# CHAPTER 21 - Gas turbines and jet propulsion¶

## EXAMPLE 21.1 PAGE 848¶

In [1]:
from __future__ import division
# Initialisation of Variables
p1=1#.........#Pressure of air while entering the turbine in bar
t1=293#........#Temperature of air entering the turbine in K
p2=4#.........#Pressure of air after compression in bar
etac=0.8#....#Efficiency of compressor
etat=0.85#.....#Efficiency of turbine
afr=90#........#Air fuel ratio
ma=3#...........#Mass of air in kg/s
ga=1.4#........#Ratio of specific heats
cp=1#.............#Specific heat at constant pressure in kJ/kgK
C=41800#.............#Calorific value of fuel in kJ/kg
#Calculations
t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
t3=round((C/((afr+1)*cp))+t21)#..............#Temperature before expansion in turbine in K
p4=p1#
p3=p2#
t4=t3*((p4/p3)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
wt=((afr+1)/afr)*cp*(t3-t41)#........#Work done by turbine  in kJ/kg of air
wc=round(1*cp*(t21-t1))#.................#Work done by compression in kJ/kg of air
wnet=wt-wc#..........#Net work done in kJ/kg
P=wnet*ma#.................#Power developed in kW/kg of air
print "Power developed = %0.2f kW/kg of air"%P
qs=(1/afr)*C#................#Heat supplied in kJ/kg of air
etath=wnet/qs#................#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)

Power developed = 250.22 kW/kg of air
Thermal efficiency = 17.96 %


## EXAMPLE 21.2 PAGE 848¶

In [2]:
from __future__ import division
# Initialisation of Variables
t1=288#........#Temperature of air entering the turbine in K
t3=883#..............#Temperature before expansion in turbine in K
etac=0.8#....#Efficiency of compressor
etat=0.82#.....#Efficiency of turbine
rp=6#...........#Pressure ratio
ma=16#...........#Mass of air in kg/s
gac=1.4#........#Ratio of specific heats for compression process
gae=1.333#............#Ratio of specific heats for expansion process
cpc=1.005#.............#Specific heat at constant pressure in kJ/kgK during compression process
cpe=1.11#.............#Specific heat at constant pressure in kJ/kgK during expansion process
C=41800#.............#Calorific value of fuel in kJ/kg
#Calculations
t2=t1*((rp)**((gac-1)/gac))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
t4=t3/((rp)**((gae-1)/gae))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
wt=cpe*(t3-t41)#........#Work done by turbine  in kJ/kg of air
wc=(1*cpc*(t21-t1))#.................#Work done by compression in kJ/kg of air
wnet=wt-wc#..........#Net work done in kJ/kg
P=wnet*ma#.................#Power developed in kW/kg of air
print "Power developed = %0.2f kW/kg of air"%P

Power developed = 770.31 kW/kg of air


## EXAMPLE 21.3 PAGE 849¶

In [3]:
from __future__ import division
# Initialisation of Variables
p1=1#.........#Pressure of air while entering the turbine in bar
t1=300#........#Temperature of air entering the turbine in K
p2=6.2#.........#Pressure of air after compression in bar
etac=0.88#....#Efficiency of compressor
etat=0.9#.....#Efficiency of turbine
far=0.017#........#Fuel air ratio
ga=1.4#........#Ratio of specific heats for compression
gae=1.333#........#Ratio of specific heats for expansion
cp=1.147#.............#Specific heat at constant pressure in kJ/kgK during expansion
cpc=1.005#.............#Specific heat at constant pressure in kJ/kgK during compression
C=44186#.............#Calorific value of fuel in kJ/kg
#Calculations
t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
t3=(((C*far)/((far+1)*cpc))+t21)#..............#Temperature before expansion in turbine in K
p4=p1#
p3=p2#
t4=t3*((p4/p3)**((gae-1)/gae))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
wt=(cp*(t3-t41))#........#Work done by turbine  in kJ/kg of air
wc=round(1*cpc*(t21-t1))#.................#Work done by compression in kJ/kg of air
wnet=wt-wc#..........#Net work done in kJ/kg
qs=(far)*C#................#Heat supplied in kJ/kg of air
etath=wnet/qs#................#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)

Thermal efficiency = 32.65 %


## EXAMPLE 21.4 PAGE 850¶

In [4]:
from __future__ import division
# Initialisation of Variables
t1=300#........#Temperature of air entering the turbine in K
t3=1148#..............#Temperature before expansion in turbine in K
etac=0.8#....#Efficiency of compressor
etat=0.852#.....#Efficiency of turbine
rp=4#...........#Pressure ratio
p1=1#...............#Pressure of air before entering compressor
ga=1.4#........#Ratio of specific heats
cp=1.0#.............#Specific heat at constant pressure in kJ/kgK
C=42000#.............#Calorific value of fuel in kJ/kg
perlcc=10#............#Percent loss of calorific value of fuel in combustion chamber
#Calculations
p2=p1*rp#.................#Pressure of air after compression in  bar
etacc=(100-perlcc)/100#.......#Efficiency of combustion chamber
t2=t1*((rp)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
afr=((C*etacc)/(cp*(t3-t21)))-1#........#Air fuel ratio
print "Air fuel ratio is %d:1"%round(afr)

