# Chapter 26 : Gas Turbines¶

## Example 26.1 Page No : 433¶

In [3]:
import math
from sympy import Symbol, solve

def horner(coeffs, x):
acc = 0
for c in reversed(coeffs):
acc = acc * x + c
return acc

#Given:
P1  =  101.325      #Pressure at the beginning(1) in kPa
T1  =  27.+273      #Temperature at the beginning(1) in K
r_p  =  6.      #pressure ratio
g  =  1.4      #Specific heat ratio(gamma)
cp  =  1.005      #Specific heat in kJ/kgK
W_TC  =  2.5      #Ratio of Turbine work and compressor work
m  =  1.      #Assume mass in kg

#Solution:
#Refer fig 26.22
T2  =  T1*r_p**((g-1)/g)      #Temperature at 2 in K
T3  =  Symbol('T3')      #Defining temperature at 3 as a unknown in K
T4  =  T3/r_p**((g-1)/g)      #Defining temperature at 4 in terms of T3 in K
W_C  =  m*cp*(T2-T1)      #Compressor work in kJ
W_T  =  m*cp*(T3-T4)      #Turbine work in kJ
T3  =  solve(W_T-W_TC*W_C)[0]      #Temperature at 3 in K
T4  =  horner([0.5993],T3)      #Temperature at 4 in K
eta  =  ((T3-T4)-(T2-T1))/(T3-T2)      #Cycle efficiency

#Results:
print " The maximum temperature in the cycle, T3   =   %.1f K"%(T3)
print " The cycle efficiency, eta   =   %.2f percent"%(eta*100)

 The maximum temperature in the cycle, T3   =   1251.4 K
The cycle efficiency, eta   =   139.88 percent


## Example 26.2 Page No : 438¶

In [2]:
import math

#Given:
T1  =  25.+273
T3  =  825.+273      #Minimum and maximum temperature in K
r_p  =  4.5      #pressure ratio
eta_C  =  85.
eta_T  =  90.      #Isentropic efficiencies of compressor and turbine in percent
P  =  1300.      #Power rating of the turbine in kW
cp  =  1.005      #Specific heat in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 26.23
T21  =  T1*r_p**((g-1)/g)      #Isentropic temperature at 2 in K
T2  =  (T21-T1)/(eta_C/100)+T1      #Temperature at 2 in K
T41  =  T3/r_p**((g-1)/g)      #Isentropic temperature at 4 in K
T4  =  T3-eta_T/100*(T3-T41)      #Temperature at 4 in K
W_C  =  cp*(T2-T1)      #Compressor work in kJ/kg
W_T  =  cp*(T3-T4)      #Turbine work in kJ/kg
Q1  =  cp*(T3-T2)      #Heat added in kJ/kg
W  =  W_T-W_C      #Work output in kJ/kg (Round off error)
eta  =  W/Q1      #Cycle efficiency
r_w  =  W/W_T      #Work ratio
HR  =  3600/(eta)      #Heat rate in kJ/kWh (Round off error)
m  =  P/W      #Mass flow rate in kg/s

#Results:
print " The specific work output,  W   =   %d kJ/kg"%(W)
print " The cycle efficiency, eta   =   %.1f percent"%(eta*100)
print " The work ratio, rw   =   %.3f"%(r_w)
print " The heat rate   =   %d kJ/kWh"%(HR)
print " The mass flow rate for 1300 kW, m   =   %.2f kg/s"%(m)
#Round off error in the values of 'W' and 'HR'

 The specific work output,  W   =   157 kJ/kg
The cycle efficiency, eta   =   25.7 percent
The work ratio, rw   =   0.455
The heat rate   =   14029 kJ/kWh
The mass flow rate for 1300 kW, m   =   8.24 kg/s


