Chapter_3_0114 (1)

3. Moist Air Properties and Conditioning Processes

3.1 Calculate values of humidity ratio, enthalpy, and specific volume for saturated air at one standard atmosphere using perfect gas relations for temperatures of (a) 70 F (20 C) and (b) 20 F (-6.7 C).

Solution:

(a) In English units, t = 70 F

Humidity Ratio: Eq. (3-14b)

s

s

a

s

s pPp

pp

W

== 6219.06219.0

at t = 70 F, ps = 0.363 psia

P = 14.696 psia

363.0696.14363.06219.0

=s

W = 0.01575 lbmv/lbma

Enthalpy: Eq. (3-20a)

( )tWti 444.02.1061240.0 ++= Btu/lbma ( ) ( ) ( )( )[ ]70444.02.106101575.070240.0 ++=i = 34.0 Btu/lbma

Specific Volume: Ra = 53.352 ft-lbf/lbm-R

s

a

a

a

pPTR

pTR

v

==

( )( )( )( )144363.0696.14

67.45970352.53

+=v = 13.69 ft3/lbma

In SI units, t = 20 C


s

s

a

s

s pPp

pp

W

== 6219.06219.0

at t = 20 C, ps = 0.00234 MPa = 2.34 kPa

P = 101.325 kPa

34.2325.10134.26219.0

=s

W = 0.01407 kgv/kga

Enthalpy: Eq. (3-20b)


( )tWti 86.13.25010.1 ++= kJ/kga ( ) ( ) ( )( )[ ]2086.13.250101407.0700.1 ++=i = 55.7 kJ/kga

Specific Volume: Ra = 287 J/kg.K

s

a

a

a

pPTR

pTR

v

==

( )( )( )( )100034.2325.101

15.27320287

+=v = 0.85 m3/kga

(b) In English units, t = 20 F


s

s

a

s

s pPp

pp

W

== 6219.06219.0

at t = 20 F < 32.02 F, use ps at 32.02 F which is nearly equal by plotting on curve = 0.089 psia

P = 14.696 psia

089.0696.14089.06219.0

=s

W = 0.0038 lbmv/lbma

Enthalpy: Eq. (3-20a)

( )tWti 444.02.1061240.0 ++= Btu/lbma ( ) ( ) ( )( )[ ]20444.02.10610038.020240.0 ++=i = 8.7 Btu/lbma

Specific Volume: Ra = 53.352 ft-lbf/lbm-R

s

a

a

a

pPTR

pTR

v

==

( )( )( )( )144089.0696.14

67.45920352.53

+=v = 12.17 ft3/lbma

In SI units, t = -6.7 C


s

s

a

s

s pPp

pp

W

== 6219.06219.0

at t = -6.7 C < 0.01 C, use ps at 0.01C which is nearly equal by plotting on curve = 0.00061 Mpa = 0.61 kPa


P = 101.325 kPa

61.0325.10161.06219.0

=s

W = 0.0038 kgv/kga

Enthalpy: Eq. (3-20b)

( )tWti 86.13.25010.1 ++= kJ/kga ( ) ( ) ( )( )[ ]7.686.13.25010038.07.60.1 ++=i = 2.8 kJ/kga

Specific Volume: Ra = 287 J/kg.K

s

a

a

a

pPTR

pTR

v

==

( )( )( )( )100061.0325.101

15.2737.6287

+=v = 0.76 m3/kga

3.2 The temperature of a certain room is 22 C, and the relative humidity is 50 percent. The barometric pressure is 100 kPa. Find (a) the partial pressures of the air and water vapor, (b) the vapor density, and (c) the humidity ratio of the mixtures.

Solution:

t = 22 C = 50 % = 0.50 P = 100 kPa

(a) ps at 22 C = 2.672 kPa

s

v

pp

= ; sv

pp = = (0.50)(2.672) = 1.336 kPa

vapPp = = 100 1.336 = 98.664 kPa

(b) v

v

pTR

v =

Rv = 462 J/kg.K ( )( )( )( )1000336.1

15.27322462 +=v = 102.065 m3/kgv

(c) v

v

pPpW

= 6219.0

336.1100336.16219.0

=W = 0.008421 kgv/kga


3.3 Compute the local atmospheric pressure at elevation ranging from sea level to 6000 ft (1830 m) in (a) inches of mercury and (b) kilopascals.

