5-56C A steady-flow system involves no changes with time anywhere within the system
or at the system boundaries
5-57C No.
5-58C It is mostly converted to internal energy as shown by a rise in the fluid
temperature.
5-59C The kinetic energy of a fluid increases at the expense of the internal energy as
evidenced by a decrease in the fluid temperature.
5-60C Heat transfer to the fluid as it flows through a nozzle is desirable since it will
probably increase the kinetic energy of the fluid. Heat transfer from the fluid will
decrease the exit velocity.
5-61 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit
temperature, and the exit area of the nozzle are to be determined.
steady-flow process since there is no change with time.
air is ideal gas with constant specific heats.
Potential energy changes are negligible.
adiabatic and so heat transfer is negligible and no work interactions.
average temperature of 450 K is Cp = 1.02 kJ/kg.C (Table A-2).
(a) There is only one inlet and one exit, and so m 1  m 2  m . Using the ideal gas relation,
the specific volume and the mass flow rate of air are determined to be
RT1
(8.314 kPa  m 3 /kg  K )( 473 K )
v1 

 0.4525 m 3 /kg
MWP1
28.8 * 300 kPa

m
1
1
A1 V1 
(0.008m 2 )(30m/s)  0.5304kg/s
3
v1
0.4525 m /kg
(b) energy balance
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
  pe  0)
m (h1  V12 / 2)  m (h2 + V22 /2) (since Q  W
0  h2  h1 
V22  V12
V 2  V12

 0  C p ,ave T2  T1   2
2
2
(180 m/s ) 2  (30 m/s ) 2
0  (1.02 kJ/kg  K )(T2  200 C) 
2

T2 = 184.6C
Check Cp at 457.6 K ~ 1.021
 1 kJ/kg

 1000 m 2 /s 2





(c)
v2 
m 
RT 2 (8.314 kPa  m 3 /kg  K )(184.6  273 K )

 1.313 m 3 /kg
P2
28.8 *100 kPa
1
1
A2 V2 
 0.5304 kg/s 
A2 180 m/s 
v2
1.313 m 3 /kg
A2 = 0.00387 m2 = 38.7 cm2
5-65 Steam is accelerated in a nozzle from a velocity of 40 m/s to 300 m/s
Potential energy changes are negligible. no work. device is adiabatic
From the steam tables (Table A-6),
3
P1  3MPa 
 v1  0.09936 m /kg

T1  400  C
 h1  3230.9 kJ/kg
(a) mass balance m 1  m 2  m .
nozzle control volume since mass crosses the boundary
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
  pe  0)
m (h1  V12 / 2)  m (h2 + V22 /2) (since Q  W
0  h2  h1 
V22  V12
2
or,
h2  h1 
V22  V12
(300 m/s ) 2  (40 m/s ) 2
 3230.9 kJ/kg 
2
2
 1 kJ/kg

 1000 m 2 /s 2


  3186.7 kJ/kg



 T2  376.7 C

h2  3186.7 kJ/kg  v 2  0.1153 m 3 /kg
P2  2.5 MPa
(b)
A
v V
(0.09936 m 3 /kg )(300 m/s )
1
1
A2 V2  A1 V1 
 1  1 2 
 6.46
v2
v1
A2 v 2 V1
(0.1153 m 3 /kg )(40 m/s )
5-72 Nitrogen is decelerated in a diffuser from 200 m/s to a lower velocity. The exit
velocity of nitrogen and the ratio of the inlet-to-exit area are to be determined.
Nitrogen is an ideal gas with variable specific heats
adiabatic and no work interactions.
The molar mass of nitrogen is M = 28 kg/kmol (Table A-1). The Cp
T1  7C = 280 K  C p  1.039kJ / kgK
T2  22C = 295 K  C p  1.039 kJ/kgK
(a) m 1  m 2  m .
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
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
  pe  0)
m (h1  V12 / 2)  m (h2 + V22 /2) (since Q  W
0  C p (T2  T1 ) 
V22  V12
2
1 ,
N2
2
Substituting,
0
It
1.039(295 - 280)
V22  200m/s 
2
2

