# Chapter 5 : Variable Specific Heat¶

## Example 5.2 Page no : 105¶

In [1]:
import math

#Input data
r = 8.					#Compression ratio
n = 1.41					#Adiabatic index of the medium
cv = 0.17					#Mean Specific heat at consmath.tant volume in kcal/kg/degree C
x = 2.					#Percentage with which spcific heat at consmath.tant volume increases
R = 29.3					#Characteristic gas consmath.tant in mkg/kg/degree C
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
e = (1-(1/r**(n-1)))					#Air standard efficiency neglecting the variation in specific heat
debye = ((x/100)*((1-e)/e)*(R/(J*cv))*math.log(r))*100					#Ratio of de and e in percent

#Output
print 'The change in air standard efficiency of the cycle is %3.3f percent'%(debye)

The change in air standard efficiency of the cycle is 1.247 percent


## Example 5.3 Page no : 106¶

In [2]:
import math

#Input data
#Cv = 0.125+0.000005T where Cv is Specific heat at consmath.tant volume and T is the temperature in K
R = 28.9					#Characteristic gas consmath.tant in mkg/kg/degree C
T = [100+273,50+273]					#Temperature in K
J = 427					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
def f(x):
return  0.125+(0.00005*x)

#Output
print 'The work done is %i m.kg/kg of gas'%(I)

The work done is 4241 m.kg/kg of gas


## Example 5.4 Page no : 109¶

In [3]:
import math

#Input data
af = 25.					#Air fuel ratio
cv = [0.17,0.00004]					#Cv = 0.17+0.00004T where Cv is Specific heat at constant volume and T is the temperature in K
r = 14.					#Compression ratio
p1 = 1.					#Pressure at the beginning of compression in kg/cm**2
T1 = 153.+273					#Temperature at the beginning of compression in K
CV = 10000.					#Heating value of fuel in kcal/kg
R = 29.					#Characteristic gas constant in mkg/kg.K
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
T2 = (T1*r**(n-1))					#Temperature at the end of compression in K
a = (cv[1]/2)					#For solving T3
b = cv[0]+(R/J)					#For solving T3
c = (-T2*cv[0])-((cv[1]/2)*T2**2)-((R/J)*T2)-(CV/(af+1))					#Foe solving T3
T3 = (-b+math.sqrt(b**2-(4*a*c)))/(2*a)					#Soving for T3 in K
pc = (((T3/T2)-1)/(r-1))*100					#Percentage cut off

#Output
print 'The percentage of stroke at which the constant pressure combustion stops is %i percent'%(pc)

The percentage of stroke at which the constant pressure combustion stops is 9 percent


## Example 5.5 Page no : 113¶

In [4]:
import math

#Input data
af = 25.					#Air fuel ratio
CV = 10000.					#Calorific value in kcal/kg
cv = [0.17,0.00004]					#Cv = 0.17+0.00004T where Cv is Specific heat at constant volume and T is the temperature in K
r = 14.					#Compression ratio
T2 = 800.+273					#Temperature at the end of compression in K
R = 29.					#Characteristic gas constant in mkg/kg/degree C
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
CVm = (CV/(af+1))					#Calorific value of mixture in kcal/kg
cpv = (R/J)					#Difference in mean specific heats in kcal/kg mol.K
a = (cv[1]/2)					#For solving T3
b = cpv+cv[0]					#For solving T3
c = (-T2*(cpv+cv[0]))-((cv[1]/2)*T2**2)-CVm					#Foe solving T3
T3 = (-b+math.sqrt(b**2-(4*a*c)))/(2*a)					#Soving for T3 in K
s = ((T3/T2)/(r-1))*100					#Percentage of the stroke

#Output
print 'The percentage of the stroke at which the combustion will be complete is %3.2f percent'%(s)

The percentage of the stroke at which the combustion will be complete is 16.70 percent


