# Chapter 10: Vapor and Combined Power Cycles¶

## Example 10-1 ,Page No.555¶

In [2]:
#given data
P1=75;#outlet pressure at compressor in kPa
P2=3000;#inlet pressure at turbine in kPa
P3=P2;
T3=350;#steam inlet temperature in C
P4=P1;

#from steam tables
#at state 1
v1=0.001037;
h1=384.44;
#at state 3
h3=3116.1;
s3=6.7450;
#at state 4
s4=s3;
sf=1.2132;
sfg=6.2426;
hf=384.44;
hfg=2278;

#calculations
win=v1*(P2-P1);
h2=h1+win;
x4=(s4-sf)/sfg;
h4=hf+x4*hfg;
qin=h3-h2;
qout=h4-h1;
nth=1-(qout/qin);
print'thermal efficency is %f'%round(nth,3)

thermal efficency is 0.260000


## Example 10-2 ,Page No.559¶

In [2]:
#given data
#from figure
P1=9;#in kPa
T1=38;#in C
P2=16000;#in kPa
P3=15.9;#in MPa
T3=35;#in C
P4=15.2;#in MPa
T4=625;#in C
P5=15;#in MPa
T5=600;#in C
#from question
nT=0.87;#isentropic efficiency of turbine
nP=0.85;#isentropic efficiency of pump
m=15;#mass flow rate in kg/s

#from steam tables
v1=0.001009;
h5=3583.1;
h6s=2115.3;
h4=3647.6;
h3=160.1;

#calculations
Win=v1*(P2-P1)/nP;
Wout=nT*(h5-h6s);
qin=h4-h3;
Wnet=Wout-Win;
nth=Wnet/qin;
print'thermal efficency is %f'%round(nth,3)
Wnet=m*Wnet;
print'power output %f MW'%round(Wnet/1000,2)

thermal efficency is 0.361000
power output 18.870000 MW


## Example 10-3 ,Page No.562¶

In [3]:
#given data
P1=10;#outlet pressure at compressor in kPa
P2=3000;#inlet pressure at turbine in kPa
P3=P2;
T3=350;#steam inlet temperature in C
P4=P1;

#from steam tables
#at state 1
h1=191.81;
v1=0.00101;
#at state 2
#s2=s1
#at state 3
h3=3116.1;
s3=6.7450;
#at state 4
s4=s3;
sf=0.6492;
sfg=7.4996;
hf=191.81;
hfg=2392.1;

#calculations
#part - a
win=v1*(P2-P1);
h2=h1+win;
x4=(s4-sf)/sfg;
h4=hf+x4*hfg;
qin=h3-h2;
qout=h4-h1;
nth=1-(qout/qin);
print'the thermal efficiency of this power plant is %f'%round(nth,3);
#part - b
#States 1 and 2 remain the same in this case, and the enthalpies at state 3 (3 MPa and 600°C) and state 4 (10 kPa and s4=s3) are determined to be
h3=3682.8;
h4=2380.3;
x4=0.915;
qin=h3-h2;
qout=h4-h1;
nth=1-(qout/qin);
print'the thermal efficiency if steam is superheated to 600° instead of 350°C is %f'%round(nth,3);
#part - c
#State 1 remains the same in this case, but the other states change. The enthalpies at state 2 (15 MPa and s2   s1), state 3 (15 MPa and 600°C),and state 4 (10 kPa and s4   s3) are determined in a similar manner to be
h2=206.95;
h3=3583.1;
h4=2115.3;
x4=0.804;
qin=h3-h2;
qout=h4-h1;
nth=1-(qout/qin);
print'the thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600°C is %f'%round(nth,3);

the thermal efficiency of this power plant is 0.334000
the thermal efficiency if steam is superheated to 600° instead of 350°C is 0.373000
the thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600°C is 0.430000


## Example 10-4 ,Page No.566¶

In [7]:
#given data
P1=10;#outlet pressure at compressor in kPa
P2=15000;#inlet pressure at turbine in kPa
P3=P2;
T3=600;#steam inlet temperature in C
P4=4000;#in kPa
T5=T3;
P6=P1;
x6=0.896;#dryness fraction

