Here are some worked problems like those I will give you in Chapter 4 on the examination tomorrow.
Let me suggest you try the following Review problems as well 4-130, 4-136, 4-143, and 4-154.
4-63 1 – kg of Oxygen is heated from 20 to 120C. Determine the amount of heat transfer required when this is done
during a (a) constant-volume process and (b) isobaric process.
Givens 1 kg of Oxygen experiences a 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 (20+120)/2=70C=343 K are cp = 0.927
kJ/kgK and cv = 0.667 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



Changein internal, kinetic,
potential,etc. energies
O2
T1 = 20°C
T2 = 120°C
Qin  U  mcv (T2  T1 )
The energy balance during a constant-pressure process (such as in a
piston-cylinder device) can be expressed as
E E
inout


Net energy transfer
by heat, work, and mass
Q
E system



Changein internal, kinetic,
potential,etc. energies
Qin  Wb,out  U
O2
T1 = 20°C
T2 = 120°C
Qin  Wb,out  U
Qin  H  mc p (T2  T1 )
Q
since U + Wb = H during a constant pressure quasi-equilibrium
process. Substituting for both cases,
Qin,V const  mcv (T2  T1 )  (1 kg)(0.667 kJ/kg  K)(120  20)K  66.7 kJ
Qin, Pconst  mc p (T2  T1 )  (1 kg)(0.927 kJ/kg  K)(120  20)K  92.7 kJ
4-74 A spring-loaded piston-cylinder device contains 1 kg of carbon dioxide. This system is heated from 100 kPa and
25C to 1000kPa and 300C. Determine the total heat transfer to and work produced by this system.
Givens Carbon dioxide contained in a spring-loaded piston-cylinder device is heated. The work done and the heat
transfer are to be determined.
Assumptions 1 CO2 is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2
The kinetic and potential energy changes are negligible, ke  pe  0 .
Properties The properties of CO2 are R = 0.1889
kJ/kgK and cv = 0.657 kJ/kgK (Table A-2a).
Analysis We take CO2 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



Changein internal, kinetic,
potential,etc. energies
Qin  Wb,out  U  mcv (T2  T1 )
The initial and final specific volumes are
V1 
mRT1 (1 kg)(0.1889 kPa  m 3 /kg  K)(298 K)

 0.5629 m 3
P1
100 kPa
V2 
mRT 2 (1 kg)(0.1889 kPa  m 3 /kg  K)(573 K)

 0.1082 m 3
P2
1000 kPa
P
(kPa)
1000
100
2
1
V (m3)
Pressure changes linearly with volume and the work done is equal to the area under the process line 1-2:
P1  P2
(V 2  V1 )
2
 1 kJ
(100  1000)kPa

(0.1082  0.5629)m 3 
 1 kPa  m 3
2

 250.1 kJ
Wb,out  Area 




Thus,
Wb,in  250.1kJ
Using the energy balance equation,
Qin  Wb,out  mcv (T2  T1 )  250.1 kJ  (1 kg)(0.657 kJ/kg  K)(300  25)K  69.4 kJ
Thus,
Qout  69.4 kJ
4-77 A piston-cylinder device contains 1.5 kg of nitrogen initially at 10 kPa and 17C. The nitrogen is now
compressed slowly in a polytropic process during which PV1.3= constant until the volume is reduced by one-half.
Determine the work done and the heat transfer for this process.
Givens A cylinder initially contains nitrogen gas at a specified state. The gas is compressed polytropically until the
volume is reduced by one-half. The work done and the heat transfer are to be determined.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The N2 is an
ideal gas with constant specific heats. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Properties The gas constant of N2 is R = 0.2968 kPa.m3/kg.K (Table A-1). The cv value of N2 at the anticipated
average temperature of 350 K is 0.744 kJ/kg.K (Table A-2b).
Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the
system boundary. The energy balance for this closed system can be expressed as
E  Eout
in



Net energy transfer
by heat, work, and mass
Esystem



Changein internal, kinetic,
potential,etc. energies
Wb,in  Qout  U  m(u2  u1 )
Wb,in  Qout  mcv (T2  T1 )
The final pressure and temperature of nitrogen are
1.3
P2V 21.3

