Chapter 26 : Jet Propulsion

Example 26.1 Page no : 447

In [1]:
import math 
					
#Input data
Ve = 2700.					#Jet exit velocity in m/s
Vf = 1350					#Forward flight velocity in m/s
m = 78.6					#Propellant consumption in kg/s

					
#Calculations
T = ((m/9.81)*(Ve-Vf))					#Thrust in kg
TH = ((T*Vf)/75)/10**5					#Thrust horse power in HP*10**5
pn = (2/(1+(Ve/Vf)))*100					#Propulsive efficiency in percent

					
#Output
print 'i) Thrust is %3.0f kg  \
\nii) Thrust horse power is %3.3f*10**5 H.P  \
\niii) Propulsive efficiency is %3.1f percent'%(T,TH,pn)

i) Thrust is 10817 kg  
ii) Thrust horse power is 1.947*10**5 H.P  
iii) Propulsive efficiency is 66.7 percent

Example 26.2 Page no : 451

In [2]:
import math 
					
#Input data
CV = 10000					#Calorific value in kcal/kg
F = 1.4					#Fuel consumption in kg per hour per kg of thrust
T = 900					#Thrust in kg
Va = 425					#Aircraft velocity in m/s
w = 19.5					#Weight of air in kg/sec

					
#Calculations
af = (w/((F*T)/3600))					#Air fuel ratio
nv = ((T*Va*3600)/(427*F*T*CV))*100					#Overall efficiency in percent

					
#Output
print 'Air fuel ratio is %3.1f  \
\nOverall efficiency is %3.1f percent'%(af,nv)

Air fuel ratio is 55.7  
Overall efficiency is 25.6 percent

Example 26.3 Page no : 451

In [4]:
import math 
					
#Input data
a = 11500.					#Altitude in m
n = 123.					#Number of passengers
c = 3.					#Cargo in tonnes
Va = 650.					#Velocity of air craft in km/hour
d = 640.					#Drag in kg
pe = 50.					#Propulsion efficiency in percent
oe = 18.					#Overall efficiency in percent
CV = 10000.					#Calorific value in kcal/kg
da = 0.0172					#Density of air at 11500 m in kg/cm**2

					
#Calculations
Vp = ((Va*1000)/3600)					#Velocity of aeroplane in m/s
Vr = ((2/(pe/100))-1)*Vp					#Velocity of working medium in m/s
nhp = ((d*Vp)/(75*(pe/100)))					#Net horse power in H.P
wf = ((nhp*75*3600)/((oe/100)*427*CV))					#Mass flow rate in kg/hr
thp = ((Va*Vp)/75)					#Thrust horse power in H.P
F = (wf/thp)					#Fuel consumption per thrust H.P hour in kg
W = ((Va*9.81)/Vr)					#Air flow in kg/sec
va = (W/da)					#Volume of air in cu.m/sec
aa = (va/(3*Vr))					#Area of jet in m**2
d = math.sqrt((4*aa)/3.14)*100					#Diameter of jet in cm
af = ((W*3600)/wf)					#Air fuel ratio

					
#Output
print 'a) Absolute velocity of the jet is %3.1f m/sec  \
\nb) Net horse power of the gas plant is %3.0f H.P  \
\nc) Fuel consumption per thrust H.P hour is %3.3f kg  \
\nd) Diameter of the jet is %3.1f cm  \
\ne) Air-fuel ratio of the engine is %3.1f'%(Vr,nhp,F,d,af)

a) Absolute velocity of the jet is 541.7 m/sec  
b) Net horse power of the gas plant is 3081 H.P  
c) Fuel consumption per thrust H.P hour is 0.692 kg  
d) Diameter of the jet is 73.2 cm  
e) Air-fuel ratio of the engine is 39.1

Example 26.4 Page no : 451

In [5]:
import math 
					
#Input data
p1 = 7.					#Pressure of gas before expansion in kg/cm**2
p2 = 5.					#Pressure of gas after expansion in kg/cm**2
T1 = 250.+273					#Temperature of gas before expansion in K
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
Cv = 0.17					#Specific heat at constant volume in kJ/kg.K
nv = 0.8					#Nozzle efficiency

					
#Calculations
R = 427*(Cp-Cv)					#Characteristic gas constant in kg.m/kg.K
g = (Cp/Cv)					#Ratio of specific heats
V1 = (R*T1)/(p1*10**4)					#Volume in cu.m per kg
V2 = (V1*(p1/p2)**(1/g))					#Volume in cu.m per kg
Wd = (g/(g-1))*(p1*V1-p2*V2)*10**4					#Work done in m.kg per kg
KE = (nv*Wd)					#Kinetic energy at exit in m.kg per kg
v3 = math.sqrt(2*9.81*nv*Wd)					#Velocity in m/s
T2 = (T1*(p2/p1)*(V2/V1))					#Temperature in K
T3 = (((1-nv)*Wd)/(427*Cp))+T2					#Temperature in K
V3 = (V2*(T3/T2))					#Volume in cu.m per kg
qa = (V3/v3)*10**4					#Discharge area unit rate of mass flow in cm**2

					
#Output
print 'Area of discharge per unit rate of mass flow is %3.2f sq.cm'%(qa)

Area of discharge per unit rate of mass flow is 10.32 sq.cm

Example 26.5 Page no : 451

In [6]:
import math 
					
#Input data
p = 3.5					#Pressure at the delivery is 3.5 times that at entrance
T = 1.15					#Temperature rise during compression is 1.15 times that for frictionless adiabatic compression. In textbook it is given wrong as 1.5
T3 = 500.+273					#Temperature of products of combustion in K
pa = 1.					#Atmospheric pressure in kg/cm**2
Ta = 15.+273					#Atmospheric temperature in K
Cp = 0.24					#Specific heat at constant pressure in kJ/kg.K
g = 1.4					#Ratio of specific heats
J = 427.					#Mechanical equivalent of heat in kg.m/kcal

					
#Calculations
p2 = p*pa					#Pressure in kg/cm**2
T2a = (Ta*(p2/pa)**((g-1)/g))					#Temperature in K
T2 = (T2a-Ta)*T+Ta					#Temperature in K
wcomp = (Cp*(T2-Ta))					#Work done by compressor in kcal/kg
T5 = T3/(p2/pa)**((g-1)/g)					#Temperature in K
dh35 = (Cp*(T3-T5))					#Change in enthalpy in kcal/kg
dhnozzle = (dh35-wcomp)					#Change in enthalpy of nozzle in kcal/kg
v5 = math.sqrt(2*9.81*J*dhnozzle)					#Velocity at the nozzle exit in m/sec
Th = (v5/9.81)					#Thrust in kg per kg of air/sec

					
#Output
print 'a) the power required to drive the compressor per kg of air per second is %3.1f kcal/kg  \
\nb) Static thrust developed per kg of air per second is %3.1f kg'%(wcomp,Th)

a) the power required to drive the compressor per kg of air per second is 34.2 kcal/kg  
b) Static thrust developed per kg of air per second is 43.4 kg
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