103

hc = 1250

hhd

= 1800

hcd

= 1800

( )( )

( )( ) ( )( ) ( )( )0.9021250

1

0.9021800

1

0.951221

112

0.049

1800

1

1750

1

U

1

o

++++=

U

o = 377.5 Btu/hr-sq ft -F

q = UoA

o∆t

oL

( )( )( )

−

−=

c2

c2

t212

152ln

60t180.11377.5q

( )

−

−=

c2

c2

t212

152ln

60t67,992q

But: q = wccc(t

c2 - t

c1)

q = (600,000)(1.0)(tc2

- 60)

q = 600,000(tc2

- 60)

Equating:

600,000

67,992

t212

152ln

c2

=

−

tc2

= 76.285 F

q = (600,000)(76.285-60) = 9,771,000 Btu/hr and q = w

hh

fg

wh = 1 / h

fg

wh = 9,771,000 / 970.3

wh = 10,070 lb/hr . . . Ans.

8_006. It is desired to design an adiabatic heat exchanger to cool continuously 110,000 pounds/hr of a solution from 150 to

103 F, using 100,000 pounds/hr of cooling water, available at a temperature of 50 F. The specific heat of the solution is 0.91, and the over-all coefficient of heat transfer may be assumed to be 400. Calculate the heat-transfer area required for each of the following proposals:

a. Parallel flow. b. Counter flow.

c. Reversed current exchanger with two shell passes and four tube passes, with the hot solution flowing through the shell and the cold water flowing through the tubes.

d. Cross flow, with one tube pass and one shell pass. Solution: w

h = 110,000 lb/hr

th1

= 150 F

th2

= 103 F


104

ch = 0.91

wc = 100,000 lb/hr

tc1

= 50 F

cc = 1.0

Uo = 400

q = w

hc

h(t

h1 - t

h2) = w

cc

c(t

c2 - t

c1)

(110,0000(0.91)(150 - 103) = (100,000)(1)(tc2

- 50)

tc2

= 97.05 F

q = (110,000)(0.91)(150 - 103) = 4,704,700 Btu/hr a. Parallel Flow

( ) ( )

−

−

−−−=

c2h2

c1h1

c2h2c1h1oL

tt

ttln

tttt∆t

( ) ( )F33.33

97.05103

50150ln

97.0510350150∆t oL =

−

−

−−−=

( )( )33.33400

4,704,700

∆tU

qA

oLoo ==

A

o = 352.9 sq ft . . . Ans.

b. Counter Flow

( ) ( )

−

−

−−−=

c1h2

c2h1

c1h2c2h1oL

tt

ttln

tttt∆t

( ) ( )F52.975

50103

97.05150ln

5010397.05150∆t oL =

−

−

−−−=

( )( )52.975400

4,704,700

∆tU

qA

oLoo ==

A

o = 222.0 sq ft . . . Ans.

c. Two shell passes and four tube passes. Figure 8-6, Letter B.

0.4750150

5097.05

tt

ttη

c1h1

c1c2h =

−

−=

−

−=

0.99895097.05

103150

tt

ttZ

c1c1

h2h1 =−

−=

−

−=

then F

G = 0.965

F52.975∆toL =


105

( )( )( )52.9750.965400

4,704,700

∆tFU

qA

oLGoo ==

A

o = 230 sq ft . . . Ans.

d. Crossflow, One tube pass, One shell pass.

0.4750150

5097.05

tt

ttη

c1h1

c1c2h =

−

−=

−

−=

0.99895097.05

103150

tt

ttZ

c1c1

h2h1 =−

−=

−

−=

then F

G = 0.92

F52.975∆toL =

( )( )( )52.9750.92400

4,704,700

∆tFU

qA

oLGoo ==

A

o = 241.3 sq ft . . . Ans.

8_007. A steam-heated oil preheater consists of standard 1-in. condenser tubes heated externally by steam condensing at a

gauge pressure of 15 lb/sq in. The oil enters the tubes at a gauge pressure of 10 lb/sq in. The volatility of the oil is such that there is no substantial vapor generation in the preheater.

The unit was recently by-passed and shut down for cleaning the inside of the tubes. When the preheater was again

returned to operation, the capacity was at first appreciably greater than that which existed before cleaning but within a few hours was far below that which existed just prior to cleaning. The preheater was again shut down, and inspection revealed that the tubes were still clean. What do these facts mean to you?

Ans. These facts means that after cleaning heat exchanger will have a capacity greater than that which existed before

cleaning but will go down below that which existed just prior to cleaning even it is still steel. Especially for this type of steam-heated oil cooler with volatile oil with no substantial vapor generation in the preheater.

8_008 It is proposed to recover waste heat from the stack flue gases in a certain plant by inserting an economizer coil in the

stack. The flue gas is available at 300 F, at the rate of 1800 pounds/hr. It is desired to produce saturated steam at 212 F by feeding water at 60 F to the economizer. It is agreed that the economizer coil can be arranged as desired an can be very extensive if necessary. The dew point of the flue gas is 50 F, and the specific heat is 0.26. Calculate the maximum rate of steam generation in the economizer. In which direction should the water flow through the coil for greatest thermal efficiency?

Solution: w = 1800 lb/hr t

h1 = 300 F

th2

= 50 F

ch = 0.26

tc2

= 212 F steam

tc1

= 60 F water

q = wc(h

c2 - h

c1)

For steam at 212 F h

c2 = 1150.4 Btu/lb


106

For water at 60 F h

c1 = 28.06 Btu/lb

w

c(h

c2 - h

c1) = w

hc

h(t

h1 - t

h2)

wc (1150.4 - 28.06) = (1800)(0.26)(300 - 50)

wc = 104.25 lb/hr . . . Ans.

If single tube pass is not passible, use two tube passes with flue gas flows across second and first passes in series.

- end -

107

CHAPTER 9

HEATING AND COOLING INSIDE TUBES

Chapter 9 Heating and Cooling Inside Tubes

108

9_001. A single-pass tubular heat exchanger is to be designed to heat water by condensing steam. The water is to pass through the horizontal tubes in turbulent flow, and the steam is to be condensed dropwise in the shell. The water flow rate, the initial and final water temperatures, the condensing temperature of the steam, and the available tube-side pressure drop (neglecting entrance and exit losses) are all specified. In order to determine the optimum exchanger design, it is desired to know how the total required area of the exchanger varies with the tube diameter selected. Assuming that the water flow remains turbulent and that the thermal resistance of the tube wall and the steam-condensate film is negligible, determine the effect of tube diameter on the total area required in the exchanger.

Solution: W = water flow rate t

1 and t

2 = initial and final temperature.

tc = condensing temperature

p1 - p

2 = available tube side pressure drop.

( )0.2ReN

0.046f =

µ

DGNRe =

D2ρρ

L4fG

r2g

νLfGpp

c

2

hc

2

21 ==−

2πND

4wG =

Equation 9-14:

( )0.2

b

0.70.14

b

w

b

p

p

L

µ

DG

LD10.023

µ

µ

k

µc

Gc

h3

2

+

=

πµND

4w

µ

πND

4wD

N2

Re =

=

0.2

0.2 w

µND0.04383

πµND

4w

0.046f

=

=

( )

D2ρρ

LπND2

4w

w

µND0.043834

ppc

20.2

21

=−

=−

3.80.8

1.80.2

c21

DN

wµ

Dρg

L0.14211pp

( )1.80.2

3.0.821c

wµ

8DNpp7.0368ρ.

D

L −=

( ) 3.80.8

21c

1.80.2

DNppρg

w0.14211µ

L

D

−=


109

( )0.2

b

0.70.14

b

w

b

p

p

L

µ

DG

LD10.023

µ

µ

k

µc

Gc

h3

2

+

=

w4c

hπND

πND

4wc

h

Gc

h

p

L2

2p

L

p

L =

=

πµND

4w

µ

πND

4wD

µ

DG

b

2

b

=

=

( )0.2

3.80.821c

1.80.2

0.14

b

w

b

p

p

L2

πµND

4w

DNppρg

w0.14211µ10.023

µ

µ

k

µc

w4c

hπND3

2

−+

=

( )

−+

=

2.660.560.7

210.7

c0.7

1.260.140.2p

0.14

b

w

b

pL

DNppgρ

w0.2552µ1

4w

πµND

πND2

w4c0.023

µ

µ

k

µch

32

( )

−+

=

2.660.560.7

210.7

c0.7

1.260.14

1.80.8

0.20.8p

0.14

b

w

b

pL

DNppgρ

w0.2552µ1

DN

µwc0.0279

µ

µ

k

µch

32

4.461.361.80.8L

DN

Bc

DN

Ach +=

q = h

LA

w∆t

L

Aw = πNDL

L = CcN0.8

D3.8

Aw = πCcN

1.8D

4.8

L

4.81.8

4.461.361.80.8∆tDπCcN

DN

Bc

DN

Acq

+=

( ) L

0.340.443 ∆tDBcNAcNDπCcq +=

1.81

4.8w

πCcD

AN

=

L0.34

4.8w3

4.8w ∆tD

πCcD

ABcD

πCcD

AAcπCcq

1.80.44

1.81

+

=

( ) L2.3267

0.2444w

'c0.33330.5556

w'c ∆t

D

ABDAAπCcq

+=

Set dA

w/dD = 0.

( ) ( )0

D

A2.3267B

D

A0.3333A3.3267

0.2444w

'c

0.6667

0.5556w

'c =−

( ) ( )3.3267

0.2444w

'c

0.6667

0.5556w

'c

D

A2.3267B

D

A0.3333A=


110

2.66'

c

'c2.27332

wDA

6.9808BA =

( )0.2444

'c

πCc

BcB =

( )0.2444

'c

πCc

AcA =

( ) 2.660.3112

2.27332w

DπCcAc

6.9808BcA =

2.660.3112

2.27332w

DAcCc

4.8887BcA =

( )0.721

0.7c

0.7

1.260.14

ppgρ

w0.2552µ

Ac

Bc

−=

( )1.80.2

21c

wµ

pp7.0368ρ.Cc

−=

( ) ( )

0.3112

21c

1.80.2

210.7

c0.7

1.260.14

0.3112 pp7.0368ρ.

wµ

ppgρ

w0.2552µ

AcCc

Bc

−

−=

( )1.01121

1.011c

1.011

1.820.20

0.3112 ppgρ

w0.13905µ

AcCc

Bc

−=

( )( )( ) 2.661.011

211.011

c1.011

1.820.202.27332

wDp-pgρ

wµ0.139054.8887A =

Answer

( )[ ] 1.170.44521c

0.800.088

wDp-pρg

w0.84384µA =

Heat Transfer area Aw increases as the diameter increases for optimum exchanger design. 9_002. A tubular air heater has been accurately designed to heat a given stream of air to 170 F. The air passes through 120

tubes, arranged in parallel, with steam condensing on the outside. From Eq. (9-17), it is calculated that the tubes should be 16 ft long.

Because of the deviation of some of the data from Eq. (9-17), it is now decided to build the apparatus with a 20 per

cent factor of safety. One engineer proposes to obtain this factor of safety by using 20 per cent more 16-ft tubes of the same diameter. Another argues that the 20 per cent factor of safety should be obtained by making the tube diameters 20 per cent greater, while keeping the same number of tubes and the same tube length.

Discuss the advantages and disadvantages of the foregoing methods. How would you build the heater to have a 20 per

cent factor of safety? Solution: Equation 9-17:

( )( )( )0.2

L

12

DG

DL0.0576

∆t

tt=

−

Equation 9-15:

0.2

0.8p

L D

G0.0144ch =


111

Equation 9-16: q = h

LπDL∆t

L

Then N = number of tubes; ∆t

L= CONSTANT

q = hLπDL∆t

L

( )( )L0.2

0.8

p ∆tπDNLD

G0.0144cq

=

( ) L

0.80.8p NL∆LDGπc0.0144q =

2πND

4wG =

( ) L0.8

0.8

2p NL∆LDπND

4wπc0.0144q

=

( )L

0.20.8

0.8

p L∆∆ND

w0.054883cq

=

( )

= 0.8

0.20.8Lp

DNwL∆∆0.054883cq

( )

= 0.8

0.2

DNCONSTANTq

Proposal (a), 20 Per Cent factor of safety using 20 per cent more 16 ft tubes:

Over all heat transfer factor = (N2

0.2 - N

1

0.2) / N

1

0.2

= ( 1.200.2

- 1 ) / 1 = 0.03714 or 3.714 Percent Proposal (b), 20 Per Cent factor of safety using 20 per cent greater tube diameter:

Over all heat transfer factor = [(1/D2

)0.8 - (1/D

1

)0.8 ] / (1/D

1

)0.8

= [(1/1.2)0.8

- 1] / 1 = -0.13572 or -13.572 Percent Therefore: 1. Increasing number of tubes by 20 per cent will increase heat transfer rate by 3.714 per cent. 2. Increasing tube diameter by 20 per cent more will decrease heat transfer rate by 13.572 per cent. Then use proposal number one, increasing number of tubes by 20 per cent is the desirable factor of safety. 9_003. A liquid metal is flowing in well-developed turbulent motion in along smooth circular pipe with an i.d. of 1.95 inches.

The Prandtl number of the liquid metal is 10-4, the kinematic viscosity µ/ρ is 0.0388 sq ft /hr, and the average velocity

of the flow is 3.33 ft/sec.

a. The following equation defines a special heat-transfer coefficient h’’:

( )axisww

w

ttA

qh'

−=

where the subscript w refers to the inside surface of the pipe wall. Estimate the numerical magnitude of the

special Nusselt number h’D/k by employing the generalized velocity distribution of Eqs. (9-4), (9-4a), and (9-


112

4b) and by making the assumption that EH and E

M are equal.

b. What value is predicted by the Martinelli analogy for the Nusselt number based on the standard heat-transfer

coefficient h, defined as:

( )mww

w

ttA

qh

−=

where t

m is the “mixing-box’ mean temperature?

Solution: a.

( )axisww

w

ttA

qh'

−=

Page 210 to 212. tw - t

axis = t

w - t

3

where t

w - t

3 is from Equation 9-5, 9-5b, and 9-5e.

Equation 9-5.

( ) 11www /yttkAq −=

5y1 =+

w

11

r

y

2

f

µ

DVρ

2

1y =

+

µ

DVρNRe =

D = 1.95 in = 0.1625 ft rw = 0.08125 ft

V = 3.33 ft/sec = 11,988 ft/hr µ/ρ = 0.0388 sq ft / hr

( )( )( )

502080.0388

119880.1625NRe ==

( )0.00528

N

0.046f

0.2Re

==

( ) 50.08125

y

2

0.0052850208

2

1y 1

1 =

=

+

15875y

1 = 5

y1 = 3.15 x 10

-4 ft

×==− −

w

w4

w

w11w

kA

q103.15

kA

qytt

Equation 9-5b


113

212

Pr

pw

w tt15

yN1ln

uρcA

5q−=

−+

+

∗

k

µc10N

p4Pr == −

30y2 =

+

µ/ρ = 0.0388 sq ft / hr

2fVu =∗

( ) 616

2

0.0052811988u ==∗

( ) 21

4

pw

w tt15

30101ln

616ρcA

5q−=

−+ −

×=− −

pw

w621

ρcA

q104.06tt

×=− −

ρ

µ

µc

k

kA

q104.06tt

pw

w621

( )0.038810

1

kA

q104.06tt

4-w

w621

×=− −

×=− −

w

w321

kA

q101.5753tt

Equation 9-5e.

−+

−+

=−∗

w

2

w

2

ww

pw

w32

r

y1

r

y5λ

r

y1

r

y5λ

lnuρcA

q1.25tt

−

−

+

−

+

−

−

−

+

∗

zr

2y1z

r

2y1

zr

2y1z

r

2y1

lnz

uρcA

q1.25

w

2

w

w

2

wpw

w

where:

==

k

ρcDV

f2

kDGc

f2

λpp

k

µc10N

p4Pr == −

µ/ρ = 0.0388 sq ft / hr V = 11988 ft / hr

µ

ρ

µ=

ρ

k

cDV

k

cDV

pp


114

( )( )( )

=

−

0.0388

1101198800.1625

k

ρcDV 4p

5.021k

ρcDV

p=

3.8762

5.021

0.00528

2

λ ==

20λ1z +=

( )3.8762201z +=

z = 8.8614

30y2 =

+

3015875yy 22 ==

+

y

2 = 1.890 x 10

-3 ft

rw = 0.08125 ft

y2 / r

w = 1.890 x 10

-3 / 0.08125 = 0.02326

at the axis (center), y / r

w = 1.0

Then:

( )[ ] ( )( )[ ]( )[ ] ( )( )[ ]8.86140.02326218.861421

8.86140.02326218.861421ln

zr

2y1z

r

2y1

zr

2y1z

r

2y1

ln

w

2

w

w

2

w

−−+−

+−−−=

−

−

+

−

+

−

−

−

[ ][ ][ ][ ]

0.44277.907927.8614

9.814889.8614ln =

−

−=

( ) ( )( )( ) ( )( )0.0232610.023263.87625

1113.87625ln

r

y1

r

y5λ

r

y1

r

y5λ

ln

w

2

w

2

ww

−+

−+=

−+

−+

3101.18603

19.404

19.381ln −×−==

Substitute in Equation 9-5e, becomes:

( ) ( )0.4427z

uρcA

q1.25

101.18603uρcA

q1.25tt

*pw

w

3

*pw

w32 +×−=− −

( )( )( )

+

×−=−

−

8.8614616

0.4427

ρcA

q1.25

616

101.18603

ρcA

q1.25tt

pw

w3

pw

w32


115

pw

w5-32

ρcA

q109.897tt ×=−

×=−

ρ

µ

µc

k

kA

q109.897tt

pw

w5-32

( )0.038810

1

kA

q109.897tt

4-w

w5-32

×=−

=−

w

w32

kA

q0.0384tt

32211w3w tttttttt −+−+−=−

( )

+×+×=− −

−w

343w

kA

q0.0384101.5753103.15tt

=−=−

w

waxisw3w

kA

q0.04029tttt

( )

( )0.04029

ttkAttAhq axisww

axisww'

w

−=−=

24.82

k

h'

=

( )( )0.162524.82

k

Dh'

=

Answer:

4.03325

k

Dh'

=

b.

