# Chapter 3 : Air Standard Cycles¶

## Example 3.1 Page no : 19¶

In [1]:
import math

#Input data
p = [1,8]					#Pressure at the beginning and end of compression in kg/m**3
g = 1.4					#Ratio of specific heats

#Calculations
r = (p[1]/p[0])**(1/g)					#Compression ratio
n = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent

#Output
print 'Air standard efficiency of an engine working on the Otto cycle is %3.1f percent'%(n)

Air standard efficiency of an engine working on the Otto cycle is 44.8 percent


## Example 3.2 Page no : 24¶

In [2]:
import math

#Input data
D = 0.25					#Bore in m
L = 0.45					#Stroke in m
Cv = 5.					#Clearance volume in litres
g = 1.4					#Ratio of specific heats
IHP = 32.					#Indicated Horse power in h.p
m = 14.					#Gas consumption in m**3/hr
CV = 4000.					#Calorific value of gas in kcal/m**3

#Calculations
Vs = (3.14/4)*D**2*L					#Stroke volume in m**3
Vc = Cv/1000					#Clearance volume in m**3
r = (Vs+Vc)/Vc					#Compression ratio
na = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent
q = (m*CV)/60					#Heat supplied in kcal/min
aI = (IHP*4500)/427					#Heat equivalent of I.H.P in kcal/min
itn = (aI/q)*100					#Indicated thermal efficiency in percent
rn = (itn/na)*100					#Relative efficiency in percent

#Output
print 'The air standard efficiency is %3.1f percent  \
\nIndicated thermal efficiency is %3.1f percent  \
\nRelative efficiency is %3.1f percent'%(na,itn,rn)


The air standard efficiency is 49.1 percent
Indicated thermal efficiency is 36.1 percent
Relative efficiency is 73.6 percent


## Example 3.3 Page no : 26¶

In [3]:
import math

#Input data
r = 6.					#Compression ratio
It = 0.6					#Indicated thermal efficiency ratio
CV = 10000.					#Calorific value in kcal/kg
g = 1.4					#Ratio of specific heats

#Calculations
an = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent
In = (It*(an/100))					#Indicated thermal efficiency
SFC = ((4500*60)/(427*CV*In))					#Specific fuel consumption in kg/I.H.P.hr

#Output
print 'Specific fuel consumption is %3.3f kg/I.H.P. hr'%(SFC)

Specific fuel consumption is 0.206 kg/I.H.P. hr


## Example 3.4 Page no : 27¶

In [4]:
import math

#Input data
T = [100.+273,473.+273]					#Temperatures at the beginning and at the end of adiabatic compression in K
g = 1.4					#Ratio of specific heats

#Calculations
an = (1-(T[0]/T[1]))*100					#Air standard efficiency in percent
r = (T[1]/T[0])**(1/(g-1))					#Compression ratio

#Output
print 'The compression ratio is %3.2f  \
\nAir standard efficiency is %i percent'%(r,an)

The compression ratio is 5.66
Air standard efficiency is 50 percent


## Example 3.5 Page no : 29¶

In [5]:
import math

#Input data
T1 = 45.+273					#Temperature at the beginning of compression in K
p1 = 1.					#Pressure at the beginning of compression in kg/cm**2
T2 = 325.+273					#Temperature at the end of compression in K
T3 = 1500.+273					#Temperature at the end of consmath.tant volume heat addition in K
g = 1.4					#Ratio of specific heats

#Calculations
r = (T2/T1)**(1/(g-1))					#Compression ratio
an = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent
p2 = (p1*r**g)					#Pressure at the end of compression in kg/cm**2
p3 = (p2*(T3/T2))					#Pressure at the end of consmath.tant volume heat addition in kg/cm**2
p4 = p3/p2					#Pressure at the end of adiabatic expansion in kg/cm**2
T4 = T3/r**(g-1)					#Temperature at the end of adiabatic expansion in K
t4 = T4-273					#Temperature at the end of adiabatic expansion in degree C

#Output
print 'The air standard efficiency is %3.1f percent  \
\nTemperature at the end of adiabatic expansion is %i degree C  \
\nPressure at the end of adiabatic expansion is %3.0f kg/cm**2'%(an,t4,p4)

The air standard efficiency is 46.8 percent
Temperature at the end of adiabatic expansion is 669 degree C
Pressure at the end of adiabatic expansion is   3 kg/cm**2


