Chapter 9 : Turbulent flow in Pipes¶

Example 9.1 Page no 308¶

In [1]:
# Determine Head loss

from math import *

# Given

S = 1.26                # specific gravity

mu = 0.826              # kinematic viscosity in Ns/m**2

# for water

rho = 998               # density of water in kg/m**3

mu1 = 1.005*10**-3      # viscosity in Ns/m**2

# for glycerine

rho1 = S*rho            # density of glycerine in kg/m**3

Q = 0.1                 # discharge in m**3/s

d1 = 0.2                # diameter in m

A = pi*d1**2/4          # area in m**2

g = 9.81                # acceleration due to gravity in m/s**2

l =100                  # length of the pipe

# Solution

V = Q/A

R = rho1*V*d1/mu

print "It is a laminar flow"

f = 64/R               # friction factor

Hf = f*l*V**2/(2*g*d1) # head loss due to friction

print "(a) Head loss due to flow for glycerine =",round(Hf,1),"m "

R1 = rho*V*d1/mu1

print "The flow is turbulent"

e = 0.025

r = e/(d1*100)

f = 0.021

hf = f*l*V**2/(2*g*d1)

print "(a) Head loss due to flow for water =",round(hf,2),"m "

It is a laminar flow
(a) Head loss due to flow for glycerine = 17.1 m
The flow is turbulent
(a) Head loss due to flow for water = 5.42 m


Example 9.2 Page no 311¶

In [2]:
# Discharge of water

from math import *

# Given

# for water

nu = 1.007*10**-6            # viscosity in m**2/s

e = 0.025                    # for cast iron in cm

L = 100                      # length of the pipe in m

D = 0.2                      # diameter in m

hf = 5.43                    # head loss due to friction

r = e/(D*100)

g = 9.81                     # acceleration due to gravity in m/s**2

# Solution

A = sqrt(2*g*D*hf/L)

B = D/nu

f = 0.021                   # from moodys diagram

V = A/sqrt(f)

print V

R = B*f

A = pi*D**2/4

Q = A*V

print "Discharge =",round(Q,2),"m**3/s"

3.1853324563
Discharge = 0.1 m**3/s


Example 9.3 Page no 314¶

In [3]:
# Size of the case iron

from math import *

# Given

Q =0.1                          # discharge in m**3/s

hf = 5.43                       # friction loss head in m

L = 100                         # length of pipe

nu = 1.00*10**-6                # viscosity in m**2/s

e = 0.025                       # for cast iron in cm

g = 9.81                        # acceleration due to gravity in m/s**2

# Solution

A = 8*L*Q**2/(hf*g*pi**2)

B = 4*Q/(pi*nu)

# for D = 0.172 ; f=0.01
D = 0.172

r = e/D

Re = B/D

f = 0.022                       # for Re and r

# for D1=0.199 ; f=0.021

D1 = 0.199

r1 = e/D1

R = B/D1

f = 0.021                       # for R and r

print "Hence the convergence is attained, D=",round(D1,1),"m"

Hence the convergence is attained, D= 0.2 m


Example 9.4 Page no 318¶

In [4]:
# Head loss

from math import *

# Given

L = 500                 # length of the pipe in ft

D= 9*2.54/100           # diameter in cm

C = 100                 # constant

S = 0.004

# Solution

Hf = S*L


Head loss = 2.0 ft


Example 9.5 Page no 319¶

In [6]:
# Head loss of water

from math import *

# Given

Q = 0.1                  # water flow rate in m**3/s

d = 30                  # diameter in m

l = 500                  # length in m

e = 0.025                # for cast iron

g = 9.81                 # acceleration due to gravity in m/s**2

# Solution

r = log(d/e,10)

K = (pi/4)*sqrt(2*g)*(2*r+1.14)*(0.3)**(2.5)

S = (Q/K)**2

hf = S*l

print "Head loss of water =",round(hf,1),"m"

Head loss of water = 3.2 m


Example 9.6 Page no 319¶

In [8]:
# Head loss by conveyance method

from math import *

# Given

Q = 0.1                  # water flow rate in m**3/s

d = 20                  # diameter in cm

l = 500                  # length in m

e = 0.025                # for cast iron

g = 9.81                 # acceleration due to gravity in m/s**2

S = 5.43

# Solution

r = log(d/e,10)

K = (pi/4)*sqrt(2*g)*(2*r+1.14)*(0.2)**2.5

Q=K*sqrt(S/100)

print "Head loss of water =",round(Q,2),"m**3/s"

Head loss of water = 0.1 m**3/s


Example 9.7 Page no 320¶

In [10]:
# Solve using the conveyence method

from math import *

# given

eps = 0.025*10**-2           # for cast iron epsilon = 0.0025 cm

# we get the value of K = 0.432 m**2/s
# we need to do trial and error to find the value of D

# we use the value of D = 0.2 m

D = 0.2               # value in m

g = 9.81

# Solution

K = (pi/4)*sqrt(2*g)*(2*log10(D/(eps))+1.14)*D**(2.5)

print "K = ",round(K,3)," from trial and error"

K =  0.432  from trial and error


Example 9.8 Page no 326¶

In [11]:
# Determine head loss

from math import *

# Given

d = 0.1                # diameter of the pipe

Q= 0.075               # discharge in m**3/s

L = 30                 # length in m

A = pi*d**2/4

g = 9.81               # acceleration due to gravity in m/s**2

# for water

nu = 1.007*10**-6       # viscosity in m**2/s

e = 0.025

r = e/(10*d)