## Example 26.3 Page No : 443¶

In [3]:
#Given:
T1  =  25.+273
T3  =  750.+273      #Minimum and maximum temperature in K
r_p  =  4.      #pressure ratio
eta_C  =  75.      #Isentropic efficiency of compressor in percent
g  =  1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 26.24
T21  =  T1*r_p**((g-1)/g)      #Isentropic temperature at 2 in K
T2  =  (T21-T1)/(eta_C/100)+T1      #Temperature at 2 in K
T41  =  T3/r_p**((g-1)/g)      #Isentropic temperature at 4 in K
#For zero efficiency of the cycle (T3-T4)   =   (T2-T1)
eta_T  =  (T2-T1)/(T3-T41)      #Turbine efficiency

#Results:
print " The turbine efficiency for zero cycle efficiency, eta_T   =   %.1f percent"%(eta_T*100)

 The turbine efficiency for zero cycle efficiency, eta_T   =   57.7 percent


## Example 26.4 Page No : 448¶

In [4]:
#Given:
P1  =  1.
P2  =  6.      #Pressure at entering and leaving of compressor in bar
T1  =  27.+273      #Temperature at entering in K
T3  =  700.+273      #Maximum temperature in K
eta_C  =  0.80
eta_T  =  0.85      #Isentropic efficiencies of compressor and turbine in percent
eta_c  =  0.98      #Combustion efficiency in percent
P3  =  P2-0.1      #Pressure at 3 after falling 0.1 bar in bar
cp_a  =  1.005      #Specific heat of air in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma)
cp_g  =  1.147      #Specific heat of gas in kJ/kgK
g1  =  1.333      #Specific heat ratio(gamma) of gas
CV  =  42700.      #Calorific value of fuel in kJ/kg

#Solution:
#Refer fig 26.25
T21  =  T1*(P2/P1)**((g-1)/g)      #Isentropic temperature at 2 in K
T2  =  (T21-T1)/(eta_C)+T1      #Temperature at 2 in K
T41  =  T3/(P3/P1)**((g1-1)/g1)      #Isentropic temperature at 4 in K
T4  =  T3-eta_T*(T3-T41)      #Temperature at 4 in K
W_C  =  cp_a*(T2-T1)      #Compressor work in kJ/kg
W_T  =  cp_g*(T3-T4)      #Turbine work in kJ/kg
W  =  W_T-W_C      #Work output in kJ/kg
Q1  =  cp_g*(T3-T2)/eta_c      #Heat added in kJ/kg
eta  =  W/Q1      #Cycle efficiency
r_w  =  W/W_T      #Work ratio
AR  =  round(3600/W)      #Air rate in kg/kWh
sfc  =  Q1*AR/CV      #Specific fuel consumption in kg/kWh
A_F  =  AR/sfc      #Air fuel ratio

#Results:
print " a)The thermal efficiency, eta   =   %.1f percent"%(eta*100)
print " b)The work ratio, rw   =   %.3f"%(r_w)
print " e)The air rate   =   %d kg/kWh"%(AR)
print " d)The specific fuel consumption, sfc   =   %.3f kg/kWh"%(sfc)
print " c)The air fuel ratio   =   %.1f"%(A_F)

 a)The thermal efficiency, eta   =   17.8 percent
b)The work ratio, rw   =   0.258
e)The air rate   =   41 kg/kWh
d)The specific fuel consumption, sfc   =   0.475 kg/kWh
c)The air fuel ratio   =   86.4


## Example 26.5 Page No : 453¶

In [5]:
#Given:
P1  =  1.
P2  =  6.20      #Pressure at entering and leaving of compressor in bar
T1  =  300.      #Temperature at entering in K
eta_C  =  88.
eta_T  =  90.      #Isentropic efficiencies of compressor and turbine in percent
CV  =  44186.      #Calorific value of fuel in kJ/kg
F_A  =  0.017      #Fuel air ratio
cp_a  =  1.005      #Specific heat of air in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma)
cp_g  =  1.147      #Specific heat of gas in kJ/kgK
g1  =  1.333      #Specific heat ratio(gamma) of gas

