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) p ps W s = 0.6219 s = 0.6219 pa P − ps at t = 70 F, ps = 0.363 psia

P = 14.696 psia Ws = 0.6219

0.363 = 0.01575 lbmv/lbma 14.696 − 0.363

Enthalpy: Eq. (3-20a) i = 0.240t + W (1061.2 + 0.444t ) Btu/lbma i = 0.240(70) + (0.01575)[1061.2 + (0.444)(70)] = 34.0 Btu/lbma

Specific Volume: Ra = 53.352 ft-lbf/lbm-R RT RT v= a = a pa P − ps (53.352)(70 + 459.67) = 13.69 ft3/lbma v= (14.696 − 0.363)(144) In SI units, t = 20 C Humidity Ratio: Eq. (3-14b) p ps W s = 0.6219 s = 0.6219 pa P − ps at t = 20 C, ps = 0.00234 MPa = 2.34 kPa P = 101.325 kPa Ws = 0.6219

2.34 = 0.01407 kgv/kga 101.325 − 2.34

Enthalpy: Eq. (3-20b)

3. Moist Air Properties and Conditioning Processes

i = 1.0t + W (2501.3 + 1.86t ) kJ/kga i = 1.0(70) + (0.01407 )[2501.3 + (1.86)(20 )] = 55.7 kJ/kga Specific Volume: Ra = 287 J/kg.K RT RT v= a = a pa P − ps (287 )(20 + 273.15) = 0.85 m3/kga v= (101.325 − 2.34)(1000) (b) In English units, t = 20 F Humidity Ratio: Eq. (3-14b) p ps W s = 0.6219 s = 0.6219 pa P − ps 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 Ws = 0.6219

0.089 = 0.0038 lbmv/lbma 14.696 − 0.089

Enthalpy: Eq. (3-20a) i = 0.240t + W (1061.2 + 0.444t ) Btu/lbma i = 0.240(20) + (0.0038)[1061.2 + (0.444)(20 )] = 8.7 Btu/lbma Specific Volume: Ra = 53.352 ft-lbf/lbm-R RT RT v= a = a pa P − ps (53.352)(20 + 459.67 ) = 12.17 ft3/lbma v= (14.696 − 0.089)(144) In SI units, t = -6.7 C Humidity Ratio: Eq. (3-14b) p ps W s = 0.6219 s = 0.6219 pa P − ps 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

3. Moist Air Properties and Conditioning Processes

P = 101.325 kPa Ws = 0.6219

0.61 = 0.0038 kgv/kga 101.325 − 0.61

Enthalpy: Eq. (3-20b) i = 1.0t + W (2501.3 + 1.86t ) kJ/kga i = 1.0(− 6.7 ) + (0.0038)[2501.3 + (1.86)(− 6.7 )] = 2.8 kJ/kga Specific Volume: Ra = 287 J/kg.K RT RT v= a = a pa P − ps (287 )(− 6.7 + 273.15) = 0.76 m3/kga v= (101.325 − 0.61)(1000) 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 φ = 50 % = 0.50 P = 100 kPa (a) ps at 22 C = 2.672 kPa pv ; pv = φ ps = (0.50)(2.672) = 1.336 kPa ps pa = P − pv = 100 – 1.336 = 98.664 kPa

φ=

RvT pv Rv = 462 J/kg.K (462)(22 + 273.15) = 102.065 m3/kgv v= (1.336)(1000)

(b) v =

pv P − pv 1.336 W = 0.6219 = 0.008421 kgv/kga 100 − 1.336

(c) W = 0.6219

3. Moist Air Properties and Conditioning Processes

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. Moist Air Properties and Conditioning Processes

