# Chapter8 - Hydraulic pumps¶

## Ex 8.1 Page 354¶

In :
from __future__ import division
from math import pi, tan
#input data
D=1.3#Diameter of the pump in m
Q=3.5/60#Discharge of water by pump in m**3/s
U2=10#Tip speed of pump in m/s
Cr2=1.6#Flow velocity of water in pump in m/s
b2=30#Outlet blade angle tangent to impeller periphery in degree
Cx1=0#Whirl velocity at inlet in m/s
U=10#Tip speed of pump in m/s
d=1000#Density of water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2

#calculations
Wx2=Cr2/tan(b2*pi/180)#Exit relative velocity in m/s
E=(U2/g)*(U2-(Wx2))#Euler head in m or W/(N/S)
m=d*Q#Mass flow rate of water in kg/s
W=E*m*g#Power delivered in W
r=D/2#Radius of the pump in m
T=W/(U/r)#Torque delivered in Nm

#output
print 'Torque delivered by the impeller is %3.1f Nm'%(T)
# Answer in the textbook is wrong.

Torque delivered by the impeller is 274.1 Nm


## Ex 8.2 Page 355¶

In :
from math import sin, tan
#input data
b2=30#Impeller blade angle to the tangent at impeller outlet in degree
N=1450#Pump rotation speed in rpm
Q=0.028#FLow rate of the pump in m**3/s
sf=0.77#Slip factor
g=9.81#Acceleration due to gravity in m/s**2

#calculations
A=3.1415*d*D#Flow area in m**2
Cr2=Q/A#Flow velocity in m/s
Wx2=Cr2/tan(b2*pi/180)#Exit relative velocity in m/s
U2=(3.14*D*N)/60#Tip speed of pump in m/s
Cx2=U2-Wx2#Absolute whirl component at exit in m/s
E=(U2*Cx2)/g#Euler head with no whirl at inlet in m
Cx21=sf*Cx2#Actual value of component of absolute value in tangential direction in m/s
Es=sf*E#Theoretical head with slip in m
Z=(3.145*sin(b2*pi/180))/((1-sf)*(1-((Cr2/U2)/tan(b2*pi/180))))#Number of blades required based on stodola slip factor

#output
print '(a)Theoretical head with slip is %3.2f m\n(b)Number of blades required is %3.f'%(Es,Z)

(a)Theoretical head with slip is 23.65 m
(b)Number of blades required is   8


## Ex 8.3 Page 356¶

In :
#input data
D2=0.4#Outer diameter of impeller in m
b2=0.05#Outlet width of impeller in m
N=800#Running speed of pump in rpm
Hm=16#Working head of pump in m
b22=40#Vane angle at outlet in degree
nm=0.75#Manometric efficiency
g=9.81#Acceleration due to gravity in m/s**2

#calculations
U2=(3.1415*D2*N)/60#Impeller tip speed in m/s
Cx2=(g*Hm)/(U2*nm)#Absolute whirl component at exit in m/s
Wx2=U2-Cx2#Exit relative velocity in m/s
Cr2=Wx2*tan(b22*pi/180)#Flow velocity of water in pump in m/s
A=3.14*D2*b2#Area of flow in m**2
Q=A*Cr2#Discharge of the pump in m**3/s

#output
print 'The discharge of the pump is %3.4f m**3/s'%(Q)

The discharge of the pump is 0.2247 m**3/s


## Ex 8.4 Page 357¶

In :
from math import tan, atan, degrees
#input data
D2D1=2#The ratio of outer and inner diameter
N=1200#The running speed of pump in rpm
Hm=75#Total head producing work in m
Cr1=3#Flow velocity through impeller at inlet in m/s
Cr2=Cr1#Flow velocity through impeller at outlet in m/s
b22=30#Vanes set back angle at outlet in degree
D2=0.6#Outlet diameter of impeller in m
d=1000#Density of water in kg/m**3
b2=0.05#Width of impeller at outlet in m
g=9.81#Acceleartion due to gravity in m/s**2

