# Chapter 25 : Gas Turbines¶

## Example 25.2 Page no : 389¶

In :
import math

#Input data
p = 4					#Pressure ratio
T3 = 1000					#Turbine inlet temperature in K
T1 = 15+273					#Inlet temperature in K
p1 = 1					#Inlet pressure in kg/cm**2
m = 11					#Mass flow rate of air in kg/s
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
R = 29.27					#haracteristic gas constant in kg.m/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
Pc = (m*R*T1*(p**((g-1)/g)-1))/75					#Power consumed by the compressor in H.P
Pt = (m*R*T3*(1-(1/p)**((g-1)/g)))/75					#Power consumed by the turbine in H.P
N = (Pt-Pc)					#Net output of the plant in H.P. In textbook answer is given wrong
T2 = (T1*(p)**((g-1)/g))					#Temperature at the end of compression in K
q = (Cp*(T3-T2))					#Heat supplied in kcal/kg of air
n = (((N*4500)/427)/(q*m*60))*100					#Overall efficiency of the plant in percent

#Output
print 'Horse power developed is %3.0f H.P  The overall efficiency of the plant is %3.2f percent'%(N,n)

Horse power developed is 3692 H.P  The overall efficiency of the plant is 42.94 percent


## Example 25.3 Page no : 391¶

In :
import math

#Input data
T1 = 15+273					#Temperature of air entering the compressor in K
rp = 5					#Pressure ratio
T3 = 700+273					#Temperature of air after heating in K
g = 1.4					#Ratio of specific heats
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K

#Calculations
T2 = (T1*rp**((g-1)/g))					#Temperature of air after compression in K
T4 = (T3/rp**((g-1)/g))					#Temperature of air after expansion in K
Wc = (Cp*(T2-T1))					#Workdone in the compressor in kcal/kg of air
Wt = (Cp*(T3-T4))					#Workdone in the turbine in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
SHP = (N*427)/75					#Shaft horse power in H.P per kg of air/sec
q = (Cp*(T3-T2))					#Heat supplied in kcal/kg of air
n = (N/q)*100					#Overall efficiency in percent

#Output
print 'Efficiency of plant is %3.1f percent  \
\nThe shaft horse-power per kg of air per second is %3.0f H.P'%(n,SHP)

Efficiency of plant is 36.9 percent
The shaft horse-power per kg of air per second is 260 H.P


## Example 25.4 Page no : 396¶

In :
import math

#Input data
g = 1.4					#Ratio of specific heats
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
m = 20.5					#Air flow rate in kg/sec
p = [5.85,1.03,1.03,5.85]					#Inlet and outlet pressure of turbine and compressor respectively in kg/cm**2
T = [20+273,250+273,600+273,360+273]					#Inlet and outlet temperatures of turbine and compressor respectively in degree C. In textbook instead of 360 degree C, 375 degree C is given

#Calculations
T2 = (T*(p/p)**((g-1)/g))					#Temperature at the outlet of compressor in ideal cycle in K
T4 = (T/(p/p)**((g-1)/g))					#Temperature at the outlet of turbine in ideal cycle in K
ic = ((T2-T)/(T-T))*100					#Isentropic efficiency of compressor in percent
it = ((T-T)/(T-T4))*100					#Isentropic efficiency of turbine in percent
Wc = (Cp*(T-T))					#Workdone in compressor in kcal/kg of air
Wt = (Cp*(T-T))					#Workdone in turbine in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
P = (N*427*m)/75					#Power output in H.P

#Output
print 'The net output is %3.0f H.P'%(P)

The net output is 280 H.P


## Example 25.5 Page no : 397¶

In :
import math

#Input data
rp = 5.					#Pressure ratio
T1 = 15.+273					#Inlet temperature in K
nc = 80.					#Adiabatic efficiency of the compressor in percent

#Calculations
T2 = (T1*rp**((n-1)/n))					#Temperature at the outlet of compressor in ideal cycle in K. The textbook answer is wrong. Instead of 456 K, it is given as 452K
T2i = (((nc/100)*T1)+T2-T1)/(nc/100)					#Temperature at the outlet of the compressor in the actual cycle in K

#Output
print 'The temperature at the end of compression is %3.0f K'%(T2i)

The temperature at the end of compression is 498 K


## Example 25.6 Page no : 400¶

In :
import math

#Input data
p1 = 5.62					#Pressure of gas entering the turbine in kg/cm**2
T1 = 1000+273					#Temperature of gas entering the turbine in K
p2 = 1.124					#Pressure of gas leaving the turbine in kg/cm**2. In textbook it is given as 1.24 instead of 1.124
n1 = 0.8					#Isotropic efficiency of the turbine in ratio
Cp = 0.25					#Specific heat at constant pressure in kJ/kg.K

#Calculations
T2 = (T1/(p1/p2)**((n-1)/n))					#Temperature at the end of adiabatic expansion in K
dt = (T1-T2)					#Isentropic temperature drop in K
adt = (n1*dt)					#Actual temperature drop in K
T2i = (T1-adt)					#Temperature at the end of actual expansion in K
W = (Cp*(T1-T2i))					#Workdone per kg of gas in kcal
q = (W*427)/4500					#H.P developed per kg of gas per minute
t2i = (T2i-273)					#Exhaust gas temperature in degree C

#Output
print '1) H.P developed per kg of gas per min is %3.2f  \
\n2) Exhaust gas temperature is %3.1f degree C'%(q,t2i)

1) H.P developed per kg of gas per min is 8.38
2) Exhaust gas temperature is 646.7 degree C


## Example 25.7 Page no : 403¶

In :
import math

#Input data
pt1 = [1.,15.+273]					#Pressure and temperature at the inlet of compressor in kg/cm**2 and K respectively
pt3 = [4.,650.+273]					#Pressure and temperature at the inlet of turbine in kg/cm**2 and K respectively
n = [85.,80.]					#Isentropic efficiencies of turbine and compressor respectively in percent
g = 1.4					#Ratio of specific heats
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K

#Calculations
T2 = (pt1*(pt3/pt1)**((g-1)/g))					#Temperature at the end of adiabatic compression in K
T2i = (pt1+((T2-pt1)/(n/100)))					#Temperature at the end of actual compression in K
T4 = (pt3/(pt3/pt1)**((g-1)/g))					#Temperature at the end of adiabatic expansion in K
T4i = (pt3-((pt3-T4)*(n/100)))					#Temperature at the end of actual expansion in K
Wt = (Cp*(pt3-T4i))					#Workdone in turbine in kcal/kg of air
Wc = (Cp*(T2i-pt1))					#Workdone in compressor in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
q = (Cp*(pt3-T2i))					#Heat supplied in kcal/kg of air
nt = (N/q)*100					#Thermal efficiency in percent

#Output
print 'Thermal efficiency of the cycle is %3.2f percent'%(nt)

Thermal efficiency of the cycle is 17.74 percent


## Example 25.8 Page no : 404¶

In :
import math

#Input data
p1 = 1.03					#Inlet air pressure in kg/cm**2
T1 = 15.5+273					#Inlet temperature of air in K
rp = 5.					#Compression ratio
nc = 85.					#Isentropic efficiency of the compressor in percent
T3 = 540.+273					#Temperature of the gas entering the turbine in K
p3 = 1.03					#Pressure of gas entering the turbine in kg/cm**2
nt = 80.					#Isentropic efficiency of the turbine in percent
O = 2500.					#Net output in H.P
fp = 0.07					#Fall of pressure through the combustion chamber in kg/cm**2
g = 1.4					#Ratio of specific heats for both air and gas
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K for both air and gas

#Calculations
T2 = (T1*rp**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
T2i = (T1+((T2-T1)/(nc/100)))					#Temperature of air at the end of actual compression in K
T4 = (T3/((rp*p3-fp)/p3)**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
T4i = (T3-((T3-T4)*(nt/100)))					#Temperature of air at the end of actual compression in K
Wt = (Cp*(T3-T4i))					#Workdone in turbine in kcal/kg of air
Wc = (Cp*(T2i-T1))					#Workdone in compressor in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
Fl = (O*4500)/(427*N*60)					#Flow rate for 2500 H.P in kg/sec