#Solution:
#Refer fig 26.26
T21  =  T1*(P2/P1)**((g-1)/g)      #Isentropic temperature at 2 in K
T2  =  (T21-T1)/(eta_C/100)+T1      #Temperature at 2 in K
m_a  =  1      #Assume mass of air in kg
m_f  =  F_A*m_a      #Mass of fuel in kg
T3  =  (cp_a*m_a*T2+m_f*CV)/(cp_g*(m_a+m_f))      #Temperature at 3 in K
r_p  =  P2/P1      #pressure ratio
T41  =  T3/r_p**((g1-1)/g1)      #Isentropic temperature at 4 in K
T4  =  T3-eta_T/100*(T3-T41)      #Temperature at 4 in K
W_C  =  m_a*cp_a*(T2-T1)      #Compressor work in kJ/kg
W_T  =  (m_a+m_f)*cp_g*(T3-T4)      #Turbine work in kJ/kg
W  =  W_T-W_C      #Work output in kJ/kg
Q1  =  m_f*CV      #Heat added in kJ/kg
eta  =  W/Q1      #Cycle efficiency

#Results:
print " The turbine work, W_T   =   %.2f kJ/kg"%(W_T)
print " The compressor work, W_C   =   %.2f kJ/kg"%(W_C)
print " The thermal efficiency, eta   =   %.2f percent"%(eta*100)

 The turbine work, W_T   =   424.03 kJ/kg
The compressor work, W_C   =   234.42 kJ/kg
The thermal efficiency, eta   =   25.24 percent


## Example 26.6 Page No : 458¶

In [6]:
from scipy.optimize import fsolve
import math

#Given:
T1  =  17.+273      #Temperature at entering in K
P1  =  1.      #Pressure at entering of compressor in bar
r_p  =  4.5      #pressure ratio
W  =  4000.      #Work output in kW
m  =  40.      #Mass flow rate in kg/s
e  =  0.6      #Thermal ratio or effectiveness of heat exchanger
eta_C  =  84.      #Isentropic efficiency of compressor in percent
eta  =  0.29      #Thermal efficiency
cp_a  =  1.005      #Specific heat of air in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma) of air
cp_g  =  1.07      #Specific heat of gas in kJ/kgK
g1  =  1.365      #Specific heat ratio(gamma) of gas

#Solution:
#Refer fig 26.27
T21  =  T1*r_p**((g-1)/g)      #Isentropic temperature at 2 in K
T2  =  (T21-T1)/(eta_C/100)+T1      #Temperature at 2 in K
W  =  W/m      #Specific work output in kJ/kg
Q1  =  W/eta      #Heat added in kJ/kg
W_C  =  cp_a*(T2-T1)      #Compressor work in kJ/kg
W_T  =  W+W_C      #Turbine work in kJ/kg
def  f(T4):
T3  =  T4-Q1/cp_g      #Defining temperature at 3 in terms of T4 in K
T5  =  T4-W_T/cp_g      #Defining temperature at 5 in terms of T4 in K
return (cp_a*(T3-T2))/(cp_g*(T5-T2))-e

#Since effectiveness from the relation must be equal to the given effectiveness
#Thus their difference must be equal to Zero, thus function, f solve for zero to get the value of variable(T4)
T4  =  fsolve(f,1000)
T5  =  T4-W_T/cp_g      #Temperature at 5 in K
T51  =  T4/r_p**((g1-1)/g1)      #Isentropic temperature at 5 in K
eta_T  =  (T4-T5)/(T4-T51)      #Isentropic efficiency of turbine

#Results:
print " The isentropic efficiency of the gas turbine, eta_T   =   %.1f percent"%(eta_T*100)

