Chapter 16:Fluid Flow Machines

Example 16.8 Page No408

In [3]:
from __future__ import division

import math

#variable initialization

Ht=H=30                            #Height

Sp=2.0                             #speed ratio

Db=0.35                            #Diameter of the boss

F=0.65                             #Flow ratio

ga=9.81                            #density

eta0=0.90                          #overall efficiency

P=15000E3

#Calculation

U1=round(Sp*math.sqrt(2*ga*H),2)

Vf1=round(F*math.sqrt(2*ga*H),2)

#using power equation,P=ga*Q*H*eta0

Q=P/(ga*998*H*eta0)                #Discharge

D=math.sqrt(Q/((math.pi/4)*(1-Db**2)*Vf1))

print "(i)D=",round(D,3),"m"       #Diameter of the runner

Db=D*Db

N=(U1*60)/(math.pi*D)              #If Speed=N rpm

print "(ii)N=",round(N,1),"rpm"    #Rotational speed

Ns=(N*math.sqrt(P/1000))/(H**(5/4))#Specific speed

print "(iii)Ns=",round(Ns,1)
(i)D= 2.285 m
(ii)N= 405.5 rpm
(iii)Ns= 707.4

Example 16.11 Page No409

In [6]:
from __future__ import division

import math

#variable initialization

H=N=300                            #Height

Cv=0.98                            #Coefficient of velocity of jet

D=2.5                              #Diameter                           

d=0.20                             #Diameter of the jet

F=0.65                             #Flow ratio

k=0.95                             #Blade friction coefficient

Beta=165                           #Blade angle

ga=9.81                            #density

g=9.79                             #gravity

etam=0.95                          #Mechanical efficiency

#Calculation

BetaD=180-Beta

V1=round(((Cv*math.sqrt(2*ga*H))),3)                 #Jet velocity

Q=round((math.pi/4)*((d**2)*V1),3)                   #Discharge

u=(math.pi*D*H)/(60)

BetaD=BetaD*0.0175                                   #converting into radians

He=round((1/ga)*(u*(V1-u))*(1+k*(math.cos(BetaD))),1)#Head extracted

etaH=He/H                                            #Hydraulic efficiency

print "(i)etaH=",round(etaH,3)

P=g*Q*H*etaH*etam                                    #Power developed

print "(ii)P=",int(round(P,0)),"kW"

Ns=(N*math.sqrt(P))/(H**(5/4))                       #Specific speed

print "(iii)Ns=",round(Ns,1)
(i)etaH= 0.919
(ii)P= 6057 kW
(iii)Ns= 18.7

Example 16.12 Page No409

In [4]:
import math

#variable initialization

u=14.0                          #man bucket diameter in m^3/s

beta=165                        #angle deflected in degree

k=1.0                           #assumed value

Q=0.8                           #discharge

H=45                            #height

etam=0.95                       #overall efficiency

Cv=0.985                        #Given value

g=9.81                          #gravity

rho=998                         #relative density

#Calculation

BetaD=180-beta                                      #in degrees

V1=((Cv*math.sqrt(2*g*H)))

BetaD=BetaD*0.0175                                  #converting into radians

P=((rho*Q*u*((V1-u)*(1+math.cos(BetaD))))/1000)     #Power produced

Ps=round(P*etam,2)

print "Ps=",round(Ps,1),"kW"

eta0=Ps/(((g*rho)/1000)*Q*H)                       #Overall efficiency

#Answer given in the book varies slightly and correct answer is,

print "eta0=",round(eta0,3)
Ps= 318.7 kW
eta0= 0.904

Example 16.13 Page No410

In [5]:
import math

#variable initialization

f=0.05                           #friction loss

h=400                            #height

N=420                            #speed in rpm

ga=9.79                          #density

eta0=0.85                        #overall efficiency

Cv=0.98                          #Given value

Ns=14                            #Specific Speed

P=500                            #Power per jet

phy=0.46                         #speed ratio

g=9.81                           #gravity

#Calculation

H=h*(1-f)                                       #Net available Head

N=int((Ns*(H**(5/4)))/(math.sqrt(P)))           #Rotational speed

print "N=",N,"rpm"

