Chapter 17:Compressible Fluid Flow

Ex17.2:pg-717

In [2]:
import math
P1 = 0.18 # Diffuser static pressure in MPa
R = 0.287 # Gas constant
T1 = 37 # Static temperature 
P0 = 0.1# Atmospheric pressure in MPa
A1 = 0.11 # intake area in m**2
V1 = 267 # Inlet velocity in m/s
w = (P1*1e3/(R*(T1+273)))*A1*V1 # mass flow rate
g = 1.4 # Heat capacity ratio
c1 = math.sqrt(g*R*(T1+273)*1000) # velocity
M1 = V1/c1 # Mach number
A1A_ = 1.0570 # A1/A* A* = A_
P1P01 = 0.68207 # pressure ratio
T1T01 = 0.89644# Temperature ratio
F1F_ = 1.0284# Impulse function ratio
A2A1 = 0.44/0.11  # Area ratio
A2A_ = A2A1*A1A_# Area ratio
M2 = 0.135 # Mach number
P2P02 = 0.987 # Pressure ratio
T2T02 = 0.996 # Temperature ratio
F2F_ = 3.46# Impulse function ratio
P2P1 = P2P02/P1P01 # Pressure ratio
T2T1 = T2T02/T1T01# Temperature ratio
F2F1 = F2F_/F1F_ # Impulse function ratio
P2 = P2P1*P1 # Outlet  pressure
T2 = T2T1*(T1+273) # Outlet temperature
A2 = A2A1*A1 # Exit area
F1 = P1*A1*(1+g*M1**2) # Impulse function
F2 = F2F1*F1 # Impulse function
Tint = F2-F1 # Internal thrust
Text = P0*(A2-A1) # External thrust
NT = Tint - Text  # Net thrust

print "\n Example 17.2 \n"
print "\n Mass flow rate of air through diffuser is ",w ," Kg/s"
print "\n  Mach number of leaving air is ",M2
print "\n  Temperature of leaving air is ",T2-273 ," degree celcius"
print "\n  Pressure of leaving air is ",P2 ," MPa "
print "\n Net thrust is ",NT*1e3 ," kN"

#The answers vary due to round off error
 Example 17.2 


 Mass flow rate of air through diffuser is  59.4200292233  Kg/s

  Mach number of leaving air is  0.135

  Temperature of leaving air is  71.4290750078  degree celcius

  Pressure of leaving air is  0.260471799082  MPa 

 Net thrust is  51.3284455434  kN

Ex17.3:pg-718

In [6]:
import math
M2 = 2.197 # Mach number
P2P0 = 0.0939 # pressure ratio
T2T0 = 0.5089 # Temperature ratio
P0 = 1 # Stagnation pressure in MPa 
T0 = 360 # Stagnation temperature in K
g = 1.4 # Heat capacity ratio
R = 0.287 # Gas constant
P2 = P2P0*P0*1e3 # Static Pressure
T2 = T2T0*T0 # Static temperature
c2 = math.sqrt(g*R*T2*1000)
V2 = c2*M2 #velocity at the exit from the nozzle
# for air
P_P0 = 0.528 # pressure ratio
T_T0 = 0.833 # Temperature ratio
P_ = P_P0*P0*1e3 # Static Pressure
T_ = T_T0*T0 #Static temperature
rho_ = P_/(R*T_) # density
V_ = math.sqrt(g*R*T_*1000) # Velocity at the exit from the nozzle 
At = 500e-06 # throat area
w = At*V_*rho_# Maximum flow rate of air

print "\n Example 17.3\n"
print "\n When divergent section act as a nozzle"
print "\n Maximum flow rate of air is ",w ," kg/s"
print "\n Static temperature is ",T2 ," K"
print "\n Static Pressure is ",P2 ," kPa"
print "\n Velocity at the exit from the nozzle is ",V2 ," m/s"
#The answers vary due to round off error

# Part (b)
Mb = 0.308  # Mach number
P2P0b = 0.936 # Pressure ratio
T2T0b = 0.9812 # Temperature ratio
P2b = P2P0b*P0*1e3#Static Pressure 
T2b = T2T0b*T0 # Static temperature
c2b = math.sqrt(g*R*T2b*1000) # Velocity 
V2b = c2b*Mb #Velocity at the exit from the nozzle
print "\n\n When divergent section act as a diffuser"
print "\n Maximum flow rate of air is ",w ," kg/s"
print "\n Static temperature is ",T2b ," K"
print "\n Static Pressure is ",P2b ," kPa"
print "\n Velocity at the exit from the nozzle is ",V2b ," m/s"
 Example 17.3


 When divergent section act as a nozzle

 Maximum flow rate of air is  1.06476372092  kg/s

 Static temperature is  183.204  K

 Static Pressure is  93.9  kPa

 Velocity at the exit from the nozzle is  596.077184351  m/s


 When divergent section act as a diffuser

 Maximum flow rate of air is  1.06476372092  kg/s

 Static temperature is  353.232  K

 Static Pressure is  936.0  kPa

 Velocity at the exit from the nozzle is  116.03411731  m/s

Ex17.4:pg-720

In [7]:
import math
Px = 16.0 # pressure in kPa
Poy = 70.0 #pressure in kPa 
Mx = 1.735 # Mach number
Pyx = 3.34 # Pressure ratio
rho_yx = 2.25 # Density ratio
Tyx = 1.483  # Temperature ratio
Poyox = 0.84 # pressure ratio
My = 0.631 # Mach number
g = 1.4 # Ratio of heat capacities
Tox = 573.0 # stagnation temperature in K 
Toy = Tox # temperature equivalence
Tx = Tox/(1+((g-1)/2.0)*Mx**2) # temperature at x
Ty = Tyx*Tx # temperature at y
Pox = Poy/Poyox  # total pressure 
# From table
Mx = 1.735

print "\n Example 17.4\n"
print "\n Mach number of the tunnel is ",Mx
 Example 17.4


 Mach number of the tunnel is  1.735

Ex17.5:pg-721

In [9]:
import math
Ax = 18.75 # cross sectional area in divergent part in m**2
A_ = 12.50 # throat area in m**2
AA_ = 1.5 # Area ratio
Pxox = 0.159 # pressure ratio from table
R = 0.287 # Gas constant
Pox = 0.21e03 # pressure in kPa
Px = Pxox*Pox # pressure calculation
# from the gas table on normal shock
Mx = 1.86 
My = 0.604 
Pyx = 3.87 
Poyx = 4.95 
Poyox = 0.786
Py = Pyx*Px
Poy = Poyx*Px
My = 0.604
Ay_ = 1.183
A2 = 25 
Ay = 18.75
A2_ = (A2/Ay)*Ay_
# From isentropic table 
M2 = 0.402
P2oy = 0.895
P2 = P2oy*Poy
syx = -R*math.log(Poy/Pox) # sy-sx

print "\n Example 17.5\n"
print "\n Exit Mach number is ",M2
print "\n Exit pressure is ",P2 ," kPa"
print "\n Exit Stagnation pressure is ",Pox-Poy ," kPa"
print "\n Entropy increase is ",syx ," kJ/kg K"
#The answers vary due to round off error
 Example 17.5


 Exit Mach number is  0.402

 Exit pressure is  147.9260475  kPa

 Exit Stagnation pressure is  44.7195  kPa

 Entropy increase is  0.068726024552  kJ/kg K