Chapter 16 : Combustion In Compresson Ignition Engines

Example 16.1 Page no : 315

In [1]:
import math 
from numpy import array
					
#Input data
s = 0.005					#Delay in sec
d = 30.					#Bore in cm
N = 600.					#Speed in r.p.m
dx = array([10,15,20])					#Bore diameters in cm

					
#Calculations
t = (s/d)*dx					#Time of delay in sec. In textbook, t[1] is given wrong as 0.00025 sec instead of 0.0025 sec

					
#Output
print 'The delay time for %i cm diameter bore is %3.5f sec  \
\nThe delay time for %i cm diameter bore is %3.5f sec  \
\nThe delay time for %i cm diameter bore is %3.5f sec'%(dx[0],t[0],dx[1],t[1],dx[2],t[2])

The delay time for 10 cm diameter bore is 0.00167 sec  
The delay time for 15 cm diameter bore is 0.00250 sec  
The delay time for 20 cm diameter bore is 0.00333 sec

Example 16.2 Page no : 317

In [3]:
import math 
					
#Input data
d = [15.,60.]					#Bore in cm
N = [1600.,400.]					#Speed in r.p.m respectively
q = 30.					#Injection of oil occupies 30 degrees of crank travel in each case
pc = 30.					#Compression pressure in kg/cm**2
d = 0.001					#Delay time in sec
rp = 5.					#Rapid combustion period is 5 degree of crank travel
pe = 60.					#Compression pressure at the end of rapid compression in kg/cm**2

					
#Calculations
					#For small engine
It1 = (60/N[0])*(q/360)					#Injection time in sec
pf1 = ((d/It1)+(rp/pc))*100					#Percent fuel
					#For large engine
It2 = (60/N[1])*(q/360)					#Injection time in sec
pf2 = ((d/It2)+(rp/pc))*100					#Percent fuel
pr = (pc*(pf2/pf1))					#Pressure rise in kg/cm**2
mp = (pc+pr)					#Maximum pressure in kg/cm**2

					
#Output
print 'Pressure in the large engine is %3.1f kg/cm**2'%(mp)

Pressure in the large engine is 45.2 kg/cm**2

Example 16.3 Page no : 320

In [5]:
import math 
					
#Input data
n = 4.					#Number of cylinders
d = 105.					#Bore in mm
l = 127.					#Stroke in mm
BHP = 63.					#Brake horse power in h.p
N = 1800.					#Speed in r.p.m
t = 15.					#Test time in min
mf = 2.75					#Mass of fuel in kg
CV = 11000.					#Calorific value in kcal/kg
af = 14.8					#Air fuel ratio
v = 0.805					#Specific volume in m**3/kg
nv = 80.					#Volumetric efficiency in percent
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

					
#Calculations
bth = ((BHP*4500)/(J*(mf/t)*CV))*100					#Brake thermal efficiency in percent
Vs = ((3.14/4)*(d/10)**2*(l/10))					#Stroke volume in c.c
Vsw = (Vs*n*(N/2)*t)					#Swept volume in c.c
Va = (Vsw*10**-6*(nv/100))					#Volume of air sucked in m**3
wa = (Va/v)					#Weight of air sucked in kg
wr = (af*mf)					#Weight of air reqired in kg
pei = (wr/wa)*100					#Percentage of air available for combustion

					
#Output
print 'Brake thermal efficiency is %3.1f percent  \
\nThe percentage of air used for combustion is %i percent'%(bth,pei)

Brake thermal efficiency is 32.9 percent  
The percentage of air used for combustion is 69 percent