#Output
print 'Flow rate of air is %3.1f kg/sec for a net output of %i H.P'%(Fl,O)

Flow rate of air is 45.8 kg/sec for a net output of 2500 H.P


## Example 25.9 Page no : 407¶

In :
import math

#Input data
p1 = 1.					#Inlet air pressure in kg/cm**2
T1 = 16.+273					#Inlet temperature of air in K
rp = 3.5					#Pressure ratio
nc = 85.					#Isentropic efficiency of the compressor in percent
T3 = 500.+273					#Temperature of the gas entering the turbine in K
nt = 80.					#Isentropic efficiency of the turbine in percent
mc = 4.					#Mass of air entering the compressor in tonnes/hour
g = 1.4					#Ratio of specific heats
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K

#Calculations
T2 = (T1*rp**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
dt = (T2-T1)					#Isentropic temperature rise in K
adt = (dt/(nc/100))					#Actual temperature rise in K
T2i = (T1+((T2-T1)/(nc/100)))					#Temperature of air at the end of actual compression in K
T4 = (T3/rp**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
T4i = (T3-((T3-T4)*(nt/100)))					#Temperature of air at the end of actual compression in K
Wt = (Cp*(T3-T4i))					#Workdone in turbine in kcal/kg of air
Wc = (Cp*(T2i-T1))					#Workdone in compressor in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
q = (Cp*(T3-T2i))					#Heat supplied in kcal/kg of air
NHP = (N*427*mc*1000)/(60*4500)					#Net Horse Power available in H.P
nt = (N/q)*100					#Thermal efficiency in percent

#Output
print 'i) The net Horse power available from this unit is %3.1f H.P  \
\nii) The thermal efficiency of the plant is %3.2f percent'%(NHP,nt)

i) The net Horse power available from this unit is 60.3 H.P
ii) The thermal efficiency of the plant is 11.77 percent


## Example 25.10 Page no : 411¶

In :
import math

#Input data
p1 = 1.02					#Inlet air pressure in kg/cm**2
T1 = 27.+273					#Inlet temperature of air in K
p2 = 4.08					#Pressure after compression in kg/cm**2
nc = 80.					#Isentropic efficiency of compressor in percent
mf = 1.					#Mass of fuel in kg
ma = 80.					#Mass of air in kg
nt = 85.					#Isentropic efficiency of the turbine in percent
CV = 10000.					#Calorific value of fuel n kcal per kg
g = 1.4					#Ratio of specific heats
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K

#Calculations
rp = (p2/p1)					#Pressure ratio
T2 = (T1*rp**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
dt = (T2-T1)					#Isentropic temperature rise in K
adt = (dt/(nc/100))					#Actual temperature rise in K
T2i = (T1+((T2-T1)/(nc/100)))					#Temperature of air at the end of actual compression in K
q = (mf/ma)*CV					#Heat supplied per kg of air in kcal
T3 = (q/Cp)+T2i					#Temperature of gas at the inlet of the turbine in K
T4 = (T3/rp**((g-1)/g))					#Temperature of air at the end of adiabatic expansion in K
T4i = (T3-((T3-T4)*(nt/100)))					#Temperature of air at the end of actual expansion in K
Wt = (Cp*(T3-T4i)*((ma+mf)/ma))					#Workdone in turbine in kcal/kg of air
Wc = (Cp*(T2i-T1))					#Workdone in compressor in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
nt = (N/q)*100					#Thermal efficiency in percent

#Output
print 'a) The net work output of installation is %3.2f kcal/kg of air  \
\nb) Overall efficiency of the plant is %3.1f percent'%(N,nt)

a) The net work output of installation is 24.02 kcal/kg of air
b) Overall efficiency of the plant is 19.2 percent