#calculations
D1=D2/D2D1#Inlet diameter of impeller in m
U1=(3.1415*D1*N)/60#Impeller tip speed at inlet in m/s
b11=degrees(atan(Cr1/U1))#Vane angle at inlet in degree
U2=(3.1415*D2*N)/60#Impeller tip speed at exit in m/s
A=3.1415*D2*b2#Area of flow in m**2
Q=A*Cr2#Discharge of the pump in m**/s
m=d*Q#Mass flow rate of water in kg/s
Wx2=Cr2/tan(b22*pi/180)#Exit relative velocity in m/s
Cx2=U2-Wx2#Absolute whirl component at exit in m/s
W=m*U2*Cx2*10**-3#Work done per second in kW
nm=Hm/((U2*Cx2)/g)#Manometric efficiency

#output
print '(a)Vane angle at inlet is %3.3f degree\n(b)Work done per second is %3.2f kW\n(c)Manometric efficiency is %0.2f %%'%(b11,W,nm*100)

(a)Vane angle at inlet is 9.043 degree
(b)Work done per second is 346.42 kW
(c)Manometric efficiency is 60.05 %


## Ex 8.5 Page 358¶

In :
#input data
Q=75#Discharge from the pump in l/s
D1=0.1#Inlet diameter of the pump in m
D2=0.29#Outlet diameter of the pump in m
Hm=30#Total head producing work in m
N=1750#Speed of the pump in rpm
b1=0.025#Width of impeller at inlet per side in m
b2=0.023#Width of impeller at outlet in total in m
a11=90#The angle made by the entering fluid to impeller in degree
b22=27#Vanes set back angle at outlet in degree
Qloss=2.25#Leakage loss in l/s
ml=1.04#Mechanical loss in kW
cf=0.87#Contraction factor due to vane thickness
n0=0.55#Overall efficiency
d=1000#Density of water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2

#calculations
A1=3.1415*D1*b1*cf*10**3#Area of flow at inlet in m**2
Qt=Q+Qloss#Total quantity of water handled by pump in l/s
Qts=Qt/2#Total quantity of water handled by pump per side in l/s
Cr1=(Qts*10**-3)/(A1*10**-3)#Flow velocity through impeller at inlet in m/s
b11=degrees(atan(Cr1/U1))#Inlet vane angle in degree
A2=3.1415*D2*(b2/2)*cf*10**3#Area of flow at outlet in m**2 here b2 is calculated per side
Cr2=Qts/A2#Velocity of flow at outlet in m/s
U2=(3.1415*D2*N)/60#Peripheral speed at outlet in m/s
Wx2=Cr2/tan(b22*pi/180)#Exit relative velocity in m/s
Cx2=U2-Wx2#Absolute whirl component at exit in m/s
a22=degrees(atan(Cr2/Cx2))#The absolute water angle at outlet in degree
C2=Cr2/sin(a22*pi/180)#Absolute velocity of water at exit in m/s
nh=Hm/((U2*Cx2)/g)#Manometric efficiency
nv=Q/Qt#Volumetric efficiency
SP=(d*g*(Q*10**-3/2)*Hm)/n0*10**-3#Shaft power in kW
nm=(SP-ml)/SP#Mechanical efficiency

#output
print '(a)Inlet vane angle is %3.2f degree\n(b)The absolute water angle is %3.2f degree\n(c)Absolute velocity of water at exit is %3.2f m/s\n(d)Manometric efficiency is %0.1f %%\n(e)Volumetric efficiency is %0.2f %%\n(f)Mechanical efficiency is %0.1f %%'%(b11,a22,C2,nh*100,nv*100,nm*100)

(a)Inlet vane angle is 31.67 degree
(b)The absolute water angle is 13.07 degree
(c)Absolute velocity of water at exit is 18.74 m/s
(d)Manometric efficiency is 60.7 %
(e)Volumetric efficiency is 97.09 %
(f)Mechanical efficiency is 94.8 %


## Ex 8.6 Page 360¶

In :
#input data
Hi=0.25#Vaccum gauge reading in m of Hg vaccum
Z01=0.5#Effective height between gauges in m
P=22#Power of electric motor in kW
Di=0.15#Inlet diameter in m
Do=0.15#Outlet diameter in m
Q=0.1#Discharge of pump in m**3/s
dHg=13600#Density of mercury in kg/m**3
dw=1000#Density of water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2