Air fuel ratio is 56:1


## EXAMPLE 21.5 PAGE 851¶

In [5]:
from __future__ import division
# Initialisation of Variables
t1=300#........#Temperature of air entering the turbine in K
t3=883#..............#Temperature before expansion in turbine in K
etac=0.8#....#Efficiency of compressor
etat=0.852#.....#Efficiency of turbine
rp=4#...........#Pressure ratio
p1=1#...............#Pressure of air before entering compressor
ga=1.4#........#Ratio of specific heats
cp=1.11#.............#Specific heat at constant pressure in kJ/kgK
C=42000#.............#Calorific value of fuel in kJ/kg
perlcc=10#............#Percent loss of calorific value of fuel in combustion chamber
#Calculations
p2=p1*rp#.................#Pressure of air after compression in  bar
etacc=(100-perlcc)/100#.......#Efficiency of combustion chamber
t2=t1*((rp)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
qs=cp*(t3-t21)#...................#Heat supplied in kJ/kg
t4=t3/((rp)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
wt=cp*(t3-t41)#........#Work done by turbine  in kJ/kg of air
wc=(1*cp*(t21-t1))#.................#Work done by compression in kJ/kg of air
wnet=wt-wc#..........#Net work done in kJ/kg
etath=wnet/qs#................#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)
wrr=wnet/wt#...................#Work ratio
print "The work ratio is %f"%wrr

Thermal efficiency = 15.92 %
The work ratio is 0.259290


## EXAMPLE 21.6 PAGE 851¶

In [6]:
from __future__ import division
# Initialisation of Variables
p1=1#.........#Pressure of air while entering the turbine in bar
t1=293#........#Temperature of air entering the turbine in K
p2=5#.........#Pressure of air after compression in bar
plcc=0.1#.....#Pressure loss in combustion chamber in bar
t3=953#............#Temperature before expansion in turbine in K
etac=0.85#....#Efficiency of compressor
etat=0.8#.....#Efficiency of turbine
etacc=0.85#......#Efficiency of combustion chamber
ga=1.4#........#Ratio of specific heats
cp=1.024#.............#Specific heat at constant pressure in kJ/kgK
P=1065#.............#Power developed by the plant in kW

#Calculations
p3=p2-plcc#........................#Pressure before expansion in turbine in bar
p4=p1#
t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
t4=t3*((p4/p3)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
wt=(cp*(t3-t41))#........#Work done by turbine  in kJ/kg of air
wc=round(1*cp*(t21-t1))#.................#Work done by compression in kJ/kg of air
wnet=wt-wc#..........#Net work done in kJ/kg
ma=P/wnet#.............#Quantity of air in circulation in kg
print "Quantity of air in circulation = %0.2f kg"%(ma)
qs=cp*(t3-t21)/etac#..................#Actual heat supplied per kg of air circulation in kJ
print "Actual heat supplied per kg of air circulation = %0.2f kJ"%qs
etath=wnet/qs#.............#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)

Quantity of air in circulation = 13.49 kg
Actual heat supplied per kg of air circulation = 552.66 kJ
Thermal efficiency = 14.28 %


## EXAMPLE 21.7 PAGE 852¶

In [7]:
from __future__ import division
# Initialisation of Variables
ma=20#..............#Air flow rate in kg/s
t1=300#........#Temperature of air entering the turbine in K
t3=1000#............#Temperature before expansion in turbine in K
rp=4#...............#Pressure ratio
cp=1#.............#Specific heat at constant pressure in kJ/kgK
ga=1.4#........#Ratio of specific heats
#Calculations
t2=t1*((rp)**((ga-1)/ga))#...............#Temperature of air after compression in K
t4=t3-t2+t1#............#Temperature after expansion in turbine in K
prlp=rp/((t3/t4)**(ga/(ga-1)))#.............#Pressure ratio of low pressure turbine
print "Pressure ratio of low pressure turbine : %0.4f"%prlp
t5=t4/((prlp)**((ga-1)/ga))#............#Temperature of the exhaust from the unit in K
print "Temperature of the exhaust from the unit = %0.2f K"%t5

Pressure ratio of low pressure turbine : 2.3042
Temperature of the exhaust from the unit = 672.95 K


## EXAMPLE 21.8 PAGE 853¶

In [8]:
from __future__ import division
# Initialisation of Variables
p1=1#.........#Pressure of air while entering the turbine in bar
t1=300#........#Temperature of air entering the turbine in K
t21=490#........#Actual temperature of air after compression in K
t3=1000#............#Temperature before expansion in turbine in K
rp=5#.............#Pressure ratio
etac=0.8#....#Efficiency of compressor
etat=0.8#.....#Efficiency of turbine
ga=1.4#........#Ratio of specific heats
cp=1.005#.............#Specific heat at constant pressure in kJ/kgK
#Calculations
t4=t3/((rp)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
t5=((t41-t21)*etac)+t21#...........#Temperature of the exhaust from the unit in K
wc=cp*(t21-t1)#.............#Work consumed by compressor in kJ/kg
wt=cp*(t3-t41)#........#Work done by turbine in kJ/kg
qs=cp*(t3-t5)#..........#Heat supplied in kJ/kg
etac=(wt-wc)/qs#.........#Cycle efficiency
print "Cycle efficiency = %0.2f %%"%(etac*100)

