# Chapter 2 : Air Standard Cycles¶

## Example 2.1 Page No : 38¶

In [1]:
#Given:
T2 = 27.+273      #Temperature of cooling pond in K
eta = 30.      #Efficiency in percent
Q2 = 200.      #Heat received by cooling pond in kJ/s

#Solution:
#Since, eta  =  (Q1-Q2)/Q1  =  (T1-T2)/T1
T1 = T2/(1-(eta/100))      #Temperature of heat source in K
Q1 = Q2/(1-(eta/100))      #Heat supplied by source in kJ/s
Power = round(Q1-Q2)      #Power of engine in kJ/s

#Results:
print " Temperature of heat source, T1  =  %.1f degreeC"%(T1-273)
print " Power of engine  =  %d kW"%(Power)

 Temperature of heat source, T1  =  155.6 degreeC
Power of engine  =  86 kW


## Example 2.2 Page No : 43¶

In [3]:
import math

#Given:
T3 = 800+273.
T1 = 15.+273      #Temperature of a hot and cold reservoir in K
P3 = 210.
P1 = 1      #Maximum and minimum pressure in bar

#Solution:
#Refer fig 2.21
eta_carnot = 1-(T1/T3)      #Efficiency of Carnot cycle
T4 = T3      #Isothermal process 3-4
g = 1.4      #Specific heat ratio(gamma)
P4 = P1*(T4/T1)**(g/(g-1))      #Initial pressure of isentropic process 4-1 in bar
R = 0.287      #Specific gas consmath.tant in kJ/kgK
Q3_4 = R*T3*math.log(P3/P4)      #Heat supplied in kJ/kg
W3_4 = Q3_4      #Work supplied in kJ/kg
Net_work = eta_carnot*Q3_4      #Net work output in kJ/kg
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
W4_1 = cv*(T4-T1)      #Work for isentropic process in kJ/kg
Gross_work = W3_4+W4_1      #Gross work supplied in kJ/kg
work_ratio = Net_work/Gross_work      #Work ratio

#Results:
print " The efficiency of the Carnot cycle, eta_carnot  =  %.1f percent"%(eta_carnot*100)
print " The work ratio of the Carnot cycle  =  %.3f"%(work_ratio)

 The efficiency of the Carnot cycle, eta_carnot  =  73.2 percent
The work ratio of the Carnot cycle  =  0.211


## Example 2.3 Page No : 48¶

In [5]:
import math

#Given:
d = 17.
l = 30.      #Bore and stroke in cm
V_c = 0.001025      #Clearance volume in m**3

#Solution:
V_s = (math.pi/4)*d**2*l      #Swept volume in cc
V_c = V_c*10**6      #Clearance volume in cc
V = V_c+V_s      #Total cylinder volume in cc
r = V/V_c      #Compression ratio
g = 1.4      #Specific heat ratio(gamma)
eta = 1-1/r**(g-1)      #Air standard efficiency

#Results:
print " The Air standard efficiency of Otto cycle, eta  =  %.1f percent"%(eta*100)

 The Air standard efficiency of Otto cycle, eta  =  55.7 percent


## Example 2.4 Page No : 53¶

In [6]:
#Given:
P1 = 97.      #Pressure at the beginning(1) in kN/m**2
T1 = 40.+273      #Temperature at the beginning(1) in K
r = 7.      #Compression ratio
Q = 1200.      #Heat supplied in kJ/kg
g = 1.4      #Specific heat ratio(gamma)
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK

#Solution:
#(a)
T2 = T1*(r)**(g-1)
T3 = round(Q/cv+T2)      #Temperature at 2, 3 in K
#(b)
eta = 1-1/r**(g-1)      #Thermal efficiency
#(c)
W = Q*eta      #Workdone per cycle in kJ/kg

#Results:
print " a)The maximum temperature attained in the cycle, T3  =  %d degreeC"%(T3-273)
print " b)The thermal efficiency of the cycle, eta  =  %.1f percent"%(eta*100)
print " c)The workdone during the cycle/kg of working fluid, W  =  %d kJ"%(W)

 a)The maximum temperature attained in the cycle, T3  =  2080 degreeC
b)The thermal efficiency of the cycle, eta  =  54.1 percent
c)The workdone during the cycle/kg of working fluid, W  =  649 kJ


## Example 2.5 Page No : 58¶

In [7]:
#Given:
r = 8.      #Compression ratio
P1 = 1.
P3 = 50.      #Pressure at 1, 3 in bar
T1 = 100.+273      #Temperature at 1 in K
m = 1.      #Air flow in kg
R = 0.287      #Specific gas consmath.tant in kJ/kgK
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.22
#Point 1
V1 = m*R*10**3*T1/(P1*10**5)      #Ideal gas equation, Volume at 1 in m**3
#Point 2
P2 = P1*r**g      #Pressure at 2 in bar
V2 = V1/r      #Volume at 2 in m**3
T2 = P2*V2*T1/(P1*V1)      #Temperature at 2 in K
#Point 3
V3 = V2      #Constant volume process, Volume at 3 in m**3
T3 = (P3/P2)*T2      #Temperature at 3 in K (Wrong in book)
#Point 4
P4 = P3*(1/r)**g      #Pressure at 4 in bar
V4 = V1      #Constant volume process, Volume at 4 in m**3
T4 = T1*(P4/P1)      #Temperature at 4 in K
cv = R/(g-1)      #Specific heat at consmath.tant volume in kJ/kgK
ratio = (cv*(T3-T2))/(cv*(T4-T1))      #Ratio of heat supplied to the heat rejected (Round off error)

#Results:
print " Point 1: Pressure  =  %d bar, Volume  =  %.4f m**3, Temperature  =  %d degreeC"%(P1,V1,T1-273)
print " Point 2: Pressure  =  %.1f bar, Volume  =  %.4f m**3, Temperature  =  %.1f degreeC"%(P2,V2,T2-273)
print " Point 3: Pressure  =  %.1f bar, Volume  =  %.4f m**3, Temperature  =  %.1f degreeC"%(P3,V3,T3-273)
print " Point 4: Pressure  =  %.2f bar, Volume  =  %.4f m**3, Temperature  =  %.1f degreeC"%(P4,V4,T4-273)
print " Ratio of heat supplied to the heat rejected  =  %.3f"%(ratio)
#Textbook answer for T3 is wrong
#Round off error in the value of 'ratio'

 Point 1: Pressure  =  1 bar, Volume  =  1.0705 m**3, Temperature  =  100 degreeC
Point 2: Pressure  =  18.4 bar, Volume  =  0.1338 m**3, Temperature  =  583.9 degreeC
Point 3: Pressure  =  50.0 bar, Volume  =  0.1338 m**3, Temperature  =  2058.2 degreeC
Point 4: Pressure  =  2.72 bar, Volume  =  1.0705 m**3, Temperature  =  741.7 degreeC
Ratio of heat supplied to the heat rejected  =  2.297


## Example 2.6 Page No : 63¶

In [8]:
#Given:
P1 = 1.      #Pressure at 1 in bar
T1 = 15.+273      #Temperature at 1 in K
r = 8.      #Compression ratio
Q1 = 1000.      #Heat added in kJ/kg
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.23
#(a)
P2 = P1*(r)**g      #Pressure at 2 in bar
T2 = T1*r**(g-1)      #Temperature at 2 in K
T3 = Q1/cv+T2      #Temperature at 3 in K (Round off error)
#(b)
eta = 1-1/r**(g-1)      #Air standard efficiency
#(c)
W = Q1*eta      #Work done in kJ/kg (Round off error)
#(d)
Q2 = Q1-W      #Heat rejected in kJ/kg

#Results:
print " a)The maximum temperature in the cycle, T3  =  %d degreeC"%(T3-273)
print " b)The air standard efficiency, eta  =  %.1f percent"%(eta*100)
print " c)The workdone per kg of air  =  %d kJ/kg"%(W)
print " d)The heat rejected  =  %d kJ/kg"%(Q2)
#Round off error in the values of 'T3' and 'W'

 a)The maximum temperature in the cycle, T3  =  1781 degreeC
b)The air standard efficiency, eta  =  56.5 percent
c)The workdone per kg of air  =  564 kJ/kg
d)The heat rejected  =  435 kJ/kg


## Example 2.7 Page No : 68¶

In [9]:
#Given:
P1 = 1.05
P2 = 13.
P3 = 35.      #Pressure at 1, 2, 3 in bar
T1 = 15.+273      #Temperature at 1 in K
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
R = 0.287      #Specific gas consmath.tant in kJ/kgK

#Solution:
r = "V1/V2"      #Compression ratio
g = R/cv+1      #Specific heat ratio(gamma)
r = (P2/P1)**(1/g)      #By adiabatic process relation
eta = 1-1/r**(g-1)      #Air standard efficiency
T2 = P2*T1/(P1*r)      #Temperature at 2 in K
T3 = (P3/P2)*T2      #Temperature at 3 in K
Q1 = cv*(T3-T2)      #Heat added in kJ/kg
W = Q1*eta      #Work done in kJ/kg
V1 = 1*R*10**3*T1/(P1*10**5)      #Ideal gas equation, Volume at 1 in m**3/kg
V2 = V1/r      #Volume at 2 in m**3/kg
V_s = V1-V2      #Swept volume in m**3/kg
mep = W*1000/(V_s*10**5)      #Mean effective pressire in bar

#Results:
print " The air standard efficiency, eta  =  %.1f percent"%(eta*100)
print " The compression ratio, r  =  %d"%(r)
print " The mean effective pressure, mep  =  %.2f bar"%(mep)

 The air standard efficiency, eta  =  51.3 percent
The compression ratio, r  =  6
The mean effective pressure, mep  =  5.60 bar


## Example 2.8 Page No : 73¶

In [10]:
#Given:
r = 8.      #Compression ratio
T1 = 20.+273      #Temperature at 1 in K
P1 = 1.      #Pressure at 1 in bar
Q1 = 1800.      #Heat added in kJ/kg
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
g = 1.4      #Specific heat ratio(gamma)

#Solution:
T2 = T1*r**(g-1)      #Temperature at 2 in K
T3 = Q1/cv+T2      #Temperature at 3 in K (printing error)
P2 = P1*(r)**g      #Pressure at 2 in bar
P3 = P2*(T3/T2)      #Pressure at 3 in bar
T4 = T3/r**(g-1)      #Temperature at 4 in K
eta = 1-1./r**(g-1)      #Air standard efficiency
W1_2 = cv*(T1-T2)      #Work done for process 1-2 in kJ/kg
W3_4 = cv*(T3-T4)      #Work done for process 3-4 in kJ/kg
W = W1_2+W3_4      #Net work done for the cycle in kJ/kg
V1 = cv*(g-1)*10**3*T1/(P1*10**5)      #Ideal gas equation, Volume at 1 in m**3/kg
V2 = V1/r      #Volume at 2 in m**3/kg
V_s = V1-V2      #Swept volume in m**3/kg
mep = W*1000/(V_s*10**5)      #Mean effective pressire in bar

#Results:
print " The maximum temperature, T3  =  %d K"%(T3)
print " The maximum pressure, P3  =  %.1f bar"%(P3)
print " The temperature at the end of the expansion process, T4  =  %d K"%(T4)
print " The air standard efficiency, eta  =  %.1f percent"%(eta*100)
print " The mean effective pressure of the cycle, mep  =  %.1f bar"%(mep)
#Answers in the book are wrong

 The maximum temperature, T3  =  3180 K
The maximum pressure, P3  =  86.8 bar
The temperature at the end of the expansion process, T4  =  1384 K
The air standard efficiency, eta  =  56.5 percent
The mean effective pressure of the cycle, mep  =  13.8 bar


## Example 2.9 Page No : 78¶

In [12]:
import math

#Given:
power = 50.      #Internal power in kW
N = 4800.      #Engine speed in rpm
l = 80.
d = 80.      #Stroke and bore of engine in mm
n = 4.      #Number of cylinders
V_c = 50000.      #Clearance volume in mm**3
delta_P = 45.      #Pressure rise during combustion in bar
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.24
V_s = (math.pi/4)*d**2*l      #Swept volume in mm**3
r = (V_c+V_s)/V_c      #Compression ratio
eta = 1-1/r**(g-1)      #Air standard efficiency
ideal_mep = eta*delta_P/((g-1)*(r-1))      #Ideal mean effective pressure in bar
W = power*60*2/(n*N)      #Actual work transfer per cycle per cylinder in kJ
V_s = V_s*1e-9      #Swept volume in m**3
actual_mep = W*1000/(V_s*10**5)      #Actual mean effective pressire in bar

#Results:
print " The mean effective pressure of the engine, actual mep  =  %.2f bar"%(actual_mep)
print " The mean effective pressure of the Otto cycle, ideal mep  =  %.2f bar"%(ideal_mep)

 The mean effective pressure of the engine, actual mep  =  7.77 bar
The mean effective pressure of the Otto cycle, ideal mep  =  8.19 bar


## Example 2.10 Page No : 83¶

In [13]:
from scipy.optimize import fsolve
import math

#Given:
CV = 42000.      #Calorific value of the fuel in kJ/kg
a = 30./100
b = 70./100      #Fraction of compression stroke at point a, b
P_a = 1.33
P_b = 2.66      #Pressure at point a, b
n = 1.33      #Polytropic index
eta_cycle = 50./100      #Air standard cycle efficiency

#Solution:
#Refer fig 2.25
#Since, compression follows PV**n  =  C
#Thus, P_a*V_a**n  =  P_b*V_b**n
#Assume a_b  =  V_a/V_b
a_b = (P_b/P_a)**(1/n)      #Ratio of volume at a to volume at b
#Defining the function, ratio of r(compression ratio)
def Volume(r):
V_a = 1+0.7*(r-1)
V_b = 1+0.3*(r-1)
return V_a/V_b-a_b

r = fsolve(Volume,1)
g = 1.4      #Specific heat ratio(gamma)
eta = round(1000*(1-1/r**(g-1)))/1000      #Air standard efficiency
eta_it = eta_cycle*eta      #Indicated thermal efficiency
#Since, 1 kWh  =  3600 kJ
Q1 = 3600/eta_it      #Heat supplied in kJ/kWh
isfc = Q1/CV      #Indicated specific fuel consumption in kg/kWh

#Results:
print " The compression ratio, r  =  %.2f"%(r)
print " The fuel consumption, isfc  =  %.3f kg/kWh"%(isfc)

 The compression ratio, r  =  4.51
The fuel consumption, isfc  =  0.378 kg/kWh


## Example 2.11 Page No : 88¶

In [14]:
#Given:
r = 14.      #Compression ratio
P1 = 1.      #Pressure at 1 in bar
T1 = 27+273.
T3 = 2500.+273      #Temperature at 1 and 3 in K

#Solution:
#Refer fig 2.26
g = 1.4      #Specific heat ratio(gamma)
T2 = T1*(r)**(g-1)      #Temperature at 2 in K
P2 = P1*(T2/T1)**(g/(g-1))      #Pressure at 2 in bar
rho = T3/T2      #Cut off ratio
T3_T4 = (r/rho)**(g-1)      #Temperature ratio T3/T4
T4 = round(T3/T3_T4)      #Temperature at 4 in K
eta = 1-((T4-T1)/(g*(T3-T2)))      #Efficiency of diesel cycle
R = 0.287
cp = 1.005
cv = 0.718      #Specific gas consmath.tant, heat capacities at consmath.tant pressure and volume in kJ/kgK
V1 = R*T1*10**3/(P1*10**5)      #Volume at 1 in m**3/kg
V_s = V1*(1-1/r)      #Stroke volume in m**3/kg
mep = (cp*(T3-T2)-cv*(T4-T1))*10**3/(V_s*10**5)      #Mean effective pressure in bar

#Results:
print " The thermal efficiency of the diesel cycle, eta  =  %.1f percent"%(eta*100)
print " The mean effective pressure of the cycle, pm  =  %.2f bar"%(mep)

 The thermal efficiency of the diesel cycle, eta  =  53.6 percent
The mean effective pressure of the cycle, pm  =  12.88 bar


## Example 2.12 Page No : 93¶

In [15]:
#Given:
P1 = 1.
P2 = 50.      #Pressure at 1, 2 in bar
V1 = 1.
V3 = 0.1      #Volume at 1, 3 in m**3
T1 = 18.+273      #Temperature at 1 in K
g = 1.4      #Specific heat ratio(gamma)

#Solution:
T2 = T1*(P2/P1)**((g-1)/g)      #Temperature at 2 in K
V2 = V1*(P1/P2)*(T2/T1)      #Volume at 2 in m**3
T3 = round(T2*(V3/V2))      #Temperature at 2 in K (printing error)
V4 = V1      #Constant volume process, volume at 4 in m**3
T4 = T3*(V3/V4)**(g-1)      #Temperature at 4 in K
eta = 1-((T4-T1)/(g*(T3-T2)))      #Efficiency of diesel cycle

#Results:
print " Temperature at 1, T1  =  %d K \
\nTemperature at 2, T2  =  %.1f K \
\nTemperature at 3, T3  =  %d K \
\nTemperature at 4, T4  =  %.1f K"%(T1,T2,T3,T4)
print " The thermal efficiency of the cycle, eta  =  %.1f percent"%(eta*100)
#Answer in the book is printed wrong

 Temperature at 1, T1  =  291 K
Temperature at 2, T2  =  889.8 K
Temperature at 3, T3  =  1455 K
Temperature at 4, T4  =  579.2 K
The thermal efficiency of the cycle, eta  =  63.6 percent


## Example 2.13 Page No : 98¶

In [16]:
#Given:
r = 18.      #Compression ratio
p = 10.      #percentage of stroke at which consmath.tant pressure process ends
P1 = 1.
T1 = 20.+273      #Pressure and temperature at 1 in bar and K
V_a = 100.      #Volume of air used per hour in m**3/hr
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.27
#Calculation of cut off ratio (rho)
V_s = r-V_c      #Swept volume in unit
V3 = V_c+V_s*p/100      #Volume at consmath.tant pressure process ends or point 3 in unit
V2 = V_c      #Volume at consmath.tant pressure process starts or point 2 in unit
rho = V3/V2      #Cut off ratio
eta = 1-((rho**g-1)/(r**(g-1)*g*(rho-1)))      #Thermal efficiency
P2 = P1*(r)**g      #Pressure at 2(maximum) in bar (printing error)
P3 = P2      #Constant pressure process, pressure at 3 in bar
T2 = T1*(r)**(g-1)      #Temperature at 2 in K
T3 = T2*rho      #Temperature at 3(maximum) in K
#Consider the cycle for 100 m**3 of swept volume with air, thus
V_s = V_a      #Swept volume in m**3/hr
V2 = V_s/(r-1)      #Volume at 2 in m**3/hr
V1 = V_s+V2      #Volume at 1 in m**3/hr
V3 = rho*V2      #Volume at 3 in m**3/hr
V4 = V1      #Constant volume process, volume at 4 in m**2
P4 = P3*(V3/V4)**g      #Pressure at 4 in bar
W = (P2*(V3-V2)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(g-1))*10**5      #Work done in cycle in Nm
ip = W/3600

#Results:
print " a)The maximum temperature, T3  =  %d degreeC and the maximum pressure, P2  =  %.1f bar"%(T3-273,P2)
print " b)The thermal efficiency of the engine, eta  =  %d percent"%(eta*100)
print " c)The indicated power of the engine, ip  =  %.2f kW"%(ip/1000)
#Answers in the book are wrong

 a)The maximum temperature, T3  =  564 degreeC and the maximum pressure, P2  =  57.2 bar
b)The thermal efficiency of the engine, eta  =  69 percent
c)The indicated power of the engine, ip  =  -2.26 kW


## Example 2.14 Page No : 103¶

In [18]:
#Given:
d = 15.
l = 20.      #Diameter and stroke of cylinder in cm
p2 = 6.      #Percentage of stroke at which cut off takes place
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.28
V_s = (math.pi/4)*d**2*l      #Stroke volume in cm**3
V_c = math.pi*V_s/100      #Clearance volume in cm**3
V1 = V_s+V_c      #Total volume at 1 in cm**3
V2 = V_c      #Volume at 2 in cm**3
V3 = V2+p2*V_s/100      #Volume at 3 in cm**3
r = V1/V2      #Compression ratio
rho = V3/V2      #Cut off ratio
eta = 1-((rho**g-1)/(r**(g-1)*g*(rho-1)))      #Thermal efficiency

#Results:
print " The air standard efficiency of the engine, eta  =  %d percent"%(eta*100)

 The air standard efficiency of the engine, eta  =  67 percent


## Example 2.15 Page No : 108¶

In [19]:
#Given:
r = 15.      #Compression ratio
P1 = 1.
T1 = 25.+273
V1 = .1      #Pressure, temperature, volume at 1 in bar, K, m**3
P4 = 65.
T4 = 1500.+273      #Pressure and temperature at 4 in bar and K
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.29
V2 = V1/r      #Volume at 2 in m**3
P2 = P1*(r)**g      #Pressure at 2 in bar
T2 = T1*(r)**(g-1)      #Temperature at 2 in K
P3 = P4      #Pressure at 3 in bar
T3 = T2*(P3/P2)      #Temperature at 3 in K
V3 = V2      #Volume at 3 in m**3
V4 = V3*(T4/T3)      #Volume at 4 in m**3
V5 = V1      #Volume at 5 in m**3
P5 = P4*(V4/V5)**g      #Pressure at 5 in bar
T5 = T4*(V4/V5)**(g-1)      #Temperature at 5 in K
eta = 1-(T5-T1)/((T3-T2)+g*(T4-T3))      #Thermal efficiency

#Results:
print " Point 1: Pressure  =  %d bar \
\nVolume  =  %.1f m**3 \
\nTemperature  =  %d degreeC"%(P1,V1,T1-273)
print " Point 2: Pressure  =  %.1f bar \
\nVolume  =  %.4f m**3 \
\nTemperature  =  %d degreeC"%(P2,V2,T2-273)
print " Point 3: Pressure  =  %d bar \
\nVolume  =  %.4f m**3 \
\nTemperature  =  %d degreeC"%(P3,V3,T3-273)
print " Point 4: Pressure  =  %d bar \
\nVolume  =  %.4f m**3 \
\nTemperature  =  %d degreeC"%(P4,V4,T4-273)
print " Point 5: Pressure  =  %.2f bar \
\nVolume  =  %.1f m**3 \
\nTemperature  =  %d degreeC"%(P5,V5,T5-273)
print " The thermal efficiency of the cycle, eta  =  %d percent"%(eta*100)
#Answers in the book are wrong

 Point 1: Pressure  =  1 bar
Volume  =  0.1 m**3
Temperature  =  25 degreeC
Point 2: Pressure  =  44.3 bar
Volume  =  0.0067 m**3
Temperature  =  607 degreeC
Point 3: Pressure  =  65 bar
Volume  =  0.0067 m**3
Temperature  =  1018 degreeC
Point 4: Pressure  =  65 bar
Volume  =  0.0092 m**3
Temperature  =  1500 degreeC
Point 5: Pressure  =  2.29 bar
Volume  =  0.1 m**3
Temperature  =  408 degreeC
The thermal efficiency of the cycle, eta  =  64 percent


## Example 2.16 Page No : 113¶

In [20]:
#Given:
r = 18.      #Compression ratio
P1 = 1.01
P3 = 69.      #Pressure at 1, 3 in bar
T1 = 20.+273      #Temperature at 1 in K
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
cp = 1.005      #Specific heat at consmath.tant pressure in kJ/kgK
g = 1.4      #Specific heat ratio(gamma)
R = 0.287      #Specific gas consmath.tant in kJ/kgK

#Solution:
T2 = T1*r**(g-1)      #Temperature at 2 in K
P2 = P1*r**g      #Pressure at 2 in bar
T3 = T2*(P3/P2)      #Temperature at 3 in K
Q_v = cv*(T3-T2)      #Heat added at consmath.tant volume in kJ/kg
#Given, Heat added at consmath.tant volume is equal to heat added at consmath.tant pressure
T4 = Q_v/cp+T3      #Temperature at 4 in K
rho = T4/T3      #Cut off ratio
T5 = T4*(rho/r)**(g-1)      #Temperature at 5 in K
Q1 = 2*Q_v      #Heat supplied in cycle in kJ/kg
Q2 = cv*(T5-T1)      #Heat rejected in kJ/kg
eta = 1-Q2/Q1      #Thermal efficiency
W = Q1-Q2      #Work done by the cycle in kJ/kg
V1 = 1*R*T1/(P1*100)      #Volume at 1 in m**3/kg
V2 = V1/r      #Volume at 2 in m**3/kg
V_s = V1-V2      #Swept volume in m**3/kg
mep = W/(V_s*100)      #Mean effective pressure in bar

#Results:
print " The air standard efficiency, eta  =  %.1f percent"%(eta*100)
print " The mean effective pressure, mep  =  %.2f bar"%(mep)

 The air standard efficiency, eta  =  68.2 percent
The mean effective pressure, mep  =  2.25 bar


## Example 2.17 Page No : 118¶

In [22]:
import math

#Given:
P1 = 1.      #Pressure at 1 in bar
T1 = 50.+273      #Temperature at 1 in K
r = 14.
rho = 2.
alpha = 2.      #Compression ratio, cut off ratio, pressure ratio
g = 1.4      #Specific heat ratio(gamma)
cp = 1.005      #Specific heat at consmath.tant pressure in kJ/kgK
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK

#Solution:
#Refer fig 2.30
T2 = T1*math.ceil(100*r**(g-1))/100      #Temperature at 2 in K
P2 = round(P1*r**g)      #Pressure at 2 in bar
P3 = alpha*P2      #Pressure at 3 in bar
T3 = T2*(P3/P2)      #Temperature at 3 in K
T4 = T3*rho      #Temperature at 4 in K
e = r/rho      #Expansion ratio
T5 = T4/e**(g-1)      #Temperature at 5 in K (Round off error)
Q1 = cv*(T3-T2)+cp*(T4-T3)      #Heat added in kJ/kg
Q2 = cv*(T5-T1)      #Heat rejected in kJ/kg
eta = 1-Q2/Q1      #Air standard efficiency

#Results:
print " The temperature\tT1  =  %d K\tT2  =  %d K\tT3  =  %d K\tT4  =  %d K\tT5  =  %d K"%(T1,T2,T3,T4,T5)
print " The ideal thermal efficiency, eta  =  %.1f percent"%(eta*100)
#Round off error in the value of 'T5'

 The temperature	T1  =  323 K	T2  =  930 K	T3  =  1860 K	T4  =  3720 K	T5  =  1708 K
The ideal thermal efficiency, eta  =  60.8 percent


## Example 2.18 Page No : 123¶

In [24]:
import math
from sympy import Symbol, solve

#Given:
r = 15.      #Compression ratio
P1 = 1.
P3 = 55.      #Pressure at 1, 3 in bar
T1 = 27.+273      #Temperature at 1 in K
g = 1.4      #Specific heat ratio(gamma)
cp = 1.005      #Specific heat at consmath.tant pressure in kJ/kgK
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK

#Solution:
#Refer fig 2.31
T2 = T1*r**(g-1)      #Temperature at 2 in K
P2 = P1*r**g      #Pressure at 2 in bar
alpha = P3/P2      #Constant volume pressure ratio
T3 = T2*(P3/P2)      #Temperature at 3 in K
Q1_v = cv*(T3-T2)      #Heat supplied at consmath.tant volume in kJ/kg
T4 = Symbol("T4")      #Defining temperature at 4 as unknown in K
#Given, heat supplied at consmath.tant volume, Q1_v is twice of heat supplied at consmath.tant pressure, Q1_p
Q1_p = cp*(T4-T3)      #Heat supplied at consmath.tant pressure in kJ/kg
T4 =solve(Q1_v-2*Q1_p)[0]      #Temperature at 4 in K
rho = T4/T3      #Cut off ratio
e = r/rho      #Expansion ratio
T5 = T4/e**(g-1)      #Temperature at 5 in K
eta = 1-(T5-T1)/((T3-T2)+g*(T4-T3))      #Thermal efficiency
eta = round(100*eta)
#Results:
print " The consmath.tant volume pressure ratio, alpha  =  %.2f"%(alpha)
print " The cut off ratio, rho  =  %.2f"%(rho)
print " The thermal efficiency of the cycle, eta  =  %d percent"%(eta)

 The consmath.tant volume pressure ratio, alpha  =  1.24
The cut off ratio, rho  =  1.07
The thermal efficiency of the cycle, eta  =  66 percent


## Example 2.19 Page No : 128¶

In [25]:
#Given:
n = 6.      #Number of cylinders
V_s = 300.      #Engine swept volume in cm**3 per cylinder
r = 10.      #Compression ratio
N = 3500.      #Engine speed in rpm
bp = 75.      #Brake power in kW
P1 = 1.      #Pressure at 1 in bar
T1 = 15.+273      #Temperature at 1 in K (misprint)
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
cp = 1.005      #Specific heat at consmath.tant pressure in kJ/kgK
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.32
#Otto cycle
eta_o = 1-1/r**(g-1)      #Cycle efficiency
Q1 = bp/eta_o      #Rate of heat addition in kW
P_o = bp/n      #Power output per cylinder in kW
W_o = P_o/(N/(2*60))      #Work output per cycle per cylinder in kJ
mep_o = W_o/V_s*10**6/100      #Mean effective pressure in bar
T2 = T1*r**(g-1)      #Temperature at 2 in K
Q1 = Q1/(n*N/(2*60))      #Heat supplied per cycle per cylinder in kJ
R = 0.287      #Specific gas consmath.tant in kJ/kgK
v1 = R*T1/(P1*100)      #Volume of air in m**3/kg
V1 = V_s/(1-1/r)*10**-6      #Volume at 1 in m**3
m = V1/v1      #Mass of the air in kg
T3 = T2+Q1/(m*cv)      #Temperature at 3 in K
#Diesel cycle
T31 = T2+Q1/(m*cp)      #Temperature at 3 in diesel cycle in K
rho = T31/T2      #Cut off ratio for diesel cycle
eta_d = 1-((rho**g-1)/(r**(g-1)*g*(rho-1)))      #The air standard efficiency
Power = eta_d*bp/(eta_o)      #Power output in kW
P_d = Power/n      #Power output per cylinder in kW
W_d = P_d/(N/(2*60))      #Work output per cycle per cylinder in kJ
mep_d = W_d/V_s*10**6/100      #Mean effective pressure in bar

#Results:
print " The rate of heat addition same for both Petrol and Diesel engine, Q1  =  %.1f kW"%(bp/eta_o)
print " For Petrol engine\t Cycle efficiency, eta  =  %.3f\t Mean effective pressure, mep  =  %.2f bar\t The maximum temperature of the cycle, Tmax  =  %.0f K"%(eta_o,mep_o,T3)
print " For Diesel engine\t Cycle efficiency, eta  =  %.2f\t Mean effective pressure, mep  =  %.2f bar\t The maximum temperature of the cycle, Tmax  =  %.0f K\t Power output  =  %.1f kW"%(eta_d,mep_d,T3,Power)

 The rate of heat addition same for both Petrol and Diesel engine, Q1  =  124.6 kW
For Petrol engine	 Cycle efficiency, eta  =  0.602	 Mean effective pressure, mep  =  14.29 bar	 The maximum temperature of the cycle, Tmax  =  3183 K
For Diesel engine	 Cycle efficiency, eta  =  0.46	 Mean effective pressure, mep  =  10.92 bar	 The maximum temperature of the cycle, Tmax  =  3183 K	 Power output  =  57.3 kW


## Example 2.20 Page No : 133¶

In [26]:
#Given:
r = 10.      #Compression ratio
P1 = 1.      #Pressure at 1 in bar
T1 = 40.+273      #Temperature at 1 in K
Q1 = 2700.      #Heat added in kJ