#calculations
Pi=dHg*g*Hi#Inlet pressure in N/m**2 vaccum
Po=P0*10**5#Outlet pressure in N/m**2
V0=Q/((3.1415*Do**2)/4)#Velocity of water in delivery pipe in m/s
Vi=V0#vleocity of water in suction pipe in m/s
n0=(dw*g*Q*Hm)/(P*10**3)#Overall efficiency

#output
print '(a)Manometric head is %3.2f m\n(b)Overall efficiency is %0.1f %%'%(Hm,n0*100)

(a)Manometric head is 19.19 m
(b)Overall efficiency is 85.6 %


## Ex 8.7 Page 361¶

In :
#input data
Hm=20#Head against which work is produced in pump in m
b22=45#Vanes set back angle at outlet in degree
N=600#Rotating speed of pump in rpm
Cr1=2#Flow velocity through impeller at inlet in m/s
Cr2=Cr1#Flow velocity through impeller at outlet in m/s
g=9.81#acceleration due to gravity in m/s**2

#calculations
Wx2=Cr2/tan(b22*pi/180)#Exit relative velocity in m/s
U2=(4+(16+(4*3*792.8))**(1/2))/(2*3)#   Blade outlet speed in m/s
#The above equation is obtained by solving
#Cx2=U2-Wx2     #Absolute whirl component at exit in m/s
#C2=(Cx2**2+Cr2**2)**(1/2)    #Absolute velocity of water at exit in m/s
#Hm=(U2*Cx2/g)-((C2**2)/(4*g))     #Total head producing work in m
#3*(U2**2)-(4*U2)-792.8=0
D2=(60*U2)/(3.1415*N)#Impeller diameter in m

#output
print 'The impeller diameter is %3.4f m'%(D2)

The impeller diameter is 0.5391 m


## Ex 8.8 Page 362¶

In :
#input data
n0=0.7#Overall efficiency
Q=0.025#Discharge of water by the pump in m**3/s
H=20#Height of supplied by the pump in m
D=0.1#Diameter of the pump in m
L=100#Length of the pipe in m
f=0.012#Friction coefficient
g=9.81#Acceleration due to gravity in m/s**2
d=1000#Density of water in kg/m**3

#calculations
V0=Q/((3.1415/4)*D**2)#Velocity of water in the pipe in m/s
hf0=(4*f*L*V0**2)/(2*g*D)#Loss of head due to friction in pipe in m
P=(d*g*Q*Hm)/(n0)*10**-3#Power required to drive the pump in kW

#output
print 'Power required to drive the pump is %3.2f kW'%(P)

Power required to drive the pump is 15.87 kW


## Ex 8.9 Page 363¶

In :
#input data
Q=0.015#Discharge of water in pump in m**3/s
D1=0.2#Internal diameter of the impeller in m
D2=0.4#External diameter of the impeller in m
b1=0.016#Width of impeller at inlet in m
b2=0.008#Width of impeller at outlet in m
N=1200#Running speed of the pump in rpm
b22=30#Impeller vane angle at outlet in degree
g=9.81#Acceleration due to gravity in m/s**2
d=1000#Density of water in kg/m**3

#calculations
print 'From velocity triangles the following values have been deduced'
a11=90#The absolute water angle at inlet in degree
Cx1=0#Absolute whirl component at inlet in m/s
A1=3.1415*D1*b1#Area of flow at inlet in m**2
Cr1=Q/A1#Flow velocity through impeller at inlet in m/s
C1=Cr1#Absolute velocity at inlet in m/s
A2=3.1415*D2*b2#Area of flow at outlet in m**2
Cr2=Q/A2#Flow velocity through impeller at outlet in m/s
Cx2=U2-(Cr2/tan(b22*pi/180))#Absolute whirl component at outlet in m/s
C2=(Cx2**2+Cr2**2)**(1/2)#Velocity at impeller exit in m/s
Ihl=((Cx2*U2)/g)-((C2**2)/(2*g))+((C1**2)/(2*g))#Pressure rise in impeller in m

#output
print '\n\nThe rise in pressure in the impeller is %3.3f m'%(Ihl)

From velocity triangles the following values have been deduced

The rise in pressure in the impeller is 31.852 m


## Ex 8.10 Page 365¶

In :
#input data
Ihl=3#Head loss in impeller in m
Cr2=4.64#Flow velocity through impeller at outlet in m/s
dPi=35.3#Difference in pressure gauge readings at impeller inlet and outlet in m of water
Pg=4.7#Pressure gain in the casing in m of water
n=0.385#Part of absolute kinetic energy converted into pressure gain
g=9.81#Acceleration due to gravity in m/s**2
d=1000#Density of water in kg/m**3
ss=0.85#Slip coefficient