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 φ = 80 % = 0.80 pv = φ ps = (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 pv 0.2048 W = 0.6219 = 0.6219 = 0.008789 lbmv/lbma P − pv 14.696 − 0.2048 Eq. (3-20a) i = 0.240t + W (1061.2 + 0.444t ) Btu/lbma i = 0.240(60) + (0.008789)[1061.2 + (0.444)(60)] = 23.96 Btu/lbma In SI units ps at 16 C = 1.836 kPa φ = 80 % = 0.80 pv = φ ps = (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 pv 1.469 W = 0.6219 = 0.6219 = 0.00915 kgv/kga P − pv 101.325 − 1.469 Eq. (3-20b) i = 1.0t + W (2501.3 + 1.86t ) kJ/kga i = 1.0(16 ) + (0.00915)[2501.3 + (1.86)(16)] = 39.16 kJ/kga

(b) English units

3. Moist Air Properties and Conditioning Processes

ps at 60 F = 0.256 psia φ = 80 % = 0.80 pv = φ ps = (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 pv 0.2048 = 0.010583 lbmv/lbma W = 0.6219 = 0.6219 P − pv 12.24 − 0.2048 Eq. (3-20a) i = 0.240t + W (1061.2 + 0.444t ) Btu/lbma i = 0.240(60) + (0.010583)[1061.2 + (0.444)(60)] = 25.91 Btu/lbma In SI units ps at 16 C = 1.836 kPa φ = 80 % = 0.80 pv = φ ps = (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 pv 1.469 W = 0.6219 = 0.6219 = 0.011045 kgv/kga P − pv 84.186 − 1.469 Eq. (3-20b) i = 1.0t + W (2501.3 + 1.86t ) kJ/kga i = 1.0(16 ) + (0.011045)[2501.3 + (1.86)(16)] = 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 φ = 0.50

3. Moist Air Properties and Conditioning Processes

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 pv = φ ps = (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 p (14.4910 )(144) = 0.0753 lb/ft3 ρ= a = RaT (53.352 )(60 + 459.67 ) Q& = 5000 ft3/min m& = ρ Q& = (0.0753)(5000) = 376.5 lb/min a

SI Units ps at 16 C = 1.836 kPa pv = φ ps = (0.80)(1.836) = 1.4688 psia At sea level, P = 101.325 kPa pa = P – pv = 101.325 – 1.4688 = 99.8562 p (99.8562)(1000) = 1.2033 kg/m3 ρ= a = RaT (287 )(16 + 273.15) Q& = 2.36 m3/s m& = ρ Q& = (1.2033)(2.36) = 2.84 kg/s a

(b) English units ps at 60 F = 0.2563 psia pv = φ ps = (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

3. Moist Air Properties and Conditioning Processes

pa = P – pv = 11.798 – 0.20504 = 11.593 psia p (11.593)(144) = 0.06021 lb/ft3 ρ= a = RaT (53.352 )(60 + 459.67 ) Q& = 5000 ft3/min m& = ρ Q& = (0.06021)(5000) = 301.05 lb/min a

SI Units ps at 16 C = 1.836 kPa pv = φ ps = (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 p (79.667 )(1000) = 0.96 kg/m3 ρ= a = RaT (287 )(16 + 273.15) Q& = 2.36 m3/s m& = ρ Q& = (0.96)(2.36) = 2.2656 kg/s a

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 ps at 80 F = 0.507 psia pv = φ ps = (0.50)(0.507) = 0.2535 psia Dew Point = tdp = 59.68 F SI units ps at 27 C = 3.602 kPa pv = φ ps = (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. Moist Air Properties and Conditioning Processes

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 pv = φ ps p 0.150 φ = v (100% ) = (100%) = 41.32 % ps 0.363 pv 0.150 W = 0.6219 = 0.6219 = 0.006413 lbmv/lbma P − pv 14.696 − 0.150 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 pv = φ ps p 1.014 φ = v (100% ) = (100%) = 40.46 % ps 2.506 W = 0.6219