Cycle efficiency = 31.04 %


## EXAMPLE 21.9 PAGE 853¶

In [2]:
from __future__ import division
# Initialisation of Variables
p1=1#.........#Pressure of air while entering the turbine in bar
t1=288#........#Temperature of air entering the turbine in K
p2=8#.........#Pressure of air after compression in bar
t3=1173#.............#Temperature before expansion in turbine in K
etac=0.76#....#Efficiency of compressor
etat=0.86#.....#Efficiency of turbine
ma=23#.........#Quantity of air circulation in kg/s
ga=1.4#........#Ratio of specific heats for compression
gag=1.34#........#Ratio of specific heats for expansion
cp=1.005#.............#Specific heat at constant pressure in kJ/kgK
cpg=1.128#.............#Specific heat at constant pressure in kJ/kgK
C=4200#.............#Calorific value of fuel in kJ/kg
etamech=0.95#........#Mechanical efficiency
etagen=0.96#.........#Generator efficiency
#Calculations
t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
p4=p1#
p3=p2#.............#Isobaric processes
t4=t3*((p4/p3)**((gag-1)/gag))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
wc=cp*(t21-t1)#................#Work dony by compressor
m1=(wc)/(cpg*(t3-t41))#.............#Flow through compressor turbine in kg
m2=1-m1#..............#Flow through power turbine in kg
wpt=m2*(cpg*(t3-t41))#.........#turbine work in kJ/kg
P=ma*wpt*etamech*etagen#.........#Power output in kW
qi=cpg*t3-cp*t21#.............#Input heat in kJ/kg of air
etath=wpt/qi#.............#Thermal efficiency of power turbine
print "Thermal efficiency of power turbine = %0.2f %%"%(etath*100)

Thermal efficiency of power turbine = 21.73 %


## EXAMPLE 21.10 PAGE 854¶

In [10]:
from __future__ import division
# Initialisation of Variables
t1=288#........#Temperature of air entering the turbine in K
t3=883#..............#Temperature before expansion in turbine in K
etac=0.82#....#Efficiency of compressor
etathp=0.85#.....#Efficiency of high pressure turbine
etatlp=0.85#.....#Efficiency of low pressure turbine
rp=7#...........#Pressure ratio
p1=1.01#...............#Pressure of air before entering compressor
ga=1.4#........#Ratio of specific heats for compression
gag=1.333#........#Ratio of specific heats for expansion
cp=1.005#.............#Specific heat at constant pressure in kJ/kgK
cpg=1.15#.............#Specific heat at constant pressure in kJ/kgK in generator
#Calculations
p2=p1*rp#
t2=t1*((p2/p1)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#..............#Actual temperature of air after compression in K
wc=cp*(t21-t1)#............#Compressor work in kJ/kg
t41=t3-(wc/cpg)#..........#Temperature of gasses entering the turbine in K
print "Temperature of gasses entering the turbine = %0.2f K"%t41
t4=round(t3-((t3-t41)/etathp))#.........#Ideal temperature of gases entering the turbine in K
p3=p2#.........#Isobaric processes
p4=p3/((t3/t4)**(1/((gag-1)/gag)))#....#Pressure of gasses entering the power turbine in bar
print "Pressure of gasses entering the power turbine = %0.2f bar"%p4
t5=t41*((((t3/t4)**(1/((gag-1)/gag)))/(rp))**((gag-1)/gag))#
t51=t41-(etatlp*(t41-t5))#
wlp=cpg*(t41-t51)#............#Net power output in kW
print "Net power output = %0.2f kW"%wlp
wr=wlp/(wlp+wc)#............#Work ratio
print "Work ratio : %0.4f"%wr
qs=cpg*(t3-t21)#...........#Heat supplied in kJ/kg
etath=wlp/qs#..........#Thermal efficiency
print "Thermal efficiency of the unit : %0.2f %%"%(etath*100)

Temperature of gasses entering the turbine = 654.75 K
Pressure of gasses entering the power turbine = 1.65 bar
Net power output = 73.95 kW
Work ratio : 0.2198
Thermal efficiency of the unit : 19.26 %


## EXAMPLE 21.11 PAGE 855¶

In [11]:
from __future__ import division
from math import sqrt
# Initialisation of Variables
rp=5.6#................#Pressure ratio
t1=303#.............#Temperature of intake air in K
p1=1#............#Pressure of intake air in bar
t5=973#............#Highest temperature of the cycle in K
etac=0.85#..........#Effeciency of compressor
etat=0.9#..........#Efficiency of turbine
ma=1.2#..........#Rate of air flow in kg/s
cp=1.02#...........#Specific heat at constant volume in kJ/kgK
ga=1.41#.............#Ratio of specific heats
#Calculations
t2=t1*((sqrt(rp))**((ga-1)/ga))#
t21=((t2-t1)/etac)+t1#
wc=2*ma*cp*(t21-t1)#............#Work input for the two stage compressor in kJ/s
t6=t5/(rp**((ga-1)/ga))#
t61=t5-etat*(t5-t6)#
wt=ma*cp*(t5-t61)#...............#Work output from turbine in kJ/s
wnet=wt-wc#....................#Net work available in kJ/s
print "Net work output = %0.2f kW"%wnet
qs=ma*cp*(t5-t21)#.................#Heat supplied in kJ/s
etath=wnet/qs#
print "Power plant efficiency = %0.2f %%"%(etath*100)

Net work output = 173.98 kW
Power plant efficiency = 25.00 %


## EXAMPLE 21.13 PAGE 856¶

In [12]:
from __future__ import division
# Initialisation of Variables
t1=288#.............#Temperature of intake air in K
rp=4#.............#Pressure ratio
etac=0.82#.........#Compressor efficiency
etahe=0.78#...........#Efficiency of heat exchanger
etat=0.7#...........#Turbine efficiency
t3=873#............#Temperature before expansion in turbine in K
R=0.287#............#Gas constant for air in kJ/kgK
ga=1.4#...........#Ratio of specific heats
#Calculations
t2=t1*((rp)**((ga-1)/ga))#...............#Ideal temperature of air after compression in K
t21=((t2-t1)/etac)+t1#...............#Actual temperature of air after compression in K
t4=t3/(rp**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K
t41=t3-etat*(t3-t4)#............#Actual temperature after expansion in turbine in K
cp=R*(ga/(ga-1))#..............#Specific heat at constant pressure in kJ/kgK
wc=cp*(t21-t1)#.............#Compressor work in kJ/kg
wt=cp*(t3-t41)#....................#Turbine work in kJ/kg
wnet=wt-wc#....................#Net work available in kJ/s
t5=(etahe*(t41-t21))+t21#
qs=cp*(t3-t5)#.................#Heat supplied in kJ/kg
etac=wnet/qs#...............#Cycle efficiency
print "Cycle efficiency = %0.2f %%"%(etac*100)

Cycle efficiency = 11.81 %


## EXAMPLE 21.14 PAGE 857¶

In [13]:
from __future__ import division
# Initialisation of Variables
etahe=0.72#.................#Efficiency of heat exchanger
p1=1.01#.........#Pressure of air while entering the turbine in bar
t1=293#........#Temperature of air entering the turbine in K
p2=4.04#.........#Pressure of air after compression in bar
etat=0.85#..........#Turbine efficiency
pdhe=0.05#............#Pressure drop on each side of heat exchanger in bar
pdcc=0.14#...........#Pressure drop in combustion chamber in bar
etac=0.8#...........#Compressor efficiency
ga=1.4#.............#Ratio of specific heats
C=41800#.............#Calorific value of fuel in kJ/kg
cp=1.024#...........#Specific heat at constant pressure in kJ/kgK
afrc=90#..............#Air fuel ratio for simple cycle
#Calculations
t2=(t1*((p2/p1)**((ga-1)/ga)))#...............#Ideal temperature of air after compression in K
t21=round(((t2-t1)/etac)+t1)#...............#Actual temperature of air after compression in K
t3=((1*C)/(cp*(afrc+1)))+t21#............#Temperature before expansion in turbine in K
p4=p1#
p3=p2-pdcc#
t4=round(t3*((p4/p3)**((ga-1)/ga)))#............#Ideal temperature after expansion in turbine in K
t41=t3-(etat*(t3-t4))#.................#Actual temperature after expansion in turbine in K
etath=(t3-t41-t21+t1)/(t3-t21)#...........#Thermal efficiency in simple cycle
print "Thermal efficiency in simple cycle = %0.2f %%"%(etath*100)
p3he=p2-pdhe-pdcc#..........#Pressure before expansion in turbine in bar in heat exchanger cycle
p4he=p1+pdhe#................#Pressure after expansion in turbine in bar in heat exchanger cycle
t4he=t3*((p4he/p3he)**((ga-1)/ga))#............#Ideal temperature after expansion in turbine in K in heat exchanger cycle
t41he=round(t3-(etat*(t3-t4he)))#.................#Actual temperature after expansion in turbine in K in heat exchanger cycle
t5=(etahe*(t41he-t21))+t21#
etathhe=(t3-t41he-t21+t1)/(t3-t5)#.............#Thermal efficiency for heat exchanger cycle
print "Thermal efficiency in heat exchanger cycle = %0.2f %%"%(etathhe*100)
inc=etathhe-etath#
print "Increase in thermal efficiency = %0.2f %%"%(inc*100)

Thermal efficiency in simple cycle = 23.25 %
Thermal efficiency in heat exchanger cycle = 20.94 %
Increase in thermal efficiency = -2.31 %


