# Chapter 9:GAS CYCLES¶

## Example 9.1, Page No:425¶

In :
import math
from __future__ import division

#Variable declaration
r=8; # Compression ratio of an engine
p1=100; # Pressure of air before compression in lPa
T1=300; # Temperature air before compression in kelvin
qH=1800; # Heat added to the air in kJ/kg
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K

#Calculation for (1)
# Otto cycle
#  (1) state 2
p2=p1*(r)**k; # Pressure at the end of compression
T2=T1*(p2/p1)**((k-1)/k);# Temperature at the end of compression
# state 3
T3=(qH/Cvo)+T2; # Temperatue after heat addition
p3=p2*(T3/T2); # Pressure after heat addition
# state 4
p4=p3*(1/r)**k; # Pressure after expansion
T4=T3*(p4/p3)**((k-1)/k);# Temperature after expansion

#Result for (1)
print "(1).state 1","\n    Pressure of air before compression = ",p1,"kPa","\n    Temperature air before compression =",T1,"K"
print "state 2","\n    Pressure of air at the end of compression = ",round(p2,0),"kPa"
print "    Temperature at the end of compression =",round(T2,1),"K"
print "state 3","\n    Pressure after heat addition = ",round(p3,0),"kPa   (round off error)"
print "    Temperature after heat addition =",round(T3,1),"K   (round off error)"
print "state 4","\n    Pressure after expansion = ",round(p4,1),"kPa   (round off error)"
print "    Temperature after expansion =",round(T4,1),"K   (round off error)"

#Calculation for (2)
#  (2).Thermal efficiency
qL=Cvo*(T4-T1); # Heat rejected
eff_th=1-qL/qH; # thermal efficiency

#Result for (2)
print "\n(2).Thermal efficienvy = ",round(eff_th*100,2),"%   (round off error)"

#Calculation for (3)
#   (3). Mean effective pressure
wnet=qH-qL; # net work
v1=R*T1/p1; # Specific volume at state 1
v2=v1/r; # Specific volume at state 2
pm=wnet/(v1-v2); # Mean effective pressure

#Result for (3)
print "\n(3).Mean effective pressure = ",round(pm,0),"kPa"

(1).state 1
Pressure of air before compression =  100 kPa
Temperature air before compression = 300 K
state 2
Pressure of air at the end of compression =  1838.0 kPa
Temperature at the end of compression = 689.2 K
state 3
Pressure after heat addition =  8537.0 kPa   (round off error)
Temperature after heat addition = 3201.4 K   (round off error)
state 4
Pressure after expansion =  464.5 kPa   (round off error)
Temperature after expansion = 1393.5 K   (round off error)

(2).Thermal efficienvy =  56.47 %   (round off error)

(3).Mean effective pressure =  1349.0 kPa


## Example 9.2, Page No:430¶

In :
#Variable declaration
r=18; # Compression ratio of an engine
p1=100; # Pressure of air before compression in lPa
T1=300; # Temperature air before compression in kelvin
qH=1800; # Heat added to the air in kJ/kg
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K

#Calculation
#   Diesel cycle
# state 2
T2=T1*(r)**(k-1); # Temperature at the end of compression
p2=p1*(r)**k; # Pressure at the end of compression
# state 3
T3=(qH/Cpo)+T2; # Temperatue after heat addition
p3=p2; #constant pressure
Tmax=T3; # maximum temperature
Pmax=p3; # Maximum pressure
# state 4
v3=R*T3/p3; # Specific volume at state 3
v4=R*T1/p1; # Specific volume at state 4
T4=T3*(v3/v4)**(k-1); # Temperature after expansion
p4=p3*(v3/v4)**k; # Pressure after expansion
qL=Cvo*(T4-T1); # Heat rejected
wnet=qH-qL; # net work
eff_th=wnet/qH; # thermal efficiency
v1=R*T1/p1; # Specific volume at state 1
v2=v1/r; # Specific volume at state 2
pm=wnet/(v1-v2); # Mean effective pressure

#Results
print "Maximum pressure = ",Pmax,"kPa","Maximum Temperature = ",Tmax,"K"
print "Thermal efficienvy = ",round(eff_th*100,1),"%"
print "Mean effective pressure =  ",round(pm,0),"kPa    (Error in textbook)"

Maximum pressure =  5719.80874166 kPa Maximum Temperature =  2747.02343004 K
Thermal efficienvy =  59.4 %
Mean effective pressure =   1315.0 kPa    (Error in textbook)


## Example 9.3, Page No: 440¶

In :
#Variable declaration
p1=0.1; # Pressure of air at inlet in MPa
T1=300; # Temperature of air at inlet in kelvin
p2=0.6; # Pressure of air at exit in MPa
T3=1200; # Maximun temperature of air in kelvin
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K

#Calculation
# Brayton cycle
rp=p2/p1; # pressure ratio
T2=T1*(p2/p1)**((k-1)/k);# Temperature at the end of compression
wc=Cpo*(T2-T1); # compressor work
T4=T3*(p1/p2)**((k-1)/k);# Temperature at the end of expansion
wT=Cpo*(T3-T4); # Turbine work
wnet=wT-wc; # net work
eff_th=wnet/qH; # thermal efficiency
rw=wnet/wT; # worh ratio

#Result
print "Thermal Efficiency = ",round(eff_th*100,1),"%"
print "Work Ratio = ",round(rw,3)

Thermal Efficiency =  40.1 %
Work Ratio =  0.583


## Example 9.4, Page No: 441¶

In :
#Variable declaration
p1=0.1; # Pressure of air at inlet in MPa
T1=300; # Temperature of air at inlet in kelvin
p2=0.6; # Pressure of air at exit in MPa
T3=1200; # Maximun temperature of air in kelvin
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K
eff_t=0.85; # Turbine efficiency
eff_c=0.8; # Compressor efficienct

#Calculation
# Brayton cycle
rp=p2/p1; # pressure ratio
T2s=T1*(p2/p1)**((k-1)/k);# Isentropic Temperature at the end of compression
T2=((T2s-T1)/eff_c)+T1; # Actual Temperature at the end of compression
p3=0.585; # as per given in MPa
p4s=0.11; # As per given in MPa
T4s=T3*(p4s/p3)**((k-1)/k); # Isentropic temperature after reversible adiabatic expansion
T4=T3-(eff_t*(T3-T4s));# Actual temperature at state 4
wc=Cpo*(T2-T1); # compressor work
wT=Cpo*(T3-T4); # Turbine work
wnet=wT-wc; # net work
eff_th=wnet/qH; # thermal efficiency
rw=wnet/wT; # worh ratio

#Result
print "Thermal Efficiency = ",round(eff_th*100,0),"%"
print "Work Ratio = ",round(rw,3)

Thermal Efficiency =  21.0 %
Work Ratio =  0.353


## Example 9.5, Page No: 448¶

In :
#Variable declaration
p1=0.1; # Pressure of air at inlet in MPa
T1=300; # Temperature of air at inlet in kelvin
p2=0.6; # Pressure of air at exit in MPa
T3=1200; # Maximun temperature of air in kelvin
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K

#Calculation
# Brayton cycle
rp=p2/p1; # pressure ratio
T2=T1*(p2/p1)**((k-1)/k);# Temperature at the end of compression
T4=T3*(p1/p2)**((k-1)/k);# Temperature at state 4
Tx=T4; Ty=T2; # regenerator temperatures
qH=Cpo*(T3-Tx); # Heat added in the cycle with regenerator
qL=Cpo*(Ty-T1);# Heat rejected in the cycle with regenerator
eff_th=1-qL/qH; # Thermal efficiency

#Result
print "Thermal efficiency with regenerator = ",round(eff_th*100,1),"%"

Thermal efficiency with regenerator =  58.3 %


## Example 9.6, Page No: 453¶

In :
import math
from __future__ import division

#Variable declaration
V1=250; # Velocoty of jet aircraft in m/s
p1=60; # Atmospheric pressure in kPa
T1=260; # Atmospheric temperature in kelvin
rp=8; # Pressure ratio of compressor
T4=1350; # Temperature of gas at turbine inlet in kelvin
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K

#Calculation for (a)
# (a).The pressure and temperature at each point of the cycle
# process 1-2 isentropic diffusion
T2=T1+(V1**2)/(2*Cpo*10**3); # Temperature at state 2
p2=p1*(T2/T1)**(k/(k-1)); # Pressure at state 2
# process 2-3 isentropic compression
p3=rp*p2; # perssure at state 3
T3=T2*(p3/p2)**((k-1)/k); # Temperature at state 3
wc=Cpo*(T3-T2); # compressor work
# process 3-4 Constant pressur heat addition
p4=p3; # constant pressure
# process 4-5 isentropic expansion in turbine
wT=wc;
T5=T4-(wT/Cpo); # Temperature at state 5
p5=p4*(T5/T4)**(k/(k-1)); # Pressure at state 5
# process 5-6 Isentropic expansion in nozzle
p6=p1;
T6=T5*(p6/p5)**((k-1)/k); # Temperature at state 6

#Result for (a)
print "(a).The pressure and temperature at each point of the cycle","\n  State 1","\n     p1 =",p1,"kPa","\n     T1 =",T1,"K"
print "\n  State 2","\n     p2 =",round(p2,1),"kPa","\n     T2 =",round(T2,1),"K",
print "\n  State 3","\n     p3 =",round(p3,0),"kPa","\n     T3 =",round(T3,1),"K"
print "\n  State 4","\n     p4 =",round(p4,0),"kPa","\n     T4 =",T4,"K"
print "\n  State 5","\n     p5 =",round(p5,0),"kPa   (roundoff error)","\n     T5 =",round(T5,1),"K    (roundoff error)"
print "  State 6","\n     p6 =",p6,"kPa","\n     T6 =",round(T6,0),"K"

