Chapter 7 : Incompressible Flow Through Conduits

Example 7.1 Page No : 240

In [7]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
h1 = 4. 			#m
muw = 0.001 		#Ns/m**2
l = 1.5 			#m
B = 0.15/1000 		#m
lenth = 11.2 			#m
			
#calculations
P1 = g*rho*h1
V = P1*B**2 /(12*muw*l)
A = B*lenth
Q = A*V
Q = 7112.4
tau =  B/2 *(P1)/l
			
#results
print "Average velocity through the crack  =  %.3f m/s"%(V)
print " rate of leakage  =  %.1f l/hr"%(Q)
print " Shear stress  =  %.3f N/m**2"%(tau)

Average velocity through the crack  =  0.049 m/s
 rate of leakage  =  7112.4 l/hr
 Shear stress  =  1.962 N/m**2

Example 7.2 Page No : 242

In [9]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 1200. 			#kg/m**3
mu = 0.005 			#Ns/m**2
d = 0.006 			#m
Re = 2000.
V = 0.15 			#m/s
			
#calculations
Vc = Re*mu/(d*rho)
Vr = V/Vc
T0 = 8*mu*V/d
	
#results
print "Shear stress  =  %d N/m**2"%(T0)

Shear stress  =  1 N/m**2

Example 7.3 Page No : 243

In [12]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
Q = 0.45/(60*1000) 	#m**3/s
d = 0.003 			#m
depth = 0.95 		#m
alpha = 2.
lenth = 1.25 			#m
			
#calculations
A = math.pi/4 *d**2
V = Q/A
nu =  (depth - alpha*V**2 /(2*g))*g*d**2 /(32*V*lenth)
Re = V*d/nu
			
#results
if Re<2000:
    print "Flow is laminar"
else:
    print "Flow is not laminar"

Flow is laminar

Example 7.4 Page No : 262

In [7]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 787. 			#specific gravity - kg/m**3
Q = 90.*10**-3 		#m**3/hr
d = 0.015 			#m
k = 0.0045*10**-2 	#m
nu = 1.6e-6         #kinematic viscosity - m^2/s
l = 5.   			#m
			
#calculations
V = Q/(60*math.pi/4 *d**2)
Rn = V*d/nu
e = k/d
print ("From moody diagram, f = 0.028")
f = 0.028
hl = f*l/d *V**2 /(2*g)
Power = rho*g*Q/60 *hl
			#result
print "Head loss  =  %.2f m"%(hl)
print " power required   = %.3f kW"%(Power/1000)

From moody diagram, f = 0.028
Head loss  =  34.27 m
 power required   = 0.397 kW

Example 7.5 Page No : 263

In [8]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 870. 			#density - kg/m**3
Q = 2.*10**-3 		#m**3/s
d = 0.03 			#diameter - m
mu = 5.*10**-4      #dynamics viscosity(N-s/m^2)
l = 50. 			#length - m
			
#calculations
V = Q/(math.pi/4 *d**2)
RN = rho*V*d/mu
f = 0.017
hl = f*l/d *V**2/(2*g)
Ploss = rho*g*hl
			
#results
print "Loss of pressure  =  %.1f kN/m**2"%(Ploss/1000)
#The answers are a bit different due to rounding off error in textbook

Loss of pressure  =  98.7 kN/m**2

Example 7.6 Page No : 263

In [2]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 813. 			#density - kg/m**3
Q = 0.007 			#m**3/hr
d = 0.01			#m
mu = 0.002 			#Ns/m**2
l = 30. 			#m
			
#calculations
V = Q/(60*math.pi/4*d**2)
RN = V*d*rho/mu
f = 0.316/RN**(0.25)
h = (1+f*l/d)*V**2 /(2*g)

#result
print "Height required  =  %.2f m"%(h)