## Example 25.11 Page no : 413¶

In :
import math

#Input data
rp = 5.					#Pressure ratio
T3 = 580.+273					#Temperature of gas at the inlet of the turbine in K
p1 = 1.03					#Inlet air pressure in kg/cm**2
T1 = 15.+273					#Inlet temperature of air in K
nc = 80.					#Isentropic efficiency of compressor in percent
no = 18.					#Overall efficiency of the plant in percent
Cpa = 0.239					#Specific heat of air at constant pressure in kJ/kg.K
Cpg = 0.261					#Specific heat of gas at constant pressure in kJ/kg.K
R = 29.27					#haracteristic gas constant in kg.m/kg.K
g = 1.4					#Ratio of specific heats for air
g1 = 1.355					#Ratio of specific heats for gas

#Calculations
T2 = (T1*rp**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
T2i = (T1+((T2-T1)/(nc/100)))					#Temperature of air at the end of actual compression in K
q = (Cpg*(T3-T2i))					#Heat supplied in kcal/kg of air
Wc = (Cpa*(T2i-T1))					#Workdone in compressor in kcal/kg of air
Wt = ((no/100)*q)+Wc					#Turbine work output in kcal/kg of air
T4i = (T3-(Wt/Cpg))					#Temperature of air at the end of actual expansion in K
T4 = (T3/rp**((g1-1)/g1))					#Temperature of air at the end of adiabatic expansion in K
nt = ((T3-T4i)/(T3-T4))*100					#Isentropic efficiency of turbine in percent

#Output
print 'Isentropic efficiency of turbine is %3.1f percent'%(nt)

Isentropic efficiency of turbine is 87.3 percent


## Example 25.12 Page no : 418¶

In :
import math

#Input data
p1 = 1.03					#Inlet air pressure in kg/cm**2
T1 = 15.+273					#Inlet temperature of air in K
rp = 5.					#Pressure ratio
nc = 85.					#Isentropic efficiency of the compressor in percent
T3 = 540.+273					#Temperature of the gas entering the turbine in K
nt = 80.					#Isentropic efficiency of the turbine in percent
NHP = 2000.					#Net horse power in H.P
fp = 0.1					#Fall of pressure through the combustion system in kg/cm**2
g = 1.4					#Ratio of specific heats for both air and gas
Cp = 0.25					#Specific heat at constant pressure in kJ/kg.K for both air and gas

#Calculations
T2i = (T1*rp**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
dt = (T2i-T1)					#Isentropic temperature rise in K
adt = (dt/(nc/100))					#Actual temperature rise in K
Wc = (Cp*adt)					#Workdone in compressor in kcal/kg of air
e = ((rp*p1-fp)/p1)					#Expansion ratio
T4i = (T3/e**((g-1)/g))					#Temperature of air at the end of adiabatic expansion in K
dt1 = (T3-T4i)					#Isentropic temperature rise in K
adt1 = (dt1/(nt/100))					#Actual temperature rise in K
Wt = (Cp*adt1)					#Workdone in turbine in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
w = (NHP*75)/(427*9.8)					#Flow rate in kg of air per sec

#Output
print 'Flow rate is %3.2f kg of air per sec'%(w)

Flow rate is 35.85 kg of air per sec


## Example 25.13 Page no : 422¶

In :
import math

#Input data
nc = 75.					#Isentropic efficiency of the compressor in percent
nt = 85.					#Isentropic efficiency of the turbine in percent
nm = 98.					#Mechanical efficiency in percent
rp = 6.					#Pressure ratio
T3 = 727.+273					#Temperature of the gas entering the turbine in K
p1 = 1.					#Inlet air pressure in kg/cm**2
T1 = 15.5+273					#Inlet temperature of air in K
m = 2.2					#Mass flow rate in kg/sec
Cpa = 0.24					#Specific heat of air at constant pressure in kJ/kg.K
Cpg = 0.276					#Specific heat of gas at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats for air
g1 = 1.33					#Ratio of specific heats for gas

#Calculations
T2 = (T1*rp**((g-1)/g))					#Temperature of air at the end of adiabatic compression in K
T2i = (T1+((T2-T1)/(nc/100)))					#Temperature of air at the end of actual compression in K
T4 = (T3/rp**((g1-1)/g1))					#Temperature of air at the end of adiabatic compression in K
T4i = (T3-((T3-T4)*(nt/100)))					#Temperature of air at the end of actual compression in K
Wt = (Cpg*(T3-T4i))					#Workdone in turbine in kcal/kg of air
Wc = (Cpa*(T2i-T1))					#Workdone in compressor in kcal/kg of air
N = (Wt-Wc)					#Net workdone in kcal/kg of air
P = (N*m*427)/(75*(nm/100))					#Power output of the plant in H.P

