Chapter 14 Air Standard Cycles¶

Example 1 Page No:302¶

In [1]:
#Input data
Tmax=477+273        #Temperature limits for the engine 477 degree celcius
Tmin=27+273         #Temperature limits for the engine 27 degree celcius
wd=150              #Carnot cycle produce in KJ

#Calculatkion
eta=(1-(Tmin/Tmax)) #Thermal efficiency of the carnot cycle in %
Qs=(wd/eta)         #Added during the process in Kj

#Output
print("thermal efficiency of the carnot cycle eta=",100*(eta),"%")

thermal efficiency of the carnot cycle eta= 60.0 %
added during the process Qs= 250.0 KJ


Example 2 Page No:302¶

In [2]:
#Input data
QR=1.5                 #tau=QS-QR
#T=Tmax-Tmin
T=300                  #temperature limit of the cycle in degree celsius

#Calculation
#QR=1.5*(QS-QR)
QR=(1.5/2.5)           #Engin work on carnot cycle
eta=(1-QR)             #Thermal effeciency
Tmax=(T/eta)-273.15    #Maximum temperataure
Tmin=(Tmax-T)          #Minimum temperataure

#Output
print("Engin work on carnot cycle=",QR,"QS")
print("Thermal effeciency=",100*(eta),"%")
print("Maximum temperataure=",round(Tmax,),"degree celsius")
print("Minimum temperataure=",round(Tmin,),"degree celsius")

Engin work on carnot cycle= 0.6 QS
Thermal effeciency= 40.0 %
Maximum temperataure= 477 degree celsius
Minimum temperataure= 177 degree celsius


Example 3 Page No:303¶

In [3]:
#Input data
#Refer figure
import math
T1=300                   #Carnot engine work in minimum temperature in kelvin
T2=750                   #Carnot engine work in maximum temperature kelvin
P2=50                    #pressure of carnot engine N/m**2
P4=1                     #pressure of carnot engine N/m**
#considering air as the working fluid therefore
R=0.287                  #Air as the working fluid in KJ/Kg K
Cp=1.005                 #KJ/Kg K
Cv=0.718                 #KJ/Kg K
K=1.4
gamma=1.4

#Calculation
#T2/T1=(P2/P1)**(gamma-1)/gamma
P1=P2*(T1/T2)**(gamma/(gamma-1)) #Pressure at intermediate salient points(1-2) in bar
P3=P4*(T2/T1)**(gamma/(gamma-1)) #Pressure at intermediate salient points(3-4) in bar
QS=R*T2*math.log(P2/P3 )         #Heat supplied and rejected per Kg of air  in KJ/Kg
QR=R*T1*math.log(P1/P4 )         #Heat supplied and rejected per Kg of air  in KJ/Kg
W=QS-QR                          #Work done in KJ/Kg
eta=(1-(T1/T2))                  #Thermal  of the carnot cycle

#Output
print("pressure at intermediate salient points(1-2)=",round(P1,2),"bar")
print("pressure at intermediate salient points(3-4)=",round(P3,1),"bar")
print("heat supplied and rejected per Kg of air(2-3)=",round(QS,1),"KJ/Kg")
print("heat supplied and rejected per Kg of air(4-1)=",round(QR,2),"KJ/Kg")
print("work done=",round(W,1),"KJ/Kg")
print("thermal  of the carnot cycle=",100*(eta),"%")

pressure at intermediate salient points(1-2)= 2.02 bar
pressure at intermediate salient points(3-4)= 24.7 bar
heat supplied and rejected per Kg of air(2-3)= 151.8 KJ/Kg
heat supplied and rejected per Kg of air(4-1)= 60.7 KJ/Kg
work done= 91.1 KJ/Kg
thermal  of the carnot cycle= 60.0 %


Example 4 Page No:304¶

In [4]:
#input data
import math
T2=377+273                  #Carnot cycle temperature in bar
P2=20*10**5                 #Carnot cycle pressure in bar
V2=1
V1=5
V3=2
#consider air as the working fluid therefore
R=0.287                     #In KJ/Kg K
Cp=1.005                    #In KJ/Kg K
Cv=0.718                    #In KJ/Kg K
K=1.4
gamma=1.4

#calculation
T1=T2*((V2/V1)**(gamma-1))                #Minimum temp in degree celsius
Qs=R*T2*math.log(V3/V2)                   #Heat supplied process in KJ/Kg
QR=R*T1*math.log((V1/V2)*(V2/V3)*((T2/T1)**(1/(gamma-1)))) #Heat Rejected Process in KJ/Kg
etath=(1-(T1/T2))*100                           #Thermal Effeiciency of the carnot cycle in %

