Chapter 17: Compressible Flow in Pipes

Example 17.1, Page 566

In [1]:
from __future__ import division
import math
import sympy
from sympy import solve,symbols

 #Initializing  the  variables
g  =  9.81;
rho  =  1000;
rhoHg  =  13.6*rho;
d1  =  0.075;
d2  =  0.025;
Pi  =  0.250;
Pt  =  0.150;
P_Hg  =  0.760;
rho1  =  1.6;
gma  =  1.4;

 #Calculations
P1  =  (Pi+P_Hg)*rhoHg*g;
P2  =  (Pt+P_Hg)*rhoHg*g;
rho2  =  rho1*(P2/P1)**(1/gma);
V0=symbols('V0')
V1=symbols('V1')
Velo  =  solve([d2**2*V1*rho2-d1**2*V0*rho1,0.5*(V1**2  -  V0**2)*((gma-1)/gma)*(rho2*rho1/(rho2*P1-rho1*P2))-1],[V0,V1])
s=(Velo[1])[0]
Flow  =  math.pi*d1**2/4*s;

print "Volume of flow (m3/s):",round(Flow,3)

Volume of flow (m3/s): 0.06

Example 17.2, Page 571

In [3]:
from __future__ import division
import math


 #Initializing  the  variables
Ma  =  4;
gma  =  1.4;
At  =  500;                                                              #  in  mm

 #Calculations
N  =  1  +  (gma-1)*Ma**2/2;
D  =  (gma+1)/2  ;
#ratio of A/At ==R
R  = round( (N/D)**((gma+1)/(2*(gma-1)))/Ma,2);
A=At*R
print "Area of test section (mm^2):",A

Area of test section (mm^2): 5360.0

Example 17.3, Page 575

In [4]:
from __future__ import division
import math


 #Initializing  the  variables
Ma1  =  2;
gma  =  1.4;
T1  =  15+273;                                                        #  In  kelvin
P1  =  105;  

 #Calculations
Ma2  =  (((gma-1)*Ma1**2  +2)/(2*gma*Ma1**2-gma+1))**0.5;
P2  =  P1*(1+gma*Ma1**2)/(1+gma*Ma2**2);
T2  =  T1*(1  +(gma-1)/2*Ma1**2)/(1  +(gma-1)/2*Ma2**2);


print "Mach No downstream shock wave                   :",round(Ma2,3)
print "Pressure (bar) of downstream shock wave         :",round(P2)
print "Temperature (Degree C) of downstream shock wave :",T2 - 273

Mach No downstream shock wave                   : 0.577
Pressure (bar) of downstream shock wave         : 473.0
Temperature (Degree C) of downstream shock wave : 213.0

Example 17.4, Page 581

In [5]:
from __future__ import division
import math


 #Initializing  the  variables
gma  =  1.4;
f  =  0.00375;
d  =  0.05;

 #Calculations
m  =  d/4;
def  x(Ma):
    A  =(1  -Ma**2  )/(gma*Ma**2);
    B  =  (gma+1)*Ma**2/(2+(gma-1)*Ma**2);  
    y  =  m/f*(A+  (gma+1)*math.log(B)/(2*gma));
    return y

X1  =  x(0.2);                                  #  At  entrance  Ma  =  0.2;
X06_X1  =x(0.6);                                #  Section(b)  Ma  =  0.6;

X06  =    X1-X06_X1;

print "The Distance X1 at which the Mach number is unity (m) :",round(X1,2)
print "Distance from the entrance (m)                        :",round(X06,2)

The Distance X1 at which the Mach number is unity (m) : 48.44
Distance from the entrance (m)                        : 46.81

Example 17.5, Page 585

In [6]:
from __future__ import division
import math
from scipy.optimize import fsolve

 #Initializing  the  variables
gma  =  1.4;
Q  =  28/60;                             #  m3/s
d  =  0.1;
p1  =  200*10**3;
f  =  0.004;
x_x1  =  60;
R  =  287;
T  =  15+273;

 #Calculations
A  =  math.pi*d**2/4;
m  =  d/4;
v1  =  Q/A;
pa  =  p1*(1-f*(x_x1)*v1**2/(m*R*T))**0.5;

def g(p):
    A  =  (v1*p1)**2/(R*T)
    B  =  f*A*x_x1/(2*m);
    y  =  (p**2  -  p1**2)/2  -A*math.log(p/p1)  +B;
    return y
    
pb=fsolve(g,pa)                        #  Guessing  solution  around  pa
pb=pb[0]
print "Pressure at the outlet, neglecting velocity changes   (kN/m2) :",round(pa/1000,1)
print "Pressure at the outlet, allowing for velocity changes (kN/m2) :",round(pb/1000,1)

Pressure at the outlet, neglecting velocity changes   (kN/m2) : 153.6
Pressure at the outlet, allowing for velocity changes (kN/m2) : 150.4