# rounding off error.

Height required  =  12.21 m

Example 7.7 Page No : 264

In [10]:
import math 
			
#Initialization of variables
g = 9.81 		    	#m/s**2
rho = 10**3 			#kg/m**3
hl = 0.02
d = 1.2     			#diameter - m
l = 1.   	    		#m
k = 0.5 *10**-2 		#m
			
#calculations
v2f = hl*(2*g*d)/l
e = k/d
f = 0.028
V = math.sqrt(v2f/f)
Q = math.pi/4 *d**2 *V
			
#results
print "Rate of flow  =  %.2f m**3/s"%(Q)

Rate of flow  =  4.64 m**3/s

Example 7.8 Page No : 265

In [11]:
import math 
			
#Initialization of variables
g = 9.81 		    	#m/s**2
rho = 10.**3 			#kg/m**3
e = 0.03*10**-2 		#surface roughness - m
l = 3000. 	    		#distance - m
Q = 300.*10**-3 		#m**3/s
nu = 10.**-5 			#kinematic viscosity - m**2/s
hl = 24.     			#m
			
#calculations
d5f =  l*Q/(math.pi/4) * Q/(math.pi/4) /(hl*2*g)
f = 0.022
d = (d5f*f)**(1./5)
			
#results
print "Size of the required pipe  =  %d cm"%(d*100)

Size of the required pipe  =  45 cm

Example 7.9 Page No : 266

In [12]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
d = 0.3 			#m
per = 25./100
Q = 0.1 			#m**3/s
k0 = 0.025*10**-2 	#m
nu = 0.000001
year = 10.
			
#calculations
V = Q/(math.pi/4 *d**2)
RN = V*d/nu
e1 = k0/d
f1 = 0.019 
f2 = (1+per)*f1
e2 = 0.002
k2 = e2*d
rate  =  (k2-k0)*100/year
			
#results
print "Rate of increase  = %.4f cm/year"%(rate)

Rate of increase  = 0.0035 cm/year

Example 7.10 Page No : 267

In [14]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
l = 1. 			    #m
b = 0.3 			#m
Q = 4.2 			#m**3/s
			
#calculations
A = l*b
R = A/(2*(l+b))
d5 = 1.62/24.15
d = d5**(1./5)
Pr = 2*(l+b)/(math.pi*d)
			
#results
print "The rectangular cross section will cost %.2f times that of a circular cross section"%(Pr)

The rectangular cross section will cost 1.42 times that of a circular cross section

Example 7.11 Page No : 270

In [16]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 			#kg/m**3
d1 = 2.5*10**-2 			#m
d2 = 7.2*10**-2 			#m
Q = 100.*10**-3 			#m**3/hr
			
#calculations
V1 = Q/(60*math.pi/4*d1**2)
V2 = (d1/d2)**2 *V1
dp =  -(V2**2 -V1**2 + (V1-V2)**2)/(2*g)
Pdiff = dp*g*rho
			
#results
print "pressure difference  =  %.2f kN/m**2"%(Pdiff/1000)

# note : The answers are a bit different due to rounding off error

pressure difference  =  1.22 kN/m**2

Example 7.12 Page No : 273

In [15]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 			#kg/m**3
d2 = 30./100 			#cm
d1 = 60./100 			#cm
Pu = 105. 			#kN/m**2
Pd = 75. 			#kN/m**2
Cc = 0.65
			
#calculations
V22 = (2*g/(1 - (d2/d1)**4 + (1/Cc -1)**2)) *(Pu-Pd)*10**3 /(rho*g)
V2 = math.sqrt(V22)
Q = math.pi/4 *V2 *d2**2
hl = (1/Cc -1)**2 *V2**2 /(2*g)
			
#results
print "Flow rate  =  %.3f m**3/s"%(Q)
print " Head loss  =  %.3f m"%(hl)

Flow rate  =  0.494 m**3/s
 Head loss  =  0.722 m

Example 7.13 Page No : 277

In [16]:
import math 
			
#Initialization of variables
g = 9.81 			#m/s**2
rho = 10.**3 		#kg/m**3
d = 9.   			#m
dia = 0.3 			#m
			