#Solution:
#Refer fig 2.33
g = 1.4      #Specific heat ratio(gamma)
R = 0.287      #Specific gas consmath.tant in kJ/kgK
cp = 1.005      #Specific heat at consmath.tant pressure in kJ/kgK
V1 = 1*R*T1/(P1*100)      #Volume at 1 in m**3/kg
V5 = V1      #Volume at 5 in m**3/kg
V2 = V1/r      #Volume at 2 in m**3/kg
V3 = V2      #Volume at 3 in m**3/kg
V_s = V1-V2      #Swept volume in m**3/kg
T2 = T1*r**(g-1)      #Temperature at 2 in K
P2 = P1*r**g      #Pressure at 2 in bar
#(a)Limited-pressure cycle
P3 = 70      #Limited maximum pressure in bar.
T3 = T2*(P3/P2)      #Temperature at 3 in K
cv = 0.718      #Specific heat at consmath.tant volume in kJ/kgK
Q_v = cv*(T3-T2)      #Heat supplied at consmath.tant volume in kJ
Q_p = Q1-Q_v      #Heat supplied at consmath.tant pressure in kJ
T4 = Q_p/cp+T3      #Temperature at 4 in K
V4 = V3*(T4/T3)      #Volume at 4 in m**3/kg
T5 = T4*(V4/V5)**(g-1)      #Temperature at 5 in K
Q2 = cv*(T5-T1)      #Heat rejected in kJ/kg
W = Q1-Q2      #Work done in kJ/kg
eta1 = W/Q1      #Efficiency of Limited pressure cycle
mep1 = W/(V_s*100)      #Mean effective pressure in bar
#(b)Constant volume cycle
#All the heat is supplied at consmath.tant volume in consmath.tant volume cycle
T6 = Q1/cv+T2      #Temperature at 6 in K
P6 = P2*T6/T2      #Pressure at 6 in bar
T7 = T6*(1/r)**(g-1)      #Temperature at 7 in K
Q2 = cv*(T7-T1)      #Heat rejected in kJ/kg
W = Q1-Q2      #Work done in kJ/kg
eta2 = W/Q1      #Efficiency of consmath.tant volume cycle
mep2 = W/(V_s*100)      #Mean effective pressure in bar
#If gases expanded isentropically to their original pressure of 1 bar, this point is named as 8
P8 = P1      #Pressure at 8 in bar
T8 = T6*(P8/P6)**((g-1)/g)      #Temperature at 8 in K
Q3 = cp*(T8-T1)      #Heat rejected at consmath.tant pressure in kJ/kg
W_inc = Q2-Q3      #Work increased if gas expanded isentropically in kJ/kg

#Results:
print " a)For Limited pressure cycle\t The mean effective pressure, mep  =  %.2f bar \
\nThe thermal efficiency, eta  =  %.1f percent"%(mep1,eta1*100)
print " a)For Constant volume cycle\t The mean effective pressure, mep  =  %.1f bar \
\nThe thermal efficiency, eta  =  %.1f percent"%(mep2,eta2*100)
print "   Additional work per kg of charge  =  %.1f kJ"%(W_inc)

 a)For Limited pressure cycle	 The mean effective pressure, mep  =  18.97 bar
The thermal efficiency, eta  =  56.8 percent
a)For Constant volume cycle	 The mean effective pressure, mep  =  20.1 bar
The thermal efficiency, eta  =  60.2 percent
Additional work per kg of charge  =  287.7 kJ


## Example 2.21 Page No : 138¶

In [27]:
#Given:
r = 6.      #Compression ratio
P1 = 1.
P3 = 20.      #Pressure at 1, 3 in bar
T1 = 27.+273      #Temperature at 1 in K
g = 1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 2.34
eta_otto = 1-1/r**(g-1)      #Efficiency of Otto cycle (printing error)
#For Atkinson cycle
e = (P3/P1)**g      #Expansion ratio
eta_atk = 1-g*(e-r)/(e**g-r**g)      #Efficiency of Atkinson cycle

#Results:
print " Efficiency of Otto cycle  =  %.2f percent"%(eta_otto*100)
print " Efficiency of Atkinson cycle  =  %.1f percent"%(eta_atk*100)
#Answer in the book is printed wrong

 Efficiency of Otto cycle  =  51.16 percent
Efficiency of Atkinson cycle  =  75.4 percent


## Example 2.22 Page No : 143¶

In [29]:
#Given:
P1 = 1.02
P2 = 6.12      #Pressure at 1, 2 in bar
T1 = 15.+273
T3 = 800+273.      #Temperature at 1, 3 in K
g = 1.4      #Specific heat ratio(gamma)
cp = 1.005      #Specific heat at consmath.tant pressure in kJ/kgK

#Solution:
#Refer fig 2.18
r_p = P2/P1      #pressure ratio
eta = 1-1/r_p**((g-1)/g)      #Thermal efficiency
r_w = 1-(T1/T3)*r_p**((g-1)/g)      #Work ratio

#Results:
print " The thermal efficiency, eta  =  %.1f percent"%(eta*100)
print " The work ratio, r_w  =  %.2f"%(r_w)

 The thermal efficiency, eta  =  40.1 percent
The work ratio, r_w  =  0.55