4-30E The heat transfer during a process that a closed system undergoes without any
internal energy change is to be determined.
Assumptions 1 The system is stationary and thus the kinetic and potential energy changes
are zero. 2 The compression or expansion process is quasi-equilibrium.
Analysis The energy balance for this stationary closed system can be expressed as
E E
inout


Net energy transfer
by heat, work, and mass
E system



Change in internal,kinetic,
potential,etc. energies
Qin  Wout  U  0
(since KE = PE = 0)
Qin  Wout
Then,
1 Btu


Qin  1.6  10 6 lbf  ft 
  2056 Btu
 778.17 lbf  ft 
4-35 An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of
water. An electric heater in the tank is turned on, and the entire liquid in the tank is
vaporized. The length of time the heater was kept on is to be determined, and the process
is to be shown on a P-v diagram.
Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes
are zero. 2 The device is well-insulated and thus heat transfer is negligible. 3 The energy
stored in the resistance wires, and the heat transferred to the tank itself is negligible.
Analysis We take the contents of the tank as the system. This is a closed system since no
mass enters or leaves. Noting that the volume of the system is constant and thus there is
no boundary work, the energy balance for this stationary closed system can be expressed
as
E  Eout
in


Net energy transfer
by heat, work, and mass

Esystem



H2O
V=
const.
Change in internal,kinetic,
potential,etc. energies
We,in  U  m(u2  u1 )
(since Q  KE = PE = 0)
VIt  m(u2  u1 )
W
The properties of water are (Tables A-4 through A-6)
e
P1  100 kPa  v f  0.001043 , v g  1.6941 m3/kg

x1  0.25  u f  417 .40 , u fg  2088.2 kJ/kg
T
v1  v f  x1v fg  0.001043  0.25  1.6941  0.001043
  0.42431
u1  u f  x1u fg  417.40  0.25  2088.2   939.4 kJ/kg
v 2  v1  0.42431 m3/kg 
sat.vapor
Substituting,
 u2  u g @ 0.42431m 3 /kg  2556.2 kJ/kg

m3/kg
2
1
v
 1000 VA
(110 V)(8 A)t  (5 kg)(2556.2  939.4)kJ/k g
 1 kJ/s
t  9186 s  153.1 min



4-69 Nitrogen in a rigid vessel is cooled by rejecting heat. The internal energy change of
the nitrogen is to be determined.
Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure
relative to its critical point values of 126.2 K and 3.39 MPa. 2 The kinetic and potential
energy changes are negligible, ke  pe  0 . 3 Constant specific heats at room
temperature can be used for nitrogen.
Analysis We take the nitrogen as the system. This is a closed system since no mass
crosses the boundaries of the system. The energy balance for this system can be
expressed as
E E
inout


Net energy transfer
by heat, work, and mass
E system



Change in internal,kinetic,
potential,etc. energies
 Qout  U  mc v (T2  T1 )
Thus,
u  q out  100 kJ/kg
Nitroge
n
Q
4-71 Oxygen is heated to experience a specified temperature change. The heat transfer is
to be determined for two cases.
Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure
relative to its critical point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential
energy changes are negligible, ke  pe  0 . 3 Constant specific heats can be used for
oxygen.
Properties The specific heats of oxygen at the average temperature of
(25+300)/2=162.5C=436 K are cp = 0.952 kJ/kgK and cv = 0.692 kJ/kgK (Table A-2b).
Analysis We take the oxygen as the system. This is a closed system since no mass crosses
the boundaries of the system. The energy balance for a constant-volume process can be
expressed as
E E
inout

Net energy transfer
by heat, work, and mass

E system



Change in internal,kinetic,
potential,etc. energies
Qin  U  mc v (T2  T1 )
The energy balance during a constant-pressure
process (such as in a piston-cylinder device) can be
expressed as
O2
T1 = 25°C
T2 =
300°C
Q
E E
inout


