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The steady-state energy balance is given by Equation (2.20):
0=\sum_{\text {in }} \dot{m}_{\text {in }} \hat{h}_{\text {in }}-\sum_{\text {out }} \dot{m}_{\text {out }} \hat{h}_{\text {out }}+\dot{Q}+\dot{W}_{s}0=∑
in
m
˙
in
h
^
in
−∑
out
m
˙
out
h
^
out
+
Q
˙
+
W
˙
s
(2.20)
0=\sum_{\text {in }} \dot{m}_{ in } \hat{h}_{ in }-\sum_{\text {out }} \dot{m}_{ out } \hat{h}_{ out }+\dot{Q}+\dot{W}_{s}0=∑
in
m
˙
in
h
^
in
−∑
out
m
˙
out
h
^
out
+
Q
˙
+
W
˙
s
(E2.5A)
For a turbine, there is one stream in and one stream out. Moreover, heat dissipation is negligible, since \dot{Q}<<\dot{W}_{s}
Q
˙
<<
W
˙
s
. If we label the inlet stream “1” and the outlet stream “2,” Equation (E2.5A) becomes:
0=\dot{m}_{1} \hat{h}_{1}-\dot{m}_{2} \hat{h}_{2}+\dot{W}_{s}0=
m
˙
1
h
^
1
−
m
˙
2
h
^
2
+
W
˙
s
(E2.5B)
At steady-state, the mass balance can be written as:
\dot{m}_{1}=\dot{m}_{2}=\dot{m}
m
˙
1
=
m
˙
2
=
m
˙
(E2.5C)
Plugging in Equation (E2.5C) into (E2.5B) gives:
\dot{W}_{s}=\dot{m}\left(\hat{h}_{2}-\hat{h}_{1}\right)
W
˙
s
=
m
˙
(
h
^
2
−
h
^
1
) (E2.5D)
We can look up values for specifi c enthalpy from the steam tables. For state 2, we use saturated steam at 1 bar (= 100 kPa) :
\hat{h}_{2}=2675.5[ kJ / kg ]
h
^
2
=2675.5[kJ/kg]
while state 1 is superheated steam at 500ºC and 100 bar (= 10 Mpa) :
\hat{h}_{1}=3373.6[ k J / kg ]
h
^
1
=3373.6[kJ/kg]
Plugging these numerical values into Equation (E2.5D) gives:
\dot{W}_{s}=10[ kg / s ](2675.5-3373.6)[ kJ / kg ]=-6981[ kW ]
W
˙
s
=10[kg/s](2675.5−3373.6)[kJ/kg]=−6981[kW]
Thus, this turbine generates approximately 7 MW of power. Note the negative sign, which indicates that we are getting useful work from the system. This is the equivalent power to that delivered by approximately 70 automobiles running simultaneously.anation: