# Chapter 17: COMPRESSIBLE FLOW¶

## Ex17.1:PG-710¶

In [5]:
#ques1
#to determine isentropic stagnation pressure and temperature

T=300;#Temperature of air in K
P=150;#Pressure of air in kPa
v=200;#velocity of air flow n m/s
Cp=1.004;#specific heat at constant pressure in kJ/kg
To=v**2/(2000*Cp)+T;#stagnation temperature in K
k=1.4;#constant
Po=P*(To/T)**(k/(k-1));#stagnation pressure in kPa
print 'Stagnation Temperature is ',round(To,1),' K \n'
print 'Stagnation Pressure is ',round(Po,2),'kPa \n'
Stagnation Temperature is  319.9  K

Stagnation Pressure is  187.85 kPa

## Ex17.2:PG-713¶

In [6]:
#ques2
#to determine the force

#initializing variables
mdot=-1 # mass flow rate out of control volume in kg/s
Vx=-1 # x component of velocity of control volume in m/s
Vy=10 # y component of velocity of control volume in m/s

Fx=mdot*Vx # Force in X direction

Fy=mdot*Vy # Force in Y direction

print "the force the man exert on the wheelbarrow",round(Fx),"N"
print "the force the floor exerts on the wheelbarrow",round(Fy),"N"
the force the man exert on the wheelbarrow 1.0 N
the force the floor exerts on the wheelbarrow -10.0 N

## Ex17.3:PG-715¶

In [11]:
#ques3
#determining the thrust acting on a control surface

#i-inlet
#e-exit
#using momentum equation on control surface in x direction
me=20.4;#mass exiting in kg
mi=20;#mass entering in kg
ve=450;#exit velocity in m/s
vi=100;#inlet velocity in m/s
Pi=95;#Pressure at inlet in kPa
Pe=125;#Pressure at exit in kPa
Po=100;#surrounding pressure in kPa
Ai=0.2;#inlet area in m^2
Ae=0.1;#exit area in m^2
Rx=(me*ve-mi*vi)/1000-(Pi-Po)*Ai+(Pe-Po)*Ae;#thrust in x direction in kN

print "Thrust acting in x direction is ",round(Rx,2),"kN"
Thrust acting in x direction is  10.68 kN

## Ex17.4:PG-717¶

In [19]:
#ques4
#to determine increase in enthalpy
#initializing variables
#i-inlet
#e-exit

T=25+273 # temperature in kelvin
v=0.001003 # specific volume of water in kg/m^3 at 25 *c from table B.1.1
ve=7;#exit velocity in m/s
vi=30;#inlet velocity in m/s
Pi=350;#Pressure at inlet in kPa
Pe=600;#Pressure at exit in kPa

#using momentum equation on control surface
Pes= (vi**2-ve**2)/(2*v*1000)+Pi # exit pressure for reversible diffuser
delH=(vi**2-ve**2)/(2*1000.0) # change in enthalpy from first law in kJ/kg
delU=delH-v*(Pe-Pi) # change in internal energy in kJ/kg
delS=delU/T # change in entropy in kJ/kg.K
print"the exit pressure for reversible diffuser is ",round(Pes),"kPa"
print"the increase in enthalpy is ",round(delH,5),"kJ/kg"
print"the increase in internal energy is ",round(delU,5),"kJ/kg"
print"the increase in entropy is ",round(delS,6),"kJ/kg.K"
the exit pressure for reversible diffuser is  774.0 kPa
the increase in enthalpy is  0.4255 kJ/kg
the increase in internal energy is  0.17475 kJ/kg
the increase in entropy is  0.000586 kJ/kg.K

## Ex17.5:PG-720¶

In [24]:
#ques5
#determining velocity of sound in air
import math
k=1.4;#constant
R=0.287;#gas constant
#at 300K
T1=300;# temperature in kelvin
c1=math.sqrt(k*R*T1*1000)
print "Speed of sound at 300 K is",round(c1,1)," m/s"
#at 1000K
T2=1000;# temperature in kelvin
c2=math.sqrt(k*R*T2*1000)
print "Speed of sound at 1000 K is",round(c2,1)," m/s"
Speed of sound at 300 K is 347.2  m/s
Speed of sound at 1000 K is 633.9  m/s