 The isentropic efficiency of the gas turbine, eta_T   =   90.3 percent


## Example 26.7 Page No : 463¶

In [9]:
import math
from sympy import Symbol,solve

#Given:
r_p  =  4.      #pressure ratio
eta_C  =  0.86
eta_HPT  =  0.84
eta_LPT  =  0.80      #Isentropic efficiencies of compressor and high and low pressure turbine in percent
e  =  70.      #Effectiveness of heat exchanger in percent
eta_d  =  0.92      #Mechanical efficiency of drive to compressor
T4  =  660.+273
T6  =  625.+273      #Temperature of gases entering H.P. turbine and L.P. turbine in K
cp_a  =  1.005      #Specific heat of air in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma)
cp_g  =  1.15      #Specific heat of gas in kJ/kgK
g1  =  1.333      #Specific heat ratio(gamma) of gas
T1  =  15.+273      #Atmospheric temperature in K
P1  =  1.      #Atmospheric pressure in bar

#Solution:
#Refer fig 26.28, 26.29
P2  =  r_p*P1;P4  =  P2      #Pressure at 2, 4 in bar
T21  =  T1*r_p**((g-1)/g)      #Isentropic temperature at 2 in K
T2  =  (T21-T1)/(eta_C)+T1      #Temperature at 2 in K
W_C  =  cp_a*(T2-T1)      #Compressor work in kJ/kg
W_HPT  =  W_C/eta_d      #Work done by H.P. turbine in kJ/kg
T5  =  T4-W_HPT/cp_g      #Temperature at 5 in K
T51  =  T4-(T4-T5)/(eta_HPT)      #Isentropic temperature at 5 in K
P5  =  P4/(T4/T51)**(g1/(g1-1))      #Pressure at 5 in bar
P6  =  P5;P7  =  P1      #Pressure at 6, 7 in bar
T71  =  T6*(P7/P6)**((g1-1)/g1)      #Isentropic temperature at 7 in K
T7  =  T6-eta_LPT*(T6-T71)      #Temperature at 7 in K
W_LPT  =  cp_g*(T6-T7)      #Work done by L.P. turbine in kJ/kg
T3  =  Symbol('T3')      #Defining temperature at 3 as a unknown in K
e1  =  (cp_a*(T3-T2))/(cp_g*(T7-T2))      #Effectiveness in terms of T3
#Effectiveness from the relation must be equal to the given effectiveness
#Thus their difference must be zero
T3  =  solve(e1-e/100)[0]      #Temperature at 3 in K
W  =  cp_g*(T6-T7)      #Work output in kJ/kg (error in book)
Q1  =  cp_g*(T4-T3)+cp_g*(T6-T5)      #Heat added in kJ/kg
eta  =  W/Q1      #Cycle efficiency

#Results:
print " The pressure of the gas entering L.P.T., P6   =   %.2f bar"%(P6)
print " The net specific power, W   =   %.2f kW/kg/s"%(W)
print " The overall efficiency, eta   =   %.4f"%(eta)

 The pressure of the gas entering L.P.T., P6   =   1.66 bar
The net specific power, W   =   98.23 kW/kg/s
The overall efficiency, eta   =   0.2738


## Example 26.8 Page No : 468¶

In [10]:
#Given:
r_p  =  6.      #pressure ratio
e  =  65.      #Effectiveness of heat exchanger in percent
T5  =  800.+273
T1  =  15.+273      #Inlet temperature to H.P. turbine and L.P. compressor in K
m  =  0.7      #Mass flow rate in kg/s
eta_C  =  0.8
eta_HPT  =  0.85
eta_LPT  =  0.85      #Isentropic efficiency of compressor and high and low pressure turbine in percent
eta_d  =  98.      #Mechanical efficiency to drive compressor in percent
eta_c  =  97.      #Combustion efficiency in percent
CV  =  42600.      #Calorific value of fuel in kJ/kg
cp  =  1.005      #Assume specific heat in kJ/kgK
g  =  1.4      #Specific heat ratio(gamma)