#from steam table
#at state 1
h1=191.81;
v1=0.00101;
#at state 3
h3=3593.1;
s3=6.6796;
#at state 4
h4=3155;
T4=375.5;
#at state 6
sf=0.6492;
sfg=7.4996;
hf=191.81;
hfg=2392.1;

#calculations
s6=sf+x6*sfg;
h6=hf+x6*hfg;
#s5 = s6
#from tables
P5=4000.0;#in kPa
h5=3674.9;
print'the pressure at which the steam should be reheated %i MPa'%(P5/1000);
#s2 = s1
win=v1*(P2-P1);
h2=h1+win;
qin=(h3-h2)+(h5-h4);
qout=h6-h1;
nth=1-(qout/qin);
print'thermal efficency is %f'%round(nth,3)

the pressure at which the steam should be reheated 4 MPa
thermal efficency is 0.451000


## Example 10-5 ,Page No.571¶

In [8]:
#given data
P1=10;#outlet pressure at compressor in kPa
P2=1200;#in kPa
P3=P2;
P4=15000;#inlet pressure at turbine in kPa
P5=P4;
T5=600;#steam inlet temperature in C
P6=P2;
P7=P1;

#from steam table
#at state 1
h1=191.81;
v1=0.00101;
#at state 3
h3=798.33;
v3=0.001138;
#at state 4
h4=3155;
T4=375.5;
#at state 5
h5=3583.1;
s5=6.6796;
#at state 6
h6=2860.2;
T6=218.4;
#at state 7
P7=10;
sf=0.6492;
sfg=7.4996;
hf=191.81;
hfg=2392.1;

#calculations
#s2 = s1
win=v1*(P2-P1);
h2=h1+win;
#s4 = s3
win=v3*(P4-P3);
h4=h3+win;
s7=s5;
x7=(s7-sf)/sfg;
h7=hf+(x7*hfg);
#y is the fraction of steam extracted from the turbine
y=(h3-h2)/(h6-h2);
qin=h5-h4;
qout=(1-y)*(h7-h1);
nth=1-(qout/qin);
print'fraction of steam extracted is %f'%round(y,4);
print'thermal efficency is %f'%round(nth,3)

fraction of steam extracted is 0.226900
thermal efficency is 0.463000


## Example 10-6 ,Page No.574¶

In [12]:
#given data
P1=10;#outlet pressure at compressor in kPa
P2=500;#steam extracted from low-pressure turbine in kPa
P3=P2;
P4=15000;#inlet pressure at turbine in kPa
P5=P4;
P6=4000;#steam extracted from turbine at pressure in kPa
P7=P5;
P8=P7;
P9=P7;
P10=P6;
P11=P10;
P12=P3;
P13=P1;

#enthalpies at the various states and the pump work per unit mass of fluid flowing through them are
h1=191.81;
h2=192.30;
h3=640.09;
h4=643.92;
h5=1087.4;
h6=h5;
h7=1101.2;
h8=1089.8;
h9=3583.1;
h10=3155;
h11=3679.9;
h12=3014.8;
h13=2335.7;
wIin=0.49;
wIIin=3.83;
wIIIin=13.77;

#calculations
y=(h5-h4)/((h10-h6)+(h5-h4));
z=(1-y)*(h3-h2)/(h12-h2);
h8=(1-y)*h5+(y*h7);
qin=(h9-h8)+(1-y)*(h11-h10);
qout=(1-y-z)*(h13-h1);
nth=1-(qout/qin);
print'fraction of steam extracted from closed feedwater is %f'%round(y,4);
print'fraction of steam extracted from open feedwater is %f'%round(z,4);
print'thermal efficency is %f'%round(nth,3)

fraction of steam extracted from closed feedwater is 0.176600
fraction of steam extracted from open feedwater is 0.130600
thermal efficency is 0.492000