( )mww

w

ttA

qh

−=

t

w - t

m = t

w - t

b

Equation 9-6.

( ) ( )

+++

−

−=

2

f

60

Nln0.5N5N1lnN5

tt

tt

2f

Gc

h

ReDRPrPr

cw

bwp

N

Pr = 10

-4

NRe

= 50,208 = 5 x 10-4

NPc

= NRe

x NPr

= 5

Use Equation 9-5f.

( )

−+

=

w

2

w

2

ReDR

r

y1

λ5r

y1

1ln

2

f

60

Nln2N


116

−−

−+

++

w

2

w

2

r

2y1z

r

2y1z

1-z

1zln

z

1

( )( )( )( )

−−

−+

−

+=

−−

−+

+

0.02326218.8614

0.02326218.8614

18.8614

18.8614ln

r

2y1z

r

2y1z

1-z

1zln

w

2

w

2

( )( )[ ] 0.4427041.241151.25441ln ==

( )( )

3

w

2

w

2

101.17154

0.0232613.87625

0.023261

1ln

r

y1

λ5r

y1

1ln −×−=

−+

=

−+

Substitute in Equation 9-5f.

( )

8.8614

0.44270410-1.17154

2

f

60

Nln2N 3-Re

DR +×=

0.024394

2

f

60

NlnN Re

DR =

Tablle 9-3, NPr

= 10-4, N

Re = 50208 = 5 x 10

4

0.562394tt

tt

cw

bw =−

−

Substitute in Equattion 9-6.

( )( ) ( )( ) ( )[ ] 1.427867

0.0243940.51051ln1050.562394

2

0.00528

Gc

h4-4

p

=+++

=−

1.427867Gc

h

ρVc

h

pp

==

1.427867ρ

µ

µc

k

kV

h

ρVc

h

pp

=

=

( )

( ) 1.4278670.038810

1

11988k

h4-

=

44.1167

k

h=

( )( )0.162544.1167

k

hD=

Answer:


117

7.1690

k

hD=

9_004. It is planned to construct a horizontal single-pass multitubular preheater to warm 6000 pounds/hr of viscous fuel oil

from 80 to 100 F, while flowing in streamline motion inside horizontal tubes jacketed by exhaust steam condensing at 220 F in the shell. For tubes having i.d. of 0.67 inch and lengths of 8, 12, 16, or 20 ft. complete heaters (both shell and tubes) can be obtained at a first cost of $1.80 per foot of each tube. The fixed charges on the heater in dollars per year per foot of length of each tube are $0.36. Mechanical energy, delivered to the fluid, will cost $0.005/million ft-lb.

Show which tube length should be selected.

Data and Notes. The oil density is 56 lb/cu ft. The specific heat cp is constant at 0.48 gm-cal/(gm)(deg C), and k may

be assumed constant at 0.08 Btu/(hr)(sq ft)(deg F per ft). Viscosity varies with temperature as shown below:

Deg F 80 90 100 110 120 130

Centipoises 400 270 185 140 110 82 It may be assumed that the pressure drop through the heater is due only to friction in the tubes. Assume tubes clean

on both sides and the Ui = h

i.

Solution: U

i = h

i

1 cp = 2.42 lb/hr-ft

D = 0.67 inch = 0.05584 ft Then at 100 F. µ = 2.42 * 185 = 447.7 lb/hr-sq ft DG/µ = 4w/πDµ = 4(6000) / [(π)(0.0584)(447.7)] DG/µ = 306 < 2100 ( 1 tube minimum) a. For Length L = 8 ft. Solve for number of tubes by trial and error. For one (1) tube

( )( )( )( )( )

5,73080.8π

0.4860004

πkL

4wc=

=

Equation 9-28a:

31

πkL

4wc

µ

µ

D

1.86kh

0.14

sa

=

asa t)(t∆t −=

ts = 215 F

1.25µ

µ0.14

s

=

t = (1/2)(80 F + 100 F) = 90 F

( )( ) ( )( ) 59.6157301.250.05584

0.081.86h 3

1

a ==

∆ta = 215 F - 90 F = 125 F

Aw = N

TπDL

NT = Number of tubes


118

p

awa12

wc

∆tAhtt =−

( ) ( )( )( )( )( )( )( )( )0.486000

12580.05584πN59.6180100 T=−

N

T = 5.5 or 6 tubes

Try 6 tubes. w = (6,000 lb/hr) / 6 = 1000 lb/hr

( )( )( )( )( )

95580.8π

0.4810004

πkL

4wc=

=

( )( ) ( )( ) 32.819551.250.05584

0.081.86h 3

1

a ==

p

awa12

wc

∆tAhtt =−

( )( )( )( )( )( )( )

( )( )0.486000

12580.05584πN32.8180100 T=−

N

T = 10 tubes

Try 10 tubes. w = (6,000 lb/hr) /10 = 600 lb/hr

( )( )( )( )( )

57388.0π

48.06004

πkL

wc4=

=

( )( ) ( )( ) 27.675731.250.05584

0.081.86h 3

1

a ==

p

awa12

wc

∆tAhtt =−

( ) ( )( )( )( )( )( )( )( )0.486000

12580.05584πN27.6780100 T=−

N

T = 12 tubes

Try 13 tubes. w = (6,000 lb/hr) /13 = 461.54 lb/hr

( )( )( )( )( )

44180.8π

0.48461.544

πkL

4wc=

=

( )( ) ( )( ) 25.3544411.250.05584

0.081.86h 3

1

a ==

p

awa12

wc

∆tAhtt =−

( ) ( )( )( )( )( )( )( )( )0.486000

12580.05584πN25.35480100 T=−

N

T = 13 tubes

Then use 13 tubes for 8 ft length.


119

hc

2

ir2g

vLfG∆p =

f = 16/N

Re

r

h = D/4 = (0.05584) / 4 = 0.01396 ft

G = w/Ai = 4w/πD

2= 4 (461.54)/[π(0.05584)

2]=188464

v = 1/ρ = 1/56 N

Re = DG/µ = 4w/πDµ = (4)(461.54)/[π(0.05584)(653.4)]

NRe

= 16.11

f = 16/NRe

= 16/16.11 = 0.9932

gc = 4.17x10

8 (lb of fluid)(ft)/(hr)(hr)(lb of fluid)

( )( ) ( )( )( )( )560.01396104.172

81884640.9932∆p

8

2

i×

=

∆p

i = 433 lb/sq ft

First Cost = $1.80 x 8 x 13 = $ 187.20 Fixed Charges = $ 0.36 x8 x13 = $ 37.44 per year. Mechanical Energy: $ 0.005 / million ft-lb ∆p

i = 433 lb/sq ft

1 year = 8640 hours Mechanical Energy = (6000 lb/hr)(433 lb/sq ft)(8640 hrs/yr)/(56 lb/cu ft) = 400.84 Million Ft-lb / yr. Mechanical Energy = ($ 0.005)(400.84) = $ 2.01 per year. b. For Length L = 12 ft. Try 9 tubes. w = (6,000 lb/hr) /9 = 666.67 lb/hr

( )( )( )( )( )

424.4120.8π

0.48666.674

πkL

4wc=

=

( )( )( )( ) 25.032424.41.25

0.05584

0.081.86h 3

1

a ==

p

awa12

wc

∆tAhtt =−

( ) ( )( )( )( )( )( )( )( )0.486000

125120.05584πN25.03280100 T=−

N

T = 8.75 tubes or 9 tubes.

Then use 9 tubes for 12 ft length. N

Re = DG/µ = 4w/πDµ = (4)(666.67)/[π(0.05584)(653.4)]

NRe

= 23.265

f = 16/NRe

= 16/23.265 = 0.6877


120

gc = 4.17x10


G = w/Ai = 4w/πD

2= 4 (666.67)/[π(0.05584)

2]=272226

( )( ) ( )( )( )( )560.01396104.172

122722260.6877∆p

8

2

i×

=

∆p

i = 938 lb/sq ft

First Cost = $1.80 x 12 x 9 = $ 194.40 Fixed Charges = $ 0.36 x12 x 9 = $ 38.88 per year. Mechanical Energy: $ 0.005 / million ft-lb ∆p

i = 938 lb/sq ft

1 year = 8640 hours Mechanical Energy = (6000 lb/hr)(938 lb/sq ft)(8640 hrs/yr)/(56 lb/cu ft) = 868.32Million Ft-lb / yr. Mechanical Energy = ($ 0.005)(868.32) = $ 4.34 per year. c. For Length L = 16 ft. Try 7 tubes. w = (6,000 lb/hr) /7 = 857.14 lb/hr

( )( )( )( )( )

409.25160.8π

0.48857.144

πkL

4wc=

=

( )( ) ( )( ) 24.73409.251.250.05584

0.081.86h 3

1

a ==

p

awa12

wc

∆tAhtt =−

( ) ( )( )( )( )( )( )( )( )0.486000

125160.05584πN24.7380100 T=−

N

T = 6.64 tubes or 7 tubes.


Re = DG/µ = 4w/πDµ = (4)(857.14)/[π(0.05584)(653.4)]

NRe

= 29.91

f = 16/NRe

= 16/29.91= 0.5349

gc = 4.17x10


G = w/Ai = 4w/πD

2= 4 (857.14)/[π(0.05584)

2]=350002

( )( ) ( )( )( )( )560.01396104.172

163500020.5349∆p

8

2

i×

=

∆p

i = 1608 lb/sq ft

First Cost = $1.80 x 16 x 7 = $ 201.60 Fixed Charges = $ 0.36 x16 x 7 = $40.32 per year. Mechanical Energy: $ 0.005 / million ft-lb ∆p

i = 1608 lb/sq ft

1 year = 8640 hours Mechanical Energy = (6000 lb/hr)(1608 lb/sq ft)(8640 hrs/yr)/(56 lb/cu ft)


121

= 1488.55 Million Ft-lb / yr. Mechanical Energy = ($ 0.005)(1488.55) = $ 7.44 per year. d. For Length L = 20 ft. Try 6 tubes. w = (6,000 lb/hr) /6 = 1000 lb/hr

( )( )( )( )( )

382200.8π

0.4810004

πkL

4wc=

=

( )( ) ( )( ) 24.173821.250.05584

0.081.86h 3

1

a ==

p

awa12

wc

∆tAhtt =−

( ) ( )( )( )( )( )( )( )( )0.486000

125200.05584πN24.1780100 T=−

N

T = 5.43 tubes or 5 or 6 tubes.

Try 5 tubes. w = (6,000 lb/hr) /5 = 1200 lb/hr

( )( )( )( )( )

458.4200.8π

0.4812004

πkL

4wc=

=

( )( ) ( )( ) 25.684458.41.250.05584

0.081.86h 3

1

a ==

p

awa12

wc

∆tAhtt =−

( ) ( )( )( )( )( )( )( )( )0.486000

125200.05584πN25.68480100 T=−

N

T = 5.11 tubes or 5


Re = DG/µ = 4w/πDµ = (4)(1200)/[π(0.05584)(653.4)]

NRe

= 41.876

f = 16/NRe

= 16/41.876= 0.382

gc = 4.17x10


G = w/Ai = 4w/πD

2= 4 (1200)/[π(0.05584)

2]=490004

( )( ) ( )( )( )( )560.01396104.172

204900040.382∆p

8

2

i×

=

∆p

i = 2814 lb/sq ft

First Cost = $1.80 x 20 x 5 = $ 180.00 Fixed Charges = $ 0.36 x 20 x 5 = $ 36.00 per year. Mechanical Energy: $ 0.005 / million ft-lb ∆p

i = 2814 lb/sq ft

1 year = 8640 hours


122

Mechanical Energy = (6000 lb/hr)(2814 lb/sq ft)(8640 hrs/yr)/(56 lb/cu ft) = 2605 Million Ft-lb / yr. Mechanical Energy = ($ 0.005)(2605) = $ 13.03 per year.

SUMMARY FOR ONE YEAR

Length First Cost Fixed Charges Mech'l Energy Total Amount

8 ft $187.20 $37.44 $2.01 $226.65

12 ft $194.40 $38.88 $4.34 $237.62

16 ft $201.60 $40.32 $7.44 $249.36

20 ft $180.00 $36.00 $13.03 $229.03 Therefore Select 8 ft length of tube as the most economical. -- Ans. 9_005. An experimental heat exchanger consists of a vertical copper tube with an i.d. of 1 inch, heated by a steam jacket for a

length of 10 ft. The steam condenses at a pressure of 1 atm, and the thermal resistance of the copper tube and steam-condensate film is negligible. A cold oil, entering at 50 F, flows upward through the copper tube.

Plot the exit temperature t of the oil vs. the oil flow rate w for the flow-rate range of 1 to 100,000 pounds/hr. In order to

simplify the calculation, the physical properties of the oil will be assumed to be independent of temperature: c

p = 0.5

k = 0.1 µ = 10 centipoises Solution: Flow rate range = 1 to 100,000 lb/hr. t

1 = 50 F

Equations: N

Re < 2100, L/D > 60, Equation formed as discussed on page 240 and 241, reference.

31

L

D

k

µc

µ

DG1.86F

k

Dh p

b1

a

=

2100 < N

Re < 10,000. Equation of Hausen, Page 241.

Re

Reb

s

b

p

pb

m

N

125N0.116

L

D1

0.14µ

µ

k

µc

Gc

h

j3

2

32

32

−=

+

=′′

10,000 < N

Re < 120,000, Equation 9-10c.

0.2

b

0.14

b

w

b

p

pb

L

µ

DG

0.023

µ

µ

k

µc

Gc

h3

2

=


123

at NRe

= 2100, D = 1 in = 0.08333 in

NRe

= DG/µb = 4w/πDµ = 2100

4w / [π(0.08333)(24.2)] = 2100 w = 3,326 lb/hr at N

Re = 10,000, D = 1 in = 0.08333 in

NRe

= DG/µb = 4w/πDµ = 10,000

4w / [π(0.08333)(24.2)] = 10,000 w = 15,838 lb/hr at N

Re = 120000, D = 1 in = 0.08333 in

NRe

= DG/µb = 4w/πDµ = 120,000

4w / [π(0.08333)(24.2)] = 120,000 w = 190,059 lb/hr Interval: 1000, 2000, 3000, 4000, 6000, 8000, 10000, 15000, 20000, 40000, 60000, 80000, 100000 lb/hr

p

awa12

wc

∆tAhtt =−

p

awa12

wc

∆tAhtt +=

@ 1 atm = 212 F, then wall is at 207 F. A

w = πDL = p(0.08333)(10) = 2.6179 sq ft

∆ta = 207 F - (1/2)(t

2 + t

1)

∆ta = 207 F - (1/2)(t

2 + 50) = 182 - 0.5t

2

N

Re < 2100, w = 1000, 2000, 3000 lb/hr:

2

2

a

2

0.5t180

50t

∆t

t1tZ

−

−=

−=

( )( )

−+

=

Z2Z2ln

ZF1

31

L

D

k

µc

µ

DG1.86F

k

Dh p

b1

a

=

c

pµ/k = (0.5)(24.2)/(0.1) = 121

D/L = 0.08333 ft / 10 ft = 0.008333 [w = 1000 lb/hr] N

Re = DG/µ

b = 4w/πDµ = 4(1000)/[(π)(0.08333)(24.2)] = 631.4

( )( )

( )( )( )[ ] 31

0.008333121631.41.86F0.1

0.08333h1

a =

h

a = 19.2F

1


124

p

awa12

wc

∆tAhtt +=

( )( )( )( )0.51000

0.5t1822.617919.2Ftt 2112

−+=

Solving trial and error:

t2 Z ln[(2+Z)/(2-Z)] F1 F2

100 F 0.3788 0.3834 0.988 63.11

70 F 0.1361 0.1363 0.9985 64.76

65 F 0.1004 0.1005 0.999 65 At w = 1000 lb/hr , t

2 = 65 F

[w = 2000 lb/hr] N

Re = DG/µ

b = 4w/πDµ = 4(2000)/[(π)(0.08333)(24.2)] = 1262.8

( )( )

( )( )( )[ ] 31

0.0083331211262.81.86F0.1

0.08333h1

a =

h

a = 24.2F

1

p

awa12

wc

∆tAhtt +=

( )( )( )( )0.52000

0.5t1822.617924.2Ftt 2112

−+=


t2 Z ln[(2+Z)/(2-Z)] F1 F2

65 F 0.1004 0.1005 0.999 59.62

60 F 0.0658 0.0658 1 59.63

59.65 F 0.0634 0.634 1 59.64 At w = 2000 lb/hr , t

2 = 59.64 F

[w = 3000 lb/hr] N

Re = DG/µ

b = 4w/πDµ = 4(3000)/[(π)(0.08333)(24.2)] = 1894.15

( )( )

( )( )( )[ ] 31

0.0083331211894.151.86F0.1

0.08333h1

a =

h

a = 27.7F

1

p

awa12

wc

∆tAhtt +=

( )( )( )( )0.53000

0.5t1822.617927.7Ftt 2112

−+=



125

t2 Z ln[(2+Z)/(2-Z)] F1 F2

59 F 0.05902 0.05904 1 57.37

57.4 F 0.04827 0.04828 1 57.4 At w = 3000 lb/hr , t

2 = 57.4 F

[w = 4000 lb/hr]

Re

Reb

s

b

p

pb

m

N

125N0.116

L

D1

0.14µ

µ

k

µc

Gc

h

j3

2

32

32

−=

+

=′′

c

pµ/k = (0.5)(24.2)/(0.1) = 121

1.25µ

µ0.14

b

s =

1.041110

0.083331

L

D1

32

32

=

+=

+

2πD

4wG =

N

Re = 4w/πDµ = 4(4000)/[(π)(0.08333)(24.2)] = 2526

0.02435N

125N

Re

Re3

2

=−

( )( )