#Solution:
#Refer fig 26.30, 26.31
P1  =  1.      #Atmospheric pressure in bar
P3  =  r_p*P1
P5  =  P3
P7  =  P1      #Pressure at 3, 5, 7 in bar
T31  =  T1*r_p**((g-1)/g)      #Isentropic temperature at 3 in K
T31  =  round(T31*10)/10
T3  =  (T31-T1)/(eta_C)+T1      #Temperature at 3 in K
W_C  =  m*cp*(T3-T1)      #Compressor work in kW
W_HPT  =  W_C*100/eta_d      #Work done by H.P. turbine in kW
T6  =  T5-W_HPT/(m*cp)      #Temperature at 6 in K
T61  =  T5-(T5-T6)/(eta_HPT)      #Isentropic temperature at 6 in K
P6  =  P5/(T5/T61)**(g/(g-1))      #Pressure at 6 in bar
T71  =  T6*(P7/P6)**((g-1)/g)      #Isentropic temperature at 7 in K
T7  =  T6-eta_LPT*(T6-T71)      #Temperature at 7 in K
W  =  m*cp*(T6-T7)      #Net power developed in kW
T4  =  e/100*(T7-T3)+T3      #Temperature at 4 in K
Q1  =  m*cp*(T5-T4)*100/eta_c      #Heat supplied in kJ/s
eta  =  W/Q1      #Overall thermal efficiency
sfc  =  Q1*3600/(CV*W)      #Specific fuel consumption in kg/kWh

#Results:
print " a)The net power developed, W   =   %.2f kW"%(W)
print " b)The overall thermal efficiency, eta   =   %.1f percent"%(eta*100)
print " c)The specific fuel consumption, sfc   =   %.3f kg/kWh"%(sfc)

 a)The net power developed, W   =   89.00 kW
b)The overall thermal efficiency, eta   =   28.4 percent
c)The specific fuel consumption, sfc   =   0.298 kg/kWh


## Example 26.9 Page No : 473¶

In [11]:
import math

#Given:
P1  =  4;P2  =  16      #Pressure at entering and leaving of compressor in bar
T1  =  320;T2  =  590      #Temperature at entering and leaving of compressor in K
e  =  70      #Effectiveness of heat exchanger in percent
P3  =  15.5;P4  =  4.2      #Pressure at entering and leaving of turbine in bar
T3  =  1400;T4  =  860      #Temperature at entering and leaving of turbine in K
P  =  100      #Net power output in MW
cp_h  =  5.2      #Specific heat of helium in kJ/kgK
g_h  =  1.67      #Specific heat ratio(gamma) for helium

#Solution:
#Refer fig 26.32, 26.33
T21  =  T1*(P2/P1)**((g_h-1)/g_h)      #Isentropic temperature at 2 in K
eta_C  =  (T21-T1)/(T2-T1)      #Compressor efficiency
T41  =  T3/(P3/P4)**((g_h-1)/g_h)      #Isentropic temperature at 4 in K
eta_T  =  (T3-T4)/(T3-T41)      #Turbine efficiency
Tx  =  T2+(T4-T2)*e/100      #Temperature at leaving of regenerator in K
Q1  =  cp_h*(T3-Tx)      #Heat supplied in kJ/kg
W_T  =  cp_h*(T3-T4)      #Turbine work in kJ/kg
W_C  =  cp_h*(T2-T1)      #Compressor work in kJ/kg
W  =  W_T-W_C      #Work output in kJ/kg
eta  =  W/Q1      #Cycle efficiency
T5  =  T4-(Tx-T2)      #Temperature at 5 in K
Qout  =  cp_h*(T5-T1)      #Heat rejected in precooler in kJ/kg
m_h  =  P*1000/W      #Helium flow rate in kg/s
#Results:

print " a)The compressor efficiency, ,eta_C   =   %.3f\tThe turbine efficiency, eta_T   =   %.3f"%(eta_C,eta_T)
print " b)The thermal efficiency of the cycle, eta   =  ; %.1f percent"%(eta*100)
print " c)The heat rejected in the cooler before compressor, Qout   =   %.1f kJ/kg"%(Qout)
print " d)The helium flow rate for the net power output of 100 MW, m   =   %.2f kg/s"%(m_h)

 a)The compressor efficiency, ,eta_C   =   0.882	The turbine efficiency, eta_T   =   0.946
b)The thermal efficiency of the cycle, eta   =  ; 43.5 percent
c)The heat rejected in the cooler before compressor, Qout   =   1825.2 kJ/kg
d)The helium flow rate for the net power output of 100 MW, m   =   71.23 kg/s