## Example 3.6 Page no : 34¶

In [6]:
import math

#Input data
T1 = 40.+273					#Temperature at the beginning of compression in K
p1 = 1.					#Pressure at the beginning of compression in kg/cm**2
p2 = 15.					#Pressure at the end of adabatic compression in kg/cm**2
T3 = 2000.+273					#Maximum temperature during the cycle in K
Cv = 0.17					#Specific heat at consmath.tant volume in kJ/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
T2 = T1*(p2/p1)**((g-1)/g)					#Temperature at the end of adabatic compression in K
na = (1-(T1/T2))*100					#Air standard efficiency in percent
q = (Cv*(T3-T2))					#Heat added in kcal/kg of air
W = ((na/100)*q)					#Workdone per kg of air in kcal
W1 = (4.28*W)					#Workdone per kg of air in kg.m
p3 = (p2*(T3/T2))					#Pressure at the end of consmath.tant volume heat addition in kg/cm**2
p4 = (p3*p1)/p2					#Pressure at the end of adiabatic expansion in kg/cm**2

#Output
print 'a) The heat supplied is %3.0f kcal/kg of air  \
\nb) The workdone is %i kcal/kg of air  \
\nc) The pressure at the end of adiabatic expansion is %3.2f kg/cm**2'%(q,W,p4)

a) The heat supplied is 271 kcal/kg of air
b) The workdone is 146 kcal/kg of air
c) The pressure at the end of adiabatic expansion is 3.35 kg/cm**2


## Example 3.7 Page no : 38¶

In [7]:
import math

#Input data
r = 16.					#Compression ratio
k = 5.					#Cut off takes place at 5% of the stroke
g = 1.4					#Ratio of specific heats

#Calculations
c = (((k/100)*(r-1))+1)					#Cut off ratio
na = (1-((1/r**(g-1))*((c**g-1)/(g*(c-1)))))*100					#Air standard efficiency in percent

#Output
print 'The air standard efficiency is %3.1f percent'%(na)

The air standard efficiency is 62.6 percent


## Example 3.8 Page no : 38¶

In [8]:
import math

#Input data
p1 = 1.05					#Inlet pressure in kg/cm**2
T1 = 15.+273					#Inlet temperature in K
p2 = 33.4					#Pressure at the end of adiabatic compression in kg/cm**2
r = 5.					#The ratio of expansion
Cp = 0.238					#Specific heat at consmath.tant pressure in kJ/kg.K
Cv = 0.17					#Specific heat at consmath.tant volume in kJ/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
r1 = (p2/p1)**(1/g)					#Compression ratio
k = r1/r					#Cutoff ratio
T2 = (p2/p1)**((g-1)/g)*T1					#Temperature at the end of adiabatic compression in K
T3 = T2*k					#Temperature at the end of consmath.tant pressure heat addition in K
T4 = T3*(1/r)**(g-1)					#Temperature at the end of adiabatic expansion in K
qa = (Cp*(T3-T2))					#Heat added in kcal/kg of air
qre = (Cv*(T4-T1))					#Heat rejected in kcal/kg of air
nt = ((qa-qre)/qa)*100					#Ideal thermal efficiency in percent

#Output
print 'The ideal thermal efficiency is %3.1f percent'%(nt)

The ideal thermal efficiency is 54.5 percent


## Example 3.9 Page no : 41¶

In [9]:

#Input data
p1 = 1.					#Pressure at the end of suction stroke in kg/cm**2
T1 = 30.+273					#Temperature at the end of suction stroke in kg/cm**2
T3 = 1500.+273					#Maximum temperature during the cycle in K
r = 16.					#Compression ratio
Cp = 0.24					#Specific heat at consmath.tant pressure in kJ/kg.K
Cv = 0.17					#Specific heat at consmath.tant volume in kJ/kg.K
g = 1.41					#Ratio of specific heats

#Calculations
T2 = T1*r**(g-1)					#Temperature at the end of adiabatic compression in K
s = (((T3/T2)-1)/(r-1))*100					#Percentage of the stroke at which cut off occurs
r1 = (r/(T3/T2))					#Expansion ratio
T4 = T3/(r1)**(g-1)					#Temperature at the end of adiabatic expansion in K
qa = (Cp*(T3-T2))					#Heat added in kcal/kg of air
qre = (Cv*(T4-T1))					#Heat rejected in kcal/kg of air
nt = ((qa-qre)/qa)*100					#Air standard efficiency in percent

#Output
print 'a) The percentage of stroke at which cut off takes place is %3.2f percent  \
\nb) The temperature at the end of expansion stroke is %3.0f K  \
\nc) The theoretical efficiency is %3.0f percent'%(s,T4,nt)

a) The percentage of stroke at which cut off takes place is 5.85 percent
b) The temperature at the end of expansion stroke is 737 K
c) The theoretical efficiency is  63 percent


