# Chapter 6 : Aircraft Propulsion¶

## Example 6.1 page : 23¶

In :
import math

#Input data
eff_com = 0.8 				#Compressor efficiency
eff_t = 0.85 				#Turbine efficiency
pr = 4. 				#Pressure ratio including combustion pressure loss(Po2s/Po1)
eff_c = 0.98 				#Combustion efficiency
eff_m = 0.99 				#Mechanical transmission efficiency
eff_n = 0.9 				#Nozzle efficiency
Tmax = 1000. 				#Maximum cycle temperature in K
To3 = Tmax 				#Stagnation temperature before turbine inlet in K
w = 220. 				#mass flow rate in N/s
Cp_air = 1005. 				#Specific heat capacity at consmath.tant pressure in J/kg-K
k = 1.4 				#Adiabatic consmath.tant for air
Cp_gas = 1153. 				#Specific heat capacity at consmath.tant pressure in J/kg-K
k_gas = 1.3 				#Adiabatic consmath.tant
To1 = 15.+273 				#Inlet Stagnation temperature of compressor in K
Po1 = 1. 				#Inlet Stagnation pressure in bar
Poe = Po1 				#Exit stagnation pressure in bar, Since exit at ambient conditions
g = 9.81 				#Acceleration due to gravity in m/s**2

#Calculation
To2s = To1*(pr)**((k-1)/k) 				#Exit Stagnation temperature of compressor at isentropic process in K
To2 = ((To2s-To1)/eff_com)+To1 				#Exit Stagnation temperature of compressor in K
Wc = (Cp_air*(To2-To1)) 				#Work given to compressor in J/kg, Cp in J/kg-K
To4 = To3-(Wc/Cp_gas*eff_m) 				#Exit Stagnation temperature of turbine in K
To4s = To3-((To3-To4)/eff_t) 				#Exit Stagnation temperature of turbine at isentropic process in K
Po2 = Po1*pr 				#Exit Stagnation pressure of compressor in bar
Po3 = Po2 				#Exit Stagnation pressure of combustion chamber in bar, Since the process takes place at consmath.tant pressure process
p1 = (To3/To4s)**(k_gas/(k_gas-1)) 				#Stagnation Pressure ratio of inlet and outlet of turbine
Po4s = Po3/p1 				#Stagnation Pressure at outlet of turbine at isentropic process in bar
pr_n = Po4s/Poe 				#Pressure ratio of nozzle
Toes = To4/((pr_n)**((k_gas-1)/k_gas)) 				#Exit Stagnation temperature of nozzle at isentropic process in K
Toe = To4-((To4-Toes)*eff_n) 				#Exit Stagnation temperature of nozzle in K
Cj = math.sqrt(2*Cp_gas*(To4-Toe)) 				#Jet velocity in m/s
m = w/g 				#Mass flow rate of air in kg/s
F = m*Cj*10**-3 				#Thrust in kN
Fs = (F*10**3)/m  				#Specific thrust in Ns/kg, F in N
Is = F/w 				#Specific impulse in sec

#Output
print 'A)Thrust is %3.3f kN \
\nB)Specific thrust is %3.2f Ns/kg'%(F,Fs)

A)Thrust is 10.180 kN
B)Specific thrust is 453.94 Ns/kg


## Example 6.2 page : 26¶

In :
import math

#Input data
u = 800.*(5./18) 				#Flight velocity in m/s
Pe = 60. 					#Ambient pressure in kPa
Pn = 300. 					#Pressure entering nozzle in kPa
Tn = 200.+273 				#Temperature entering nozzle in K
m = 20. 					#Mass flow rate of air in kg/s
Cp = 1005. 					#Specific heat capacity at consmath.tant pressure in J/kg-K
k = 1.4 					#Adiabatic consmath.tant for air

#Calculation
Te = Tn*(Pe/Pn)**((k-1)/k) 				#Exit temperature of nozzle in K
Cj = math.sqrt(2*Cp*(Tn-Te)) 				#Jet velocity in m/s
F = m*(Cj-u) 				#Thrust in N
P = F*u*10**-3 				#Thrust power in kW
eff = ((2*u)/(Cj+u))*100 				#Propulsive efficiency in percent

#Output
print 'A)Thrust developed is %3.1f N \
\nB)Thrust developed is %3.2f kW \
\nC)Propulsive efficiency is %3.3f percent'%(F,P,eff)

A)Thrust developed is 7395.4 N
B)Thrust developed is 1643.42 kW
C)Propulsive efficiency is 54.586 percent


## Example 6.3 page : 26¶

In :
import math

#Input data
Mi = 0.8 				#Inlet mach number
h = 10000. 				#Altitude in m
pr_c = 8. 				#Pressure ratio of compressor
To3 = 1200. 				#Stagnation temperature at turbine inlet in K
eff_c = 0.87 				#Compressor efficiency
eff_t = 0.9 				#Turbine efficiency
eff_d = 0.93 				#Diffuser efficiency
eff_n = 0.95  				#Nozzle efficiency
eff_m = 0.99 				#Mechanical transmission efficiency
eff_cc = 0.98 				#Combustion efficiency
pl = 0.04 				#Ratio of combustion pressure loss to compressor delivery pressure
k = 1.4 				#Adiabatic consmath.tant of air
R = 287. 				#Specific gas consmath.tant in J/kg-K
k_g = 1.33 				#Adiabatic consmath.tant of gas
Cp_a = 1005. 				#Specific heat capacity at consmath.tant pressure of air in J/kg-K
Cp_g = 1100. 				#Specific heat capacity at consmath.tant pressure of gas in J/kg-K
CV = 43000000. 				#Calorific value in J/kg (AssumE)