#output
print("Minimum temp= ",round(T1,1),"degree celsius")
print("Heat supplied process= ",round(Qs,1),"KJ/Kg")
print("Heat Rejected Process= ",round(QR,1),"KJ/Kg")
print("Thermal Effeiciency of the carnot cycle= ",round(etath,1)," %")

Minimum temp=  341.4 degree celsius
Heat supplied process=  129.3 KJ/Kg
Heat Rejected Process=  247.5 KJ/Kg
Thermal Effeiciency of the carnot cycle=  47.5  %


Example 5 Page No:308¶

In [5]:
#Input data
P1=1                          #Isentropic Compression in bar
P2=20                         #Isentropic Compression in bar
#consider air as the working fluid therefore
gamma=1.4

#Calculation
r=(P2/P1)**(1/gamma)          #Isentropic process
eta=100*(1-(1/(r**(gamma-1))))#Otto cycle air standard effeciency in %

#Output
print("compression ratio=",round(r,2),)
print("standard efficiency=",round(eta,1),"%")

compression ratio= 8.5
standard efficiency= 57.5 %


Example 6 Page No:308¶

In [6]:
#Input data
T1=27+273                      #Initial temp in degree celsius
T2=450+273                     #Final temp in degree celsius

#calculation
r=(T2/T1)**(1/(gamma-1))       #Isentropic process
eta=100*(1-(1/(r**(gamma-1)))) #Otto cycle air standard effeciency in %

#output
print("compression ratio=",round(r),)
print("standard efficiency=",round(eta,1),"%")

compression ratio= 9
standard efficiency= 58.5 %


Example 7 Page No:309¶

In [7]:
#Input data
D=200*10**-3                   #Otto cycle cylindrical bore in mm
L=450*10**-3                   #Otto cycle Stroke in mm
vc=2*10**-3                    #Clearance volume in mm**3
gamma=1.4
pi=3.142

#calculation
vs=(pi/4)*(D**2*L)             #Swept volume
r=((vs+vc)/vc)                 #Compression ratio
eta=100*(1-(1/(r**(gamma-1)))) #Standard efficiency

#output
print("Swept volume=",round(vs,6),"m**3")
print("compression ratio=",round(r,3),)
print("standard efficiency=",round(eta,1),"%")

Swept volume= 0.014139 m**3
compression ratio= 8.07
standard efficiency= 56.6 %


Example 8 Page No:309¶

In [8]:
#Input data
P1=0.1*10**6                 #Otto cycle air
T1=35+273                    #Otto cycle temp degree celsius
r=9                          #Compression ratio
Qs=1800                      #Supplied heat in kJ/kg
v1=9
v2=1
R=0.287*10**3
gamma=1.4
Cv=0.718

#calculation
T2=(T1*((v1/v2)**(gamma-1))) #Temperature at point 2 in K
P2=(P1*((v1/v2)**1.4))*10**-6         #pressure at point 2 in MPa
T3=((Qs/Cv)+(T2))            #Max temp of cycle in degree celsius
P3=(T3/T2*P2)        #Max pressure of cycle in MPa
eta=100*(1-(1/(r**(gamma-1))))#Otto cycle thermal efficiency in %
WD=(Qs*eta)*10**-2           #Work done during the cycle in KJ/Kg
v1=((R*T1)/P1)               #Char gass equation in m**3/Kg
v2=v1/r                      #Char gass equation in m**3/Kg
Sv=v1-v2                     #Swept volume in m**3/Kg
Pme=(WD/Sv)*10**-3           #Mean effective pressure in MPa
alpha=P3/P2                  #Explosion ratio
Pm=(((P1*r)/((r-1)*(gamma-1)))*(((r**(gamma-1))-1)*(alpha-1)))*10**-6#Mean effective pressure in MPa

#Output
print("Temperature at point=",round(T2,1),"K")
print("pressure at point=",round(P2,3),"MPa")
print("Max temp of cycle=",round(T3,3),"K")
print("Max pressure= ",round(P3,1),"MPa")
print("Otto cycle thermal efficiency=",round(eta,1),"%")
print("Work done during the cycle=",round(WD,),"J/Kg")
print("Char gass equation=",round(v1,3),"m**3/Kg")
print("Char gass equation=",round(v2,4),"m**3/Kg")
print("Swept volume=",round(Sv,4),"m**3/Kg")
print("Mean effective pressure=",round(Pme,2),"MPa")
print("Explosion ratio=",round(alpha,2))
print("Mean effective pressure=",round(Pm,2),"MPa")