7334430.08333π

40004

πD

4wG

22===

( )( )

( ) ( ) ( )( )0.024350.1161.0411

1.25121

7334430.5

h 32

m =

h

m = 35.266

−

−=

2

2m

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.54000

2.617935.266Ftt

t

2 = 57.082 F

[w = 4000 lb/hr] N

Re = 4w/πDµ = 4(4000)/[(π)(0.08333)(24.2)] = 3788

0.03115N

125N

Re

Re3

2

=−


126

( )( )

11001650.08333π

60004

πD

4wG

22===

( )( )

( ) ( ) ( )( )0.031150.1161.0411

1.25121

11001650.5

h 32

m =

h

m = 67.67

−

−=

2

2m

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.56000

2.617967.67Ftt

t

2 = 59.002 F

[w = 8000 lb/hr]

NRe

= 4w/πDµ = 4(8000)/[(π)(0.08333)(24.2)] = 5051

0.03354N

125N

Re

Re3

2

=−

( )( )

14668860.08333π

80004

πD

4wG

22===

( )( )

( ) ( ) ( )( )0.033540.1161.0411

1.25121

14668860.5

h 32

m =

h

m = 97.15

−

−=

2

2m

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.58000

2.617997.15Ftt

t

2 = 59.671F

[w =10000 lb/hr] N

Re = 4w/πDµ = 4(10000)/[(π)(0.08333)(24.2)] = 6316

0.03431N

125N

Re

Re3

2

=−

( )( )

18336080.08333π

100004

πD

4wG

22===


127

( )( )

( ) ( ) ( )( )0.034310.1161.0411

1.25121

18336080.5

h 32

m =

h

m = 124.23

−

−=

2

2m

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.510000

2.6179124.23Ftt

t

2 = 59.888 F

[w = 15000 lb/hr] N

Re = 4w/πDµ = 4(15000)/[(π)(0.08333)(24.2)] = 9471

0.03407N

125N

Re

Re3

2

=−

( )( )

27504110.08333π

150004

πD

4wG

22===

( )( )

( ) ( ) ( )( )0.034070.1161.0411

1.25121

27504110.5

h 32

m =

h

m = 185.04

−

−=

2

2m

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.515000

2.6179185.04Ftt

t

2 = 59.820 F

[w = 20000 lb/hr]

0.2

b

0.14

b

w

b

p

pb

L

µ

DG

0.23

µ

µ

k

µc

Gc

h3

2

=

c

pb = 0.5

c

pµ/k = 121

(cpµ/k )

2/3 = 24.46378


128

1.25µ

µ0.14

b

s =

N

Re = 4w/πDµ = 4(20000)/[(π)(0.08333)(24.2)] = 12628

( )( )

36672150.08333π

200004

πD

4wG

22===

( )( )

( )( )( )0.2

L

12628

0.0231.2524.46378

36672150.5

h=

h

L = 208.6

−

−=

2

2L

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.520000

2.6179208.6Ftt

t

2 = 58.340 F

[w = 40000 lb/hr] N

Re = 4w/πDµ = 4(40000)/[(π)(0.08333)(24.2)] = 25264

( )( )

73344300.08333π

400004

πD

4wG

22===

( )( )

( )( )( )0.2

L

25264

0.0231.2524.46378

73344300.5

h=

h

L = 363.19

−

−=

2

2L

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.540000

2.6179363.19Ftt

t

2 = 57.29 F

[w = 60000 lb/hr] N

Re = 4w/πDµ = 4(60000)/[(π)(0.08333)(24.2)] = 37883

( )( )

110016440.08333π

600004

πD

4wG

22===

( )( )

( )( )( )0.2

L

37883

0.0231.2524.46378

110016440.5

h=

h

L = 502.38

−

−=

2

2L

t207

157ln

50t∆t


129

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.560000

2.6179502.38Ftt

t

2 = 56.735 F

[w = 80000 lb/hr] N

Re = 4w/πDµ = 4(80000)/[(π)(0.08333)(24.2)] = 50511

( )( )

146688590.08333π

800004

πD

4wG

22===

( )( )

( )( )( )0.2

L

50511

0.0231.2524.46378

146688590.5

h=

h

L = 632.39

−

−=

2

2L

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.580000

2.6179632.39Ftt

t

2 = 56.366 F

[w = 100000 lb/hr] N

Re = 4w/πDµ = 4(100000)/[(π)(0.08333)(24.2)] = 63138

( )( )

183360740.08333π

1000004

πD

4wG

22===

( )( )

( )( )( )0.2

L

63138

0.0231.2524.46378

183360740.5

h=

h

L = 756

−

−=

2

2L

t207

157ln

50t∆t

( )( )( )

−

−⋅+=

2

2112

t207

157ln

50t

0.5100000

2.6179756Ftt

t

2 = 56.096 F

SUMMARY: W, LB/HR t

2, F

1000 65 2000 59.64 3000 57.4 4000 57.082


130

6000 59.002 8000 59.671 10000 59.888 15000 59.82 20000 58.34 40000 57.29 60000 56.735 80000 56.366 100000 56.096 Plot of temperature and flow rate:

9_006 A single-pass heat exchanger is being used to preheat the oil fed to a distillation column. Heat is supplied by saturated


131

steam condensing on the outside of the tubes while the liquid flows inside the tubes. A “promoter” is added to the steam, which causes dropwise condensation. The tube pass consists of 100 tubes, each 20 ft long with an i.d. of 1 inch.

It is proposed to install an emergency “stand-by” heat exchanger in parallel with the existing exchanger described

above. Because of space limitations, the stand-by exchanger will employ tubes only 10 ft in length. The tubes will have an i.d. of 1/2 inch, and a single tube pass will be used.

How many tubes are required in the stand-by exchanger for it to preheat the feed stock to the same temperature as the

original exchanger? Note: It is known that the flow in the original exchanger is in the turbulent range and that the thermal resistance of the

tube and the steam-condensate film is negligible. Solution: Single pass heat exchanger data (old): 100 tubesm 29 ft long, ID = 1 inch = 0.08333 ft New standby heat exchanger L 10 ft, Single pass, ID = 1/2 inch =0.0417 ft Turbulent type. wc

p(t

2 - t

1) = h

LA

wDt

L

w’cp’(t

2’ - t

1’) = h

L’A

w’Dt

L’

h

LA

w = h

L’A

w’

Existing (old): h

L, A

w

Stand By (new); hL’, A

w’

Aw = (100)(π)(0.0833)(20) = 523.58 sq ft

Aw’ = (N

T)(π)(0.0417)(10) = 1.31N

T

where N

T = number of tubes required.

Equation 9-10a.

bb

0.4

b

p0.8

bb

L

πDµ

4w

µ

DG;

k

µc

µ

DG0.023

k

Dh=

=

hLD = (Constant) / D

0.8

Constant = hLD

1.8

hL’(D’)

1.8 = h:(D)

1.8

hL’(0.5 in)

1.8 = hL (1.0 in)

1.8

hL’ / h

L = 3.4822

Then:


132

hLA

w = h

L’A

w’

hL (523.58) = 3.4822h

L(1.31NT)

N

T = No. of tubes = 115 tubes. . . Ans.

- end -

133

CHAPTER 10

HEATING AND COOLING OUTSIDE

TUBES

Chapter 10 Heating and Cooling Outside Tubes

134

10_001. An air heater consists of a bank of vertical 2-inch standard steel pipes, each 6 ft long. Each row, normal to the air flow, consists of 30 tubes with center-to-center spacings of 4 inches. The bank is 40 rows deep, and the row centers are 3 inches apart. Successive rows are staggered, every other row being displaced 1 1/2 inches to the side. Exhaust steam, available at 11 lb/sq in. gauge, will be condensed inside the tubes, and air will be heated from 65 F to 220 F as it flows past the tube bank.

Estimate the capacity of this heater, expressed as pounds of air per hour. Solution: Equation 10-11.

n

f

maxo2

f

om

µ

GDb

k

Dh

=

Using Table 10-5. t

f = t

s - (t

s - t )

m / 2

sT = 4 inches

sL = 3 inches

Do = 2.375 inches = 0.1979 ft

xL = s

L/D

o = 3.0 / 2.375 = 1.263

xT = s

T/D

o = 4.0 / 2.375 = 1.684

Therefore: b

2 = 0.507

n = 0.555 Then;

0.555

f

maxo

f

om

µ

GD0.507

k

Dh

=

@ 11 psig , t

s = 241.51 F - 5 F = 236.51 F

t = (65 + 220) / 2 = 142.6 F t

f = 236.51 - (236.51 - 142.5) / 2 = 190 F

Minimun cross section = Smin S

min = (L)(center-to-center - diameter)(No. of tubes per row)

Smin

= (6 ft)(4.0 in - 2.375 in)(1 ft / 12 in)(30)

Smin

= 24.375 sq ft

Gmax = w / Smin @ 190 F. for air. k

f = 0.01798 Btu/(hr)(sq ft)(deg F per ft)

µf = 0.0515 lb/(hr)(ft)

cp = 0.24 Btu/lb-F


135

( )( )

0.555

m

0.0515

24.375

w0.1979

0.5070.01791

0.1979h

=

0.555

m 0.01644wh =

( ) ( )

−

−

−−−=

220241.51

65241.51ln

220241.5165241.51∆t m

∆t

m = 73.64 F

wcp(t

2 - t

1) = h

mA

w∆t

m

A

w = p(0.1979 ft)(6 ft)(30)(40)

Aw = 4476.4 sq ft

w(0.24)(220 - 65) = (0.01644w0.555

)(4476.4)(73.64) w = 72,707 lb/hr . . . Ans. 10_002. Hot air is flowing inside a steel pipe with an i.d. of 18 inches. Two thermocouples are used to measure the air

temperature, one at the center of the pipe and one embedded in the pipe wall. The thermocouple in the center of the pipe is shielded by a cylindrical metal tube, concentric with the large pipe, which is centered around the junction of the thermocouple. The shield is 1/2 inch in diameter and 12 inches long, and its emissivity is 0.3. The emissivity of the thermocouple junction may be assemed to be 0.7.

The center-line thermocouple indicates a temperature of 917 F, and the temperature of the pipe wall is 631 F. Estimate

the true temperature of the gas. What temperature would the center-line thermocouple indicate if the shield were removed? It is agreed to neglect the open ends of the shield and to neglect conduction in the thermocouple leads. Assume h

c = 8 for convection to the shield and to the thermocouple.

Solution: Equation 10-4b t

g - t

p = (t

p - t

s)(h

r)/h

c

t

p = 917 F

ts = 631 F

hc = 8

Pipe ID = 18 in

Shield = 1/2 in dia. x 12 in, ε = 0.3

Thermocouple juction, ε = 0.7 Equation 4-44a.

( ) 3

avgavgr Tσεn4h +=

with n = 0.

3

avgavgr T4σσh =

σ = 0.1713 x 10-8


136

εavg

= (1/2)(0.3 + 0.7) = 0.5

Tavg

= (1/2)(917 F + 631 F) / 2 + 460 F = 1234 R

hr = 4(0.1713 x 10

-8)(0.5)(1234)

3 = 6.438

t

g -917 = (917 - 631)(6.438)/8

True temperature of gas. t

g = 1147 F ...Ans.

When shield was removed.

εavg

= 0.7

hr = 4(0.1713 x 10

-8)(0.7)[(1/2(t

p+t

s) + 460]

3

hr = (1/2)(t

p + 631 + 920)

3(0.1713 x 10

-8)(0.7)

hr = (1/2)(0.1713 x 10

-8)(0.7)(t

p + 1551)

3

tg - 917 = (917 - 631)(h

r) / (8)

but tg = 1147 F

hr = 8 (1147 - 917) / (917 - 631) = 6.4336

hr = 6.4336 = (1/2)(0.1713 x 10

-8)(0.7)(t

p + 1551)

3

Center-Line Thermocouple indication, t

p = 655 F... Ans.

10_003. An air heater involves condensation of steam inside horizontal tubes and flow of air at right angles to the tubes in a

shell containing a number of vertical segmental baffles. When using exhaust steam condensing at a gauge pressure of 5 lb/sq in. and an air rate of 30,000 lb/hr, the apparatus will heat the air from 80 to 180 F.

If this exchanger were used to heat hydrogen from 80 to 180 F using the same steam pressure, what would be the

capacity of the exchanger, expressed as pounds of hydrogen per hour? Solution: Equation 10-11a:

0.6

f

mao

f

p

f

om

µ

xGD

k

µc0.33

k

Dh3

1

=

G

max = w / S

min

0.6

hf

mao

hf,

p

0.6

af

mao

af,

p

hf

om

af

om

µ

xGD

k

µc

µ

xGD

k

µc

k

Dh

k

Dh

31

31

=

Where subscript a is for air and h is for hydrogen.


137

0.6

hfhf,

p

0.6

afaf,

p

hf

m

af

m

µw

kµc

µw

kµc

kh

kh

31

31

=

wcp( t

2 - t

1) = h

mA

w∆t

m

@ 5 psig (19.7 psia), ts = 227 F - 5 F = 222 F t

f = t

s - (t

s - t)

m/2

tf = 222 - (222 - 130) / 2 = 176 F

Properties of air at 176 F c

p = 0.2410

µ = 0.0506 k = 0.0175

NPr

= (cpµ/k)

f,a = 0.697

Properties of hydrogen at 176 F c

p = 3.5

µ = 0.0242 k = 0.1189

NPr

= (cpµ/k)

f,h = [(3.5)(0.0242)/(0.1189)] = 0.7124

( ) ( )F82.1

180222

80222ln

18022280222∆t m =

−

−

−−−=

( )[ ]( )[ ]

[ ][ ]

hmwm

amwm

h12p

a12p

∆tAh

∆tAh

ttwc

ttwc=

−

−

[ ][ ] mh

ma

hp

ap

h

h

wc

wc=

[ ] [ ]ap

ma

mh

hp wch

hwc =

Then:

0.6

phh

mh

0.6

paa

ma

1/3hPr,

1/3aPr,

0.6

hmph

mhapa

0.6

a

a

1/3hPr,

1/3aPr,

fh

mh

fa

ma

cµ

h

cµ

h

N

N

aµhc

hcw

µ

w

N

N

kh

kh

⋅=

⋅=

( )( )

( )

( )

( )( )( )( )

0.60.4

mh

ma

0.2400.0506

3.50.0242

0.7124

0.697

0.11891

0.01751

h

h

31

31

⋅=

( ) ( )( )3.207120.992756.7943h

h0.4

mh

ma =


138

0.1503h

h

mh

ma =

0.1503h

h

cw

cw

mh

ma

phh

paa==

(30,000)(0.2410) = (0.1503)(w

h)(3.5)

wh = 13,744 lb/hr . . . Ans.

10_004. Air is heated in the annulus of a concentric-tube heat exchanger by steam condensing in the center tube. The o.d. of

the inner tube is 1 inch, and its outer surface is finned with 32 longitudinal fins, each of which extends the length of the tube. The fins extend out radially 1 inch from the tube, and each fin is 0.06 inch thick. The fins are made of an alloy steel (k = 16). The saturation temperature of the condensing steam is 240 F, the air-side heat-transfer coefficient is estimated to be 4, and the thermal resistance of the pipe wall and steam-condensate film are considered negligible.

Calculate the heat transfer rate, expressed as Btu per hour per foot of pipe length at a point in the exchanger at which

the air temperature of 140 F, by the following procedures: a. By using Eq. (10-8b).

b. By using Eq. (10-8a) and adding S/b to xf to allow for the bare tips.

c. By using Eq. (10-8a) for the sides of the fins and evaluating the heat transfer from the tips by using Eq. (10-8).

Solution: a. Equation 10-8b.

( )

−

⋅+

−+

= 1

e

1

1m'

1m'1

2

a

∆tbhq

f2ax

om

kb/hSka/hm' ==

( )0.5

kShba =

then: x

f = 1 inch = 0.08333 ft

h = 4.0 k = 16

(∆T)o = 235 F - 140 F = 95 F

2Lb ≈

t = 0.06 in = 0.005 ft

tLS ⋅=

( )0.5

kShba =

( )( )( )( )

10005.016

42=

=

0.5

a

( )( )2.82843

4

1622k/hkb/hSka/hm' =====

2ax

f = 2(10)(0.08333) = 1.6666


139

0.1889e

1

f2ax=

0.47763.82843

1.82843

1m'

m'-1==

+ b = 2L

( )( )

−

⋅+

−+

= 1

e

1

1m'

1m'1

2

a

∆t2h

L

q

f2ax

om

( )( )( )( )( )

−

+= 1

0.18890.47761

2

10

9524

L

q

q/L = 63.422 Btu/hr-ft pipe length per fins For 32 fins: q

T/L = 32 (63.422) = 2029.5 Btu/hr-ft pipe length . . . Ans.

b. Equation 10-8a, add S/b to x

f

( )( )

( )( )f

f

o

m

ax

axtanh

∆t

∆tη ==

q = h

mA

f(∆t)

m

Af = [(0.08333)(2) + 0.005] (L)(32) = 5.4931L sq ft

S/b = t/2 x

f = 0.08333 ft + 0.005 ft/2 = 0.08583 ft

axf = (10)(0.08583) = 0.8583

( )( )

( )( )

( )0.81

0.8583

0.8583tanh

ax

axtanh

∆t

∆tη

f

f

o

m ====

q = h

mA

fη(∆t)

o=(4)(5.4931L)(0.81)(85)

q/L = 1691 Btu/hr-ft pipe length ... Ans. c. Equation 10-8a for the sides Equation 10-8 for the fin tips Equation 10-8.

( )( )

( )[ ]( )f

f

o

x

axcosh

xxacosh

∆t

∆t −=

at fin tip;

( )( ) ( )fo

x

axcosh

1

∆t

∆t=

x

f = 0.08333 ft

a = 10, axf = 0.8333

( )( ) ( )

0.731120.8333cosh

1

∆t

∆t

o

x ==


140

Equation 10-8a

( )( )

( )( )f

f

o

m

ax

axtanh

∆t

∆tη ==

( )( )

( )( )

0.81880.8333

0.8333tanh

∆t

∆tη

o

m ===

A

f = 32tL = 32(0.005)L = 0.16L

As = (32)(2x

fL) = 32(2)(0.0833)L = 5.3331L

q = hmA

s(∆t)

m + h

tA

t(∆t)

t

say hm = h

t =4

q/L = [(4)(5.3331)(0.8188) + (4)(0.16)(0.73112)](95) q/L = 1704 Btu/hr-ft pipe length... Ans. 10_005. An experimental apparatus consists of a solid horizontal polished copper cylinder spanning a duct. The ends of the

cylinder, which are embedded in the walls of the duct, are maintained at a temperature of 150 F by electrical heaters, and heat is transferred from the copper cylinder to air flowing in the duct. The width of the duct is 4 ft, and the diameter of the cylinder is 1/2 inch. The approach velocity of the air is 15.4 ft/sec, and the air is at a temperature of 130 F and a pressure of 1 atm. a. Calculate the rate of heat transfer from the cylinder to the air. b. Plot the cylinder temperature vs. position along the length of the cylinder.

NOTE: It is to be assumed that radial temperature gradients in the cylinder are negligible and that the effect of the walls of the duct on the heat-transfer coefficient from the air to the cylinder near the walls may be neglected.

Solution: Equarion 10-9.

21

32

f

maxp

f

p

maxp

m

µ

Gz1.0

k

µc

Gc

h−

=

Equation 10-8.

( )( )

( )[ ]( )f

f

o

x

axcosh

xxacosh

∆t

∆t −=

Equation 10-8a

( )( )

( )( )f

f

o

m

ax

axtanh

∆t

∆tη ==

Properties of air at 140 F average.

cpµ/k = N

Pr = 0.702

µf = 0.04844

cp = 0.2406

( )( )( )( )

0.06646014053.34

14414.696

RT

pρ =

+==

ρ = 0.066 lb/cu ft z

p = 1/2 inch = 0.0417 ft

Gmax

= ρV = (0.066)(25.4)(3600) = 6126 lb/hr-sq ft

( )( )

( )( )( ) 2

1

32

0.04844

61260.04171.00.702

61260.2406

hf

m

−

=


141

h

m = 25.7 Btu/hr-sq ft-F

xf = (1/2)(4 ft) = 2 ft

a = (hb/kS)0.5

Copper at 145 F, k =220

b = πD

S = πD2/4

b/S = 4/D = 4/0.0417 = 95.9236

( )( )3.3475

220

95.923625.7a

0.5

=

=

ax

f = (3.3475)(2) = 6.695

a.

( )( )

( )( )

( )( )

0.156.695

6.695tanh

ax

axtanh

∆t

∆tη

f

f

o

m ====

q = h

mA

f(∆t)

m = h

mA

fη(∆t)

o

Af = πDL = π(0.04170(4) = 0.524 sq ft

(∆t)o = 150 F - 130 F = 20 F

q = (25.7)(0.524)(0.15)(20) q = 40.4 Btu/hr . . . Ans. b.

( )( )

( )[ ]( )f

f

o

x

axcosh

xxacosh

∆t

∆t −=

( )[ ]( )6.695cosh

x23.3475cosh

130-150

130-t s −=

( )[ ]( )6.695cosh

x23.347520cosh130t s

−+=

x ts

0 150

0.5 133.75

1 130.7

1.5 130.14

2 130.05 Plot of temperature vs position:


142

10_006. A multipass heat exchanger contains 1200 tubes, each 16 ft long, having an o.d. of 1.25 in. and an i.d. of 1.12 in.

These tubes are arranged as a 12-6 exchanger in which the fluid in the tubes makes 12 equal passes and that in the shell makes 6 corresponding passes. Each tube pass contains 100 tubes, and each shell pass is well baffled. It is desired to predict the performance of this apparatus if 693,000 lb/hr of a hot oil, c = 0.540 gm-cal/(gm)(deg C), were

introduced continuously into the shell at a temperature of 360 F and cold oil at 60 F, c = 0.470 gm-cal/(gm)(deg C), ρ = 56 lb / cu ft, were fed to the first pass of the tubes at a rate of 796,000 lb/hr. From the data and notes given below, calculate the temperature to which the hot oil will be cooled.

Data and Notes. It is estimated that the average coefficient from hot oil to the outer surface of the tubes, including

suitable allowance for scale deposit on both sides, will be 100 Btu/(hr)(sq ft)(deg F), based on the outside surface. The thermal conductivity of the tubes will be taken as 26 Btu/(hr)(sq ft)(deg F per ft). The coefficient h on the inside of the tubes is given by Eq. (9-10b). The average absolute viscosity of the oil in the tubes may be taken as 0.0416 gm/(sec)(cm); and the thermal conductivity of the oil in the tubes is 0.080 Btu/(hr)(sq ft)(deg F per ft).

Solution: Inside Tube (Cold Oil) Cold oil. Equation 9-10b


143

0.2

f

f

p

pb

L

µDG

0.023

k

µc

Gc

h3

2

=

w = 796,000 lb/hr

µ = 0.0416 gm/(sec)(cm) = 10.0672 lb/hr-ft k = 0.080 Btui/(hr)(sq ft)(deg F per ft) cp = 0.47 gm-cal/(gm)(deg C) = 0.47 Btu/lb-F

( )

1163452

10012

1.12

4

π

796000

A

wG

2=

==

( )( )( )( )

( )0.2

f

L

10.0672

116345212

1.12

0.023

0.08

10.06720.47

11634520.47

h 32

=

h

L=h

c = 129.34Btu/(hr)(sq ft)(deg F)

Outside Tube (Hot Oil) w = 693,000 lb/hr cp = 0.54 gm-cal/(gm)(deg C) = 0.54 Btu/lb-F h

h = 100 Btu.(hr)(sq ft)(deg F) including scale deposits on both side;

Equation 8-7B

+

+=

1D

OD

h

1

D

OD

k

x

h

1

U

1

cmw

w

h neglecting tube resistance, etc.

+=

ID

OD

h

1

h

1

U

1

ch OD = 1.25 in ID = 1.12 in

+=

1.12

1.25

129.34

1

100

1

U

1

U = 53.68 Btu/(hr)(sq ft)(deg F) Solving for th2 (temperature of hot oil leaving the heater).

q = UAFG(∆T)

L

Equation 8-19

( )( ) ( )

−

−

−−−=

c1h2

c2h1

c1h2c2h1L

tt

ttln

tttt∆T

t

h1 = 360 F, t

c1 = 60 F

( )( ) ( )

−

−

−−−=

60t

t360ln

60tt360∆T

h2

c2

h2c2L


144

Then, q = whc

h(t

h1 - t

h2) = w

cc

c(t

c2 - t

c1)

(693,000)(0.54)(360-t

h2) = (796,000)(0.47)(t

c2 - 60)

360 - th2

= tc2

- 60

tc2

= 420 - th2

( )( )( ) ( )

( )

−

−−

−−−−=

60t

t420360ln

60tt420360∆T

h2

h2

h2h2L

( )( ) ( )

−

−

−−−=

60t

60tln

60t60t∆T

h2

h2

h2h2L

Therefore use

(∆t)L = (1/2)(360 + t

h2) - (1/2)(60 + t

c2)

(∆t)L = 150 + 0.5(t

h2 - t

c2)

(∆t)L = t

h2 - 60

q = UAFG(∆T)

L

For FG; Using Figure 8-6: 6 shell passes, 12 tub passes Z = “fall” / “rise” = 1.0

( )c1h1

c

c1h1

c1c2H

ttZ

fall""

Z

η

tt

ttη

−==

−

−=

( )( )( ) 300

t360

603601

t - 360η h2h2

H

−=

−=

A = heat transfer area

A = (π)(1.25 / 12)(1200)(16) = 6283.3 sq ft

q = UAFG(∆T)

L

q = (693,000)(0.54)(360 - th2

) = (53.68)(6283.2)(FG)(t

h2 - 60)

( )300

t - 360η h2

H =

1.10952(360 - th2

) = FG(t

h2 - 60)

1.10952F

60F399.43t

G

Gh2

+

+=

F

G from Fig. 8-6

Solving by trial and error method with the derived equations above:

Trial th2

ηH F

G Result t

h2

210 0.5 1.0 217.79 217.79 0.974 1.0 217.79 Therefore: t

h2 = 217.79 F . . . Ans.

- end -

145

CHAPTER 11

COMPACT EXCHANGERS, PACKED AND

FLUIDIZED SYSTEMS

Chapter 11 Compact Exchangers, Packed and Fluidized Systems

146

11_001. A fluidized bed, 3 in. in diameter, is being heated through the external walls. At a given time during the heating operation it is found that the wall temperature is 266 F and the temperature of the fluidized bed is 150 F. The air flow rate was 914 lb/(hr)(sq ft), and the heat flux was 3970 Btu/(hr)(sq ft). The solid being fluidized was glass beads having

an average diameter of 14.8 x 10-4

ft. The absolute density of the glass beads was found to be 159 lb/cu ft, and the dense-phase density of the fluidized bed was 51.7 lb/cu ft. Using these data, calculate the heat-transfer coefficient from the wall to the fluidized bed.

Solution: With the data presents the only computation required is as follows,

( )bw ttA

qh

−=

where q/A = 3970 Btu/(hr)(sq ft) t

w = 266 F

tb = 150 F

( )150266

3970h

−=

h = 34.22 Btu/(hr)(sq ft)(deg F) . . . Ans. 11_002 A carbon-steel ball, 0.975 in. in diameter, is heated and then plunged into a fluidized bed. The temperature at the

center of the steel ball is measured with a thermocouple. In one test after the ball had been in the fluidized bed for 4 min., the center temperature was found to be 96 F. The original temperature of the steel ball was 210 F, and the temperature of the fluidized bed remained constant at 78 F. The fluidized bed was 3 in. in diameter and the solid employed was finely divided activated alumina. In this particular test the air flow rate was 4.0 cu ft/min, measured at standard temperature and pressure. Using these data, calculate the average heat transfer coefficient between the steel ball and the fluidized bed.

Solution: Diameter of carbon steell = 0.975 in

Volume of carbon steel = (4/3)(π)(0.975/12)3 = 3.88243 cu in

Specific heat of steel = 0.12 Btu/lb-F Density of steel = 0.283 lb/cu in Weight of steel = (0.283)(3.88243) = 1.10 lb

q = wcp∆t / θ

q = (1.1)(0.12)(210 - 96) / (4/60) = 225.72 Btuh Average temperature = (1/2)(210 F + 96 F) = 153 F

Surface area of sphere =A = 4π(0.975 / 12)2 = 0.083 sq ft

( )bs

avettA

qh

−=

( )( )961530.083

225.72have

−=

h

ave = 47.71 Btu/(hr)(sq ft)(deg F) . . . Ans.


147

11_003. In order to measure the ability of a fluidized bed to transmit heat from one region to another, measurements were made in which heat was added at one level and removed at some higher level. Temperature gradients were measured in the section between the heater and the cooler. They were found to be essentially linear with distance, and the results for the intermediate section could be expressed as an apparent thermal conductivity by using the usual Fourier conduction equation. In one such investigation, a 3-in. column was employed fluidizing microspherical cracking catalyst with particle sizes from 74 to 160 microns. The superficial air velocity at the average fluidized-bed conditions was 0.230 ft/sec. The average bed pressure was 30.8 in. Hg abs. The air entered the fluidized bed at 75 F and left the top of the unit at 144 F. The water rate to the cooler was 186 lb/hr, and it entered at a temperature of 14.4 C and left at a temperature of 24.0 C. The temperature in the bed between the heater and cooler was measured at two levels 29 in. apart and found to be 209 and 167 F. The pressure drop per foot through the dense phase was 4.25 in. of water. Neglecting the heat losses from the column,

a. Calculate the apparent thermal conductivity.

b. Assuming that the heat is carried by solid at 209 F flowing up and being cooled to 167 F, estimate the pounds of solid that must flow per second.

c. For the density prevailing, estimate the solid velocity corresponding to the answer for (b). Solution: 3-in column, particle size = 74 to 160 microns, use 117 microns = 0.00461 in = 0.000384 ft Vo = 0.230 ft / sec Average bed pressure = 30.8 in Hg abs. Air temperature: 75 F Bottom, 144 F Top Water Cooler data: 186 lb/hr, 14.4 C to 24 C (57.92 F to 75.2 F) Two levels considered: 29 in apart, 209 F and 167 F.

∆p / foot of dense phase = 4.25 inch of water Equation 11-39.

( ) sf

c ρε1Lg

∆pq−=

Equation 11-38.

2m

mf

12m

mfmf

o

ε

ε

ε1

ε1

V

V

−

−=

+

Equation 11-36.

( ) gρρρDV

0.1g

0.9gs

2pmf −=

Solving for ρ

s:

gc = 32.2 (ft)(pounds matter)/(sec)(sec)(pounds force)

g = 32.17 (ft)/(sec)(sec)

∆p/Lf = 4.25 in of water = (4.25/12)(62.4) = 22.1 lb/sq ft per foot

Equation 11-39.

( )( )( )

( ) sρε132.17

22.1032.2−=

Equation No. 1

( ) 22.12ρε1 s =−

Equation 11-36

( ) gρρρDV

0.1g

0.9gs

2pmf −=

RT

pρg =

p = (30.8 / 29.92)(14.6960(144) lb/sq ft = 2178.5 lb/sq ft


148

T = (1/2)(75 + 144) + 460 = 569.5 R

( )( )569.553.34

2178.5ρg =

ρ

g = 0.07172 lb/cu ft

( ) ( ) ( ) ( )( )20.10.9

s2

mf 360032.170.071720.07172ρ0.000389V −= ( )0.9smf 0.07172ρ1.50683V −=

Figure 11-14, Dp = 0.00461 in; m = -1.8

Figure 11-15, iron-oxide catalyst, Dp = 0.00461 in; ε

mf = 0.588

Equation 11-38, Vo = 0.23 ft/sec = 828 ft/hr

2m

mf

12m

mfmf

o

ε

ε

ε1

ε1

V

V

−

−=

+

( )

2m12m

0.9s

ε

0.588

0.5881

ε1

0.07172-ρ1.50683

828

−

−=

+

Substituting Equation no. 1

( )

3.6

s

2.6

s

0.9s

ρ

22.121

0.588

0.412ρ

22.20.07172ρ1.50683828

−

−

−

−=

By trial and error method,

ρs = 90.5 lb/cu ft.

Table A-14, at (1/2)(209 F + 167 F) = 188 F. cs = 0.1151

Equation 11-43, ρb’ = ρ

s

0.45

op

0.5

g

p

0.36

p

s

0.18

g

b

g

pm

µ

GBD

k

µc

c

c'0.58

k

Dh

ρ

ρ=

5.0to0.1µ

GBD op=

For air at (1/2)(75 F + 144 F) = 109/5 F c

p = 0.2402

µ = 0.04658 k

g = 0.01593

cpµ/k

g = 0.7024

Go = ρ

gVo = (0.07172)(828) = 59.384

B =1

( )( )( )0.48956

0.04658

59.3840.0003841

µ

GBD op==

within 0.1 to 5.0 range.

( )( ) ( )0.450.5

0.360.18

m 0.489560.70240.2402

0.1151

0.07172

90.50.58

0.01593

0.000384h

=


149

hm = 40.53

q = hmA∆t

m

Parallel Flow:

( ) ( )62.984

144167

75209ln

14416775209∆t m =

−

−

−−−=

A = π(0.000384 ft)(2.41667 ft) = 0.0029155 sq ft q = (40.53)(0.0029155)(62.984) = 7.4425 Btu/hr a. Apparent Thermal Conductivity

KF = (q/s)(∆x / ∆t)

KF = (7.4425 / 0.04909)(2.41667 / 42)

KF = 8.7236 Btu/(hr)(sq ft)(deg F per ft) . . . ans.

b. Pounds of solid that must flow.

q = wsc

s∆t

7.4425 = ws(0.1151)(209 - 167)

ws = 1.53955 lb/hr = 0.000428 lb/sec . . . ans.

c. Solid velocity:

ρsAV

s = w

s

(90.5)(0.0029155)(Vs) = 1.53955 lb/hr V

s = 5.8349 ft/hr = 0.001621 ft/sec . . . ans.

11_004. The thermal conductivity of a packed bed was measured in the annular space between a center heater and the outer

cooling surface. The internal core, which was 0.84 in. in diameter, was heated with steam at atmospheric pressure. The outer tube, which served as a cooling system, had an internal diameter of 3.06 in. The length of the test section was 24.1 in. In one experiment the annular space between the heater and the cooler was packed with glass beads

0.158 in. in diameter. During a 10-min. run, 14.4 cm2 of condensate was obtained. The temperature of the cooling

surface was 69.8 F. a. Neglecting any temperature drop through the metal wall of the heater, calculate the thermal conductivity of

the packed section. b. During the experiment given above, the voids around the glass beads were filled with air at atmospheric

pressure. The same bed was tested when the absolute pressure was 0.4 mm Hg and the cubic centimeters of condensate obtained in 10 min. decreased to 5.4. Estimate the thermal conductivity in this case.

Solution: Diameter of internal core = 0.84 in Diameter of outer tube = 3.06 in i.d. Length of section = 24.1 in

10 minutes run, 14.4 cm3 of condensate

Temperature of cooling surface = 69.8 F a.


150

Steam properties at 212 F, atmospheric pressure. h

fg = 970.3 Btu/lb

νf = 0.01672 cu ft / lb

νg = 26.80 cu ft / lb

Volume rate of condensate:

= ( 14.4 cm3 / 10 min)(1 in / 2.54 cm)

3(1 ft / 12 in)

3(60 min / hr)

= 0.003054 cu ft / hr. Steam rate = Condensate rate w

s = (0.003054 cu ft / hr) / (0.01672 cu ft / lb)

ws = 0.18266 lb/hr

q = ws∆h

fg = (0.18266)(970.3) = 117.235 Btu/hr

Equation 2-8

∆π=

1

2

m

x

xln

t2kq

L = 24.1 in,

∆t = 212 F - 69.8 F = 142.2 F x

2 = 3.06 in

x1 = 0.84 in

( )( )( )( )( )

π==

0.84

3.06ln

142.22.00832k177.235q m

k

m = 0.1277 Btu/(hr)(sq ft)(deg F per ft) . . . Ans.

b. Steam properties at 212 F, atmospheric pressure. h

fg = 970.3 Btu/lb

νf = 0.01672 cu ft / lb

νg = 26.80 cu ft / lb

Volume rate of condensate:

= ( 5.4 cm3 / 10 min)(1 in / 2.54 cm)

3(1 ft / 12 in)

3(60 min / hr)

= 0.001144 cu ft / hr. Steam rate = Condensate rate w

s = (0.001144 cu ft / hr) / (0.01672 cu ft / lb)

ws = 0.06842 lb/hr

q = ws∆h

fg = (0.06842)(970.3) =66.388 Btu/hr

Equation 2-8


151

∆π=

1

2

m

x

xln

t2kq

L = 24.1 in,

∆t = 212 F - 69.8 F = 142.2 F x

2 = 3.06 in

x1 = 0.84 in

( )( )( )( )( )

π==

0.84

3.06ln

142.22.00832k66.388q m

k

m = 0.04783 Btu/(hr)(sq ft)(deg F per ft) . . . Ans.

- end -

152

CHAPTER 12

HIGH VELOCITY FLOW RAREFIED

GASES

Chapter 12 High Velocity Flow Rarified Gases

153

12_001. It is desired to heat air from a stagnation temperature of 400 R to a stagnation temperature of 2600 R while at the maximum possible mass velocity through a single smooth horizontal tube having an actual i.d. of 0.10 ft, with the temperature of the inner wall maintained at a uniform temperature of 4000 R. The air is to have an absolute pressure of 1 atm at the outlet. Calculate the following items: initial pressure, final temperature of the moving stream, value of 2f(L/D)

max, and the tube length in feet.

Solution: z

1 = T

s1 / T

w = 400 R / 4000 R = 0.1

z2 = T

s2 / T

w = 2600 R / 4000 R = 0.65

p2 = 1 atm

γ = 1.4 a. Equation 12-11b

1

2

2Ma,1

2Ma,1

1

2Ma,2

2Ma,2

2

1

2

21

21

N2

11N

z

N2

11N

z

p

p

ψ

ψ=

−γ+

−γ+

=

Figure 12-5, z

1 = 0.1, z

2 = 0.65, k = 1.4 or Equation 12-13.

Maximum possible mass velocity

z2 = 0.65, N

Ma = 1.0 . . .ψ

2=0.736

z1 = 0.10, N

Ma = 0.16 . . . ψ

1 = 1.971

0.37341.971

0.736

ψ

ψ

p

p

1

2

1

2 ===

p

1 = (1 atm) / (0.3734)

p1 = 2.6781 atm . . . Ans.

b. Equation 12-11a

22 m

2Ma,2s TN

2

11T

−γ+=

(2600) = [1 + 0.2(1)2]T

m2

Tm2

= 2167 R or 1707 F.

c.

max2

1m

D

L2f

z1

z1ln

D

L2f

=

−

−=

0.94450.651

0.11ln

z1

z1ln

D

L2f

2

1

max

=

−

−=

−

−=


154

d. L = tube length in feet

0.9445z1

z1ln

D

L2f

2

1m =

−

−=

D = 0.10 ft

11 m

2Ma,1s TN

2

11T

−γ+=

(400) = [1 + 0.2(0.16)2]T

m1

Tm1

= 399 R

Average Temperature = (1/20(2167 R + 399 R) = 1283 R or 823 F

µ = 0.08224

−γ+γ=

2Ma

2Ma2

c

sG2

N2

11N

pg

TRG

( )( )

( ) ( )( )( )[ ]( )( ) ( )

−+=

22

2

2

12

11.4111.4

14414.7 0.373432.2

260053.34G

G = 15.61 lb/sec-sq ft = 56196 lb/hr-sq ft

GD/µNRe =

N

Re = (56196)(0.1) / (0.08224) = 68332

( )0.32

ReN0.1250.00140f +=

f = 0.00495

( )0.9445

0.1

L0.004952=

L = 9.54 ft. . . . Ans. 12_002. Dry air is to be heated from a stagnation temperature of 500 R to a stagnation temperature of 1000 R, while flowing

continuously through a straight smooth tube heated by steady flow of d-c electricity from one end to the other. The tube will have a very thin wall of uniform thickness and an i.d. of 1.2 inches. The tube is to be constructed of a special alloy which has an electrical resistivity which is independent of temperature over the range involved. At the inlet, the temperature of the wall is to be 1100 R, the static pressure of the air is to be 3.83 atm, and the Mach number of the air is to be 0.25.

Assuming air to be a perfect gas with a molecular weight of 19 and a specific-heat ratio γ of 1.4, calculate the tube length, and estimate the final static pressure.

Solution: p

1 = 3.83 atm, N

Ma1 = 0.25

Ts1

= 500 R

Tw = 1100 R

M = 29, γ = 1.4 T

s2 = 1000 R

z

1 = T

s1 / T

w = 500 R / 1100 R = 0.4545

−γ+γ=

2Ma

2Ma2

c

sG2

N2

11N

pg

TRG


155

RG = 1545 / 29 = 53.3

p = 3.83 atm x 14.696 x 144 = 8105 lb/sq ft g

c = 32.2 (ft)(pound mass)/(sec)(sec)(pound force)

( )( )

( )( )( )( ) ( )

−+=

22

2

2

0.252

11.410.251.4

810532.2

50053.3G

G = 83.856 lb / sec-sq ft = 301882 lb / hr-sq ft z

2 = T

s2/ T

w = 1000 R / 1100 R = 0.9091

at z1 = 0.4545, N

Ma1 = 0.25, ψ

1 = 2.68

at NMa2

= 1, z2 = 0.9091, ψ

2 = 0.8704

Equation 12-11b.

1

2

1

2

p

p

ψ

ψ=

Final static pressure. p

2 = (3.83 atm)(0.8704) / (2.68) = 1.244 atm ... Ans.

Equation 12-12.

D

L2f

z1

z1ln m

2

1 =−

−

µ=

DGNRe

−γ+=

2Mas N

2

11T

( )1m

2T0.25

2

11.41500

−+=

T

m1 = 494 R

( )2m

2T1

2

11.411000

−+=

T

m2 = 833 R

Average T

m = (1/2)(494 + 833) R = 664 R = 204 F

µ = 0.05233

( )( )( )

5768820.05233

3018820.1NRe ==

( )0.32

ReN0.1250.00140f +=

( )0.0032

5768820.1250.00140f 0.32 =+=

( )0.1

L0.00322

0.90911

0.45451ln =

−

−

Tube Length = 28 ft . . . . Ans.


156

12_003. Let Φm be defined as the ratio of the rate of isothermal flow of a gas through a pipe to that flow rate which would exist,

under the same conditions of temperature, inlet and exit pressures, pipe diameter, and pipe length, if the gas obeyed the Hagen-Poiseuille law.

Show that Φm is equal to Φ evaluated at the Knudsen number corresponding to the arithmetic mean of the inlet and

outlet pressures. Solution: Isothermal Flow:

2

ccc

2

c Dg

32µ2µ

g

VdV

D2g

dL4fV

g

VdVdp +=+=ν−

−

+=

dLdpConstant

32µ

DgρV

2c

i

Hagen-Poiseuille Law,

2

cDg

32µ2dL

dp=

−

−

=

dLdp

32µ

DgρV

2c

h

Ratio:

ConstantDg

32µ1

32µ

Dg

Constant32µ

Dg

2c

2c

2c

m

+=

+

=φ

Then

Kn11N1+=φ

Kn2c

11NConstantDg

32µ=

ρDV

µ

2N

N

2N

aRe

aMKn ⋅

πγ=⋅

πγ=

pV

TDRg

232

11

ρV

Dg

232

11Constant

a

Gc

a

c ⋅πγ

=⋅πγ

=

p

TRgD

232

11Constant

Gc⋅

π=

ConstantDg

32111N1

c

Knm

µ+=+=φ

2

Where

ConstantD11g

32µ

ρDV

µ

2N

2ca

Kn =⋅πγ

=

p

TRg

D

µ

2ρDV

µ

2N

Gc

aKn ⋅⋅

π=⋅

πγ=


157

Φm is equal to Φ evaluated at average pressure p. T is constant for isothermal flow.

p

TRg

D

µ

2N

GcKn ⋅⋅

π=

12_004. It is desired to maintain an experimental apparatus at a pressure of 2 x 10-2 mm Hg by means of a vacuum pump. It is

estimated that the rate of air leakage into the apparatus will be 10-5 pounds of air/hr, and this air must be pulled through

10 ft of 1/2-inch i.d. copper tubing to the intake of the vacuum pump. The temperature of the air will be 70 F. Calculate the required intake pressure of the vacuum pump. Solution: Solving p

2 by trial and error.

Equation 12-26

φ

µ=

− 1

Dg

V32

dL

dP2

c

p1 = 2 x 10

-2 mm Hg = 0.0557 lb/sq ft

w = 10-5 pounds of air / hr

∆L = 10 ft D = 1/2 in = 0.04167 ft T = 70 F

At 70 F, µ = 0.0441

ρ = (0.0557) / [(53.34)(70+460)] = 0.000002 lb/cu ft

ρ = 2.0 x 10-6 lb/cu ft

( )( )( )( )

0.0070.04410.04167

104

D

4wN

5

Re =π

=µπ

=−

aMa

V

VN =

( )( )( )( )53053.3432.21.4TRgV Gca =γ=

V

a = 1129 ft/sec

( )( )( )( )26

5

0.04167102.0

104

D2

4wV

π×=

ρπ=

−

−

V = 3666 ft/hr

( )( )0.0009

36001129

3666NMa ==

( )0.191

0.007

0.0009

2N

N

2N

Re

MaKn =

⋅

π=⋅

πγ=

4.1

( ) 3.100.0191111.011N1.0 Kn =+=+=φ

φ

µ=

− 1

Dg

V32

dL

dP2

c

( )( )

( )( ) ( )

=

−

3.10

1

0.04167360032.2

36660.044132

dL

dP22

-dp/dL = 0.0023 lb/sq ft-ft -dp/dL = 0.000826 mmHg/ft


158

For ∆L = 10 ft

-∆p = 0.00826 mm Hg

∆p = -0.00826 mm Hg

p2 = 2 x 10

-2- 8.26 x 10

-3 = 0.01174 mm Hg

Average p = (1/2)(2 x 10-2 + 1.174 x 10

-2) = 1.587 x 10

-2 mm Hg

p = 0.0442 lb/sq ft

ρ = 1.5635 x 10-6 lb/cu ft N

Re = 0.007

V = 4690 ft/hr N

Ma = 0.001154

NKn

= 0.2445

φ = 3.69 -dP/dL = 0.00248 lb/sq ft-ft = 0.000891 mm Hg/ft

∆p = -0.00891 mm Hg

p2 = 2 x 10

-2 - 8.91 x 10

-3 = 0.01109 mm Hg

Average p = (1/2)(2 x 10-2 + 1.109 x 10

-2) = 1.5545 x 10

-2 mm Hg

p = 0.0433 lb/sq ft

ρ = 1.532 x 10-6 lb/cu ft N

Re = 0.007

V = 4786 ft/hr N

Ma = 0.001178

NKn

= 0.2496

φ = 3.75 -dP/dL = 0.002486 lb/sq ft-ft = 0.000893 mm Hg/ft

∆p = -0.00893 mm Hg

p2 = 2 x 10

-2 - 8.93 x 10

-3 = 0.01107 mm Hg

Therefore p

2 = 0.01108 mm Hg . . . Ans.

∞∞∞

159

CHAPTER 13

CONDENSING VAPORS

Chapter 13 Condensing Vapors

160

13_001. It is proposed to heat 10,000 pounds/hr of process water from 60 to 80 F by direct countercurrent contact with air discharged from a drier. The hot air is available at the rate of 10,000 pounds/hr of bone-dry air with a dry-bulb temperature of 120 F and a wet-bulb temperature of 99 F.

For a tower with a cross-sectional area of 10 sq ft, the superficial mass velocities are 1000 lb bone-dry air/(hr)(sq ft) and 1000 lb water/(hr)(sq ft). For the proposed packing and the above superficial mass velocities, available data indicate that the following coefficients should apply: h

Ga = 120 Btu/(hr)(cu ft)(deg F); h

La = 460 Btu/(hr)(cu ft)(deg F).

a. Calculate the local rate of evaporation of water at the bottom of the tower in pounds of water per hour per

cubic foot of tower volume. b. What are the dry- and wet-bulb temperatures of the air leaving the top of the tower? c. Estimate the depth of the packing required. d. Calculate the maximum amount of process water which can be heated from 60 to 80 F by countercurrent

contact with the air from the drier if the tower height were infinite. Solution: t

G1 = 120 F, w

G = 10,000 lb/hr, t

wb = 99 F

wL = 10,000 lb/hr

tL1 = 60 F, tL2

= 80 F

So = 10, G

G = 1000, G

L = 1000, h

Ga

H = 120. h

La

H = 460

at 99 F wet bulb, i

G1 = 70 Btu/lb

Total Volume of tower = S

oz

a = A / Soz

Evaporated water = w

G(H

G1 - H

G2)

Rate of evaporation:

( )zS

HHw

o

G21GG −=

Graph:


161

wG∆i

G = W

Lc

L∆t

L

(10000)(iG1

- iG2

) = (10000)(10)(80 - 60)

iG1

- iG2

= 20

iG2

= 70 - 20 = 50 Btu/lb

Wet Bulb = 85 F Slope of the line:

'G

L

Li

Gi

K

h

tt

ii−=

−

−

aK

ah

K

h'G

L

'G

L =

but

s'GG cKh =

s

G

L

'G

L cah

ah

K

h=

c

s = 0.24 + 0.45H

G

say cs = 0.25

( )( )

( ) 0.960.25120

460

K

h'G

L ==

0.96tt

ii

Li

Gi −=−

−

∫∫ −

=−

=Gi

GG

Gi

G

'G

G

ii

diz

ii

di

aK

Gz

( )

aiG

G21G

Gi

G

ii

ii

ii

di

−

−=

−∫

(iG - i

i)a = (1/3)[(50-33.3) + (60-41.7) + (70-53.3)] = 17.23 Btu/lb

1.160817.23

5070

ii

di

Gi

G =−

=−∫

( )( )( )

2.083120

0.251000

ah

G

aK

Gz

G

G

'G

GG ====

( )( ) ft2.4181.16082.083ii

dizz

Gi

GG ==

−= ∫

To solve for H

G1 , H

G2 and H

G.

iG1

= (1061.4 + 0.45tG1

)HG1

+ 0.24tG1

iG1

= 70 = (1061.4 + 0.45(120))HG1

+ 0.24(120)

HG1

= 0.03694 lb/lb

i

G2 = (1061.4 + 0.45t

G2)H

G2 + 0.24t

G2

From graph, tG2 = 90 F

50 = (1061.4 + 0.45(90))HG2

+ 0.24(90)

HG2

= 0.02577 lb/lb


162

HG = (1/2)(H

G1 +H

G2) = (1/2)(0.03694 + 0.02577) = 0.03136 lb/lb

cs = 0.24 + 0.45(0.03136) = 0.2541( ~ 0.25)

zG = (1000)(0.2541)/(120) = 2.1175

( )( )

( ) 0.9740.2541120

460c

ah

ah

K

hs

G

L

'G

L ===

0.974K

h

tt

ii'G

L

Li

Gi −=−=−

−

( )( ) ft2.45801.16082.1175ii

dizz

Gi

GG ==

−= ∫

a. Rate of evaporation

( )zS

HHw

o

G21GG −=

( )( )( )( )2.458010

0.025770.0369410000 −=

= 4.544 lb/(hr)(cu ft) b. t

G2 = 90 F, t

wb = 85 F

c. z = 2.4580 ft d. If tower height is infinite i

G2 = 26.7 Btu/lb

( )L2L1L

G1G2

G

L

ttc

ii

w

W

−

−=

( )( )( )( )80-601

7026.710000WL

−=

W

L = 21,650 lb/hr . .Ans.

13_002. The following data were obtained in tests on a packed column being used to cool water by countercurrent contact with

atmospheric air. Water was fed to the top of the tower at a rate of 10,000 lb/hr and a temperature of 103 F and left the tower at a temperature of 82 F. Air was fed to the tower at a rate of 21,300 lb of wet air/hr, with an entering temperature of 105 F and a dew point of 65 F. The temperature of the exit air was 90.5 F.

a. Assuming that the flow rates of dry air and water remain the same and that the water still enters at 103 F,

calculate the temperature to which the water will be cooled in this tower on a day when the atmospheric air is at temperature of 100 F and has a dew point of 81.5 F. What would be the temperature of the exit air?

b. Estimate the quantity of water evaporated per hour under the test conditions and under the new conditions. Solution: Solving for wet bulb temperature of air at 105 F DB, DP = 65 F @ DP = 65 F, ps = 0.3097 psia

0.0133928.97

18.02

0.309714.696

0.3097

28.97

18.02

pP

pH

s

si =

−=

−=

t

G1 = 105 F, H

G1 = H

i = 0.01339

psG1

= 1.112 psia

iG1

= 1061.4HG1

+ cstG1

cs = 0.24 + 0.45H

G1

cs = 0.24 + 0.45(0.01339)

cs = 0.24603


163

iG1

= 1061.4(0.01339) + (0.24603)(105)

iG1

= 50.05 Btu/lb

Table A-26 t

wbG1 = 76.20 F

tL1

= 103 F, tL2

= 82 F

tG2

= 90.5 F

Dry air flow w

G = 21,300 - 0.01339w

G

wG = 21,019 lb/hr dry air.