 1kJ/kg 


2 2 
 1000m /s 
V2 = 93.9 m/s
(b) The ratio of the inlet to exit area is determined from the conservation of mass
relation,
 RT / P  V
A
v V
1
1
A2 V2  A1 V1 
 1  1 2   1 1  2
v2
v1
A2 v 2 V1  RT 2 / P2  V1
or,
A1  T1 / P1  V2 280 K/60 kPa 93.9 m/s 



 0.625
A2  T2 / P2  V1 295 K/85 kPa 200 m/s 
5-76C The volume flow rate at the compressor inlet will be greater than that at the
compressor exit.
5-77C Yes. Because energy (in the form of shaft work) is being added to the air.
5-78C No.
5-83 Steam expands in a turbine. The exit temperature of the steam for a power output of
2 MW is to be determined.
Kinetic and potential energy changes are negligible.
adiabatic and so heat transfer is negligible.
From the steam tables (Tables A-4 through 6)
P1  10 MPa 

 h1  3373.7 kJ/kg

T1  500 C 

1  m
2  m
.
m
E  E out
in



Rate of net energy transfer
by heat, work, and mass
E system 0 (steady)



1
0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
H2O
(since Q  ke  pe  0)
 1  W out  mh
 2
mh
W out  m (h1  h2 )
Substituting,
2
2000 kJ/s  3kg/s 3373.7  h2 kJ/kg
h2  2707 kJ/kg
Then the exit temperature becomes
P2  20 kPa


T2  110.8 C
h2  2707 kJ/kg 
5-84 Argon gas expands in a turbine. The exit temperature of the argon for a power
output of 250 kW is to be determined.
adiabatic and so heat transfer is negligible
Argon is an ideal gas with constant specific heats.
Ar is Cp = 0.5203 kJ/kg·C (Tables A-2a)
There is only one inlet and one exit, and so m 1  m 2  m . The inlet specific volume of
argon and its mass flow rate are
v1 


RT1
8.314kPa  m3 /k - mol  K 723K 

 0.167 m3 /kg
MWP1
40 * 900kPa
So,
m 


1
1
A1V1 
0.006 m 2 80 m/s   2.874 kg /s
v1
0.167 m 3 /kg
E  E out
in



Rate of net energy transfer
by heat, work, and mass
E system 0 (steady)



0
ARGON
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
  pe  0)
m (h1  V12 / 2)  W out  m (h2 + V22 /2) (since Q

V 2  V12
W out  m  h2  h1  2

2

Substituting,
A1 = 60 cm2
P1 = 900 kPa
T1 = 450C
V1 = 80 m/s




250 kW
P2 = 150 kPa
V2 = 150 m/s

(150 m/s) 2  (80 m/s) 2


250 kJ/s  (2.874 kg/s ) (0.5203 kJ/kg C)(T2  450 C) 
2

It
 1 kJ/kg

 1000 m 2 /s 2





T2 = 267.3C
5-91 CO2 is compressed by a compressor. The volume flow rate of CO2 at the compressor
inlet and the power input to the compressor are to be determined.
Kinetic and potential energy changes are negligible
Helium is an ideal gas with constant specific heats.
adiabatic and so heat transfer is negligible.
molar mass is M = 44 kg/kmol (Table A-1).
Tavg 
T1  T2
 375K  C pavg  0.906kJ/kgK
2
(a) There is only one inlet and one exit, and so m 1  m 2  m . The inlet specific volume of
air and its volume flow rate are
v1 


RT1
8.314kPa  m 3 /k - mol  K 300K 

 0.5667m 3 /kg
MWP1
44 *100kPa
 1  (0.5 kg / s)(0.5667 m3 / kg)  0.283 m3 / s
V  mv
(b) We take the compressor control volume since mass crosses the boundary. The energy
balance for this steady-flow system can be expressed in the rate form as
2
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
W in  m h1  m h2 (since Q  ke  pe  0)
W in  m * C p (T2  T1 )
0.5kg/s 0.906 *150  67.95kW
W in 
CO2
1
5-92C Because usually there is a large temperature drop associated with the throttling
process.
5-93C Yes.
5-94C No. Because air is an ideal gas and h = h(T) for ideal gases. So if h remains
constant, so does the temperature.
5-98 Steam is throttled by a well-insulated valve. The temperature drop of the steam after
the expansion is to be determined.
Kinetic and potential energy changes are negligible.
Heat transfer to or from the fluid is negligible.
There are no work interactions involved.
The inlet enthalpy of steam is (Tables A-6),
P1 = 8 MPa
T1 = 500C
P1  8MPa 

h  3398.3 kJ/kg
  1
T1  500 C 

There is only one inlet and one exit, and so m 1  m 2  m .
H2O
0
E in  E out  E system
(steady)
0
 1  mh
 2  h1  h2
 E in  E out  mh
since Q  W  ke  pe  0 . Then the exit temperature of steam becomes
P2 = 6 MPa
P2  6MPa 