## Example 3.10 Page no : 46¶

In [10]:
import math

#Input data
p1 = 1.					#Pressure at the beginning of compression in kg/cm**2
T1 = 80.+273					#Temperature at the beginning of compression in K
r = 14.					#Compression ratio
p4 = 2.7					#Pressure at the end of expansion in kg/cm**2
Cp = 0.24					#Specific heat at consmath.tant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
p2 = p1*r**g					#Pressure at the end of compression in kg/cm**2
s = (((r*(p4/p2)**(1/g))-1)/(r-1))*100					#Percentage of stroke when the fuel is cut off in percent
T2 = (T1*(p2/p1))/r					#Temperature at the end of compression in K
T3 = (T2*r*(p4/p2)**(1/g))					#Temperature at the end of adiabatic expansion in K
q = (Cp*(T3-T2))					#Heat supplied in kcal/kg

#Output
print 'a) The maximum pressure attained during the cycle is %3.1f kg/cm**2  \
\nb) The percentage of working stroke at which the heat supply to the working fluid ceases is %3.2f percent  \
\nc) The heat received per kg of woring substance during the cycle is %3.0f kcal/kg'%(p2,s,q)

a) The maximum pressure attained during the cycle is 40.2 kg/cm**2
b) The percentage of working stroke at which the heat supply to the working fluid ceases is 7.95 percent
c) The heat received per kg of woring substance during the cycle is 251 kcal/kg


## Example 3.11 Page no : 47¶

In [11]:
import math

# variables
d = 0.25					#Diameter of the cylinder in m
L = 0.35					#Stroke in m
Cv = 1500.					#Clearance volume in c.c
s = 5.					#cut off ratio takes place at 5 percent of stroke
a = 1.4					#Explosion ratio
g = 1.4					#Ratio of specific heats for air

#Calculations
Vs = (3.14/4)*d**2*L					#Stroke volume in m**3
r = (Vs*10**6+Cv)/Cv					#Compression ratio
k = (Cv+((s/100)*Vs*10**6))/Cv					#Cut off ratio
na = (1-((1/(r**(g-1)))*((a*k**g-1)/((a-1)+a*g*(k-1)))))*100					#Air standard efficiency in percent

#Output
print 'The air standard efficiency of the engine is %3.1f percent'%(na)

The air standard efficiency of the engine is 60.7 percent


## Example 3.12 Page no : 48¶

In [12]:
import math

#Input data
d = 0.2					#Diameter of the cylinder in m
L = 0.4					#Stroke in m
r = 13.5					#Compression ratio
a = 1.42					#Explosion ratio
s = 5.1					#Cut off occurs at 5.1 percent of the stroke
g = 1.4					#Ratio of specific heats for air

#Calculations
Vs = (3.14/4)*d**2*L*10**-6					#Stroke volume in c.c
Vc = Vs/r					#Clearance volume in c.c
k = (((s/100)*Vs)+Vc)/Vc					#Cut off ratio
ASE = (1-((1/(r**(g-1)))*((a*k**g-1)/((a-1)+a*g*(k-1)))))*100					#Air standard efficiency in percent

#Output
print 'The air standard efficiency of the engine is %3.1f percent'%(ASE)

The air standard efficiency of the engine is 61.4 percent


## Example 3.13 Page no : 53¶

In [13]:
import math

#Input data
x = [2./3,1./3]					#The dual cycle atkes two-thirds of its total heat supply at consmath.tant volume and one-third at consmath.tant pressure
r = 13.					#Compression ratio
p3 = 43.					#Maximum pressure of the cycle in kg/cm**2
p1 = 1.					#Pressure at intake in kg/cm**2
T1 = 15.+273					#Intake temperature in K
Cp = 0.24					#Specific heat at consmath.tant pressure in kJ/kg.K
Cv = 0.17					#Specific heat at consmath.tant volume in kJ/kg.K
g = 1.41					#Ratio of specific heats