# Solution

V = Q/A

Re = V*d/nu

f = 0.025              # firction factor from moodys diagram

hf = f*L*V**2/(2*g*d)

K= 0.5                 # contraction constant

hc = K*V**2/(2*g)

K1 =10                 # loss of the globe valve

hg = K1*V**2/(2*g)

Th = hf+hc+hg


Total head loss = 83.7 m


Example 9.9 Page no 328¶

In [12]:
# discharge through the pipe

from math import *

# Given

# for water

nu = 1.007*10**-6                # viscosity in m**2/s

d1 = 0.3                         # diameter of pipe 1 in m

d2 = 0.15                        # diameter of pipe 2 in m

d3 = 0.08                        # diameter of pipe 3 in m

g = 9.81                         # acclelration due to gravity in m/s**2

e = 0.025                        # for cast iron

f1 = 0.019                       # foe e/d1

f2 = 0.022                       # foe e/d2

# Solution

V3 = sqrt(2*g*100/((8.4*(f1)+268.85*(f2)+4.85)))

V1 = (d3/d1)**2*V3

V2 = (d3/d2)**2*V3

# reynolds number for pipe BC

R1 = V1*d1/nu

R2 = V2*d2/nu

Q = V3*pi*d3**2/4

print "Discharge through the pipe =",round(Q,3),"m**3/s"

Discharge through the pipe = 0.067 m**3/s


Example 9.10 Page no 332¶

In [13]:
# Replace the flow system

from math import *

# Given

D = 0.2                      # diameter of pipe 1

D1 = 0.15                    # diameter of pipe 2

Q = 0.1                     # discharge in m**3/s

nu = 1.007*10**-6           # viscosity in m**2/s

e = 0.025                   # e for cast iron

r = e/(100*D)

# Solution

V = Q/(pi*(0.2)**2/4)

R = V*D/nu

f = 0.021                 # from moodys law

r2 = e/D1

V1 = Q/(pi*D1**2/4)

R1 = V*D1/nu

f2 = 0.023                # from moodys law

L2 = 28562*D1**5/f2

print "Replacement of the flow system =",round(L2,2),"m"

Replacement of the flow system = 94.3 m


Example 9.11 Page no 335¶

In [14]:
# Discharge through each branch

from math import *

# Given

e = 0.025                # in cm

nu = 1.007*10**-6        # viscosity in m**2/s

Q1 = 0.5                 # discharge in m**3/s

D1 = 50

L1 = 500                 # length in m

g = 9.81

# Pipe 1

r1 = e/D                 # r1 for pipe 1

V1 = Q1/(pi*(0.5)**2/4)

R = V*(0.5)/nu

f1 = 0.018               # for the reynolds no

hf1 = f*L1*V1**2/(2*g*D1)

# pipe 2

hf2 = hf1

L2 =200                 # length in m

D2 = 0.3                # diameter in m

r2 = e/D2

f2 =  0.02

V2 = 0.419/sqrt(f2)

R2 = V2*D2/nu

Q2 = V2*(pi*D2**2/4)

#pipe 3

hf3=hf1

L3 = 300              # length of pipe 3 in m

D3  =0.15             # diameter of pipe 3 in m

r3 = e/D3

f = 0.022            # from moody's law

V3 = 0.242/sqrt(f2)

R3 = V3*D3/nu

Q3 = V3*(pi*D3**2/4)

Td = Q1+Q2+Q3

q1 = Q1*(2.0/Td)

q2 = Q2*(2.0/Td)

q3 = Q3*(2.0/Td)

print "Discharge through branch 1 =",round(q1,2),"m**3/s"

print "Discharge through branch 2 =",round(q2,3),"m**3/s"

print "Discharge through branch 3 =",round(q3,3),"m**3/s"

d = 0.5

v1 = q1/(pi*(d)**2/4)

R4 = v1*d/nu

r4 = 0.0005                 # ratio of e/D

f = 0.018

Hf1 = f*L1*v1**2/(2*g*d)

print "It is found that hf1=hf2=hf3 =",round(Hf1,1),"The distribution od discharge is correct"

Discharge through branch 1 = 1.35 m**3/s
Discharge through branch 2 = 0.566 m**3/s
Discharge through branch 3 = 0.082 m**3/s
It is found that hf1=hf2=hf3 = 43.5 The distribution od discharge is correct


Example 9.14 Page no 349¶

In [15]:
# Find minimum depth below the ridge

from math import *

# Given

e = 0.00015             # from moody's chart

D = 2                   # depth in ft

r = e/D

z1 = 100                # elevation in ft

mu = 1.084*10**-5       # viscosity in Ns/ft**2

p1 = 34                 # pressure head in ft

p2 = 10                 # pressure head in ft

g = 32.2                # acclelration due to gravity in ft/s**2

L = 1000                # length in ft

# Solution

f = 0.011           # assume

V = sqrt(100/(10000/(2*2*g)))/sqrt(f)

R = V*D/mu

V1 = 10.15

f1 = 0.0125

Q = V1*pi*D**2/4

x = p1-p2-(V1**2/(2*g))-(f1*L*V1**2/(2*g*D))

Dp = 30 - x

print "Minimum depth =",round(Dp,2),"ft"

Minimum depth = 17.6 ft

In [ ]: