# Chapter 2 Test¶

## Example 2.1 Page No: 44¶

In :
#Input data
d=20.0            #Cylinder bore diameter in cm
L=25.0            #Stroke length in cm
Vc=1570.0         #The clearance volume in cm**3
P1=1.0            #Pressure at the beginning of the compression in bar
T1=300.0          #Temperature at the beginning of the compression in K
T3=1673           #The maximum temperature of the cycle in K
Cv=0.718          #specific heat at constant volume for air in kJ/kgK
R=0.287           #Real gas constant in kJ/kgK
g=1.4             #Isentropic index
c=500.0           #Number of cycles per minute

#Calculations
import math
Vs=(math.pi/4.0)*d**2*L
V1=Vs+Vc
V2=Vc                                     #cm**3
r=V1/V2                                   #Compression ratio
T2=T1*r**(g-1)
P2=P1*r**g
P3=P2*(T3/T2)                             #In constant volume, process Pressure at point 3 in bar
T4=T3*(1/r)**(g-1)                        #In isentropic process, Temperature at point 4 in degree centigrade
P4=P3*(1/r)**(g)                          #In isentropic process, Pressure at point 4 in bar
no=(1-(1/r)**(g-1))*100                   #Air standard efficiency of otto cycle
Q1=Cv*(T3-T2)                                    #Heat supplied in kJ/kg
Q2=Cv*(T4-T1)                                    #Heat rejected in kJ/kg
W=Q1-Q2                                          #Work done per unit mass in kJ/kg
m=((P1*10**5*V1*10**-6)/(R*T1))/1000.0           #The amount of mass in kg
W1=W*m                                           #Work done in kJ
pm=((W1*10**3)/(Vs*10.0**-6))/10.0**5            #Mean effective pressure in N/m**2
P=W1*(c/60.0)                                    #Power developed in kW

#Output
print"Temperature at point 2 = ",round(T2-273.15,1),"C"
print"Pressure at point 2 =" ,round(P2,2)," bar"
print"Pressure at point 3 = ",round(P3,2),"bar"
print"Temperature at point 4 = ",round(T4-273.15)," C \nPressure at point 4 = ",round(P4,3),"bar"
print"Air standard efficiency of otto cycle = ",round(no,2)," percent "
print"Work done = ",round(W1,2)," kJ"
print"Mean effective pressure = ",round(pm,2),"bar "
print"Power developed = ",round(P,1),"kW "

Temperature at point 2 =  341.3 C
Pressure at point 2 = 12.29  bar
Pressure at point 3 =  33.47 bar
Temperature at point 4 =  544.0  C
Pressure at point 4 =  2.723 bar
Air standard efficiency of otto cycle =  51.17  percent
Work done =  4.26  kJ
Mean effective pressure =  5.42 bar
Power developed =  35.5 kW


## Example 2.2 Page No: 46¶

In :
#Input data
CV=42000.0          #The calorific value of the fuel in kJ/kg
pa=5.0            #Percentage of compression
Pa=1.2            #Pressure in the cylinder at 5% compression stroke
pb=75             #Percentage of compression
Pb=4.8            #Pressure in the cylinder at 75% compression stroke
g=1.3             #polytropic index
g1=1.4            #Isentropic index
n=0.6             #Air standard efficiency

#Calculations
V=(Pb/Pa)**(1/1.3)#Ratio of volumes
r=(V*(pb/100.0)-(pa/100.0))/((1-(pa/100.0))-(V*(1-(pb/100.0))))               #Compression ratio
n1=((1-(1/r)**(g1-1)))*100                                                    #Relative efficiency
nthj=n*(n1/100.0)                                                             #Indicated thermal efficiency
x=(1/(CV*nthj))*3600                                                          #Specific fuel consumption in kg/kW.h

#Output
print"The compression ratio of the engine is ",round(r,1)
print"The specific fuel consumption is ",round(x,3),"kg/kwh"

The compression ratio of the engine is  9.5
The specific fuel consumption is  0.241 kg/kwh


