# Chapter 3 : Internal Combustion Engines¶

## Example 3.1 Page no : 139¶

In :
# Variables
d = 200.;			#diameter of cylinder in mm
L = 300.;			#stroke of cylinder in mm
Vc = 1.73;			#Clearance volume in litres
imep = 650.;			#indicated mean effective pressure in kN/(m**2)
g = 6.2;			#gas consumption in (m**3)/h
CV = 38.5;			#Calorific value in MJ/(m**3)
y = 1.4;			#Ratio of specific heats
N = 150.;			#No. of firing cycles per minute

# Calculations
Vs = ((3.1415/4)*(d**2)*L)*(10**-6);			#Stroke volume in litres
Vt = Vs+Vc;			#Total volume in litres
rv = (Vt/Vc);			#Compression ratio
n = (1-(1/rv**(y-1)))*100;			#Air standard efficiency
IP = imep*(Vs*10**-3)*(N/60);			#Indicated power in kW
F = (g*CV*1000)/3600;			#Fuel energy input in kW
nT = (IP/F)*100;			#Indicated thermal efficiency

# Results
# 1st answer is wrong in book
print 'Air Standard Efficiency is %3.1f percent  \
\nIndicated Power is %3.1f kW \
\nIndicated thermal efficiency is %3.0f percent'%(n,IP,nT)

Air Standard Efficiency is 52.5 percent
Indicated Power is 15.3 kW
Indicated thermal efficiency is  23 percent


## Example 3.2 Page no : 140¶

In :
# Variables
Vs = 0.0008;			#Swept volume in m**3
Vc = 0.00015;			#Clearance volume in m**3
CV = 38.;			#Calorific value in MJ/(m**3)
v = 0.45;			#volume in m**3
IP = 81.5;			#Indicated power in kW
y = 1.4;			#Ratio of specific heats

# Calculations
rv = (Vs+Vc)/Vc;			#Compression ratio
n = (1-(1/rv**(y-1)));			#Air standard efficiency
Ps = (v*CV*1000.)/60;			#Power supplied in kW
nact = IP/Ps;			#Actual efficiency
nr = (nact/n)*100;			#Relative efficiency

# Results
print 'Relative Efficiency is %3.2f percent'%(nr)

# rounding error in book answer. please check.

Relative Efficiency is 54.77 percent


## Example 3.3 Page no : 141¶

In :
# Variables
n = 6.;			    #No. of cylinders
d = 0.61;			#Diameter in m
L = 1.25;			#Stroke in m
N = 2.;		    	#No.of revolutions per second
m = 340.;			#mass of fuel oil in kg
CV = 44200.;		#Calorific value in kJ/kg
T = 108.;			#Torque in kN-m
imep = 775.;		#Indicated mean efective pressure in kN/(m**2)

# Calculations
IP = (imep*L*3.1415*(d**2)*N)/(8);			#Indicated power in kW
TotalIP = (n*IP);			    #Total indicated power in kW
BP = (2*3.1415*N*T);			#Brake power in kW
PI = (m*CV)/3600.;			    #Power input in kW
nB = (BP/PI)*100.;		    	#Brake thermal efficiency
bmep = (BP*8)/(n*L*3.1415*(d**2)*2);			#Brake mean effective pressure in kN/(m**2)
nM = (BP/TotalIP)*100;			#Mechanical efficiency
bsfc = m/BP;	        		#Brake specific fuel consumption in kg/kWh

# Results
print 'Total Indicated Power is %3.1f kW  \
\nBrake Power is %3.1f kW \
\nBrake thermal efficiency is %3.1f percent  \
\nBrake mean effective pressure is %3.1f kN/m**2 \
\nMechanical efficiency is %3.1f percent  \
\nBrake specific fuel consumption is %3.3f kg/kW.hr'%(TotalIP,BP,nB,bmep,nM,bsfc)

Total Indicated Power is 1698.6 kW
Brake Power is 1357.1 kW
Brake thermal efficiency is 32.5 percent
Brake mean effective pressure is 619.2 kN/m**2
Mechanical efficiency is 79.9 percent
Brake specific fuel consumption is 0.251 kg/kW.hr


## Example 3.4 Page no : 142¶

In :
# Variables
Hm = 21.;			#Mean height of indicator diagram in mm
isn = 27.;			#indicator spring number in kN/(m**2)/mm
Vs = 14.;			#Swept volume in litres
N = 6.6;			#Speed of engine in rev/s
Pe = 77.;			#Effective brake load in kg
Re = 0.7;			#Effective vrake radius in m
mf = 0.002;			#fuel consumed in kg/s
CV = 44000.;			#Calorific value of fuel in kJ/kg
mc = 0.15;			#cooling water circulation in kg/s
Ti = 311.;			#cooling water inlet temperature in K
To = 344.;			#cooling water outlet temperature in K
C = 4.18;			#specific heat capacity of water in kJ/kg-K
Ee = 33.6;			#Energy to exhaust gases in kJ/s
g = 9.81;			#Acceleration due to geravity in m/(s**2)

# Calculations
imep = isn*Hm;			#Indicated mean efective pressure in kN/(m**2)
IP = (imep*Vs*N)/(2000);			#Indicated Power in kW
BP = (2*3.1415*N*g*Pe*Re)/1000;			#Brake Power in kW
nM = (BP/IP)*100;			#Mechanical efficiency
Ef = mf*CV;			#Eneergy from fuel in kJ/s
Ec = mc*C*(To-Ti);			#Energy to cooling water in kJ/s
Es = Ef-(BP+Ec+Ee);			#Energy to surroundings in kJ/s
p = (BP*100)/Ef;			#Energy to BP in %
q = (Ec*100)/Ef;			#Energy to coolant in %
r = (Ee*100)/Ef;			#Energy to exhaust in %
w = (Es*100)/Ef;			#Energy to surroundings in %

# Results
print 'Indicated Power is %3.1f kW  \
\nBrake Power is %3.0f kW  \
\nMechanical Efficiency is %3.0f percent  \
\nENERGY BALANCE                     kJ/s     Percentage \
\nEnergy from fuel                  %3.0f        100 \
\nEnergy to BP                      %3.0f        %3.0f \
\nEnergy to coolant                 %3.01f       %3.1f \
\nEnergy to exhaust                 %3.1f        %3.1f \
\nEnergy to surroundings, etc       %3.1f        %3.1f'%(IP,BP,nM,Ef,BP,p,Ec,q,Ee,r,Es,w)

Indicated Power is 26.2 kW
Brake Power is  22 kW
Mechanical Efficiency is  84 percent
ENERGY BALANCE                     kJ/s     Percentage
Energy from fuel                   88        100
Energy to BP                       22         25
Energy to coolant                 20.7       23.5
Energy to exhaust                 33.6        38.2
Energy to surroundings, etc       11.8        13.4


## Example 3.5 Page no : 143¶

In :
# Variables
t = 30.;			#duration of trial in minutes
N = 1750.;			#speed in rpm
T = 330.;			#brake torque in Nm
m = 9.35;			#mass of fuel in kg
CV = 42300.;			#Calorific value in kJ/kg
mj = 483.;			#jacket cooling water circulation in kg
Ti = 290.;			#inlet temperature in K
T0 = 350.;			#outlet temperature in K
ma = 182.;			#air consumption in kg
Te = 759.;			#exhaust temperature in K
Ta = 256.;			#atmospheric temperature in K
nM = 0.83;			#Mechanical efficiency
ms = 1.25;			#mean specific heat capacity of exhaust gas in kJ/kg-K
Cw = 4.18;			#specific heat capacity of water in kJ/kg-K

# Calculations
BP = (2*3.1415*T*N)/(60*1000);			#Brake power in kW
sfc = (m*2)/BP;			#specific fuel consumption in kg/kWh
IP = BP/nM;			#Indicated power in kW
nIT = IP/(m*2/3600*CV)*100    			#Indicated thermal efficiency
Ef = (m/t*CV)     			#Eneergy from fuel in kJ/min
EBP = BP*60;			#Energy to BP in kJ/min
Ec = (mj*Cw*(T0-Ti))/t;			#Energy to cooling water in kJ/min
Ee = ((ma+m)*ms*(Te-Ti))/30;			#Energy to exhaust in kJ/min
Es = Ef-(EBP+Ec+Ee);			#Energy to surroundings in kJ/min

# Results
print 'Break power is %3.1f kW  \
\nSpecific fuel consumption is %3.3f kg/kWh  \
\nIndicated thermal efficiency is %3.1f percent  \
\nEnergy from fuel is %3.0f kJ/min  \
\nEnergy to BP is %3.0f kJ/min  \
\nEnergy to cooling water is %3.0f kJ/min  \
\nEnergy to exhaust is %3.0f kJ/min  \
\nEnergy to surroundings is %d kJ/min'%(BP,sfc,nIT,round(Ef,-2),EBP,Ec,Ee,Es)

# rounding off error

Break power is 60.5 kW
Specific fuel consumption is 0.309 kg/kWh
Indicated thermal efficiency is 33.2 percent
Energy from fuel is 13200 kJ/min
Energy to BP is 3628 kJ/min
Energy to cooling water is 4038 kJ/min
Energy to exhaust is 3739 kJ/min
Energy to surroundings is 1777 kJ/min


## Example 3.6 Page no : 144¶

In :
# Variables
BP0 = 12.;			#Brake Power output in kW
BP1 = 40.5;			#Brake Power in trial 1 in kW
BP2 = 40.2;			#Brake Power in trial 2 in kW
BP3 = 40.1;			#Brake Power in trial 3 in kW
BP4 = 40.6;			#Brake Power in trial 4 in kW
BP5 = 40.7;			#Brake Power in trial 5 in kW
BP6 = 40.0;			#Brake Power in trial 6 in kW

# Calculations
BPALL = BP0+BP6;			#Total Brake Power in kW
IP1 = BPALL-BP1;			#Indicated Power in trial 1 in kW
IP2 = BPALL-BP2;			#Indicated Power in trial 2 in kW
IP3 = BPALL-BP3;			#Indicated Power in trial 3 in kW
IP4 = BPALL-BP4;			#Indicated Power in trial 4 in kW
IP5 = BPALL-BP5;			#Indicated Power in trial 5 in kW
IP6 = BPALL-BP6;			#Indicated Power in trial 6 in kW
IPALL = IP1+IP2+IP3+IP4+IP5+IP6;			#Total Indicated Power in kW
nM = (BPALL/IPALL)*100;			#Mechanical efficiency

# Results
print 'Indicated Power of the engine is %3.1f kW  \
\nMechanical efficiency of the engine is %3.1f percent'%(IPALL,nM)

Indicated Power of the engine is 69.9 kW
Mechanical efficiency of the engine is 74.4 percent


## Example 3.7 Page no : 145¶

In :
import math

# Variables
n = 2.;			#No. of cylinders
N = 4000.;			#speed of engine in rpm
nV = 0.77;			#Volumetric efficiency
nM = 0.75;			#Mechanical efficiency
m = 10.;			#fuel consumed in lit/h
g = 0.73;			#spcific gravity of fuel
Raf = 18.;			#air-fuel ratio
Np = 600.;			#piston speed in m/min
imep = 5.;			#Indicated mean efective pressure in bar
R = 281.;			#Universal gas constant in J/kg-K
T = 288.;			#Standard temperature in K
P = 1.013;			#Standard pressure in bar

# Calculations
L = Np/(2*N);			#Piston stroke in m
mf = m*g;			#mass of fuel in kg/h
ma = mf*Raf;			#mass of air required in kg/h
Va = (ma*R*T)/(P*60*(10**5));			#volume of air required in (m**3)/min
D = math.sqrt((2*Va)/(nV*L*N*3.1415));			#Diameter in m
IP = (2*imep*100*L*3.1415*(D**2)*N)/(4.*60);			#Indicated Power in kW
BP = nV*IP;			#Brake Power in kW

# Results
print 'Piston Stroke is %3.3f m  \
\nBore diameter is %3.4f m  \
\nBrake power is %3.1f kW'%(L,D,BP)

# rounding off error

Piston Stroke is 0.075 m
Bore diameter is 0.0694 m
Brake power is 14.6 kW