# CHAPTER 3 - Air Standard Cycles¶

## EXAMPLE 3.1 PAGE 87¶

In [5]:
from __future__ import division
# Initialisation of Variables
t1=673#.....................#Max temp in Kelvin
t3=313##...................#Min temp in Kelvin
W=130#.................#Work produced in kJ
#calculations
etath=(t1-t3)/t1#................#Engine thermal efficiency
print "Engine thermal efficiency = %0.2f  %%"%(etath*100)
dels=(ha-W)/t3#...........#Change in entropy
print "Change in entropy = %0.3f kJ/K"%dels

Engine thermal efficiency = 53.49  %
Change in entropy = 0.361 kJ/K


## EXAMPLE 3.2 PAGE 88¶

In [6]:
from __future__ import division
# Initialisation of Variables
m=0.5#.....................#Mass of air in kg
etath=0.5#.................#Thermal efficiency of engine
hie=40#...................#Heat transferred during isothermal expansion in kJ
p1=7#....................#Pressure in bar at the beginning of expansion
v1=0.12#..................#Volume in m**3 at the beginning of expansion
cv=0.721#...................#Specific heat at constant volume in kJ/kgK
cp=1.008#..................#Specific heat at constant pressure in kJ/kgK
R=287#......................#Gas constant in J/kgK
#Calculations
t1=(p1*10**5*v1)/(R*m)#....................#Max temp in K
t2=t1*(1-etath)#.......................#Min temp in K
print "The maximum temperature = %0.2f Kelvin"%t1
print "The minimum temperature = %0.2f Kelvin"%t2
from math import exp
v2=(exp((hie*1000)/(m*R*t1)))*v1#..................#Volume at the end of isothermal expansion in m**3
print "Volume at the end of isothermal expansion = %0.2f m**3"%v2
print "\n\n"
print "Process                                     Heat transfer\n"
print "_______________________________________________________________\n"
print "Isothermal expansion                               %d kJ\n"%(hie)
print "Adiabatic reversible expansion                     %d kJ\n"%(0)
print "Isothermal compression                            %d kJ\n"%(-hie)
print "Adiabatic reversible compressions                  %d kJ"%(0)

The maximum temperature = 585.37 Kelvin
The minimum temperature = 292.68 Kelvin
Volume at the end of isothermal expansion = 0.19 m**3

Process                                     Heat transfer

_______________________________________________________________

Isothermal expansion                               40 kJ

Isothermal compression                            -40 kJ



## EXAMPLE 3.3 PAGE 89¶

In [7]:
from __future__ import division
from math import log
# Initialisation of Variables
p1=18#..................#Maximum pressure in bar
t1=410+273#.............#Maximum temperature in Kelvin
ric=6#.....................#Ratio of isentropic compression
rie=1.5#.................#Ratio of isothermal expansion
v1=0.18#..................#Volume of air at the beginning of expansion
ga=1.4#...................#Degree of freedom of gas
R=287#.....................#Gas constant in J/kgK
nc=210#..................#no of working cycles
#Calculations

t4=t1/(ric**(ga-1))#.............#Min temp in K
t3=t4#
p4=p1/(ric**ga)#..................#Min pressure in bar
p2=p1/rie#.......................#pressure of gas before isentropic expansion in bar
p3=p2*((1/6)**ga)#.................#Pressure of gas after isentropic expansion in bar
print "p1=%.f bar \np2=%.f bar \np3=%.3f bar \np4=%.3f bar \nt1=t2=%0.f Kelvin \nt3=t4=%.f Kelvin \n"%(p1,p2,p3,p4,t1,t3)
dels=(p1*10**5*v1*log(rie))/(1000*t1)#....................#Change in entropy
print "Change in entropy = %0.2f kJ/K"%dels
qs=t1*dels#.......................#Heat supplied in kJ
Qr=t4*dels#.......................#Heat rejected in kJ
eta=(qs-Qr)/qs#............#Efficiency of the cycle
v3byv1=ric*rie#
Vs=(v3byv1-1)*v1#.................#Stroke volume
pm=((qs-Qr)*10**3)/(Vs*10**5)#........#Mean effective pressure of the cycle in bar
print "Mean effective pressure of the cycle = %0.2f bar"%pm
P=(qs-Qr)*(nc/60)#.........................#Power of engine
print "Mean effective pressure of the cycle = %0.2f bar"%P

p1=18 bar
p2=12 bar
p3=0.977 bar
p4=1.465 bar
t1=t2=683 Kelvin
t3=t4=334 Kelvin

Change in entropy = 0.19 kJ/K
Mean effective pressure of the cycle = 0.47 bar
Mean effective pressure of the cycle = 235.25 bar


## EXAMPLE 3.4 PAGE 92¶

In [1]:
from __future__ import division
# Initialisation of Variables
eta=1/6#...................#Efficiency of the engine
rts=70#.................#The amount of temp which is reduced in the sink in C
#Calculation
t1byt2=1/(1-eta)#
t2=(rts+273)/((2*eta*t1byt2)-t1byt2+1)#............#Temperature of the sink in K
print "Temperature of the sink = %0.f Celsius"%(t2-273)
t1=t1byt2*t2#...............#Temperature of source in K
print "temperature of source = %0.f Celsius:"%(t1-273)

Temperature of the sink = 1442 Celsius
temperature of source = 1785 Celsius:


## EXAMPLE 3.5 PAGE 92¶

In [2]:
from __future__ import division
# Initialisation of Variables
t1=1990#....................#Temperature of the heat source in K
t2=850#..................#Temperature of the sink in K
Q=32.5#...................#Heat supplied in kJ/min
P=0.4#....................#Power developed by the engine in kW
#Calculations
eta=1-(t2/t1)#..........#Efficiency of carnot engine
etath=P/(Q/60)#..................#Efficiency of the given engine
if (etath>eta):
print "Since the efficiency of the given engine is more than efficiency of carnot engine, the claim is not true."

Since the efficiency of the given engine is more than efficiency of carnot engine, the claim is not true.