V 

 P2   1  P1  21.3 (100 kPa)  246.2 kPa
V 2 
P V
246.2 kPa

 T2  2 2 T1 
 0.5 (290 K)  357.0 K
P1 V1
100 kPa
P1V11.3
P1V1 P2V 2

T1
T2
N2
100 kPa
17C
PV1.3 = C
Q
Then the boundary work for this polytropic process can be determined from
P2V 2  P1V1
mR (T2  T1 )

1 n
1 n
(1.5 kg)(0.2968 kJ/kg  K)(357.0  290)K

 99.5 kJ
1  1.3
Substituting into the energy balance gives
Wb,in  
2
 PdV  
1
Qout  Wb,in  mcv (T2  T1 )
 99.5 kJ  (1.5 kg)(0.744 kJ/kg.K)(357.0  290)K
= 24.7 kJ
4-90 Carbon steel balls ( = 7833 kg/m3 and cp = 0.465 kJ/kgC) 8 mm in diameter are annealed by heating them
first 900C in a furnace, and then allowing them to cool slowly to 100C in ambient air at 35C. If 2500 ball are to
be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air.
Givens Carbon steel balls are to be annealed at a rate of 2500/h by heating them first and then allowing them to cool
slowly in ambient air at a specified rate. The total rate of heat transfer from the balls to the ambient air is to be
determined.
Assumptions 1 The thermal properties of the balls are constant. 2 There are no changes in kinetic and potential
energies. 3 The balls are at a uniform temperature at the end of the process
Properties The density and specific heat of the balls are given to be  = 7833 kg/m3 and cp = 0.465 kJ/kg.C.
Analysis We take a single ball as the system. The
energy balance for this closed system can be
expressed as
Furnace
E  Eout

Esystem
Air, 35C
in





Net energy transfer
Changein internal, kinetic,
Steel
balls,
900C
by heat, work, and mass
potential,etc. energies
 Qout  U ball  m(u2  u1 )
Qout  mc (T1  T2 )
(b) The amount of heat transfer from a single ball is
m  V  
D3
 (7833 kg/m3 )
 (0.008 m)3
 0.00210 kg
6
6
Qout  mc p (T1  T2 )  (0.0021 kg)( 0.465 kJ/kg.C)(900  100)C  0.781 kJ (per ball)
Then the total rate of heat transfer from the balls to the ambient air becomes
Q  n Q  (2500 balls/h)  (0.781 kJ/ball)  1,953 kJ/h  542 W
out
ball out
Here are some worked problems like those I will give you in Chapter 5 on the examination tomorrow.
Let me suggest you try the following Review problems as well 5-147, 5-152, 5-156, and 5-196.
5-30 Air enters an adiabatic nozzle steadily at 300 kPa, 200C, and 45 m/s and leaves at 100 kPa and 180 m/s. The
inlet area of the nozzle is 110 cm2. Determine (a) the mass flow rate through the nozzle, (b) the exit temperature of
the air and (c) the exit area of the nozzle.
Givens Air is accelerated in a nozzle from 45 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 of air at the anticipated average
P1 = 300 kPa
temperature of 450 K is cp = 1.02 kJ/kg.C (Table A-2).
P2 = 100 kPa
T1 = 200C
AIR
Analysis (a) There is only one inlet and one exit, and thus
V2 = 180 m/s
V1 = 45 m/s
1  m
2  m
 . Using the ideal gas relation, the specific volume
m
A1 = 110 cm2
and the mass flow rate of air are determined to be
RT1 (0.287 kPa  m 3 /kg  K)( 473 K)

 0.4525 m 3 /kg
P1
300 kPa
v1 
m 
1
v1
A1V1 
1
3
(0.0110 m 2 )( 45 m/s)  1.094 kg/s
0.4525 m /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

E system0 (steady)
0
in



Rate of net energy transfer
by heat, work, and mass
Rate of changein internal, kinetic,
potential,etc. energies
E in  E out
  pe  0)
m (h1  V12 / 2)  m (h2 + V 22 /2) (since Q  W
0  h2  h1 
V22  V12
V 2  V12

 0  c p ,ave T2  T1   2
2
2
Substituting,
0  (1.02 kJ/kg  K)(T2  200  C) 
(180 m/s) 2  (45 m/s) 2
2
 1 kJ/kg

 1000 m 2 /s 2





It yields
T2 = 185.2C
(c) The specific volume of air at the nozzle exit is
v2 
RT2 (0.287 kPa  m 3 /kg  K)(185.2  273 K)