pv 1.014 = 0.006287 kgv/kga = 0.6219 P − pv 101.325 − 1.014

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, pv 2 = ps 2 = 0.1013 psia

3. Moist Air Properties and Conditioning Processes i ∗fg 2 = 1973.3 Btu/lbm

iw∗ = 3.0 Btu/lbm t 2∗ = 35 F at 40 F , iv1 = 1078.5 Btu/lbm P = 29 in Hg = 14.244 psia 0.1013 Ws∗2 = 0.6219 = 0.004454 lbmv/lbma 14.244 − 0.1013 Then c (t ∗ − t ) + W ∗ i ∗ W1 = pa 2 1 ∗ s 2 fg 2 iv1 − iw 0.24(35 − 40 ) + (0.004454 )(1073.3) W1 = = 0.003283 lbmv/lbma 1078.5 − 3 i = 0.24t + W (1061.2 + 0.444t ) Btu/lbma i1 = 0.24(40) + (0.003283)[1061.2 + 0.444(40 )] = 13.14 Btu/lbma Solving for W2 and i2 at point 2 At 70 F, ps = 0.363 psia pv = φ ps = (0.40)(0.363 psia) = 0.1452 psia P = 14.244 psia pv 0.1452 W2 = 0.6219 = 0.6219 = 0.006408 lbmv/lbma P − pv 14.244 − 0.1452 i2 = 0.240t 2 + W2 (1061.2 + 0.444t 2 ) Btu/lbma i2 = 0.240(70) + (0.006408)[1061.2 + 0.444(70)] = 23.8 Btu/lbma (a) Mass of water vapor added: m& w = W2 − W1 = 0.006408 – 0.003283 = 0.003125 lbmv/lbma m& a

(b) At 50 F, iw = 18.1 Btu/lb q& m& = (i2 − i1 ) − w iw = (23.8 – 13.14) – (0.003125)(18.1) = 10.3434 Btu/lbma m& a m& a 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:

3. Moist Air Properties and Conditioning Processes

t1 = 70 F t 2∗ = 65 F Solving for Ws∗2 , Eq. (3-14b) pv 2 = ps 2 at 65 F = 0.3095 psia P2 = P1 = 29.92 in Hg = 14.696 psia pv 2 0.3095 Ws∗2 = 0.6219 = 0.6219 = 0.013379 lbmv/lbma P2 − pv 2 14.696 − 0.3095 Solving for W1 , Eq. (3-21c) c (t ∗ − t ) + W ∗ i ∗ W1 = pa 2 1 ∗ s 2 fg 2 iv1 − iw i ∗fg 2 = i fg at 65 F = 1056.5 Btu/lbm iw∗ = i f at 65 F = 33 Btu/lbm iv1 = i g at 70 F = 1091.7 Btu/lbm W1 =

0.24(65 − 70 ) + (0.013379 )(1056.5) = 0.012218 lbmv/lbma 1091.7 − 33

Solving for pv1 , Eq. (3-14b) p v1 W1 = 0.6219 P1 − pv1 p v1 0.012218 = 0.6219 14.696 − pv1 pv1 = 0.2832 psia at 70 F, ps1 = 0.363 psia (a) Relative Humidity p 0.2832 φ = v1 = = 0.78 or 78 % p s1 0.363 (b) Vapor Density p (0.2832)(144 ) = 0.000898 lbmv/ft3 ρ= v = RvT (85.78)(70 + 459.67 ) (c) Dew Point At p v1 = 0.2832 psia t dp = 62.54 F (d) Humidity Ratio

3. Moist Air Properties and Conditioning Processes

W=

m& v = W1 = 0.012218 lbmv/lbma m& a

(e) Volume occupied by mixture per pound of mass of dry air. R T (53.352 )(70 + 459.67 ) v= a = = 13.62 ft3/lbma pa (14.696 − 0.2832)(144) 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 t1 = 75 F t wb = t 2∗ = 70 F pv 2 = ps 2 at 70 F = 0.363 psia P2 = 14.696 psia pv 2 0.363 Ws∗2 = 0.6219 = 0.6219 = 0.01575 lbmv/lbma P2 − pv 2 14.696 − 0.363 c (t ∗ − t ) + W ∗ i ∗ W1 = pa 2 1 ∗ s 2 fg 2 iv1 − iw ∗ i fg 2 = i fg at 70 F = 1053.7 Btu/lbm iw∗ = i f at 70 F = 38Btu/lbm iv1 = i g at 75F = 1093.85 Btu/lbm