## Example 5.6 Page no : 115¶

In [5]:
import math

#Input data
T = [500,2000]					#Change in temperature in K
x = [11.515,-172,1530]					#Cp = 11.515-172/math.sqrt(T)+1530/T in kcal/kg mole.K
mO2 = 32					#Molecular weight of oxygen

#Calculations

def f(T):
return (x[0]+(x[1]/math.sqrt(T))+(x[2]/T))

dh = (I/mO2)					#Change in enthalpy in kcal/kg

#Output
print 'The change in enthalpy is %3.1f kcal/kg'%(dh)

The change in enthalpy is 365.7 kcal/kg


## Example 5.7 Page no : 117¶

In [6]:
import math

#Input data
r = 14.					#Compression ratio
s = 5.					#Fuel injection stops at 5% stroke after inner head centre
pm = 50.					#Maximum pressure in kg/cm**2
p4 = 1.					#Pressure at the end of suction stroke in kg/cm**2
T4 = 90.+273					#Temperature at the end of suction stroke in K
R = 29.3					#Characteristic gas constant in mkg/kg/degree C
cv = [0.171,0.00003]					#Cv = 0.171+0.00003T where Cv is Specific heat at constant volume and T is the temperature in K
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
a = (R/J)+cv[0]					#a value in kcal/kg.mole.K
g = (a+cv[1]*T4)/(cv[0]+cv[1]*T4)					#Adiabatic index of compression
z = 1.3					#Rounding off 'z' value to one decimal.
T5 = (T4*r**(z-1))					#Temperature in K
p5 = (p4*r**g)					#Pressure in kg/cm**2
T1 = T5*(pm/p5)					#Tmperature in K
T2 = (T1*(1+(s/100)*(r-1)))					#Temperature in K
T3 = (T2*((1+(s/100)*(r-1))/r)**(g-1))					#Temperature in K
p3 = (p4*(T3/T4))					#Pressure in kg/cm**2
def f1(T):
return  cv[0]+(cv[1]*T)

def f2(T):
return (a+(cv[1]*T))

qs = (I1+I2)					#Heat supplied per kg of air in kcal/kg

def f3(T):
return  a+(cv[1]*T)

qre = quad(f3,T4,T3)[0]					#Heat required per kg of air in kcal/kg

nth = ((qs-qre)/qs)*100					#Thermal efficiency in percent

#Output
print 'The tempertautes and pressures at salient points of the cycle are :  T1  =  %3.0f K  \
\np1  =  %3.1f kg/cm**2  \
\nT2  =  %3.0f K  \
\np2  =  %3.1f kg/cm**2  \
\nT3  =  %3.0f K  \
\np3  =  %3.1f kg/cm**2  \
\nT4  =  %3.0f K  \
\np4  =  %3.1f kg/cm**2  \
\nT5  =  %3.0f K  \
\np5  =  %3.1f kg/cm**2  \
\nHeat supplied per kg of air is %3.1f kcal/kg  \
\nThe thermal efficiency of the cycle is %3.1f percent'%(T1,pm,T2,pm,T3,p3,T4,p4,T5,p5,qs,nth)


The tempertautes and pressures at salient points of the cycle are :  T1  =  1057 K
p1  =  50.0 kg/cm**2
T2  =  1745 K
p2  =  50.0 kg/cm**2
T3  =  779 K
p3  =  2.1 kg/cm**2
T4  =  363 K
p4  =  1.0 kg/cm**2
T5  =  801 K
p5  =  37.9 kg/cm**2
Heat supplied per kg of air is 244.5 kcal/kg
The thermal efficiency of the cycle is 56.4 percent


