#   5-31

```5-31
5-49 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be
determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible.
3 The device is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4 through 6)
P1 = 6 MPa
T1 = 400C
V1 = 80 m/s
P1  6 MPa  v 1  0.047420 m 3 /kg

T1  400C  h1  3178.3 kJ/kg
and
P2  40 kPa 
 h2  h f  x 2 h fg  317.62  0.92  2392.1  2318.5 kJ/kg
x 2  0.92 
&middot; STEAM
m = 12 kg/s
&middot;
W
Analysis (a) The change in kinetic energy is determined from
ke 
V22  V12 50 m/s 2  (80 m/s) 2

2
2
 1 kJ/kg

 1000 m 2 /s 2


  1.95 kJ/kg


1  m
2  m
 . We take the turbine as the
(b) There is only one inlet and one exit, and thus m
system, which is a 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 out
in


Rate of net energy transfer
by heat, work, and mass

P2 = 40 kPa
x2 = 0.92
V2 = 50 m/s
0
Rate of change in internal, kinetic,
potential, etc. energies
E in  E out
  pe  0)
m (h1  V12 / 2)  W out  m (h2 + V 22 /2) (since Q

V 2  V12
W out  m  h2  h1  2

2





Then the power output of the turbine is determined by substitution to be
W out  (20 kg/s)(2318.5  3178.3  1.95)kJ/kg  14,590 kW  14.6 MW
(c) The inlet area of the turbine is determined from the mass flow rate relation,
m 
1
v1
A1V1 
 A1 
m v 1 (20 kg/s)(0.047420 m 3 /kg )

 0.0119 m 2
V1
80 m/s
PROPRIETARY MATERIAL. &copy; 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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