In [1]:

```
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.
```

In [2]:

```
from math import sin, tan
#input data
b2=30#Impeller blade angle to the tangent at impeller outlet in degree
d=0.02#Blade depth in m
D=0.25#Blade diameter in m
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)
```

In [3]:

```
#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)
```

In [4]:

```
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)
```

In [5]:

```
#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
U1=(3.1415*D1*N)/60#Blade inlet speed in m/s
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)
```

In [6]:

```
#input data
Hi=0.25#Vaccum gauge reading in m of Hg vaccum
P0=1.5#Pressure gauge reading in bar
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
Hm=((Po+Pi)/(dw*g))+((V0**2-Vi**2)/(2*g))+(Z01)#Manometric head in m
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)
```

In [7]:

```
#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)
```

In [8]:

```
#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
Hm=H+hf0+(V0**2/(2*g))#Manometric head 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)
```

In [9]:

```
#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
U2=(3.1415*D2*N)/60#Blade outlet speed 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)
```

In [10]:

```
#input data
Ihl=3#Head loss in impeller in m
Cr2=4.64#Flow velocity through impeller at outlet in m/s
U2=30#Blade outlet speed 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)
```

In [11]:

```
#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
b=0.064#Blade depth in m
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
E=(U2/g)*(U2-Wx2)#Theoretical head developed in m
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)
```

In [12]:

```
from math import pi, tan
from __future__ import division
#input data
Q=0.04#Discharge of the pump design in m**3/s
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
TH=TL+H#Total required head 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[2] # 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)
```

In [13]:

```
#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)
```

In [14]:

```
from math import ceil
#input data
D1=0.6#Inlet impeller diameter in m
D2=1.2#Outlet impeller diameter in m
Cr2=2.5#Radial flow velocity in m/s
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))
```

In [15]:

```
#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)
```

In [16]:

```
#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)
```

In [17]:

```
#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)
```

In [18]:

```
#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)
```

In [19]:

```
#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)
```

In [20]:

```
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
U1=(3.1415*Dh*N)/60#Blade inlet speed in m/s
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)
```

In [21]:

```
#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)
```

In [22]:

```
#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)
```