V1=round((Cv*math.sqrt(2*g*H)),2)

U=phy*(math.sqrt(2*g*H))

D=round((U*60)/(math.pi*N),3)

print "D=",D,"m"                               #mean diameter of bucket circle

Q=P*1000/(ga*998*H*eta0)                       #Discharge

d=math.sqrt(Q/((math.pi/4)*V1))

print "d=",round((d*100),2),"cm"
N= 1050 rpm
D= 0.722 m
d= 4.88 cm

Example 16.16 Page No411

In [6]:
import math

#variable initialization

H=270                            #Height

D=1.5                            #Diameter

N=400                            #speed in rpm

ga=9.81                          #density

eta0=0.90                        #overall efficiency

I=3000                           #Impulse

Cv=0.95                          #Given value

#Calculation

P=I/2

#using power equation,P=ga*Q*H*eta0

Q=P*1000/(ga*998*H*eta0)                #Discharge

V1=round(Cv*math.sqrt(2*ga*H),2)

d=math.sqrt(Q/((math.pi/4)*V1))

print "(i)d=",round(d*100,2),"cm"       #Diameter of the nozzle

U=round(((math.pi*D*N)/60),2)           #Peripheral velocity of the bucket

phy=U/(math.sqrt(2*ga*H))

print "(ii)phy=",round(phy,3)           #Speed ratio

Ns=(N*math.sqrt(P))/(H**(5/4))          #Specific speed

print "(iii)Ns=",round(Ns,2)
(i)d= 10.78 cm
(ii)phy= 0.432
(iii)Ns= 14.15

Example 16.17 Page No411

In [8]:
import math

#variable initialization

H=500                   #Head in m

Cv=0.98                 #specific heat

g=9.81                  #gravity

fi=0.45                 #assume

eta_o=0.85              #assume

d=18                    #diameter in cm

gamma=9.79              #specific weight

N=420 

#solution

V1=Cv*(math.sqrt(2*g*H)) #Velocity

d=d/100                  #diameter in m

Q=(math.pi/4)*(d**2)*V1  #Discharge

P=eta_o*gamma*Q*H        #Power developed

Ns=(N*math.sqrt(P))/(math.pow(H,5/4)) #Specific speed

print "Ns=",int(Ns)
Ns= 18

Example 16.19 Page No412

In [9]:
import math

#variable initialisation

N1=100         #speed of turbine in rpm

H1=30          #head on turbine in m

H2=18          #head reduced in m

P1=8000        #p in kW

#solution

#For geometrically similar turbines,the unit speed,Nu=N/math.sqrt(H)

N2=round(N1*(math.sqrt(H2/H1)),2)  #speed 

print "N2=",N2,"rpm"

P2=P1*(math.pow((H2/H1),3/2))      #power developed

print "P2=",int(P2),"kW"
N2= 77.46 rpm
P2= 3718 kW

Example 16.20 Page No412

In [10]:
import math

#variable initialisation

P1=6750                #p in kW

N1=300                 #speed in rpm

H1=45                  #net head in m

H2=60                  #net head under homologus conditions in m

ete_o=85               #efficiency in percentage

ga=9.81*998            #density in kg/m^3

#solution

#using unit relationships,

eta_o=85/100

Q=round(P1/((eta_o)*(ga/1000)*H1),2) 

N2=round(N1*(math.sqrt(H2/H1)),1)    #revolutions per minute

print "N2=",N2,"rpm"

Q1=18.03                             #Q value

Q2=Q1*(math.sqrt(H2/H1))             #Discharge

print "Q2=",round(Q2,2),"m^3/s"