## EXAMPLE 3.7 PAGE 96¶

In [3]:
from __future__ import division
# Initialisation of Variables
etaotto=0.6#............#Efficiency of otto engine
ga=1.5#.................#Ratio of specific heats
#Calculations
r=(1/(1-etaotto))**(1/(ga-1))#................#Compression ratio
print "The compression ratio of the engine is:",r

The compression ratio of the engine is: 6.25


## EXAMPLE 3.8 PAGE 96¶

In [4]:
from __future__ import division
from math import pi
# Initialisation of Variables
D=0.25#......................#Engine bore in m
L=0.375#...................#Engine stroke in m
Vc=0.00263#................#Clearence volume in m**3
p1=1#..................#Initial pressure in bar
t1=323#...................#Initial temperature in K
p3=25#....................#Max pressure in bar
ga=1.4#....................#Ratio of specific heats
#Calculations
Vs=(pi/4)*D*D*L#................#Swept volume in m**3
r=round((Vs+Vc)/Vc)#..........................#Compression ratio
etaotto=1-(1/(r**(ga-1)))#..................#Air standard efficiency of otto cycle
print "The air standard efficiency of otto cycle = %0.2f %%"%(etaotto*100)
p2=p1*((r)**ga)#
rp=p3/p2#..........................#Pressure ratio
pm=(p1*r*((r**(ga-1))-1)*(rp-1))/((ga-1)*(r-1))#................#Mean effective pressure in bar
print "Mean effective pressure = %0.2f bar"%pm

The air standard efficiency of otto cycle = 56.47 %
Mean effective pressure = 1.34 bar


## EXAMPLE 3.9 PAGE 98¶

In [12]:
from __future__ import division
# Initialisation of Variables
p1=1#.....................#Pressure in bar
t1=300#......................#Temperature in K
r=8#.......................#Compression ratio
Cv=0.72#....................#Specific heat at constant volume
ga=1.4#......................#Ratio of specific heats
#Calculations
t2=t1*(r)**(ga-1)#..........#Temperature after adiabatic compression in K
p2=p1*(r**ga)#..............#Pressure after adiabatic compression in bar
t3=(Q/Cv)+t2#.................#Temperature after isochoric compression in K
p3=(p2*t3)/t2#................#Pressure after isochoric compression in bar
t4=t3/(r**(ga-1))#.......................#Temperature after adiabatic expansion in K
p4=p3*(1/(r**(ga)))#................#Pressure after adiabatic expansion in bar
Ws=Cv*(t3-t2-t4+t1)#.........#Specific work in kJ/kg
etath=1-(1/(r**(ga-1)))#............#Thermal efficiency
print "Temperature after adiabatic compression = %0.2f K"%t2
print "Pressure after adiabatic compression = %0.2f bar"%p2
print "Temperature after isochoric compression = %0.2f K"%t3
print "Pressure after isochoric compression = %0.2f bar"%p3
print "Temperature after adiabatic expansion = %0.2f K"%t4
print "Pressure after adiabatic expansion = %0.2f bar"%p4
print "Specific work = %0.2f kJ/kg"%Ws
print "Thermal efficiency = %0.2f %%"%(etath*100)

Temperature after adiabatic compression = 689.22 K
Pressure after adiabatic compression = 18.38 bar
Temperature after isochoric compression = 2772.55 K
Pressure after isochoric compression = 73.93 bar
Temperature after adiabatic expansion = 1206.82 K
Pressure after adiabatic expansion = 4.02 bar
Specific work = 847.09 kJ/kg
Thermal efficiency = 56.47 %


## EXAMPLE 3.10 PAGE 99¶

In [13]:
from __future__ import division
# Initialisation of Variables
r=6#..............#Compression ratio
p1=1#................#Pressure after isochoric expansion in bar
t1=300#................#Temperature after isochoric expansion in K
t3=1842#...............#Temperature after isochoric compression in K
ga=1.4#...............#Ratio of specific heats
#Calculations
p2=p1*(r**ga)#...............#Pressure after adiabatic compression in bar
t2=t1*(r**(ga-1))#.............#Temperature after adiabatic compression in K
p3=p2*(t3/t2)#..................#pressure after isochoric compression in bar
t4=t3/(r**(ga-1))#..............#Temperature after adiabatic expansion in K
p4=p3*(1/(r**(ga)))#...........#Pressure after adiabatic expansion in bar
etaotto=1-(1/(r**(ga-1)))#............#Efficiency of otto cycle
p5=p1#
t5=((p5/p3)**((ga-1)/ga))*t3#................#Atkinson cycle temp after further adiabatic expansion in K
etatk=1-((ga*(t5-t1))/(t3-t2))#...........#Efficiency of atkinson cycle
print "Temperature after adiabatic compression = %0.2f K"%t2
print "Pressure after adiabatic compression = %0.2f bar"%p2
print "Temperature after isochoric compression = %0.2f K"%t3
print "Pressure after isochoric compression = %0.2f bar"%p3
print "Temperature after adiabatic expansion = %0.2f K"%t4
print "Pressure after adiabatic expansion = %0.2f bar"%p4
print "Thermal efficiency = %0.2f %%"%(etath*100)

Temperature after adiabatic compression = 614.30 K
Pressure after adiabatic compression = 12.29 bar
Temperature after isochoric compression = 1842.00 K
Pressure after isochoric compression = 36.84 bar
Temperature after adiabatic expansion = 899.56 K
Pressure after adiabatic expansion = 3.00 bar
Thermal efficiency = 56.47 %


## EXAMPLE 3.11 PAGE 101¶

In [1]:
from __future__ import division
# Initialisation of Variables
p1=1#...................#Initial pressure in bar
t1=343#..................#Initial temperature in K
Qs=465#...............#Heat addition at constant volume in kJ/kg
cp=1#.....................#Specific heat at constant pressure in kJ/kg
cv=0.706#..................#Specific heat at constant volume in kJ/kg
ga=cp/cv#.................#Ratio of specific heats
#Calculations
r=(p2/p1)**(1/ga)#...............#Compression ratio
t2=t1*(r**(ga-1))#.....................#Temperature at the end of compression in K
t3=t2+(Qs/cv)#.............#Temperature at the end of heat addition in K
print "Compression ratio:",round(r,2)
print "Temperature at the end of compression = %0.2f K"%t2
print "Temperature at the end of heat addition = %0.2f K"%t3

Compression ratio: 3.95
Temperature at the end of compression = 607.79 K
Temperature at the end of heat addition = 1266.43 K


## EXAMPLE 3.12 PAGE 102¶

In [2]:
from __future__ import division
# Initialisation of Variables
ga=1.4#..............#Ratio of specific heats
p2byp1=15#...............#Ratio pressure at the end of compression to that of pressure at the start
t1=311#................#Initial temperature in K
t3=2223#...............#Maximum temperature in K
R=0.287#...............#Gas constant in kJ/kg K
#Calculations
r=p2byp1**(1/ga)#...............#Compression ratio
etath=1-(1/(r**(ga-1)))#.............#Thermal efficiency
t2=t1*(r**(ga-1))#............#Temperature at the end of compression in K
t4=t3/(r**(ga-1))#...........#Temperature at the end of isothermal expansion in K
cv=R/(ga-1)#................#Specific heat at constant volume in kJ/kg
Q=cv*(t3-t2)#..............#Heat supplied in kJ/kg of air
Qr=cv*(t4-t1)#.................#Heat rejected in kJ/kg of air
W=Q-Qr#.................#Work done
print "Compression ratio:",round(r,2)
print "Thermal efficiency = %0.2f %%"%(etath*100)
print "Work done = %0.2f kJ"%W