#Calculations
T2 = T1*r**(g-1)					#Temperature at the end of compression in K
p2 = (p1*r**g)					#Pressure at the end of compression in kg/cm**2
T3 = T2*p3/p2					#Temperature at the end of consmath.tant volume heat addition in K
q23 = Cv*(T3-T2)					#Heat added at consmath.tant volume in kcal/kg
q34 = (1./2)*q23					#Heat added at consmath.tant pressure in kcal/kg
T4 = (q34/Cp)+T3					#Temperature at the end of consmath.tant pressure heat supply in K
T5 = (T4*((p1*(T4/T3))/r)**(g-1))					#Temperature at the end of expansion in K
na = (1-((Cv*(T5-T1))/((Cv*(T3-T2))+(Cp*(T4-T3)))))*100					#Efficiency in percent
T = [T1-273,T2-273,T3-273,T4-273,T5-273]					#Temperature at the five cardinal points in degree C

#Output
print 'a) The temperature at the five cardinal points of the cycle are :  point 1 is %3.0f degree C  \
\npoint 2 is %3.0f degree C  \
\npoint 3 is %3.0f degree C  \
\npoint 4 is %3.1f degree C  \
\npoint 5 is %3.0f degree C  \
\nb) The ideal thermal efficiency of the cycle is %3.1f percent'%(T[0],T[1],T[2],T[3],T[4],na)

a) The temperature at the five cardinal points of the cycle are :  point 1 is  15 degree C
point 2 is 551 degree C
point 3 is 680 degree C
point 4 is 725.0 degree C
point 5 is  82 degree C
b) The ideal thermal efficiency of the cycle is 65.0 percent


## Example 3.14 Page no : 56¶

In [14]:
import math

#Input data
p1 = 1.					#Pressure at intake in kg/cm**2
T1 = 100.+273					#Intake temperature in K
r = 10.					#Compression ratio
p3 = 70.					#Maximum pressure of the cycle in kg/cm**2
q = 400.					#Amount of heat added in kcal/kg of air
Cp = 0.24					#Specific heat at consmath.tant pressure in kJ/kg.K
Cv = 0.17					#Specific heat at consmath.tant volume in kJ/kg.K
g = 1.41					#Ratio of specific heats

#Calculations
T2 = (T1*r**(g-1))					#Temperature at the end of compression in K
p2 = (p1*r**g)					#Pressure at the end of compression in kg/cm**2
T3 = T2*(p3/p2)					#Temperature at the end of consmath.tant volume heat addition in K
qv = (Cv*(T3-T2))					#Heat added at consmath.tant volume in kcal/kg
qp = (q-qv)					#Heat added at consmath.tant pressure in kcal/kg
T4 = (qp/Cp)+T3					#Temperature at the end of consmath.tant pressure heat supply in K
k = (T4/T3)					#Cut off ratio
T5 = T4/(r/k)**(g-1)					#Temperature at the end of expansion in K
qv2 = Cv*(T5-T1)					#Heat added at consmath.tant volume in kcal/kg
W = q-qv2					#Workdone in kcal/kg of air
na = (W/q)*100					#Air standard efficiency in percent

#Output
print 'The temperature at the five cardinal points of the cycle are :  \
\npoint 1 is %3.0f K  \
\npoint 2 is %3.0f K  \
\npoint 3 is %3.0f K  \
\npoint 4 is %3.0f K  \
\npoint 5 is %3.0f K  \
\nThe air standard efficiency of the engine is %3.1f percent'%(T1,T2,T3,T4,T5,na)

The temperature at the five cardinal points of the cycle are :
point 1 is 373 K
point 2 is 959 K
point 3 is 2611 K
point 4 is 3107 K
point 5 is 1298 K
The air standard efficiency of the engine is 60.7 percent


## Example 3.15 Page no : 59¶

In [15]:
import math

#Input data
r = 12.					#Compression ratio
p1 = 0.955					#Pressure at the start of compression in kg/cm**2
T1 = 85.+273					#Temperature at the start of compression in K
p3 = 55.					#Maximum pressure of the cycle in kg/cm**2
x = (1./30)					#Consmath.tant pressure heat reception contnues for 1/30 of the stroke
Cp = 0.238					#Specific heat at consmath.tant pressure in kJ/kg.K
Cv = 0.17					#Specific heat at consmath.tant volume in kJ/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
T2 = T1*r**(g-1)					#Temperature at the end of compression in K
p2 = (p1*r**g)					#Pressure at the end of compression in kg/cm**2
T3 = T2*(p3/p2)					#Temperature at the end of consmath.tant volume heat addition in K
T4 = (T3*((p1+x*(r-1))/p1))					#Temperature at the end of consmath.tant pressure heat supply in K
T5 = T4*((p1+x*(r-1))/r)**(g-1)					#Temperature at the end of expansion in K
qs = (Cv*(T3-T2))+(Cp*(T4-T3))					#Heat supplied in kcal/kg of air
qre = (Cv*(T5-T1))					#Heat rejected in kcal/kg of air
W = (qs-qre)					#Workdone in kcal/kg of air
an = ((qs-qre)/qs)*100					#Air standard efficiency in percent