Solution:

(a) H = 6000 ft

Eq. (3-4) P = a + bH Table 3-2: H > 4000 ft a = 29.42 b = -0.0009 P = 29.42 + (-0.0009)(6000) = 24.02 in. Hg.

(b) H = 1830 m

Eq. (3-4) P = a + bH Table 3-2: H > 1220 m a = 99.436 b = -0.010 P = 99.436 + (-0.010)(1830) = 81.136 kPa.

3.4 Rework Problem 3.1 for an atmospheric pressure corresponding to an elevation of (a) 5280 ft and (b) 1600 m.

Solution:

(a) H = 5280 ft

Eq. (3-4) P = a + bH Table 3-2: H > 4000 ft a = 29.42 b = -0.0009 P = 29.42 + (-0.0009)(5280) = 24.668 in. Hg.

(b) H = 1830 m

Eq. (3-4) P = a + bH Table 3-2: H > 1220 m a = 99.436 b = -0.010 P = 99.436 + (-0.010)(1600) = 83.346 kPa.


3.5 Compute the enthalpy of moist air at 60 F (16 C) and 80 percent relative humidity for an elevation of (a) sea level and (b) 5000 ft (1525 m).

Solution:

(a) English units

ps at 60 F = 0.256 psia = 80 % = 0.80

svpp = = (0.80)(0.256) = 0.2048 psia

At sea level, H = 0 Eq. (3-4) P = a + bH Table 3-2: H < 4000 ft a = 29.92 P = a = 29.92 in. Hg = 101.325 kPa = 14.696 psia

2048.0696.142048.06219.06219.0

=

=

v

v

pPpW = 0.008789 lbmv/lbma

Eq. (3-20a) ( )tWti 444.02.1061240.0 ++= Btu/lbma

( ) ( ) ( )( )[ ]60444.02.1061008789.060240.0 ++=i = 23.96 Btu/lbma

In SI units

ps at 16 C = 1.836 kPa = 80 % = 0.80

svpp = = (0.80)(1.836) = 1.469 psia

At sea level, H = 0 Eq. (3-4) P = a + bH Table 3-2: H < 4000 ft a = 101.325 P = a = 101.325 kPa

469.1325.101469.16219.06219.0

=

=

v

v

pPpW = 0.00915 kgv/kga

Eq. (3-20b) ( )tWti 86.13.25010.1 ++= kJ/kga

( ) ( ) ( )( )[ ]1686.13.250100915.0160.1 ++=i = 39.16 kJ/kga

(b) English units


ps at 60 F = 0.256 psia = 80 % = 0.80

svpp = = (0.80)(0.256) = 0.2048 psia

At H = 5000 ft > 4000 ft Eq. (3-4) P = a + bH Table 3-2: H > 4000 ft a = 29.42 b = - 0.0009 P = 29.42 + (-0.0009)(5000) = 24.92 in. Hg = 12.24 psia

2048.024.122048.06219.06219.0

=

=

v

v


Eq. (3-20a) ( )tWti 444.02.1061240.0 ++= Btu/lbma

( ) ( ) ( )( )[ ]60444.02.1061010583.060240.0 ++=i = 25.91 Btu/lbma

In SI units

ps at 16 C = 1.836 kPa = 80 % = 0.80

svpp = = (0.80)(1.836) = 1.469 psia

At H = 1525 m > 1220 , Eq. (3-4) P = a + bH Table 3-2: H < 4000 ft a = 99.436 b = - 0.010 P = 99.436 + (-0.010)(1525) = 84.186 kPa

469.1186.84469.16219.06219.0

=

=

v

v


Eq. (3-20b) ( )tWti 86.13.25010.1 ++= kJ/kga

( ) ( ) ( )( )[ ]1686.13.2501011045.0160.1 ++=i = 43.96 kJ/kga

3.6 The condition within a room is 70 F db, 50 percent relative humidity, and 14.696 psia pressure. The inside surface temperature of the window is 40 F. Will moisture condense on the window glass?