## EXAMPLE 21.15 PAGE 858¶

In [14]:
from __future__ import division
from math import sqrt
# Initialisation of Variables
t1=293#........#Temperature of air entering the turbine in K
rp=9#............#Overall pressure ratio
etac=0.8#........#Efficiency of compressor
t6=898#..........#Reheat remperature
t8=t6#
etat=0.85#.......#Efficiency of turbine
etamech=0.95#..........#Mechanical efficiency
etahe=0.8#...............#Heat exchanger thermal efficiency
cpg=1.15#.............#Specific heat capacity for gases in heat exchanger in kJ/kgK
cpa=1.005#............#Specific heat capacity for normal air in kJ/kgK
gag=1.333#.............#Ratio of specific heats for gases in heat exchanger
ga=1.4#...............#Ratio of specific heats for normal gases
P=4500#.................#Power output of turbine in kW
#Calculations
t2=t1*((sqrt(rp))**((ga-1)/ga))#
t21=((t2-t1)/etac)+t1#
wc=cpa*(t21-t1)#............#Work input per compressor stage
whp=(2*wc)/etamech#.........#Work output of HP turbine in kJ/kg
t71=t6-(whp/cpg)#
t7=round(t6-((t6-t71)/etat))#
k=(rp/((t6/t7)**((gag)/(gag-1))))**((gag-1)/gag)#
k1=((round((k/2)*100))*2)/100#..............#Rounding off upto 2 decimals
t9=t8/(k1)#
t91=t8-((t8-t9)*etat)#
wout=cpg*(t8-t91)*etamech#..............#Net work output in kJ/kg
t5=etahe*(t91-t21)+t21#
qs=cpg*(t6-t5)+cpg*(t8-t71)#...............#Heat supplied
etath=wout/qs#.................#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)
wgross=whp+(wout/etamech)#.........#Gross work output in kJ/kg
wr=wout/wgross#................#Work ratio
print "The workk ratio is %0.3f"%wr
m1=P/wout#...............#Mass flow in kg/s
print "Mass flow = %0.2f kg/s"%m1

Thermal efficiency = 23.44 %
The workk ratio is 0.283
Mass flow = 39.12 kg/s


## EXAMPLE 21.16 PAGE 859¶

In [15]:
from __future__ import division
from math import sqrt
# Initialisation of Variables
#Conditions of the closed gas turbine
t1=293#.............#Temperature at the inlet of first stage compressor in K
t5=1023#.................#Maximum temperature in K
p1=1.5#................#Inlet pressure in bar
p2=6#.................#Pressure in bar
etac=0.82#..............#Compressor efficiency
etat=0.82#..............#Turbine efficiency
etare=0.70#................#Regenerator efficiency
P=350#....................#Power developed by the plant in kW
ga=1.4#................#Ratio of specific heats
cp=1.005#..............#Specific heat at constant pressure in kJ/kgK
t3=t1#
#Calculations
t2=t1*((sqrt(p2/p1))**((ga-1)/ga))#
t21=((t2-t1)/etac)+t1#t41=t21#
t6=t5/((p2/sqrt(p1*p2))**((ga-1)/ga))#
t61=t5-(etat*(t5-t6))#
t81=t61#
t7=t5#
ta=(etare*(t81-t41))+t41#.......#Temperature of air coming out of regenerator in K
wnet=2*cp*(t5-t61-t21+t1)#........#Net work done in kJ/kg of air
qs=cp*(t5-t41+t7-t61)#...........#Heat supplied without regenerator in kJ/kg of air
qsr=cp*(t5-ta+t7-t61)#............#Heat supplied with regenerator in kJ/kg of air
etath=wnet/qs#.............#Thermal efficiency (without regenerator)
etathr=wnet/qsr#.........#Thermal efficiency (with regenerator)
mfl=P/wnet#..........#mass of fluid circulated in kg/s
print "Thermal efficiency of the turbine without regenerator = %0.2f %%"%(etath*100)
print "Thermal efficiency of the turbine with regenerator = %0.2f %%"%(etathr*100)
print "Mass of fluid circulated = %0.2f kg/s"%mfl

Thermal efficiency of the turbine without regenerator = 27.01 %
Thermal efficiency of the turbine with regenerator = 38.96 %
Mass of fluid circulated = 2.40 kg/s


## EXAMPLE 21.17 PAGE 860¶

In [16]:
from __future__ import division
# Initialisation of Variables
t1=293#............#Temperature of inlet air into low pressure compressor in K
p1=1.05#.........#Pressure of inlet air into low pressure compressor in bar
t3=300#...........#Temperature of air after passing it through intercooler in K
t6=1023#..........#temperature of air in combustion chamber in K
rp=2#...........#Pressure ratio of each compressor
etac=0.82#........#Compressor efficiency
etat=0.82#..........#Turbine efficiency
etaht=0.72#............#Heat exchanger efficiency
ma=16#...........#Air flow in kg/s
ga=1.4#...........#Ratio of specific heats for air
gag=1.33#..........#Ratio of specific heats for gases
cpa=1.0#...........#Specific heat at constant pressure in kJ/kgK for air
cpg=1.15#.........#Specific heat at constant pressure in kJ/kgK for gases
C=42000#.........#Calorific value of fuel in kJ/kg
#Calculations
t2=round(t1*(rp**((ga-1)/ga)))#
t21=round(((t2-t1)/etac)+t1)#
t4=t3*(rp**((ga-1)/ga))#
t41=round(((t4-t3)/etac)+t3)#
t71=round(((cpg*t6)-cpa*(t21-t1+t41-t3))/cpg)#
t7=t6-((t6-t71)/etat)#
p6=p1*rp*rp#
p7=p6/((t6/t7)**((gag)/(gag-1)))#
t8=round(t71/((p7/p1)**((gag-1)/gag)))#
t81=round(t71-(etat*(t71-t8)))#
P=cpg*(t71-t81)#...........#Net power output in kJ/kg
print "Net power output = %0.2f kW "%(P*ma)
t5=etaht*(t81-t41)+t41#
qs=ma*cpg*(t6-t5)#......#Heat supplied in combustion chamber in kJ/s
etath=P*ma/qs#.........#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)
afr=C/(cpg*(t6-t5))#......#Air fuel ratio
mf=ma*3600/afr#..............#Fuel supplied per hour in kg
sfc=mf/(P*ma)#...........#Specific fuel consumption in kg/kWh
print "Specific fuel consumption = %0.2f kg/kWh"%sfc