Compression ratio: 6.92
Thermal efficiency = 53.87 %
Work done = 598.65 kJ


## EXAMPLE 3.13 PAGE 104¶

In [3]:
from __future__ import division
# Initialisation of Variables
v1=0.45#.............#Initial volume in m**3
p1=1#...............#Initial pressure in bar
t1=303#.............#Initial temperature in K
p2=11#...................#Pressure at the end of compression stroke in bar
Q=210#...................#heat added at constant volume in kJ
N=210#.................#No of working cycles per min
ga=1.4#.............#Ratio of specific heats
R=287#.................#Gas constant in kJ/kgK
cv=0.71#.................#Specific heat at constant volume in kJ/kg
#Calculations
r=(p2/p1)**(1/ga)#...................#Compression ratio
t2=t1*(r**(ga-1))#...................#Temperature at the end of adiabatic compression
v2=(t2*p1*v1)/(t1*p2)#.................#Volume at the end of adiabatic compression in m**3
m=(p1*v1*10**5)/(R*t1)#................#Mass of engine fluid in kg
t3=(Q/(m*cv))+t2#...................#Temperature at the end of isochoric compression in K
p3=(t3/t2)*p2#................#Pressure at the end of isochoric compression in bar
v3=v2#
t4=t3*(1/r)**(ga-1)#...................#Temperature at the end of adiabatic expansion in K
p4=p3*(1/r)**ga#......................#Pressure at the end of adiabatic expansion in bar
v4=v1#
pc=(v2*100)/(v1-v2)#..................#Percentage clearence
etaotto=1-(1/(r**(ga-1)))#........................#Efficiency of otto cycle
Qr=m*cv*(t4-t1)#...............................#Heat rejected in kJ/kg
pm=((Q-Qr)*1000)/((v1-v2)*100000)#......#Mean effective pressure in bar
P=(Q-Qr)*(N/60)#........................#Power developed in kW
print "Temperature after adiabatic compression: %0.2f K\n"%(t2)
print "Pressure after adiabatic compression: %0.2f bar\n"%(p2)
print "Volume after adiabatic compression: %0.2f m**3\n"%(v2)
print "Temperature after isochoric compression: %0.2f K\n"%(t3)
print "Pressure after isochoric compression: %0.2f bar\n"%(p3)
print "Volume after isochoric compression: %0.2f m**3\n"%(v3)
print "Temperature after adiabatic expansion: %0.2f K\n"%(t4)
print "Pressure after adiabatic expansion: %0.2f bar\n"%(p4)
print "Volume after adiabatic expansion: %0.2f m**3\n"%(v4)
print "Percentage clearance: %0.2f\n"%(pc)
print "Efficiency of otto cycle: %0.2f\n"%(etaotto*100)
print "Mean effective pressure: %0.2f bar:\n"%(pm)
print "Power developed: %0.2f kW\n"%(P)

Temperature after adiabatic compression: 601.15 K

Pressure after adiabatic compression: 11.00 bar

Volume after adiabatic compression: 0.08 m**3

Temperature after isochoric compression: 1172.73 K

Pressure after isochoric compression: 21.46 bar

Volume after isochoric compression: 0.08 m**3

Temperature after adiabatic expansion: 591.09 K

Pressure after adiabatic expansion: 1.95 bar

Volume after adiabatic expansion: 0.45 m**3

Percentage clearance: 22.01

Efficiency of otto cycle: 49.60

Mean effective pressure: 2.82 bar:

Power developed: 364.54 kW



## EXAMPLE 3.14 PAGE 106¶

In [4]:
from __future__ import division
# Initialisation of Variables
t1=310#................#Min temperature in K
t3=1220#................#Max temperature in K
ga=1.4#................#Ratio of specific heats for air
cph=5.22#............#Specific heat at constant volume for helium in kJ/kg
cvh=3.13#...............#Specific heat at constant pressure for helium in kJ/kg
#Calculations
r=(t3/t1)**(1/((ga-1)*2))#..............#Compression ratio
etaotto=1-(1/(r**(ga-1)))#................#Air standard efficiency
gah=cph/cvh#................#Ratio of specific heats for Helium
rh=(t3/t1)**(1/((gah-1)*2))#..............#Compression ratio when Helium is used
etaottoh=1-(1/(rh**(gah-1)))#................#Air standard efficiency when Helium is used
print "Air standard efficiency of the engine = %0.2f %%"%(etaotto*100)
if ((round (etaotto)- round (etaottoh)) == 0):
print "There is no change in efficiency when Helium is used as working fluid instead of air"

Air standard efficiency of the engine = 49.59 %
There is no change in efficiency when Helium is used as working fluid instead of air


## EXAMPLE 3.15 PAGE 108¶

In [5]:
from __future__ import division
from math import sqrt
# Initialisation of Variables
t1=310#.........#Minimum temperature in K
t3=1450#............#maximum temperature in K
m=0.38#...........#Mass of working fluid in kg
cv=0.71#...........#Specific heat at constant volume in kJ/kg
#Calculations
t4=sqrt(t1*t3)#............#Temperature at the end of adiabatic expansion in K
t2=t4#
W=cv*(t3-t2-t4+t1)#..................#Work done in kJ/kg
P=W*(m/60)#.................#Power developed in kW
print "Power developed = %0.2f kW"%P

Power developed = 1.88 kW


## EXAMPLE 3.17 PAGE 113¶

In [6]:
from __future__ import division
# Initialisation of Variables
r=15#...................#Compression ratio
ga=1.4#..............#Ratio os fpecific heats for air
perQ=6#................#Heat addition at constant pressure takes place at 6% of stroke
#Calculations
rho=1+((perQ/100)*(r-1))#.............#Cut off ratio
print "Efficiency of diesel engine = %0.2f %%"%(etad*100)

Efficiency of diesel engine = 61.19 %


## EXAMPLE 3.18 PAGE 114¶

In [7]:
from __future__ import division
from math import sqrt,pi
# Initialisation of Variables
L=0.25#...............#Engine stroke in m
D=0.15#..................#Engine bore in m
v2=0.0004#...............#Clearance volume in m**3
pers=5#...............#Percentage of stroke when fuel injection occurs
ga=1.4#..............#Ratio of specific heats
#Calculations
Vs=(pi/4)*D*D*L#..............#Swept volume in m**3
Vt=Vs+v2#....................#Total cylinder volume in m**3
v3=v2+((pers/100)*Vs)#..............#Volume at point of cut off
rho=v3/v2#............#Cut off ratio
r=1+(Vs/v2)#.............#Compression ratio
print "Efficiency of diesel engine = %0.2f %%"%(etad*100)