Heat Balance

wG∆iG = W

cL∆t

L

W = 10,000 lb/hr c

L = 1.0

∆tL = 104 F - 82 F = 21 F

(21,019)(iG2

- iG1

) = (10,0000)(1.0)(21)

(21,019)(iG2

-40.05) = (10,0000)(1.0)(21)

iG2

= 50.04 Btu/lb

iG2

= 1061.4HG2

+ cstG2

cs = 0.24 + 0.45H

G2

iG2

= (1061.4 + 0.45tG2)HG2 + 0.24tG2

50.04 = [1061.4 + 0.45(90.5)]HG2

= 0.24(90.5)

HG2

= 0.02570

Assume water recirculation; Equilibrium temperature equal to t

wb.

ti = t

wb @ t

G2 = 50.04

twb

= 85.5 F

Solving for Hi. i

i = (1061.4 + 0.45t

i)H

i + 0.24t

i

50.04 = [1061.4 + 0.45(85.5)]Hi + 0.24(85.5)

Hi = 0.02684

Equation 13-34.

iG2

iG1

t tt

ttln

z

z

−

−=

3.985.590.5

85.5105

tt

tt

iG2

iG1 =−

−=

−

−

Equation 13-35.

G2i

G1i

G HH

HHln

z

z

−

−=

11.80.025700.02684

0.013390.02684

HH

HH

G2i

G1i =−

−=

−

−


164

a. For new condition: w

G = 21,019 lb/hr

W = 10,000 lb/hr Solving for wet bulb temperature of air at 100 F DB, DP = 81.5 F @ DP = 81.5 F, p

s = 0.5356 psia

0.0235328.97

18.02

0.535614.696

0.5356

28.97

18.02

pP

pH

s

si =

−=

−=

t

G1 = 100 F, H

G1 = H

i = 0.02353

psG1

= 0.9492 psia

iG1

= 1061.4HG1

+ cstG1

cs = 0.24 + 0.45H

G1

cs = 0.24 + 0.45(0.02353)

cs = 0.25059

iG1

= 1061.4(0.02353) + (0.2353)(100)

iG1

= 50.03 Btu/lb

Table A-26 t

wbG1 = 85.16 F

tL1

= 103 F

tL2

= unknown

tG2

= unknown

wG∆i

G = W

Lc

L∆t

L

(21,019)(iG2

- 50.03) = (10,000)(1.0)(103 - tL2

)

tL2

= 208.158 - 2.1027iG2

Other Equations. i

G2 = (1061.4 + 0.45t

G2)H

G2 + 0.24tG2

ii = (1061.4 + 0.45ti)Hi + 0.24ti

−=

28.97

18.02

pP

pH

s

si

3.9tt

tt

iG2

iG1 =−

−

11.8HH

HH

G2i

G1i =−

−

w

G(i

G2 - i

G1) = Wc

L(t

L1 - t

L2)

For t

i (Wet Buld at i

G2)

t

G1 - t

i = 3.9(t

G2 - t

i)

Equation 1. t

G2 = (t

G1 - t

i) / 3.9 + t

i

Equation 2. ps @ ti


165

−=

28.97

18.02

pP

pH

s

si

Equation 3. H

G2 = H

i - (H

i - h

G1) / 11.8

Equation 4. i

G2 = (1061.4 + 0.45t

G2)H

G2 + 0.24t

G2

Equation 5.

( )

LL

G1G2GL1L2

cW

iiwtt

−−=

t

L2 = 208.158 - 2.1027i

G2

Equation 6. t

i is the wet bulb of i

G2 (Table A-26)

Equation 7. i

i = i

G2

t

G1 = 100 F, t

L1 = 103 F, i

G1 = 50.0, H

G1 = 0.02353

By trial and error method.

ti tG2 ps Hi HG2 iG2=ii ti

90 92.56 0.6982 0.03103 0.03039 55.74 89.86

89.86 92.46 0.6955 0.0309 0.03028 55.59 89.75

89.75 92.38 0.6934 0.0308 0.03018 55.44 89.64

89.64 92.3 0.6913 0.037 0.03009 55.34 89.56

89.56 92.24 0.6898 0.03064 0.03004 55.27 89.51

89.51 92.2 0.6888 0.03059 0.03 55.21 89.47

89.47 92.17 0.6881 0.03056 0.02996 55.16 89.43

89.4 92.12 0.6867 0.03049 0.0299 55.08 89.37

89.35 92.08 0.6858 0.03045 0.02986 55.03 89.34

89.34 92.07 0.6856 0.03044 0.02985 55.02 89.33

89.33 92.07 0.6854 0.03043 0.02985 55.02 89.33 Exit air temperature = t

G2 = 92.07 F . . Ans.

Exit water temperature = t

L2 = 208.158 - 2.1027 (55.02) = 92.47 F . . Ans.

b. Water Evaporated per hour

∆W = wG(H

G2 - H

G1)

Test condition:

∆W = (21,019)(0.02570 - 0.01339) = 259 lb/hr . . Ans. New Condition:


166

∆W = (21,019)(0.02985 - 0.02353) = 133 lb/hr . . Ans. 13_003. It is proposed to use am existing tubular heat exchanger to condense exhaust steam. The steam, saturated at a gauge

pressure of 5 lb/sq in., is to condense on the outside of the 3/4-inch. 18 BWG copper condenser tubes. The water, available at a temperature of 75 F, is to flow through the tubes in a single pass. The tubes in the exchanger are 12 ft long, and the exchanger is presently installed in a horizontal position. The exchanger could be installed in a vertical position if extra piping were provided. Assuming the tube to be clean and neglecting any effect of vapor velocity expressed as pounds of steam condensed per hour, for:

a. The horizontal position. b. The vertical position. Note: The tube layout for the horizontal position is given in the table below, where n is the number of vertical tiers

having N tubes per tier (thus the total number of tubes is 385):

N 3 9 13 15 17 19 21 23

n 2 2 2 2 2 4 46 63 Solution: a. Horizontal position

hm/h

i = 1/N

0.25

N 3 9 13 15 17 19 21 23

hm/hi 0.7598 0.5774 0.5266 0.5081 0.4925 0.4925 0.4671 0.4566

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )23

0.456630.467160.479040.492520.508120.526620.577420.75982

h

h

i

m +++++++=

h

m/h

i = 0.5138

Equation 13-16.

1/32

f3

f

1/4

fo

2f

3f

w

gLk0.95

tND

gk0.725h

µ

ρ=

∆µ

λρ=

f

m

1/4

fo

2f

3f

1tD

gk0.725h

∆µ

λρ=

at p = 5 psig, t

sv = 227 F,

Do = 0.75 in = 0.0625 ft

at infinite water flow rate, ts = t2, t

2=t

1, t

w = t

1

ts = (1/2)(227 F+ 75 F) = 151 F

tf = t

sv - (3/4)(∆t) = 227 - (3/4)(227 - 75) = 113 F

Properties: k

f = 0.369

ρf = 61.82

µf = 1.465

at 5 psig, λ = 960.72 Btu/lb


167

( )1/3

2f

3f

1/4

fo

2f

3f

w

gLk0.95

tD

gk0.51380.725h

µ

ρ=

∆µ

λρ=

f

m

∆t = 225 F - 75 F = 152 F

( ) ( ) ( ) ( )( )( )( )( )

( ) ( ) ( )( )( )

1/3823

1/4823

w1.465

12104.1761.820.3690.95

1521.4650.0625

960.72104.1761.820.3690.51380.725

×=

×

w = 3,020 lb/hr . . . Ans. b. Vertical position Equation 13-3

1/4

f

2f

3f

mtL

gk0.943h

∆µ

λρ=

Equation 13-4

3/1

f

2

f 447.1

−

µ

Γ=

ρ

µ1/3

2f

3f

mgk

h

D

w

π=Γ

3/1

f

2

f 447.1

−

µπ=

ρ

µ

D

w

gkh

1/3

2f

3f

m

1/3

2f

2f

3f

1/3

fm

gk

4w

D1.47h

µ

ρ

µπ=

1/3

f

2f

3f

1/3

mw

gDk

41.47h

µ

ρ

π=

1/3

f

2f

3f

mw

gDk1.3563h

µ

ρ=

Then:

1/3

f

2f

3f

1/4

f

2f

3f

mw

gDk1.3563

tL

gk0.943h

µ

ρ=

∆µ

λρ=

( ) ( ) ( )( )( )( )( )

( )( ) ( ) ( )( )

1/3823

1/4823

1.465w

104.1761.820.3690.06251.3563

152121.465

960.72104.1761.820.3690.943

×=

×

w = 145 lb/hr per tube For 23 tubes w = 23 x 145 w = 3,335 lb/hr . . . Ans.


168

13_004. Tests have been made to determine whether or not heat-transfer coefficients for organic liquids with high surface

tension would follow the prediction of Nusselt when condensing on horizontal finned tubes, described below. The over-all heat-transfer coefficients were measured at a series of water velocities and were plotted by the graphical method of Wilson to obtain the condensing-film coefficients.

Calculate the experimentally determined heat-transfer coefficient for acetone condensing at 154 F. Compare the

calculated coefficient with the coefficient predicted by the Nusselt equations. Physical dimensions of copper condenser tube. Fifteen fins per in. Root diameter, 0.621 inch. Length of finned section, 34 3/8 inches. Diameter over fins,0.746 inch. Length of cooled section, 5 7/8 inches. Diameter of plain ends, 0.75 inch. Actual outside area, 1.452 sq ft. Physical properties of acetone: Latent heat of vaporization at 154 F = 121.4 cal/gm Viscosity at 94 F = 0.30 centipoise Thermal conductivity = 0.1 Btu/(hr)(sq ft)(deg F per ft) Test Data. Avg. bulk-water temperature = 74.0 F Avg. condensing-vapor temperature = 154.0 F Avg. mean temperature difference = 80.0 F

Run

number

Avg water

velocity,

ft/sec

U, Btu/(hr)(sq

ft)(deg F)(based

on outside area)

Water rate,

lb/hr

Temperature

rise of water,

deg F

48 4.6 254.4 1709 17.28

51 6.43 289.6 2394 14.08

53 8.6 324.3 3208 11.75

55 10.68 349.1 3982 10.17

57 12.4 367.5 4620 9.25 Solution: Data for Wilson Plot

V Uo 1/V (1/V)^0.8 1/Uo

4.6 254.5 0.21739 0.29498 0.003929

6.43 289.6 0.15552 0.22565 0.003453

8.6 324.3 0.11627 0.1788 0.003084

10.68 349.1 0.09363 0.15036 0.002865

12.4 367.5 0.08065 0.13344 0.002721

( )0.8

1

wvo VC

1RR

U

1

′++=

From curve-fitting technique R

v + R

w = 0.001740

1/C1 = 0.007478

C1 = 133.726


169

then

w

ww

vv

k

xR:

h

1R ==

For copper condenser tube , 154 F. k = 220 Btu/hr-sq ft-deg F per ft For 3/4 in o.d. x

w = 0.746 inch - 0.621 inch = 0.125 inch

Rw = (0.125 / 12) / 220 = 0.00004735

0.000047350.001740R)R(Rh

1R wwv

vv −=−+==

R

v = 1/h

v = 0.001693

hv = 591 Btu/hr-sq ft- deg F . . . Ans.

By Nusselt Equation:

1/4

fo

2f

3f

m∆tµD

gλρk0.725h

=

λ = 121.4 cal / gm = 218.5 Btu/llb

µf = 3 x 2.42 = 0.726

Do = 0.75 inch = 0.0625 ft

ρf of acetone = 49.4 lb/cu ft

( ) ( ) ( )( )( )( )( )

1/4823

m800.7260.0625

218.5104.1749.40.10.725h

×=

h

m = 361 Btu/hr-sq ft-deg F . . . Ans. (< 591)

Nusselt equation is for bare horizontal tubes and is lesser value than finned tube. 13_005. Ultimately it will be necessary to design a multitubular total condenser for an ammonia-refrigeration plant to condense

8000 pounds of ammonia/hr, but because of uncertainties in some of the design data, it is planned to construct a pilot unit, consisting of a single tube of steel, having an o.d. of 1.000 inch and a wall thickness of 0.049 inch, provided with a suitable external water jacket. In the pilot unit, ammonia vapor at an absolute pressure of 300 pounds/sq inch and a temperature of 300 F is to be introduced continuously at the top of the vertical tube at a rate of 20 pounds/hr. The condensate will leave at the bottom, and brackish clean river water at 80 F will flow vertically upward through the jacket, leaving at 105 F. Because of limitations of headroom, it is desired to attain complete condensation at the bottom, thus using the shortest possible length of condenser tube. Other data are given below.

a. Estimate the tube length.

b. Estimate the temperature of the condensate leaving the bottom of the tube. c. Should the pilot plant differ from the one proposed? Enthalpy of liquid ammonia:

t, deg F 80 90 100 110 123

i, Btu/lb 131.5 143 154.5 166 181 Properties of NH3, vapor at 300 psia:

t, deg F Saturated

vapor at 123 F

Superheated

vapor at 300 F

i, Btu/lb 634 758

v, cu ft/lb 1.003 1.496


170

The critical pressure is 1638 lb/sq in. abs, and critical temperature is 270 F. It will be assumed that the ammonia side will remain clean, but the water side will become fouled. The coefficient on

the water side when clean will be 2000 Btu/(hr)(sq ft)(deg F). Film-type condensation will be obtained. The saturation

temperature is 123 F; ρL = 35.2 lb per cu ft.

Solution:

mowpwwcpscsspdss tAUtcwtcwtcwwq ∆=∆=∆+∆+λ= s

( ) ( ) ( )

3mo2mo1momo tAUtAUtAUtAU ∆+∆+∆=∆

ccdw

w

hdmo h

1

h

1

k

x

h

1

h

1

U

1++++=

1/h

dh = 0

hc = 2000

xw = 0.049 inch = 0.004083 ft

kw = 26

For Brackish Clean River Water, 80 F to 105 F, Table 8-2 h

dc = 750

2000

1

750

1

26

0.0040830

h

1

U

1

mo

++++=

0.002h

1

U

1

mo

+=

w

s = 20 lb/hr

1. For temperature of condensate = 80 F i = 131. 5 Btu/lb Water flow rate, w

w.

ww(1.0)(105 - 80) = (20)(758 - 131.5)

ww = 501.2 lb/hr

Let t1 = temperature of water at saturated liquid ammonia

and t2 = temperature of water at saturated vapor ammonia.

For t

1:

(501.2)(1.0)(t1 - 80) = (20)(181 -131.5)

t1 = 82 F

For t

2:

(501.2)(1.0)(105 - t2) = (20)(758 - 634)

t2 = 100 F

Condensing section: Water from t

1 = 82 F to t

2 = 100 F

Equation 13-4.

31

31

f2

f3

f

2f

mµ

4Γ1.47

gρk

µh

−

=

πD

wΓ =


171

ρf =ρ

L = 35.2

ρf

2 = ρ

f(ρ

f - ρ

v)

ρv = 1/ 1.003 = 0.997

ammonia at 123 F.

µf = 0.6 x 2.42 = 1.452 (Table A-18, Figure A-4)

kf = 0.283 (Table A-10)

D = 1.0 in = 0.08333 ft

( )( )( )

210.51.4520.08333π

204

πDµ

4w

µ

4Γ

ff

===

ρf

2 = ρ

f(ρ

f - ρ

v) = 35.2 ( 35.2 - 0.997) = 1204

( )( ) ( )( )

( ) 31

31

210.51.47104.1712040.283

1.452h

83

2

m−

=

×

hm = 433.5 Btu/hr-sq ft-deg F

Table 13-2.

1.7826100123

82123

∆t

∆t

ot

ob =−

−=

F

c = 1.047

Fch

m = (1.047)(433.5) = 453.9

002.0+=mco hF

1

U

1

( )

002.0+=453.9

1

U

1

o U

o = 237.9 Btu/hr-sq ft-deg F

q = UoA∆t

om

( ) ( )31.14

100123

82123ln

10012382123∆t om =

−

−

−−−=

A = πDL q = w

wcp

w(t

2 - t

1)

Then:

(501.2)(1.0)(100 - 82) = (237.9)(π)(0.08333)(L)(31.14) L

c = 4.65 ft of condensing section

Desuperheating section: Equarion 9-10a

0.4

p0.8

bb

L

k

µc

µ

DG0.023

k

Dh

=


172

bb πDµ

4w

µ

DG=

t

ave = (1/2)(123 + 300) = 211.5 F

TR = (211.5 + 460) / (270 + 460) = 0.92

pR = p/p

c = 300 / 1638 = 0.183

Viscosity ratio

1.05µ

µ

1,t

p,t=

Table A-19, Figure A-6

µ1,t

= 0.0126 x 2.42 = 0.03050

µp,t

= 1.05 x 0.0305 = 0.03203

( )( )( )

95410.032030.08333π

204

πDµ

4wN

bRe ===

Table A-12 k

b = 0.0192

Figure A-3. c

p = 0.25

( )( )

( ) ( )( )( )

0.40.8L

0.01920

0.032030.2595410.023

0.0192

0.08333h

=

h

L = h

m = 5.70

0.002h

1

U

1

mo

+=

0.0025.70

1

U

1

o

+=

U

o = 5.636

( ) ( )80.47

100123

105300ln

100123105300∆t om =

−

−

−−−=

q = w

wc

pw(105 - t

2)

Then

(501.2)(1.0)(105 - 100) = (5.636)(π)(0.08333)(L)(80.47) L

ds = 21.1 ft of desuperheating section

Subcooling section: Equation 9-37a.