#Output
print 'Power output of the plant is %3.0f H.P'%(P)

Power output of the plant is 287 H.P


## Example 25.14 Page no : 425¶

In :
import math

#Input data
T1 = 15.+273					#Inlet temperature of air in K
rp = 4.					#Pressure ratio
T4 = 560.+273					#Maximum temperature of the cycle in K
nc = 83.					#Isentropic efficiency of the compressor in percent
nt = 86.					#Isentropic efficiency of the turbine in percent
x = 75.					#Heat exchanger making use of heat available in percent
g = 1.4					#Ratio of specific heats

#Calculations
T5i = (T4*(1/rp)**((g-1)/g))					#Temperature in K
dt = (T4-T5i)					#Isometric temperature drop through turbine in degree C
ta = ((nt/100)*dt)					#Actual temperature drop in degree C
T5 = (T4-ta)					#Temperature in K
T2i = (T1*rp**((g-1)/g))					#Temperature in K
tc = (T2i-T1)					#Temperature change in degree C
T2 = (tc/(nc/100))+T1					#Temperature in K
q = (T5-T2)					#Available heat in exchanger in kcal per kg *Cp
T3 = ((q*(x/100))+T2)					#Temperature in K
#Without heat exchanger
qw = (T4-T2)					#Heat supplied *Cp in kcal/kg
tw = (T4-T5)					#Turbine work *Cp in kcal/kg
cw = (T2-T1)					#Compressor work *Cp in kcal/kg
nw = (tw-cw)					#Net workdone *Cp in kcal/kg
no = (nw/qw)*100					#Overall efficiency in percent
#With heat exchanger
qs = (T4-T3)					#Heat supplied *Cp in kcal/kg
no1 = (nw/qs)*100					#Overall efficiency in percent

#Output
print 'The overall efficiency  \
\na) without heat exchanger is %3.1f percent  \
\nb) with heat exchanger making use of %i percent of heat available is %3.1f percent'%(no,x,no1)

The overall efficiency
a) without heat exchanger is 17.4 percent
b) with heat exchanger making use of 75 percent of heat available is 24.3 percent


## Example 25.15 Page no : 427¶

In :
import math

#Input data
p1 = 1.					#Initial pressure in kg/cm**2
T1 = 15.+273					#Initial temperature in K
p2 = 5.5					#Pressure after compression in kg/cm**2
T3 = 750.+273					#Temperature at the entrance of turbine in K
v = 225.					#Speed in m/s
x = 70.					#Percentage
in1 = 75.					#Isentropic efficiency of compressor in percent
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
T2 = (T1*(p2/p1)**((g-1)/g))					#Temperature in K
at = (T2-T1)/(in1/100)					#Actual temperature rise in the compressor in K
T2i = (T1+at)					#Temperature in K
T4 = (T3/(p2/p1)**((g-1)/g))					#Temperature in K
to = (Cp*(T3-T4))					#Theoritical turbine output in kcal/kg of air
ci = (Cp*(T2i-T1))					#Actual compressor input in kcal/kg of air
ke = (v**2/(2*9.81*427))					#Kinetic energy in gas leaving the exhaust annulus in kcal/kg
dT34 = (ci+ke)/((x/100)*Cp)					#Change in temperature in K
r = 1/(1-(dT34/T3))**(g/(g-1))					#Ratio of pressures
p4 = (r/p2)					#Pressure in kg/cm**2

#Output
print 'The pressure of the gases in the turbine exhaust annulus is %3.1f kg/cm**2'%(p4)

The pressure of the gases in the turbine exhaust annulus is 0.9 kg/cm**2


## Example 25.16 Page no : 431¶

In :
import math

#Input data
p = [1.,5]					#Pressures in atm
T1 = 288					#Temperature in K
T3 = 650+273					#Temperature in K
er = 0.85					#Efficiency ratio
x = 0.72					#Effectiveness of heat exchanger
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
T2 = (T1*(p/p)**((g-1)/g))					#Temperature in K
T2i = (T1+((T2-T1)/er))					#Temperature in K
T4 = (T3/(p/p)**((g-1)/g))					#Temperature in K
T4i = (T3-(er*(T3-T4)))					#Temperature in K
Tc = ((x*(T4i-T2i))+T2i)					#Temperature in K
W = ((Cp*((T3-T4i)-(T2i-T1))))					#Workdone in kcal/kg
q = (Cp*(T3-Tc))					#Heat supplied in kcal/kg
n = (W/q)*100					#Efficiency in percent

#Output
print 'The heat efficiency of the plant is %3.1f percent'%(n)

The heat efficiency of the plant is 27.6 percent


## Example 25.17 Page no : 435¶

In :
import math

#Input data
T1 = 15.+273					#Inlet temperature of air in K
p1 = 1.03					#Inlet pressure of air in kg/cm**2
rp = 5.					#Pressure ratio
T3 = 815.+273					#Temperature of air entering the turbine in K
nc = 0.83					#Adiabatic efficiency of the compressor
nt = 0.92					#Internal engine efficiency of the turbine
nr = 0.65					#Effectiveness of regenerator
p2 = 2.45					#Pressure in kg/cm**2
T6 = T1					#Temperature in K
T9 = T3					#Temperature in K
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heat

#Calculations
T2 = (T1*rp**((g-1)/g))					#Temperature in K
T4 = (T3/rp**((g-1)/g))					#Temperature in K
Wt = (Cp*(T3-T4))					#Isentropic work done in the turbine in kcal/kg of air
Wc = (Cp*(T2-T1))					#Isentropic work done in the compressor in kcal/kg of air
Wr = (Wt/Wc)					#Work ratio
qa = (Cp*(T3-T2))					#Heat added in kcal/kg of air
nth = ((Wt-Wc)/qa)*100					#Thermal efficiency in percent
T2i = (T1+((T2-T1)/nc))					#Temperature in K
T4i = (T3-(nt*(T3-T4)))					#Temperature in K
Wti = (Cp*(T3-T4i))					#work done in the turbine in kcal/kg of air
Wci = (Cp*(T2i-T1))					#work done in the compressor in kcal/kg of air
Wri = (Wti/Wci)					#Work ratio
qai = (Cp*(T3-T2i))					#Heat added in kcal/kg of air
nthi = ((Wti-Wci)/qai)*100					#Thermal efficiency in percent
T2ii = (T2i+((T4i-T2i)*nr))					#Temperature in K
qaii = (Cp*(T3-T2ii))					#Heat added in kcal/kg of air
nthii = ((Wti-Wci)/qaii)*100					#Thermal efficiency in percent
T5 = (T1*(p2/p1)**((g-1)/g))					#Temperature in K
T5i = (T1+((T5-T1)/nc))					#Temperature in K
T7 = (T1*((rp*p1)/p2)**((g-1)/g))					#Temperature in K
T7i = (T6+((T7-T6)/nc))					#Temperature in K
T7ii = (T7i+((T4i-T7i)*nr))					#Temperature in K
Wcomp = (Cp*((T5i-T1)+(T7i-T6)))					#Compressor work in kcal/kg of air
Wratio = (Wti/Wcomp)					#Work ratio
qaa = (Cp*(T3-T7ii))					#Heat added in kcal/kg of air
nthe = ((Wti-Wcomp)/qaa)*100					#Thermal efficiency in percent
T8 = (T3*(p2/(rp*p1))**((g-1)/g))					#Temperature in K
T8i = (T3-((T3-T8)*nt))					#Temperature in K
T10 = (T9/(p2/p1)**((g-1)/g))					#Temperature in K
T10i = (T9-((T9-T10)*nt))					#Temperature in K
T2iii = (T2i+((T10i-T2i)*nr))					#Temperature in K
Wturb = (Cp*((T3-T8i)+(T3-T10i)))					#Compressor work in kcal/kg of air
Wratioi = (Wturb/Wci)					#Work ratio
qaai = (Cp*((T3-T2iii)+(T9-T8i)))					#Heat added in kcal/kg of air
nthei = ((Wturb-Wci)/qaai)*100					#Thermal efficiency in percent
T7iii = (T7i+((T10i-T7i)*nr))					#Temperature in K
Wratioii = (Wturb/Wcomp)					#Work ratio
qaaii = (Cp*((T3-T7iii)+(T9-T8i)))					#Heat added in kcal/kg of air
ntheii = ((Wturb-Wcomp)/qaaii)*100					#Thermal efficiency in percent

