# Chapter 8 : Two Marks Questions and Answers¶

## Example 8.34 page : 8¶

In :
import math

#Input data
C = 500. 				#Airplane velocity in m/s
T = 20.+273 				#Temperature in K
k = 1.4 				#Adiabatic consmath.tant
R = 287 				#Specific gas consmath.tant in J/kg-K

#Calculation
a = math.sqrt(k*R*T) 				#Sound velocity in m/s
M = C/a 				#Mach number
alp = math.degrees(math.asin((1/M))) 				#Mach angle in degree

#Output
print 'Mach angle is %3.3f degree'%(alp)

Mach angle is 43.332 degree


## Example 8.35 page : 8¶

In :
import math

#Input data
a1 = 2.2 				#Area ratio (A/At)
Po = 10 				#Stagnation Pressure in bar

#Calculation
#Two values of mach number at a1 from gas tables

M1 = 0.275 				#Mach number from gas tables
p1 = 0.949 				#Presure ratio (P/Po)
P1 = Po*p1 				#back pressure in bar

M2 = 2.295 				#Mach number from gas tables
p2 = 0.0806 				#Presure ratio (P/Po)
P2 = Po*p2 				#back pressure in bar

#Output
print 'A)When M = %3.3f, back pressure is %3.2f bar \
\nB)When M = %3.3f, back pressure is %3.3f bar'%(M1,P1,M2,P2)

A)When M = 0.275, back pressure is 9.49 bar
B)When M = 2.295, back pressure is 0.806 bar


## Example 8.37 page : 9¶

In :
import math

#Input data
M = 0.8 				#Mach number
T = 20+273 				#Temperature in K
k = 1.4 				#Adiabatic consmath.tant

#Calculation
To = T*(1+(((k-1)/2)*M**2)) 				#Temperature of air at nose of aircraft in K
To1 = To-273 				#Temperature of air at nose of aircraft in degree Centigrade

#Output
print 'Temperature of air at nose of aircraft is %3.1f degree Centigrade'%(To1)

Temperature of air at nose of aircraft is 57.5 degree Centigrade


## Example 8.38 page 9¶

In :
import math

#Input data
P = 1. 				#Pressure in bar
T = 400. 				#Temperature in K
C = 400. 				#Air velocity in m/s
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Specific gas consmath.tant in J/kg-K
Cp = 1005. 				#Specific heat capacity at constnat pressure in J/kg-K

#Calculation
To = T+(C**2/(2*Cp)) 				#Stagnation Temperature in K
Poi = P+((P*C**2)/(R*T*2)) 				#Stagnation Pressure (if it is incompressible) in bar
Poc = P*(To/T)**(k/(k-1)) 				#Stagnation Pressure (if it is compressible) in bar

#Output
print 'Stagnation Temperature is %3.1f K \
\nC)Stagnation Pressure:    \
\nIf it is incompressible is %3.4f bar    \
\nIf it is compressible is %3.4f bar'%(To,Poi,Poc)

Stagnation Temperature is 479.6 K
C)Stagnation Pressure:
If it is incompressible is 1.6969 bar
If it is compressible is 1.8874 bar


## Example 8.39 page : 9¶

In :
import math

#Input data
v1 = 8 				#Intial volume in litres
P1 = 0.7 				#Intial pressure in MPa
v2 = 7.8 				#Final volume in litres
P2 = 2.7 				#Final pressure in MPa

#Calculation
k = (P2-P1)/(math.log(v1/v2)) 				#Bulk modulus of elasticity of a liquid in MPa

#Output
print 'Bulk modulus of elasticity of a liquid is %3.3f MPa'%k

Bulk modulus of elasticity of a liquid is 78.996 MPa


## Example 8.40 page : 9¶

In :
import math

#Input data
To = 15+273 				#Air Temperature in K
Cp = 1005 				#Specific heat capacity at constnat pressure in J/kg-K

#Calculation
Cmax = math.sqrt(2*Cp*To) 				#Highest possible velocity in m/s

#Output
print 'Highest possible velocity is %3.2f m/s'%Cmax

Highest possible velocity is 760.84 m/s


## Example 8.3.10 page : 12¶

In :
import math

#Input data
M = 0.25 				#mach number
D = 0.04 				#Diamter in m
f = 0.002 				#frictional factor

#Calculation
X = 8.483 				#fanno parameter from gas tables at M
Lmax = (X*D)/(4*f) 				#Lenggth of the pipe in m

#Output
print 'Length of the pipe is %3.3f m'%Lmax

Length of the pipe is 42.415 m


## Example 8.3.15 page : 13¶

In :
import math

#Input data
M = 3. 				#mach number
D = 0.04 				#Diamter in m
f = 0.002 				#frictional factor

#Calculation
X = 0.522 				#fanno parameter from gas tables at M
L = (X*D)/(4*f) 				#Lenggth of the pipe in m