Solution:

At 70 F db, ps = 0.363 psia = 0.50


pv = 0.50 ( 0.363 psia ) = 0.1815 psia

at 0.1815 psia, t = 50.45 F

Since 40 F < 50.45 F , the moisture will condense on the window glass.

3.7 A duct has moist air flowing at a rate of 5000 ft3/min (2.36 m3/s). What is the mass flow rate of the dry air, where the dry bulb temperature is 60 F (16 C), the relative humidity is 80 percent and the pressure inside the duct corresponds to (a) sea level, and (b) 6000 ft (1830 m).

Solution:

(a) English units

ps at 60 F = 0.2563 psia sv

pp = = (0.80)(0.2563) = 0.20504 psia At sea level, P = 29.92 in. Hg = 14.696 psia pa = P pv = 14.696 0.20504 = 14.4910 psia ( )( )

( )( )67.45960352.531444910.14

+==

TRp

a

a = 0.0753 lb/ft3

Q& = 5000 ft3/min Qm

a&& = = (0.0753)(5000) = 376.5 lb/min

SI Units ps at 16 C = 1.836 kPa

svpp = = (0.80)(1.836) = 1.4688 psia

At sea level, P = 101.325 kPa pa = P pv = 101.325 1.4688 = 99.8562 ( )( )

( )( )15.2731628710008562.99

+==

TRp

a

a = 1.2033 kg/m3

Q& = 2.36 m3/s Qm

a&& = = (1.2033)(2.36) = 2.84 kg/s

(b) English units


pp = = (0.80)(0.2563) = 0.20504 psia At H = 6000 ft > 4000 ft P = a + bH a =29.42 b = - 0.0009 P = 29.42+ (-0.0009)(6000) = 24.02 in. Hg = 11.798 psia


pa = P pv = 11.798 0.20504 = 11.593 psia ( )( )( )( )67.45960352.53

144593.11+

==

TRp

a

a = 0.06021 lb/ft3

Q& = 5000 ft3/min Qm

a&& = = (0.06021)(5000) = 301.05 lb/min

SI Units ps at 16 C = 1.836 kPa

svpp = = (0.80)(1.836) = 1.4688 kPa

At H = 1830 m > 1220 m P = a + bH a = 99.436 b = - 0.010 P = 99.436 + (-0.010)(1830) = 81.136 kPa pa = P pv = 81.136 1.4688 = 79.667 kPa ( )( )

( )( )15.273162871000667.79

+==

TRp

a

a = 0.96 kg/m3

Q& = 2.36 m3/s Qm

a&& = = (0.96)(2.36) = 2.2656 kg/s

3.8 Compute the dew point for moist air at 80 F (27 C) and 50 percent relative humidity for pressures corresponding to (a) sea level and (b) 5000 ft (1225 m).

Solution:

(a) English units


pp = = (0.50)(0.507) = 0.2535 psia Dew Point = tdp = 59.68 F

SI units

ps at 27 C = 3.602 kPa sv

pp = = (0.50)(3.602) = 1.801 kPa Dew Point = tdp = 15.72 C

(b) H = 5000 ft (1225 m)

Since elevation does not affect dew point, the answers are the same as in (a).


3.9 A space is to be maintained at 70 F (21 C) dry bulb. It is estimated that the inside wall surface temperature could be as low as 45 F (7 C). What maximum relative and specific humidity can be maintained without condensation on the walls?

Solution:

English units

At 45 F, pv = 0.150 psia At 70 F, ps = 0.363 psia

svpp = ( ) ( )%100

363.0150.0%100 ==

s

v

pp = 41.32 %

150.0696.14150.06219.06219.0

=

=

v

v


Maximum relative humidity = 41.32 % Maximum specific humidity = 0.006413 lbmv/lbma

SI units

At 7 C, pv = 1.014 kPa At 21 C, ps = 2.506 kPa

svpp = ( ) ( )%100

506.2014.1%100 ==

s

v

pp = 40.46 %

014.1325.101014.16219.06219.0

=

=

v

v


Maximum relative humidity = 40.46 % Maximum specific humidity = 0.006287 kgv/kga

3.10 Outdoor air with a temperature of 40 F db and 35 F wb and with a barometric pressure of 29 in. Hg is heated and humidified under steady-flow conditions to a final temperature of 70 F db and 40 percent relative humidity. (a) Find the mass of water vapor added to each pound mass of dry air. (b) If the water is supplied at 50 F, how much heat is added per pound mass of dry air?