#calculations
Kei=Pg/n#Kinetic energy at impeller exit in m/s
C2=((Kei)*2*g)**(1/2)#Velocity at impeller exit in m/s
Cx22=(C2**2-Cr2**2)**(1/2)#Absolute whirl component at outlet with fliud slip in m/s
Cx2=Cx22/ss#Ideal absolute whirl velocity in m/s
b22=degrees(atan(Cr2/(U2-Cx2)))#Blade angle at exit in degree
Wm=ss*U2*Cx2#Euler work input in J/kg
nm=dPi/(U2*Cx22/g)#Manometric efficiency
dP=(U2*Cx22/g)-(Ihl)-(C2**2/(2*g))#Pressure rise in impeller in m

#output
print '(a)\n    Exit blade angle is %3.2f degree\n    Euler work input is %3.2f J/kg\n(b)Manometric efficiency is %0.2f %%\n(c)Pressure rise in the impeller is %3.3f m'%(b22,Wm,nm*100,dP)

(a)
Exit blade angle is 20.17 degree
Euler work input is 442.93 J/kg
(b)Manometric efficiency is 78.18 %
(c)Pressure rise in the impeller is 29.943 m


## Ex 8.11 Page 366¶

In :
#input data
r1=0.051#Eye radius of the impeller in m
D2=0.406#Outer diameter of the impeller in m
b11=(90-75)#Inlet blade angle measured from tangential flow direction in degree
b22=(90-83)#Outlet blade angle measured from tangential flow direction in degree
Cx1=0#Inlet whirl velocity in m/s
nh=0.89#Hydraulic efficiency
g=9.81#Acceleration due to gravity in m/s**2
d=1000#Density of water in kg/m**3
N=900#Rotating speed of impeller in rpm

#calculations
w=(2*3.1415*N)/60#Angular velocity at inlet in rad/s
U1=(w*r1)#Inlet tangential impeller velocity in m/s
C1=U1*tan(b11*pi/180)#Velocity at impeller inlet in m/s
A=2*3.1415*r1*b#Area of flow through the pump in m**2
Cr1=C1#Flow velocity through impeller at inlet in m/s
Q=A*Cr1#Volume flow through the pump in m**3/s
r2=D2/2#Outer radius of the impeller in m
Cr2=(r1*Cr1)/r2#Flow velocity through impeller at outlet in m/s
U2=w*r2#Outlet tangential impeller velocity in m/s
Wx2=Cr2/tan(b22*pi/180)#Exit relative velocity in m/s
Hm=nh*E#Total stagnation head developed by the pump in m
dP021=Hm*d*g*10**-3#Total pressure head coefficient in kPa
Cx2=U2-(Cr2/tan(b22*pi/180))#Absolute whirl velocity in m/s
C2=(Cr2**2+Cx2**2)**(1/2)#Velocity at impeller exit in m/s
dP21=(Hm-(((C2**2)-(C1**2))/(2*g)))*d*g*10**-3#The static pressure head in kPa
P=d*g*Q*Hm*10**-3#Power given to the fluid in kW
Ps=P/nh#Input power to impeller in kW

#output
print '(a)Volume flow rate through the impeller is %3.4f m**3/s\n(b)\n    stagnation pressure rise across the impeller is %3.1f kPa\n    Static pressure rise across the impeller is %3.1f kPa\n(c)Power given to fluid is %3.2f kW\n(d)Input power to impeller is %3.2f kW'%(Q,dP021,dP21,P,Ps)

(a)Volume flow rate through the impeller is 0.0264 m**3/s
(b)
stagnation pressure rise across the impeller is 280.9 kPa
Static pressure rise across the impeller is 145.6 kPa
(c)Power given to fluid is 7.42 kW
(d)Input power to impeller is 8.34 kW


