5-85 Ethylene glycol is cooled by water in a heat exchanger. The rate of heat transfer in
the heat exchanger and the mass flow rate of water 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
Cold
water and ethylene glycol are
Water
given to be 4.18 and 2.56
20C
kJ/kg.C, respectively.
Hot
Analysis (a) We take the ethylene glycol
Glycol
tubes as the system, which is a control
volume. The energy balance for this
40C
steady-flow system can be expressed in
80C
the rate form as
2 kg/s
E  E out
in



Rate of net energy transfer
by heat, work, and mass
E system0 (steady)

0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
m h1  Q out  m h2 (since ke  pe  0)
Q out  m c p (T1  T2 )
Then the rate of heat transfer becomes
Q  [m c p (Tin  Tout )] glycol  (2 kg/s)(2.56 kJ/kg. C)(80 C  40 C) = 204.8 kW
(b) The rate of heat transfer from glycol must be equal to the rate of heat transfer to the
water. Then,
 c p (Tout  Tin )]water 
 water 
Q  [m
 m
Q
c p (Tout  Tin )

204 .8 kJ/s
= 1.4 kg/s
(4.18 kJ/kg. C)(55 C  20 C)
5-96 Two streams of water are mixed in an insulated chamber. The temperature of the
exit stream 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 There are no work interactions. 4 The
device is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Table A-4),
h1  hf @ 90C = 377.04 kJ/kg
h2  hf @ 50C = 209.34 kJ/kg
Analysis We take the mixing chamber as the system, which is a control volume since
mass crosses the boundary. The mass and energy balances for this steady-flow system
can be expressed in the rate form as
m in  m out  m system0 (steady)  0
m in  m out

m1  m 2  m 3
Mass balance:
Water
90C
30 kg/s
Energy balance:
E  E
inout

E system0 (steady)




Rate of net energy transfer
by heat, work, and mass
3
0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
m 1 h1  m 2 h2  m 3 h3 (since Q  W  ke  pe  0)
1
Water
50C
200 kg/s
2
Solving for the exit enthalpy,
h3 
m 1 h1  m 2 h2 (30 )(377 .04 )  (200 )( 209 .34 )

 231 .21 kJ/kg
m 1  m 2
30  200
The temperature corresponding to this enthalpy is
T3 =55.2C (Table A-4)
5-124 A rigid tank initially contains air at atmospheric conditions. The tank is connected
to a supply line, and air is allowed to enter the tank until mechanical equilibrium is
established. The mass of air that entered and the amount of heat transfer are to be
determined.
Assumptions 1 This is an unsteady process since the conditions within the device are
changing during the process, but it can be analyzed as a uniform-flow process since the
state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific
heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions
involved. 5 The direction of heat transfer is to the tank (will be verified).
Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The properties of
air are (Table A-17)
Ti  295 K 
 hi  295 .17 kJ/kg
T1  295 K 
 u1  210 .49 kJ/kg
T2  350 K 
 u 2  250 .02 kJ/kg
Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.
Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and
internal energy u, respectively, the mass and energy balances for this uniform-flow system can be
expressed as
Mass balance:
min  mout  msystem  mi  m2  m1
Energy balance:
E  Eout
in


Net energy transfer
by heat, work, and mass

Pi = 600 kPa
Ti = 22C
Esystem



Change in internal,kinetic,
potential,etc. energies
Qin  mi hi  m2u2  m1u1 (since W  ke  pe  0)
The initial and the final masses in the tank are
V1 = 2 m
P1 = 100 kPa
T1 = 22C
3
·
Q
m1 
P1V
(100 kPa )( 2 m3 )

 2.362 kg
RT1 (0.287 kPa  m3/kg  K )( 295 K )
m2 
P2V
(600 kPa )( 2 m3 )

