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4.11
(a) This is a Joule-Thomson expansion
⇒=
()
==
°
≈=
=
±
?
±
.
±
.
HTH
T
H
T
70 bar,
10133
400
1
400
32782
bar,
C
bar,
C
kJ kg
°
and
T
=
,
°
447 C
±
.
S
=
6619 kJ kg K
(b) If turbine is adiabatic and reversible
, then
²
S
gen
=
0
ch
±±
.
SS
out
in
kJ kg K
6619
and
P
=
1013
.
bar.
This suggests that a two-phase mixture is leaving the turbine
Let
fraction vapor
kJ kg K
kJ kg K
V
L
x
S
S
=
=
=
±
.
±
.
73594
13026
Then
xx
7 3594
1
13026
..
.
+−
(
)
=
kJ kg K or
x
=
08778
.
.
Therefore the enthalpy of
fluid leaving turbine is
±
H
HH
=×
+
−
×=
08788
26755
1
08778
417 46
2399 6
VL
sat’d, 1 bar
sat’d, 1 bar
kJ
kg
.
²
±
²
±
MH
M H
in
in
out
out
Energy balance
0
=+
+
²
Q
0
²
WP
s
dV
dt
0
but
²²
MM
in
out
=−
⇒
−
=−=
²
²
.
W
M
s
in
kJ
kg
2399 6
8786
(c) Saturated vapor at 1 bar
±
.;
±
.
²
.
%
.
.
.
²
²
.
SH
W
M
S
M
s
−=
−
=
() =
×
=
7 3594
26755
26755
602 7
602 7
100
686%
7 3594
0740
kJ kg K
kJ kg
kJ kg
Efficiency
kJ Kh
in Actual
gen
in

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** preview**
has intentionally

(d)
0
0
0
12
2
1
11
2
2
=+⇒ =
−
=−
+
+
−
+
+
±± ±
±
±
²² ±±
±
²²
±
±
MM M
M
MH H
W Q P
dV
dt
MS S
Q
T
S
s
ch

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