#Calculation
Ti = 223.15 				#Inlet temperature in K from gas tables
Pi = 26.4 				#Inlet pressure in kPa from gas tables
ai = math.sqrt(k*R*Ti) 				#Sound velocity in m/s
Ci = ai*Mi 				#Velocity of air in m/s,
u = Ci 				#Flight velocity in m/s, Since it is reaction force of Ci
t1 = 0.886 				#Ratio of static to stagnation temperature a entry from gas tables at M = 0.8
To1s = Ti/t1 				#Stagnation temperature at inlet of compressor at isentropic process in K
To1 = ((To1s-Ti)/eff_d)+Ti 				#Stagnation temperature at inlet of compressor in K
p1 = (To1s/Ti)**(k/(k-1)) 				#Pressure ratio i.e. (Po1s/Pi)
Po1s = Pi*p1 				#inlet Stagnation pressure of compressor at isentropic process in kPa
Po1 = Po1s 				#Inlet Stagnation pressure of compressor in kPa
Po2 = pr_c*Po1 				#Exit Stagnation pressure of compressor in kPa
To2s = To1s*(Po2/Po1)**((k-1)/k) 				#Exit Stagnation temperature of compressor at isentropic process in K
To2 = ((To2s-To1)/eff_c)+To1 				#Exit Stagnation temperature of compressor in K
P_los = pl*Po2 				#combustion pressure loss in kPa
Po3 = Po2-P_los 				#Exit Stagnation pressure of combustion chamber in kPa
To4 = To3-((Cp_a*(To2-To1))/(eff_m*Cp_g)) 				#Exit Stagnation temperature of turbine in K
To4s = To3-((To3-To4)/eff_t) 				#Exit Stagnation temperature of turbine at isentropic process in K
p1 = (To3/To4s)**(k_g/(k_g-1)) 				#Pressure ratio i.e. (Po3/Po4s)
Po4s = Po3/p1 				#Stagnation Pressure at outlet of turbine at isentropic process in kPa
Poe = Pi 				#Exit stagnation pressure in kPa, Since exit is at ambient conditions
pr_n = Po4s/Poe 				#Pressure ratio of nozzle
Toes = To4/((pr_n)**((k_g-1)/k_g)) 				#Exit Stagnation temperature of nozzle at isentropic process in K
Toe = To4-((To4-Toes)*eff_n) 				#Exit Stagnation temperature of nozzle in K
Cj = math.sqrt(2*Cp_g*(To4-Toe)) 				#Jet velocity in m/s
Fs = Cj-u 				#Specific thrust in Ns/kg
f = ((Cp_g*To3)-(Cp_a*To2))/((eff_cc*CV)-(Cp_g*To3)) 				#Fuel-air ratio
TSFC = (f/Fs)#*10**5 				#Thrust specific fuel consumption in kg/s-N x10**-5

#Output
print 'A)Specific thrust is %3.2f Ns/kg \
\nB)Thrust specific fuel consumption is %.3e kg/s-N'%(Fs,TSFC)

A)Specific thrust is 575.62 Ns/kg
B)Thrust specific fuel consumption is 3.537e-05 kg/s-N


## Example 6.4 page : 29¶

In :
import math

#Input data
u = 300. 				#Flight velocity in m/s
Pi = 35. 				#Inlet pressure in kPa
Ti = -40.+273 				#Inlet temperature in K
pr_c = 10. 				#Pressure ratio of compressor
T3 = 1100.+273 				#Inlet turbine temperature in K
m = 50. 				#Mass flow rate of air in kg/s
k = 1.4 				#Adiabatic consmath.tant of air
Cp = 1005. 				#Specific heat capacity at consmath.tant pressure of air in J/kg-K
R = 287. 				#Specific gas consmath.tant in J/kg-K

#Calculation
ai = math.sqrt(k*R*Ti) 				#Sound velocity at diffuser in m/s
C1 = u 				#Velocity of air in m/s, Since it is reaction force of u
T1 = Ti+(C1**2/(2*Cp)) 				#Temperature at inlet of compressor in K
P1 = Pi*((T1/Ti)**(k/(k-1))) 				#Inlet pressure of compressor in kPa
P2 = pr_c*P1 				#Exit pressure of compressor in kPa
P3 = P2 				#Exit pressure of combustion chamber in kPa, Since the process takes place at consmath.tant pressure process
T2 = T1*(P2/P1)**((k-1)/k) 				#Exit temperature of compressor in K
T4 = T3-(T2-T1) 				#Exit temperature of turbine in K
P4 = P3/((T3/T4)**(k/(k-1))) 				#Pressure at outlet of turbine in kPa
Pe = Pi 				#Exit pressure in kPa, Since exit is at ambient conditions
pr_n = P4/Pe 				#Pressure ratio of nozzle
Te = T4/((pr_n)**((k-1)/k)) 				#Exit temperature of nozzle in K
Cj = math.sqrt(2*Cp*(T4-Te)) 				#Jet velocity in m/s
sig = u/Cj 				#Jet speed ratio
eff_prop = ((2*sig)/(1+sig))*100 				#Propulsive efficiency of the cycle in %

#Output
print 'A)Temperature and pressure of gases at turbine exit is %3.2f K and %3i kPa \
\nB)Velocity of gases is %3.2f m/s \
\nC)Propulsive efficiency of the cycle is %3.2f percent'%(T4,P4,Cj,eff_prop)

A)Temperature and pressure of gases at turbine exit is 1114.47 K and 311 kPa
B)Velocity of gases is 1020.35 m/s
C)Propulsive efficiency of the cycle is 45.44 percent


## Example 6.5 page : 30¶

In :
import math

#Input data
n = 2 				#Number of jets
D = 0.25 				#Diameter of turbojet in m
P = 3000 				#Net power at turbojet in W
mf_kWh = 0.42 				#Fuel consumption in kg/kWh
CV = 49000 				#Calorific value in kJ/kg
u = 300 				#Flight velocity in m/s
d = 0.168 				#Density in kg/m**3
AFR = 53 				#Air fuel ratio

#Calculatioon
mf = mf_kWh*P/3600 				#Mass flow rate of fuel in kg/s
ma = AFR*mf 				#Mass flow rate of air in kg/s
m = ma+mf 				#Mass flow rate of gas in kg/s
Q = m/d 				#Volume flow rate in m**3/s
Cj = (Q*4)/(2*math.pi*D**2) 				#Jet velocity in m/s
Ca = Cj-u 				#Absolute Jet velocity in m/s
F = ((m*Cj)-(ma*u))*10**-3 				#Thrust in kN
eff = ((F*u)/(mf*CV))*100 				#Overall efficiency in %
eff_prop = ((2*u)/(Cj+u))*100 				#Propulsive efficiency of the cycle in %
eff_ther = (eff/eff_prop)*100 				#Efficiency of turbine in %

#Output
print 'A)Absolute velocity of jet is %3.3f m/s \
\nB)Resistance of the plane is %3.4f kN \
\nC)Overall efficiency is %3.2f percent \
\nD)Efficiency of turbine is %3.3f percent'%(Ca,F,eff,eff_ther)

A)Absolute velocity of jet is 845.916 m/s
B)Resistance of the plane is 16.0928 kN
C)Overall efficiency is 28.15 percent
D)Efficiency of turbine is 67.839 percent


## Example 6.6 page : 31¶

In :
import math

#Input data
u = 900*(5./18) 				#Flight velocity in m/s
ma = 3000./60 				#Mass flow rate of air in kg/s
dh = 200. 				#Enthalpy drop of nozzle in kJ/kg
eff_n = 0.9 				#Nozzle efficiency
AFR = 85 				#Air fuel ratio
eff_cc = 0.95 				#Combustion efficiency
CV = 42000 				#Calorific value in kJ/kg

#Calculation
mf = ma/AFR 				#Mass flow rate of fuel in kg/s
m = ma+mf 				#Mass flow rate of gas in kg/s
Cj = math.sqrt(2*eff_n*dh*10**3) 				#Jet velocity in m/s
sig = u/Cj 				#Jet speed ratio
F = ((m*Cj)-(ma*u))*10**-3 				#Thrust in kN
Pt = F*u 				#Thrust power in kW
Pp = 0.5*((m*Cj**2)-(ma*u**2))*10**-3 				#Propulsive power in kW
HS = eff_cc*mf*CV 				#Heat supplied in kW
eff_ther = (Pp/HS)*100 				#Efficiency of turbine in %
eff_prop = (Pt/Pp)*100 				#Propulsive efficiency of the cycle in %
eff = (Pt/HS)*100 				#Overall efficiency in %

#Output
print 'A)Propulsive power is %3.2f kW \
\nB)Thrust power is %3.1f kW \
\nC)Propulsive efficiency is %3.3f percent \
\nD)Thermal efficiency is %3.2f percent \
\nE)Total fuel consumption is %3.3f kg/s F)Overall efficiency is %3.3f percent'%(Pp,Pt,eff_prop,eff_ther,mf,eff)

A)Propulsive power is 7543.38 kW
B)Thrust power is 4463.2 kW
C)Propulsive efficiency is 59.168 percent
D)Thermal efficiency is 32.14 percent
E)Total fuel consumption is 0.588 kg/s F)Overall efficiency is 19.016 percent


## Example 6.7 page : 32¶

In :
import math

#Input data
M = 0.8 				#Mach number
CV = 42800. 				#Calorific value in kJ/kg
h = 10. 				#Altitude in km
F = 50. 				#Thrust in kN
ma = 45. 				#Mass flow rate of air in kg/s
mf = 2.65 				#Mass flow rate of fuel in kg/s

#Calculation
m = ma+mf 				#Mass flow rate of gas in kg/s
a = 299.6 				#Sound velocity in m/s, from gas tables
T = 233.15 				#Inlet temperature in K
u = a*M 				#Flight velocity in m/s
Cj = ((F*10**3)+(ma*u))/m 				#Jet velocity in m/s
sig = u/Cj 				#Jet speed ratio
Fs = F*10**3/m 				#Specific thrust in Ns/kg, F in N
TSFC = mf*3600/(F*10**3) 				#Thrust specific fuel consumption in kg/N-hr, F in N
Pt = F*u 				#Thrust power in kW
Pp = 0.5*((m*Cj**2)-(ma*u**2))*10**-3 				#Propulsive power in kW
HS = mf*CV 				#Heat supplied in kW
eff_ther = (Pp/HS)*100 				#Efficiency of turbine in %
eff_prop = (Pt/Pp)*100 				#Propulsive efficiency of the cycle in %
eff = (Pt/HS)*100 				#Overall efficiency in %

