# Chapter 4:Fuel Air Cycles and their analysis¶

## Example 4.1 Page No 117¶

In [3]:
#given
r=8.5                                #The compression ratio
sv=1.4                               #The specific heat at constant volume in percent

#Calculations
import math
n=1-(1/r)**(sv-1)
ef=(((1-n)/n)*(sv-1)*(math.log(r))*(sv/100.0))*100

#Output
print"The efficiency decreases by ",round(ef,3),"percent"

The efficiency decreases by  0.885 percent


## Example 4.2 Page No 118¶

In [5]:
#given
r=18.0                                      #The compression ratio
l=6.0                                       #The cut off taking place corresponding of the stroke in percent
sc=2.0                                      #The specific heat at constant volume increases in percent
cv=0.717                                  #The specific heat at constant volume in kJ/kgK
R=0.287                                   #Gas constant in kJ/kgK

#Calculations
import math
Vs=(r-1)
B=((l/100.0)*Vs)+1
cp=cv+R
R1=cp/cv
n=1-(((((1/r)**(R1-1))*(B**R1-1))/(R1*(B-1))))
dn=(((1-n)/n)*((R1-1)*((math.log(r))-(((B**R1)*math.log(B))/(B**R1-1))+(1/B)))*(sc/100.0))*100

#Output
print"The efficiency decreases by ",round(dn,1),"percent"

The efficiency decreases by  1.1 percent


## Example 4.3 Page No 120¶

In [4]:
#given
r=8                                    #The compression ratio
af=15                                  #Air/fuel ratio
p1=1                                   #The pressure at the beginning of a compression stroke in bar
t=60                                   #The temperature at the beginning of a compression stroke in degree centigrade
cv=44000                               #The calorific value of the fuel in kJ/kg
n=1.32                                 #The index of the compression
Cv=0.717                               #specific heat at constant volume in kJ/kgK

#Calculations
T1=t+273
p2=p1*(r)**n
T2=T1*r**(n-1)
f=(1/(af+1))
a=(af/(af+1))
q23=cv/(af+1)
T3=((-10430+((10430)**2+(4*494.8*10**5))**(1/2.0))/2.0)
p3=(T3/T1)*(r)*p1
T31=(q23/Cv)+T2
p31=(T31/T1)*r*p1

#Output
print"(a) The Maximum temperature in the cylinder  = ",round(T3,0),"K"
print"The Maximum pressure in the cylinder P3 = ",round(p3,0),"bar"
print"(b)With constant value of Cv "
print"The Maximum temperature in the cylinder  = ",round(T31,0),"K"
print"The Maximum pressure in the cylinder P3 = ",round(p31,1),"bar"

(a) The Maximum temperature in the cylinder  =  3541.0 K
The Maximum pressure in the cylinder P3 =  85.0 bar
(b)With constant value of Cv
The Maximum temperature in the cylinder  =  4483.0 K
The Maximum pressure in the cylinder P3 =  107.7 bar


## Example 4.4 Page No 121¶

In [10]:
#given
r=21                                     #The compression ratio
af=29                                    #Air/fuel ratio
T=1000                                   #The temperature at the end of compression in K
cv=42000                                 #The calorific value of the in kJ/kg
R=0.287                                  #Gas constant in kJ/kgK

#Calculations
q23=cv/(af+1)
T3=(-0.997+(((0.997)**2)+(4*2411*14*10**-6))**(1/2.0))/(28.0*10.0**-6)
V3=(T3/T)
Vs=(r-1)
V=V3-1
pc=(V/Vs)*100

#Output
print"The percentage of stroke at which combustion is complete = ",round(pc,3),"percent"

The percentage of stroke at which combustion is complete =  6.706 percent


## Example 4.5 Page No 122¶

In [32]:
#given
r=16.0                                         #The compression ratio
l=6.0                                          #The cut-off of the stroke in percent
p3=70.0                                        #The maximum pressure obtained in bar
p1=1.0                                         #The pressure at the beginning of compression in bar
T1=(100.0+273.0)                                 #The temperature at the beginning of compression in K
R=0.287                                      #Gas constant in kJ/kgK
g=1.4                                        #Assume the isentropic index