Solution:

Solving for for W1 and i1 at Point 1 Using eq. (3-21d) and (3-14b) with its symbols. At 35 F, 22 sv pp = = 0.1013 psia


2fgi = 1973.3 Btu/lbm

wi = 3.0 Btu/lbm

2t = 35 F at 40 F , 1vi = 1078.5 Btu/lbm P = 29 in Hg = 14.244 psia

1013.0244.141013.06219.02

=

sW = 0.004454 lbmv/lbma

Then ( )

+=

wv

fgspa

iiiWttc

W1

22121

( ) ( )( )35.1078

3.1073004454.0403524.01

+=W = 0.003283 lbmv/lbma

( )tWti 444.02.106124.0 ++= Btu/lbma ( ) ( ) ( )[ ]40444.02.1061003283.04024.01 ++=i = 13.14 Btu/lbma

Solving for W2 and i2 at point 2

At 70 F, ps = 0.363 psia sv

pp = = (0.40)(0.363 psia) = 0.1452 psia P = 14.244 psia

1452.0244.141452.06219.06219.02

=

=

v

v


( )2222 444.02.1061240.0 tWti ++= Btu/lbma ( ) ( ) ( )[ ]70444.02.1061006408.070240.02 ++=i = 23.8 Btu/lbma

(a) Mass of water vapor added:

12 WWm

m

a

w=

&

& = 0.006408 0.003283 = 0.003125 lbmv/lbma

(b) At 50 F, iw = 18.1 Btu/lb ( )

w

a

w

a

im

miim

q&

&

&

&= 12 = (23.8 13.14) (0.003125)(18.1) = 10.3434 Btu/lbma

3.11 Air with a dry bulb temperature of 70 F and wet bulb temperature of 65 F is at a barometric pressure of 29.92 in. Hg. Without making use of psychrometric chart, find (a) the relative humidity of the air, (b) the vapor density, (c) the dew point, (d) the humidity ratio, and (e) the volume occupied by the mixture associated with a pound mass of dry air.

Solution:


1t = 70 F

2t = 65 F

Solving for 2sW , Eq. (3-14b) 2vp = 2sp at 65 F = 0.3095 psia

12 PP = = 29.92 in Hg = 14.696 psia

3095.0696.143095.06219.06219.0

22

22

=

=

v

v

s pPpW = 0.013379 lbmv/lbma

Solving for 1W , Eq. (3-21c) ( )

+=

wv

fgspa

iiiWttc

W1

22121

2fgi = fgi at 65 F = 1056.5 Btu/lbm

wi = fi at 65 F = 33 Btu/lbm

1vi = gi at 70 F = 1091.7 Btu/lbm ( ) ( )( )

337.10915.1056013379.0706524.0

1

+=W = 0.012218 lbmv/lbma

Solving for 1vp , Eq. (3-14b)

11

11 6219.0

v

v

pPpW

=

1

1

696.146219.0012218.0

v

v

pp

=

1vp = 0.2832 psia

at 70 F, 1sp = 0.363 psia

(a) Relative Humidity

363.02832.0

1

1==

s

v

pp = 0.78 or 78 %

(b) Vapor Density ( )( )

( )( )67.4597078.851442832.0

+==

TRp

v

v = 0.000898 lbmv/ft3

(c) Dew Point At 1vp = 0.2832 psia

dpt = 62.54 F (d) Humidity Ratio


1Wm

mWa

v==

&

&= 0.012218 lbmv/lbma

(e) Volume occupied by mixture per pound of mass of dry air. ( )( )( )( )1442832.0696.14

67.45970352.53

+==

a

a

pTR

v = 13.62 ft3/lbma

3.12 Air is cooled from 75 F db and 70 F wb until it is saturated at 55 F. Find (a) the moisture removed per pound of dry air, (b) the heat removed to condense the moisture, (c) the sensible heat removed, and (d) the total amount of heat removed.