 T2  490.1 C
h2  h1  
5-100E High-pressure air is throttled to atmospheric pressure. The temperature of air
after the expansion is to be determined.
Kinetic and potential energy changes are negligible.
Heat transfer to or from the fluid is negligible. no work interactions involved.
Air is an ideal gas.
There is only one inlet and one exit, and so m 1  m 2  m .
0
E in  E out  E system
(steady)
0
 1  mh
 2  h1  h2
 E in  E out  mh
P1 = 200 psia
T1 = 90F
since Q  W  ke  pe  0 . For an ideal gas, h = h(T).
Therefore,
T2 = T1 = 90°F
5-101C Yes, if the mixing chamber is losing heat to the surrounding medium.
Air
P2 = 14.7 psia
5-102C Under the conditions of no heat and work interactions between
the mixing
chamber and the surrounding medium.
5-106 Feedwater is heated in a chamber by mixing it with superheated steam. If the
mixture is saturated liquid, the ratio of the mass flow rates of the feedwater and the
superheated vapor is to be determined.
Kinetic and potential energy changes are negligible.
There are no work interactions
adiabatic and so heat transfer is negligible.
T < Tsat @ 800 kPa = 170.43C, the cold water stream and the mixture exist as a compressed
liquid, which can be approximated as a saturated liquid at the given temperature. So,
h1  hf @ 50C = 209.33 kJ/kg
h3  hf @ 800 kPa = 721.11 kJ/kg
and
P2  800 kPa 

 h2  2839.3 kJ/kg
T2  200  C 

 in  m
 out  m
 system 0
m
Mass balance:
(steady)
0

 in  m
 out
m
1  m
2  m
3
 m
Energy balance:
E  E out
in



Rate of net energy transfer
by heat, work, and mass
E system 0 (steady)



0
T1 = 50C
·1
m
Rate of change in internal, kinetic,
potential, etc. energies
H2O
(P = 800 kPa)
Sat. liquid
E in  E out
 1h1  m
 2 h2  m
 3h3
m
Combining the two,
Dividing by m 2 yields
(since Q  W  ke  pe  0)
 1h1  m
 2h2  m
1  m
 2 h3
m
y h1  h2   y  1h3
y
Solving for y:
T2 = 200C
m·2
h3  h2
h1  h3
where y  m 1 / m 2 is the desired mass flow rate ratio. Substituting,
y
721.11  2839.3
 4.14
209.33  721.11
5-118 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.
The heat exchanger is well-insulated so that heat loss to the surroundings is negligible
and so heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.
Changes in the kinetic and potential energies of fluid streams are negligible.
The specific heats of water and ethylene glycol are given to be 4.18 and 2.56 kJ/kg.C,
respectively.
(a) ethylene glycol tubes control volume.
E  E out
in

0 (steady)
E system




Rate of net energy transfer
by heat, work, and mass
0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
 1  Q out  mh
 2 (since ke  pe  0)
mh

 p (T1  T2 )
Qout  mC
Cold Water
20C
Hot Glycol
40C
80C
2 kg/s
Then the rate of heat transfer becomes
 p (Tin  Tout )]glycol  (2 kg / s)(2.56 kJ / kg.  C)(80 C  40 C) = 204.8 kW
Q  [mC
(b) The rate of heat transfer from water must be equal to the rate of heat transfer to the
glycol. Then,
Q  [m C p (Tout  Tin )] water 
 m water 

Q
C p (Tout  Tin )
204 .8 kJ/s
= 1.4 kg/s
(4.18 kJ/kg. C)(55 C  20 C)
5-122 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.
The heat exchanger is well-insulated so that heat loss to the surroundings is negligible
and so heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.
Changes in the kinetic and potential energies of fluid streams are negligible
The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg.C,
respectively.
exhaust pipes as control volume.
E  E out
in

0 (steady)

Rate of net energy transfer
by heat, work, and mass
E system



0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
 1  Q out  mh
 2 (since ke  pe  0)
mh

 p (T1  T2 )
Qout  mC
Air
95 kPa
20C
0.8 m3/s
Then the rate of heat transfer from the exhaust gases becomes
 p (Tin  Tout )]gas.  (11
Q  [mC
. kg / s)(1.1 kJ / kg.  C)(180 C  95 C) = 102.85 kW
The mass flow rate of air is

m
Exhaust gases
1.1 kg/s, 95C
PV
(95 kPa)(0.8 m3 / s)

 0.904 kg / s
RT (0.287 kPa.m3 / kg.K)  293 K
heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature
of the air becomes
Q
 p (Tc,out  Tc,in ) 
Q  mC
 Tc,out  Tc,in 
 p
mC
 20 C 
102.85 kW
 133.2 C
(0.904 kg / s)(1.005 kJ / kg.  C)
5-125 A desktop computer is to be cooled safely by a fan in hot environments and high
elevations. The air flow rate of the fan and the diameter of the casing are to be
determined.
Air is an ideal gas
Kinetic and potential energy changes are negligible.
The specific heat of air at the average temperature of Tave = (45+60)/2 = 52.5C = 325.5
K is Cp = 1.0065 kJ/kg.C (Table A-2b)
The fan selected must be able to meet the cooling requirements of the computer at worst conditions.
Therefore, we assume air to enter the computer at 66.63 kPa and 45C, and leave at 60C.
E  E out
in


Rate of net energy transfer
by heat, work, and mass
0 (steady)
E system



0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
 1  mh
 2 (since ke  pe  0)
Q in  mh
 p (T2  T1 )
Q in  mC
Then the required mass flow rate of air to absorb heat at a rate of 60 W is determined to be
Q
60 W
Q  m C p (Tout  Tin )  m 

C p (Tout  Tin ) (1006.5 J/kg.C)(60 - 45) C
 0.00397 kg/s = 0.238 kg/min
The density of air entering the fan at the exit and its volume flow rate are
P
66 .63 kPa

 0.6972 kg/m 3
RT (0.287 kPa.m 3 /kg.K)(60 + 273)K
0.238 kg/min
m
V  
 0.341 m 3 /min
 0.6972 kg/m 3

For an average exit velocity of 110 m/min, the diameter of the casing of the fan is
determined from
(4)(0.341 m 3 /min)
D 2
4V
V  Ac V 
VD

 0.063 m = 6.3 cm
4
V
 (110 m/min)
5-128 A sealed electronic box is to be cooled by tap water flowing through channels on
two of its sides. The mass flow rate of water and the amount of water used per year are to
be determined.
Entire heat generated is dissipated by water.
Water is an incompressible substance with constant specific heats at room temperature.
Kinetic and potential energy changes are negligible.
The specific heat of water at room temperature is Cp = 4.18 kJ/kg.C (Table A-3).
E  E out
in

0 (steady)
E system




Rate of net energy transfer
by heat, work, and mass
0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
Water
1nlet
 1  mh
 2 (since ke  pe  0)
Q in  mh
 p (T2  T1 )
Q in  mC
1
Then the mass flow rate of tap water flowing through the electronic box becomes
 p T 

Q  mC
 m
Q
2 kJ / s

 0.1196 kg / s
C p T (4.18 kJ / kg.  C)(4 C)
Electronic
box
2 kW
Therefore, 0.11962 kg of water is needed per second to cool this electronic box. Then the amount of
cooling water used per year becomes
Water
exit
 t  (0.1196 kg/s)(365 days/yr 24 h/day 3600 s/h)
mm
= 3,772,000 kg/yr = 3,772 tons/yr
2
5-143 The ducts of a heating system pass through an unheated area. The rate of heat loss
from the air in the ducts is to be determined.
Air is an ideal gas
Kinetic and potential energy changes are negligible
There are no work interactions involved.
The constant pressure specific heat of air at room temperature is Cp = 1.005 kJ/kg·K
(Table A-2)
We take the heating duct as the system
1  m
2  m
.
There is only one inlet and one exit, and so m
E  E out
in


Rate of net energy transfer
by heat, work, and mass
0 (steady)
E system



0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
 1  Q out  mh
 2 (since W  ke  pe  0)
mh
 p (T1  T2 )
Q out  m (h1  h2 )  mC
120 kg/min AIR
·
Q
Substituting,
Q out  (120 kg / min)(1.005 kJ / kg C)(4  C)  482 kJ / min
5-144E The ducts of an air-conditioning system pass through an unconditioned area. The
inlet velocity and the exit temperature of air are to be determined.
Air is an ideal gas with constant specific heats at room temperature.
Kinetic and potential energy changes are negligible
There are no work interactions involved.
The constant pressure specific heat of air at room temperature is Cp = 0.240 Btu/lbm.R
(Table A-2E)
(a) The inlet velocity of air through the duct is
V1 
V1
V
450 ft 3 /min
 12 
 825 ft/min
A1  r
 (5/12 ft ) 2
Then the mass flow rate of air becomes
450 ft3/min
RT1 1.986psia  ft 3 /lbm  R 510R 
3
v1 

 12.6ft /lbm
P1
28.815psia 
V1 450ft 3 /min
m 

 35.7lbm/min  0.595lbm /s
v1 12.6ft 3 /lbm


AIR
D = 10 in
2 Btu/s
(b) We take the air-conditioning duct control volume since mass crosses the boundary. There is only one
1  m
2  m
 . The energy balance for this steady-flow system can be expressed in
inlet and one exit, and so m
the rate form as
E  E out
in


Rate of net energy transfer
by heat, work, and mass
0 (steady)
E system



0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
 1  mh
 2 (since W  ke  pe  0)
Q in  mh
 p (T2  T1 )
Q in  m (h2  h1 )  mC
Then the exit temperature of air becomes
T2  T1 
Q in
2 Btu/s
 50  F 
 64.0 F
m C p
(0.595 lbm/s )(0.24 Btu/lbm  F)
5-145 Water is heated by a 7-kW resistance heater as it flows through an insulated tube.
The mass flow rate of water is to be determined.
Water is an incompressible substance with constant specific heats at room temperature.
Kinetic and potential energy changes are negligible.
The tube is adiabatic and so heat losses are negligible.
The specific heat of water at room temperature is C = 4.18 kJ/kg·°C (Table A-3).
We take the water pipe as control volume since mass crosses the boundary. There is only one inlet and one
1  m
2  m
.
exit, and so m
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
W e,in  m h1  m h2 (since Q out  ke  pe  0)
WATER
20C
75C
W e,in  m (h2  h1 )  m [C (T2  T1 )  vP 0 ]  m C T2  T1 
7 kW
Substituting, the mass flow rates of water is determined to be
m 
W e,in
C (T2  T1 )

7 kJ/s
(4.184 kJ/kg C)(75  20)  C
 0.0304kg/s
5-146 Steam pipes pass through an unheated area, and the temperature of steam drops as
a result of heat losses. The mass flow rate of steam and the rate of heat loss from are to be
determined. 
Kinetic and potential energy changes are negligible
There are no work interactions involved.
From the steam tables (Table A-6),
P1  1MPa  v1  0.2327 m 3 /kg

T1  250  C h1  2942.6 kJ/kg
and
1 MPa
250C
·
P2  800 kPa 
 h2  2839.3 kJ/kg
T2  200  C 
Q
(a) The mass flow rate of steam is determined directly from
m 


1
1
A1 V1 
 0.06 m 2 2m/s   0.0972 kg/s
v1
0.2327 m 3 /kg
(b) We take the steam pipe as the system
1  m
2  m
.
There is only one inlet and one exit, and so m
E  E out
in


Rate of net energy transfer
by heat, work, and mass
0 (steady)
E system



out
0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
 1  Q out  mh
 2
mh

Q  m (h  h )
1
800 kPa
200C
STEAM
(since W  ke  pe  0)
2
Substituting, the rate of heat loss is determined to be
Q loss  (0.0972 kg/s )(2942.6 - 2839.3) kJ/kg  10.04 kJ/s