## Example 10-7 ,Page No.577¶

In [4]:
#given data
T0=290;#temperature at heat which is rejected in K
Tsource=1600;#temperature of furnace in K
Tsink=T0;
#from Ex 10.1
qin=2728.6;#in kJ/Kg
qout=2018.6;#in kJ/Kg
h4=2403;#in kJ/Kg

#from steam tables
s1=1.2132;
s3=6.7450;

#calculations
s2=s1;s4=s3;#isentropic processes
xdest12=0;
xdest34=0;
xdest23=T0*(s3-s2-(qin/Tsource));
xdest41=T0*(s1-s4+(qout/Tsink));
print'exergy destruction in process 1-2 %i kJ/kg'%xdest12;
print'exergy destruction in process 2-3 %i kJ/kg'%round(xdest23);
print'exergy destruction in process 3-4 %i kJ/kg'%xdest34;
print'exergy destruction in process 4-1 %i kJ/kg'%round(xdest41);
xdestcy=xdest12+xdest23+xdest34+xdest41;
print'exergy destruction %i cycle in kJ/kg'%round(xdestcy);
#from steam tables
#at 290 K and 100 kPa
h0=71.355;
s0=0.2533;
X4=(h4-h0)-T0*(s4-s0);
print'exergy of the leaving steam %i kJ/kg'%(round(X4))

exergy destruction in process 1-2 0 kJ/kg
exergy destruction in process 2-3 1110 kJ/kg
exergy destruction in process 3-4 0 kJ/kg
exergy destruction in process 4-1 414 kJ/kg
exergy destruction 1524 cycle in kJ/kg
exergy of the leaving steam 449 kJ/kg


## Example 10-8 ,Page No.581¶

In [14]:
#given data
m1=15;#mass flow rate in kg/s
P1=7000;#inlet pressure at turbine in kPa
P2=P1;
P3=P1;
P4=500;#throttled to pressure in kPa
P5=P4;
P6=5;#steam expanded at a pressure in kPa
P7=P4;
P8=P6;
P9=P1;
P10=P1;

#from steam tables
v7=0.001005;
v8=0.001093;
h1=3411.4;
h2=h1;
h3=h1;
h4=h1;
h5=2739.3;
h6=2073.0;
h7=640.09;
h8=137.75;
h11=144.78;

#calculations
wIin=v8*(P9-P8);
wIIin=v7*(P10-P7);
h9=h8+wIin;
h10=h7+wIIin;
Qmax=m1*(h1-h7);
print'the maximum rate %i kW'%round(Qmax);
Wtout=m1*(h3-h6);#turbine
Wpin=m1*wIin;#pump
Wnet=Wtout-Wpin;
print'the power produced %i MW'%(round(Wnet/1000));
Qp=0;
Qin=m1*(h1-h11);
Eu=(Wnet+Qp)/Qin;
print'the utilization factor is %f'%round(Eu,3);
m4=0.1*m1;
m5=0.7*m1;
m7=m4+m5;
Qout=m4*h4+m5*h5-m7*h7;
print'the rate of process heat supply %f MW'%round(Qout/1000,1)

the maximum rate 41570 kW
the power produced 20 MW
the utilization factor is 0.407000
the rate of process heat supply 26.200000 MW


## Example 10-9 ,Page No.585¶

In [5]:
#given data
P1=5;#intial pressure in kPa
P2=7000;#final pressure ink Pa
P3=P2;
T3=500;#inlet temperature in K
P4=P1;

#gas cycle from Ex9-6
#d stands for '
h4d=880.36;
T4d=853;
qin=790.58;
wnetg=210.41;
nth=0.266
h5d=451.80;
#steam cycle
h2=144.78;
T2=33;
h3=3411.4;
T3=500;
wnets=1331.4;
nth=0.408;

#calculations
#Ein = Eout
#y is the ratio of ms/mg
y=(h4d-h5d)/(h3-h2);
print'the ratio of the mass flow rates of the steam and the combustion gasesis %f'%round(y,3);
wnet=wnetg+y*wnets
nth=wnet/qin;
print'thermal efficency is %f'%round(nth,3)

the ratio of the mass flow rates of the steam and the combustion gasesis 0.131000
thermal efficency is 0.487000