#Ouptut
print 'The wordone per kg of air is %3.2f kcal  \
\nThe ideal thermal efficiency is %3.1f percent'%(W,an)

The wordone per kg of air is 178.24 kcal
The ideal thermal efficiency is 62.6 percent


## Example 3.16 Page no : 63¶

In [16]:
import math

#Input data
p1 = 1.					#Pressure of air intake in kg/cm**2
T1 = 50.+273					#Temperature of air intake in K
v = (1./14)					#Volume compresses by it adiabatically of its original volume
r = (1/v)					#Compression ratio
Cp = 0.237					#Specific heat at consmath.tant pressure in kJ/kg.K
Cv = 0.169					#Specific heat at consmath.tant volume in kJ/kg.K
g = 1.4					#Ratio of specific heats for air

#Calculations
T2 = T1*r**(g-1)					#Temperature at the end of compression in K
p2 = (p1*r**g)					#Pressure at the end of compression in kg/cm**2
p3 = x*p2					#Pressure at the end of the heat addition at consmath.tant volume in kg/cm**2
T3 = T2*(p3/p2)					#Temperature at the end of consmath.tant volume heat addition in K
T4 = (T3*x)					#Temperature at the end of consmath.tant pressure heat supply in K
T5 = T4/(r/x)**(g-1)					#Temperature at the end of expansion in K
qs = (Cv*(T3-T2))+(Cp*(T4-T3))					#Heat supplied in kcal/kg of air
qre = (Cv*(T5-T1))					#Heat rejected in kcal/kg of air
na = ((qs-qre)/qs)*100					#Air standard efficiency in percent
T = [T1-273,T2-273,T3-273,T4-273,T5-273]					#Temperature at the five key points in degree C

#Output
print 'a) The temperature at the five key points of the cycle are :  \
\npoint 1 is %3.0f K  =  %3.0f degree C  \
\npoint 2 is %3.0f K  =  %3.0f degree C  \
\npoint 3 is %3.0f K  =  %3.0f degree C  \
\npoint 4 is %3.0f K  =  %3.0f degree C  \
\npoint 5 is %3.0f K  =  %3.0f degree C  \
\nb) The ideal thermal efficiency of the cycle is %3.2f percent'%(T1,T[0],T2,T[1],T3,T[2],T4,T[3],T5,T[4],na)

a) The temperature at the five key points of the cycle are :
point 1 is 323 K  =   50 degree C
point 2 is 928 K  =  655 degree C
point 3 is  31 K  =  -242 degree C
point 4 is   1 K  =  -272 degree C
point 5 is   0 K  =  -273 degree C
b) The ideal thermal efficiency of the cycle is 65.62 percent


## Example 3.18 Page no : 68¶

In [17]:
import math

#Input data
n = 6.					#Six cylinder engine
r = 5.					#Compression ratio
Vc = 110.					#Clearance volume in c.c
a = 0.66					#Efficiency ratio referred to the air standard cycle
N = 2400.					#Speed in r.p.m
m = 9.9					#Mass of petrol in kg
CV = 10600.					#Calorific value of fuel in kcal/kg
g = 1.4					#Ratio of specific heats

#Calculations
Vs = (r*Vc-Vc)					#Swept Volume in c.c
na = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent
nt = (na/100)*a					#Thermal efficiency
IHP = (nt*CV*m*427)/(4500*60)					#Indicated Horse Power in h.p
pm = (((IHP/n)*4500*100*2)/(Vs*N))					#Average indicated mean effective pressure in kg/cm**2

#Output
print 'The average indicated mean effective pressure in each cylinder is %3.3f kg/cm**2'%(pm)

The average indicated mean effective pressure in each cylinder is 7.386 kg/cm**2


## Example 3.19 Page no : 68¶

In [18]:
import math

#Input data
n = 4.					#Four cylinder engine
BHP = 40.					#Brake horse power in h.p
N = 3000.					#Speed in r.p.m
nm = 70.					#Mechanical efficiency in percent
pm = 13.5					#Indicated mean effective pressure in kg/cm**2
#Bore is equal to stroke

#Calculations
#case(i)
d1 = ((BHP*100*4500*n*2)/(n*(nm/100)*pm*N*3.14))**(1/3)					#Cylinder bore or stroke length in cm

