5-58 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input required is
to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential
energy changes are negligible. 3 Helium is an ideal gas with constant specific heats.
Properties The constant pressure specific heat of helium is cp = 5.1926 kJ/kg·K (Table A-2a).
1  m
2  m
.
Analysis There is only one inlet and one exit, and thus m
We take the compressor as the system, which is a control volume since
P2 = 700 kPa
mass crosses the boundary. The energy balance for this steady-flow
T2 = 430 K
system can be expressed in the rate form as
E  E out

E system0 (steady)
0
in



Rate of net energy transfer
·
Rate of change in internal,kinetic,
by heat, work, and mass
Q
potential,etc. energies
He
E in  E out
90 kg/min
·
W
W  m h  Q  m h (since ke  pe  0)
in
1
out
2
W in  Q out  m (h2  h1 )  m c p (T2  T1 )
Thus,
Win  Q out  m c p T2  T1 
P1 = 120 kPa
T1 = 310 K
 (90/6 0 kg/s)(20 kJ/kg) + (90/60 kg/s)(5.19 26 kJ/kg  K)(430  310)K
 965 kW
5-89 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the
outlet temperature of the air are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to
the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the
cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties
are constant.
Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg.C,
respectively.
Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this
steady-flow system can be expressed in the rate form as
E  E out
in


E system0 (steady)


Rate of net energy transfer
by heat, work, and mass
 0  E in  E out
Rate of change in internal,kinetic,
potential,etc. energies
m h1  Q out  m h2 (since ke  pe  0)
Q out  m c p (T1  T2 )
Then the rate of heat transfer from the
exhaust gases becomes
Q  [m c (T  T )]
p
in
out
gas
 (1.1 kg/s)(1.1 kJ/kg. C)(180 C  95 C)
= 102.85 kW
Air
95 kPa
20C
0.8 m3/s
The mass flow rate of air is
m 
(95 kPa)(0.8 m 3 /s)
PV

 0.904 kg/s
RT (0.287 kPa.m 3 /kg.K)  293 K
Exhaust gases
1.1 kg/s, 95C
Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the
air becomes
Q
102 .85 kW
 c p (Tc,out  Tc,in ) 
Q  m
Tc,out  Tc,in 
 20 C 
 133.2C
 cp
m
(0.904 kg/s)(1.00 5 kJ/kg. C)
5-62E Air is expanded in an adiabatic turbine. The mass flow rate of the air and the power produced are to
be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is wellinsulated, and thus there is no heat transfer. 3 Air is an ideal gas with constant specific heats.
Properties The constant pressure specific heat of air at the average temperature of (800+250)/2=525°F is cp
= 0.2485 Btu/lbm·R (Table A-2Eb). The gas constant of air is R = 0.3704 psiaft3/lbmR (Table A-1E).
1  m
2  m
 . We take the turbine as the system,
Analysis There is only one inlet and one exit, and thus m
which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system
can be expressed in the rate form as
E  E

E system0 (steady)
0
inout



Rate of net energy transfer
Rate of change in internal,kinetic,
500 psia
by heat, work, and mass
potential,etc. energies
800°F
E in  E out

V2
m  h1  1

2

2 


  m  h2  V2   W out


2 



V 2  V22
W out  m  h1  h2  1

2

Turbine
2
2


  m  c p (T1  T2 )  V1  V2


2






60 psia
250°F
50 ft3/s
The specific volume of air at the exit and the mass flow rate are
v2 

m
RT2 (0.3704 psia  ft 3 /lbm  R)(250  460 R)

 4.383 ft 3 /lbm
P2
60 psia
V2
50 ft 3 /s

 11.41 kg/s
v 2 4.383 ft 3 /lbm
V2 
 v 2 (11 .41 lbm/s)(4.3 83 ft 3 /lbm)
m

 41 .68 ft/s
A2
1.2 ft 2
Similarly at the inlet,
v1 
RT1 (0.3704 psia  ft 3 /lbm  R)(800  460 R)

 0.9334 ft 3 /lbm
P1
500 psia
V1 
 v 1 (11.41 lbm/s)(0.9 334 ft 3 /lbm)
m

 17 .75 ft/s
A1
0.6 ft 2
Substituting into the energy balance equation gives

V 2  V 22 
W out  m  c p (T1  T2 )  1


2



(17 .75 ft/s) 2  (41 .68 m/s) 2
 (11 .41 lbm/s) (0.2485 Btu/lbm  R)(800  250 )R 
2

 1559 kW
 1 Btu/lbm

 25,037 ft 2 /s 2





5-68 Refrigerant-134a is throttled by a capillary tube. The quality of the refrigerant at the exit is to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential
energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work
interactions involved.
1  m
2  m
 . We take the throttling valve as the
Analysis There is only one inlet and one exit, and thus m
system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as
E in  E out  E system0 (steady)  0
E in  E out
m h1  m h2
50°C
Sat. liquid
h1  h2
since Q  W  ke  Δpe  0 .
R-134a
The inlet enthalpy of R-134a is, from the refrigerant tables (Table A-11),
T1  50 C 
 h1  h f  123 .49 kJ/kg
sat. liquid 
The exit quality is
h2  h f 123 .49  35 .92
T2  12 C 

 0.422
 x2 
h2  h1
h fg
207 .38

-12°C