#Output
print 'A)Specific thrust is %3.2f N/kg \
\nB)Thrust specific fuel consumption is %3.4f kg/N-hr \
\nC)Jet velocity is %3.3f m/s \
\nD)Thermal efficiency is %3.2f percent \
\nE)Propulsive efficiency is %3.3f percent F)Overall efficiency is %3.2f percent'%(Fs,TSFC,Cj,eff_ther,eff_prop,eff)

A)Specific thrust is 1049.32 N/kg
B)Thrust specific fuel consumption is 0.1908 kg/N-hr
C)Jet velocity is 1275.668 m/s
D)Thermal efficiency is 33.04 percent
E)Propulsive efficiency is 31.976 percent F)Overall efficiency is 10.57 percent


## Example 6.8 page : 34¶

In :
import math

#Input data
Mi = 0.8 				#Inlet mach number
h = 10. 				#Altitude in km
To3 = 1200. 				#Stagnation temperature before turbine inlet in K
dTc = 175. 				#Stagnation temperature rise through the compressor in K
CV = 43000. 				#Calorific value in kJ/kg
eff_c = 0.75 				#Compressor efficiency
eff_cc = 0.75 				#Combustion efficiency
eff_t = 0.81 				#Turbine efficiency
eff_m = 0.98 				#Mechanical transmission efficiency
eff_n = 0.97 				#Nozzle efficiency
Is = 25. 				#Specific impulse in sec
k = 1.4 				#Adiabatic consmath.tant of air
R = 287. 				#Specific gas consmath.tant in J/kg-K
Cp = 1005. 				#Specific heat capacity at consmath.tant pressure of air in J/kg-K
g = 9.81 				#Acceleration due to gravity in m/s**2

#Calculation
Ti = 223.15 				#Inlet temperature in K from gas tables
ai = math.sqrt(k*R*Ti) 				#Sound velocity in m/s
Toi = (1+((0.5*(k-1)*Mi**2)))*Ti 				#Stagnation temperature at diffuser inlet in K
To1 = Toi 				#Inlet Stagnation temperature of compressor in K, math.since hoi = ho1
To2 = dTc+To1 				#Exit Stagnation temperature of compressor in K
pr_c = (1+(eff_c*((To2-To1)/To1)))**(k/(k-1)) 				#Compressor pressure ratio
f = ((Cp*To3)-(Cp*To2))/((eff_cc*CV*10**3)-(Cp*To3)) 				#Fuel-air ratio, calculation mistake in textbook
dTt = dTc/(eff_m*(1+f)) 				#Temperature difference across turbine
pr_t = 1/((1-(dTt/(To3*eff_t)))**(k/(k-1))) 				#Turbine pressure ratio
To4 = To3-dTc 				#Exit Stagnation temperature of turbine in K
u = ai*Mi 				#Flight velocity in m/s
sig = 1/(((Is*g)/u)+1) 				#Jet speed ratio
Ce = u/sig 				#Exit velocity in m/s
Cj = Ce 				#Jet velocity in m/s, Since Cj is due to exit velociy
Te = To4-(Ce**2/(2*Cp)) 				#Exit temperature in K
Tes = To4-((To4-Te)*eff_n) 				#Exit temperature in K, (At isentropic process)
pr_n = (To4/Te)**(k/(k-1)) 				#Nozzle pressure ratio
ae = math.sqrt(k*R*Te) 				#Exit Sound velocity in m/s
Me = Ce/ae 				#Exit mach number

print 'A)Fuel-air ratio is %3.5f  \
\nB)Compressor, turbine, nozzle pressure ratio are %3.3f, %3.3f, %3.2f respectively \
\nC)Mach number at exhaust jet is %3.3f'%(f,pr_c,pr_t,pr_n,Me)

A)Fuel-air ratio is 0.02503
B)Compressor, turbine, nozzle pressure ratio are 4.344, 1.996, 1.53 respectively
C)Mach number at exhaust jet is 0.803


## Example 6.9 page : 36¶

In :
import math

#Input data
D = 2.5 				#Diameter in m
u = 500.*(5./18) 				#Flight velocity in m/s
h = 8000. 				#Altitude in m
sig = 0.75 				#Jet speed ratio
g = 9.81 				#Acceleration due to gravity in m/s**2

#Calculation
d = 0.525 				#from gas tables
A = math.pi*D**2*0.25 				#Area of flow in m**2
Cj = u/sig 				#Jet velocity in m/s
Vf = (u+Cj)/2 				#Velocity of flow in m/s
ma = d*A*Vf 				#Mass flow rate of air in kg/s
F = ma*(Cj-u)*10**-3 				#Thrust in kN
P = F*u 				#Thrust power in kW
Fs = F*10**3/ma 				#Specific thrust in Ns/kg
Is = Fs/g 				#Specific impulse in sec

#Output
print 'A)Flow rate of air through the propeller is %3.3f m/s \
\nB)Thrust produced is %3.3f kN \
\nC)Specific thrust is %3.2f N-s/kg \
\nD)Specific impulse is %3.3f sec \
\nE)Thrust power is %3.1f kW'%(ma,F,Fs,Is,P)

A)Flow rate of air through the propeller is 417.584 m/s
B)Thrust produced is 19.333 kN
C)Specific thrust is 46.30 N-s/kg
D)Specific impulse is 4.719 sec
E)Thrust power is 2685.1 kW


## Example 6.10 page : 37¶

In :
import math

#Input data
h = 3000. 				#Altitude in m
Pi = 0.701 				#Inlet pressure in bar
Ti = 268.65 				#Inlet temperature in K
u = 525*(5./18) 				#Flight velocity in m/s
eff_d = 0.875 				#Diffuser efficiency
eff_c = 0.79 				#Compressor efficiency
C1 = 90. 				#Velocity of air at compressor in m/s
dTc = 230. 				#Temperature rise through compressor
k = 1.4 				#Adiabatic consmath.tant of air
Cp = 1005. 				#Specific heat capacity at consmath.tant pressure of air in J/kg-K
R = 287. 				#Specific gas consmath.tant in J/kg-K

#Calculation
ai = math.sqrt(k*R*Ti) 				#Sound velocity in m/s
Mi = u/ai 				#Inlet mach number
Toi = (1+((0.5*(k-1)*Mi**2)))*Ti 				#Stagnation temperature at diffuser inlet in K
To1 = Toi 				#Inlet Stagnation temperature of compressor in K, math.since hoi = ho1
T1 = To1-(C1**2/(2*Cp)) 				#Temperature at inlet of compressor in K
P1 = Pi*((1+(eff_d*((T1/Ti)-1)))**(k/(k-1))) 				#Inlet pressure of compressor in bar
dPc = P1-Pi 				#Pressure rise through inlet diffuser in bar
pr_c = (((eff_c*dTc)/To1)+1)**(k/(k-1)) 				#Pressure ratio of compressor
P = Cp*(dTc) 				#Power required by the compressor in kW/(kg/s)
eff = 1-(1/pr_c**((k-1)/k)) 				#Air standard efficiency

#Output
print 'A)Pressure rise through diffuser is %3.4f bar \
\nB)Pressure developed by compressure is %3.4f bar \
\nC)Air standard efficiency of the engine is %3.4f'%(dPc,P1,eff)

A)Pressure rise through diffuser is 0.0538 bar
B)Pressure developed by compressure is 0.7548 bar
C)Air standard efficiency of the engine is 0.3942


## Example 6.11 page : 38¶

In :
import math

#Input data
h = 9500. 				#Altitude in m
u = 800*(5./18) 				#Flight velocity in m/s
eff_prop = 0.55 				#Propulsive efficiency of the cycle
eff_o = 0.17 				#Overall efficiency
F = 6100. 				#Thrust in N
d = 0.17 				#Density in kg/m**3
CV = 46000. 				#Calorific value in kJ/kg

#Calculation
mf = (F*u)/(eff_o*CV*10**3) 				#Mass flow rate of fuel in kg/s
Cj = ((2*u)/(eff_prop))-u 				#Jet velocity in m/s, wrong calculation in textbook
Ca = Cj-u 				#Absolute Jet velocity in m/s
ma = (F-(mf*Cj))/(Ca) 				#Mass flow rate of air in kg/s
m = ma+mf 				#Mass flow rate of gas in kg/s
f = ma/mf 				#Air fuel ratio
Q = m/d 				#Volume flow rate in m**3/s
Dj = math.sqrt((4*Q)/(math.pi*Cj))*10**3 				#Diameter of jet in mm, Cj value wrong in textbook
P = ((F*u)/eff_prop)*10**-3 				#Power output of engine in kW

#Output
print 'A)Diamter of the jet is %3.1f mm \
\nB)Power output is %3.1f kW \
\nC)Air-fuel ratio is %3.3f \
\nD)Absolute velocity of the jet is %3i m/s'%(Dj,P,f,Ca)

A)Diamter of the jet is 461.6 mm
B)Power output is 2464.6 kW
C)Air-fuel ratio is 95.161
D)Absolute velocity of the jet is 363 m/s


## Example 6.12 page : 39¶

In :
import math

#Input data
u = 960*(5./18) 				#Flight velocity in m/s
ma = 40. 				#Mass flow rate of air in kg/s
AFR = 50. 				#Air fuel ratio
sig = 0.5 				#Jet speed ratio, for maximum thrust power
CV = 43000. 				#Calorific value in kJ/kg

#Calculation
mf = ma/AFR 				#Mass flow rate of fuel in kg/s
m = ma+mf 				#Mass flow rate of gas in kg/s
Cj = u/sig 				#Jet velocity in m/s
F = ((m*Cj)-(ma*u))*10**-3 				#Thrust in kN
Fs = F*10**3/m 				#Specific thrust in Ns/kg, F in N
Pt = F*u 				#Thrust power in kW
eff_prop = ((2*sig)/(1+sig))*100 				#Propulsive efficiency of the cycle in %
eff_ther = ((0.5*m*(Cj**2-u**2))/(mf*CV*10**3))*100 				#Efficiency of turbine in %
eff = (eff_prop/100)*(eff_ther/100)*100 				#Overall efficiency in %
TSFC = mf*3600/(F*10**3) 				#Thrust specific fuel consumption in kg/Nhr

#Output
print 'A)Jet velocity is %3.1f m/s \
\nB)Thrust is %3.3f kN \
\nC)Specific thrust is %3.2f N-s/kg \
\nD)Thrust power is %3.2f kW \
\nE)propulsive, thermal and overall efficiency is %3.2f, %3.2f and %3.3f respectively \
\nF)Thrust specific fuel consumption is %3.4f kg/Nhr'%(Cj,F,Fs,Pt,eff_prop,eff_ther,eff,TSFC)