#Case(ii)
d2 = ((BHP*100*4500*n)/(n*(nm/100)*pm*N*3.14))**(1/3)*10					#Cylinder bore or stroke length in cm

#Output
print 'The cylinder sizes for a bore equal to stroke of a four cylinder in case of  \
\ni)Four stroke engine is %3.1f cm  \
\nii)Two stroke engine is %3.0f mm'%(d1,d2)

The cylinder sizes for a bore equal to stroke of a four cylinder in case of
i)Four stroke engine is 1.0 cm
ii)Two stroke engine is  10 mm


## Example 3.20 Page no : 72¶

In [19]:
import math

#Input data
T = [50.+273,345.+273]					#Temperatures at the beginning and end of compression in K
g = 1.4					#ratio of specific heats
IHP = 25.					#Indicated horse power in h.p
m = 5.44					#Mass of fuel consumed per hour in kg
CV = 10300.					#Calorific value in kcal/kg

#Calculations
na = (1-(T[0]/T[1]))*100					#Air standard efficiency in percent
r = (T[1]/T[0])**(1/(g-1))					#Compression ratio
qIHP = (IHP*4500)/427					#Heat equivalent of I.H.P in kcal/min
q = (m*CV)/60					#Heat supplied per minute in kcal/min
Ith = (qIHP/q)*100					#Indicated thermal efficiency in percent
nr = (Ith/na)*100					#Efficiency ratio

#Output
print 'The air standard efficiency is %3.1f percent  \
\nThe compression ratio is %3.2f  \
\nIndicated thermal efficiency is %3.1f percent  \
\nEfficiency ratio is %3.1f percent'%(na,r,Ith,nr)

The air standard efficiency is 47.7 percent
The compression ratio is 5.06
Indicated thermal efficiency is 28.2 percent
Efficiency ratio is 59.1 percent


## Example 3.21 Page no : 74¶

In [20]:
import math

#Input data
CV = 10000.					#Calorific value of petrol in kcal/kg
pe = [30.,70.]					#Percentage of compression strokes in percent
p = [1.33,2.66]					#Pressures in the cylinder corresponding to the compression strokes in kg/cm**2
n = 1.33					#Polytropic consmath.tant
rn = 50.					#Relative efficiency in percent
g = 1.4					#ratio of specific heats

#Calculations
v = (p[1]/p[0])**(1/n)					#Ratio of specific volumes
r = ((pe[1]/100)*v-(pe[0]/100))/((pe[1]/100)-((pe[0]/100)*v))					#Compression ratio
na = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent
ith = (rn*na)/100					#Indicated thermal efficiency in percent
q = (4500*60)/(427*(ith/100))					#Heat supplied in kcal/i.h.p.hr
Sc = (q/CV)					#Specific consumption in kg/i.h.p.hr

#Output
print 'Compression ratio is %3.2f  \
\nSpecific consumption is %3.3f kg/i.h.p.hr'%(r,Sc)

Compression ratio is 4.51
Specific consumption is 0.279 kg/i.h.p.hr


## Example 3.22 Page no : 76¶

In [21]:
import math

#Input data
n = 4.					#Four cylinder four stroke
d = 7.5					#Bore in cm
L = 8.75					#Stroke in cm
r = 6.					#Compression ratio
n1 = 55.					#Efficiency in percent
g = 1.4					#ratio of specific heats
N = 2400.					#Speed in r.p.m
pm = 7.					#Brake mean effective pressure in kg/cm**2
m = 9.					#Mass of fuel per hour in kg
CV = 10500.					#Calorific Value in kcal/kg

#Calculations
an = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent
In = (an*n1)/100					#Indicated thermal efficiency in percent. In textbook, answer is wrong
BHP = (pm*(3.14/4)*d**2*(L/100)*(N/2)*n)					#Brake horse power in kg.m/min
Bth = ((BHP*60)/(427*CV*m))*100					#Brake thermal efficiency in percent
nm = (Bth/In)*100					#Mechanical efficiency in percent
Sc = ((4500*60)/(427*(Bth/100)*CV))					#Specific consumption in g/i.h.p.hr

#Output
print 'Indicated thermal efficiency is %3.1f percent  \
\nBrake thermal efficiency is %3.1f percent  \
\nMechanical efficiency is %3.1f percent  \
\nSpecific fuel consumption is %3.3f kg/i.h.p.hr'%(In,Bth,nm,Sc)

Indicated thermal efficiency is 28.1 percent
Brake thermal efficiency is 19.3 percent
Mechanical efficiency is 68.6 percent
Specific fuel consumption is 0.312 kg/i.h.p.hr