P2=P1*(math.pow((H2/H1),3/2))        #brake power

print "P2=",int(round(P2,5)),"kW"
N2= 346.4 rpm
Q2= 20.82 m^3/s
P2= 10392 kW

Example 16.38 Page No420

In [12]:
import math

#variable initialization

r=0.15                            #radius in m

N=60                              #speed at rpm

rho=9800                          #density

Ht=15                             #Height

Qa=310                            #actual discharge in l/min

#calculation

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

L=2*r

Qt=round(((A*L*N)/60),4)

Qt1=Qt*60*1000                    #converting into l/min

slip=(Qt1-Qa)*100/Qt1

print "slip=",round(slip,2),"%"

Cd=round(Qa/Qt1,3)                #Coefficient of discharge

print "Coefficient of discharge=",Cd

Pt=rho*(Qt)*Ht                    #Power

print "Answer in the book for power is wrong.It should be as,"

print "Power,Pt=",round(Pt/1000,3),"kW"
slip= 2.52 %
Coefficient of discharge= 0.975
Answer in the book for power is wrong.It should be as,
Power,Pt= 0.779 kW

Example 16.40 Page No420

In [13]:
import math

#variable initialization

r=0.40                          #radius in cm

Ld=45.0                         #length in m

g=9.81                          #gravity

A=20                            #size of cylinder in cm

Ad=10                           #suction pipe in cm

Hv=2.5                          #in m water (abs)

Hatmo=10                        #atmospheric pressure of water in m

Hd=40                           #height of water

#Calculation

Had=int(-(Hv-Hd-Hatmo))#At incipient caviation in the delivery pipe

N=Had/((Ld/g)*((A/Ad)**2)*(((2*math.pi)/60)**2)*r)

print "N=",round(N**(1/2),2),"rpm"
N= 24.17 rpm

Example 16.41 Page No421

In [14]:
import math

#variable initialization

N=30                            #speed in rpm

D=0.30                          #radius of cylinder

Ls=5.0                          #length in m

g=9.81                          #gravity

A=15                            #size of cylinder in cm

As=5                            #suction pipe in cm

Hv=2.0                          #in m water (abs)

Hatmo=10                        #atmospheric pressure of water in m

Hs=2.5                          #Suction head

#Calculation

omega=(2*math.pi*N)/60

r=D/2

Hasm=(-Hv-Hs+Hatmo)

N=Hasm/((Ls/g)*((A/As)**2)*(((2*math.pi)/60)**2)*r)#At limiting condition for a suction pipe,

print "N=",round(N**(1/2),0),"rpm"
N= 27.0 rpm

Example 16.42 Page No421

In [15]:
import math

#variable initialization

N=30                            #speed in rpm

S=40                            #stroke in cm

Ls=5.0                          #length in m

g=9.81                          #gravity

A=20                            #size of cylinder in cm

As=10                           #suction pipe in cm

Hv=2.5                          #in m water (abs)

Hatmo=10                        #atmospheric pressure of water in m

#Calculation

omega=(2*math.pi*N)/60

r=(S/100)/2

Hasm=(Ls/g)*((A/As)**2)*(omega**2)*r

Hs=Hatmo-Hv-Hasm

print "Hs=",round(Hs,3),"m"
Hs= 3.476 m

Example 16.45 Page No422

In [16]:
import math

#variable initialization

N=50                            #speed in rpm

Ls=5.0                          #length in m

g=9.81                          #gravity

r=0.20                          #size of cylinder in cm

Hv=2.5                          #in m water (abs)

Hatmo=10.0                      #atmospheric pressure of water in m

Hs=3.0                          #in m

d=25                            #diameter in cm

#Calculation

N=N/2                           #since there are 2 strokes per revolution

omega=round((2*math.pi*N)/60,3)

Hasm=round((Ls/g)*(omega**2)*r,2)#Maximum acceleration head in suction pipe

Ds=(Hatmo-Hs-Hv)/(Hasm)          #At limiting condition for a suction pipe

print "ds=",round(d/Ds**(1/2),2),"cm"
ds= 9.86 cm