Chapter 11:Pipe Flow Systems

Example 11.1 Page No238

In [2]:
from __future__ import division

import math

#variable initialisation

Q=450                #discharge in L/s

D1=6                 #diameter in cm

D2=9                 #diameter in cm

g=9.81               #gravity

p1=20                #pressure at upstream of the expansion in kN/m^3

ga=9.79              #relative density in kg/m^3

#Calculation

Q=(Q/60)/1000                              #coverting into m^3/s

V1=round(Q/((math.pi/4)*((D1/100)**2)),3)  #Velocity before expansion

V2=round(V1*(((D1/100)/(D2/100))**2),3)    #Velocity downstream of the expansion

H_L=round(((V1-V2)**2)/(2*g),3)

#By energy equation:

#(p1/ga)+(V1**2)/(2*g)+Z1)=(p2/ga)+((V2**2)/(2*g))+Z2

#As Z1=Z2

p2=round(((p1/ga)+(V1**2)/(2*g)-((V2**2)/(2*g))-H_L)*ga,2)

print "p2=",p2,"kPa"
p2= 21.73 kPa

Example 11.4 Page No239

In [3]:
import math

#variable initialisation

L1=1800                      #length of pipe1

L2=1200                      #length of pipe 2

L3=600                       #length of pipe 3

D1=50                        #diameter of pipe 1 in cm

D2=40                        #diameter of pipe 2 in cm

D3=30                        #diameter of pipe 3 in cm

De=40                        #Equivalent length of pipe in cm

#Calculation

#It's assumed that f is same for all pipes.

D1=D1/100

D2=D2/100                    #Converting into m

D3=D3/100

De=De/100

Le=round(((L1/(D1**5))+(L2/(D2**5))+(L3/(D3**5)))*(De**5),1)

print "(i)Le=",Le,"m"

Le=3600                      #assumed equivalent length of pipe in m

De=round(((Le/((L1/(D1**5))+(L2/(D2**5))+(L3/(D3**5))))**(1/5))*100,2)

print "(ii)De=",De,"cm"

#when three pipes in parallel

#h_f1=h_f2=h_f3=h_fe

#assume a=(V2/V1),b=(V3/V1),c=Ve/V1

a=round(math.sqrt((L1*D2)/(L2*D1)),4)

b=round(math.sqrt((L1*D3)/(L3*D1)),4)

#Total discharges is given by,Q0=(0.546)*(math.pi/4)*V1

#Qe=(math.pi/4)*(D1/100)**2)*Ve

#for an equivalent pipe,Q0=Qe,so equating,

c=((0.546)/((D1/100)**2))/10000

#Ve/V1=math.sqrt(L1/Le)

Le=L1/c**2

print "(iii)Le=",round(Le,2),"m"

print "Using Alternative method,"

De=0.50                     #discharge in m

a=round(((((D1**5)/L1)**(1/2))+(((D2**5)/L2)**(1/2))+(((D3**5)/L3)**(1/2))),4)

Le=((De**5)**(1/2))/a

print "Le=",round(Le**2,2),"m"
(i)Le= 4318.2 m
(ii)De= 38.57 cm
(iii)Le= 377.37 m
Using Alternative method,
Le= 377.37 m

Example 11.5 Page No240

In [4]:
import math

#variable initialisation

D1=10                                     #Diameter of pipe AB in cm

D2=16                                     #Diameter of pipe BC in cm

g=9.81                                    #gravity

f=0.02                                    #value of pipe AB

f1=0.018                                  #Value of pipe BC

#Calculation
 
#(i)entrance loss h_L1=0.5 V^2/2g

#(ii)loss at expansion:

#Equating, math.pi/4 *D1^2*V1=math.pi/4 *D2^2*V2

#we get,h_eL=0.3714(V1^2)/(2*g)

#(iii)he=V2^2/(2*g)

#Total loss H_L=6.300(V1**2)/(2*g)

H_L=10                                    #taking H_L as 10m,

V1=math.sqrt((H_L*2*g)/(6.300))

V2=(((D1/100)/(D2/100))**2)*(V1)

Q=(math.pi/4)*((D1/100)**2)*(V1)          #Calculating discharge

print "Q=",round(Q*1000,1),"L/s"          #Converting from m^3/s into L/s

#taking Q=15 L/s

Q=15/1000                                 #Converting into m^3/s

V1=Q/((math.pi/4)*(D1/100))

H_L=6.300*((V1**2)/(2*g))                 #Difference in reservoir elevations 

print "H_L=",round(H_L*100,3),"m"
Q= 43.8 L/s
H_L= 1.171 m

Example 11.9 Page No243

In [5]:
import math

#variable initialisation

fb=0.018                        #friction factor

Lb=100                          #length of pipe b in m

D2=12                           #diameter of the pipe b in cm

D1=15                           #diameter of the pipe 1 in cm

fa=0.018                        #friction factor

La=150                          #length of pipe 1 in cm

#Calculation

#Since the discharges are same in both the pipes,

#AaVa=AbVb

#math.pi/4*(D1**2)*Va=math.pi/4*(D2**2)*(Vb)

