  extracted for regeneration and the thermal efficiency of the cycle...

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10-106
10-102 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam
extracted for regeneration and the thermal efficiency of the cycle are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h1  h f
@ 15 kPa
 225.94 kJ/kg
v1  v f
@ 15 kPa
 0.001014 m 3 /kg
wpI,in  v 1 P2  P1 


 1 kJ
 0.001014 m 3 /kg 600  15 kPa 
 1 kPa  m 3

 0.59 kJ/kg
h2  h1  wpI,in  225.94  0.59  226.53 kJ/kg
P3  0.6 MPa
sat. liquid
5




7

P II
h4  h3  wpII,in  670.38  10.35  680.73 kJ/kg
1
3
PI
1 MPa
7
5
T
10 MPa
P6  1.0 MPa 
 h6  2783.8 kJ/kg
s6  s5

P8  0.6 MPa 
 h8  3310.2 kJ/kg
s8  s7

Condens.
2

P7  1.0 MPa  h7  3479.1 kJ/kg
T7  500C  s7  7.7642 kJ/kg  K
9
Open
fwh
4
 1 kJ 

 0.001101 m3/kg 10,000  600 kPa 
1 kPa  m3 

 10.35 kJ/kg
P5  10 MPa  h5  3375.1 kJ/kg
T5  500C  s5  6.5995 kJ/kg  K
1-y
8
y
 h3  h f @ 0.6 MPa  670.38 kJ/kg
v  v
3
f @ 0.6 MPa  0.001101 m /kg
 3
wpII,in  v 3 P4  P3 
Turbine
6
Boiler
4
3
6
8
0.6 MPa
2
15 kPa
1
9
s
s9  s f
7.7642  0.7549

 0.9665
P9  15 kPa  x9 
s
7.2522

fg
s9  s7

h9  h f  x9 h fg  225.94  0.96652372.3  2518.8 kJ/kg
The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater
heaters. Noting that Q  W  Δke  Δpe  0 ,
E in  E out  E system0 (steady)  0  E in  E out
 m h   m h
i i
e e

 m 8 h8  m 2 h2  m 3 h3 
 yh8  1  y h2  1h3 
where y is the fraction of steam extracted from the turbine (  m 8 / m 3 ). Solving for y,
y
h3  h2 670.38  226.53

 0.144
h8  h2 3310.2  226.53
(b) The thermal efficiency is determined from
q in  h5  h4   h7  h6   3375.1  680.73  3479.1  2783.8  3389.7 kJ/kg
q out  1  y h9  h1   1  0.14402518.8  225.94   1962.7 kJ/kg
and
 th  1 
q out
1962.7 kJ/kg
 1
 42.1%
3389.7 kJ/kg
q in
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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