```8-106
8-114 Steam expands in a two-stage adiabatic turbine from a specified state to another specified state. Steam is reheated
between the stages. For a given power output, the reversible power output and the rate of exergy destruction are to be
determined.
Assumptions 1 This is a steady-flow process since there is no change
Heat
with time. 2 Kinetic and potential energy changes are negligible. 3 The
turbine is adiabatic and thus heat transfer is negligible. 4 The
2 MPa
2 MPa
environment temperature is given to be T0 = 25C.
350C
500C
Properties From the steam tables (Tables A-4 through 6)
P1  8 MPa  h1  3399.5 kJ / kg

T1  500C  s1  6.7266 kJ / kg  K
Stage I
Stage II
5 MW
P2  2 MPa  h2  3137.7 kJ / kg

T2  350C  s 2  6.9583 kJ / kg  K
8 MPa
500C
P3  2 MPa  h3  3468.3 kJ / kg

T3  500C  s 3  7.4337 kJ / kg  K
30 kPa
x = 97%
P4  30 kPa  h4  h f  x 4 h fg  289.27  0.97  2335.3  2554.5 kJ/kg

x 4  0.97  s 4  s f  x 4 s fg  0.9441  0.97  6.8234  7.5628 kJ/kg  K
Analysis We take the entire turbine, excluding the reheat section, 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

0
inout



Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
m h1  m h3  m h2  m h4  W out
W out  m [(h1  h2 )  (h3  h4 )]
Substituting, the mass flow rate of the steam is determined from the steady-flow energy equation applied to the actual
process,
W out
5000 kJ/s
m 

 4.253 kg/s
h1  h2  h3  h4 (3399.5  3137.7  3468.3  2554.5)kJ/kg
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 destroyed0 (reversible)  X system0 (steady)  0
X  X out
in

 
Rate of net exergy transfer
by heat, work,and mass
Rate of exergy
destruction
Rate of change
of exergy
X in  X out
m 1  m  3  m  2  m  4  Wrev,out
Wrev,out  m (1  2 )  m ( 3  4 )
 m [(h1  h2 )  T0 (s2  s1)  Δke0  Δpe0 ]
 m [(h3  h4 )  T0 (s4  s3 )  Δke0  Δpe0 ]
Then the reversible power becomes
W rev,out  m h1  h2  h3  h4  T0 ( s 2  s1  s 4  s 3 )
 (4.253 kg/s)[(3399.5  3137.7  3468.3  2554.5)kJ/kg
(298 K)(6.9583  6.7266  7.5628  7.4337)kJ/kg  K]
 5457 kW
Then the rate of exergy destruction is determined from its definition,
X
 W
 W  5457  5000  457 kW
destroyed
rev,out
out
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
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