Net power output = 2042.40 kW
Thermal efficiency = 30.95 %
Specific fuel consumption = 0.28 kg/kWh


## EXAMPLE 21.18 PAGE 861¶

In [17]:
from __future__ import division
# Initialisation of Variables
t1=293#............#Temperature of inlet air into low pressure compressor in K
p1=1.1#.........#Pressure of inlet air into low pressure compressor in bar
p2=3.3#..........#Pressure of air in the low pressure compressor in bar
t3=300#.............#Intercooled temperature in K
pli=0.15#..........#Loss in pressure due to intercooling in bar
p3=p2-pli#...........#Pressure after intercooling in bar
p4=9.45#............#Pressure of air after high pressure compressor in bar
p6=p4#
t6=973#.........#Temperature of gases supplied to high pressure turbine in K
t8=943#.........#Reheat temperature in K
plr=0.12#...........#Loss of pressure after reheating in bar
p7=3.62#............#Pressure of gases at the end of expansion in high pressure turbine in bar
p8=p7-plr#...........#Pressure of outlet gases in bar
ga=1.4#...........#Ratio of specific heats for air
gag=1.33#..........#Ratio of specific heats for gases
cpa=1.005#...........#Specific heat at constant pressure in kJ/kgK for air
cpg=1.15#.........#Specific heat at constant pressure in kJ/kgK for gases
etac=0.82#........#Compressor efficiency
etat=0.85#..........#Turbine efficiency
etaht=0.65#.........#Efficiency of heat exchanger
P=6000#..................#Power generated in kW
p9=p1#
#Calculations
t2=round(t1*((p2/p1)**((ga-1)/ga)))#
t21=round(((t2-t1)/etac)+t1)#
t4=round(t3*((p4/p3)**((ga-1)/ga)))#
t41=round(((t4-t3)/etac)+t3)#
t7=round(t6/((p6/p7)**((gag-1)/gag)))#
t71=round(t6-(etat*(t6-t7)))#
t9=round(t8/((p8/p9)**((gag-1)/gag)))#
t91=round(t8-(etat*(t8-t9)))#
t5=round(etaht*(t91-t41)+t41)#
wthp=cpg*(t6-t71)#.......#Work done by high pressure turbine in kJ/kg of gas
wtlp=cpg*(t8-t9)#.......#Work done by low pressure turbine in kJ/kg of gas
wchp=cpg*(t21-t1)#.......#Work done by high pressure compressor in kJ/kg of gas
wclp=cpg*(t41-t3)#.......#Work done by low pressure compressor in kJ/kg of gas
qs=cpg*(t6-t5+t8-t71)#.........#Heat supplied in kJ/kg of gas
etath=(wthp+wtlp-wchp-wclp)/qs#..#Overall efficiency
print "Overall efficiency = %0.2f %%"%(etath*100)
wr=(wthp+wtlp-wchp-wclp)/(wthp+wtlp)#......#Work ratio
print "Work ratio  = %0.3f "%wr,
m=P/(wthp+wtlp-wchp-wclp)#.....#Mass flow rate
print "Mass flow rate = %0.2f kg/s"%m

Overall efficiency = 30.89 %
Work ratio  = 0.363  Mass flow rate = 34.32 kg/s


## EXAMPLE 21.19 PAGE 862¶

In [18]:
from __future__ import division
from math import sqrt
# Initialisation of Variables
ma=60.2#...........#Rate of air consumption in kg/s
delh=230#.......#Enthalpy change for nozzle in kJ/kg
z=0.96#..........#Velocity co efficient
afr=70#............#Air fuel ratio
etaco=0.92#...............#Combustion eficiency
CV=42000#..............#Calorific value of fuel in kJ/kg
v=1000#............#Velocity of aircraft in km/h
Ca=v*(5/18)#............#Aircraft velocity in m/s
#Calculations
Cj=z*sqrt(2*delh*v)#...........#Exit velocity of jet
print "Exit velocity of jet = %0.2f m/s"%Cj
mf=ma/afr#.........#Rate of fuel consumption
print "Rate of fuel consumption = %0.2f kg/s"%mf
tp=ma*(Cj-Ca)#......#Thrust produced in N
tsfc=mf/tp#.........#Thrust specific fuel consumption in kg/N
print "Thrust specific fuel consumption = %0.2f kg/N"%tsfc
etath=((Cj**2)-(Ca**2))/(2*(1/afr)*CV*etaco*1000)#.........#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)
pp=(ma/1000)*((Cj**2)-(Ca**2))/2#................#Propulsive power in kW
print "Propulsive power = %0.2f kW"%pp
etapp=(2*Ca)/(Cj+Ca)#......................#Propulsive efficiency
print "Propulsive efficiency = %0.2f %%"%(etapp*100)
etao=((Cj-Ca)*Ca)/((1/afr)*CV*etaco*1000)#............#Overall efficiency
print "Overall efficiency = %0.2f %%"%(etao*100)