## Ex 8.12 Page 368¶

In :
from math import pi, tan
from __future__ import division
#input data
Ns=0.075#Specific speed in rev
b22=(180-120)#Outlet angle with the normal in degree
H=35#Distance to which pumping of water is done in m
Dp=0.15#Diameter of suction and delivery pipes in m
L=40#Combined length of suction and delivery pipes in m
WD=1/10#Width to diameter ratio at outlet of impeller
f=0.005#Friction factor
g=9.81#Acceleration due to gravity in m/s**2
nh=0.76#Hydraulic effficiency neglecting the slip
n=0.06#Percentage occupied by blades on circumference area

#calculations
A=(pi/4)*(Dp**2)#Area of flow in pipe in m**2
V=Q/A#Velocity in the pipes in m/s
OL=3*V**2/(2*g)#Other loses in the pipes in m
TL=(4*f*L*V**2/(2*g*Dp))+(OL)#Total loses in a pipe in m
N=(Ns*((g*TH)**(3/4)))/((Q)**(1/2))#The speed of the pump in rev/s
from sympy import symbols, solve
from sympy import N as NN
D = symbols('D')
Ao=pi*WD*(1-n)*D**2#Flow area perpendicular to impeller outlet periphery
Cr2=Q/Ao#Flow velocity through impeller at outlet in m/s
U2=pi*D*N#Outlet tangential impeller velocity in m/s
Cx2=(g*TH)/(U2*nh)#Absolute whirl velocity in m/s
expr = tan(b22*pi/180)-(Cr2/(Cx2-U2)) # polynomial of D
D = solve(expr, D) # discarding -ve values
D = D # Now discard imaginary part as negligible(in powers of e**-23)
D = NN(abs(D),3) # in meters # rounding off
#output
print 'The pump impeller diameter is %3.3f m'%(D)

The pump impeller diameter is 0.214 m


## Ex 8.13 Page 370¶

In :
#input data
N=2875#Speed of the pump in rpm
Q=57.2/3600#Discharge of the pump in m**3/s
Hm=42.1#Total head developed by the pump in m
d=1000#Density of the water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2
n=0.76#Efficiency of the pump

#calculations
Ns=(N*Q**(1/2))/(Hm**(3/4))#Specific speed of the pump
P=((d*g*Q*Hm)/n)*10**-3#Power input in kW

#calculations
print '(a)Specific speed of the pump is %3.f\n(b)Power input is %3.3f kW'%(Ns,P)

(a)Specific speed of the pump is  22
(b)Power input is 8.634 kW


## Ex 8.14 Page 371¶

In :
from math import ceil
#input data
D1=0.6#Inlet impeller diameter in m
D2=1.2#Outlet impeller diameter in m
N=200#Running speed of the pump in rpm
Q=1.88#Discharge of the pump in m**3/s
Hm=6#Head which the pump has to overcome in m
b22=26#Vane angle at exit at tangent to impeller in degree
d=1000#Density of the water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2

#calculations
U2=(3.1415*D2*N)/60#Outlet tangential impeller velocity in m/s
Wx2=Cr2/tan(b22*pi/180)#Exit relative velocity in m/s
Cx2=U2-Wx2#Absolute whirl velocity in m/s
nm=(Hm/(U2*Cx2/g))#Manometric efficiency
Nls=((2*g*Hm*60**2)/((3.1415**2)*((1.2**2)-(0.6**2))))**(1/2)#Least starting speed of the pump in rpm

#output
print '(1)Manometric efficiency is %0.1f %%\n(2)Least speed to start the pump is %3.2f rpm, rounding off = %0.f rpm'%(nm*100,Nls, ceil(Nls))

(1)Manometric efficiency is 63.0 %
(2)Least speed to start the pump is 199.40 rpm, rounding off = 200 rpm


## Ex 8.15 Page 372¶

In :
#input data
D2=1.25#External diameter of the impeller in m
D1=0.5#Internal diameter of the impeller in m
Q=2#Discharge of the pump in m**3/s
Hm=16#Head over which pump has to operate in m
N=300#Running speed of the pump in rpm
b22=30#Angle at which vanes are curved back in degree
Cr1=2.5#Flow velocity through impeller at inlet in m/s
Cr2=Cr1#Flow velocity through impeller at outlet in m/s
d=1000#Density of the water in kg/m**3
g=9.81#Acceleration due to gravity in m/s**2