#Output
print 'Condition     Work ratio     Thermal efficiencyin percent)   \
\na)           %3.3f                %3.1f   \
\nb)           %3.2f                 %3.1f   \
\nc)           %3.2f                 %3.1f   \
\nd)           %3.2f                 %3.1f   \
\ne)           %3.3f                %3.1f   \
\nf)           %3.3f                %3.1f'%(Wr,nth,Wri,nthi,Wri,nthii,Wratio,nthe,Wratioi,nthei,Wratioii,ntheii)

Condition     Work ratio     Thermal efficiencyin percent)
a)           2.385                36.9
b)           1.82                 27.9
c)           1.82                 37.1
d)           2.06                 38.6
e)           2.029                38.4
f)           2.289                39.5


## Example 25.18 Page no : 435¶

In :
import math

#Input data
p = [1.,9.]					#Pressures in ata
T = [25.+273,1250.+273]					#Minimum and maximum temperatures in K
n = 0.83					#Compressor and turbine efficiencies
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats
x = 0.65					#Cycle with 65% regeneration

#Calculations
#(a)Without regeneration
ip = math.sqrt(p*p)					#Intermediate pressure in ata
T2 = (T*(ip/p)**((g-1)/g))					#Temperature in K
T3 = (T+((T2-T)/n))					#Temperature in K
T4 = T					#Temperature in K
T5 = T2					#Temperature in K
T6 = T3					#Temperature in K
T7 = T					#Temperature in K
T8 = T7/(ip/p)**((g-1)/g)					#Temperature in K
T9 = (T7-((T7-T8)*n))					#Temperature in K
T10 = T7					#Temperature in K
T11 = T8					#Temperature in K
T12 = T9					#Temperature in K
Wc = (2*Cp*(T3-T))					#Work of compression in kcal/kg of air
We = (2*Cp*(T7-T8))					#Work of expansion in kcal/kg of air
NW = (We-Wc)					#Net output in kcal/kg of air
qi = (Cp*((T7-T6)+(T10-T9)))					#Heat input in kcal/kg of air
nth = (NW/qi)*100					#Thermal efficiency in percent

#(b)Cycle efficiency with 65% regeneration
Tg = (T6+(x*(T12-T6)))					#Temperature in K
q = (Cp*((T7-Tg)+(T10-T9)))					#Heat input in kcal/kg of air
nthi = (NW/q)*100					#Thermal efficiency in percent

#(c)Cycle efficiency with ideal regeneration
Eg = T12					#Temperature in K
qa = (2*Cp*(T7-Eg))					#Heat added in kcal/kg of air
nthii = (NW/qa)*100					#Thermal efficiency in percent

#Output
print 'a)Cycle efficiency without regeneration is %3.1f percent  \
\nb)Cycle efficiency with 65 percent regeneration is %3.1f percent  \
\nc)Cycle efficiency with ideal regeneration is %3.0f percent'%(nth,nthi,nthii)

a)Cycle efficiency without regeneration is 38.8 percent
b)Cycle efficiency with 65 percent regeneration is 58.9 percent
c)Cycle efficiency with ideal regeneration is  82 percent