Exit velocity of jet = 651.10 m/s
Rate of fuel consumption = 0.86 kg/s
Thrust specific fuel consumption = 0.00 kg/N
Thermal efficiency = 31.41 %
Propulsive power = 10437.94 kW
Propulsive efficiency = 59.81 %
Overall efficiency = 18.79 %


## EXAMPLE 21.20 PAGE 863¶

In [19]:
from __future__ import division
from math import pi,sqrt
# Initialisation of Variables
v=800#.............#Speed of the turbojet in km/h
etapp=0.55#......#Propulsive efficiency
etao=0.17#.........#Overall efficiency
al=9500#...............#Altitude in m
rhoa=0.17#............#Density of air at the given altitude in kg/m**3
dr=6100#...........#Drag on the plane in N
CV=46000#.........#Calorific value of fuel in kJ/kg
#Calculations
Ca=v*(1000/3600)#.........#Velocity of jet in m/s
Cj=((2*Ca)/etapp)-Ca#........#Velocity of gases at nozzle exit relative to the aircraft in m/s
print "Absolute velocity of jet = %0.2f m/s"%(Cj-Ca)
ma=dr/(Cj-Ca)#............#Rate of air flow in kg/s
Va=(ma/rhoa)*60#..........#Volume of air compresssed per min in kg
print "Volume of air compressed = %0.2f kg/min"%Va
d=sqrt((Va*4)/(60*pi*Cj))#..........#Diameter of the jet in m
print "Diameter of the jet = %0.2f mm"%(d*1000)
tp=dr*(Ca/1000)#...........#Thrust power in kW
wt=tp/etapp#................#Turbine output in kW
print "Turbine output = %0.2f kW"%wt
mf=wt/(etao*CV)#...........#Rate of fuel consumption in kg/s
afr=ma/mf#..........#Air fuel ratio
print "\nAir fuel ratio is %0.3f:1"%(afr)

Absolute velocity of jet = 363.64 m/s
Volume of air compressed = 5920.59 kg/min
Diameter of the jet = 463.09 mm
Turbine output = 2464.65 kW

Air fuel ratio is 53.225:1


## EXAMPLE 21.21 PAGE 864¶

In [20]:
from __future__ import division
from math import pi,sqrt
# Initialisation of Variables
t1=288#..........#Temperature of the inlet air into compressor in K
p1=1.01#......#Pressure of the inlet air into compressor in bar
t3=1023#.........#Maximum temperature in K
p2=4.04#.........#Pressure of air at the end of compression in bar
etac=0.82#.......#compressor efficiency
etat=0.78#......#Turbine efficiency
etan=0.88#........#Nozzle efficiency
R=0.287#.........#Gas constant for air in kJ/kgK
ga=1.4#............#Ratio of specific heats
C=42000#..........#Calorific value of fuel in kJ/kg
#Calculations
t2=t1*((p2/p1)**((ga-1)/ga))#........#Ideal temperature at the end of compression in K
t21=((t2-t1)/etac)+t1#...........#Actual temperature at the end of compression in K
cp=R*(ga/(ga-1))#..............#Specific heat at constant pressure in kJ/kgK
Pc=cp*(t21-t1)#.............#Power required to drive the compressor in kW
print "Power required to drive the compressor = %0.2f kW"%Pc
afr=((C)/(cp*(t3-t21)))-1#....#Air fuel ratio
print "\n\nAir fuel ratio %0.3f:1\n"%(afr)
t41=t1+t3-t21#......#Actual temperatur of gases leaving the turbine in K
t4=t3-((t3-t41)/etat)#......#Ideal temperature of gases leaving the turbine in K
p3=p2#
p4=p3*((t4/t3)**(ga/(ga-1)))#.......#Pressure of gases leaving the turbine in bar
print "Pressure of gases leaving the turbine = %0.2f bar"%p4
p5=p1#
t5=t41/((p4/p5)**((ga-1)/ga))#
t51=t41-(etan*(t41-t5))#
Cj=sqrt(2*cp*(t41-t51)*1000)#..............#Jet velocity in m/s
th=Cj*1#..................#Thrust per kg per second in N
print "Thrust per kg of air per second = %0.2f N"%th

Power required to drive the compressor = 171.46 kW

Air fuel ratio 73.094:1

Pressure of gases leaving the turbine = 1.74 bar
Thrust per kg of air per second = 465.69 N