Efficiency of diesel engine = 59.34 %


## EXAMPLE 3.19 PAGE 114¶

In [8]:
from __future__ import division
# Initialisation of Variables
r=14#....................#Compression ratio
pers1=5#...............#Percentage of stroke when fuel cut off occurs
pers2=8#...............#Percentage of stroke when delayed fuel cut off occurs
v2=1#.....................#Clearance volume in m**3
ga=1.4#..................#Ratio of specific heats
#Calculations
#When the fuel is cut off at 5 %
rho1=((pers1/100)*(r-1))+1#.............#Cut off ratio
#When the fuel is cut off at 8 %
rho2=((pers2/100)*(r-1))+1#.............# Delayed Cut off ratio
etad2=1-((((rho2**ga)-1)/(rho2-1))*(1/(ga*(r**(ga-1)))))#..................#Efficiency of diesel engine when cut off ratio is deyaled
print "Percentage loss in efficiency due to delay in cut off : %0.2f"%((etad1-etad2)*100)

Percentage loss in efficiency due to delay in cut off : 2.10


## EXAMPLE 3.20 PAGE 115¶

In [9]:
from __future__ import division
from scipy.optimize import fsolve
# Initialisation of Variables
pm=7.5#.................#Mean effective pressure in bar
r=12.5#..................#Compression ratio
p1=1#....................#Initial pressure in bar
ga=1.4#.................#Ratio of specific heats
#Calculations
k=(pm*(ga-1)*(r-1))/(p1*(r**ga))#
c1=(r**(1-ga))/k#
c2=(-ga)/k#
c=1+(ga/k)-((r**(1-ga))/k)#
def F(rho):
f=c1*(rho**ga)+c2*rho+c#
return f
#Initial guess
rho=2#
#Derivative
def D(rho):
z=c1*ga*(rho**(ga-1))+c2#
return z
y=fsolve(F,rho)
perc=((y-1)/(r-1))*100#..................#Percentage of cutoff
print "Cut off Percentage : %0.2f"%perc

Cut off Percentage : 11.17


## EXAMPLE 3.21 PAGE 115¶

In [1]:
from __future__ import division
from math import pi
# Initialisation of Variables
D=0.2#.................#Engine bore in m
L=0.3#.............#Engine stroke in m
p1=1#................#Initial pressure in bar
N=380#.................#No of working cycles per min
t1=300#..............#Initial temperature in K
co=8#................#Cut off percentage
r=15#..................#Compression ratio
R=287#.................#gas constant in J/kg
ga=1.4#................#Ratio of specific heats
#Calculations
Vs=(pi/4)*D*D*L#.............#Stroke volume in m
v1=(r/(r-1))*Vs#................#Volume at the end of isochoric compression in m**3
m=(p1*v1*10**5)/(R*t1)#................#Mass of air in cylinder in kg/cycle
p2=p1*(r**ga)#.......................#Pressure at the end of isentropic compression in bar
t2=t1*(r**(ga-1))#....................#Temperature at the end of isentropic compression in K
v2=Vs/(r-1)#..................#Volume at the end of isentropic compressionin m**3
p3=p2#
rho=((r-1)*(co/100))+1#................#Cut off ratio
v3=rho*v2#.......................#Volume at the end of isobaric expansion in m**3
t3=t2*(v3/v2)#..................#Temperature at the end of isobaric expansion in K
p4=((rho/r)**ga)*p3#..............#Pressure at the end of adiabatic expansion in bar
t4=((rho/r)**(ga-1))*t3#..............#Temperature at the end of adiabatic expansion in K
v4=v1#
print "Temperature after adiabatic compression: %f K\n"%(t2)
print "Pressure after adiabatic compression: %f bar\n"%(p2)
print "Volume after adiabatic compression: %f m**3\n"%(v2)
print "Temperature after isobaric compression: %f K\n"%(t3)
print "Pressure after isobaric compression: %f bar\n"%(p3)
print "Volume after isobaric compression: %f m**3\n"%(v3)
print "Temperature after adiabatic expansion: %f K\n"%(t4)
print "Pressure after adiabatic expansion: %f bar\n"%(p4)
print "Volume after adiabatic expansion: %f m**3\n"%(v4)
print "Efficiency of diesel engine = %0.2f %%"%(etad*100)
pm=p1*(r**ga)*(ga*(rho-1)-((r**(1-ga))*((rho**ga)-1)))*(1/(ga-1))*1/(r-1)#.......#Mean effective pressure
print "Mean effective pressure : %0.2f"%pm
Wdc=(pm*Vs*10**5)/1000#..................#Work done per cycle in kJ/cycle
P=(Wdc*N)/60#...........................#Power developed in kW
print "Power developed = %0.2f kW "%P

Temperature after adiabatic compression: 886.253082 K

Pressure after adiabatic compression: 44.312654 bar

Volume after adiabatic compression: 0.000673 m**3

Temperature after isobaric compression: 1878.856533 K

Pressure after isobaric compression: 44.312654 bar

Volume after isobaric compression: 0.001427 m**3

Temperature after adiabatic expansion: 858.996630 K

Pressure after adiabatic expansion: 2.863322 bar

Volume after adiabatic expansion: 0.010098 m**3

Efficiency of diesel engine = 59.77 %
Mean effective pressure : 7.42
Power developed = 44.27 kW


## EXAMPLE 3.22 PAGE 118¶

In [2]:
# Initialisation of Variables
rc=15.3#....................#Compression ratio
re=7.5#...................#Expansion ratio
cp=1.005#.................#Specific heat at constant pressure in kJ/kg K
cv=0.718#..................#Specific heat at constant volume in kJ/kgK
ga=1.4#....................#Ratio of specific heats
p1=1#....................#Initial pressure in bar
t1=300#..................#Initial temperature in K
etamech=0.8#..................#Mechanical efficiency
C=42000#...........................#Calorific value of fuel in kJ/kg
rita=0.5#.........................#Ratio of indicated thermal efficiency to air standard efficiency
R=287#..........................#Gas constant in kJ/kgK
#Calculations
t2=t1*(rc**(ga-1))#.................#Temperature at the end of adiabatic compression in K
p2=p1*(rc**ga)#...................#Pressure at the end of adiabatic compression in bar
t3=(rc*t2)/re#....................#Temperature at the end of constant pressure process in K
v2=1#..................#Volume at the end of adiabatic process in m**3
m=(p2*v2*10**5)/(R*t2)#..................#Mass of working fluid in kg
t4=t3*((1/re)**(ga-1))#...................#Temperature at the end of adiabatic expansion in K
W=(m*(cp*(t3-t2)))-(m*(cv*(t4-t1)))#........#Work done in kJ
pm=W/(rc-1)#..............................#Mean effective pressure in kN/m**2
print "Mean effective pressure = %0.2f bar "%(pm/100)
print "Ratio of maximum pressure to mean effective pressure %0.2f"%((p2*100)/(pm))
etacy=W/(m*cp*(t3-t2))#...............#Cycle efficiency
print "Cycle efficiency = %0.2f %%"%(etacy*100)
etaith=rita*etacy#..................#Indicated thermal efficiency
etabth=etaith*etamech#...............#Brake thermal efficiency
mf=3600/(etabth*C)#................#Fuel consumption per kWh
print "Fuel consumption = %0.2f kg/kWh"%mf