#Calculations
T2=T1*(r)**(g-1)
Cv=(1/(T2-T1))*(0.716+125*10**-6*(T2**2-T1**2))
Cp=Cv+R
g1=Cp/Cv
T21=T1*(r)**(g1-1)
Cv1=(1/(T21-T1))*((0.716+125*10**-6*(T21**2-T1**2)))
Cp1=Cv1+R
g2=Cp1/Cv1
gm=1.358                       #mean value
T22=T1*(r)**(gm-1)
p2=(T22/T1)*r*p1
T3=(p3/p2)*T22
V=((l/100.0)*(r-1))+1
T4=(V)*T3
p4=p3
g3=1.3
V5=r/V
T5=T4*(1/V5)**(g3-1)
Cv2=((0.716*(T5-T4))+(62.5*10**-6*(T5**2-T4**2)))/(T5-T4)
Cp2=Cv2+R
g4=Cp2/Cv2
T51=T4*(1/V5)**(g4-1)
Cv3=((0.716*(T51-T4))+(62.5*10**-6*(T51**2-T4**2)))/(T51-T4)
Cp3=Cv3+R
g5=Cp3/Cv3
T52=T4*(1/V5)**(g5-1)
p5=(T52/T1)*p1

#Output
print"The pressure and temperature at all points of the cycle \nat point 2: Temperature T2 = ",round(T22,0),"K and Pressure P2 = ",round(p2,2)," bar"
print"at point 3 :Temperature T3 = ",round(T3,1)," K  and Pressure P3 = ",p3," bar"
print"at point 4 : Temperature T4 = ",T4," K    and Pressure P4 = ",p4,"bar"
print"at point 5 :Temperature T5 = ",round(T52,0)," K  and Pressure P5 = ",round(p5,2),"bar"

The pressure and temperature at all points of the cycle
at point 2: Temperature T2 =  1006.0 K and Pressure P2 =  43.17  bar
at point 3 :Temperature T3 =  1631.9  K  and Pressure P3 =  70.0  bar
at point 4 : Temperature T4 =  3100.5625  K    and Pressure P4 =  70.0 bar
at point 5 :Temperature T5 =  1698.0  K  and Pressure P5 =  4.55 bar


## Example 4.6 Page No 125¶

In [8]:
#given
r=8                                                #Compression ratio
lcv=44000                                          #The lower heating value of the fuel in kJ/kg
af=15                                              #The air/fuel ratio
Cv=0.71                                            #The specific heat at constant volume in kJ/kgK
p=1                                                #The pressure at the beginning of the compression in bar
t=60                                               #The temperature at the beginning of the compression in degree centigrade
Mo=32                                              #Molecular weight of oxygen
Mn=28.161                                          #Molecular weight of nitrogen
Mh=18                                              #Molecular weight of water
n=1.3                                              #Polytrpic index

#Calculations
T1=(t+273)
sa=(12.5*(Mo+(3.76*Mn)))/((12*8)+(1*Mh))
Y=af*(((12*8)+(1*Mh))/(Mo+(3.76*Mn)))
x=(12.5-Y)*2
nb=1+Y+(Y*3.76)
na=x+7.8+9+46.624
Me=((na-nb)/nb)*100
T2=T1*(r)**(n-1)
T3=(lcv/(af+1))*(1/Cv)+(T2)
p3=r*(T3/T1)*p
p31=p3*(na/nb)

#Output
print"The percentage molecular expansion is ",round(Me,0),"percent"
print"(a) Without considering the molecular expansion "
print" The maximum temperature is ",round(T3,0),"K"
print"The maximum pressure is ",round(p3,0),"bar"
print"(b) With molecular expansion "
print"The maximum temperature is ",round(T3,0),"K"
print"The maximum pressure is ",round(p31,1),"bar"