A)Jet velocity is 533.3 m/s
B)Thrust is 11.093 kN
C)Specific thrust is 271.90 N-s/kg
D)Thrust power is 2958.22 kW
E)propulsive, thermal and overall efficiency is 66.67, 12.65 and 8.434 respectively
F)Thrust specific fuel consumption is 0.2596 kg/Nhr


## Example 6.13 page : 40¶

In :
import math

#Input data
u = 960*(5./18) 				#Flight velocity in m/s
ma = 54.5 				#Mass flow rate of air in kg/s
dh = 200. 				#Change of enthalpy for nozzle in kJ/kg
Cv = 0.97 				#Velocity coefficient
AFR = 75. 				#Air fuel ratio
eff_cc = 0.93 				#Combustion efficiency
CV = 45000. 				#Calorific value in kJ/kg

#Calculation
mf = ma/AFR 				#Mass flow rate of fuel in kg/s
Cj = Cv*math.sqrt(2*dh*10**3) 				#Jet velocity in m/s
F = ma*(Cj-u) 				#Thrust in kN
TSFC = mf*3600/(F) 				#Thrust specific fuel consumption in kg/Nhr
HS = mf*eff_cc*CV 				#Heat supplied in kJ/s
Pp = 0.5*ma*(Cj**2-u**2)*10**-3 				#Propulsive power in kW
Pt = F*u 				#Thrust power in kW
eff_p = Pt/(Pp*10**3) 				#Propulsive efficiency of the cycle
eff_t = Pp/HS 				#Efficiency of turbine
eff_o = Pt*10**-3/HS  				#Overall efficiency

#Output
print 'A)Exit velocity of the jet is %3.2f m/s \
\nB)Fuel rate is %3.4f kg/s \
\nC)Thrust specific fuel consumption is %3.5f kg/Nhr \
\nD)Thermal efficiency is %3.3f \
\nE)Propulsive power is %3.2f kW \
\nF)Propulsive efficiency is %3.4f \
\nG)Overall efficiency is %3.5f'%(Cj,mf,TSFC,eff_t,Pp,eff_p,eff_o)

A)Exit velocity of the jet is 613.48 m/s
B)Fuel rate is 0.7267 kg/s
C)Thrust specific fuel consumption is 0.13840 kg/Nhr
D)Thermal efficiency is 0.274
E)Propulsive power is 8318.03 kW
F)Propulsive efficiency is 0.6060
G)Overall efficiency is 0.16574


## Example 6.14 page : 41¶

In :
import math

#Input data
u = 750*(5./18) 				#Flight velocity in m/s
h = 10000. 				#Altitude in m
eff_p = 0.5 				#Propulsive efficiency of the cycle
eff_o = 0.16 				#Overall efficiency
d = 0.173 				#Density in kg/m**3
F = 6250. 				#Thrust in N
CV = 45000. 				#Calorific value in kJ/kg

#Calculation
sig = eff_p/(2-eff_p) 				#Jet speed ratio
Cj = u/sig 				#Jet velocity in m/s
Ca = Cj-u 				#Absolute Jet velocity in m/s
ma = F/Ca 				#Mass flow rate of air in kg/s
Q = ma*60/d 				#Volume flow rate in m**3/min
A = Q/(Cj*60) 				#Area of flow in m**2
D = math.sqrt((4*A)/(math.pi))*10**3 				#Diameter in mm
Pt = F*u 				#Thrust power in W
Pp = (Pt/eff_p)*10**-3  				#Propulsive power in kW
eff_t = eff_o/eff_p 				#Efficiency of turbine
HS = Pp/eff_t 				#Heat supplied in kJ/s
mf = HS/CV 				#Mass flow rate of fuel in kg/s
AFR = ma/mf 				#Air fuel ratio

#Output
print 'A)Absolute velocity of the jet is %3.2f m/s \
\nB)Volume of air compressed per minute is %3.2f m**3/min \
\nC)Diameter of the jet is %3i mm \
\nD)Power unit of the unit is %3.3f kW \
\nE)Air fuel ratio is %3.1f'%(Ca,Q,D,Pp,AFR)

A)Absolute velocity of the jet is 416.67 m/s
B)Volume of air compressed per minute is 5202.31 m**3/min
C)Diameter of the jet is 420 mm
D)Power unit of the unit is 2604.167 kW
E)Air fuel ratio is 82.9


## Example 6.15 page : 42¶

In :
import math

#Input data
P1 = 0.56 				#Inlet pressure of compressor in bar
T1 = 260 				#Temperature at inlet of compressor in K
pr_c = 6 				#Pressure ratio of compressor
eff_c = 0.85 				#Compressor efficiency
u = 360*(5./18) 				#Flight velocity in m/s
D = 3 				#Propeller diameter in m
eff_p = 0.8 				#Efficiency of propeller
eff_g = 0.95 				#Gear reduction efficiency
pr_t = 5 				#Expansion ratio
eff_t = 0.88 				#Turbine efficiency
T3 = 1100 				#temperature at turbine inlet in K
eff_n = 0.9 				#Nozzle efficiency
Cp = 1005. 				#Specific heat capacity at consmath.tant pressure of air in J/kg-K
CV = 40000 				#Calorific value in kJ/kg
k = 1.4 				#Adiabatic consmath.tant of air
R = 287 				#Specific gas consmath.tant in J/kg-K

