Chapter 6 : Combustion Charts

Example 6.1 Page no : 123

In [2]:
import math 
					
#Input data
r = 7.5					#Compression ratio
					#Data from combustion chart
p = [1,15.1,26.95,1.95]					#Pressure of air fuel mixture in kg/cm**2
T = [60,460,1150,435]					#Temperature of air fuel mixture in K
V = [16.98,2.264,2.264,16.98]					#Volume in m**3/kg
U = [17,78.8,212,80]					#Internal energy in kcal/kg
S = [0.07,0.07,0.22,0.22]					#Entropy in kcal/kg.degree C
g = 1.4					#Ratio of specific heats

					
#Calculations
n = (((U[2]-U[3])-(U[1]-U[0]))/(U[2]-U[1]))*100					#Thermal efficiency in percent
na = (1-(1/r)**(g-1))*100					#Air standard efficiency in percent

					
#Output
print 'Thermal efficiency is %3.1f percent  \
\nAir standard efficiency is %3.1f percent'%(n,na)

Thermal efficiency is 52.7 percent  
Air standard efficiency is 55.3 percent

Example 6.2 Page no : 127

In [4]:
import math 
					
#Input data
					#Data from combustion chart
p = [1,33,33,1]					#Pressure of air fuel mixture in kg/cm**2
T = [65,600,1450,725]					#Temperature of air fuel mixture in K
V = [16,1.23,3.45,16]					#Volume in m**3/kg
U = [11.8,110,295,140]					#Internal energy in kcal/kg
H = [22.7,150,395,225]					#Enthalpy in kcal/kg
S = [0.068,0.068,0.264,0.264]					#Entropy in kcal/kg.degree C

					
#Calculations
r = (V[0]/V[1])					#Compression ratio
q = (H[2]-H[1])					#Heat supplied in kcal/kg
qre = (U[3]-U[0])					#Heat rejected in kcal/kg
nt = ((q-qre)/q)*100					#Thermal efficiency in percent

					
#Output
print 'a)Compression ratio is %3.0f  \
\nb) Heat supplied to the cycle is %3.0f kcal/kg  \
\nc) Heat rejected by the cycle is %3.2f kcal/kg  \
\nd) Thermal efficiency is %3.2f percent'%(r,q,qre,nt)

a)Compression ratio is  13  
b) Heat supplied to the cycle is 245 kcal/kg  
c) Heat rejected by the cycle is 128.20 kcal/kg  
d) Thermal efficiency is 47.67 percent

Example 6.3 Page no : 130

In [5]:
import math 
					
#Input data
					#Data from combustion chart
p = [1,51.5,77.25,77.25,3.75]					#Pressure of air fuel mixture in kg/cm**2
T = [16,1,1,1.5,16]					#Temperature of air fuel mixture in K
V = [65,745,1400,2200,1030]					#Volume in m**3/kg
U = [14.7,135,275,475,197]					#Internal energy in kcal/kg
H = [21.9,85,372,625,280]					#Enthalpy in kcal/kg
S = [0.068,0.068,0.19,0.32,0.32]					#Entropy in kcal/kg.degree C

					
#Calculations
nth = (((U[2]-U[1])+(H[3]-H[2])-(U[4]-U[0]))/((U[2]-U[1])+(H[3]-H[2])))*100					#Thermal efficiency in percent

					
#Output
print 'Thermal efficiency is %3.2f percent'%(nth)

Thermal efficiency is 53.61 percent