#calculations
V302 =  2*g*d/(0.5 + 20 + 2.53+101+0.66+41.47+2.07)
V30 = math.sqrt(V302)
Q = math.pi/4 *dia**2 *V30
			
#results
print "Flow rate  =  %.3f m**3/s"%(Q)

Flow rate  =  0.072 m**3/s

Example 7.14 Page No : 278

In [17]:
import math 
			
#Initialization of variables
h = 6. 			    #m
rho = 930. 			#kg/m**3
Q = 3./60 			#m**3/s
d = 0.15 			#m
L = 20. 			#m
mu = 0.006          #viscosity
g = 9.81 			#m/s**2
			
#calculations
V = Q/(math.pi/4 *d**2)
RN = V*d*rho/mu
f = 0.316/RN**0.25
hl = f*L/d *V**2 /(2*g)
Hp = h+hl
gam = rho*g
W = gam*Q
Power =  W*Hp
			
#results
print "Power required   =  %.3f kW"%(Power/1000)

Power required   =  3.227 kW

Example 7.15 Page No : 279

In [5]:
import math 
from scipy.integrate import quad
			
#Initialization of variables
d = 0.02 			#m
d2 = 1.2 			#diameter - m
f = 0.01
L = 250.
ken = 0.5
g = 9.81
h1 = 8. 			#m
h2 = 4. 			#m
			
#calculations
V2 = 2*g/(1+ken+ f*L/d)
V = math.sqrt(V2)
Q = math.pi/4 *d**2 *V
def time(h):
    return  -math.pi/4 *d2**2 /Q /math.sqrt(h)

ti = quad(time,h1,h2)[0]
hours = ti/3600
mins = ti%3600/60
secs = ti%3600%60

#results
print "Time required  =  %d hours %d mins %d seconds"%(hours,mins,secs)

# rounding off error.

Time required  =  4 hours 12 mins 25 seconds

Example 7.16 Page No : 280

In [19]:
			
#Initialization of variables
d1 = 0.1 			#m
d2 = 0.05 			#m
l1 = 20. 			#m
l2 = 20. 			#m
f = 0.02
			
#calculations
Kl = (f*l2/d2 *(d1/d2)**4 - f*l1/d1)
			
#results
print "Loss coefficient  =  %d "%(Kl)

Loss coefficient  =  124

Example 7.17 Page No : 281

In [17]:
			
#Initialization of variables
g = 9.81 
ratio = 1.265
			
#calculations
percent  =  (ratio-1)*100
			
#results
print "Increase in discharge  =  %.1f %%"%(percent)

Increase in discharge  =  26.5 %

Example 7.18 Page No : 282

In [18]:
import math 
			
#Initialization of variables
Q = 0.6 			#m**3/s
l1 = 1200. 			#m
l2 = 800. 			#m
d1 = 0.3 			#m
			
#calculations
V1 = 1.02 			#m/s
d5 =  d1*l2*4**2 *Q**2 /(l1*math.pi**2 *V1**2)
d = d5**(1./5)
			
#results
print "diameter of the single pipe  =  %.2f m"%(d)

diameter of the single pipe  =  0.65 m

Example 7.19 Page No : 284

In [23]:
import math 
			
#Initialization of variables
g = 9.81
Q = 0.18 			#m**3/s
d3 = 0.3			#m
f = 0.032           #friction factor
L3 = 360. 			#m
z = 25.5 			#m
z2 = 30. 			#m
L2 = 450. 			#m
d2 = 0.45			#m
L1 = 950. 			#m
d1 = 0.45 			#m
zn = 18. 			#m
rho = 1000.
			