Temperature at point= 741.7 K
pressure at point= 2.167 MPa
Max temp of cycle= 3248.697 K
Max pressure=  9.5 MPa
Otto cycle thermal efficiency= 58.5 %
Work done during the cycle= 1053 J/Kg
Char gass equation= 0.884 m**3/Kg
Char gass equation= 0.0982 m**3/Kg
Swept volume= 0.7857 m**3/Kg
Mean effective pressure= 1.34 MPa
Explosion ratio= 4.38
Mean effective pressure= 1.34 MPa


Example 9 Page No:311¶

In [9]:
#Input data
P1=0.1                    #Beginning compression in MPa
T1=40+273                 #Beginning temp in degree celsius
eta=0.55                  #Standard effeciency in %
QR=540                    #Rejected heat in KJ/Kg
r=7.36                    #Compression ratio

#calculation
#eta=(1-(1/(r**(gamma-1))))
QS=(-QR/(eta-1))          #Heat supplied/unit mass in KJ/Kg
WD=QS-QR                  #Work done per Kg of air in KJ/Kg
T2=T1*(r**(gamma-1))      #Temp at end of compression in K
P2=P1*((r)**gamma)        #pressure at point 2 in MPa
T3=(QS/Cv)+T2             #max temp of the cycle in K
P3=(T3/T2)*P2             #max pressure of the cycle in MPa

#output
print("Heat supplied/unit mass=",round(QS,),"KJ/Kg")
print("Work done per Kg of air= ",round(WD,),"KJ/Kg")
print("Temp at end of compression=",round(T2,1),"K")
print("pressure at point two=",round(P2,3)," MPa")
print("max temp of the cycle=",round(T3,1),"K")
print("max pressure of the cycle=",round(P3,3)," MPa")

Heat supplied/unit mass= 1200 KJ/Kg
Work done per Kg of air=  660 KJ/Kg
Temp at end of compression= 695.5 K
pressure at point two= 1.635  MPa
max temp of the cycle= 2366.8 K
max pressure of the cycle= 5.565  MPa


Example 10 Page No:312¶

In [10]:
#Input data
T1=300                            #Initial temp in K
T3=2500                           #Final temp in K
P1=1                              #Initial pressure in N/m**2
P3=50                             #Final pressure in N/m**2
gamma=1.4
Cv=0.718

#calculation
r=(P3*T1)/(P1*T3)                    #Compression ratio
eta=(1-(1/r**(gamma-1)))             #Standard effeciency in %
T2=T1*((P3/P1)**((gamma-1)/gamma))   #Middle temperature in K
Qs=Cv*(T3-T2)                        #Heat supplied in KJ/Kg
WD=eta*Qs                            #Work done KJ/Kg

#output
print("Compression ratio=",r,"")
print("Standard effeciency=",round(eta,4),"%")
print("Middle temperature=",round(T2,2),"K")
print("Heat supplied=",round(Qs,2),"KJ/Kg")
print("Work done=",round(WD,1),"KJ/Kg")

Compression ratio= 6.0
Standard effeciency= 0.5116 %
Middle temperature= 917.36 K
Heat supplied= 1136.33 KJ/Kg
Work done= 581.4 KJ/Kg


Example 11 Page No:316¶

In [11]:
#input data
r=18   #compression ratio of diesel engine
K=6    #cut-off ratio of the stroke in%
rho=2.02

#calculation
#diesel engine air standard efficiency
eta=100*((1-(1/r**(gamma-1)))*(1/gamma*(rho**(gamma-1)/(rho-1))))

#output
print("diesel engine air standard efficiency",round(eta,1),"%")

diesel engine air standard efficiency 63.6 %


Example 12 Page No:317¶

In [12]:
#Input Data
r=22              #compression ratio of diesel engine r=v1/v2
r1=11             #expansion ratio r1=v4/v3
gamma=1.4
rho=1.4

#calculation
rho=r/r1          #cut-off ratio
#diesel engine air standard efficiency
eta=100*((1-(1/r**(gamma-1)))*(1/gamma*(rho**(gamma-1)/(rho-1))))

#output
print("cut-off ratio=",rho,)
print("diesel engine air standard efficiency=",round(eta,2),"%")

cut-off ratio= 2.0
diesel engine air standard efficiency= 66.88 %


Example 13 Page No:317¶

In [13]:
#Input data
Vc=10/100              #Clearance volume in %
Vs=Vc/0.1
K=0.05                 #Cut-off of the strok in
gamma=1.4

#Calculation
r=((Vs+Vc)/(Vc))       #Compression ratio
rho=1+K*(r-1)          #Cut-off ratio
#Effeciency in %
eta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100

#output
print("Compression ratio=",r,"Vs")
print("Cut-off ratio=",rho,)
print("Effeciency=",round(eta,2),"%")