 11.946 kg
RT2 (0.287 kPa  m3/kg  K )(350 K )
Then from the mass balance,
mi  m2  m1  11.946  2.362  9.584 kg
(b) The heat transfer during this process is determined from
Qin  mi hi  m2u2  m1u1
 9.584 kg 295.17 kJ/kg   11.946 kg 250.02 kJ/kg   2.362 kg 210.49 kJ/kg 
 339 kJ  Qout  339 kJ
Discussion The negative sign for heat transfer indicates that the assumed direction is
wrong. Therefore, we reversed the direction.
6-19E The work output and heat rejection of a heat engine that propels a ship are given.
The thermal efficiency of the engine is to be determined.
Assumptions 1 The plant operates steadily. 2 Heat
losses from the working fluid at the pipes and
Furnac
other components are negligible.
qeH
Analysis Applying the first law to the heat engine gives
HE
q H  wnet  q L  500 Btu/lbm  300 Btu/lbm  800 Btu/lbm
wnet
qL
Substituting this result into the definition of the thermal efficiency, sink
 th 
wnet 500 Btu/lbm

 0.625
qH
800 Btu/lbm
6-40 The COP and the refrigeration rate of a refrigerator are given. The power
consumption and the rate of heat rejection are to be determined.
Assumptions The refrigerator operates steadily.
Analysis (a) Using the definition of the coefficient of
performance, the power input to the refrigerator is
Kitchen air
determined to be
COP=1.2
Q L
60 kJ/min
Wnet,in 

 50 kJ/min  0.83 kW
R
COP
1.2
R
(b) The heat transfer rate to the kitchen air is determined
from the energy balance,
Q L
cool space
Q H  Q L  Wnet,in  60  50  110 kJ/min
6-50 A house is heated by resistance heaters, and the amount of electricity consumed
during a winter month is given. The amount of money that would be saved if this house
were heated by a heat pump with a known COP is to be determined.
Assumptions The heat pump operates steadily.
Analysis The amount of heat the resistance heaters supply to the house is equal to he
amount of electricity they consume. Therefore, to achieve the same heating effect, the
house must be supplied with 1200 kWh of energy. A heat pump that supplied this much
heat will consume electrical power in the amount of
Wnet,in 
Q H
1200 kWh

 500 kWh
COP HP
2.4
which represent a savings of 1200 – 500 = 700 kWh. Thus the homeowner would have
saved
(700 kWh)(0.085 $/kWh) = $59.50
6-54 The rate of heat loss, the rate of internal heat gain, and the COP of a heat pump are
given. The power input to the heat pump is to be determined.
Assumptions The heat pump operates steadily.
Analysis The heating load of this heat pump system is the
60,00
difference between the heat lost to the outdoors and the
0
kJ/h
House
heat generated in the house from the people, lights, and
Q H
appliances,
HP
Q H  60,000  4,000  56,000 kJ / h
COP =
Using the definition of COP, the power input to the heat
pump is determined to be
Outside2.5
Wnet,in 
Q H
56,000 kJ/h  1 kW 

  6.22 kW

COP HP
2.5
 3600 kJ/h 
6-58 A commercial refrigerator with R-134a as the working fluid is considered. The
evaporator inlet and exit states are specified. The mass flow rate of the refrigerant and the
rate of heat rejected are to be determined.
QH
Assumptions 1 The refrigerator operates
steadily. 2 The kinetic and potential energy
Condenser
changes are zero.
Properties The properties of R-134a at the
Expansion
evaporator inlet and exit states are (Tables
Win
valve
A-11 through A-13)
Compressor
P1  120 kPa 
h1  65 .38 kJ/kg
x1  0.2

P2  120 kPa 
h2  238 .84 kJ/kg
T2  20C 
Evaporator
120 kPa
x=0.2
Analysis (a) The refrigeration load is
Q L  (COP)W in  (1.2)(0.45 kW)  0.54 kW
The mass flow rate of the refrigerant is determined from
QL
120 kPa
-20C
m R 
Q L
0.54 kW

 0.0031 kg/s
h2  h1 (238 .84  65 .38) kJ/kg
(b) The rate of heat rejected from the refrigerator is
Q H  Q L  W in  0.54  0.45  0.99 kW