# Chapter 1 : Introduction¶

## Example 1.1 Page No : 3¶

In [1]:
import math

#Given:
n  =  4.      #Number of cylinders
d  =  68./10      #Bore in cm
l  =  75./10      #Stroke in cm
r  =  8.      #Compression ratio

#Solution:
V_s  =  (math.pi/4)*d**2*l      #Swept volume of one cylinder in cm**3
cubic_capacity  =  n*V_s      #Cubic capacity in cm**3
#Since, r   =   (V_c + V_s)/V_c
V_c  =  V_s/(r-1)      #Clearance volume in cm**3

#Results:
print " The cubic capacity of the engine   =   %.1f cm**3"%(cubic_capacity)

 The cubic capacity of the engine   =   1089.5 cm**3


## Example 1.2 Page No : 8¶

In [3]:
import math

#Given:
ip  =  10.      #Indicated power in kW
eta_m  =  80.      #Mechanical efficiency in percent

#Solution:
#Since, eta_m   =   bp/ip
bp  =  (eta_m/100)*ip      #Brake power in kW
fp  =  ip-bp      #Friction power in kW

#Results:
print " The brake power delivered, bp   =   %d kW"%(bp)
print " The friction power, fp   =   %d kW"%(fp)

 The brake power delivered, bp   =   8 kW
The friction power, fp   =   2 kW


## Example 1.3 Page No : 13¶

In [5]:
import math

#Given:
bp  =  100.      #Brake power at full load in kW
fp  =  25.      #Frictional power in kW (printing error)

#Solution:
eta_m  =  bp/(bp+fp)      #Mechanical efficiency at full load
bp  =  bp/2      #Brake power at half load in kW
eta_m1  =  bp/(bp+fp)      #Mechanical efficiency at half load
bp  =  bp/2      #Brake power at quarter load in kW
eta_m2  =  bp/(bp+fp)      #Mechanical efficiency at quarter load

#Results:
print " The mechanical efficiency at full load, eta_m   =   %d percent"%(eta_m*100)
print " The mechanical efficiency, \
\na)At half load, eta_m   =   %.1f percent \
\nb)At quarter load, eta_m   =   %d percent"%(eta_m1*100,eta_m2*100)

#Data in the book is printed wrong

 The mechanical efficiency at full load, eta_m   =   80 percent
The mechanical efficiency,
a)At half load, eta_m   =   66.7 percent
b)At quarter load, eta_m   =   50 percent


## Example 1.4 Page No : 18¶

In [6]:
import math
#Calculations on four stroke petrol engine
#Given:
bp  =  35.      #Brake power in kW
eta_m  =  80.      #Mechanical efficiency in percent
bsfc  =  0.4      #Brake specific fuel consumption in kg/kWh
A_F  =  14./1      #Air-fuel ratio
CV  =  43000.      #Calorific value in kJ/kg

#Solution:
#(a)
ip  =  bp*100/eta_m      #Indicated power in kW
#(b)
fp  =  ip-bp      #Frictional power in kW
#(c)
#Since, 1 kWh   =   3600 kJ
eta_bt  =  1/(bsfc*CV/3600)      #Brake thermal efficiency
#(d)
eta_it  =  eta_bt/eta_m*100      #Indicated thermal efficiency
#(e)
m_f  =  bsfc*bp      #Fuel consumption in kg/hr
#(f)
m_a  =  A_F*m_f      #Air consumption in kg/hr

#Results:
print " a)The indicated power, ip   =   %.2f kW \
\nb)The friction power, fp   =   %.2f kW"%(ip,fp)
print " c)The brake thermal efficiency, eta_bt   =   %.1f percent \
\nd)The indicated thermal efficiency, eta_it   =   %.1f percent"%(eta_bt*100,eta_it*100)
print " e)The fuel consumption per hour, m_f   =   %.1f kg/hr \
\nf)The air consumption per hour, m_a   =   %d kg/hr"%(m_f,m_a)

 a)The indicated power, ip   =   43.75 kW
b)The friction power, fp   =   8.75 kW
c)The brake thermal efficiency, eta_bt   =   20.9 percent
d)The indicated thermal efficiency, eta_it   =   26.2 percent
e)The fuel consumption per hour, m_f   =   14.0 kg/hr
f)The air consumption per hour, m_a   =   196 kg/hr