Compression ratio= 11.0 Vs
Cut-off ratio= 1.5
Effeciency= 58.17 %


Example 14 Page No:¶

In [14]:
#Input data
T1=50+273          #Temperature at the beginning of the compression
T2=700+273         #Temperature at the end of the compression
T3=2000+273        #Temperature at the beginning of the expansion

#Calculation
r=((T2/T1)**(1/(gamma-1))) #Compression ratio
rho=(T3/T2)                #Cut-off ratio
K=((rho-1)/(r-1))          #Also cut-off ratio
#Air standard efficiency
eta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100

#Output
print("compression ratio=",round(r,2),"")
print("cut-off ratio=",round(rho,3),)
print("also cut-off ratio=",round(K,2),"")
print("air standard efficiency=",round(eta,2),"%")

compression ratio= 15.75
cut-off ratio= 2.336
also cut-off ratio= 0.09
air standard efficiency= 59.54 %


Example 15 Page No:317¶

In [15]:
#Input data
P1=0.1            #Diesel cycle is supplied# with air in MPa
T1=40+273         #Diesel cycle is supplied with temperature in degree celsius
r=18              #Compression ratio
Qs=1500           #Heat supplied
v1=18
v2=1
Cp=1.005

#Calculation
T2=T1*((v1/v2)**(gamma-1)) #For isentropic process the temperature is
P2=P1*((v1/v2)**(gamma))   #For isentropic process the pressure is
T3=(Qs/Cp)+T2              #Maximum temperatureof the cycle
rho=T3/T2                  #Cut-off ratio
#Air standard efficiency
eta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100
NWD=(Qs*eta)*10**-2                  #Net work done

#Output
print("for isentropic process the temperature=",round(T2,1),"K")
print("for isentropic process the pressure=",round(P2,2),"MPa")
print("maximum temperatureof the cycle=",round(T3,2),"K")
print("cut-off ratio=",round(rho,1),"MPa")
print("air standard efficiency=",round(eta,2),"%")
print("net work done=",round(NWD,),"KJ/Kg")

for isentropic process the temperature= 994.6 K
for isentropic process the pressure= 5.72 MPa
maximum temperatureof the cycle= 2487.15 K
cut-off ratio= 2.5 MPa
air standard efficiency= 60.93 %
net work done= 914 KJ/Kg


Example 16 Page No:317¶

In [16]:
#Input data
r=14               #compression ratio of standard diesel cycle
P1=1               #compression stroke in bar
T1=300             #temperature of air in k
T3=2774            #temperature rises in k
CP=1.005
v1=14
v2=1
gamma=1.4
Qs=1921.43
R=0.287*10**3

#calculation
T2=T1*((v1/v2)**(gamma-1))      #constant pressure
rho=T3/T2                       #cut-off ratio
eta=(1-(1/r**(gamma-1))*((1/gamma)*(((rho**(gamma))-1)/(rho-1))))*100 #air standard efficiency
HS=(CP*(T3-T2))                 #heat supplied
WD=(Qs*eta)*10**-2                     #Net work done
v1=(R*T1/P1) *10**-5                     #characteristics gas equation
v2=(v1/r )                       #characteristics gas equation
Sv=(v1-v2)                       #Swept volume
Pme=(WD/Sv )*10**-2                     #Mean effective pressur
Pm=((P1*r)/((r-1)*(gamma-1)))*((gamma*(r**(gamma-1)))*(rho-1)-((rho**(gamma))-1))# mean effective pressure

#output
print("constant pressure=",round(T2,2),"K")
print("cut-off ratio= ",round(rho,2),)
print("air standard efficiency=",round(eta,2),"%")
print("heat supplied= ",round(HS,2),"KJ/Kg")
print("Net work done= ",round(WD,2),"KJ/Kg")
print("characteristics gas equation= ",round(v1,3),"m**3/Kg")
print("characteristics gas equation= ",round(v2,4),"m**3/Kg")
print("Swept volume= ",round(Sv,4),"m**3/Kg")
print("Mean effective pressure= ",round(Pme,1),"bar")
print("Mean effective pressure= ",round(Pm,1),"bar")

constant pressure= 862.13 K
cut-off ratio=  3.22
air standard efficiency= 53.65 %
heat supplied=  1921.43 KJ/Kg
Net work done=  1030.91 KJ/Kg
characteristics gas equation=  0.861 m**3/Kg
characteristics gas equation=  0.0615 m**3/Kg
Swept volume=  0.7995 m**3/Kg
Mean effective pressure=  12.9 bar
Mean effective pressure=  12.9 bar