## Example 26.10 Page No : 478¶

In [12]:
import math
#Calculations on closed cycle gas turbine
#Given:
r_p  =  9.      #Overall pressure ratio
eta_LPC  =  85.;eta_HPC  =  85.      #Isentropic efficiency of L.P. and H.P. compressors in percent
eta_LPT  =  90.;eta_HPT  =  90.      #Isentropic efficiency of L.P. and H.P. turbine in percent
T1  =  300.;T5  =  1100.      #Inlet temperature to turbine and compressor in K
cp_ar  =  0.5207      #Specific heat of Argon in kJ/kgK
g_ar  =  1.667      #Specific heat ratio(gamma) for Argon
R_ar  =  0.20813      #Specific gas consmath.tant for Argon in kJ/kgK

#Solution:
#Refer fig. 26.34; 26.35
m_ar  =  1.      #Assume mass flow rate in kg/s
P1  =  1.      #Assume pressure at entering to L.P. compressor in bar
P2  =  math.sqrt(r_p)*P1      #Pressure at leaving to L.P. compressor in bar
P3  =  P2      #Pressure at entering to H.P. compressor in bar
P4  =  r_p*P1      #Pressure at leaving to H.P. compressor in bar
T21  =  T1*(P2/P1)**((g_ar-1)/g_ar)      #Isentropic temperature at 2 in K
T2  =  (T21-T1)/(eta_LPC/100)+T1      #Temperature at ;2 in K
W_LPC  =  m_ar*cp_ar*(T2-T1)      #Work required by L.P. compressor in kJ/kg/s
T3  =  T1      #Temperature at 3 in K
T41  =  T3*(P4/P3)**((g_ar-1)/g_ar)      #Isentropic temperature at 4 in K
T4  =  (T41-T3)/(eta_HPC/100)+T3      #Temperature at 4 in K
#Work required is same for both L.P.C. and H.P.C. as pressure ratio is same for both
W_HPC  =  W_LPC      #Work required by H.P. compressor in kJ/kg/s
P5  =  P4;P6  =  P2;P7  =  P6;P8  =  P1      #Pressure at 5; 6; 7; 8 in bar
T61  =  T5/(P5/P6)**((g_ar-1)/g_ar)      #Isentropic temperature at 6 in K
T6  =  T5-eta_HPT/100*(T5-T61)      #Temperature at 6 in K
W_HPT  =  m_ar*cp_ar*(T5-T6)      #Work done by H.P. turbine in kJ/kg/s
#Work done is same for both L.P.T. and H.P.T. as pressure ratio is same for both
W_LPT  =  W_HPT      #Work done by L.P. turbine in kJ/kg/s
T7  =  T5      #Temperature at 7 in K
#(a)
W  =  (W_HPT+W_LPT)-(W_HPC+W_LPC)      #Net work done in kW/kg
#(b)
r_w  =  W/(W_HPT+W_LPT)      #Work ratio
#(c)
Q1_c  =  m_ar*cp_ar*(T5-T4)      #Heat supplied in combustion chamber in kJ/kg/s
Q1_r  =  m_ar*cp_ar*(T7-T6)      #Heat supplied in reheater in kJ/kg/s
eta  =  W/(Q1_c+Q1_r)      #Overall efficiency

#Results:
print " a)The work done per kg of fuel flow, W   =   %.1f kW/kg"%(W)
print " b)The work ratio, r_w   =   %.3f"%(r_w)
print " c)The overall efficiency, eta   =   %.3f"%(eta)

 a)The work done per kg of fuel flow, W   =   163.8 kW/kg
b)The work ratio, r_w   =   0.447
c)The overall efficiency, eta   =   0.329