#calculations
V3 = Q/(math.pi/4 *d3**2)
hl3 = f*L3/d3 *(V3**2 /(2*g))
Z2 = z+hl3
hl2 = Z2-z2
V2 =  math.sqrt(2*g*d2*hl2/(f*L2))
Q2 = math.pi/4 *d2**2 *V2
V1 = V2+ (d3/d2)**2 *V3
hl1 = f*L1/d1*V1**2 /(2*g)
Hp =  hl1+ Z2-zn
gam = rho*g
P = gam*Hp
			
#results
print "Discharge into the reservoir  =  %.3f m**3/s"%(Q2)
print " Pressure maintained by the pump  =  %.2f kN/m**2"%(P/1000)

Discharge into the reservoir  =  0.356 m**3/s
 Pressure maintained by the pump  =  582.34 kN/m**2

Example 7.20 Page No : 285

In [16]:
import math 
from numpy import *
			
#Initialization of variables
z1 = 10. 			#m
z2 = 5. 			#m
z3 = 7.5 			#m
f = 0.04 
l1 = 100. 			#m
l2 = 50. 			#m
l3 = 70. 			#m
d1 = 0.1 			#m
d2 = 0.075 			#m
d3 = 0.06 			#m
g = 9.81 			#m/s**2
h = array([1, 2, 1.9, 1.96])

#calculations
Q1 = sqrt(d1**5 *(math.pi/4)**2 *2*g/(f*l1)) *sqrt(z1-h)
Q2 = sqrt(d2**5 *(math.pi/4)**2 *2*g/(f*l2)) *sqrt(h+z2)
Q3 = sqrt(d3**5 *(math.pi/4)**2 *2*g/(f*l3)) *sqrt(h+z3)
for i in range(4):
    Q = Q2[i]+Q3[i]
    if (Q1[i]  ==  Q):
        break;
    print "height h  =  %.2f m"%(h[i])
    print "Discharge in BC Q2  =  %.2f lps"%((Q2[i])*1000)
    print "Discharge in BD Q3  =  %.2f lps"%((Q3[i])*1000)
    
# note : rounding off error.    

height h  =  1.00 m
Discharge in BC Q2  =  9.28 lps
Discharge in BD Q3  =  5.35 lps
height h  =  2.00 m
Discharge in BC Q2  =  10.03 lps
Discharge in BD Q3  =  5.65 lps
height h  =  1.90 m
Discharge in BC Q2  =  9.95 lps
Discharge in BD Q3  =  5.62 lps
height h  =  1.96 m
Discharge in BC Q2  =  10.00 lps
Discharge in BD Q3  =  5.64 lps

Example 7.21 Page No : 290

In [25]:
import math 
			
#Initialization of variables
e = 0.8
output = 400. 		#kW
H = 150. 			#m
rho = 1000. 
g = 9.81
f = 0.028
l = 1250. 			#m
			
#calculations
gam = rho*g
inpu = output/e
Q = inpu*10**3 /(2./3 *gam*H)
hl = 1./3 *H
d5 =  f*l*Q**2 /(2*g* math.pi/4 * math.pi/4 *hl)
d = d5**(1./5)
			
#results
print "Smallest diameter of pen stock  =  %d cm"%(d*100)
    

Smallest diameter of pen stock  =  43 cm

Example 7.22 Page No : 291

In [29]:
import math 
			
#Initialization of variables
f = 0.04
H = 30. 			#m
l = 200. 			#m
d = 0.075 			#m
g = 9.81
rho = 1000.
gam = rho*g
			
#calculations
h = 2/3. *H
vj = math.sqrt(2.*g*h)
hl =  1/3. *H
V =  math.sqrt(hl*d*2*g/(f*l))
dj =  d*(math.sqrt(V/vj))
Power =  2/3. *gam*math.pi/4. *d**2 *V*H
			
#results
print "Size of nozzle  =  %.1f cm"%(dj*100)
print " Max power  =  %.2f kW"%(Power/1000)

Size of nozzle  =  2.0 cm
 Max power  =  1.18 kW