Mean effective pressure = 7.02 bar
Ratio of maximum pressure to mean effective pressure 6.49
Cycle efficiency = 60.48 %
Fuel consumption = 0.35 kg/kWh


## EXAMPLE 3.23 PAGE 123¶

In [3]:
# Initialisation of Variables
Vs=0.0053#................#Swept volume in m**3
Vc=0.00035#...............#Clearance volume in m**3
v3=Vc#
v2=Vc#
p3=65#..................#Max pressure in bar
co=5#...................#Cut off percentage
p4=p3#ga=1.4#...............#Ratio of specific heats
t1=353#....................#Temperature at the start of compression in K
p1=0.9#...................#Pressure at the start of compression in bar
#Calculations
r=1+(Vs/Vc)#...................#Compression ratio
rho=(((co/100)*Vs)/Vc)+1#...................#Cut off ratio
p2=p1*(r**ga)#
Beta=p3/p2#.............................#Explosion ratio
print "Efficiency of dual cycle : %0.2f"%(etadual*100)

Efficiency of dual cycle : 63.76


## EXAMPLE 3.24 PAGE 124¶

In [4]:
# Initialisation of Variables
r=14#......................#Compression ratio
Beta=1.4#................#Explosion ratio
co=6#..................#Cut off percentage
ga=1.4#.................#Ratio of specific heats
#Calculation
rho=((co/100)*(r-1))+1#...............#Cut off ratio
print "Efficiency of dual cycle : %0.2f"%(etadual*100)

Efficiency of dual cycle : 61.42


## EXAMPLE 3.25 PAGE 124¶

In [5]:
from math import pi
# Initialisation of Variables
D=0.25#.................#Engine bore in m
L=0.3#.............#Engine stroke in m
p1=1#................#Initial pressure in bar
N=3#...............#No of cycles per second
p3=60#................#Maximum pressure in bar
t1=303#..............#Initial temperature in K
co=4#................#Cut off percentage
r=9#..................#Compression ratio
R=287#.................#gas constant in J/kg
cv=0.71#...............#Specific heat at constant volume in kJ/kgK
cp=1.0#.................#Specific heat at constant pressure in kJ/kgK
ga=1.4#...............#Ratio of specific heats
#Calculations
p4=p3#
Vs=(pi/4)*D*D*L#.............#Stroke volume in m**3
Vc=Vs/(r-1)#..................#Clearance volume in m**3
rho=((r-1)*(co/100))+1#................#Cut off ratio
v1=Vc+Vs#.................#Volume after isochoric compression in m**3
p2=p1*(r**ga)#................#Pressure after adiabatic compression in bar
t2=t1*(r**(ga-1))#..............#Temperature after adiabatic expansion in K
t3=(p3*t2)/p2#..............#Temperature after isochoric compression in K
t4=t3*rho#.....................#Temperature after isobaric expansion in K
t5=t4*((rho/r)**(ga-1))#.........#Temperature after adiabatic expansion in K
p5=p4*(rho/r)**ga#...............#Pressure after adiabatic expansion in bar
Qs=(cv*(t3-t2)+cp*(t4-t3))#.....#Heat supplied in kJ/kg
Qr=cv*(t5-t1)#...................#Heat rejected in kJ/kg
etast=1-(Qr/Qs)#.................#Air standard efficiency
print "Air standard efficiency = %0.2f %%"%(etast*100)
m=(p1*v1*10**5)/(R*t1)#...............#Mass of air in cycle
W=m*(Qs-Qr)#....................#Work done per cycle in kJ
P=W*N#............................#Power developed in kW
print "Power developed = %0.2f kW"%P

Air standard efficiency = 57.55 %
Power developed = 51.39 kW


## EXAMPLE 3.26 PAGE 127¶

In [1]:
from __future__ import division
# Initialisation of Variables
p1=1#................#Initial pressure in bar
t1=363#.............#Initial temperature in K
r=9#.................#Compression ratio
p3=68#...............#Max pressure
p4=p3#
Qs=1750#..............#Total heat supplied
ga=1.4#...............#Ratio of specific heats
R=287#................#Gas constant in kJ/kgK
cv=0.71#..............#Specific heat at constant volume in kJ/kgK
cp=1#................#Specific heat at constant pressure in kJ/kgK
#Calculations
p2=p1*((r)**ga)#............#Pressure at the end of adiabatic compression in bar
t2=t1*((r)**(ga-1))#..........#Temperature at the end of adiabatic compression in K
t3=t2*(p3/p2)#............#Temperature at the end of isochoric compression in K
Qv=cv*(t3-t2)#.............#Heat added at constant volume in kJ/kg
Qp=Qs-Qv#.....................#Heat added at constant pressure in kJ/kg
t4=(Qp/cp)+t3#................#Temperature at the end of isobaric expansion in kJ/kg
rho=t4/t3#.....................#Cut off ratio
p5=p4*((rho/r)**ga)#................#Pressure at the end of adiabatic expansion in kJ/kg
t5=t4*((rho/r)**(ga-1))#...........#Temperature at the end of adiabatic expansion in kJ/kg
print "Temperature after adiabatic compression: %0.2f K\n"%(t2)
print "Pressure after adiabatic compression: %0.2f bar\n"%(p2)
print "Temperature after isochoric compression: %0.2f K\n"%(t3)
print "Pressure after isochoric compression: %0.2f bar\n"%(p3)
print "Temperature after isobaric expansion: %0.2f K\n"%(t4)
print "Pressure after isobaric expansion: %0.2f bar\n"%(p4)
print "Temperature after adiabatic expansion: %0.2f K\n"%(t5)
print "Pressure after adiabatic expansion: %0.2f bar\n"%(p5)
Qr=cv*(t5-t1)#....................#Heat rejected in kJ
etast=1-(Qr/Qs)#.................#Air standard efficiency
print "Air standard efficiency = %0.2f %%"%(etast*100)
pm=(1/(r-1))*((68*(rho-1))+(((p4*rho)-(p5*r))/(ga-1))-((p2-r)/(ga-1)))#................#Mean effective pressure in bar
print "Mean effective pressure = %0.2f bar"%pm