## Example 5.8 Page no : 119¶

In [7]:
import math

#Input data
r = 14.					#Compression ratio
c = 5.					#Cut off takes place at 5% of the stroke
p1 = 1.					#Pressure at the beginning of compression in kg/cm**2. In texbook, it is given wrong as 10
T1 = 90.+273					#Temperature at the beginning of compression in K
p3 = 50.					#Maximum pressure in kg/cm**2
R = 29.3					#Characteristic gas constant in mkg/kg/degree C
cv = [0.171,0.00003]					#Cv = 0.171+0.00003T where Cv is Specific heat at constant volume and T is the temperature in K
g1 = 1.4					#Ratio of specific heats
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
T2x = (T1*r**(g1-1))					#Temperature in K
def  f1(T):
return  cv[0]+(cv[1]*T)

Cv = (1/(T2x-T1))*I1					#Mean value of Cv in kJ/kg.K
Cp = (Cv+(R/J))					#Mean value of Cp in kJ/kg.K
g = 1.35					#(Cp/Cv) value and rounded off to 2 decimal places for calculation purpose. Ratio of specific heats
T2 = (T1*r**(g-1))					#Temperature in K
CV = (1/(T2-T1))*I2					#Maen value of Cv in kJ/kg.K
CP = (Cv+(R/J))					#Mean value of Cp in kJ/kg.K
g2 = 1.36					#(Cp/Cv) value and rounded off to 2 decimal places for calculation purpose.Ratio of specific heats
T2a = (T1*r**(g2-1))					#Temperature in K
p2 = (p1*r*(T2a/T1))					#Pressure in kg/cm**2
T3 = (T2a*(p3/p2))					#Temperature in K
T4 = (((r-1)*(c/100))+1)*T3					#Temperature in K
g3 = 1.3					#Assuming gamma as 1.3 for process 4-5
T5 = (T4/(r/(((r-1)*(c/100))+1))**(g3-1))					#Temperature in K
cV = cv[0]+(cv[1]/2)*(T5+T4)					#Mean value of Cv in kJ/kg.K
cP = cV+(R/J)					#Mean value of Cp in kJ/kg.K
g4 = (cP/cV)					#Ratio of specific heats
T5a = (T4/(r/(((r-1)*(c/100))+1))**(g4-1))

def f2(T):
return cv[0]+(R/J)+(cv[1]*T)
q = I3+I4					#Heat supplied per kg of working substance in kcal/kg

#Output
print 'a) Temperatures at all the points of the cycle are:  \
\nT1  =  %i K  T2  =  %3.0f K  T3  =  %3.0f K  T4  =  %3.0f K  T5  =  %i K  \
\nb) heat supplied per kg of the working substance is %3.1f kcal/kg'%(T1,T2a,T3,T4,T5a,q)

a) Temperatures at all the points of the cycle are:
T1  =  363 K  T2  =  939 K  T3  =  1296 K  T4  =  2139 K  T5  =  1097 K
b) heat supplied per kg of the working substance is 318.5 kcal/kg


## Example 5.10 Page no : 119¶

In [8]:
import math

#Input data
r = 20.					#Compression ratio
c = 5.					#Cut off at 5%
dc = 1.					#Specific heat at constant volume increases by 1%
Cv = 0.171					#pecific heat at constant volume in kJ/kg.K
R = 29.3					#Characteristic gas constant in mkg/kg/degree C
k = 1.95					#k can be obtained from relation de/e = -dcv/cv*(1-e/e)*(g-1)*((1/g)+ln(r)-(k**g*lnk)/(k**g-1))
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
g = (R/(J*Cv))+1					#Ratio of specific heats
e = (1-((1/g)*(1/r**(g-1))*((k**g-1)/(k-1))))					#Air standard efficiency of the cycle
dee = ((-(dc/100)*((1-e)/e)*(g-1)*((1/g)+math.log(r)-((k**g*math.log(k))/(k**g-1))))*100)					#Change in efficiency due to 1% change in cv

#Output
print 'Percentage change in air standard efficiency is %3.3f percent This indicates that there is a decrease in efficiency'%(dee)

Percentage change in air standard efficiency is -0.564 percent This indicates that there is a decrease in efficiency