# Chapter 9: Gas Power Cycles¶

## Example 9-2 ,Page No.498¶

In [2]:
#given data
T1=17+273;#temperature of air in K
P1=100;#pressure of air in kPa
r=8.0;#compression ratio i.e v1/v2
qin=800;#heat transfer rate in kJ/kg

#constants used
R=0.287;#in kPa-m^3/kg-K

#from Table A-17
#at T1
u1=206.91;
vr1=676.1;

#calculations
#Process 1-2
vr2=vr1/r;
#at this vr2
T2=652.4;
u2=475.11;
P2=P1*(T2/T1)*(r);
#Process 2-3
u3=qin+u2;
#at this u3
T3=1575.1;
vr3=6.108;
P3=P2*(T3/T2)*1;#factor of 1 as v3=v2
print'maximum temperature in the cycle %f K'%T3;
print'maximum pressure %f MPa'%round((P3/1000),3);#factor of 1000 to convert into MPa
#Process 3-4
vr4=r*vr3;
#at this vr4
T4=795.6;
u4=588.74;
#Process 4-1
qout=u4-u1;
Wnet=qin-qout;
print'net work output %f kJ/kg'%round(Wnet,2);
nth=Wnet/qin;
print'thermal efficiency is %f'%round(nth,3);
v1=R*T1/P1;
MEP=Wnet/(v1*(1-1/r));
MEP=round(MEP);
print'mean effective pressure %i kPa'%(MEP)

maximum temperature in the cycle 1575.100000 K
maximum pressure 4.345000 MPa
net work output 418.170000 kJ/kg
thermal efficiency is 0.523000
mean effective pressure 574 kPa


## Example 9-3 ,Page No.501¶

In [10]:
#given data
V1=117.0;#intial volumne in in^3
T1=80+460.0;#intial temperature in R
P1=14.7;#intial pressure in psia
r=18.0;#compression ratio
rc=2.0;#cut-off ratio

#constants used
R=0.3704;#in psia ft^3/lbm R
cp=0.240;#in Btu/lbm R
cv=0.171;#in Btu/lbm R

#from Table A-2Ea
k=1.4;

#calculations
V2=V1/r;
V3=rc*V2;
V4=V1;
#Process 1-2
T2=T1*(V1/V2)**(k-1);
P2=P1*(V1/V2)**k;
T2=round(T2);
P2=round(P2);
print('Process 1-2');
print'temperature %i R'%T2;
print'pressure %i psia'%P2;
#Process 2-3
P3=P2;
T3=T2*(V3/V2);
T3=round(T3);
P3=round(P3);
print('Process 2-3');
print'temperature %i R'%T3;
print'pressure %i psia'%P3;
#Process 3-4
T4=T3*(V3/V4)**(k-1);
P4=P3*(V3/V4)**k;
T4=round(T4);
print('Process 3-4');
print'temperature %i R'%T4;
print'pressure %f psia'%round(P4,1);
m=P1*V1/(R*T1)/1728;#factor of 1728 to covert to ft^3 from in^3
Qin=m*cp*(T3-T2);
Qout=m*cv*(T4-T1);
Wnet=Qin-Qout ;
print'work output %f Btu'%round(Wnet,3);
nth=Wnet/Qin;
print'thermal efficiency is %f'%round(nth,3);
MEP=Wnet/(V1-V2)*778.17*12;#factor of 778.17 and 12 to convert to lbf ft and in from Btu and ft respectively
MEP=round(MEP);
print'mean effective pressure %i psia'%MEP

Process 1-2
temperature 1716 R
pressure 841 psia
Process 2-3
temperature 3432 R
pressure 841 psia
Process 3-4
temperature 1425 R
pressure 38.800000 psia
work output 1.296000 Btu
thermal efficiency is 0.633000
mean effective pressure 110 psia


## Example 9-5 ,Page No.511¶

In [1]:
#given data
T1=300;#compressor inlet temperature in K
r=8;#pressure ratio
T3=1300;#turbine inlet temperature in K

#calcualtions
#Process 1-2
#at T1
h1=300.19;
Pr1=1.386;
Pr2=r*Pr1;
#at Pr2
T2=540;
h2=544.35;
print'temperature at exit of compressor %i K'%T2;
#Process 3-4
#at T3
h3=1395.17;
Pr3=330.9;
Pr4=Pr3/r;
#at Pr4
T4=770;
h4=789.37;
print'temperature at turbine exit %i K'%T4;
Win=h2-h1;
Wout=h3-h4;
rbw=Win/Wout;
print'back work ratio is %f'%round(rbw,3);
qin=h3-h2;
Wnet=Wout-Win;
nth=Wnet/qin;
print'thermal efficeincy is %f'%round(nth,3)

temperature at exit of compressor 540 K
temperature at turbine exit 770 K
back work ratio is 0.403000
thermal efficeincy is 0.425000


## Example 9-6 ,Page No.513¶

In [13]:
#from 9.5
Wsc=244.16;#work input of compressor in kJ/kg
Wst=606.60;#work output of turbine in kJ/kg
h1=300.19;#process 1-2 data
h3=1395.17;#process 3-4 data

#given data
nC=0.8;#compressor effciency
nT=0.85;#turbine efficiency

#calculations
Win=Wsc/nC;
Wout=nT*Wst;
rbw=Win/Wout;
print'back work ratio is %f'%round(rbw,3);
h2a=h1+Win;
qin=h3-h2a;
Wnet=Wout-Win;
nth=Wnet/qin;
print'thermal efficeincy is %f'%round(nth,3)
h4a=h3-Wout;
#from A-17 at h4a
T4a=853;
print'turbine exit temperature %i K is'%T4a

back work ratio is 0.592000
thermal efficeincy is 0.266000
turbine exit temperature 853 K is