#Calculation for (b)
# (b).Exit velocity of jet
V6=math.sqrt (2*Cpo*10**3*(T5-T6)); # Exit velocity of jet

#Result for (b)
print "\n\n(b).Exit velocity of jet =",round(V6,0),"m/s   (roundoff error)"

#Calculation for (c)
#   (c).Specific thrust and work output
F_mair=(V6-V1); # Specific thrust
w=F_mair*V1/1000; # Work output

#Result for (c)
print "\n\n(c).Specific thrust and work output","\nSpecific thrust =",round(F_mair,0),"N   (roundoff error)"
print "Work output = ",round(w,1),"kJ/kg"

#Calculation for (d)
#   (d).Propulsion efficiency
eff_p=w/(w+(V6**2-V1**2)/2000);# Propulsion efficiency

#Result for (d)
print "\n\n(d).Propulsion efficiency =",round(eff_p*100,1),"%"

#Calculation for (e)
#    (e).Overall thermal efficiency
eff_th=w/qH; # Overall thermal efficiency

#Result for (e)
print "(e).Overall thermal efficiency =",round(eff_th*100,1),"%"

(a).The pressure and temperature at each point of the cycle
State 1
p1 = 60 kPa
T1 = 260 K

State 2
p2 = 89.1 kPa
T2 = 291.1 K
State 3
p3 = 713.0 kPa
T3 = 527.4 K

State 4
p4 = 713.0 kPa
T4 = 1350 K

State 5
p5 = 364.0 kPa   (roundoff error)
T5 = 1113.8 K    (roundoff error)
State 6
p6 = 60 kPa
T6 = 666.0 K

(b).Exit velocity of jet = 948.0 m/s   (roundoff error)

(c).Specific thrust and work output
Specific thrust = 698.0 N   (roundoff error)
Work output =  174.6 kJ/kg

(d).Propulsion efficiency = 29.4 %
(e).Overall thermal efficiency = 21.2 %


## Example 9.7, Page No:458¶

In :
#Variable declaration
p1=100; # Pressure of air at inlet in kPa
T1=288; # Temperature of air at inlet in kelvin
rp=12; # Pressure ratio of the compressor
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K
T3=T1; # From figure

#Calculation
# process 1-2
p2=12*p1; # Pressure at state 2
T2=T1*(p2/p1)**((k-1)/k); # Temperature at state 2
wc=Cpo*(T2-T1); # Compressor work
# process 2-3
# process 3-4
T4=T3*(1/rp)**((k-1)/k); # Temperature at state 4
# process 4-1 Refrigerating coil
qL=Cpo*(T1-T4); # heat rejected
wnet=qH-qL; # net work
cop=qL/wnet; # Cop of plant
pc=wnet/qL; # Power consumption per kW of refrigeration

#Result
print "COP of the cycle =",round(cop,3),"\nPower consumption per kW of refrigeration =",round(pc,3),"kW/kW"

COP of the cycle = 0.967
Power consumption per kW of refrigeration = 1.034 kW/kW


## Example 9.8, Page No:459¶

In :
#Variable declaration
p1=100; # Pressure of air at inlet in kPa
T1=288; # Temperature of air at inlet in kelvin
rp=12; # Pressure ratio of the compressor
T4=223; # Temperature at state 4
k=1.4; # Index of reversible adiabatic process
Cvo=0.7165; # Specific heat at constant volume in kJ/kg K
Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K
R=0.287; # characteristic gas constant of air in kJ/kg K
T3=T1; # From figure

#Calculation
# process 1-2
p2=12*p1; # Pressure at state 2
T2=T1*(p2/p1)**((k-1)/k); # Temperature at state 2
wc=Cpo*(T2-T1); # Compressor work
# process 2-3
# process 3-4 cooling in regenerative heat exchanger
qregen=Cpo*(T3-T4); # cooling in regenerative heat exchanger
# process 4-5 Expander
T5=T4*(1/rp)**((k-1)/k); # Temperature at state 5
wE=Cpo*(T4-T5); # Expander work
# process 5-6 Refrigerating coil
T6=T4; # From figure 9.32
qL=Cpo*(T6-T5); # Heat rejected
# process 6-1 Heating in regenerative heat exchanger
qregen=Cpo*(T1-T6); # Heat supplied
wnet=qH-qL; # net work
cop=qL/wnet; # Cop of plant

#Result
print "COP of the modified cycle =",round(cop,3)

COP of the modified cycle = 0.615