## Example 3.26 Page no : 79¶

In [22]:
import math

#Input data
r = 7.					#Compression ratio
v = 1.					#Specific heat at consmath.tant volume increases by 1 percent
g = 1.4					#Ratio of specific heats

#Calculations
e = (1-(1/r**(g-1)))					#Air standard efficiency
dee = -(((1-e)*(g-1)*math.log(r)*(v/100))/e)*100					#Change in efficiency to the original efficiency
x = -(dee)					#For Output purpose

#Output
print 'Percentage change is efficiency is %3.2f percent i.e. a decrease of %3.2f percent'%(dee,x)

Percentage change is efficiency is -0.66 percent i.e. a decrease of 0.66 percent


## Example 3.30 Page no : 84¶

In [23]:
import math

#Input data
N = 210.					#Speed in r.p.m
d = 0.3					#Diameter of the piston in m
L = 0.4					#Stroke in m
v = 2.5					#Clearance volume is 2.5% of the swept volume. But in textbook it is given wrong as 25%
CO = 19.7					#Percentage of CO gas
H2 = 28.8					#Percentage of H2 gas
CO2 = 14.4					#Percentage of CO2 gas
N2 = 37.1					#Percentage of N2 gas
x = 0.875					#Total mixture at N.T.P admitted per suction stroke is 0.875 of the total volume behind the piston at the end of the stroke
tn = 35.					#Thermal efficiency in percent
CVH2 = 13200.					#Calorific value of H2 per kg in kcal
CVC = 2540.					#Calorific value of carbon burning from CO to CO2 in kcal/kg
de = 1.293					#Density of air in kg/m**3
mC = 12.					#Molecular weight of carbon
mO2 = 32.					#Molecular weight of O2
mH2 = 2.					#Molecular weight of H2
mCO = 28.					#Molecular weight of CO

#Calculations
a = ((100./21)*((CO2/100)+((CO/2)/100)))					#Air per cu.m of gas in cu.m
Vm = (a+1)					#Volume of mixture per cu.m of gas in cu.m
Vs = ((3.14/4)*d**2*L)					#Swept volume in cu.m
Vc = (Vs*v)/100					#Clearance volume in cu.m
V = Vc+Vs					#Total volume in cu.m
VC = V*x					#Volume of charge admitted per stroke in cu.m
VM = VC*(N/2)					#Charge volume per minute in cu.m
VG = (VM/Vm)					#cu.m of gas per minute
vH2 = (VG*(H2/100))					#Volume of H2 per minute in cu.m
vCO = (VG*(CO/100))					#Volume of CO per minute in cu.m
CVH2cum = (mH2*CVH2)/(vH2*1000)					#Calorific value of H2 per cu.m in kcal
CVCO = (CVC*(2*mC)/(2*mCO))					#Calorific value of CO per kg in kcal
CVCOcum = (mCO*CVCO)/(vH2*1000)					#Calorific value of CO per cu.m in kcal
qH2 = (16.09*CVH2cum)					#Heat in charge due to H2 in kcal
qCO = (11*CVCOcum)					#Heat in charge due to CO in kcal
qt = (qH2+qCO)					#Heat supplied per minute in kcal
qu = (qt*(tn/100))					#Heat utilised in kcal
hp = (qu*427)/4500					#H.P developed

#Output
print 'Maximum horse power that can be developed is %3.1f H.P'%(hp)

Maximum horse power that can be developed is 71.0 H.P


## Example 3.31 Page no : 84¶

In [24]:
import math

#Input data
vCH4 = 65.					#Composition by volume of CH4
vH2 = 2.					#Composition by volume of H2
vN2 = 2.					#Composition by volume of N2
vCO2 = 31.					#Composition by volume of CO2
O2 = 5.3					#Composition of O2 in dry exhaust gases when analysed in orsat apparatus
N2 = 83.					#Composition of N2 in dry exhaust gases when analysed in orsat apparatus
CO = 0.3					#Composition of CO in dry exhaust gases when analysed in orsat apparatus
CO2 = 11.4					#Composition of CO2 in dry exhaust gases when analysed in orsat apparatus
an = 79.					#Air contains 79% by volume of nitrogen

