Chapter_08.pdf

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1. Calculate the specific volume of an air-vapor mixture in cubic meters per kilogram of dry air at the following conditions t₌32C_,W₌0_.016kg kg_,p_t₌100kPa_.

Solution: s t s p p p W − =0.622 s s p p − = 100 622 0 016 0_. . kPa p_s ₌2_.508 kPa p p p_a ₌ _t₋ _s ₌100₋2_.508₌97_.492

(

)(

)

kg m p RT v a a a 3 899 0 492 97 273 32 287 0 . . . = + = =

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, p_d ₌3_.169kPa_,_h _kJ _kg g =2547.2 , % 50 = φ (a) p_s =φp_d =

(

0.50

)(

3.169

)

=1.5845kPa kPa p p p_a ₌ _t₋ _s ₌101_.4₋1_.5845₌99_.82 (b) t_dp₌t_sat atp_d ₌1_.5845₌13_.7 C (c)

(

)

kg kg p p p W s t s ₀₀₀₉₈₇ 82 99 5845 1 622 0 622 0 . . . . . = = − = (d)

(

)(

)

m kg p RT v a a a 3 857 0 82 99 273 25 287 0 . . . = + = = (e) h=c_pt+Wh_g =

(

1.0

)( ) (

25 + 0.00987

)(

2547.2

)

=50.14kJ 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

2

Solution:

From psychrometric chart, at 33 C, RH = 50% (a) t_wb ₌24_.5C

(b) t_dp=21.2C (c) W=0.0158kg kg (h) h=73.8kJ kg, (e) v=0.889m3 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 kg kJ h₁=41.9 kg m v1=0.84 3 s kg v V m 0595 84 0 50 0 1 1 . . . = = = at 36 C, W₂ =W₁=0.0090kg kg kg kJ h₂=59.4

(

h h

)

(

)

kW m Q= ₁− ₂ =0.59559.4−41.9 =10.4

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3

C

t

t_dp₌ _sat₌12_.5

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

kg kJ h₁=67.3 kg m v1=0.888 3

(

)( )

kg s v V m 0563 60 888 0 30 1 1 . . = = = & & at 19 C, W₂ =W₁=0.0129kg kg kg kJ h₂=54.0 % 82 2= RH

(

h h

)

(

)

kW m Q= ₁− ₂ =0.56467.3−54 =7.5 % 82 2= RH

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?

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CHAPTER 8

4

From psychrometric chart,

At 21 C , 30 % RH kg kJ h₁=32.8 kg kg W₁₌0_.0046 kg m v1=0.84 3

(

)( )

kg s v V m 0298 60 84 0 15 1 1 . . = = = & & at 37 C, 40 % RH kg kJ h₂₌77_.8 kg kg W₂ ₌0_.0158 heat added,

(

h h

)

(

)

kW m Q= & ₂− ₁ =0.29877.8−32.8 =13.41

moisture added=m&

(

W₂−W₁

)

=0.298

(

0.0158−0.0046

)

=0.00334kg s

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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5

At 29 C DB and 21 C WB kg kJ h₁=60.6 kg kg W₁=0.0123 kg m v1=0.873 3 At 16 C DB and 14 C WB kg kg W₂=0.0091 kg kJ h₂₌39_.2

(

)( )

kg s v V m 0955 60 873 0 50 1 1 . . = = = & &

heat removed =m&

(

h₁−h₂

)

=0.955

(

60.6−39.2

)

=20.44kW

moisture removed =m&

(

W₁−W₂

)

=0.955

(

0.0123−0.0091

)

=0.00306kg s

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?

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CHAPTER 8

6

At 1, 32 C and 20% RH kg kg W₁=0.0059 At 2, 90% RH kg kg W₂ =0.0116 moisture added =W₂−W₁=0.0116−0.0059=0.0057kg kg

air washer efficiency =

wb db db db t t t t − − 1 2 1 C t_wb₂=17.0 , t_db 180C 2 = .

air washer efficiency =

(

100%

)

93.3% 17 32 18 32 = − −

Dew point of leaving air = t_dp 164C 2 = .

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.

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7

At 35 C DB, 25 C WB kg kJ h_o =75.9 kg kg W_o=0.0159 at 24 C, 50% RH kg kJ h_r =47.8 kg kg W_r=0.0093 (a)

( )(

) ( )(

)

kJ kg m h m h m h m r r o o m 590 5 8 47 3 9 75 2 . . . = + = + = (b)

( )(

) ( )(

)

kg kg m W m W m W m r r o o m 00119 5 0093 0 3 0159 0 2 . . . = + = + = (c)

( )( ) ( )( )

C m h m h m t m r o m db db r db o db 284 5 24 3 35 2 . = + = + =

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.