The percentage molecular expansion is  6.0 percent
(a) Without considering the molecular expansion
The maximum temperature is  4495.0 K
The maximum pressure is  108.0 bar
(b) With molecular expansion
The maximum temperature is  4495.0 K
The maximum pressure is  114.4 bar


## Example 4.7 Page No 128¶

In [11]:
#given
f=0.03                          #The residual fraction of an engine
e=1.2                           #The equivalence ratio
F=0.0795                        #Fuel/air ratio for corresponding equivalence ratio

#Calculations

T=1+F
fa=1-f
ff=F*(fa)
ra=f
rf=ra*F

#Output
print"Fresh air = ",fa,"kg"
print"Fresh fuel = ",ff,"kg"
print"Air in residual = ",ra,"kg"
print"Fuel in residual = ",rf,"kg"

Fresh air =  0.97 kg
Fresh fuel =  0.077115 kg
Air in residual =  0.03 kg
Fuel in residual =  0.002385 kg


## Example 4.8 Page No 128¶

In [14]:
#given
T=800                          #The given temperature in K
e=1                            #The equivalence ratio
hi=154.723                     #Sensible Enthalpy for isooctane at 800 K in MJ/kmol
ho=15.841                      #Sensible Enthalpy for oxygen at 800 K in MJ/kmol
hn=15.046                      #Sensible Enthalpy for nitrogen at 800 K in MJ/kmol
nc=0.00058                     #Number of kmoles of C8H18 for equivalence ratio for 1 kg of air
no=0.00725                     #Number of kmoles of oxygen for equivalence ratio for 1 kg of air
nn=0.0273                      #Number of kmoles of nitrogen for equivalence ratio for 1 kg of air
R=8.314                        #Gas constant in kJ/kgK

#Calculations

Hs=(nc*hi)+(no*ho)+(nn*hn)
Hs1=Hs*1000
n=nc+no+nn
Us=Hs-(n*R*10**-3*(T-298))
Us1=Us*1000

#Output
print"Total sensible enthalpy of reactants = ",round(Hs1,3),"kJ/kg air"
print"Sensible internal energy of reactants = ",round(Us1,3),"kJ/kg air"

Total sensible enthalpy of reactants =  615.342 kJ/kg air
Sensible internal energy of reactants =  468.723 kJ/kg air


## Example 4.9 Page No 131¶

In [8]:
#given
T=500.0                                 #The given temperature in K
e=1.0                                   #Equivalence ratio
Ai=0.0662                             #The amount of isooctane for 1 kg of air in kg
Ta=298.0                                #Consider the ambient temperature in K
R=8.314                               #Gas constant in kJ/kgK

#Calculations
import math
E=((0.0662*((0.44*math.log(T/Ta))+(3.67*10**-3*(T-Ta))))+((0.921*math.log(T/Ta))+(2.31*10**-4*(T-Ta))))*1000
Ri=Ri/114.0
W=(0.5874-(0.662*Ri*math.log(T/Ta))-(0.287*math.log(T/Ta)))*1000

#Output
print"The isentropic compression functions at 500 K for the unburned"
print"isooctsne-air mixture are ",round(E,1),"J/kg air and",round(W,1),"J/kg air"

The isentropic compression functions at 500 K for the unburned
isooctsne-air mixture are  587.4 J/kg air and 438.9 J/kg air


## Example 4.10 Page No 133¶

In [13]:
#given
r=7.8                               #Compression ratio
p=1.0                               #The pressure at the start of compression in atm
T1=335.0                              #The temperature at the start of compression in K
W1=100                              #Isentropic compression function for T1 in J/kg air K
T2=645                              #The temperature corresponding to isentropic compression function in J/kg air K
U1=35                               #Internal energy corresponding to temp T1 in kJ/kg air
U2=310                              #Internal energy corresponding to temp T2 in kJ/kg air
E1=120                              #Isentropic compression function at T1
E2=910                              #Isentropic compression function at T2