## Example 2.4 Page No: 48¶

In :
#Input data
d=0.2           #The diameter of the cylinder bore in m
L=0.3           #The length of the stroke in m
P1=1            #The pressure at the beginning of the compression in bar
T1=300.0        #The temperature at the beginning of the compression in K
r=16.0          #Compression ratio
V=0.08          #Cutt off takes place at 8& of the stroke
R=0.287         #Real gas constant in kJ/kgK
g=1.4           #Isentropic index
Cp=1.005        #Specific heat at constant prassure in kJ/kgK
Cv=0.718        #specific heat at constant volume for air in kJ/kgK

#Calculations
import math
Vs=(math.pi/4.0)*d**2*L  #Swept volume in m**3
Vc=Vs/(r-1)              #Clearance volume in m**3
V2=Vc                    #Volume at point 2 in m**3
V1=Vs+Vc                 #Volume at point 1 in m**3
m=(P1*10**5*V1)/(R*T1)   #The amount of mass in kg
P2=P1*(r**g)             #Pressure at point 2 in bar
P3=P2                    #Pressure at point 3 in bar
T2=T1*r**(g-1)           #Temperature at point 2 in K
V3=(V*Vs)+V2             #Volume at point 3 in m**3
C=V3/V2                  #Cut off ratio
T3=C*T2                  #Temperature at point 3 in K
P4=P3*(C/r)**g           #Pressure at the point 4 in bar
T4=T3*(C/r)**(g-1)       #Temperature at point 4 in K
V4=V1                    #Volume at point 4 in m**3
Q1=(m*Cp*(T3-T2))/1000.0 #Heat supplied in kJ
Q2=(m*Cv*(T4-T1))/1000.0 #Heat rejected in kJ
W=(Q1-Q2)                #Work done per cycle in kJ
na=(W/Q1)*100            #Air standard efficiency
Pm=(W*1000/Vs)/10.0**5   #Mean effective pressure in bar

#Output
print"(a) Volume at point 2 = ",round(V2,5)," m**3 \nVolume at point 1 = ",round(V1,5),"m**3 "
print"Pressure at point 2 = ",round(P2,1)," bar"
print"Temperature at point 2 = ",round(T2,1),"K"
print"beeta is",C,"\nTemperature at point 3 = ",round(T3,0)," K \nPressure at point 4 =",round(P4,3)," bar"
print"Temperature at point 4 = ",round(T4,1)," K \nVolume at point 4 = ",round(V4,5),"m**3"
print"(b) cut off ratio = ",round(c,1)
print"(c) Work done per cycle = ",round(W,3),"kJ"
print"(d) air smath.tandard efficiency = ",round(na,1)," percent"
print"(e)Mean effective pressure = ",round(Pm,2)," bar "

(a) Volume at point 2 =  0.00063  m**3
Volume at point 1 =  0.01005 m**3
Pressure at point 2 =  48.5  bar
Temperature at point 2 =  909.4 K
beeta is 2.2
Temperature at point 3 =  2001.0  K
Pressure at point 4 = 3.016  bar
Temperature at point 4 =  904.7  K
Volume at point 4 =  0.01005 m**3
(b) cut off ratio =  500.0
(c) Work done per cycle =  7.736 kJ
(d) air smath.tandard efficiency =  60.4  percent
(e)Mean effective pressure =  8.21  bar


## Example 2.5 Page No:50¶

In :
#given
P1=7
g=1.4
r=12
import numpy as np
from scipy.optimize import fsolve

def f(b):
return P1-1/((g-1)*(r-1))*(g*r**g*(b-1)-r*(b**1.4-1))
b = fsolve(f, 1)
f(b)
na=(1-(1/(r**(g-1)))*(((b**g)-1)/(g*(b-1))))*100  #Air standard efficiency

#Output
print"The cut off ratio = ",round(b,1)," \n The air standard efficiency =",na,"percent"
#NOTE:In the book Answer is wrong for Air standard efficiency .