## Example 1.5 Page No : 23¶

In [7]:
#Given:
F_A  =  0.07/1      #Fuel-air ratio
bp  =  75.      #Brake power in kW
eta_bt  =  20.      #Brake thermal efficiency in percent
rho_a  =  1.2      #Density of air in kg/m**3
rho_f  =  4*rho_a      #Density of fuel vapour in kg/m**3
CV  =  43700.      #Calorific value of fuel in kJ/kg

#Solution:
m_f  =  bp*3600/(eta_bt*CV/100)      #Fuel consumption in kg/hr
m_a  =  m_f/F_A      #Air consumption in kg/hr
V_a  =  m_a/rho_a      #Volume of air in m**3/hr
V_f  =  m_f/rho_f      #Volume of fuel in m**3/hr
V_mixture  =  V_f+V_a      #Mixture volume in m**3/hr

#Results:
print " The air consumption, m_a   =   %.1f kg/hr"%(m_a)
print " The volume of air required, V_a   =   %.1f m**3/hr"%(V_a)
print " The volume of mixture required   =   %.1f m**3/hr"%(V_mixture)      #printing error)
#Answer in the book is printed wrong

 The air consumption, m_a   =   441.3 kg/hr
The volume of air required, V_a   =   367.8 m**3/hr
The volume of mixture required   =   374.2 m**3/hr


## Example 1.6 Page No : 28¶

In [9]:
#Given:
bp  =  5.      #Brake power in kW
eta_it  =  30.      #Indicated thermal efficiency in percent
eta_m  =  75.      #Mechanical efficiency in percent (printing error)

#Solution:
ip  =  bp*100/eta_m      #Indicated power in kW
CV  =  42000.      #Calorific value of diesel(fuel) in kJ/kg
m_f  =  ip*3600/(eta_it*CV/100)      #Fuel consumption in kg/hr
#Density of diesel(fuel)   =   0.87 kg/l
rho_f  =  0.87      #Density of fuel in kg/l
V_f  =  m_f/rho_f      #Fuel consumption in l/hr
isfc  =  m_f/ip      #Indicated specific fuel consumption in kg/kWh
bsfc  =  m_f/bp      #Brake specific fuel consumption in kg/kWh

#Results:
print " The fuel consumption of engine, m_f in, \
\na)kg/hr   =   %.3f kg/hr \
\nb)litres/hr   =   %.2f l/hr"%(m_f,V_f)
print " c)Indicated specific fuel consumption, isfc   =   %.3f kg/kWh"%(isfc)
print " d)Brake specific fuel consumption, bsfc   =   %.3f kg/kWh"%(bsfc)
#Data in the book is printed wrong

 The fuel consumption of engine, m_f in,
a)kg/hr   =   1.905 kg/hr
b)litres/hr   =   2.19 l/hr
c)Indicated specific fuel consumption, isfc   =   0.286 kg/kWh
d)Brake specific fuel consumption, bsfc   =   0.381 kg/kWh


## Example 1.7 Page No : 33¶

In [11]:
#Given:
bp  =  5000.      #Brake power in kW
fp  =  1000.      #Friction power in kW
m_f  =  2300.      #Fuel consumption in kg/hr
A_F  =  20./1      #Air-fuel ratio
CV  =  42000.      #Calorific value of fuel in kJ/kg

#Solution:
#(a)
ip  =  bp+fp      #Indicated power in kW
#(b)
eta_m  =  bp/ip      #Mechanical efficiency
#(c)
m_a  =  A_F*m_f      #Air consumption in kg/hr
#(d)
eta_it  =  ip*3600/(m_f*CV)      #Indicated thermal efficiency
#(e)
eta_bt  =  eta_it*eta_m      #Brake thermal efficiency

#Results:
print " a)The indicated power, ip   =   %d kW"%(ip)
print " b)The mechanical efficiency, eta_m   =   %d percent"%(eta_m*100)
print " c)The air consumption, m_a   =   %d kg/hr"%(m_a)
print " d)The indicated thermal efficiency, eta_it   =   %.1f percent \
\ne)The brake thermal efficiency, eta_bt   =   %.1f percent"%(eta_it*100,eta_bt*100)

 a)The indicated power, ip   =   6000 kW
b)The mechanical efficiency, eta_m   =   83 percent
c)The air consumption, m_a   =   46000 kg/hr
d)The indicated thermal efficiency, eta_it   =   22.4 percent
e)The brake thermal efficiency, eta_bt   =   18.6 percent