Chapter 08 (1)

CHAPTER 8 1. Calculate the specific volume of an air-vapor mixture in cubic meters per kilogram of dry air at the following conditions t = 32 C , W = 0.016 kg kg , pt = 100 kPa .

Solution: 0.622p s W= pt − ps 0.622ps 0.016 = 100 − ps ps = 2.508 kPa pa = pt − ps = 100 − 2.508 = 97.492 kPa RT (0.287 )(32 + 273) = 0.899 m 3 kg va= a = pa 97.492 2. Moist air at a dry bulb temperature of 25 C has a relative humidity of 50% when the barometric pressure is 101.4 kPa. Determine (a) the partial pressures of water vapor and dry air, (b) the dew point temperature, (c) the specific humidity, (d) the specific volume, and (e) the enthalpy. Solution: At 25 C, pd = 3.169 kPa , hg = 2547.2 kJ kg ,

φ = 50% (a) ps = φ pd = (0.50 )(3.169) = 1.5845 kPa pa = pt − ps = 101.4 − 1.5845 = 99.82 kPa

(b) t dp = t sat at pd = 1.5845 = 13.7 C (c) W =

0.622 ps 0.622(1.5845) = = 0.00987 kg kg pt − p s 99.82

(d) v a =

RTa (0.287 )(25 + 273) = = 0.857 m 3 kg pa 99.82

(e) h = c p t + Whg = (1.0 )(25) + (0.00987 )(2547.2 ) = 50.14 kJ kg 3. Air at a temperature of 33 C has a relative humidity of 50%. Determine (a) the wet bulb temperature, (b) the dew point temperature, (c) the humidity ratio, (d) the enthalpy, and (e) the specific volume.

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CHAPTER 8 Solution: From psychrometric chart, at 33 C, RH = 50% (a) t wb = 24.5 C (b) t dp = 21.2 C (c) W = 0.0158 kg kg (h) h = 73.8 kJ kg , (e) v = 0.889 m 3 kg 4. How much heat is required to raise the temperature of 0.50 m3/s of air from 19 C dry bulb and 15 C wet bulb to 36 C? What is the final dew point temperature? Solution:

From psychrometric chart, At 19 C DB and 15 C WB h1 = 41.9 kJ kg v1 = 0.84 m 3 kg V 0.50 m= 1 = = 0.595 kg s v1 0.84

at 36 C, W2 = W1 = 0.0090 kg kg h2 = 59.4 kJ kg Q = m(h1 − h2 ) = 0.595(59.4 − 41.9 ) = 10.4 kW

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CHAPTER 8 t dp = t sat = 12.5 C

5. How much heat must be removed to cool 30 cu m per minute of air from 34 C dry bulb and 18 C dew point to a wet bulb temperature of 19 C? What is the final relative humidity? Solution: From psychrometric chart, At 34 C DB and 18 C Dew Point h1 = 67.3 kJ kg v1 = 0.888 m 3 kg V& 30 &= 1= m = 0.563 kg s v1 (0.888)(60)

at 19 C, W2 = W1 = 0.0129 kg kg

h2 = 54.0 kJ kg RH2 = 82% Q = m(h1 − h2 ) = 0.564(67.3 − 54) = 7.5 kW RH2 = 82%

6. How much heat and moisture must be added to 5 m3/minute of air at 21 C dry bulb and 30% relative humidity to raise it to 37 C and 40% relative humidity? Solution:

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CHAPTER 8

From psychrometric chart, At 21 C , 30 % RH h1 = 32.8 kJ kg W1 = 0.0046 kg kg v1 = 0.84 m 3 kg V& 15 &= 1= m = 0.298 kg s v1 (0.84)(60 )

at 37 C, 40 % RH h2 = 77.8 kJ kg W2 = 0.0158 kg kg heat added, & (h2 − h1 ) = 0.298(77.8 − 32.8) = 13.41 kW Q=m & (W2 − W1 ) = 0.298(0.0158 − 0.0046) = 0.00334 kg s moisture added = m

7. How much heat must be removed to cool 50 m3/min of air at 29 C dry bulb and 21 C wet bulb temperatures to 16 C dry bulb and 14 C wet bulb temperatures? How much moisture was removed? Solution:

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CHAPTER 8

From psychrometric chart, At 29 C DB and 21 C WB h1 = 60.6 kJ kg W1 = 0.0123 kg kg v1 = 0.873 m 3 kg

At 16 C DB and 14 C WB W2 = 0.0091 kg kg h2 = 39.2 kJ kg &= m

V&1 50 = = 0.955 kg s v1 (0.873)(60)

& (h1 − h2 ) = 0.955(60.6 − 39.2 ) = 20.44 kW heat removed = m & (W1 − W2 ) = 0.955(0.0123 − 0.0091) = 0.00306 kg s moisture removed = m