Temperature after adiabatic compression: 874.19 K

Pressure after adiabatic compression: 21.67 bar

Temperature after isochoric compression: 2742.67 K

Pressure after isochoric compression: 68.00 bar

Temperature after isobaric expansion: 3166.05 K

Pressure after isobaric expansion: 68.00 bar

Temperature after adiabatic expansion: 1392.38 K

Pressure after adiabatic expansion: 3.84 bar

Air standard efficiency = 58.24 %
Mean effective pressure = 11.09 bar


## EXAMPLE 3.27 PAGE 129¶

In [2]:
from __future__ import division
# Initialisation of Variables
t1=300#...............#Initial temperature
rmami=70#....................#Ratio of max pressure and min pressure
r=15#....................#Compression ratio
ga=1.4#.................#Ratio of specific heats
R=287#....................#Gas constant in kJ/kgK
t2=t1*(r**(ga-1))#.................#Temperature at the end of adiabatic compression in K
t3=t2*(rmami/(r**ga))#............#Temperature at the end of isochoric compression in K
t4=t3+((t3-t2)/ga)#..............#Temperature at the end of isobaric process in K
t5=t4/((1/(t4/(t3*r)))**(ga-1))#..........#Temperature at the end of adiabatic expansion in K
etast=1-((t5-t1)/((t3-t2)+ga*(t4-t3)))#..............#Air standard efficiency
print "Air standard efficiency = %0.2f %%"%(etast*100,)

Air standard efficiency = 65.30 %


## EXAMPLE 3.28 PAGE 131¶

In [3]:
from __future__ import division
# Initialisation of Variables
t1=373#.............#Initial temperature in K
p1=1#...............#Initial pressure in bar
p3=65#..............#Maximum pressure in bar
R=287#.................#Gas constant in kJ/kg
p4=p3#
ga=1.41#.................#Ratio of specific heats
Vs=0.0085#............#Swept volume in m**3
afr=21#...............#Air fuel ratio
r=15#.................#Compression ratio
C=43890#..............#Calorific value of fuel in kJ/kg
cp=1#................#Specific heat at constant pressure in kJ/kgK
cv=0.71#..............#Specific heat at constant volume in kJ/kgK
#Calculations
Vc=Vs/(r-1)#...............#Clearance volume in m**3
v2=Vc
v1=Vs+v2#
v3=Vc
v5=v1#
p2=p1*(r**ga)#.....................#Pressure at the end of adiabatic compression in bar
t2=t1*(r**(ga-1))#................#Temperature at the end of adiabatic compression in K
t3=(t2*p3)/p2#...................#Temperature at the end of isochoric compression in K
m=(p1*v1*10**5)/(R*t1)#............#Mass of air in the cycle in kg
Qv=m*cv*(t3-t2)#.....................#Heat added during constant volume process in kJ
fv=Qv/C#.............................#Fuel added during constant volume process in kg
mf=m/afr#..................#Total amount of fuel added in kg
mfib=mf-fv#....................#Total amount of fuel added in isobaric process in kg
Qib=mfib*C#....................#Total amount of heat added in isobaric process in kJ
t4=(Qib/((m+mf)*cp))+t3#........#Temperature at the end of isobaric process in K
v4=(v3*t4)/t3#..................#Volume at the end of isobaric process in m**3
t5=t4/((v5/v4)**(ga-1))#.........#Temperature at the end of isochoric expansion in K
Qrv=(m+mf)*cv*(t5-t1)#...............#Heat rejected during constant volume process in kJ
W=(Qib+Qv)-Qrv#................#Work done in kJ
etath=W/(Qib+Qv)#..................#Thermal efficiency
print "Thermal efficiency = %0.2f %%"%(etath*100)

Thermal efficiency = 61.80 %


## EXAMPLE 3.29 PAGE 133¶

In [4]:
from __future__ import division
from math import pi
# Initialisation of Variables
D=0.25#.............#Engine bore in m
L=0.4#..............#Engine stroke in m
t1=303#.............#Initial temperature in K
R=287#...............#Gas constant in kJ/kgK
p1=1#...............#Initial pressure in bar
N=8#................#No of working cycles per sec
cv=0.71#.............#Specific heat at constant volume in kJ/kgK
cp=1#.................#Specific heat at constant pressure in kJ/kgK
rc=9#...............#Compression ratio
re=5#...............#Expansion ratio
rqptqe=2#...........#Ratio of heat liberated at constant pressure to heat liberated at constant volume
#Calculations
p2=p1*(rc**n)#.......................#Pressure at the end of adiabatic compression in bar
t2=t1*(rc**(n-1))#...................#Temperature at the end of adiabatic compression in K
rho=rc/re#..........................#Cut off ratio
t3=(2*cv*t2)/((2*cv)-(cp*(rho-1)))#...............#Temperature at the end of isochoric compression in K
p3=p2*(t3/t2)#....................................#Pressure at the end of isochoric compression in bar
p4=p3#t4=rho*t3#.................................#Temperature and pressure at the end of isobaric process
p5=p4*(1/(re**n))#.................................#Pressure at the end of adiabatic expansion in bar
t5=t4*(1/(re**(n-1)))#.............................#Temperature at the end of adiabatic expansion in K
pm=(1/(rc-1))*((p3*(rho-1))+(((p4*rho)-(p5*rc))/(n-1))-((p2-(p1*rc))/(n-1)))#...............#Mean effective pressure
print "Temperature after adiabatic compression: %0.2f K\n"%(t2)
print "Pressure after adiabatic compression: %0.2f bar\n"%(p2)
print "Temperature after isochoric compression: %0.2f K\n"%(t3)
print "Pressure after isochoric compression: %0.2f bar\n"%(p3)
print "Temperature after isobaric expansion: %0.2f K\n"%(t4)
print "Pressure after isobaric expansion: %0.2f bar\n"%(p4)
print "Temperature after adiabatic expansion: %0.2f K\n"%(t5)
print "Pressure after adiabatic expansion: %0.2f bar\n"%(p5)
print "Mean effective pressure = %0.2f bar"%pm
Vs=(pi/4)*D*D*L#....................#Swept volume in m**3
W=(pm*(10**5)*Vs)/1000#.................#Work done per cycle in kJ
m=(p1*(10**5)*(rc/(rc-1))*Vs)/(R*t1)#.....................#Mass of air per cycle in kg
Qs=m*(cv*(t3-t2)+cp*(t4-t3))#.....................#Heat supplied per cycle in kJ
eta=W/Qs#....................#Engine efficiency
print "Engine efficiency = %0.2f %%"%(eta*100)
P=W*N#.................#Power of the engine in kW
print "Power of the engine = %0.2f kW"%P