## Ex17.6:PG-727¶

In [30]:
#ques6
#determining mass flow rate through control volume
import math
k=1.4;#constant
R=0.287;#gas constant
To=360;#stagnation Temperature in K
T=To*0.8333;#Temperature of air in K, 0.8333 stagnation ratio from table
v=math.sqrt(k*R*T*1000);#velocity in m/s
P=528;#stagnation pressure in kPa
d=P/(R*T);#stagnation density in kg/m^3
A=500*10**-6;#area in m^2
ms=d*A*v;#mass flow rate in kg/s
print" Mass flow rate at the throat section is",round(ms,4),"kg/s"
#e-exit state
Te=To*0.9381;#exit temperature in K, ratio from table
ce=math.sqrt(k*R*Te*1000);#exit velocity of sound in m/s
Me=0.573;#Mach number
ve=Me*ce;
Pe=800;#exit pressure in kPa
de=Pe/R/Te;
mse=de*A*ve;
print" Mass flow rate at the exit section is",round(mse,4)," kg/s"
Mass flow rate at the throat section is 1.0646 kg/s
Mass flow rate at the exit section is 0.8711  kg/s

## Ex17.7:PG-728¶

In [39]:
#ques7
#determining exit properties in a control volume
import math
Po=1000;#stagnation pressure in kPa
To=360;#stagnation temperature in K

#when diverging section acting as nozzle

Pe1=0.0939*Po;#exit pressure of air in kPa
Te1=0.5089*To;#exit temperature in K
k=1.4;#constant
R=0.287;#gas constant for air
ce=math.sqrt(k*R*Te1*1000);#velocity of sound in exit section in m/s
Me=2.197;#mach number from table
ve1=Me*ce;#velocity of air at exit section in m/s
print "When diverging section act as a nozzle :-"
print "Exit pressure is",round(Pe1,4)," kPa"
print "Exit Temperature",round(Te1,1)," K"
print "Exit velocity is",round(ve1,1)," m/s "

#when diverging section act as diffuser

Me=0.308;
Pe2=0.0936*Po;#exit pressure of air in kPa
Te2=0.9812*To;#exit temperature in K
ce=math.sqrt(k*R*Te2*1000);#velocity of sound in exit section in m/s
ve2=Me*ce;
print "When diverging section act as a diffuser :-"
print "Exit pressure is",round(Pe2,1)," kPa"
print "Exit Temperature",round(Te2,2)," K"
print "Exit velocity is",round(ve2,)," m/s "

# The value of Exit pressure when diverging section acts as diffuser is wrong
When diverging section act as a nozzle :-
Exit pressure is 93.9  kPa
Exit Temperature 183.2  K
Exit velocity is 596.1  m/s
When diverging section act as a diffuser :-
Exit pressure is 93.6  kPa
Exit Temperature 353.23  K
Exit velocity is 116.0  m/s

## Ex17.8:PG-731¶

In [2]:
#ques8
#Determine the static pressure and temperature of supersonic nozzle
#x-inlet
#y-exit
# using  the data from previous example
Mx=2.197 # Mach number at x
Px=93.9 # IN kPa
Tx=183.2 # IN K
P0x=1000 # IN kPa
T0x = 360 # in K
My=0.547 # Mach number at y From Table A.13
Py = 5.46*Px # From Table A.13
Ty = 1.854 * Tx # From Table A.13
P0y= 0.630 * P0x # From Table A.13
print "Exit pressure after shock is",round(Py,1)," kPa"
print "Exit Temperature after shock is",round(Ty,1)," K"
print "Exit stagnation pressure after shock is",round(P0y,1)," kPa"
Exit pressure after shock is 512.7  kPa
Exit Temperature after shock is 339.7  K
Exit stagnation pressure after shock is 630.0  kPa

## Ex17.9:PG-733¶

In [50]:
#ques9
#determining exit plane properties in control volume

#x-inlet
#y-exit
Mx=1.5;#mach number for inlet
My=0.7011;#mach number for exit
Px=272.4;#inlet pressure in kPa
Tx=248.3;#inlet temperature in K
Tox=360 # stagnation temperature in Kelvin
Pox=1000.0;#stagnation pressure for inlet
Py=2.4583*Px;# Pressure at 1.5 mach in kPa
Ty=1.320*Tx;# temperature at 1.5 mach in K
Poy=0.9298*Pox;# pressure at 1.5 mach in kPa

Toy=Tox # constant
Me=0.339 # from table with A/A*=1.860 and M < 1
Pe=0.9222*Py;#Exit Pressure in kPa
Te=0.9771*Toy;#Exit temperature in K
Poe=0.9222*Poy;#Exit pressure in kPa

print "Exit Mach no.=",Me
print "Exit temperature =",round(Te,2),"K "
print "Exit pressure =",round(Poe,1),"kPa"
Exit Mach no.= 0.339
Exit temperature = 351.76 K
Exit pressure = 857.5 kPa