## Example 25.19 Page no : 435¶

In :
import math

#Input data
p1 = 1.					#Inlet pressure of compressor in atm
T1 = 27.+273					#Inle temperature of compressor in K
ic = 0.8					#Isentropic efficiency of compressor
ma = 20.5					#Mass flow rate of air in kg/sec
T3 = 650.+273					#Inlet temperatures of both turbines in K
p2 = 5.					#Inlet pressure of turbine in atm
it = 0.92					#Internal engine efficiency for both the turbines
CV = 10000.					#Calorific value in kcal/kg
Cpa = 0.24					#Specific heat at constant pressure of air in kJ/kg.K
ga = 1.4					#Ratio of specific heats for air
Cpg = 0.276					#Specific heat at constant pressure of gas in kJ/kg.K
gs = 1.33					#Ratio of specific heats for gas

#Calculations
T2 = (T1*(p2/p1)**((ga-1)/ga))					#Temperature in K
T2i = (T1+((T2-T1)/ic))					#Temperature in K
T4 = (T3/(p2/p1)**((gs-1)/gs))					#Temperature in K
T4i = (T3-((T3-T4)*it))					#Temperature in K
Wc = (Cpa*(T2i-T1))					#Work of compression in kcal/kg of air
We = (Cpg*(T3-T4i))					#Work of expansion in kcal/kg of air
mx = (Wc/We)					#Gas required per kg of air compressed in kg
F = ((Cpa*T2i)-(Cpg*T3))/(Cpg*T3-CV)					#Amount of fuel supplied per kg of air in kg
Wg = 1+F					#Weight of gases per kg of air in kg
Wt = (Wg-mx)					#Gases supplied to turbine in kg
Ot = ((Wt*ma*427*We)/75)
#Output of turbine in H.P
nth = ((Wt*We)/(CV*F))*100					#Thermal efficiency in percent

#Output
print 'Output is %3.0f H.P  \
\nThermal efficiency is %3.2f percent'%(Ot,nth)

Output is 2994 H.P
Thermal efficiency is 19.20 percent


## Example 25.20 Page no : 435¶

In :
import math

#Input data
q = 2250.					#Heat supplied per sec in kcal

#Input data from Fig. 25.34 from page no. 652
T1 = 200.					#Temperature in K
T2 = 100.					#Temperature in K
T3 = 625.					#Temperature in K
T4 = 550.					#Temperature in K
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats

#Calculations
Wc = (2*Cp*(T1-T2))					#Work of compression in kcal/kg of air
We = (2*Cp*(T3-T4))					#Work of expansion in kcal/kg of air
NW = (Wc-We)					#Net output in kcal/kg of air
qi = (2*Cp*(T3-T4))					#Heat input in kcal/kg of air
nth = (NW/qi)*100					#Thermal efficiency in percent
rf = (q/qi)					#Rate of flow of working substance in kg/sec
O = (NW*rf*427)/75					#Total output in H.P

#Output
print 'Output is %3.0f H.P  \
\nThermal efficiency is %3.1f percent'%(O,nth)

Output is 4270 H.P
Thermal efficiency is 33.3 percent


## Example 25.22 Page no : 436¶

In :
import math

#Input data
p = [5.,20.]					#Pressure limits in atm
T3 = 650.+273					#Temperature in K
T1 = 60.+273					#Temperature in K
T2 = T1					#Temperature in K
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats
R = 29.27					#Characteristic gas constant in kg.m/kg.K
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

#Calculations
T4 = T3/(p/p)**((g-1)/g)					#Temperature in K
Wc = ((R*T1)/J)*math.log(p/p)					#Compression work in kcal/kg
qs = (Cp*(T3-T2))					#Heat supplied at constant pressure in kcal/kg
qre = (Cp*(T4-T1))					#Heat ejected during process 4-1 in kcal/kg
nth = ((qs-Wc-qre)/(qs-qre))*100					#Thermal efficiency in percent
nc = ((T3-T1)/T3)*100					#Carnot efficiency in percent
r = (nth/nc)*100					#Ratio of air standard efficiency to carnot efficiency in percent

#Output
print 'a) air standard efficiency of the cycle is %3.1f percent  \
\nb) carnot efficiency is %3.0f percent  \
\nc) Ratio of air standard efficiency to carnot efficiency is %3.1f percent'%(nth,nc,r)

a) air standard efficiency of the cycle is 56.3 percent
b) carnot efficiency is  64 percent
c) Ratio of air standard efficiency to carnot efficiency is 88.1 percent