Solution:

Use Figure 3-7

Determine state condition 1, 75 F db, 70 F wb 1t = 75 F

= 2ttwb = 70 F

22 sv pp = at 70 F = 0.363 psia

2P = 14.696 psia

363.0696.14363.06219.06219.0

22

22

=

=

v

v


( )

+=

wv

fgspa

iiiWttc

W1

22121


wi = fi at 70 F = 38Btu/lbm

1vi = gi at 75F = 1093.85 Btu/lbm


( ) ( )( )3885.1093

7.105301575.0757024.01

+=W = 0.014581 Btu/lbma


Determine state condition2 2t = 55 F

2sp = 0.217 psia

22

22 6219.0

s

s

pPpW

=

217.0696.14217.06219.02

=W = 0.009321 lbmv/lbma


Determine state condition 3 13 tt = = 75 F

23 WW = = 0.009321 lbmv/lbma


(a) Moisture removed, Eq. (3-29) 21 WW

m

m

a

w=

&

&= 0.014581 0.009321 = 0.00526 Btu/lbma

(b) Heat removed to condense the moisture, Eq. (3-33) 31 ii

m

q

a

l=

&

&= 33.96 28.20 = 5.76 Btu/lbma

(c) Sensible heat removed 23 ii

m

q

a

s=

&

&= 28.20 23.32 = 4.88 Btu/lbma

(d) Total amount of heat removed

a

l

a

s

am

qm

qm

q&

&

&

&

&

&+= = 4.88 + 5.76 = 10.64 Btu/lbma

3.13 The dry bulb and thermodynamic wet bulb temperature are measured to be 75 F and 62 F, respectively, in a room. Compute the humidity ratio relative humidity for the air at (a) sea level and (b) 5000 ft (1225 m).


Solution:

Use only English units as temperature are given in English units.

(a) At sea level, P = 29.92 in Hg = 14.696 psia

Eq. (3-14b)

2t = 62 F

22

22 6219.0

v

v

s pPpW

=

2vp = 2sp at 62 F = 0.2774 psia

2774.0696.142774.06219.02

=

sW = 0.0119865 lbmv/lbma

Eq. (3-21d) ( )

+=

wv

fgspa

iitWttc

W1

22121


wi = fi at 62 F = 30 Btu/lbm

1vi = gi at 75F = 1093.85 Btu/lbm ( ) ( )( )

3085.109318.1058011965.0756224.0

1

+=W = 0.008969 lbmv/lbma ans.

Solving for 1vp :

1

11 696.14

6219.0v

v

ppW

=

1

1

696.146219.0008969.0

v

v

pp

=

1vp = 0.20893 psia

1sp = vp at 75 F = 0.435 psia

435.020893.0

1

11 ==

s

v

pp = 0.48 or 48 % - ans.

(b) H = 5000 ft = 1225 m

P = a + bH

Table 3-2. H > 4000 ft a = 29.42 b = - 0.0009 P = 29.42 + (-0.0009)(4000) = 25.82 in Hg = 12.682 psia


2774.0682.122774.06219.06219.0

22

22

=

=

v

v


( )

+=

wv

fgspa

iitWttc

W1

22121

( ) ( )( )3085.1093

18.1058013907.0756224.01

+=W = 0.010900 lbmv/lbma ans.

Solving for 1vp :

1

11 696.14

6219.0v

v

ppW

=

1

1

682.126219.0010900.0

v

v

pp

=

1vp = 0.218448 psia

1sp = vp at 75 F = 0.435 psia

435.0218448.0

1

11 ==

s

v

pp = 0.5022 or 50.22 % - ans.

3.14 To what temperature must atmospheric air at standard sea level pressure be cooled to be saturated with a humidity ratio of 0.001 lbv/lba ? What is the temperature if the pressure is 5 atmospheres?

Solution:

At standard sea level pressure W = 0.001 lbmv/lbma

s

s

ppW

=

696.146219.0

s

s

pp

=

696.146219.0001.0

sp = 0.0236 psia

Use Table A-1a, t 32.02 F ans.

At P = 5 atm = 73.48 psia

Solving for 1vp :

s

s

ppW

==

48.736219.0001.0

sp = 0.118 psia

Use Table A-1a, interpolation, t = 39 F ans.

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Chapter_3_0114 (1)