The cut off ratio =  2.2
The air standard efficiency = [ 55.47110058] percent


## Example 2.6 Page no 51¶

In :
#given
m=30.0                                  #The air fuel ratio by mass
T1=300                                  #The temperature of air at the beginning of the compression in K
r=16                                    #The compression ratio
CV=42000                                #The calorific value of the fuel in kJ/kg
g=1.4                                   #Isentropic index
Cp=1.005                                #Specific heat at constant prassure in kJ/kgK

#Calculations
T2=T1*(r**(g-1))                        #Temperature at point 2 in K
T3=((1/m)*(CV/Cp))+T2                   #Temperature at point 3 in K
C=T3/T2                                 #The cut off ratio
n=(1-((1/r**(g-1))*(((C**g)-1)/(g*(C-1)))))*100#The ideal efficiency of the engine based on the air standard cycle

#Output
print" The ideal efficiency of the engine based on the air standard cycle = ",round(n,1)

 The ideal efficiency of the engine based on the air standard cycle =  58.9


## Example 2.7 Page No: 52¶

In :
#given
p1=1.0      #Inlet pressure in bar
p2=32.425   #Pressure at the end of isentropic compression in bar
r=6.0       #Ratio of expansion
r1=1.4      #Isentropic index

#Calculations
rc=(p2/p1)**(1/r1)     #Compression ratio
b=(rc/r)               #cut-off ratio
n=(1-((b**r1-1)/(rc**(r1-1)*r1*(b-1))))*100
pm=((p1*rc**r1*n/100.0*r1*(b-1))/((r1-1)*(rc-1)))

#Output
print"Air-smath efficiency is ",round(n,3),"percent"
print"Mean effective pressure is ",round(pm,3),"bar"

Air-smath efficiency is  56.671 percent
Mean effective pressure is  5.847 bar


## Example 2.8 Page No: 52¶

In :
#Input data
rc=15.0         #Compression ratio
p1=1            #Pressure at which compression begins in bar
T1=27.0+273.0   #Temperature in K
pm=60           #Maximum pressure in bar
h=2             #Heat transfered to air at constant volume is twice that at consmath.tant pressure
g=1.4           #Isentropic index
Cv=0.718        #specific heat at constant volume for air in kJ/kgK
Cp=1.005        #specific heat at constant pressure for air in kJ/kgK
R=0.287         #Real gas constant in kJ/kgK

#Calculations
T2=(T1*rc**(g-1))     #Temperature in K
p2=(p1*rc**g)         #Pressure in bar
T3=(T2*(pm/p2))       #Temperature in K
T4=(Cv*(T3-T2))/(2*Cp)+T3   #Temperature in K
b=(T4/T3)                   #Cut-off ratio
T5=(T4*(b/rc)**(g-1))       #Temperature in K
p5=(p1*(T5/T1))             #Pressure in bar
Q1=(Cv*(T3-T2))+(Cp*(T4-T3))#Heat supplied per unit mass in kJ/kg
Q2=Cv*(T5-T1)               #Heat rejected per unit mass in kJ/kg
W=(Q1-Q2)                   #Workdone in kJ/kg
n=(W/Q1)*100                #Air standard efficiency
Vs=((1*R*1000*T1)/(p1*10**5))*(1-1/rc)   #Swept volume in m**3/kg
pmean=((W*1000)/Vs)/10.0**5                #Mean-effective pressure in bar

#Output
print"(a) The pressures and temperatures at the cardinal points of the cycle are "
print"T2 =",round(T2,1)," K \np2 =", round(p2,1)," bar \nT3 = ",round(T3,1), "K  \np3 =",round(p3,1),"bar"
print"T4 =",round(T4,1),"K  \nT5 = ",round(T5,1), "K  \np5 = ",round(p5,1),"bar"
print"(b) The cycle efficiency is",round(n,0),"percent"
print"(c) The mean effective pressure of the cycle is ",round(pmean,1),"bar"

