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1
HW 10 Solution
1.
7108
Saturated water vapor is compressed in a reversible steadyflow device. The work required is to be determined.
Assumptions
1
This is a steadyflow process since there is no change with time.
2
There is no heat transfer associated with
the process.
3
Kinetic and potential energy changes are negligible.
Analysis
The properties of water at the inlet state are
4)

A
(Table
/kg
m
39248
.
0
kPa
16
.
476
1
C
150
3
1
1
1
1
=
=
⎭
⎬
⎫
=
°
=
v
P
x
T
Noting that the specific volume remains constant, the reversible steadyflow work
expression gives
kJ/kg
205.6
=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
=
−
=
=
∫
3
3
1
2
1
2
1
in
m
kPa
1
kJ
1
476.16)kPa

/kg)(1000
m
(0.39248
)
(
P
P
dP
w
Compressor
1 MPa
Water
150°C
sat. vap.
2.
7112
A steam power plant operates between the pressure limits of 5 MPa and 10 kPa. The ratio of the turbine work to the
pump work is to be determined.
Assumptions
1
Liquid water is an incompressible substance.
2
Kinetic and potential energy changes are negligible.
3
The
process is reversible.
4
The pump and the turbine are adiabatic.
Analysis
Both the compression and expansion processes are reversible and adiabatic, and thus isentropic,
s
1
=
s
2
and
s
3
=
s
4
.
Then the properties of the steam are
kJ/kg
4608.1
MPa
5
K
kJ/kg
.1488
8
kJ/kg
2583.9
.
kPa
10
3
4
3
3
kPa
10
@
4
kPa
10
@
4
4
=
⎭
⎬
⎫
=
=
⋅
=
=
=
=
⎭
⎬
⎫
=
h
s
s
P
s
s
h
h
vapor
sat
P
g
g
Also,
1
=
f @
10 kPa
= 0.00101 m
3
/kg.
The work output to this isentropic turbine is determined from
the steadyflow energy balance to be
)
(
0
4
3
out
out
4
3
out
in
energies
etc.
potential,
kinetic,
internal,
in
change
of
Rate
(steady)
0
system
mass
and
work,
heat,
by
nsfer
energy tra
net
of
Rate
out
in
h
h
m
W
W
h
m
h
m
E
E
E
E
E
−
=
+
=
=
=
Δ
=
−
±
±
±
±
±
±
±
²
²³
²
²´
µ
±
µ
±
±
©
H
2
O
3
4
H
2
O
2
1
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Substituting,
kJ/kg
.2
2024
9
.
2583
1
.
4608
4
3
out
turb,
=
−
=
−
=
h
h
w
The pump work input is determined from the steadyflow work relation to be
()
kJ/kg
.041
5
m
kPa
1
kJ
1
kPa
)
0
1
5000
)(
/kg
m
0.00101
(
3
3
1
2
1
0
0
2
1
in
pump,
=
⎟
⎠
⎞
⎜
⎝
⎛
⋅
−
=
−
=
Δ
+
Δ
+
=
∫
P
P
pe
ke
dP
w
v
©
©
Thus,
402
=
=
5.041
2024.2
in
pump,
out
turb,
w
w
3.
7122E
Steam is compressed in an adiabatic closed system with an isentropic efficiency of 80%. The work produced and
the final temperature are to be determined.
Assumptions
1
Kinetic and potential energy changes are negligible.
2
The device is adiabatic and thus heat transfer is
negligible.
Analysis
We take the steam as the system. This is a closed system since no mass enters or leaves. The energy balance for
this stationary closed system can be expressed as
2
1
out
1
2
out
energies
etc.
potential,
kinetic,
internal,
in
Change
system
mass
and
work,
heat,
by
nsfer
energy tra
Net
out
in
)
(
u
u
w
u
u
m
U
W
E
E
E
−
=
−
=
Δ
=
−
Δ
=
−
±²
±³
´
´
From the steam tables (Tables A5 and A6),
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 Fall '07
 RAMUSSEN
 Heat Transfer

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