8. Air at 32 C and 20 percent relative humidity is cooled and humidified by means of an air washer until the relative humidity becomes 90%. How much moisture was added per kg of dry air. What was the air washer efficiency and the dew point temperature of the leaving air? Solution:

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CHAPTER 8

From psychrometric chart, At 1, 32 C and 20% RH W1 = 0.0059 kg kg At 2, 90% RH W2 = 0.0116 kg kg moisture added = W2 − W1 = 0.0116 − 0.0059 = 0.0057 kg kg air washer efficiency =

t db1 − t db2 t db1 − t wb

twb 2 = 17.0 C , t db2 = 18.0 C

air washer efficiency =

32 − 18 (100% ) = 93.3% 32 − 17

Dew point of leaving air = t dp2 = 16.4 C 9. A stream of outdoor air is mixed with a stream of return air in an air conditioning system that operates at 101 kPa pressure. The flow rate of outdoor air is 2 kg/s, and its condition is 35 C dry bulb temperature and 25 C wet bulb temperature. The flow rate of return air is 3 kg/s, and its condition is 24 C and 50 percent relative humidity. Determine (a) the enthalpy of the mixture, (b) the humidity ratio of the mixture, and (c) the dry bulb temperature of the mixture. Solution:

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CHAPTER 8

From psychrometric chart, At 35 C DB, 25 C WB ho = 75.9 kJ kg Wo = 0.0159 kg kg at 24 C, 50% RH hr = 47.8 kJ kg Wr = 0.0093 kg kg mo ho + mr hr (2 )(75.9 ) + (3)(47.8) = = 59.0 kJ kg mm 5 m W + mrWr (2 )(0.0159) + (3)(0.0093) = = 0.0119 kg kg (b) Wm = o o mm 5 mo hdbo + mr hdbr (2)(35) + (3)(24) (c) t dbm = = = 28.4 C mdbm 5

(a) hm =

10. An auditorium is to be maintained at 25 C dry bulb temperature and 50% relative humidity. The supply air enters the auditorium at 17 C. The sensible and latent heat loads are 150 kW and 61 kW, respectively. Determine the wet bulb temperature, relative humidity, and volume flow rate of the supply air. Solution:

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CHAPTER 8

QS = 150 kW , QL = 61 kW

At 25 C, 50% RH h2 = 50.3 kJ kg W2 = 0.0099 kg kg

(

& c p t db − t db QS = m 2 1

)

& (1.0062)(25 − 17 ) 150 = m & = 18.63 kg s m h1 = h2 −

QS + QL 150 + 61 = 50.3 − = 39.0 kJ kg & m 18.63

& (W2 − W1 ) QL = 2500m 61 = 2500(18.63)(0.0099 − W1 ) W1 = 0.0086 kg kg

From psychrometric chart, At h1 = 39.0 kJ kg , W1 = 0.0086 kg kg Then, Wet bulb at 1, t wb1 = 13.9 C Relative humidity, RH1 = 71.5%

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CHAPTER 8 v1 = 0.8334 m3 kg & v1 = (18.63)(0.8334) = 15.53 m 3 s V&1 = m

11. In a certain space to be air conditioned the sensible and latent heat loads are 20.60 kW and 6.78 kW, respectively. Outside air is at 33 C dry bulb and 24 C wet bulb temperatures. The space is to be maintained at 25 C with a relative humidity not exceeding 50%. All outside air is supplied with reheater. The conditioned air enters at 18 C. Determine (a) the refrigeration load required, (b) the capacity of the supply fan, and (c) the heat supplied in the reheater. Solution:

From pyschrometric chart. At 1, t db1 = 33 C , t wb1 = 24 C h1 = 71.9 kJ kg

At 4, t db4 = 25 C , RH 4 = 50% h4 = 50.3 kJ kg W4 = 0.0099 kg kg

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CHAPTER 8

At 3, t db3 = 18 C

(

& c p t db − t db QS = m 4 3

)

& (1.0062)(25 − 18) 20.60 = m & = 2.925 kg s m & (h4 − h3 ) QS + QL = m 20.60 + 6.78 = 2.925(50.3 − h3 ) h3 = 40.9 kJ kg

at tdb3 = 18 C , h3 = 40.9 kJ kg

W3 = 0.0090 kg kg v 3 = 0.837 m 3 kg

At 2, with W2 = W3 = 0.0090 kg kg h2 = 35.1 kJ kg & (h1 − h2 ) = 2.925(71.9 − 35.1) = 107.6 kW (a) Refrigeration load required = m & v 3 = (2.925)(0.837 ) = 2.45 m 3 s (b) Capacity m & (h3 − h2 ) = 2.925(40.9 − 35.1) = 16.97 kW (c) Heat supplied in the reheater = m