#calculations
U2=(3.1415*D2*N)/60#Outlet tangential impeller velocity in m/s
Wx2=Cr2/tan(b22*pi/180)#Exit relative velocity in m/s
Cx2=U2-Wx2#Absolute whirl velocity in m/s
nm=(Hm*g)/(U2*Cx2)#Manometric or hydraulic efficiency
m=d*Q#Mass flow rate of water in kg/s
W=m*U2*Cx2*10**-3#Fluid power developed by the impeller in kW
Ps=W#Power required by the pump in kW neglecting mechanical loses
Nls=((2*g*Hm)/(((3.1415/60)**2)*(D2**2-D1**2)))**(1/2)#Minimum starting speed of the pump in rpm

#output
print '(a)Manometric or hydraulic efficiency is %0.1f %% \n(b)Power required by the pump is %3.2f kW\n(c)Minimum starting speed of the pump is %3.1f rpm'%(nm*100,Ps,Nls)

(a)Manometric or hydraulic efficiency is 52.2 %
(b)Power required by the pump is 600.98 kW
(c)Minimum starting speed of the pump is 295.4 rpm


## Ex 8.16 Page 373¶

In :
#input data
n=3#Number of stages
D2=0.4#Outlet impeller diameter in m
b2=0.02#Outlet impeller width in m
b22=45#Backward vanes angle at outlet in degree
dA=0.1#Reduction in circumferential area
nm=0.9#Manometric efficiency of the pump
Q=0.05#Discharge of the pump in m**3/s
N=1000#Running speed of the pump in rpm
n0=0.8#Overall efficiency of the pump
g=9.81#Acceleration due to gravity in m/s**2
d=1000#Density of water in kg/m**3

#calculations
A2=(1-dA)*3.1415*D2*b2#Area of flow at outlet in m**2
Cr2=Q/A2#Flow velocity through impeller at outlet in m/s
U2=(3.1415*D2*N)/60#Outlet impeller tangential velocity in m/s
Wx2=Cr2#Exit relative velocity in m/s as tand(b22)=1
Cx2=U2-Wx2#Absolute whirl velocity in m/s
Hm=(nm*U2*Cx2)/g#Head over which pump has to operate in m
H=n*Hm#Total head generated by the pump in m
P=d*g*Q*Hm*n#Power output from the pump in W
Ps=P/n0*10**-3#Shaft power input in kW

#output
print '(1)The head generated by the pump is %3.2f m\n(2)Shaft power input is %3.3f kW'%(H,Ps)

(1)The head generated by the pump is 107.98 m
(2)Shaft power input is 66.205 kW


## Ex 8.17 Page 374¶

In :
#input data
H=156#Total head operated by the pumps in m
N=1000#Running speed of the pump in rpm
Ns=20#Specific speed of each pump
Q=0.150#Discharge of the pump in m**3/s

#calculations
Hm=((N*(Q)**(1/2))/(Ns))**(4/3)#Head developed by each pump in m
n=H/Hm#Number of pumps

#output
print 'The number of pumps are %3.f'%(n)

The number of pumps are   3


## Ex 8.18 Page 375¶

In :
#input data
Q1=120#Discharge of each of the multi stage pump in parallel in first case in m**3/s
Q2=450#Discharge of the multi stage pump in second case in m**3/s
H1=16#Head of each stage in first case in m
D1=0.15#Diameter of impeller in first case in m
H=140#Total head developed by all pumps in second case in m
N1=1500#Running speed of the pump in rpm in first case
N2=1200#Running speed of the pump in rpm in second case
#calculations
H2=H1*((Q2/Q1)*((N2/N1)**2))**(4/6)#Head of each stage in second case in m
n=H/H2#Number of stages in second case
D2=D1*(((N1/N2)**(2))*(H2/H1))**(1/2)#Diameter of impeller in second case in m

#output
print '(a)number of stages required is %3.f\n(b)Diameter of impeller in the second case is %3.2f m or %0.f mm'%(n,D2, D2*1000)

(a)number of stages required is   5
(b)Diameter of impeller in the second case is 0.25 m or 251 mm