#Calculations
import math
W2=W1-(292*math.log(1/r))
V1=(292*T1)/(p*10.0**5)
p2=p*(T2/T1)*r
V2=V1/r
W=U2-U1
p21=(math.exp((E2-E1)/292.0))

#Output
print"(a)At the end of the compression stroke"
print"The temperature is ",T2,"K"
print"The pressure is ",round(p2,0),"atm"
print"The volume per unit mass of air is ",round(V2,3),"m**3/kg air"
print"The pressure is ",round(p21,0),"atm"
print"(b)The work input during compression is ",W,"kJ/kg air"

(a)At the end of the compression stroke
The temperature is  645 K
The pressure is  15.0 atm
The volume per unit mass of air is  0.125 m**3/kg air
The pressure is  15.0 atm
(b)The work input during compression is  275 kJ/kg air


## Example 4.11 Page No 134¶

In [22]:
#given
p=65                      #The pressure in the cylinder in bar
r=10                      #The compression ratio
V3=0.1                    #The volume per unit mass of air at the start of expansion in m**3/kg air
p3=p*100                  #The pressure in the cylinder after the completion of combustion in kN/m**2
T3=2240                   #The temperature from the chart corresponding to p3,V3 in K
u3=-1040                  #The energy from the chart in kJ/kg air
s3=8.87                   #The entropy from the chart in kJ/kg air K
T4=1280                   #The temperature from the chart corresponding to p4,V4 in K
u4=-2220                  #The energy from the chart in kJ/kg air
p4=4.25                   #The pressure from the chart in bar

#Calculations
s4=s3
V4=r*V3
W=-(u4-u3)

#Output
print"(a)At the end of expansion stroke"
print"The pressure is ",p4,"bar"
print"The temperature is ",T4,"K"
print"The volume is ",V4,"m**3/kg air"
print"(b)The work during the expansion stroke is ",W,"kJ/kg air"

(a)At the end of expansion stroke
The pressure is  4.25 bar
The temperature is  1280 K
The volume is  1.0 m**3/kg air
(b)The work during the expansion stroke is  1180 kJ/kg air


## Example 4.12 Page no 137¶

In [31]:
#Given
#From Table 4.4
H1=-224.1*1000     #MJ/mol, Enthalpy of C8H18
H2=-393.52*1000    #Enthalpy of CO2
H3=-241.82*1000    #Enthalpy of H2O
U1=-204.1*1000     #MJ/mol, Internal energyof C8H18
U2=-393.52*1000    #Internal energy of CO2
U3=-240.6*1000    #Internal energy of H2O

M1=114.0       #g, molecular wt of C8H18
M2=32.0       #g, molecular wt of O2
M3=28.0       #g, molecular wt of N2
M4=44.0       #g, molecular wt of CO2
M5=18.0       #g, molecular wt of H2O
#For 1 kg air, from the eq.,  the fraction of wt are
x1=0.0661
x2=0.232
x3=0.768
x4=0.204
x5=0.094

#Calculation
import sympy
f=sympy.Symbol("f")
n1=(x1/M1)*(1-f)        #No. of kmoles of C8H18
n2=x2/M2*(1-f)
n3=x3/M3
n4=x4/M4*f
n5=x5/M5*f
N1=(n1*H1+n4*H2+n5*H3)
N2=n1*U1+n4*U2+n5*U3

#Result
print "Standard Enthalpy of formation is",N1
print "Internal energy of formation is",N2

Standard Enthalpy of formation is -2957.40091174907*f - 129.938684210526
Internal energy of formation is -2962.62629186603*f - 118.342192982456


