Chem E 260
HW #7
(Due: Friday - 11/16)
7-63 A cylinder is initially filled with saturated water vapor mixture at a specified
temperature. Steam undergoes a reversible heat addition and an isentropic process. The
processes are to be sketched and heat transfer for the first process and work done during
the second process are to be determined.
Assumptions 1 The kinetic and potential energy changes are negligible. 2 The thermal
energy stored in the cylinder itself is negligible. 3 Both processes are reversible.
Analysis (b) From the steam tables (Tables A-4 through A-6),
T1  100 C
 h1  h f  xh fg  419 .17  (0.5)( 2256 .4)  1547 .4 kJ/kg
x  0.5

h  h g  2675.6 kJ/kg
T2  100 C 2
 u 2  u g  2506.0 kJ/kg
x2  1
 s  7.3542 kJ/kg  K
2
P3  15 kPa 
u 3  2247.9 kJ/kg
s3  s 2

H2 O
100C
x = 0.5
Q
We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves.
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
Qin  Wb,out  U  m(u2  u1 )
For process 1-2, it reduces to
Q12,in  m(h2  h1 )  (5 kg)(2675.6 - 1547.4)kJ/ kg  5641 kJ
(c) For process 2-3, it reduces to
W23,b,out  m(u2  u3 )  (5 kg)(2506.0 - 2247.9)kJ/ kg  1291 kJ
SteamIAPWS
700
600
T [°C]
500
400
101.42 kPa
300
200
15 kPa
2
1
100
0
0.0
3
1.1
2.2
3.3
4.4
5.5
6.6
7.7
8.8
9.9
11.0
s [kJ/kg-K]
7-67 A hot copper block is dropped into water in an insulated tank. The final equilibrium
temperature of the tank and the total entropy change are to be determined.
Assumptions 1 Both the water and the copper block are incompressible substances with
constant specific heats at room temperature. 2 The system is stationary and thus the
kinetic and potential energies are negligible. 3 The tank is well-insulated and thus there is
no heat transfer.
Properties The density and specific heat of water at 25C are  = 997 kg/m3 and cp = 4.18
kJ/kg.C. The specific heat of copper at 27C is cp = 0.386 kJ/kg.C (Table A-3).
Analysis We take the entire contents of the tank, water + copper block, as the system.
This is a closed system since no mass crosses the system boundary during the process.
The energy balance for this system can be expressed as
E  Eout
in



Net energy transfer
by heat, work, and mass
Esystem



Change in internal,kinetic,
potential,etc. energies
WATER
0  U
or,
U Cu  U water  0
[mc (T2  T1)]Cu  [mc (T2  T1)]water  0
Coppe
r
50 kg
120 L
where
mwater  V  (997 kg/m3 )(0.120 m3 )  119.6 kg
Using specific heat values for copper and liquid water at room temperature and
substituting,
(50 kg)(0.386 kJ/kg  C) (T2  80) C  (119.6 kg)(4.18 kJ/kg  C) (T2  25) C  0
T2 = 27.0C
The entropy generated during this process is determined from
T 
 300.0 K 
  3.140 kJ/K
Scopper  mcavg ln  2   50 kg 0.386 kJ/kg  K  ln 
 353 K 
 T1 
T 
 300.0 K 
  3.344 kJ/K
S water  mcavg ln  2   119.6 kg 4.18 kJ/kg  K  ln 
T
 298 K 
 1
Thus,
Stotal  Scopper  Swater  3.140  3.344  0.204 kJ/K
7-75C For an ideal gas, dh = cp dT and v = RT/P. From the second Tds relation,
ds 
dh vdP c p dP RT dP
dT
dP



 cp
R
T
T
T
P T
T
P
Integrating,
T 
P
s 2  s1  c p ln  2   R ln  2
 T1 
 P1



Since cp is assumed to be constant.
7-92 One side of a partitioned insulated rigid tank contains an ideal gas at a specified
temperature and pressure while the other side is evacuated. The partition is removed, and
the gas fills the entire tank. The total entropy change during this process is to be
determined.
Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relations
apply.
Analysis Taking the entire rigid tank as the system, the energy balance can be expressed
as
E  Eout
in