## Example 9-7 ,Page No.516¶

In [14]:
#from 9.6
h2a=605.39;#entalpy of air leaving the compressor in kJ/kg
h4a=880.36;#entalpy of air leaving the turbinein kJ/kg
h3=1395.97;#in kJ/kg
Wnet=210.41;#net woek done in kJ/kg

#given data
n=0.80;#thermal efficiency

#calculations
# n = (h5 - h2a) / (h4a - h2a)
h5=(h4a - h2a)*n+h2a;
qin=h3-h5;
nth=Wnet/qin;
print'thermal efficeincy is %f'%round(nth,3)

thermal efficeincy is 0.369000


## Example 9-8 ,Page No.519¶

In [16]:
from math import sqrt

#given data
T1=300;#inlet temperature of compressor in K
T6=1300;#inlet temperature of turbine in K
r=8;#overall compression ratio

#calculations
#as it is case of intercooling
ri=sqrt(r);#ri stands for P2/P1 = P4/P3 = P6/P7 = P8/P9
#from A-17 at T1
h1=300.19;
Pr1=1.386;
Pr2=ri*Pr1;
#from A-17 at Pr2
T2=403.3;
h2=403.31;
#from A-17 at T6
h6=1395.97;
Pr6=330.9;
Pr7=Pr6/ri;
#from A-17 at Pr7
T7=1006.4;
h7=1053.33;
#at inlets
T3=T1;h3=h1;T8=T6;h8=h6;
#et exits
T4=T2;h4=h2;T9=T7;h9=h7;h5=h7;
Win=2*(h2-h1);
Wout=2*(h6-h7);
Wnet=Wout-Win;
qin=(h6-h4)+(h8-h7);
rbw=Win/Wout;
print'back work ratio is %f'%round(rbw,3);
nth=Wnet/qin;
print'thermal efficeincy is %f'%round(nth,3)
#part - b
print('part - b');
qin=(h6-h5)+(h8-h7);
nth=Wnet/qin;
print'thermal efficeincy is %f'%round(nth,3)

back work ratio is 0.301000
thermal efficeincy is 0.359000
part - b
thermal efficeincy is 0.699000


## Example 9-9 ,Page No.523¶

In [4]:
from math import sqrt

#given data
m=100;
P1=5;#air pressue in psia
T1=-40+460;#air temperature in R
T4=2000+460;#turbine inlet temperature in R
V1=850;#aircraft velocity in ft/sec
rp=10;#pressure ratio

#constants used
cp=0.240;#in Btu/lbm F
k=1.4;

#calculations
#Process 1-2
T2=T1+V1**2/(2*cp)/25037;#factor of 25037 to covert to Btu/lbm
P2=P1*(T2/T1)**(k/(k-1));
#Process 2-3
P3=rp*P2;
P4=P3;
T3=T2*(P3/P2)**((k-1)/k);
#Win=Wout
T5=T4-T3+T2;
P5=P4*(T5/T4)**(k/(k-1));
T5=round(T5);
print'temperature at turbine exit %i R'%T5;
print'pressure at turbine exit %f psia'%round(P5,1);
#Process 5-6
P6=P1;
T6=T5*(P6/P5)**((k-1)/k);
T6=round(T6);#round off
V6=sqrt(2*cp*(T5-T6)*25037);#factor of 25037 to covert to (ft/s)^2
print'the velocity of nozzle exit %i ft/s'%(round(V6));
Wp=m*(V6-V1)*V1/25037;#factor of 25037 to covert to Btu/lbm
Qin=m*cp*(T4-T3);
nP=Wp/Qin;
print'propulsive efficiency is %f percent'%round(nP*100,1)

temperature at turbine exit 2013 R
pressure at turbine exit 39.600000 psia
the velocity of nozzle exit 3285 ft/s
propulsive efficiency is 22.500000 percent


## Example 9-10 ,Page No.529¶

In [5]:
from math import log

#from 9.2
r=8;#compression ratio i.e v1/v2
T0=290.0;#temperature of air in K
T1=290.0;#temperature in the begining of the isentropic compression process in K
T2=652.4;#temperature in the end of the isentropic compression process in K
T3=1575.1;#temperature after process 2-3 in K
P2=1.7997;#pressure in the end of the isentropic compression process in MPa
P3=4.345;#pressure after process 2-3 in MPa
qin=800.0;#heat transfer rate in kJ/kg
qout=381.83;#heat rejection in kJ/kg
wnet=418.17;#net work in kJ/kg

#given data
Tsource=1700;#temperature of working fluid at source in K

#constants used
R=0.287;#in kPa-m^3/kg-K

#calculations
#s1=s2 ; s3=s4
s03=3.5045;
s02=2.4975;
s32=(s03-s02)-R*log(P3/P2);#s32 stands for s3-s2
xdest23=T0*(s32-qin/Tsource);
Tsink=T1;
xdest41=T0*(-s32+qout/Tsink);
xdestcycle=xdest23+xdest41;
print'exergy destrustion associated with Otto cycle %f kJ/kg'%round(xdestcycle,1);
# X4 = (u4 - u0 )- T0*(s4 - s0) + P0(v4 - v0)
# s4 - s0 = s4 - s1 = s32
# u4 - u0 = u4 - u1 = qout
# v4 - v0 = v4 - v1 = 0
#hence x4 is
X4=qout-T0*s32;
print'exergy destruction of purge stream %f kJ/kg'%round(X4,1)

exergy destrustion associated with Otto cycle 245.400000 kJ/kg
exergy destruction of purge stream 163.200000 kJ/kg