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CHAPTER 8

8

kW Q_S ₌150 _,Q_L ₌61kW At 25 C, 50% RH kg kJ h₂=50.3 kg kg W₂=0.0099

(

db2 db1

)

p S mc t t Q = & −

(

10062

)(

25 17

)

150₌m& . ₋ s kg m& =18.63 kg kJ m Q Q h h S L ₃₉₀ 63 18 61 150 3 50 2 1 . . . − + = = + − = &

(

2 1

)

2500mW W Q_L = & −

(

1863

)

(

00099 1

)

2500 61= . . −W kg kg W₁=0.0086

At h₁₌39_.0kJ kg_,W₁₌0_.0086kg kg Then,

Wet bulb at 1, t_wb 139C 1= . Relative humidity, RH₁₌71.5%

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9

kg m v1=0.8334 3

(

)(

)

m s v m V&1= & 1= 18.63 0.8334 =15.53 3

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_db 33C 1 = , twb C 24 1 = kg kJ h₁₌71_.9 At 4, t_db 25C 4 = , % 50 4 = RH kg kJ h₄₌50_.3 kg kg W₄ ₌0_.0099

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CHAPTER 8

10

At 3, t_db 18C 3 =

(

db4 db3

)

p S mc t t Q = & −

(

10062

)(

25 18

)

60 20_. ₌_m_& _. ₋ s kg m_& ₌2_.925

(

h4 h3

)

m Q Q_S+ _L = & −

(

503 3

)

925 2 78 6 60 20. + . = . . −h kg kJ h₃₌40_.9 at t_db 18C 3 = , h kJ kg 9 40 3= .

kg

W

₃

₌

0 .

0090

kg m v3=0.837 3 At 2, with W₂ ₌W₃ ₌0_.0090kg kg kg kJ h₂=35.1

(a) Refrigeration load required ₌m_&

(

h₁₋h₂

)

₌2_.925

(

71_.9₋35_.1

)

₌107_.6kW (b) Capacity m&v3=

(

2.925

)(

0.837

)

=2.45m3 s

(c) Heat supplied in the reheater =m&

(

h₃−h₂

)

=2.925

(

40.9−35.1

)

=16.97kW

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.

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11

m m&_o₌0_.40& m m&_r =0.60& At 4, t_db 24C 4 = , % 55 3 = φ kg kJ h₄=50.2 kg kg W₄ =0.0102 kg m v4 =0.856 3 At 1, t_db 34C 1 = , % 50 1= φ kg kJ h₁=77.2

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CHAPTER 8

12

At 5, t_db 040

( )

34 060

( )

24 28C 5 = . + . =

(

)

(

)

kJ kg h₅ ₌0_.4077_.2 ₊0_.6050_.2 ₌61

(

db4 db3

)

p S mc t t Q = & −

(

10062

)(

24 17

)

115₌m& _. ₋ s kg m_& ₌16_.33

(

4 3

)

2500mW W Q_L = & −

(

1633

)(

00102 3

)

2500 42₌ _. _. ₋_W kg kg W₃ =0.0092 At 2, t_db 128C 2 = . , W W kg kg 0092 0 3 2 = = . kg kJ h₂₌36_.1

(a) Volume flow rate of recirculated air = V&_r =m&_rv4 =

(

0.60

)(

16.33

)(

0.856

)

=8.39m3 s (b) Apparatus dew point = t_db 126C

2 = .

(c) Refrigeration Load = =m&

(

h₅−h₂

)

=16.33

(

61−36.1

)

=406.6kW

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: C t_db 25 4 = , % 50 = φ

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13

kW Q_S ₌18_.90 kW Q_L =6.30 C t_db 17 3 = C t_db 32 1 = C t_wb 23 1 = m m_o ₌0_.30 m m m_r₊ _b₌0_.70

From psychrometric chart At 4, t_db 25C 4 = , % 50 = φ kg kJ h₄₌50_.3 kg kg W₄=0.0099 kg m v4 =0.858 3

(

db4 db3

)

p S mc t t Q = & −

(

10062

)(

25 17

)

90 18_. ₌m& _. ₋ s kg m_& ₌2_.348

(

h4 h3

)

m Q Q_S+ _L = & −

(

503 3

)

348 2 30 6 90 18. + . = . . −h kg kJ h₃₌39_.6 At 3, t_db₃₌17C_,h₃₌39_.6kJ kg kg m v3 =0.834 3

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CHAPTER 8

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at 1, t_db 32C 1 = , twb C 23 1 = kg kJ h₁₌68_.0 To solve for 2 db t : Let m m b b = , m m c c = (1) _b_{+ c}₌1

(

4

)

(

2

)

3 db db db bt c t t = + (2) 17=25b+c

(

tdb2

)

(

4

)

(

2

)

3 wb wb wb bt ct t = + C t_wb 14 3 = , twb C 8 17 4 = . 2 2 db wb t t = (3) 14=17.8b+c

(

tdb2

)

(3) – (2)

17 ₋

14 ₌

(

25 ₋

17 .

8 )

b

417 . 0 = b 583 0 417 0 1 1₋ ₌ ₋ _. ₌ _. = b c Substitute in (2)

(

2

)

25 17= b+ct_db C t t_wb _db 113 2 2 = = . kg kJ h₁₌32_.4 ∴ m&_c =cm& =0.583

(

2.348

)

=1.367kg s

(

)

kg s m b m&_b = & =0.4172.348 =0.979

(

)

kg s m m&_o =0.30& =0.302.348 =0.704

(

2348

)

70 0 70 0. = . . = +m m

m&_r &_b &

(

2348

)

70 0 979 0. = . . + r m& s kg m&_r =0.665

∴ at 5,

(

)

(

)

kg kJ m m h m h m h o r o r ₅₉₄ 704 0 665 0 0 68 704 0 3 50 665 0 1 4 5 . . . . . . . = + + = + + = & & & &

(a) Refrigeration load ₌m_&_c

(

h₅₋h₂

)

₌1_.367

(

59_.3₋32_.4

)

₌36_.8kW (b) Volume of the bypass air at supply condition

(

)

m s

v

m_b 3 =0.9790.834 =0.816 3

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