# Chapter 3:FLOW OF LIQUIDS IN PIPES AND OPEN CHANNELS¶

## Example no:3.1,Page no:70¶

In [110]:
#Variable declaration
sap=1.25  #Sulphuric acid pumped
d=25e-3  #Diameter of pipe
l=30  #length of pipe
meu=25e-3  #Viscosity of acid
rho_a=1840  #Density of acid

#Calculation
import math
Re=4*sap/(math.pi*meu*d)
u=sap/(rho_a*math.pi/4*d**2)
#calculating pressure drop from the energy balance equation and equation 3.19
Dp=rho_a*((0.5+4*0.006*30/0.025)*u**2+9.81*12)

#Result
print"\n Pressure drop =",round((Dp/10**3)),"kN/m**2"

 Pressure drop = 320.0 kN/m**2


## Example no:3.2,Page no:70¶

In [111]:
#Variable declaration
d=50e-3  #Diameter of pipe
l=100.0  #length of pipe
e=0.013  #Roughness of pipe
DPf=50e3  #Maximum pressure drop
rho=1000.0  #density of water
meu=1e-3  #viscosity of water

#Calculatiom
phi_re2=(DPf)*d**3*rho/(4*l*meu**2)
e_d=e/(d*1e3)
Re=7.9e4
u=Re*meu/(rho*d)

#Result
print"The maximum allowable velocity is =",round(u,1),"m/s"

The maximum allowable velocity is = 1.6 m/s


## Example no:3.3,Page no:71¶

In [112]:
#Variable declaration
Dia_tank=5  #Diameter of the tank
len_pipe=100.0  #Length of pipe
dia_pipe=225e-3  #Diameter of pipe

#Calculation
X=.0020
def f(D):
return(111.5*(1+(3552*X))**0.5*D**-0.5)

print"The time taken for the level to fall is therefore about %d"%t[0],"s or %.1f"%(t[0]/60.0),"min"

The time taken for the level to fall is therefore about 751 s or 12.5 min


## Example no:3.4,Page no:72¶

In [113]:
#Variable declaration
d1=0.3  #diameter of pipe from junction A to D or B to D
l1=1.5e3  #length of pipe from junction A to D or B to D
d2=0.5  # diameter of pipe from junction D to C
l2=0.75e3  # length of pipe from junction D to C
h_A=10  # height of tank A above C
h_B=h_A+6  # height of tank A above C
rho=870  # density of liquid
Meu_l=0.7e-3  # viscosity of liquid

#Calculation
import math
import sympy
import numpy as np
x=Symbol('x')
u2=sympy.solve(x**4-(7.38*x**2)+13.57)
u1=(np.square(u2)-1.47)**0.5
u3=(u1+u2)/2.78
#taking the positive values and which satisfy equation 7
U1=u1[2]
U2=u2[2]
U3=u3[2]
Q=math.pi/4*d2**2*U3

#Result
print"\n The volumetric flow rate =%.2f"%Q,"m**3/s"

 The volumetric flow rate =0.23 m**3/s


## Example no:3.5,Page no:86¶

In [114]:
#Variable declaration
r=50.0

#Calculation
from scipy.optimize import fsolve
a=Symbol('a')
def f(a):
return(105*a**(8.0/7.0)-56*a**(15.0/7.0)-24.5)
p=fsolve(f,0.5)
y=p[0]*r

#Result
print"a =",round(p[0],2)
print"y =",round(y,1),"mm"

a = 0.33
y = 16.6 mm


## Example no:3.6,Page no:88¶

In [115]:
#Variable declaration
Q=7.2 #Water flow rate
d1=40e-3  #initial pipe diameter
d2=50e-3  #diameter of pipe after enlargement
g=9.81

#Calculation
import math
u1=(Q/3600.0)/(math.pi/4*d1**2) #Velocity in 40 mm pipe
u2=(Q/3600.0)/(math.pi/4*d2**2) #Velocity in 50 mm pipe
# The head lost is given by equation 3.77 as:
hf=(u1-u2)**2.0/(2.0*g)

#Result

Head lost = 16.7 mm of water


## Example no:3.7,Page no:92¶

In [116]:
#Variable declaration
Q_h=2.27  # flow rate of water in m**3/h
T=320  #Temperature of water to be pumped
id=40e-3  #internal diameter of pipe
l_h=150.0  #length of pipe horizontally
l_v=10.0  #length of pipe vertically
e=0.2e-3
g=9.81
rho=1000.0

#Calculation
rel_rough=e/id  #Relative roughness
meu=0.65e-3  #Viscosity at 320 K
Q_s=Q_h/3600.0  #flow rate of water in m**3/s
area=math.pi/4*id**2  # Area for flow
u=Q_s/area  #Velocity
Re=(id*u*rho)/meu
#X=R/(rho*u**2)=0.004 (from Figure 3.7)
X=.004
equi_len=l_h+l_v+(260*id)  # Equivalent length of pipe
hf=4*X*equi_len*u**2/(id*g)
mass_thr=Q_s*rho  #Mass throughput