(a) The pressures and temperatures at the cardinal points of the cycle are
T2 = 886.3  K
p2 = 44.3  bar
T3 =  1200.0 K
p3 = 52.2 bar
T4 = 1312.1 K
T5 =  460.3 K
p5 =  1.5 bar
(b) The cycle efficiency is 66.0 percent
(c) The mean effective pressure of the cycle is  2.8 bar


## Example 2.9 Page No: 55¶

In :
#Input data
r=12.0     #Compression ratio
B=1.615    #Cut off ratio
p3=52.17   #Maximum pressure in bar
p4=p3      #Maximum pressure in bar
p1=1       #Initial pressure in bar
T1=(62+273)    #Initial temperature in K
n=1.35         #Indices of compression and expansion
mR=0.287       #Real gas constant in kJ/kgK
Cv=0.718       #specific heat at constant volume for air in kJ/kgK
Cp=1.005       #specific heat at constant pressure for air in kJ/kgK

#Calculations
T2=T1*r**(n-1)     #The temperature at point 2 in K
p2=p1*(r)**n       #The pressure at point 2 in bar
T3=T2*(p3/p2)      #The temperature at point 3 in K
T4=T3*B            #The temperature at point 4 in K
T5=T4*(B/r)**(n-1) #The temperature at point 5 in K
Q12=((g-n)/(g-1))*mR*((T1-T2)/(n-1))     # kJ/kg
Q23=Cv*(T3-T2)
Q34=Cp*(T4-T3)
Q45=((g-n)/(g-1))*mR*((T4-T5)/(n-1))
Q51=Cv*(T1-T5)
Q1=Q23+Q34+Q45
Q2=-Q12+(-Q51)
W=Q1-Q2
E=(W/Q1)*100
Vs=((mR*T1)/p1)*(r-1)/r                   # m**3/kg
pm=(W*1000/Vs)/10.0**3                      #Mean effective pressure in bar

#Output
print"(a)The temperature at cardinal points \nT2 =",round(T2,0)," K\nT3 = ",round(T3,0),"K \nT4 = ",round(T4,0),"K \nT5 = ",round(T5,0),"K "
print"(b) The cycle efficiency = ",round(E,1)," percent"
print"(c) The mean effective pressure of the cycle = ",round(pm,3),"bar"

(a)The temperature at cardinal points
T2 = 799.0  K
T3 =  1456.0 K
T4 =  2352.0 K
T5 =  1166.0 K
(b) The cycle efficiency =  56.9  percent
(c) The mean effective pressure of the cycle =  9.638 bar


## Example 2.10 Page No: 57¶

In :
#Input data
p1=1.0         #Inlet pressure in bar
T1=27.0+273.0    #Temperature in K
p2=4.0         #pressure at point 2 in bar
p3=16.0        #Maximum pressure in bar
Cv=0.573     #specific heat at constant volume for gas in kJ/kgK
Cp=0.761     #specific heat at constant pressure for gas in kJ/kgK

#Calculations
g=(Cp/Cv)
T2=(T1*(p2/p1)**((g-1)/g))    # K
T3=(p3/p2)*T2
T4=T3*(p1/p3)**((g-1)/g)
Q1=Cv*(T3-T2)                 #kJ/kg
Q2=Cp*(T4-T1)
W=Q1-Q2
n=(W/Q1)*100
r=(p2/p1)**(1/g)
R=(Cp-Cv)                     #kJ/kg.K
Vs=(R*1000*T1*(r-1))/(p1*10.0**5*r)       #m**3/kg
pm=(W/(Vs*100.0))

#Output
print"(a) The work done per kg of gas is ",round(W,1),"kJ/kg"
print"(b) The efficiency of the cycle is ",round(n,1),"percent "
print"(c) Mean effective pressure is ",round(pm,1),"bar"

(a) The work done per kg of gas is  306.2 kJ/kg
(b) The efficiency of the cycle is  42.2 percent
(c) Mean effective pressure is  8.4 bar