Chapter 13:Flow Measurement

Example 13.1 Page No320

In [2]:
from __future__ import division

import math

#variable initialization

g=9.81                               #Gravity constant

H=2                                  #height in m

hL=0.2                               #headloss

Cc=0.63                              #Coefficient of contraction

D=4                                  #Diameter in cm

#Calculation

D=D/100                              #Converting into m

V=round(math.sqrt(2*g*H),3)          #Ideal Velocity

Va=math.sqrt(2*g*(H-hL))

Cv=round(Va/V,3)

print "(i)Coefficient of velocity Cv=",Cv

Cd=round(Cc*Cv,3)

print "Coefficient of discharge Cd=",Cd

a=math.pi/4*(D**2)                   #Discharge through the orifice

Q=Cd*a*(V)

print "(ii)Q=",round(Q*1000,3),"L/s"

y=0.50                              #Given value

x=Cv*(4*y*H)**(1/2)

print "(iii)x=",x,"m"
(i)Coefficient of velocity Cv= 0.949
Coefficient of discharge Cd= 0.598
(ii)Q= 4.707 L/s
(iii)x= 1.898 m

Example 13.2 Page No321

In [3]:
import math

#variable initialization

d=25                             #diameter im mm

H=5.5                            #head in m

Q=3                              #rate in L/s

x=1.5                            #horizontal distance

y=0.12                           #vertical distance

g=9.81                           #gravity in m/s^2

#solution

Cv=round(x/(math.sqrt(4*y*H)),3)    #Coefficient of velocity

print "Cv=",Cv                  

Q=Q/1000 

a=((math.pi)/4)*((d/1000)**2)

Cd=round(Q/(a*(math.sqrt(2*g*H))),3) #Coefficient of contadiction

print "Cd=",Cd

Cc=Cd/Cv                             #Coefficient of discharge

print "Cc=",round(Cc,3)
Cv= 0.923
Cd= 0.588
Cc= 0.637

Example 13.5 Page No322

In [4]:
import math

#variable initialisation

D=10                 #diameter in cm

V1=2.5               #Velocity in pipe in m/s

P1=50                #pressure in kPa

Cv=0.98              #coefficient of velocity

ga=9.79              #Relative density

g=9.81               #gravity in m/s^2

#solution

H=round((P1/ga)+((V1**2)/(2*g)),3)    #Total head

Vj=round(Cv*(math.sqrt(2*g*H)),2)     #Velocity of jet

print "Vj=",Vj,"m/s"

A1=(math.pi/4)*((D/100)**2)           #Discharge

aj=(A1*V1)/Vj

Dj=math.sqrt(aj*4/(math.pi))

print "Dj=",round((Dj*100),2),"cm"

Hl=((1/(Cv**2))-1)*((Vj**2)/(2*g))    #Headloss

print "HL=",round(Hl,3),"m" 
Vj= 10.11 m/s
Dj= 4.97 cm
HL= 0.215 m

Example 13.7 Page No323

In [5]:
import math

#variable initialization

D=20                         #Diameter in cm

del_H=0.5                    #pressure head differential in m

mu=0.001                     #Coefficient of viscosity in Pa.s

rho=998                      #density in kg/m63

g=9.81                       #Gravity constant

#Calculation

D=D/100                      #Coverting into m

#Q=K0*A2*math.sqrt(2*g*del_H)

A2=(math.pi/4)*((D/2)**2)

Q=A2*math.sqrt(2*g*del_H)

#As K0 is not known use trial and error method,

K0=0.62

Q=K0*Q

print "Q=",round(Q,5),"m^3/s"

V=Q/(math.pi*D**2/4)         #Velocity

v=mu/rho

Re=V*D/v                     #Reynold's number

#For this value of Re,K0 is same as assumed.Hence no iterations required.

print "Q=",round(Q*1000,2),"L/s"
Q= 0.01525 m^3/s
Q= 15.25 L/s

Example 13.8 Page No323

In [6]:
import math

#variable initialisation

Q=20                     #Discharge in L/s

g=9.81                   #gravity in m/s^2

Kf=0.99                  #flow coefficient of nozzle

D=3.0                    #diameter in cm

ga=0.70*9.79             #relative density

#solution

A2=math.pi/4*(D/100)**2      #calculating area

Q=Q/10000                    #converting into m^3/s

del_H=((Q/(Kf*A2))**2)/(2*g) #Head difference across nozzle

#For a Horizontal nozzle Z1-Z2=0

del_P=ga*del_H               #calculating del_p

print "del_p=", round(del_P,3),"kPa"
del_p= 2.853 kPa

Example 13.18 Page No328

In [7]:
import math

#variable initialisation

D2=10                  #diameter in cm

D1=20                  #diameter in cm

Sm=13.6                #relative density constant

Sp=0.9                 #relative density

Cd=0.99                #coefficient of discharge

g=9.81                 #gravity in m/s^2

y=9                    #manometer reading in cm

#solution

#for a differential monometer

r=(D1/2)/100            #radius

A2=(math.pi/4)*((r)**2) #area

D=round(math.sqrt(1-(D2/D1)**4),3) 

delh=(Sm/Sp)-1          #calculating delh

Q=round(((Cd*A2)/D)*(math.sqrt(2*g*(delh*(y/100)))),2) #calculating flow

print "Q=",Q*1000,"L/s"