## EXAMPLE 21.22 PAGE 865¶

In [21]:
from scipy import sqrt
# Initialisation of Variables
Ca=216#................#Speed of aircraft in m/s
t1=265.8#...............#Intake air temperature in K
p1=0.78#...............#Intake air pressure in bar
rp=5.8#..................#Pressure ratio in compressor
t4=1383#.................#Temperature of gases entering the gas turbine in K
pd=0.168#...............#Pressure drop in combustion chamber in bar
etan=0.9#............#Nozzle efficiency
etac=0.9#............#Compressor efficiency
etat=0.8#.............#Turbine efficiency
C=44150#............#Calorific value of fuel in kJ/kg
cp=1.005#.............#Specific heat at constant pressure in kJ/kgK
ga=1.4#...............#Ratio of specific heats
cin=0.12#...............#Inlet cross sectio of the diffuser in m**3
R=0.287#............#Gas constant in kJ/kgK
#Calculations
t2=t1+((Ca*Ca)/(2*cp*1000))#......#For ideal diffuser
p2=p1*((t2/t1)**(ga/(ga-1)))#
t3=t21*(rp**((ga-1)/ga))#
t31=t21+((t3-t21)/etac)#
afr=(C-(cp*t4))/(cp*(t4-t31))#............#Air fuel ratio
print "Air fuel ratio : %0.3f"%afr
p3=p2*rp#
p4=p3-pd#...............#Pressure of gases entering the turbine in bar
t51=t4-(t31-t21)#
t5=round(t4-((t4-t51)/etat))#
p5=p4/((t4/t5)**(ga/(ga-1)))#
p6=p1#
t6=t51/((p5/p6)**((ga-1)/ga))#
t61=t51-(etac*(t51-t6))#
Cj=44.72*sqrt(cp*(t51-t61))#........#Velocity at the exit of the nozzle in m/s
st=(1+(1/afr))*Cj#............#Specific thrust in N/kg
print "Specific thrust = %0.2f N/kg"%abs(st)
v1=Ca*cin#...........#Volume of flowing air in m**3/s
ma=(p1*v1*10**5)/(R*t1*1000)#.........#Mass flow of air
tt=ma*st#..............#Total thrust in N
print "Total thrust = %0.2f N"%abs(tt)

Air fuel ratio : 48.348
Specific thrust = 822.74 N/kg
Total thrust = 21805.03 N


## EXAMPLE 21.23 PAGE 866¶

In [22]:
from math import sqrt
# Initialisation of Variables
al=9000#..........#Altitude in m
Ca=215#...........#Speed of aircraft in m/s
TP=750#.............#Thrust power developed in kW
p1=0.32#...........#Inlet pressure of air in bar
t1=231#.............#Inlet temperature of air in K
t3=963#.............#Temperature of gases leaving the combustion chamber in K
rpc=5.2#............#Pressure ratio
C=42500#..........#Calorific value of fuel in kJ/kg
C41=195#.........#Velocity in ducts
etac=0.86#..........#Compressor efficiency
ga=1.4#............#Ratio of specific heats for air
gag=1.33#............#Ratio of specific heats for gases
etat=0.86#..........#Turbine efficiency
etajt=0.9#..........#Jet tube efficiency
cp=1.005#............#Specific heat at constant pressure in kJ/kgK for air
cpg=1.087#............#Specific heat at constant pressure in kJ/kgK for gases
R=0.29#..................#Gas constant for exhaust gases in kJ/kgK
#Calculations
t2=t1*(rpc**((ga-1)/ga))#
t21=t1+((t2-t1)/etac)#
mf=(cpg*(t3-t21))/(C-(cpg*(t3-t21)))#
afr=1/mf#..........#Air fuel ratio
t41=round(t3-((cp*(t21-t1))/(cpg*(1+mf))))#
t4=t3-((t3-t41)/etat)#
p4=rpc#
rpt=(t3/t4)**(gag/(gag-1))#.............#Expansion pressure ratio in turbine
rpj=p4/rpt#....................#Expansion pressure ratio in jet tube
t5=t41/(rpj**((gag-1)/gag))#
Cj=sqrt(etajt*2*((cpg*1000*(t41-t5))+((C41*C41)/2)))#
etao=((((1+mf)*Cj)-Ca)*Ca)/(1000*mf*C)#......#Overall efficiency
print "Overall efficiency = %0.2f %%"%(etao*100)
ma=(TP*1000)/((((1+mf)*Cj)-Ca)*Ca)#........#Rate of air consumption in kg/s
print "Rate of air consumption = %0.2f kg/s"%ma
P=ma*(1+mf)*cpg*(t3-t41)#..............#Power developed by the turbine in kW
print "Power developed by turbine = %0.2f kW"%P
t51=t41-(((Cj**2)-(C41**2))/(2*1000*cpg))#
rhoe=(p1*10**5)/(R*1000*t51)#..........#Density of exhaust gases
Ajt=(ma*(1+mf))/(Cj*rhoe)#.......#Discharge of jet area in m**2
print "The outlet area of jet tube = %0.2f m**2"%Ajt
sfc=(mf*ma*3600)/(1000*(TP/Ca))#..........#Specific fuel consumption in kg/thrust-hour
print "Specific fuel consumption = %0.2f kg per kg of thrust"%sfc

Overall efficiency = 12.91 %
Rate of air consumption = 9.23 kg/s
Power developed by turbine = 1496.76 kW
The outlet area of jet tube = 0.10 m**2
Specific fuel consumption = 0.14 kg per kg of thrust