12. An air conditioned auditorium with a capacity of 1000 persons is to be maintained at 24 C dry bulb temperature and 55% relative humidity. The sensible and latent heat loads are 115 kW and 42 kW, respectively. The conditioned air enters the auditorium at 17 C. For proper ventilation, 40% of the supply air is fresh air and the rest is recirculated air. Outside air is at 34 C and 50% relative humidity. Determine (a) the volume flow rate of recirculated air, (b) the apparatus dew point, and (c) the refrigeration load. Solution:

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CHAPTER 8

& o = 0.40m & m & r = 0.60m & m

At 4, t db4 = 24 C , φ3 = 55% h4 = 50.2 kJ kg W4 = 0.0102 kg kg v 4 = 0.856 m 3 kg

At 1, t db1 = 34 C , φ1 = 50% h1 = 77.2 kJ kg

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CHAPTER 8

At 5, t db5 = 0.40(34 ) + 0.60(24) = 28 C h5 = 0.40(77.2) + 0.60(50.2 ) = 61 kJ kg

(

& c p t db − t db QS = m 4 3

)

& (1.0062)(24 − 17 ) 115 = m & = 16.33 kg s m & (W4 − W3 ) QL = 2500m 42 = 2500(16.33)(0.0102 − W3 ) W3 = 0.0092 kg kg

At 2, t db2 = 12.8 C , W2 = W3 = 0.0092 kg kg h2 = 36.1 kJ kg & r v 4 = (0.60 )(16.33)(0.856) = 8.39 m 3 s (a) Volume flow rate of recirculated air = V&r = m (b) Apparatus dew point = t db2 = 12.6 C

& (h5 − h2 ) = 16.33(61 − 36.1) = 406.6 kW (c) Refrigeration Load = = m

13. A store to be maintained at 25 C and 50% relative humidity has a sensible heat load of 18.90 kW and a latent heat load of 6.30 kW. Outside air is at 32 C dry bulb and 23 C wet bulb temperatures. The conditioned air enters at 17 C. If 30% of the supply air is fresh air and the bypass system is used, determine (a) the refrigeration required, and (b) the volume of the bypass air at supply condition. Solution:

t db4 = 25 C , φ = 50%

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CHAPTER 8 QS = 18.90 kW QL = 6.30 kW t db3 = 17 C t db1 = 32 C t wb1 = 23 C

mo = 0.30m mr + mb = 0.70m

From psychrometric chart At 4, t db4 = 25 C , φ = 50% h4 = 50.3 kJ kg W4 = 0.0099 kg kg v 4 = 0.858 m 3 kg

(

& c p t db − t db QS = m 4 3

)

& (1.0062)(25 − 17 ) 18.90 = m & = 2.348 kg s m & (h4 − h3 ) QS + QL = m 18.90 + 6.30 = 2.348(50.3 − h3 ) h3 = 39.6 kJ kg

At 3, t db3 = 17 C , h3 = 39.6 kJ kg v 3 = 0.834 m 3 kg

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CHAPTER 8

at 1, t db1 = 32 C , t wb1 = 23 C h1 = 68.0 kJ kg

To solve for t db2 : Let b =

mb m , c= c m m

(1)

b + c =1 tdb 3 = b(tdb 4 ) + c (tdb 2 )

(2)

17 = 25b + c(t db 2 ) t wb3 = b(t wb 4 ) + c (t wb 2 ) t wb3 = 14 C , t wb4 = 17.8 C t wb2 = t db2

14 = 17.8b + c (t db 2 )

(3) (3) – (2)

17 − 14 = (25 − 17.8)b b = 0.417 c = 1 − b = 1 − 0.417 = 0.583

Substitute in (2) 17 = 25b + c(t db 2 ) t wb2 = t db2 = 11.3 C h1 = 32.4 kJ kg

& c = cm & = 0.583(2.348) = 1.367 kg s ∴ m & b = bm & = 0.417(2.348) = 0.979 kg s m & o = 0.30m & = 0.30(2.348) = 0.704 kg s m &r +m & b = 0.70m & = 0.70(2.348) m & r + 0.979 = 0.70(2.348) m & r = 0.665 kg s m

∴ at 5, h5 =

& r h4 + m & o h1 0.665(50.3) + 0.704(68.0 ) m = = 59.4 kJ kg &r +m &o m 0.665 + 0.704

& c (h5 − h2 ) = 1.367(59.3 − 32.4) = 36.8 kW (a) Refrigeration load = m (b) Volume of the bypass air at supply condition & bv 3 = 0.979(0.834) = 0.816 m 3 s =m

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CHAPTER 8

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