#when Q=50L/s

Q=50/1000

s=(((Cd*A2)/D)*(math.sqrt(2*g*(delh))))

y=round((Q/s)**2,2)     #manometer reading

print "y=",int(y*100),"cm"
Q= 40.0 L/s
y= 14 cm

Example 13.22 Page No330

In [9]:
import math

#variable initialisation

D1=20                    #diameter in cm

Cd=0.985                 #coefficient of discharge

Sm=0.6                   #relative density of liquid

Sp=1.0                   #relative density

y=15                     #U-tube reading in cm

D2=10                    #diameter in cm

g=9.81                   #gravityi in m/s^2

#solution

#from the inverted differential manometer

delh=(y/100)*(1-(Sm/Sp))

#By the venturimeter equation,

A2=(math.pi/4)*(((D1/100)/2)**2)  #area

s=math.sqrt((1-((D2/D1)**4)))

Q=((Cd*A2)/s)*(math.sqrt((2*g*delh)))

print "Q=",round(Q*1000,2),"L/s"

HLi=(1-(Cd**2))*delh              #Head loss in the intel section

print "HLi=",(round(HLi,4)*1000),"mm of water"
Q= 8.67 L/s
HLi= 1.8 mm of water

Example 13.24 Page No331

In [10]:
import math

#variable initialization

Sp=0.85              #Relative density

y=4                  #pitot static tube reading in cm

C=0.99               #coefficient of pitot tube

Sm=13.6              #relative density

g=9.81               #gravity

#solution

#For the differential monometer

del_h=(y/100)*((Sm/Sp)-1)    #del_h

#For the pitot tube

Vo=C*(math.sqrt(2*g*del_h))  #Velocity

print "Velocity at M=",round(Vo,3),"m/s"
Velocity at M= 3.397 m/s

Example 13.26 Page No332

In [11]:
import math

#variable initialization

Ps=3.0                     #Stagnation Pressure in kPa

Po=-3.0                    #Static pressure in kPa

rho=1.20                   #mass density in kg/m^3

C=0.98                     #instrument coefficient

#solution

#In a pitot static tube,

#Velocity of flow

Vo=C*math.sqrt(((2*(Ps-Po))/rho)*1000)

print "Vo=",int(Vo),"m/s"
Vo= 98 m/s

Example 13.27 Page No332

In [12]:
import math

#variable initialization

D=30                     #diameter in cm

Ps=-10                   #static pressure in cm

Psn=1.0                  #stagnation pressure in N/cm^2

C=0.98                   #coefficient of tube
 
ga=9790                  #density

g=9.81                   #gravity

#solution

Po=(Ps/100)*13.6                 #Calculating Po/ga

Ps=Psn*(10**4)/ga                #Calculating ps/ga

del_h=(Ps-Po)

Vm=C*math.sqrt(2*g*del_h)        #Centreline velocity

V=0.85*Vm                        #Mean velocity

Q=(math.pi/4)*((D/100)**2)*V     #Discharge

print "Q=",round(Q,3),"m^3/s"
Q= 0.402 m^3/s

Example 13.31 Page No333

In [13]:
import math

#variable initialization

L=0.40              #Width in m

Q=25                #discharge in L/s

H1=10               #head in cm

g=9.81              #gravity in m/s^2

#solution

#By the weir formula,

Cd=(Q/1000)/((2/3)*(math.sqrt(2*g))*L*(math.pow((H1/100),3/2)))

print "Cd=",round(Cd,3) #Coefficient of discharge
Cd= 0.669

Example 13.35 Page No334

In [14]:
import math

#variable initialization

L=2.5              #length in m
  
Cd=0.62            #Coefficient of discharge

H1=0.7             #head in m

t=0.15             #thickness in m

g=9.81             #gravity in m/s^2

#solution

L=L-2*t                #crest length

n=2+(2*2)              #no of end contractions

Le=L-(0.1*n*H1)        #effective crest length

#The discharge from Francis Formula by neglecting the velocity of approach,

Q=(2/3)*(Cd)*(math.sqrt(2*g))*(Le)*(math.pow(H1,3/2)) 

print "Q=",round(Q,2),"m^3/s"
Q= 1.91 m^3/s