#Result
print"Power required =%.1f"%power_reqd,"W =%.3f"%(power_reqd*1e-3),"kW"

Power required =136.6 W =0.137 kW


## Example no:3.8,Page no:92¶

In [117]:
#Variable declaration
d=0.15  # diameter of pipe
g=9.81
phi=.0045

#Calculation
import sympy
import math
x=Symbol('x')
u=sympy.solve((7.6+4*phi*(105/.15))*x**2/(2*g)-10)
rate_dis=u[1]*math.pi*d**2/4

#Result
print"\n Rate of discharge =%.d"%(rate_dis*1e3),"kg/s"

 Rate of discharge =55 kg/s


## Example no:3.9,Page no:103¶

In [118]:
#Variable declaration
u1=1.5  # velocity
D1=75e-3  #depth
g=9.81

#Calculation
#The depth of fluid in the channel after the jump is given by:
D2=0.5*(-D1+(D1**2+(8*u1**2*D1/g)**0.5))  #equation 3.113
#If the channel is of uniform cross-sectional area, then:
u2=u1*D1/D2

#Result
print"\n The velocity of fluid in the channel after the jump is =",round(u2,2),"m/s"

 The velocity of fluid in the channel after the jump is = 0.75 m/s


## Example no:3.10,Page no:127¶

In [119]:
#Variable declaration
k=10
n=0.2
Ucl=1  # centre line velocity
l=200  # length of pipe
dux_dy_1=10
dux_dy_2=50
Ry_1=k*dux_dy_1**0.2
Ry_2=k*dux_dy_2**0.2

#Calculation
#Using the Bingham-plastic model (equation 3.125):
import numpy as np
A=np.array([[1,10],[1,50]])
B=np.array([15.85,21.87])
C=np.linalg.inv(A)*B
Ry=C[0]
Meu_p=C[1]
# Using Equation 3.131
DP=2*k*l*Ucl**n*((n+1)/n)**n*r**(-n-1)
# For a Bingham-plastic fluid:
# The centre line velocity is given by equation 3.145:
X=(l*2*Ry)/(r*DP)
Up=(DP*r**2*(2-4*X+2*X**2))/(8*Meu_p*l)

#Result
print"Plastic viscosity (Meu_p) =",C[1][0]+C[1][1],"N s/m**2"
print"\nYeild stress (Ry) =",C[0][0]+C[0][1],"N s/m**2"
print"\nPressure drop (Bingham plastic model)=%.2e"%DP,"N/m**2"
print"\ncentre line velocity (Bingham plastic model) = ",round(Up[0]+Up[1],2),"m/s (APPROX taken in book)"

Plastic viscosity (Meu_p) = 0.1505 N s/m**2

Yeild stress (Ry) = 14.345 N s/m**2

Pressure drop (Bingham plastic model)=6.26e+05 N/m**2

centre line velocity (Bingham plastic model) =  0.68 m/s (APPROX taken in book)


## Example no:3.11,Page no:129¶

In [1]:
#Variable declaration
Meu=0.1  # Viscosity of liquid
d=25e-3  # Diameter of pipe
l=20.0  # length of pipe
DP=1e5  # Pressure drop
n=1/3.0  # flow index of polymer solution
dux_dy=1000.0
k=Meu

#Calculation
Meu_a=Meu
k_poly_sol=Meu_a/(dux_dy)**(n-1)
Ry=10.0*(dux_dy)**n
#From equation 3.136:
#For a power-law fluid:
u2=((DP/(4*k_poly_sol*l))**3)*(n*(d**((n+1)/n)))/(2*(3*n+1))
u1=(DP/(4*k*l))*(d**2)/8.0
ratio=u2/u1

#Result
print"Ratio of the volumetric flow rates of the two liquids =",round(ratio,3)

Ratio of the volumetric flow rates of the two liquids = 0.065


## Examople no:3.12,Page no:129¶

In [15]:
#Variable declaration
n=0.5
du_dy=0.01
import sympy
mu=sympy.Symbol('mu')

#Calculation
k=mu/(du_dy)**(n-1)
Ry=k*du_dy**(n-1)
#ux=(rho*g/k)**(1/n)*((s-y)**(n+1)/n)*(-n/(n+1)]+constant
#At the surface
y=0
ux=0
#At free surface:
#y=s
#us=(rho*g/k)**(1/n)*(n/(n+1))**(s*(n+1)/n)
s=sympy.Symbol('s')
sn=sympy.Symbol('sn')
#For non-newtonian:
us1=0.00592*sn**6/s**5
#For newtonian:
us2=0.0067*sn**5/s**4
sn_s=1/(us1/us2)

#Result
print" sN/s=",round(sn_s*sn/s,3)

 sN/s= 1.132