# Chapter 7 : Flow of a Real Fluid¶

## Example 7.1 Page No : 225¶

In [1]:
import math

# variables
nu = 0.00001;		# sqft/sec
d = 1.;		#in
R_c = 2100.;

# calculations
V = R_c*nu/(d/12);
Q = V*0.25*math.pi*(d/12)**2;

# results
print 'Q = %.6f cfs'%(Q);

Q = 0.001374 cfs


## Example 7.2 pageno : 230¶

In [1]:
import math

# calculations
e = 0.2/(1./0.33)
l2 = 0.2/(1.94 * (1/0.33)**2)
l = math.sqrt(l2)
k2 = 0.2/(1.94 * (1/0.33)**2 / (-1/(0.33**2)**0.5)**2)
k = math.sqrt(k2)

# results
print "E = %.3f lb-sec/sqft"%e
print "l = %.3f ft"%l
print "k = %.2f"%k

E = 0.066 lb-sec/sqft
l = 0.106 ft
k = 0.32


## Example 7.4 Page No : 240¶

In [3]:
import math

# variables
G = 240.;		#lb/sec
A1 = 4.;		#sqft
A2 = 2.;		#sqft
z1 = 30.;		#ft
z2 = 80.;		#ft
V1 = 600.;		# fps
V2 = 800.;		#fps
p1 = 20.;		#psia
p2 = 35.;		# psia

# calculations
gam1 = G/(A1*V1);
gam2 = G/(A2*V2);
T1 = p1*144/(53.3*gam1);
T2 = p2*144/(53.3*gam2);
del_H = 186.5*(T2-T1);
E_H1 = (V2**2)/(2*32.2) - (V1**2)/(2*32.2) +del_H+z2-z1;
E_H2 = (V2**2)/(2*32.2) - (V1**2)/(2*32.2) +del_H;
Q = G*E_H2/550.;

# results
print 'T1 = %d degreeR, T2 = %d degreeR'%(T1,T2);
print ' The net heat energy added = %d hp'%(round(Q,-1));

#answer differs due to rounding-off errors

T1 = 540 degreeR, T2 = 630 degreeR
The net heat energy added = 9230 hp


## Example 7.5 Page No : 240¶

In [4]:
import math

# variables
G = 50.;		#cfs
Q = 400.;		#hp
A1 = 4.;		#sqft
A2 = 2.;		#sqft
z1 = 30.;		#ft
z2 = 80.;		#ft
p1 = 20.;		#psi
p2 = 10.;		#psi

# calculations
V1 = G/A1;
V2 = G/A2;
E_p = Q*(550/62.4)/G;
h_L = (p1-p2)*144/62.4 + (V1**2 - V2**2)/(2*32.2) +(z1-z2)+E_p;

# results
print 'Head lost = %.1f ft'%(h_L);

Head lost = 36.3 ft


## Example 7.6 Page No : 243¶

In [9]:
import math

# variables
b = 3.;		#ft
d = 2.;		#ft
l = 200.;		#ft
h_L = 30.;		#ft

# calculations
tau_0 = h_L*62.4*b*d/(10*l);		#0.00694

# results
print 'The resistance stress exerted between fluid and conduit walls = %.2f psf = %.3f psi'%(tau_0,tau_0*0.00694);

The resistance stress exerted between fluid and conduit walls = 5.62 psf = 0.039 psi


## Example 7.7 Page No : 244¶

In [10]:
import math

# variables
h_L = 30.;		#ft
l = 200.;		#ft
d = 2.;		#ft
r = 8.;		#in

# calculations
#part (a)
tau_0 = h_L*62.4/(d*l);
#part(b)
tau = (0.5*r/12)*(tau_0*0.00694);

# results
print 'Parta): Shear stress = %.2f psf = %.4f psi '%(tau_0,tau_0*0.00694);
print 'Partb): Shear stress = %.4f psi '%(tau);

Parta): Shear stress = 4.68 psf = 0.0325 psi
Partb): Shear stress = 0.0108 psi