Temperature after adiabatic compression: 524.81 K

Pressure after adiabatic compression: 15.59 bar

Temperature after isochoric compression: 1201.99 K

Pressure after isochoric compression: 35.70 bar

Temperature after isobaric expansion: 3283.19 K

Pressure after isobaric expansion: 35.70 bar

Temperature after adiabatic expansion: 2195.60 K

Pressure after adiabatic expansion: 4.78 bar

Mean effective pressure = 10.92 bar
Engine efficiency = 32.95 %
Power of the engine = 171.53 kW


## EXAMPLE 3.31 PAGE 140¶

In [5]:
from __future__ import division
# Initialisation of Variables
cp=0.92#..................#Specific heat at constant pressure in kJ/kgK
cv=0.75#..................#Specific heat at constant volume in kJ/kgK
p1=1#...................#Pressure at the end of adiabatic expansion in bar
p2=p1#...................#Pressure at the end of isobaric compression in bar
p3=4#....................#Pressure at the end of isobaric compression in bar
p4=16#...................#Final pressure after heat addition in bar
t2=300#.....................#Temperature at the end of isobaric compression in K
ga=1.22#................#Ratio of specific heats
#Calculations
t3=t2*((p3/p2)**((ga-1)/ga))#............#Temperature at the end of isobaric compression in K
t4=(p4*t3)/p3#........................#Final temperature after heat addition in K
t1=t4/((p4/p1)**((ga-1)/ga))#...................#Temperature at the end of adiabatic compression in K
Qs=cv*(t4-t3)#.........................#Heat supplied in kJ/kg
Qr=cp*(t1-t2)#.........................#Heat rejected in kJ/kg
W=Qs-Qr#.......................#Work done per kg of gas in kJ
print "Work done = %0.2f kJ/kg"%W
eta=W/Qs#......................#Efficiency of cycle
print "Efficiency of cycle = %0.2f %%"%(eta*100)

Work done = 282.90 kJ/kg
Efficiency of cycle = 32.64 %


## EXAMPLE 3.32 PAGE 145¶

In [6]:
from __future__ import division
# Initialisation of Variables
p1=101.325#....................#Pressure of intake air in kPa
t1=300#.......................#Temperature of intake air in kPa
rp=6#.........................#Pressure ratio in the cycle
ga=1.4#.........................#Ratio of specific heats
rtc=2.5#...........................#Ratio of turbine work and compressor work
#Calculations
t2=t1*(rp**((ga-1)/ga))#..................#Temperature at the end of isentropic expansion in K
t3=(rtc*(t2-t1))/(1-(1/(rp**((ga-1)/ga))))#........#Temperature at the end of isobaric expansion in K
t4=t3/(rp**((ga-1)/ga))#.......................#Temperature at the end of isentropic compression in K
eta=(t3-t4-t2+t1)/(t3-t2)#...................#Cycle efficiency
print "Cycle efficiency = %0.2f %%"%(eta*100)

Cycle efficiency = 40.07 %


## EXAMPLE 3.33 PAGE 147¶

In [7]:
from __future__ import division
# Initialisation of Variables
p1=1#....................#Intake pressure in bar
p2=5#....................#Supply pressure in bar
t3=1000#..................#Supply temperature in Kelvin
cp=1.0425#................#Specific heat at constant pressure in kJ/kgK
cv=0.7662#.................#Specific heat at constant volume in kJ/kgK
ga=cp/cv#..................#Ratio of specific heats
#Calculations
t4=t3*((p1/p2)**((ga-1)/ga))#
P=cp*(t3-t4)#.....................#Power developed per kg of gas per second in kW
print "Power developed per kg of gas per second = %0.2f kW "%P

Power developed per kg of gas per second = 362.01 kW


## EXAMPLE 3.34 PAGE 147¶

In [8]:
from __future__ import division
# Initialisation of Variables
ma=0.1#...................#Air supplied in kg/s
p1=1#.....................#Supply pressure in bar
t4=285#.................#Temperature of air when supplied to cabin in K
p2=4#...................#Pressure at inlet to turbine in bar
cp=1.0#..................#Specific heat at constant pressure in kJ/kgK
ga=1.4#..................#Ratio of specific heats
#Calculations
t3=t4*((p2/p1)**((ga-1)/ga))#................#Temperature at turbine inlet in K
print "Temperature at turbine inlet = %0.2f K"%t3
P=ma*cp*(t3-t4)#...........................#Power developed in kW
print "Power developed = %0.2f kW"%P

Temperature at turbine inlet = 423.51 K
Power developed = 13.85 kW


## EXAMPLE 3.35 PAGE 148¶

In [9]:
from __future__ import division
# Initialisation of Variables
p1=1#......................#Pressure of air entering the compressor in bar
p2=3.5#.................#Pressure of air while leaving the compressor in bar
t1=293#..................#Temperature of air at the onlet of the compressor in K
t3=873#.................#Temperature of air at the turbine inlet in K
cp=1.005#...............#Specific heat at constant pressure in kJ/kgK
ga=1.4#...................#Ratio of specific heats
#Calculations
rp=p2/p1#....................#Pressure ratio of the cycle
eta=1-(1/(rp**((ga-1)/ga)))#..............#Efficiency of the cycle
print "Efficiency of the cycle : %0.2f"%(eta*100)
t2=t1*((rp**((ga-1)/ga)))#................#Temperature of air while leaving the compressor in K
q1=cp*(t3-t2)#................#Heat supplied to the air in kJ/kg
print "Heat supplied to the air = %0.2f kJ/kg "%(q1)
W=eta*q1#........................#Work available at the shaft in kJ/kg
print "Work available at the shaft = %0.2f kJ/kg "%W
q2=q1-W#................#Heat rejected in the cooler in kJ/kg
print "Heat rejected in the cooler = %0.2f kJ/kg "%q2
t4=t3/(rp**((ga-1)/ga))#.......................#Temperature of air leaving the turbine in K
print "Temperature of air leaving the turbine = %0.2f K "%t4

Efficiency of the cycle : 30.09
Heat supplied to the air = 456.17 kJ/kg
Work available at the shaft = 137.25 kJ/kg
Heat rejected in the cooler = 318.92 kJ/kg
Temperature of air leaving the turbine = 610.33 K


## EXAMPLE 3.36 PAGE 149¶

In [10]:
from __future__ import division
# Initialisation of Variables
p1=1#...................#Pressure of air entering the compressor in bar
t1=300#.................#Temperature of air entering the compressor in bar
rp=6#...................#Pressure ratio
rtc=2.5#.................#Ratio of turbine work to compressor work
ga=1.4#............#Ratio of specific heats
#calculations
t2=t1*(rp**((ga-1)/ga))#..................#Temperature at the end of isentropic expansion in K
t3=(rtc*(t2-t1))/(1-(1/(rp**((ga-1)/ga))))#........#Temperature at the end of isobaric expansion in K
t4=t3/(rp**((ga-1)/ga))#.......................#Temperature at the end of isentropic compression in K
eta=(t3-t4-t2+t1)/(t3-t2)#...................#Cycle efficiency
print "Maximum temperature = %0.2f K "%t3
print "Cycle efficiency = %0.2f %%"%(eta*100)