#Vb=((D1/D2)**2)*Va=1.5625Va

#let the loss in the value be Kl*Va**2/2*g

#Head losses in both pipes are the same

D1=D1/100                      #Converting into m

D2=D2/100

#(fa*La*Va^2/2gDa)+K_L*(Va^2/2g)=(fbLbVb^2/2gDb)

#calculating value of the valve coefficient

K_L=(fb*Lb*(((D1/D2)**2)**2)/(D2))-((fa*La)/(D1))

print "K_L=",round(K_L,2)
K_L= 18.62

Example 11.17 Page No248

In [6]:
import math

#variable initialisation

f=0.02                           #friction factor

L=1000                           #length of pipe in m

h_f=16                           #water surface elevation in m

g=9.81                           #gravity

D=15                             #diameter in cm

L1=600                           #length in m

p2=10.3                          #atmospherric pressure in m(abs)

Z2=100                           #elevation of the upstream in m

Z1=98                            #elevation in m

ga=9.79                          #density in kg/m^3

p=20.00                          #pressure in kPa

#Calculation

D=D/100

V=round(math.sqrt((h_f*(2*g*D))/(f*L)),3)

Q=round(((math.pi/4)*(D**2)*(V))*1000,1)

#Head loss upto the summit=h_f1

h_f1=round(((f*L1)/(2*g*D))*(V**2),1)

#By energy equation between upstream reservoir water surface and the summit.

#(p1/ga)+(V1**2)/(2*g)+Z1)=(p2/ga)+((V**2)/(2*g))+Z2

p1=((p2+Z2)-((V**2)/(2*g)+Z1+h_f1))*ga

print "(a)p1=",round(p1,2),"kPa"

#By energy equation,

a=p2+Z2

b=(p/ga)+Z1                     #Calculating volume V1

V1=math.sqrt((a-b)/(1+((f*L1)/(2*g*D))))

Q1=(math.pi/4)*(D**2)*(V1)       #Calculating discharge

print "(b)Q1=",round(Q1*1000,1),"L/s"
(a)p1= 25.26 kPa
(b)Q1= 25.1 L/s

Example 11.28 Page No259

In [7]:
import math

#variable initialisation

D=50                          #diameter in cm

f=0.019                       #friction factor

L=1000                        #length in m

H=100                         #head at the inlet in m

g=9.81                        #gravity

ga=9.79                       #density in kg/m^3

#Calculation

D=D/100

#For the condition of maximum power,the diameter of the nozzle,

d=round(((D**5)/(2*f*L))**(1/4)*100,2)

h_f=round(H/3,2)                      #Head loss

V=math.sqrt(((h_f*2*g*D))/(f*L))

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

Pm=ga*Q*(2/3)*H                       #Maximum power

print "Pm=",round(Pm,1),"kW"
Pm= 531.6 kW

Example 11.31 Page No260

In [8]:
import math

#variable initialisation

EL_p=100                                  #length of pipe

EL_a=95                                   #Length of pipe a

EL_c=110                                  #Length of pipe c

D1=15                                     #Diameter in cm

D2=12                                     #Diameter in cm

Q=0.02                                    #discharge

f1=f2=0.02                                #force

L1=20                                     #Length in m

g=9.81                                    #gravity

L2=300                                    #Length in m

ga=9.79                                   #density

Ph=10                                     #atmospheric pressure head in m

#Calculation

r=(D1/2)/100                              #Calculating radius

Hs=EL_c-EL_a                              #Static head

V1=Q/((math.pi/4)*((D1/100)**2))          #Velcoity in the suction pipe

V2=V1*(D1/D2)**2                          #Velocity in the delivery pipe

hf1=(f1*L1*(V1**2))/(D1/100*2*g)          #Head loss in the suction pipe

hL1=0.5*((V1**2)/(2*g))                   #Inlet loss

hf2=(f2*L2*(V2**2))/(D2/100*2*g)          #friction loss in the delivery pipe

hL2=((V2**2)/(2*g))                       #Loss at exit

TL=hf1+hf2+hL1+hL2                        #Total loss

Hf=Hs+TL                                  #Heat delivered by pump

P=ga*Q*Hf                                 #Power delivered

print "P=",round(P,2),"kW"

#(b)By energy equation between resorviour A and pump B

#Ps=pressure at the suction side of the pump

a=(V1**2)/(2*g)

Ps=(Ph+(EL_a-(EL_p+a+hf1+hL1)))*ga

print "Ps=",round(Ps,2),"kPa(abs)"
P= 4.57 kW
Ps= 46.29 kPa(abs)