8-49
8-61 Steam expands in a turbine from a specified state to another specified state. The actual power output of the turbine is
given. The reversible power output and the second-law efficiency are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy change is negligible. 3
The temperature of the surroundings is given to be 25C.
Properties From the steam tables (Tables A-4 through A-6)
P1  6 MPa  h1  3658.8 kJ/kg

T1  600C  s1  7.1693 kJ/kg  K
P2  50 kPa  h2  2682.4 kJ/kg

T2  100C  s 2  7.6953 kJ/kg  K
1  m
2  m
 . We take the turbine as the system, which is a
Analysis (b) There is only one inlet and one exit, and thus m
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


Rate of net energy transfer
by heat, work, and mass
E system 0 (steady)


0
80 m/s
6 MPa
600C
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
m (h1  V12 / 2)  W out  m (h2  V 22 / 2)
W out
STEAM

V 2  V 22 
 m h1  h2  1

2


5 MW
Substituting,

(80 m/s) 2  (140 m/s) 2  1 kJ/kg
5000 kJ/s  m  3658.8  2682.4 


2
 1000 m 2 / s 2

m  5.156 kg/s
 


50 kPa
100C
140 m/s
The reversible (or maximum) power output is determined from the rate form of the exergy balance applied on the turbine
and setting the exergy destruction term equal to zero,
X  X
inout

Rate of net exergy transfer
by heat, work,and mass
 X destroyed0 (reversible)  X system0 (steady)  0
 


Rate of change
of exergy
Rate of exergy
destruction
X in  X out
   W
m
1
rev,out
2
m
 ( 1  2 )  m
 [(h1  h2 )  T0 ( s1  s2 )  Δke  Δpe0 ]
W rev,out  m
Substituting,
W rev,out  W out  m T0 ( s1  s2 )
 5000 kW  (5.156 kg/s)(298 K)(7.1693  7.6953) kJ/kg  K  5808 kW
(b) The second-law efficiency of a turbine is the ratio of the actual work output to the reversible work,
W out
 II  
W
rev, out

5 MW
 86.1%
5.808 MW
Discussion Note that 13.9% percent of the work potential of the steam is wasted as it flows through the turbine during this
process.
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