81

31

f

p

3 2f

2f

3f

m

µ

4Γ

k

µc0.01

µgρk

h

=

t

f = t

m = t

sv - 3(t

sv - t

s)/8

tf = 123 - 3(123 - 81)/8 = 107 F

Then at 107 F


173

kf = 0.277 (Table A-10)

µf = 0.7 x 2.42 = 1.694 (Figure A-4, Table A-18)

cp = 1.57 (Table A-16)

( )( )( )

180.41.6940.08333π

204

πDµ

4w

µ

4Γ

ff

===

( )( )9.6014

0.277

1.6941.57

k

µcp==

( ) ( ) ( )[ ] ( ) 15641.694104.1735.20.277µgρk 3 28233 2

f2

f3

f =×=

( ) ( ) 8

13

1

180.49.60140.011564

hm =

hm = 63.64

0.002h

1

U

1

mo

+=

0.00263.64

1

U

1

o

+=

U

o = 56.45

q = wwcp

w(t

1 - 80)

( ) ( ) ( ) ( )[ ] 20.5808082123

8080

82123ln

808082123∆t

21

om =−+−=

−

−

−−−=

Then:

(501.2)(1.0)(82 - 80) = (56.45)(π)(0.08333)(L)(20.5) L

sc = 3.31 ft of subcooling section

Total Length = L

sc + L

c + L

ds = 3.31 ft + 4.65 ft + 21.1 ft = 29.06 ft

2. For temperature of condensate = 90 F i = 143 Btu/lb Water flow rate, w

w.

ww(1.0)(105 - 80) = (20)(758 - 143)

ww = 492 lb/hr



For t

1:

(492)(1.0)(t1 - 80) = (20)(181 -143)

t1 = 81.55 F

For t

2:

(492)(1.0)(105 - t2) = (20)(758 - 634)

t2 = 99.96 F


1 = 81.55 F to t

2 = 99.96 F


174

Equation 13-4.

31

31

f2

f3

f

2f

mµ

4Γ1.47

gρk

µh

−

=

πD

wΓ =

ρ

f =ρ

L = 35.2

ρf

2 = ρ

f(ρ

f - ρ

v)

ρv = 1/ 1.003 = 0.997

ammonia at 123 F.


kf = 0.283 (Table A-10)

D = 1.0 in = 0.08333 ft

( )( )( )

210.51.4520.08333π

204

πDµ

4w

µ

4Γ

ff

===

ρf

2 = ρ

f(ρ

f - ρ

v) = 35.2 ( 35.2 - 0.997) = 1204

( )( ) ( )( )

( ) 31

31

210.51.47104.1712040.283

1.452h

83

2

m−

=

×


Table 13-2.

1.79999.96123

81.55123

∆t

∆t

ot

ob =−

−=

F

c = 1.048

Fch

m = (1.048)(433.5) = 454.3

002.0+=mco hF

1

U

1

( )

002.0+=454.3

1

U

1

o U

o = 238 Btu/hr-sq ft-deg F

q = UoA∆t

om

( ) ( )31.35

99.96123

81.55123ln

99.9612381.55123∆t om =

−

−

−−−=

A = πDL q = w

wcp

w(t

2 - t

1)

Then:

(492)(1.0)(99.96 - 81.55) = (238)(π)(0.08333)(L)(31.35) L



175


0.4

p0.8

bb

L

k

µc

µ

DG0.023

k

Dh

=

bb πDµ

4w

µ

DG=

t

ave = (1/2)(123 + 300) = 211.5 F

TR = (211.5 + 460) / (270 + 460) = 0.92

pR = p/p

c = 300 / 1638 = 0.183

Viscosity ratio

1.05µ

µ

1,t

p,t=


µ1,t

= 0.0126 x 2.42 = 0.03050

µp,t

= 1.05 x 0.0305 = 0.03203

( )( )( )

95410.032030.08333π

204

πDµ

4wN

bRe ===

Table A-12 k

b = 0.0192

Figure A-3. c

p = 0.25

( )( )

( ) ( )( )( )

0.40.8L

0.01920

0.032030.2595410.023

0.0192

0.08333h

=

h

L = h

m = 5.70

0.002h

1

U

1

mo

+=

0.0025.70

1

U

1

o

+=

U

o = 5.636

( ) ( )80.51

99.96123

105300ln

99.96123105300∆t om =

−

−

−−−=

q = w

wc

pw(105 - t

2)

Then

(492)(1.0)(105 - 99.96) = (5.636)(π)(0.08333)(L)(80.51) L




176

81

31

f

p

3 2f

2f

3f

m

µ

4Γ

k

µc0.01

µgρk

h

=

t

f = t

m = t

sv - 3(t

sv - t

s)/8

tf = 123 - 3(123 - 80.78)/8 = 107 F

Then at 107 F k

f = 0.277 (Table A-10)


cp = 1.57 (Table A-16)

( )( )( )

180.41.6940.08333π

204

πDµ

4w

µ

4Γ

ff

===

( )( )9.6014

0.277

1.6941.57

k

µcp==

( ) ( ) ( )[ ] ( ) 15641.694104.1735.20.277µgρk 3 28233 2

f2

f3

f =×=

( ) ( ) 8

13

1

180.49.60140.011564

hm =

hm = 63.64

0.002h

1

U

1

mo

+=

0.00263.64

1

U

1

o

+=

U

o = 56.45

q = wwcp

w(t

1 - 80)

( ) ( )22.12

8090

81.55123ln

809081.55123∆t om =

−

−

−−−=

Then:

(492)(1.0)(81.55 - 80) = (56.45)(π)(0.08333)(L)(22.12) L


Total Length = L

sc + L

c + L

ds = 2.33 ft + 4.64 ft + 20.88 ft = 27.85 ft

3. For temperature of condensate = 100 F i = 154.5 Btu/lb Water flow rate, w

w.

ww(1.0)(105 - 80) = (20)(758 - 154.5)

ww = 482.8 lb/hr



For t

1:

(482.8)(1.0)(t1 - 80) = (20)(181 -154.5)

t1 = 81.1 F


177

For t

2:

(482.8)(1.0)(105 - t2) = (20)(758 - 634)

t2 = 99.86 F


1 = 81.1 F to t

2 = 99.86 F

Equation 13-4.

31

31

f2

f3

f

2f

mµ

4Γ1.47

gρk

µh

−

=

πD

wΓ =

ρ

f =ρ

L = 35.2

ρf

2 = ρ

f(ρ

f - ρ

v)

ρv = 1/ 1.003 = 0.997

ammonia at 123 F.


kf = 0.283 (Table A-10)

D = 1.0 in = 0.08333 ft

( )( )( )

210.51.4520.08333π

204

πDµ

4w

µ

4Γ

ff

===

ρf

2 = ρ

f(ρ

f - ρ

v) = 35.2 ( 35.2 - 0.997) = 1204

( )( ) ( )( )

( ) 31

31

210.51.47104.1712040.283

1.452h

83

2

m−

=

×


Table 13-2.

1.81199.86123

81.1123

∆t

∆t

ot

ob =−

−=

F

c = 1.049

Fch

m = (1.049)(433.5) = 454.74

002.0+=mco hF

1

U

1

( )

002.0+=454.74

1

U

1

o U


q = UoA∆t

om

( ) ( )31.60

99.86123

81.1123ln

99.8612381.1123∆t om =

−

−

−−−=

A = πDL


178

q = wwcp

w(t

2 - t

1)

Then:

(482.8)(1.0)(99.86 - 81.1) = (238)(π)(0.08333)(L)(31.60) L



0.4

p0.8

bb

L

k

µc

µ

DG0.023

k

Dh

=

bb πDµ

4w

µ

DG=

t

ave = (1/2)(123 + 300) = 211.5 F

TR = (211.5 + 460) / (270 + 460) = 0.92

pR = p/p

c = 300 / 1638 = 0.183

Viscosity ratio

1.05µ

µ

1,t

p,t=


µ1,t

= 0.0126 x 2.42 = 0.03050

µp,t

= 1.05 x 0.0305 = 0.03203

( )( )( )

95410.032030.08333π

204

πDµ

4wN

bRe ===

Table A-12 k

b = 0.0192

Figure A-3. c

p = 0.25

( )( )

( ) ( )( )( )

0.40.8L

0.01920

0.032030.2595410.023

0.0192

0.08333h

=

h

L = h

m = 5.70

0.002h

1

U

1

mo

+=

0.0025.70

1

U

1

o

+=

U

o = 5.636

( ) ( )80.63

99.86123

105300ln

99.86123105300∆t om =

−

−

−−−=

q = w

wc

pw(105 - t

2)

Then


179

(482.8)(1.0)(105 - 99.86) = (5.636)(π)(0.08333)(L)(80.63) L



81

31

f

p

3 2f

2f

3f

m

µ

4Γ

k

µc0.01

µgρk

h

=

t

f = t

m = t

sv - 3(t

sv - t

s)/8

tf = 123 - 3(123 - 81.55)/8 = 107 F

Then at 107 F k

f = 0.277 (Table A-10)


cp = 1.57 (Table A-16)

( )( )( )

180.41.6940.08333π

204

πDµ

4w

µ

4Γ

ff

===

( )( )9.6014

0.277

1.6941.57

k

µcp==

( ) ( ) ( )[ ] ( ) 15641.6941084.1735.20.277µgρk 3 2233 2

f2

f3

f =×=

( ) ( ) 8

13

1

180.49.60140.011564

hm =

hm = 63.64

0.002h

1

U

1

mo

+=

0.00263.64

1

U

1

o

+=

U

o = 56.45

q = wwc

pw(t

1 - 80)

( ) ( )29.61

80100

81.1123ln

8010081.1123∆t om =

−

−

−−−=

Then:

(482.8)(1.0)(81.1 - 80) = (56.45)(π)(0.08333)(L)(29.61) L


Total Length = L

sc + L

c + L

ds = 1.21 ft + 4.60 ft + 20.86 ft = 26.67 ft

4. For temperature of condensate = 110 F i = 166 Btu/lb Water flow rate, w

w.

ww(1.0)(105 - 80) = (20)(758 - 166)

ww = 473.6 lb/hr



180


For t

1:

(473.6)(1.0)(t1 - 80) = (20)(181 -166)

t1 = 80.63 F

For t

2:

(473.6)(1.0)(105 - t2) = (20)(758 - 634)

t2 = 99.76 F


1 = 80.63 F to t

2 = 99.76 F

Equation 13-4.

31

31

f2

f3

f

2f

mµ

4Γ1.47

gρk

µh

−

=

πD

wΓ =

ρ

f =ρ

L = 35.2

ρf

2 = ρ

f(ρ

f - ρ

v)

ρv = 1/ 1.003 = 0.997

ammonia at 123 F.


kf = 0.283 (Table A-10)

D = 1.0 in = 0.08333 ft

( )( )( )

210.51.4520.08333π

204

πDµ

4w

µ

4Γ

ff

===

ρf

2 = ρ

f(ρ

f - ρ

v) = 35.2 ( 35.2 - 0.997) = 1204

( )( ) ( )( )

( ) 31

31

210.51.47104.1712040.283

1.452h

83

2

m−

=

×


Table 13-2.

1.82399.76123

80.63123

∆t

∆t

ot

ob =−

−=

F

c = 1.049

Fch

m = (1.049)(433.5) = 454.74

002.0+=mco hF

1

U

1

( )

002.0+=454.74

1

U

1

o U



181

q = UoA∆t

om

( ) ( )31.85

99.76123

80.63123ln

99.7612380.63123∆t om =

−

−

−−−=

A = πDL q = w

wc

pw(t

2 - t

1)

Then:

(473.6)(1.0)(99.76 - 80.63) = (238.15)(π)(0.08333)(L)(31.85) L



0.4

p0.8

bb

L

k

µc

µ

DG0.023

k

Dh

=

bb πDµ

4w

µ

DG=

t

ave = (1/2)(123 + 300) = 211.5 F

TR = (211.5 + 460) / (270 + 460) = 0.92

pR = p/p

c = 300 / 1638 = 0.183

Viscosity ratio

1.05µ

µ

1,t

p,t=


µ1,t

= 0.0126 x 2.42 = 0.03050

µp,t

= 1.05 x 0.0305 = 0.03203

( )( )( )

95410.032030.08333π

204

πDµ

4wN

bRe ===

Table A-12 k

b = 0.0192

Figure A-3. c

p = 0.25

( )( )

( ) ( )( )( )

0.40.8L

0.01920

0.032030.2595410.023

0.0192

0.08333h

=

h

L = h

m = 5.70

0.002h

1

U

1

mo

+=

0.0025.70

1

U

1

o

+=

U

o = 5.636


182

( ) ( )80.75

99.76123

105300ln

99.76123105300∆t om =

−

−

−−−=

q = w

wc

pw(105 - t

2)

Then

(473.6)(1.0)(105 - 99.76) = (5.636)(π)(0.08333)(L)(80.75) L



81

31

f

p

3 2f

2f

3f

m

µ

4Γ

k

µc0.01

µgρk

h

=

t

f = t

m = t

sv - 3(t

sv - t

s)/8

tf = 123 - 3(123 - 80.78)/8 = 107 F

Then at 107 F k

f = 0.277 (Table A-10)


cp = 1.57 (Table A-16)

( )( )( )

180.41.6940.08333π

204

πDµ

4w

µ

4Γ

ff

===

( )( )9.6014

0.277

1.6941.57

k

µcp==

( ) ( ) ( )[ ] ( ) 15641.694104.1735.20.277µgρk 3 28233 2

f2

f3

f =×=

( ) ( ) 8

13

1

180.49.60140.011564

hm =

hm = 63.64

0.002h

1

U

1

mo

+=

0.00263.64

1

U

1

o

+=

U

o = 56.45

q = wwc

pw(t

1 - 80)

( ) ( )35.83

80110

80.63123ln

8011080.63123∆t om =

−

−

−−−=

Then:

(473.6)(1.0)(80.63 - 80) = (56.45)(π)(0.08333)(L)(35.83) L


Total Length = L

sc + L

c + L

ds = 0.56 ft + 4.56 ft + 20.83 ft = 25.95 ft

5. For temperature of condensate = 123 F i = 188 Btu/lb


183

Water flow rate, w

w.

ww(1.0)(105 - 80) = (20)(758 - 188)

ww = 456 lb/hr



For t

1:

(456)(1.0)(t1 - 80) = (20)(181 -188)

t1 = 80 F

For t

2:

(456)(1.0)(105 - t2) = (20)(758 - 634)

t2 = 99.56 F


1 = 80 F to t

2 = 99.56 F

Equation 13-4.

31

31

f2

f3

f

2f

mµ

4Γ1.47

gρk

µh

−

=

πD

wΓ =

ρ

f =ρ

L = 35.2

ρf

2 = ρ

f(ρ

f - ρ

v)

ρv = 1/ 1.003 = 0.997

ammonia at 123 F.


kf = 0.283 (Table A-10)

D = 1.0 in = 0.08333 ft

( )( )( )

210.51.4520.08333π

204

πDµ

4w

µ

4Γ

ff

===

ρf

2 = ρ

f(ρ

f - ρ

v) = 35.2 ( 35.2 - 0.997) = 1204

( )( ) ( )( )

( ) 31

31

210.51.47104.1712040.283

1.452h

83

2

m−

=

×


Table 13-2.

1.83599.56123

80123

∆t

∆t

ot

ob =−

−=

F

c = 1.050

Fch

m = (1.050)(433.5) = 455.18


184

002.0+=mco hF

1

U

1

( )

0.002455.18

1

U

1

o

+=

U


q = UoA∆t

om

( ) ( )32.24

99.56123

80123ln

99.5612380123∆t om =

−

−

−−−=

A = πDL q = w

wc

pw(t

2 - t

1)

Then:

(456)(1.0)(99.56 - 80) = (238.27)(π)(0.08333)(L)(32.24) L



0.4

p0.8

bb

L

k

µc

µ

DG0.023

k

Dh

=

bb πDµ

4w

µ

DG=

t

ave = (1/2)(123 + 300) = 211.5 F

TR = (211.5 + 460) / (270 + 460) = 0.92

pR = p/p

c = 300 / 1638 = 0.183

Viscosity ratio

1.05µ

µ

1,t

p,t=


µ1,t

= 0.0126 x 2.42 = 0.03050

µp,t

= 1.05 x 0.0305 = 0.03203

( )( )( )

95410.032030.08333π

204

πDµ

4wN

bRe ===

Table A-12 k

b = 0.0192

Figure A-3. c

p = 0.25

( )( )

( ) ( )( )( )

0.40.8L

0.01920

0.032030.2595410.023

0.0192

0.08333h

=

h

L = h

m = 5.70


185

0.002h

1

U

1

mo

+=

0.0025.70

1

U

1

o

+=

U

o = 5.636

( ) ( )80.98

99.56123

105300ln

99.56123105300∆t om =

−

−

−−−=

q = w

wc

pw(105 - t

2)

Then

(456)(1.0)(105 - 99.56) = (5.636)(π)(0.08333)(L)(80.98) L


Subcooling section: L


Total Length = L

sc + L

c + L

ds = 0 ft + 4.44 ft + 20.76 ft = 25.20 ft

Summary Temperature of condensate Total Length 80 F 29.06 ft 90 F 27.85 ft 100 F 26.67 ft 110 F 25.95 ft 123 F 25.20 ft Therefore it is proven that the shortest possible length is at 123 F temperature of condensate. a. Length of tube = 25.20 ft b. Temperature of condensate = 123 F c. The pilot plant should not differ from the one proposed. Proposed unit shall have 8000 / 20 - 400 tubes of

the same height as the pilot plant.

- end -

186

CHAPTER 14

BOILING LIQUIDS

Chapter 14 Boiling Liquids

187

14_001 An experimental evaporator shell consists of a horizontal cylinder approximately 1 ft in length by 8 in. in diameter, provided with an overhead water-cooled total condenser and a submerged horizontal chromium-plated copper tube that is heated by steam condensing on the inside surface. The submerged tube is 12.1 in. long and has an o.d. of 0.840 in. and a wall thickness of 0.109 in. Thermocouples are installed at points 4 in. from each end of the tube and 0.025 in. below the outer surface.

In a typical run, the heating surface is covered with water to a depth of 1.25 in., steam is passed through a vapor-liquid

separator, thence through a glass-wool filter, and is throttled to the desired condensing pressure before it enters the heating tube. Cooling water is supplied to the condenser.

When steady-state conditions have been obtained, the following readings are taken with the water boiling at 212 F:

Run number 1 2 3 4

Steam gauge pressure, lb/sq in 99 13 43 78

Average temperature of thermocouples, deg F 329 234 258.5 317

Temperature of inlet condenser water, deg C 16.2 17.2 12.8 18

Temperature of outlet condenser water, deg C 54.5 58.5 57 58.5

Condenser-water rate, lb/hr 322 305 1132 317

From these data, compute the over-all coefficient and the two surface coefficients. Solution: Run 1, steam p = 99 psig = 113.7 psia = 337.12 F

A = π(0.840)(12.1)/144 = 0.22175 sq ft q = (322)(1.0)(54.5 - 16.2))(1.8) = 22,199 q/A = 22,199 / 0.22175 = 100,108

∆t = 329-212 = 117 F

∆to = 337.12 - 212 = 125.12 F

ho = 100,108 / 117 = 856

Uo = 100,108 / 125.12 = 800

Do = 0.840 in, Di = 0.840 - 2(0.109) = 0.622 in

ii

o

oo hD

D

h

1

U

1+=

( ) ih0.622

0.840

856

1

800

1+=

h

i = 16,510

Run 2, steam p = 13 psig = 27.7 psia = 245.62 F

A = π(0.840)(12.1)/144 = 0.22175 sq ft q = (305)(1.0)(58.5 - 17.2))(1.8) = 22,674 q/A = 22,674 / 0.22175 = 102,250

∆t = 234-212 = 22 F

∆to = 245.62- 212 = 33.62 F

ho = 102,250 / 22 = 4,648

Uo = 102,250 / 33.62 = 3,042

Do = 0.840 in, Di = 0.840 - 2(0.109) = 0.622 in

ii

o

oo hD

D

h

1

U

1+=


188

( ) ih0.622

0.840

4648

1

3042

1+=

h

i = 11,888


A = π(0.840)(12.1)/144 = 0.22175 sq ft q = (1132)(1.0)(57 - 12.8))(1.8) = 90,062 q/A = 90,062 / 0.22175 = 406,142

∆t = 258.5-212 = 46.5 F

∆to = 290.15- 212 = 78.15 F

ho = 406,142 / 46.5 = 8,734

Uo = 406,142 / 78.15 = 5,197

Do = 0.840 in, Di = 0.840 - 2(0.109) = 0.622 in

ii

o

oo hD

D

h

1

U

1+=

( ) ih0.622

0.840

8734

1

5197

1+=

h

i = 17,335


A = π(0.840)(12.1)/144 = 0.22175 sq ft q = (317)(1.0)(58.5 - 18))(1.8) = 23,109 q/A = 23,109 / 0.22175 = 104,212

∆t = 317-212 = 105 F

∆to = 322.27- 212 = 115.27 F

ho = 23,109 / 105 = 993

Uo = 23,109 / 115.27 = 904

Do = 0.840 in, Di = 0.840 - 2(0.109) = 0.622 in

ii

o

oo hD

D

h

1

U

1+=

( ) ih0.622

0.840

993

1904 +=

h

i = 13,622

14_002. It is desired to boil methanol at 151 F in a natural-convection evaporator heated by steam condensing inside horizontal

submerged tubes. In view of the data below, what should be the condensing temperature of the steam?

tsv, deg F 221 226 236 246 256 266 272 286 296

U 1450 1510 1540 1500 1420 1300 1200 1040 950 Solution: t

w = condensing temperature of the steam


189

U tw - tsv q/A

1450 70 101500

1510 75 113250

1540 85 130900

1500 95 142500

1420 105 149100

1300 115 149500

1200 121 145200

1040 135 140400

950 145 137750 Maximum flux for nucleate boiling is 149,500 Btu/hr

U = 1300, ∆to = 115 F

tc = 151F + 115 F = 266 F condensing steam temperature

14_003. Estimate the pressure drop for isothermal co-current flow of water and air in a horizontal standard 1-inch iron pipe 10 ft

long. The water rate will be 1200 lb/hr and the air rate 30 lb/hr. The temperature is 70 F, and the discharge pressure is 1 atm.

Repeat for co-current upflow of the same mixture. Solution: ID = 1.049 in = 0.08742 ft at 70 F

µL = 2.36 lb/(hr)(ft)

µG = 0.0441 lb/(hr)(ft)

( )( )( )

002074062.360.08742π

12004

πDµ

4w

L

L >==

( )( )( )

002099080.04410.08742π

304

πDµ

4w

G

G >==

Table 14-9, Flow Type = t-t Eq. 14-10

( )( )L

TPL

∆L∆p

∆L∆p=φ

Eq. 14-11.

( )( )G

L

∆L∆p

∆L∆pX =

a. Horizontal For liquid. Equation 6-9b

hc

m2

m21

r2g

LGfpp

ν=−

hc

m2

m

L r2g

Gf

L

p ν=

∆

∆

at 70 F v

m = 0.01606 cu ft/lb

NRe

= 7,406


190

fm = 0.009

( )( )

199,9260.08742π

12004

πD

4wG

22

L ===

r

h = D/ 4= 0.08742 / 4 = 0.02186 ft

( )( ) ( )( )( )

0.31690.02186104.172

0.016061999260.009

∆L

∆p8

2

L

=×

=

For Gas, v

1=v

2

hc

m2

m21

r2g

LGfpp

ν=−

( )( )( )( )14414.696

4607053.34

p

RTm

+==ν

v

m = 13.36 cu ft/lb

α = 1.0

( )( )

49980.08742π

304

πD

4wG

22

L ===

hc

m2

m

G r2g

Gf

L

p ν=

∆

∆

f

m = 0.0085

( )( ) ( )( )( )

0.15560.02186104.172

13.3649980.009

∆L

∆p8

2

G

=×

=

( )( )

1.430.1556

0.3169

∆L∆p

∆L∆pX

G

L ===

Figure 14-23

( )( )

3.5∆L∆p

∆L∆p

L

TPL ==φ

( )3.5

0.3169

∆L∆p TP =

( ) 3.882∆L∆p TP =

∆p = 3.882 x 10 = 38.82 lb/sq ft b. Up-flow

L

zz

g

g

r2g

Gf

L

p

m

12

chc

m2

m

L ν

−+

ν=

∆

∆

L = z

2 - z

1

( )( )0.01606

1

1

10.3169

∆L

∆p

L

+=

62.584∆L

∆p

L

=

( )( )

200.1556

62.584

∆L∆p

∆L∆pX

G

L ===

Figure 14-23


191

( )( )

1.5∆L∆p

∆L∆p

L

TPL ==φ

( ) 140.8∆L∆p TP =

∆p = 140.8 x 10 = 1408 lb/sq ft 14_004 The outer surface of a long horizontal tube with an o.d. of 0.5 inch is maintained at a constant temperature of 1000 F

by electrical heat input. The tube is immersed in boiling water at 212 F. What is the average heat flux from the tube to the water? If the tube is 16 BWG and has a thermal conductivity of 15 Btu/(hr)(sq ft)(deg F per ft), what is the

temperature inside the tube? Use ε = 0.7. Solution: For 16 BWG tube, k = 15 Btu/(hr)(sq ft)(deg F per ft)

Use ε = 0.7

( ) 41

∆tDµ

ρρρgk0.62h

v

vLv3

vco

−λ=

For water at 212 F

ρL = 1/ 0.01672 = 59.8086 lb/cu ft

ρv = 1/ 26.80 = 0.0373 lb/cu ft

D = 0.04167 ft

∆t = 1000 F - 212 F = 788 F

λ = 970.3 Btu/lb k

v = 0.0136 Btu/(hr)(sq ft)(deg F per ft)

µv = 0.012 x 2.42 = 0.02904 lb/hr-ft

g = 4.17 x 108

( )( ) ( )( )( )( )( )( )

24.357880.029040.04167

0.037359.80860.0373970.30.136104.170.62h

41

38

co =

−×=

rco

corc hh

hhhhh

31

+

=+=

( ) 3

avgavgr Tσεn4h +=

( )( )( )( ) 5.8110660.7100.17134h

38r =×=

( ) 5.81h

24.3524.35h

31

+

=

h = 28.8

q/A = h∆∆∆∆t = (28.8)(1000-212) = 22,695 Btu/hr-sq ft For 16 BWG, 1/2 in o.d. xw = 0.065 in = 0.005417 ft ID = 0.370 in = 0.030833 ft

q = k∆t2 / xw 22,695 = (15)(t

i - 1000) / (0.005417)

ti = 1008 F

- end -

192

CHAPTER 15

APPLICATIONS TO DESIGNS

Chapter 15 Application to Designs

193

15_001. For process use, 50,000 lb/hr of air is to be heated from 70 to 300 F. In order to utilize exhaust steam, two heaters in series will be used. The air will flow through 1-in. o.d. steel tubes (18 BWG, thickness = 0.049 in.) at a mass velocity of 10,000 lb/(hr)(sq ft). the over-all coefficient of heat transfer in both heaters may be assumed constant at 8 Btu/(hr)(sq ft)(deg D) based on the outside surface area of the tubes.

The exhaust steam condenses at 212 F and is valued at 20 cents/106 Btu of latent heat. The high-pressure steam has

a condensing temperature of 350 F and costs 40 cents/106 Btu of latent heat.

The energy cost for pumping the air through the heaters at the given mass velocity is $0.40/(sq ft of heating

surface)/year, and the fixed charges on both heaters are $1 / (year)(sq ft of heating surface). (By heating surface is meant the outside surface area of the tubes.)

If the heaters operate 8000 hr/year, to what temperature should the air be heated with the exhaust steam in order to

give the lowest over-all cost of heating for air from 70 to 300 F? The specific heat of air is 0.24 Btu/(lb)(deg F). Solution:

CH’

Exhaust steam = $ 0.20 x 10-6 / Btu of latent heat

CH’’

High pressure steam = $ 0.40 x 10-6 / Btu of latent heat

Ca = Fixed charges = ($ 1.00)(1/8000) = $ 1.25 x 10

-4 / (hr)(sq ft of heating surface).

Cp = Power = ($ 0.40)/(1/8000) = $ 5.0 x 10

-5 / (hr)(sq ft of heating surface)

K2 = C

a + C

p

Equation 15-41

( )( )( )( )

( )H'H"

pa

CCU

t't"CCtt"tt'

−

−+=−−

t’ = 212 F t” = 350 F

( )( )( )( )

( )( )15094

100.20100.408

212350105.0101.25t350t212

66

54

=×−×

−×+×=−−

−−

−−

t = 140 F air heated by exhaust steam. 16_002. It is necessary to heat 24,000 lb/hr of air from 70 to 350 F while it is flowing under pressure at the optimum mass

velocity of 7200 lb/(hr)(sq ft of cross section), inside tubes having an actual i.d. of 0.870 in. Low-pressure exhaust steam (220 F saturation temperature) now being discarded would be available at a cost of $0.05/million Btu of latent heat, and high-pressure steam (370 F saturation temperature) is available at a cost of $0.20/million Btu of latent heat. The heaters must run 8400 hr/year. The annual fixed charges, expressed in dollars/(year)(ft. of each tube), will be assumed as 0.15 for the low-pressure heater and 0.25 for the high-pressure heater, independent of the length of tube or the number of tubes. The air is to leave the high-pressure heater at an absolute pressure of 10 atm and a temperature of 350 F. It is agreed to use Eq. (9-15). Calculate the minimum yearly costs and the tube length required for each heater.

Solution:

0.2

0.8

pL DG0.0144ch =

G

io = 7200

D = 0.870 / 12 = 0.0725 ft

( )( )

( )0.2

0.8

L0.0725

72000.240.0144Uh ==


194

U = 7.118

( )( )( )( )

( )H'H"

pa

CCU

t't"CCtt"tt'

−

−+=−−

t’ = 220 F, C

H’ = $ 0.05 x 10

-6

t” = 350 F, CH”

= $0.20 x 10-6

( )( )

5a 107.84

0.0725π8400

1$0.15C −

×==1

( )( )

6a2 101.307

0.0725π8400

1$0.25C −

×==

ii

i

a

p

nm3

n

C

C

−−=

n

i = 0.8 , m

i = 0.2

Cp/C

a = 0.4

Cp1

= 0.4Ca1

= 3.136 x 10-5

Cp2

= 0.4Ca2

= 5.228 x 10-5

average Ca = (1/2)(7.84 x 10

-5 + 1.307 x 10

-4) = 10.455 x 10

-5

average Cp = 0.4Ca = 4.182 x 10

-5

K2 = 10.455 x 10

-5 + 4.182 x 10

-5 = 14.637 x 10

-5

( )( )( )( )

( )H'H"

pa

CCU

t't"CCtt"tt'

−

−+=−−

( )( )( )( )

( )( )66

5

100.05100.207.118

2203701014.637t370t220

−−

−

×−×

−×=−−

( )( ) 20564t370t220 =−−

t = 133.2 F For Low Pressure Heater: q = wc(T - T

1) = UA

1∆t

m

F115.53

133.2220

70220ln

70133.2∆t m =

−

−

−=

q = (24,000)(0.24)(133.2 - 70) = (7.118)(A

1)(115.63)

A1 = 442.3 sq ft

For High Pressure Heater q = wc(T

2 - T) = UA

2∆t

m

F87.72

350370

133.2370ln

133.2350∆t m =

−

−

−=

q = (24,000)(0.24)(350 -133.2) = (7.118)(A

2)(87.72)

A2 = 2000 sq ft

Σy=wc(t - t1)C

H’ + A

1K

2 + wc(t

2 - t)C

H”+A

2K

2

Σy=(24,000)(0.24)(133.2-70)(0.05 x 10-6) + (442.3)(14.637 x 10

-5) + (24,000)(0.24)(350-133.2)(0.20 x 10

-6) +


195

(2000)(14.637 x 10-5)

ΣΣΣΣy= $ 0.6254 per hour minimum Lengths of tubes No. of tubes

( )( )( )808

0.0725π7,200

24,0002

==

( )( )

( )( )ft10.87

8080.0725π

2000

πDn

AL

ft2.48080.0725π

442.3

πDn

AL

22

11

===

===

16_003. An air cooler consisting of a bundle of 1-in. 18 BWG copper tubes, enclosed in a well-baffled shell , is being built to

cool 45,000 lb of air/hr from 200 to 90 F. The air flows under pressure in a single pass through the tubes, and cooling water at 80 F, under sufficient pressure to force it through at any desired rate, flows countercurrently through the shell. The air flows inside the pipes at a amass velocity of 8600 lb/(hr)(sq ft), which is the optimum velocity; the corresponding U

i will be 7.8. Cooling water costs 20 cents/1000 cu ft, and the annual fixed charges on the cooler are

50 cents/sq ft of heating surface. It is proposed to operate 8400 hr/year. From the standpoint of the lowest total yearly cost, calculate:

a. The optimum pounds of cooling water per pound of air. b. The over-all temperature difference at the hot end. c. The length of each tube. d. The number of tubes in parallel e. The total annual cost in dollars. Solution: For 1 in , 18 BWG, x

w = 0.049 in, ID = 0.902 in, OD = 1.00 in

q = w’c’(t1’- t

2’) = w”c”(t

2” - t

1”)

For air: t

1’ = 200 F, t

2’ = 90 F, c’ = 0.24, w’ =45,000 lb/hr

Inside area = (π/4)(0.902 / 12)2 = 0.004438 sq ft

Gio = 8,600 lb/hr-sq ft

N= No. of tubes = (45,000) / [(8,600)(0.004438)] = 1179 tubes t1” = 80 F c” = 1.0 C

w = cooling water cost = ($0.20) / [(1000)(62.3)]

Cw = $ 3.21 x 10

-6 per lb

Annual fixed charges = C

AF

A = $ 0.50 / sq ft of heating surface

Equation 15-39

oo

o

o

2

o

o XlnYY

Y1

1Y

Y1Z=

+

−

−

−+


196

wherein:

( )

( )"t't

"t't

∆t

∆tY

12

o21

C

Ho

−

−==

c"FC

θCUFX

AA

wGo =

"t't

't'tZ

12

21

−

−=

U

i = 7.8

( )( )( )( )( )( )

0.4221.00.5

8400103.211.07.8X

-6

o =×

=

11.0

8090

90200Z =

−

−=

From Figure 15-12, Y

o = 7.9

( )( )8090

"t2007.9 2

−

−=

t

2” = 121 F

a.

( )( )

( )( )( )( )801211

902000.24

"t"tc"

't'tc'

w'

w"

12

21

−

−=

−

−=

airofpoundperwatercoolingoflb0.6439

w'

w"=

b. ∆t

H = t

1’ - t

2’ = 200 F - 121 F = 79 F

c. Length of each tube: q = UAF

G∆t

L

∆tc = 10 F

F33.384

10

79ln

1079∆tL =

−=

q = w’c’(t

1’ - t

2’) = U

iA

iF

G∆t

L = U

iπD

iNL∆t

L

(45000)(0.24)(200-90) = (7.8)(π)(0.902 / 12)(1179)(L)(33.384) L = 16.4 ft d. No. of tubes in parallel = 1179 tubes e. Total Annual Cost = w”θC

w + AC

AF

A

A = π(0.902 / 12)(1179)(16.4) = 4566 sq ft

Total annual cost = (0.6439)(45000)(84000)(3.21 x 10-6) + (4566)(0.5)

Total annual cost = $ 3,064.30

- end -

Engineering

Solution Manual - Heat Transmission - 3rd Edition - William McAdams