14-54
14-93 Air is cooled and dehumidified at constant pressure. The cooling required is provided by a simple ideal vaporcompression refrigeration system using refrigerant-134a as the working fluid. The exergy destruction in the total system per
1000 m3 of dry air is to be determined.
Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire
process (m a1  m a 2  m a ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are
negligible.
Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of
the air at various states are determined from the psychrometric chart (Figure A-31) to be
h1  106.8 kJ/kg dry air
Condenser
1  0.0292 kg H 2 O/kg dry air
3
v 1  0.905 m 3 /kg dry air
2
E
x
and
h2  52.7 kJ/kg dry air
 2  0.0112 kg H 2 O/kg dry air
Compressor
4
1
We assume that the condensate leaves this system at the
average temperature of the air inlet and exit. Then, from
Table A-4,
h w  h f @ 28C  117.4 kJ/kg
Analysis The amount of moisture in the air decreases
due to dehumidification ( 2 <  1). The mass of air is
ma 
Evaporator
T2 = 24C
 2 = 60%
T1 = 32C
 1 = 95%
1 atm
V1
1000 m 3

 1105 kg
v 1 0.905 m 3 / kg dry air
Condensate
Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section,
Water Mass Balance:
 m w,i   m w ,e 
 m a1 1  m a 2 2  m w
m w  m a (1   2 )  (1105 kg)(0.0292  0.0112)  19.89 kg
Energy Balance:
E in  E out  E system 0 (steady)  0
E in  E out
 m i hi  Q out   m e he
Q out  m a1 h1  (m a 2 h2  m w h w )  m a (h1  h2 )  m w h w
Qout  m a (h1  h2 )  m w hw
Qout  (1105 kg)(106.8  52.7)kJ/kg  (19.89 kg)(117.4 kJ/kg)  57,450 kJ
We obtain the properties for the vapor-compression refrigeration cycle as follows (Tables A-11,through A-13):
T1  4C  h1  h g @ 4C  252.77 kJ/kg

sat. vapor  s1  s g @ 4C  0.92927 kJ/kg  K
P2  Psat @ 39.4C  1 MPa 
 h2  275.29 kJ/kg
s 2  s1

T
·
QH
P3  1 MPa  h3  h f @ 1 MPa  107.32 kJ/kg

sat. liquid  s 3  s f @ 1 MPa  0.39189 kJ/kg  K
h4  h3  107.32 kJ/kg ( throttling)
T4  4C
 x 4  0.2561

h4  107.32 kJ/kg  s 4  0.4045 kJ/kg  K
2
3 39.4C
·
Win
4 C
4s
4
·
QL
1
s
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
14-55
The mass flow rate of refrigerant-134a is
mR 
QL
57,450 kJ

 395.0 kg
h1  h4 (252.77  107.32)kJ/kg
The amount of heat rejected from the condenser is
Q H  m R (h2  h3 )  (395.0 kg )(275.29  107.32) kJ/kg  66,350 kg
Next, we calculate the exergy destruction in the components of the refrigeration cycle:
X destroyed,12  m R T0 ( s 2  s1 )  0 (since the process is isentropic)

Q 
X destroyed, 23  T0  m R ( s 3  s 2 )  H 
TH 

66,350 kJ 

 (305 K ) (395 kg)(0.39189  0.92927) kJ/kg  K 
  1609 kJ
305 K 

X destroyed, 34  m R T0 ( s 4  s 3 )  (395 kg)(305 K )(0.4045  0.39189) kJ/kg  K  1519 kJ
The entropies of water vapor in the air stream are
s g1  s g @ 32C  8.4114 kJ/kg  K
s g 2  s g @ 24C  8.5782 kJ/kg  K
The entropy change of water vapor in the air stream is
S vapor  m a ( 2 s g 2  1 s g1 )  (1105)(0.0112  8.5782  0.0292  8.4114)  165.2 kJ/K
The entropy of water leaving the cooling section is
S w  m w s f @ 28C  (19.89 kg )(0.4091 kJ/kg  K)  8.14 kJ/K
The partial pressures of water vapor and dry air for air streams are
Pv1  1 Pg1  1 Psat @ 32C  (0.95)(4.760 kPa)  4.522 kPa
Pa1  P1  Pv1  101.325  4.522  96.80 kPa
Pv 2   2 Pg 2   2 Psat @ 24C  (0.60)(2.986 kPa)  1.792 kPa
Pa 2  P2  Pv 2  101.325  1.792  99.53 kPa
The entropy change of dry air is

P 
T
S a  m a ( s 2  s1 )  m a  c p ln 2  R ln a 2 
Pa1 
T
1

297
99.53 

 (1105) (1.005) ln
 (0.287) ln
 38.34 kJ/kg dry air
305
96
.80 

The entropy change of R-134a in the evaporator is
S R, 41  m R ( s1  s 4 )  (395 kg)(0.92927  0.4045)  207.3 kJ/K
An entropy balance on the evaporator gives
S gen,evaporator  S R,41  S vapor  S a  S w  207.3  (165.2)  (38.34)  8.14  11.90 kJ/K
Then, the exergy destruction in the evaporator is
X dest  T0 S gen, evaporator  (305 K)(11.90 kJ/K)  3630 kJ
Finally the total exergy destruction is
X dest, total  X dest, compressor  X dest, condenser  X dest, throttle  X dest, evaporator
 0  1609  1519  3630
 6758 kJ
The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from humid air and rejected to the
ambient air at 32C (305 K), which is also taken as the dead state temperature.
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.