Net energy transfer
by heat, work, and mass
Esystem



Change in internal,kinetic,
potential,etc. energies
0  U  m(u2  u1 )
u2  u1
T2  T1
IDEA
L GAS
5 kmol
40C
since u = u(T) for an ideal gas. Then the entropy change of the gas becomes

T 0
V 
V
S  N  cv ,avg ln 2  Ru ln 2   NRu ln 2


T1
V1 
V1

 5 kmol 8.314 kJ/kmol  K  ln 2
 28.81 kJ/K
This also represents the total entropy change since the tank does not contain anything
else, and there are no interactions with the surroundings.
7-100 Oxygen is expanded in an adiabatic nozzle. The maximum velocity at the exit is to
be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 The
process is adiabatic, and thus there is no heat transfer. 3 Oxygen is an ideal gas with
constant specific heats.
Properties The properties of oxygen at room temperature are cp = 0.918 kJ/kg·K and k =
1.395 (Table A-2a).
Analysis For the maximum velocity at the exit, the process must be reversible as well as
adiabatic (i.e., isentropic). This being the case, the exit temperature will be
P
T2  T1  2
 P1



( k 1) / k
 120 kPa 
 (363 K)

 300 kPa 
0.395 / 1.395
 280.0 K
There is only one inlet and one exit, and thus m1  m 2  m . 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
inout

E system0 (steady)



Rate of net energy transfer
by heat, work, and mass
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out

V2
m  h1  1

2

h1 
0
300 kPa
90C
3 m/s
2 


  m  h2  V2 )


2 


Solving for the exit velocity,

 V
2
1

0.5
 2c p (T1  T2 )
120 kPa
T
V12
V2
 h2 + 2
2
2
V 2  V12  2(h1  h2 )
O2

300
kPa
1
120
kPa
2
s
0.5

 1000 m 2 /s 2
 (3 m/s) 2  2(0.918 kJ/kg  K)(363  280 )K 
 1 kJ/kg


 390 m/s




0.5
7-106 An ideal gas is compressed in an isentropic compressor. 10% of gas is compressed
to 400 kPa and 90% is compressed to 600 kPa. The compression process is to be
sketched, and the exit temperatures at the two exits, and the mass flow rate into the
compressor are to be determined.
Assumptions 1 The compressor operates steadily. 2 The process is reversible-adiabatic
(isentropic)
Properties The properties of ideal gas are given to be cp = 1.1 kJ/kg.K and cv = 0.8
kJ/kg.K.
Analysis (b) The specific heat ratio of the gas is
k
cp
cv

1.1
 1.375
0.8
P3 = 600 kPa
COMPRESSO
R
32 kW
P2 = 400 kPa
P1 = 100 kPa
T1 = 300 K
The exit temperatures are determined from ideal gas
isentropic relations to be,
P
T2  T1  2
 P1



P
T3  T1  3
 P1



( k 1) / k
( k 1) / k
 400 kPa
 27  273 K 
 100 kPa



 600 kPa
 27  273 K 
 100 kPa



0.375/1.375
 437.8 K
0.375/1.375
 489.0 K
(c) A mass balance on the control volume gives
1  m
2 m
3
m
where
m 2  0.1m 1
m 3  0.9m 1
T
We take the compressor 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


Rate of net energy transfer
by heat, work, and mass

E system0 (steady)

0
Solving for the inlet mass flow rate, we obtain
Win
c p 0.1(T2  T1 )  0.9(T3  T1 )
32 kW
(1.1 kJ/kg  K) 0.1(437.8 - 300)  0.9(489.0 - 300) 
 0.158 kg/s

P2
P1
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
m 1h1  Win  m 2 h2  m 3h3
m 1c pT1  Win  0.1m 1c pT2  0.9m 1c pT3
m 1 
P3
s