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)
```

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)
```

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)
```

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)
```

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)
```

In [ ]:

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