 1.315 m 3 /kg
P2
100 kPa
m 
1
v2
A2V2 
 1.094 kg/s 
1
1.315 m 3 /kg
A2 180 m/s → A2 = 0.00799 m2 = 79.9 cm2
5-82 A heat exchanger is to cool ethylene glycol (cp = 2.56 kJ/kgC) flowing at a rate of 3.2 kg/s from 80C to 40C
by water (cp = 4.18 kJ/kgC) that enters at 20C and leaves at 70C. Determine (a) the rate of heat transfer and (b)
the mass flow rate of water.
Givens 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 water and
ethylene glycol are given to be 4.18 and 2.56
kJ/kg.C, respectively.
Analysis (a) We take the ethylene glycol
tubes as the system, which is a control
volume. The energy balance for this steadyflow 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
Cold Water
20C
Hot Glycol
40C
80C
3.2 kg/s
Rate of changein internal, kinetic,
potential,etc. energies
E in  E out
m h  Q  m h
1
out
2 (since ke  pe  0)

Qout  m c p (T1  T2 )
Then the rate of heat transfer becomes
 c (T  T )]
Q  [m
 (3.2 kg/s)(2.56 kJ/kg.C)(80C  40C) = 327.7kW
p
in
out
glycol
(b) The rate of heat transfer from glycol must be equal to the rate of heat transfer to the water. Then,
Q
327.7 kJ/s
Q  [m c p (Tout  Tin )] water 
 m water 

= 1.57 kg/s
c p (Tout  Tin ) (4.18 kJ/kg.C)(70C  20C)
5-114 Refrigerant-134a enters the condenser of a refrigerator at 900 kPa and 60C, and leaves as a saturated liquid at
the same pressure. Determine the heat transfer from the refrigerant per unit mass.
Givens R-134a is condensed in a condenser. The heat transfer per unit mass 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.
Analysis We take the pipe in which R-134a is condensed 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
inout


Rate of net energy transfer
by heat, work, and mass
E system0 (steady)


0
qout
Rate of changein internal, kinetic,
potential,etc. energies
E in  E out
m h1  m h2  Q out
Q out  m (h1  h2 )
900 kPa
60C
R134a
900 kPa
sat. liq.
q out  h1  h2
The enthalpies of R-134a at the inlet and exit of the condenser are (Table A-12, A-13).
P1  900 kPa 
 h1  295.13 kJ/kg
T1  60C

P2  900 kPa 
 h2  h f @ 900kPa  101.61 kJ/kg
x0

Substituting,
qout  295.13  101.61  193.5kJ/kg
5-135 An insulated 0.15m3 tank contains helium at 3MPa and 130C. A valve is now opened, allowing some helium
to escape. The valve is closed when one-half of the initial mass has escaped. Determine the final temperature and
pressure in the tank.
Givens An insulated rigid tank initially contains helium gas at high pressure. A valve is opened, and half of the mass
of helium is allowed to escape. The final temperature and pressure in the tank 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 by using constant average properties for the helium leaving the
tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The tank is
insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats.
Properties The specific heat ratio of helium is k =1.667 (Table A-2).
Analysis 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 non-flowing 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
min  mout  msystem  me  m1  m2
Mass balance:
m2  12 m1 (given) 
 me  m2  12 m1
He
0.15 m3
3 MPa
130C
Energy balance:
E  Eout
in


Net energy transfer
by heat, work, and mass

Esystem



Changein internal, kinetic,
potential,etc. energies
 me he  m2u2  m1u1 (since W  Q  ke  pe  0)
Note that the state and thus the enthalpy of helium leaving the tank is changing during this process. But for
simplicity, we assume constant properties for the exiting steam at the average values.
Combining the mass and energy balances:
0  21 m1he  21 m1u2  m1u1
Dividing by m1/2
0  he  u 2  2u1 or 0  c p
Dividing by cv:
0  k T1  T2   2T2  4T1
Solving for T2:
T2 
T1  T2
 cv T2  2cv T1
2
since k  c p / cv
4  k  T  4  1.667 403 K   257 K
2  k  1 2  1.667
The final pressure in the tank is
P1V
m RT
mT
1 257
3000 kPa  956 kPa
 1 1 
 P2  2 2 P1 
P2V m2 RT2
m1T1
2 403