## Example 4.13 Page No 138¶

In [14]:
#given
Tu=645.0                                #The temperature at the end of compression process in K
usu=310.0                               #The internal energy at the end of compression process in kJ/kg air
pu=(15.4*1.013)                       #The pressure at the end of the compression process in bar
Vu=0.124                              #The volume at the end of the compression process in m**3/kg air
e=1.0                                   #Equivalence ratio
f=0.065                               #Burned gas fraction
Tb=2820.0                               #The temperature for constant volume
pb=6500.0                               #The pressure for constant volume
hsu=440.0                               #The enthalpy from chart corresponding to temp Tu in kJ/kg air
pb1=1560.0                              #The pressure for constant pressure adiabatic combustion in kN/m**2
ub1=-700.0                              #Trail and error along the pb internal energy in kJ/kg air
Tb1=2420.0                              #The temperature for constant pressure

#Calculations
ufu=-118.5-(2963*f)
ub=usu-ufu
Vb=Vu
hfu=-129.9-(2958*f)
hb=hsu+hfu
vb1=(118-ub1)/pb

#Output
print"The temperature is ",Tb,"K"
print"The pressure is ",pb,"kN/m**2"
print"The temperature is ",Tb1,"K"
print"The pressure is ",pb1,"kN/m**2"

(a)For constant volume adiabatic combustion
The temperature is  2820.0 K
The pressure is  6500.0 kN/m**2
The temperature is  2420.0 K
The pressure is  1560.0 kN/m**2


## Example 4.14 Page No 139¶

In [4]:
#given
r=8.0                               #The compression ratio
T1=350.0                            #The given temperature at the start of compression in K
p=1.0                               #The given pressure at the start of compression in bar
f=0.08                              #The exhaust residual fraction
cv=44000                            #The calorific value in kJ/kg
W1=150                              #Isentropic compression functions for corresponding temp T1 in J/kg air K
T2=682                              #The temperature corresponding to isentropic compression function in K
us2=350                             #The internal energy corresponding to temp T2 in K
us1=40                              #The internal energy corresponding to temp T1 in K
T3=2825                             #The temperature at point 3 corresponding to u3,V3 on the burned gas chart in K
p3=7100                             #The pressure at point 3 in kN/m**2
s3=9.33                             #Entropy at point 3 in kJ/kg air K
u4=-1540                            #The internal energy at point 4 corresponding to V4,s4 in kJ/kg air
p4=570                              #The pressure at point 4 in kN/m**2
T4=1840                             #The temperature at point 4 in K

#Calculations
import math
W2=W1-(292*math.log(1/r))
V1=(292*T1)/(p*10.0**5)
p2=p*(T2/T1)*r
V2=V1/r
Wc=us2-us1
ufu=-118.5-(2963*f)
u3=us2+ufu
V3=V2
s4=s3
V4=V1
We=u3-u4
Wn=We-Wc
nth=((Wn)/((1-f)*0.0662*cv))*100
imep=((Wn*1000)/(V1-V2))/10.0**5
nv=(((1-f)*287*298)/(1.013*10**5*(1-0.125)))*100

#Output
print"(a)At point 2, \nThe temperature is ",T2," K \nThe pressure is ",round(p2,1),"atm"
print"At point 3, \nThe temperature is ",T3," K \nThe pressure is ",p3,"kN/m**2"
print"At point 4, \nThe temperature is ",T4," K \nThe pressure is ",p4,"kN/m**2"
print"(b)The indicated thermal efficiency is ",round(nth,1),"percent"
print"(c)The indicated mean effective pressure is ",round(imep,0)," bar"
print"(d)The volumetric efficiency is ",round(nv,2), "percent"

(a)At point 2,
The temperature is  682  K
The pressure is  15.6 atm
At point 3,
The temperature is  2825  K
The pressure is  7100 kN/m**2
At point 4,
The temperature is  1840  K
The pressure is  570 kN/m**2
(b)The indicated thermal efficiency is  45.7 percent
(c)The indicated mean effective pressure is  14.0  bar
(d)The volumetric efficiency is  88.77 percent