3. Moist Air Properties and Conditioning Processes 0.24(70 − 75) + (0.01575)(1053.7 ) = 0.014581 Btu/lbma 1093.85 − 38 i1 = 0.240t1 + W1 (1061.2 + 0.444t1 ) Btu/lbma i1 = 0.240(75) + (0.014581)[1061.2 + 0.444(75)] = 33.96 Btu/lbma

W1 =

Determine state condition2 t 2 = 55 F ps 2 = 0.217 psia ps 2 W2 = 0.6219 P2 − ps 2 0.217 W2 = 0.6219 = 0.009321 lbmv/lbma 14.696 − 0.217 i2 = 0.240t 2 + W2 (1061.2 + 0.444t 2 ) Btu/lbma i2 = 0.240(55) + (0.009321)[1061.2 + 0.444(55)] = 23.32 Btu/lbma Determine state condition 3 t3 = t1 = 75 F W3 = W2 = 0.009321 lbmv/lbma i3 = 0.240t3 + W3 (1061.2 + 0.444t3 ) Btu/lbma i3 = 0.240(75) + (0.009321)[1061.2 + 0.444(75)] = 28.20 Btu/lbma (a) Moisture removed, Eq. (3-29) m& w = W1 − W2 = 0.014581 – 0.009321 = 0.00526 Btu/lbma m& a (b) Heat removed to condense the moisture, Eq. (3-33) q&l = i1 − i3 = 33.96 – 28.20 = 5.76 Btu/lbma m& a

(c) Sensible heat removed q& s = i3 − i2 = 28.20 – 23.32 = 4.88 Btu/lbma m& a

(d) Total amount of heat removed q& q& q& = s + l = 4.88 + 5.76 = 10.64 Btu/lbma m& a m& a m& a 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).

3. Moist Air Properties and Conditioning Processes

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) t 2∗ = 62 F pv 2 P2 − pv 2 pv 2 = ps 2 at 62 F = 0.2774 psia 0.2774 Ws∗2 = 0.6219 = 0.0119865 lbmv/lbma 14.696 − 0.2774 Eq. (3-21d) c pa (t 2∗ − t1 ) + Ws∗2t ∗fg 2 W1 = iv1 − iw∗ i ∗fg 2 = i fg at 62 F = 1058.18 Btu/lbm Ws∗2 = 0.6219

iw∗ = i f at 62 F = 30 Btu/lbm iv1 = i g at 75F = 1093.85 Btu/lbm W1 =

0.24(62 − 75) + (0.011965)(1058.18) = 0.008969 lbmv/lbma – ans. 1093.85 − 30

Solving for pv1 : W1 = 0.6219

pv 1 14.696 − pv1

0.008969 = 0.6219

pv1 14.696 − pv1

pv1 = 0.20893 psia ps1 = pv at 75 F = 0.435 psia p 0.20893 φ1 = v1 = = 0.48 or 48 % - ans. p s1 0.435 (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

3. Moist Air Properties and Conditioning Processes 0.2774 pv 2 = 0.6219 = 0.013907 lbmv/lbma 12.682 − 0.2774 P2 − pv 2 c pa (t 2∗ − t1 ) + Ws∗2t ∗fg 2

Ws∗2 = 0.6219

W1 =

iv1 − iw∗ 0.24(62 − 75) + (0.013907 )(1058.18) W1 = = 0.010900 lbmv/lbma – ans. 1093.85 − 30 Solving for pv1 : pv 1 W1 = 0.6219 14.696 − pv1 pv1 0.010900 = 0.6219 12.682 − pv1 pv1 = 0.218448 psia ps1 = pv at 75 F = 0.435 psia p 0.218448 φ1 = v1 = = 0.5022 or 50.22 % - ans. p s1 0.435 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 ps W = 0.6219 14.696 − ps ps 0.001 = 0.6219 14.696 − ps ps = 0.0236 psia Use Table A-1a, t ≈ 32.02 F – ans. At P = 5 atm = 73.48 psia Solving for pv1 : W = 0.001 = 0.6219

ps 73.48 − ps

ps = 0.118 psia Use Table A-1a, interpolation, t = 39 F – ans.