Net energy transfer
by heat, work, and mass
E system



Change in internal,kinetic,
potential,etc. energies
Qin  Wb,out  U
O2
T1 = 25°C
T2 =
300°C
Qin  Wb,out  U
Qin  H  mc p (T2  T1 )
since U + Wb = H during a constant pressure
quasi-equilibrium process. Substituting for both
cases,
Q
Qin,V const  mc v (T2  T1 )  (1 kg) (0.692 kJ/kg  K)(300  25)K  190.3 kJ
Qin, Pconst  mc p (T2  T1 )  (1 kg)(0.952 kJ/kg K)(300 25)K  261.8kJ
4-94 An egg is dropped into boiling water. The amount of heat transfer to the egg by the
time it is cooked is to be determined.
Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 The thermal
properties of the egg are constant. 3 Energy absorption or release associated with any
chemical and/or phase changes within the egg is negligible. 4 There are no changes in
kinetic and potential energies.
Properties The density and specific heat of the egg are given to be  = 1020 kg/m3 and cp
= 3.32 kJ/kg.C.
Analysis We take the egg as the system. This is a closes system since no mass enters or
leaves the egg. The energy balance for this closed system can be expressed as
E  Eout
in



Net energy transfer
by heat, work, and mass
Esystem



Change in internal,kinetic,
potential,etc. energies
Boiling
Water
Qin  U egg  m(u2  u1 )  mc (T2  T1 )
Then the mass of the egg and the amount of heat transfer become
m  V  
D3
 (0.055 m) 3
 (1020 kg/m )
 0.0889 kg
6
6
Qin  mc p (T2  T1 )  (0.0889 kg )(3.32 kJ/kg. C)(80  8)C  21.2 kJ
3
Egg
8C
5-31 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.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is
an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 The
device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions.
Properties The gas constant of air is 0.287
kPa.m3/kg.K (Table A-1). The specific heat
P1 = 300
of air at the anticipated average temperature
P2 = 100
AIR
kPa
of 450 K is cp = 1.02 kJ/kg.C (Table A-2).
kPa
T1 =
V2 = 180
200C
m/s
V1 = 30
m/s
A1 = 80
cm2
Analysis (a) There is only one inlet and one
exit, and thus m 1  m 2  m . Using the ideal
gas relation, the specific volume and the
mass flow rate of air are determined to be
v1 
RT1 (0.287 kPa  m 3 /kg  K)( 473 K)

 0.4525 m 3 /kg
P1
300 kPa
m 
1
v1
A1V1 
1
(0.008 m2 )(30 m/s )  0.5304 kg/s
0.4525 m3/kg
(b) We take nozzle as the system, 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 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 
Substituting,
V22  V12
V 2  V12

 0  c p,ave T2  T1   2
2
2
0  (1.02 kJ/kg  K)(T2  200  C) 
(180 m/s ) 2  (30 m/s ) 2
2
 1 kJ/kg

 1000 m 2 /s 2





It yields
T2 = 184.6C
(c) The specific volume of air at the nozzle exit is
v2 
RT2 (0.287 kPa  m 3 /kg  K)(184.6  273 K)

 1.313 m 3 /kg
P2
100 kPa
m 
1
v2
A2V 2 
 0.5304 kg/s 
1
1.313 m 3 /kg
A2 180 m/s 
→ A2 = 0.00387 m2 = 38.7
cm2
5-34 Air is decelerated in an adiabatic diffuser. The velocity at the exit is to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is
an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4
There are no work interactions. 5 The diffuser is adiabatic.
Properties The specific heat of air at the average temperature of (20+90)/2=55°C =328 K
is cp = 1.007 kJ/kgK (Table A-2b).
Analysis There is only one inlet and one exit, and thus m1  m 2  m . We take diffuser as
the system, 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
inout

E system0 (steady)



Rate of net energy transfer
by heat, work, and mass
0
Rate of change in internal,kinetic,
potential,etc. energies
100 kPa
20C
500 m/s
E in  E out
m (h1  V12 / 2)
h1  V12 / 2