#Output
print 'Lenggth of the pipe is %3.2f m'%L

Lenggth of the pipe is 2.61 m


## Example 8.3.31 page : 16¶

In :
import math

#Input data
M = 0.2 				#Mach number
To = 120.+273 				#Stagnation Temperature in K
Cp = 1005. 				#Specific heat capacity at constnat pressure in J/kg-K

#Calculation
t1 = 0.174 				#Temperature ratio (To/Tot) from Rayleigh gas tables
Tot = To/t1 				#Critical stagnation temperature in K
q = Cp*(Tot-To)*10**-3 				#Maximum amount of heat transfer in kJ/kg

#Output
print 'Maximum amount of heat transfer is %3.2f kJ/kg'%q

Maximum amount of heat transfer is 1874.95 kJ/kg


## Example 8.3.32 page : 17¶

In :
import math

#Input data
p1 = 0.75 				#Pressure ratio (Po2/Po1) Since Stagnation pressure drop is 25%
Cp = 1150. 				#Specific heat capacity at constnat pressure in J/kg-K
k = 1.33 				#Adiabatic consmath.tant

#Calculation
ds = ((k-1)/k)*Cp*math.log(1/p1) 				#Increase in entropy in J/kg-K

#Output
print 'Increase in entropy is %3.2f J/kg-K'%ds

Increase in entropy is 82.09 J/kg-K


## Example 8.3.33 page : 17¶

In :
import math

#Input data
Mi = 2.2 				#Inlet Mach number
T = 100.+273 				#Temperature in K
Cp = 1005. 				#Specific heat capacity at constnat pressure in J/kg-K

#Calculation
t1 = 0.508 				#Temperature ratio (To/Tot) from isentropic gas tables @Mi
To = T/t1 				#Stagnation Temperature in K
t2 = 0.756 				#Temperature ratio (To/Tot) from Rayleigh gas tables @Mi
Tot = To/t2 				#Critical stagnation temperature in K
q = Cp*(Tot-To)*10**-3 				#Maximum amount of heat transfer in kJ/kg

#Output
print 'Maximum amount of heat transfer is %3.4f kJ/kg'%q

Maximum amount of heat transfer is 238.1657 kJ/kg


## Example 8.5.16 page: 22¶

In :
import math

#Input data
Mx = 1.5 				#Mach number
P = 40. 				#Static pressure in kPa

#Calculation
p1 = 3.413 				#Pressure ratio in (Poy/Px) from normal shock gas tables @Mx
Poy = p1*P 				#Pressure acting on front of the body in kPa

#Output
print 'Pressure acting on front of the body is %3.1f kPa'%Poy

Pressure acting on front of the body is 136.5 kPa


## Example 8.5.17 page : 22¶

In :
import math

#Input data
M = 2. 				#Mach number at shock

#Calculation
p1 = 4.5 				#Pressure ratio (Py/Px) from normal shock gas tables @M
e = p1-1 				#Strength of shock wave

#Output
print 'Strength of shock wave is %3.1f'%e

Strength of shock wave is 3.5


## Example 8.5.20 page : 23¶

In :
import math

#Input data
Mx = 7 				#mach number upstream of shock
P = 2 				#pressure @Mx in bar
T = 57+273 				#Temperature @Mx in K
R = 287 				#Specific gas consmath.tant in J/kg-K

#Calculation
p1 = 0.72 				#Pressure ratio (Poy/Pox) from normal shock gas tables @Mx
ds = R*math.log(1/p1) 				#Irreversibility in J/kg-K

#Output
print 'Irreversibility is %3.2f J/kg-K'%ds

Irreversibility is 94.28 J/kg-K


## Example 8.5.21 page : 23¶

In :
import math

#Input data
Px = 45. 				#Static pressure in kPa
T = -20.+273 				#Static temperature in K
Poy = 395. 				#Stagnation pressure in kPa
k = 1.4 				#Adiabatic consmath.tant
R = 287 				#Specific gas consmath.tant in J/kg-K

#Calculation
p1 = Poy/Px 				#Pressure ratio
Mx = 2.536 				#Mach number from normal shock gas tables @p1
Cx = Mx*math.sqrt(k*R*T) 				#Air velocity in m/s

#Output
print 'Mach number is %3.3f \
\nAir velocity is %.f m/s'%(Mx,Cx)

Mach number is 2.536
Air velocity is 809 m/s


## Example 8.5.22 page : 23¶

In :
import math

#Input data
Cx = 750. 				#velocity upstream of shock in m/s
Px = 1. 				#Pressure upstream of shock in bar
Tx = 10.+273 				#Temperature upstream of shock in K
k = 1.4 				#Adiabatic consmath.tant
R = 287. 				#Specific gas consmath.tant in J/kg-K

#Calculation
Mx = Cx/math.sqrt(k*R*Tx) 				#Mach number upstream of shock
My = 0.545 				#Mach number downstream of shock from normal shock gas tables, Mistake in textbook
t1 = 1.875 				#Temperature ratio (Ty/Tx)
Ty = Tx*t1 				#Static temperature downstream of shock in K
p1 = 5.583 				#Pressure ratio (Py/Px)
Py = Px*p1 				#Static pressure downstream of shock in bar
Cy = My*math.sqrt(k*R*Ty) 				#velocity downstream of shock in m/s

#Output
print 'Downstream of shock:    Velocity is %3.3f m/s    Pressure is %3.3f bar    Temperature is %3.3f K'%(Cy,Py,Ty)

Downstream of shock:    Velocity is 251.649 m/s    Pressure is 5.583 bar    Temperature is 530.625 K


## Example 8.6.41 page : 31¶

In :
import math

#Calculation

#Differentiating P = m*(Cj-u)*u and equating it to zero we get jet speed ratio as 0.5
sig = 0.5 				#Jet speed ratio
eff_max = ((2*sig)/(1+sig)) 				#Propulsive efficiency for optimum thrust power, wrong notation in textbook.