Maximum temperature = 1251.38 K
Cycle efficiency = 40.07 %


## EXAMPLE 3.37 PAGE 150¶

In [11]:
from __future__ import division
from math import sqrt
# Initialisation of Variables
t1=303#........................#Min temperature in K
t3=1073#........................#Max temperature in K
C=45000#.....................#Calorific value of fuel in kJ/kg
cp=1#....................#Specific heat at constant pressure in kJ/kgK
ga=1.4#........................#Ratio os specific heats
diftc=100#..................#Difference between work done by turbine and compressor in kW
#Calculations
t2=sqrt(t1*t3)# t4 = t2#.....#Assumed
mf=diftc/(C*(1-((t4-t1)/(t3-t2))))#................#Fuel used in kg per second
print "Rate of fuel consumption = %0.2f kg/s "%mf
ma=(diftc-(mf*(t3-t4)))/((t3-t4-cp*(t2-t1)))#............#Rate of air consumption in kg/s
print "Mass flow rate of air = %0.2f kg/s "%(ma)

Rate of fuel consumption = 0.02 kg/s
Mass flow rate of air = 1.68 kg/s


## EXAMPLE 3.38 PAGE 151¶

In [12]:
from __future__ import division
# Initialisation of Variables
t1=300#.................#Inlet temperature in K
p1=1#....................#Inlet pressure in bar
ma=1#....................#Mass of air in kg
rp=6.25#.............#Pressure ratio
t3=1073#.....#Maximum temperature in K
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
t2=t1*(rp**((ga-1)/ga))#...........#Ideal Temperature of air while leaviing the compressor in K
t21=((t2-t1)/etac)+t1#............#Actual Temperature of air while leaviing the compressor in K
Wcomp=ma*cp*(t21-t1)#.............#Compressor work in kJ/kg
t4=t3/(rp**((ga-1)/ga))#........#Ideal temperature of air while leaving the turbine in K
t41=t3-(etat*(t3-t4))#..........#Actual temperature of air while leaving the turbine in K
Wtur=ma*cp*(t3-t41)#..............#Turbine work in kJ/kg
Wnet=Wtur-Wcomp#.................#Net work produced in kJ/kg
Qs=ma*cp*(t3-t21)#.................#Heat supplied in kJ/kg
print "Compressor work = %0.2f kJ/kg "%(Wcomp)
print "Turbine work = %0.2f kJ/kg "%Wtur
print "Heat supplied = %0.2f kJ/kg "%Qs
print "Cycle efficiency = %0.2f %%"%((Wnet/Qs)*100)
print "Actual exhaust temperature of turbine = %0.2f K"%t41

Compressor work = 259.32 kJ/kg
Turbine work = 351.64 kJ/kg
Heat supplied = 517.54 kJ/kg
Cycle efficiency = 17.84 %
Actual exhaust temperature of turbine = 723.11 K


## EXAMPLE 3.39 PAGE 153¶

In [13]:
# Initialisation of Variables
etat=0.85#..............#Turbine efficiency
etac=0.8#...............#Compressor efficiency
t3=1148#................#Max temperature in K
t1=300#................#Temperature of working fluid when entering the compressor in Kelvin
cp=1#...................#specific heat at constant pressure in kJ/kgK
ga=1.4#................#ratio of specific heats
p1=1#...................#Pressure of working fluid while entering the compressor in bar
rp=4#...................#Pressure ratio
C=42000#...............#Calorific value of fuel used in kJ/kgK
perlcc=10#.............#Percentage loss of calorific value in combustion chamber

#calculations
p2=p1*rp#.................#pressure of air while leaving the compressor in bar
etacc=1-(perlcc/100)#............#efficiency of combustion chamber
t2=t1*(rp**((ga-1)/ga))#...........#Ideal Temperature of air while leaviing the compressor in K
t21=((t2-t1)/etac)+t1#............#Actual Temperature of air while leaviing the compressor 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 3.40 PAGE 154¶

In [14]:
from math import log
# Initialisation of Variables
p1=1#...........#pressure before isothermal compression in bar
t1=310#.........#temperature before isothermal compression in K
p3=16#.........#pressure before isothermal expansion in bar
t3=930#.........#temperature before isothermal expansion in K
R=287#.............#Gas constant in kJ/kgK
#Calculations
v1=(R*t1)/(p1*10**5)#...............#Volume before isothermal compression in m**3
v3=(R*t3)/(p3*10**5)#...............#Volume before isothermal expansion in m**3
v2=v3#v4=v1#.................#2-3 and 1-4 are isochoric processes
r=v1/v2#...................#Compression ratio
q12=R*t1*log(r)#...............#Work done and heat rejected in process 1-2
w12=q12#
print "Work done in process 1-2 = %0.2f kJ/kg "%(q12/1000)
print "Heat rejected in process 1-2 = %0.2f kJ/kg "%(w12/1000)
q23=0#w23=q23#..................#COnstant volume process and hence work done is zero
print "Work done in process 2-3 = %0.2f kJ/kg "%(q23/1000)
print "Heat rejected in process 2-3 = %0.2f kJ/kg "%(q23/1000)
q34=R*t3*log(r)#...............#Work done and heat rejected in process 1-2
w34=q34#
print "Work done in process 3-4 = %0.2f kJ/kg "%(q34/1000)
print "Heat rejected in process 3-4 = %0.2f kJ/kg"%(w34/1000)
q41=q34-q12
w41=q41#
print "Work done in process 4-1 = %0.2f kJ/kg "%(q41/1000)
print "Heat rejected in process 4-1 = %0.2f kJ/kg "%(w41/1000)
etath=w41/q34#.....................#Thermal efficiency
print "Thermal efficiency of the cycle = %0.2f %%"%(etath*100)

Work done in process 1-2 = 148.93 kJ/kg
Heat rejected in process 1-2 = 148.93 kJ/kg
Work done in process 2-3 = 0.00 kJ/kg
Heat rejected in process 2-3 = 0.00 kJ/kg
Work done in process 3-4 = 446.80 kJ/kg
Heat rejected in process 3-4 = 446.80 kJ/kg
Work done in process 4-1 = 297.87 kJ/kg
Heat rejected in process 4-1 = 297.87 kJ/kg
Thermal efficiency of the cycle = 66.67 %