#Calculation
P2 = pr_c*P1 				#Exit pressure of compressor in bar
T2s = T1*(pr_c)**((k-1)/k) 				#Exit temperature of compressor at isentropic proces in K
T2 = T1+((T2s-T1)/eff_c) 				#Exit temperature of compressor in K
Wc = Cp*(T2-T1)*10**-3 				#Power input to compressor in kJ/kg of air
C1 = u 				#Air velocity in m/s, math.since C1 is resulmath.tant of u
C = C1/eff_p 				#Average velocity in m/s
C2 = (2*C)-C1 				#Exit velocity from compressor in m/s
Ap = 0.25*math.pi*D**2 				#Area of propeller passage in m**2
Q = Ap*C 				#Quantity of air inducted in m**3/s
mf = ((T3-T2)*Cp)/((CV*10**3)-(Cp*T3)) 				#Mass flow rate of fuel in kg/s
f = mf 				#Fuel consumption in kg/kg of air
AFR = 1/mf 				#Air fuel ratio
P3 = P2 				#Exit pressure of combustion chamber in bar, Since process is at consmath.tant pressure
P4 = P3/pr_t 				#Exit pressure of turbine in bar
T4s = T3/((pr_t)**((k-1)/k)) 				#Exit temperature of turbine at isentropic proces in K, wrong calculation
T4 = T3-(eff_t*(T3-T4s)) 				#Exit temperature of turbine in K
Po = (1+f)*Cp*(T3-T4)*10**-3 				#Power output per kg of air in kJ/kg of air
Pa = Po-Wc 				#Power available for propeller in kJ/kg of air
Pe = P1 				#Exit pressure in bar, Since exit is at ambient conditions
Tes = T4/((P4/Pe)**((k-1)/k)) 				#Exit temperature of nozzle at isentropic proces in K
Cj = math.sqrt(2*Cp*eff_n*(T4-Tes)) 				#Jet velocity in m/s
Fs = ((1+f)*Cj)-u 				#Specific thrust in Ns/kg, F in N
Pp = ((0.5*P1*10**5*Q*(C2**2-C1**2))/(R*T1))*10**-3 				#Propulsive power by propeller in kJ/s
Ps = Pp/eff_g 				#Power supplied by the turbine in kW
ma = Ps/Pa 				#Air flow rate in kg/s
Fj = ma*Cj*10**-3 				#Jet thrust in kN, calculation mistake
Fp = (Pp*eff_p)/u 				#Thrust produced by propeller in kN

#Output
print 'A)Air fuel ratio is %3.2f \
\nB)Thrust produced by the nozzle is %3.3f kN \
\nC)Thrust by the propeller is %3.3f kN \
\nD)mass flow rate through the compressor is %3.2f kg/s'%(AFR,Fj,Fp,ma)

A)Air fuel ratio is 60.90
B)Thrust produced by the nozzle is 7.168 kN
C)Thrust by the propeller is 33.155 kN
D)mass flow rate through the compressor is 27.44 kg/s


## Example 6.16 page : 45¶

In :
import math

#Input data
M1 = 1.5 				#Mach number
h = 6500. 				#Altitude in m
D = 0.5 				#Diameter in m
To4 = 1600. 				#Stagnation temperature at nozzle inlet in K
CV = 40000. 				#Calorific value in kJ/kg
k = 1.4 				#Adiabatic consmath.tant of air
R = 287. 				#Specific gas consmath.tant in J/kg-K
eff_d = 0.9 				#Diffuser efficiency
eff_cc = 0.98 				#Combustion efficiency
eff_n = 0.96 				#Nozzle efficiency
pr_l = 0.02 				#Pressure ratio i.e. Stagnation pressure loss to Exit presure of compressor
Cp = 1005. 				#Specific heat capacity at consmath.tant pressure of air in J/kg-K

#Calculation
P1 = 0.44 				#Inlet pressure of compressor in bar
T1 = 245.9  				#Temperature at inlet of compressor in K
a1 = 314.5 				#Sound velocity at compressor in m/s
d1 = 0.624 				#Density at compressor in kg/m**3
A1 = 0.25*math.pi*D**2 				#Area at diffuser inlet in m**2
u1 = M1*a1 				#Flight velocity in m/s
ma = d1*A1*u1 				#Mass flow rate of air in kg/s
To2 = T1*(1+(((k-1)/2)*M1**2)) 				#Stagnation temperature at commpressor inlet in K
To1 = To2 				#Stagnation temperature at commpressor outlet in K, (It is in case of diffuser)
pr_d = ((eff_d*(((k-1)/2)*M1**2))+1)**(k/(k-1)) 				#Pressure ratio of diffuser
P2 = pr_d*P1 				#Exit pressure of compressor in bar
Po2 = P2 				#Stagnation pressure at exit of compressor in bar
Po3 = (Po2-(pr_l*Po2)) 				#Stagnation pressure at exit of combustion chamber in bar
Poe = P1 				#Exit stagnation pressure in kPa, Since exit is at ambient conditions
pr_n = Po3/Poe 				#Pressure ratio of nozzle
p1 = 1/pr_n 				#Inverse of pr_n to find in gas tables
M4s = 1.41 				#Mach number at turbine exit  from gas tables
T4s = To4/(1+((0.5*(k-1)*M4s**2))) 				#Exit temperature of turbine at isentropic process in K
To3 = To4 				#Stagnation temperature at inlet turbine in K,
T4 = To3-(eff_n*(To3-T4s)) 				#Exit temperature of turbine in K
C4 = math.sqrt(2*Cp*(To4-T4)) 				#Flight velocity of air in m/s
a4 = math.sqrt(k*R*T4) 				#Sound velocity in m/s
Me = C4/a4 				#Nozzle jet mach number
f = (Cp*(To3-To2))/(eff_cc*CV*10**3) 				#Fuel air ratio
mf = ma*f 				#Mass flow rate of fuel in kg/s
m = ma+mf 				#Mass flow rate of gas in kg/s
eff_i = (1/(1+((2/(k-1))*(1/M1**2))))*100 				#Efficiency of the ideal cycle in %
sig = u1/C4 				#Jet speed ratio
eff_p = ((2*sig)/(1+sig)) 				#Propulsive efficiency in %
F = ((m*C4)-(ma*u1))*10**-3 				#Thrust in kN