## Ex 8.19 Page 376¶

In :
#input data
H=36#Initial total head of the pump in m
Q1=0.05#Initial discharge of the pump in m**3/s
H2=3.5#Sum of static pressure and velocity head at inlet in m
P01=0.75#Atmospheric pressure initially in m of Hg
Pvap1=1.8*10**3#Vapour pressure of water initially in Pa
Pvap2=830#Vapour pressure of water finanlly in Pa
P02=0.62#Atmospheric pressure finally in m of Hg
g=9.81#Acceleration due to gravity in m/s**2
dW=1000#Density of water in kg/m**3
dHg=13.6#Density of mercury in kg/m**3

#calculations
NPSH=H2-((Pvap1)/(dW*g))#Net positive suction head in m
s=NPSH/H#Cavitation parameter when pump dvelops same total head and discharge
dH1=(P01*dHg)-(s*H)-(Pvap1/(dW*g))#The height reduced in initial condition above supply in m
dH2=(P02*dHg)-(s*H)-(Pvap2/(dW*g))#The height reduced in final condition above supply in m
Z=dH1-dH2#The total height which the pump must be lowered at new location in m

#output
print '(a)The cavitation parameter is %3.4f\n(b)\n    The height reduced in initial condition above supply is %3.1f m\n    The height reduced in final condition above supply is %3.2f m\n    The total height which the pump must be lowered at new location is %3.2f m'%(s,dH1,dH2,Z)

(a)The cavitation parameter is 0.0921
(b)
The height reduced in initial condition above supply is 6.7 m
The height reduced in final condition above supply is 5.03 m
The total height which the pump must be lowered at new location is 1.67 m


## Ex 8.20 Page 377¶

In :
from math import degrees, atan
#input data
Dt=1#Impeller outlet diameter in m
Dh=0.5#Diameter of the boss in m
Ns=38#Specific speed of the pump
Ca=2#Velocity of the flow in m/s
H=6#Head which the pump has to drive in m

#calculations
A=(3.1415/4)*(Dt**2-Dh**2)#Area of flow in m**2
Q=A*Ca#Discharge of the pump in m**3/s
N=(Ns*H**(3/4))/(Q**(1/2))#Pump speed in rpm
b1=degrees(atan(Ca/U1))#Vane angle at the entry of the pump when the flow is axial at inlet in degree

#output
print '(a)Pump speed is %0.2f rpm\n(b)Vane angle at the entry of the pump when the flow is axial at inlet is %3.2f degree'%(N,b1)

(a)Pump speed is 134.22 rpm
(b)Vane angle at the entry of the pump when the flow is axial at inlet is 29.65 degree


## Ex 8.21 Page 378¶

In :
#input data
Q=0.180#Discharge of the pump in m**3/s
H=2#Head developed by the pump in m
Ns=250#Specific speed of the pump
SR=2.4#Speed ratio of the pump
FR=0.5#Flow ratio of the pump
g=9.81#Acceleration due to gravity in m/s**2

#calculations
N=(Ns*(H**(3/4)))/(Q**(1/2))#Pump speed in rpm
U=SR*(2*g*H)**(1/2)#Peripheral velocity in m/s
D=(60*U)/(3.1415*N)#Runner diameter of the pump in m
Ca=FR*(2*g*H)**(1/2)#Velocity of flow in m/s
Dh=((D**2)-(Q*4/(Ca*3.14)))**(1/2)#Boss diameter of the pump in m

#output
print '(a)Pump speed is %3.i rpm\n(b)Runner diameter of the pump is %3.2f m\n(c)Boss diameter of the pump is %3.2f m\n'%(N,D,Dh)

(a)Pump speed is 991 rpm
(b)Runner diameter of the pump is 0.29 m
(c)Boss diameter of the pump is 0.10 m



## Ex 8.22 Page 379¶

In :
#input data
Hs=2.5#Height of the pipe above suction reservoir in m
H1=18#Height of the pipe below supply reservoir in m
H=2.7#Total height through which the pump lifts water in m
Q1=2.75#Discharge of water used from supply reservoir in l/s
Qt=7.51#Discharge of water totally delivered in l/s

#calculations
Hd=H-Hs#Height of the pipe from discharge reservoir in m
Qs=Qt-Q1#Discharge of water in delivery reservoir in l/s
nj=(Qs/Q1)*((Hs+Hd)/(H1-Hd))#Jet pump efficiency

#output
print 'The efficiency of the jet pump is %0.1f'%(nj*100)

The efficiency of the jet pump is 26.3