Solving for exit velocity,

V 2  V12  2(h1  h2 )
m (h2 + V22 /2)
h2 + V22 /2

0.5

 V12  2c p (T1  T2 )
AIR
200 kPa
90C

0.5

 1000 m 2 /s 2
 (500 m/s) 2  2(1.007 kJ/kg  K)(20  90 )K 
 1 kJ/kg


 330.2 m/s




0.5
5-38E Air is decelerated in a diffuser from 600 ft/s to a low velocity. The exit
temperature and the exit velocity of air are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is
an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The
device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions.
Properties The enthalpy of air at the inlet temperature of 20F is h1 = 114.69 Btu/lbm
(Table A-17E).
Analysis (a) There is only one inlet and one exit, and thus m 1  m 2  m . We take diffuser
as the system, 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 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
m (h1  V12
/ 2) 
0
1
m (h2 + V22 /2) (since Q
V 2  V12
h2  h1  2
AIR
  pe  0)
W
,
2
or,
h2  h1 
V22  V12
0  600 ft/s 2  1 Btu/lbm 


 114.69 Btu/lbm 
 25,037 ft 2 /s2   121.88 Btu/lbm
2
2


From Table A-17E,
T2 = 510.0 R
(b) The exit velocity of air is determined from the conservation of mass relation,
1
v2
A2V2 
1
v1
A1V1 

1
1
A2V2 
A1V1
RT2 / P2
RT1 / P1
Thus,
V2 
A1T2 P1
1 (510 R )(13 psia)
V1 
(600 ft/s)  114.3 ft/s
A2T1 P2
5 (480 R )(14.5 psia)
2
5-44 Heat is lost from the steam flowing in a nozzle. The velocity and the volume flow
rate at the nozzle exit are to be determined.
Assumptions 1 This is a steady-flow process
since there is no change with time. 2 Potential
energy change is negligible. 3 There are no
work interactions.
400C STEAM
300C
800 kPa
200 kPa
Analysis We take the steam as the system,
10
m/s
which is a control volume since mass crosses
Q
the boundary. The energy balance for this
steady-flow system can be expressed in the
rate form as
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


V2 
V2 
m  h1  1   m  h2  2   Q out


2 
2 


h1 
or
since W  pe  0)
V12
V 2 Q
 h2  2  out
2
2
m
The properties of steam at the inlet and exit are (Table A-6)
P1  800 kPa v1  0.38429 m3/kg

T1  400 C  h1  3267 .7 kJ/kg
P2  20 0 kPa v 2  1.31623 m3/kg

T1  300 C  h2  3072 .1 kJ/kg
The mass flow rate of the steam is
m 
1
v1
A1V1 
1
(0.08 m2 )(10 m/s)  2.082 kg/s
0.38429 m3/s
Substituting,
3267.7 kJ/kg 
(10 m/s) 2  1 kJ/kg 
V22  1 kJ/kg 
25 kJ/s

3072
.
1
kJ/kg





2 2
2
2  1000 m 2 /s2  2.082 kg/s
 1000 m /s 

V2  606 m/s
The volume flow rate at the exit of the nozzle is
V2  m v 2  (2.082kg/s)(1.31623 m3/kg)  2.74 m3/s
5-80 Two streams of refrigerant-134a are mixed in a chamber. If the cold stream enters at
twice the rate of the hot stream, the temperature and quality (if saturated) of the exit
stream are 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 R-134a tables (Tables A-11 through A-13),
h1  hf @ 12C = 68.18 kJ/kg
h2 = h @ 1 MPa, 60C = 293.38 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
Mass balance:
m in  m out  m system0 (steady)  0
m in  m out
m 1  m 2  m 3  3m 2 since m 1  2m 2
T·1 = ·
12C
m1 = 2m2
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
T2 = 60C
R-134a
(P = 1
MPa)
T3, x3
E in  E out
m 1h1  m 2 h2  m 3h3 (since Q  W  ke  pe  0)
Combining the two gives
 2 h1  m
 2 h2  3m
 2 h3 or h3  2h1  h2  / 3
2m
Substituting,
h3 = (268.18 + 293.38)/3 = 143.25 kJ/kg
At 1 MPa, hf = 107.32 kJ/kg and hg = 270.99 kJ/kg. Thus the exit stream is a saturated
mixture since
hf < h3 < hg. Therefore,
T3 = Tsat @ 1 MPa = 39.37C
and
x3 
h3  h f
h fg

143 .25  107 .32
 0.220
163 .67