#Output
print 'A)Efficiency of the ideal cycle is %3i percent \
\nB)Flight speed is %3.3f m/s \
\nC)Air flow rate is %3.3f kg/s \
\nD)Diffuser pressure ratio is %3.4f \
\nE)Fuel air ratio is %3.5f \
\nF)Nozzle pressure ratio is %3.2f \
\nG)Nozzle jet mach number is %3.3f \
\nH)Propulsive efficiency is %3.4f percent \
\nI)Thrust is %3.3f kN'%(eff_i,C4,ma,pr_d,f,pr_n,Me,eff_p,F)

A)Efficiency of the ideal cycle is  31 percent
B)Flight speed is 937.202 m/s
C)Air flow rate is 57.800 kg/s
D)Diffuser pressure ratio is 3.2875
E)Fuel air ratio is 0.03188
F)Nozzle pressure ratio is 3.22
G)Nozzle jet mach number is 1.371
H)Propulsive efficiency is 0.6696 percent
I)Thrust is 28.630 kN


## Example 6.17 page : 47¶

In :
import math

#Input data
ma = 18. 				#Mass flow rate of air in kg/s
Mi = 0.6 				#Inlet mach number
h = 4600. 				#Altitude in m
Pi = 55. 				#Inlet pressure in
Ti = -20.+273 				#Inlet temperature in K
eff_d = 0.9 				#Diffuser efficiency
pr_d = 5. 				#Diffuser pressure ratio
T3 = 1000.+273 				#Inlet turbine temperature in K
Pe = 60. 				#Exit pressure in kPa
eff_c = 0.81 				#Compressor efficiency
eff_t = 0.85 				#Turbine efficiency
eff_n = 0.915 				#Nozzle efficiency
CV = 46520. 				#Calorific value in kJ/kg
Cp = 1005. 				#Specific heat capacity at consmath.tant pressure of air in J/kg-K
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Specific gas consmath.tant in J/kg-K

#Calculation
Ci = Mi*math.sqrt(k*R*Ti) 				#Velocity of air in m/s,
u = Ci 				#Flight velocity in m/s, Since it is reaction force of Ci
T1 = Ti+(Ci**2/(2*Cp)) 				#Temperature at inlet of compressor in K
P1s = Pi*(T1/Ti)**(k/(k-1)) 				#Inlet pressure of compressor at isentropic process in kPa
P1 = Pi+(eff_d*(P1s-Pi)) 				#Inlet pressure of compressor in kPa
P2 = P1*pr_d 				#Outlet pressure of compressor in kPa
T2s = T1*(pr_d)**((k-1)/k) 				#Outlet temperature of compressor at isentropic process in K
T2 = T1+((T2s-T1)/eff_c) 				#Exit temperature of compressor in K
Wc = Cp*(T2-T1)*10**-3 				#Workdone on compressor in kJ/kg of air
Pc = ma*Wc 				#Power input in kW
Pt = Pc 				#Power out put of turbine for isentropic process in kW
f = (T3-T2)/((CV*10**3/Cp)-T3) 				#Fuel air ratio
Wt = Wc 				#Workdone by the turbine in kJ/kg of air
T4 = T3-(Wt*10**3/Cp) 				#Exit temperature of turbine in K
T4s = T3-((T3-T4)/eff_t) 				#Exit temperature of turbine at isentropic process in K
P3 = P2 				#Exit pressure of combustion chamber in kPa, Since the process takes place at consmath.tant pressure process
P4 = P3*(T4s/T3)**(k/(k-1)) 				#Pressure at outlet of turbine in kPa
pr_n = P4/Pe 				#Pressure ratio of nozzle
Tes = T4/(pr_n)**((k-1)/k) 				#Exit temperature of nozzle  at isentropic process in K
Te = T4-(eff_n*(T4-Tes)) 				#Exit temperature of nozzle in K
Cj = math.sqrt(2*Cp*(T4-Te)) 				#Jet velocity in m/s
Ce = Cj 				#Flight velocity in m/s
ae = math.sqrt(k*R*Te) 				#Sound velocity at nozzle in m/s
Me = Ce/ae 				#Nozzle jet mach number
F = ma*(((1+f)*Cj)-u) 				#Thrust in N
P = F*u*10**-3 				#Thrust power in kW

#Output
print 'A)Power input of compressor is %3.2f kW \
\nB)Power output of turbine is %3.2f kW \
\nC)F/A ratio on mass basis is %3.4f \
\nD)Exit mach number is %3.3f \
\nE)Thrust is %3.2f N \
\nF)Thrust power is %3.1f kW'%(Pc,Pt,f,Me,F,P)

A)Power input of compressor is 3536.17 kW
B)Power output of turbine is 3536.17 kW
C)F/A ratio on mass basis is 0.0179
D)Exit mach number is 1.245
E)Thrust is 9668.83 N
F)Thrust power is 1849.7 kW