#Calculations
a = (100./(100-an))*(((vCH4/100)*2)+((vN2/100)*(1./2)))					#Total air required for complete combustion of 1 cu.m of gas in cu.m
xCO = (CO/2)					#O2 required to burn the CO in cu.m
xCO2 = CO					#CO2 formed in cu.m
tO2 = O2-xCO					#Total O2 in cu.m
tN2 = N2					#Total N2 in cu.m
tCO2 = CO2+xCO2					#Total CO2 in cu.m
T = tO2+tN2+tCO2					#Total mixture in cu.m
pCO2 = (tCO2*100)/T					#Percentage of CO2 in percent
mm = (a*100)					#Minimum air supply required for complete combustion of 100 cu.m of the gas in cu.m
an2 = (an/100)*mm					#N2 for this air in cu.m
tn2 = (an2+vN2)					#Total N2 in cu.m
v = (((vCH4+vCO2)*100)/pCO2)-(vCH4+vCO2+tn2)					#Increase in air supply for reduction in percentage of CO2 in cu.m
pea = (v*100)/mm					#Percentage of excess air. In textbook it is given wrong as 26.7 percent

#Output
print 'a) the air-fuel ratio by volume to give complete combustion is %3.3f  \
\nb) the percentage of excess air actually used in the test is %3.1f percent'%(a,pea)

a) the air-fuel ratio by volume to give complete combustion is 6.238
b) the percentage of excess air actually used in the test is 36.6 percent


## Example 3.32 Page no : 86¶

In [25]:
import math

#Input data
Vs = 9.45					#Swept volume in litres
Vc = 2.32					#Clearance volume in litres
m = 4.25					#Consumption of gas per hour in cu.m
N = 165.					#Speed in r.p.m
bhp = 5.62					#Brake horse power in h.p
nm = 73.4					#Mechanical efficiency in percent
CV = 3500.					#Calorific value in kcal per cubic meter
vn = 0.87					#Volumetric efficiency
g = 1.4					#Ratio of specific heats

#Calculations
tV = (Vs+Vc)*1000					#Total volume in c.c
rc = (tV/Vc)					#Compression ratio
na = (1-(1/rc**(g-1)))*100					#Air math.radians(numpy.arcmath.tan(ard efficiency in percent
W = (bhp*4500)/427					#Workdone per minute in kcal
Iw = (W/(nm/100))					#Indicated work in kcal/min
q = (m/60)*CV					#Heat supplied in kcal/min
ith = (Iw/q)*100					#Indicated thermal efficiency in percent
rn = (ith/na)*100					#Relative efficiency in percent
Vm = (Vs*1000)*vn					#Volume of mixture taken in per stroke in c.c
Vg = (m*2*10**6)/(60*N)					#Volume of gas taken in per stroke in c.c
Va = (Vm-Vg)					#Volume of air taken in per stroke in c.c
agr = (Va/Vg)					#Air gas ratio
CVc = (CV/(agr+1))					#Calorific value of charge in kcal

#Output
print 'Ratio of air to gas used is %3.2f  \
\nCalorific value of 1 cu.m of the mixture in the cylinder is %3.1f kcal'%(agr,CVc)

Ratio of air to gas used is 8.58
Calorific value of 1 cu.m of the mixture in the cylinder is 365.5 kcal


## Example 3.33 Page no : 87¶

In [26]:
import math

#Input data
d = 18.					#Bore in cm
l = 37.5					#Stroke in cm
N = 220.					#Speed in r.p.m
#Mean effective pressure in kg/cm**2
#Firing
pp = 5.9					#Positive loop
pn = 0.248					#Negative loop
#Mismath.sing
nn = 0.432					#Negative loop
bhp = 8.62					#Brake horse power in h.p
ex = 100.					#Explosions per minute
vg = 0.101					#Gas used in cu.m per minute

#Calculations
tc = (N/2)					#The number of cycles
nw = ex					#Number of working cycles
nm = (tc-nw)					#Number of mismath.sing cycles
ihp = ((l/100)*(3.14/4)*(d**2/4500))*((pp-pn)*(100-nn))					#Net I.H.P in h.p
fhp = (ihp-bhp)					#Friction horse power in h.p
W = ((pp-pn)*(3.14/4)*(d**2*(l/100)))					#Workdone per firing done in kg.m
Wp = (nn*(3.14/4)*d**2*(l/100))					#Workdone per pumping stroke in kg.m
n = ((fhp*4500)+(Wp*tc))/(W+Wp)					#Number of strokes
gf = (vg/nw)					#Gas per firing stroke in cu.m
gl = (n*gf)					#Gas per minute at no load in cu.m

#Output
print 'Friction horse power of the engine is %3.2f  Gas consumption at no load is %3.3f cu.m/min'%(fhp,gl)

Friction horse power of the engine is 3.31  Gas consumption at no load is 0.034 cu.m/min