#Output
print 'Propulsive efficiency for optimum thrust power is %3.3f'%(eff_max)

Propulsive efficiency for optimum thrust power is 0.667


## Example 8.6.42 page : 31¶

In :
import math

#Input data
u = 1200*(5./18) 				#Flight velocity in m/s
Cj = 800. 				#Effective jet velocity in m/s

#Calculation
sig = u/Cj 				#jet speed ratio
eff = ((2*sig)/(1+sig))*100 				#Propulsive efficiency in %

#Output
print 'Propulsive efficiency is %3.1f percent'%eff

Propulsive efficiency is 58.8 percent


## Example 8.7.42 page : 39¶

In :
import math

#Input data
m = 5. 				#Propellent rate in kg/s
Pamb = 1.013 				#Ambient pressure in bar
Pe = 1.02 				#Nozzle exit pressure in bar
D = 0.1 				#Nozzle exit diameter in m
Ce = 1400. 				#Exit jet velocity in m/s

#Calculation
Ae = math.pi*D**2/4 				#Exit area in m**2
F = (m*Ce)+((Pe-Pamb)*Ae) 				#Thrust in N

#Output
print 'Thrust is %3i N'%F

Thrust is 7000 N


## Example 8.7.43 page : 39¶

In :
import math

#Input data
Is = 230. 				#Specific Impulse in sec
m = 1. 				#Propellent flow in kg/s
g = 9.81 				#Acceleration due to gravity in m/s**2

#Calculation
F = m*Is*g 				#Thrust in N

#Output
print 'Thrust is %3.1f N'%F

Thrust is 2256.3 N


## Example 8.7.45 page : 39¶

In :
import math

#Input data
u = 1500. 				#Flight velocity in m/s
eff = 0.75 				#Propulsive efficiency

#Calculation
#Converting relation eff = (2*sig)/(1+sig**2) into 2nd degree polynomial of sig
sig = ((2-(math.sqrt(4-(4*eff*eff))))/(2*eff)) 				#Jet speed ratio
Cj = u/sig 				#Jet velocity in m/s

#Output
print 'Jet velocity is %3.2f m/s'%Cj

Jet velocity is 3322.88 m/s


## Example 8.7.46 page : 40¶

In :
import math

#Input data
Cj = 2700. 				#Jet velocity in m/s
u = 1350. 				#Flight velocity in m/s
m = 78.6 				#Propellent flow in kg/s

#Calculation
F = m*Cj*10**-3 				#Thrust in kN
P = F*u*10**-3 				#Thrust power in MW
sig = u/Cj 				#Jet speed ratio
eff = ((2*sig)/(1+sig**2))*100 				#Propulsive efficiency in %

#Output
print 'Thrust is %3.1f kN \
\nThrust power is %3.2f MW \
\nPropulsive efficiency is %3i percent'%(F,P,eff)

Thrust is 212.2 kN
Thrust power is 286.50 MW
Propulsive efficiency is  80 percent


## Example 47 page : 40¶

In :
import math

#Input data
D = 12683.*1000 				#Diameter of Earth in m
g = 9.81 				#Acceleration due to gravity in m/s
h = 500.*1000 				#Altitude in m

#Calculation
Uorb = (D/2)*math.sqrt(g/((D/2)+h)) 				#Orbital velocity in m/s
Uesc = math.sqrt(2)*Uorb 				#Escape velocity in m/s

#Output
print 'Orbital velocity is %3.2f m/s \
\nEscape velocity is %3.2f m/s'%(Uorb,Uesc)

Orbital velocity is 7593.65 m/s
Escape velocity is 10739.05 m/s


## Example 48 page : 40¶

In :
import math

#Input data
u = 10080*(5./18) 				#Flight velocity in m/s
Cj = 1400. 				#Jet velocity in m/s
m = 5. 				#Propellent flow in kg/s

#Calculation
F = m*Cj*10**-3 				#Thrust in kN
P = F*u*10**-3 				#Thrust power in MW
sig = u/Cj 				#Jet speed ratio
eff = ((2*sig)/(1+sig**2)) 				#Propulsive efficiency

#Output
print 'Propulsive power is %3.1f MW \
\nPropulsive efficiency is %3.1f'%(P,eff)

Propulsive power is 19.6 MW
Propulsive efficiency is 0.8