In [1]:

```
# Example 12.1
from math import *
from __future__ import division
# Given
Q = 0.25 # discharge from the pump in m**3/s
gma= 0.8*9810 # specific weight in kg/m**3
H=25 # elevation head in m
T = 350 # Torque to drive the shaft in Nm
N = 1800 # Speed in RPM
w = 2*pi*N/60 # angular velocity
# Solution
Eff = gma*Q*H*100/(T*w) # efficiency
print "Efficiency of th pump =",round(Eff,0),"%"
```

In [3]:

```
# Example 12.2
from math import *
from __future__ import division
# Given
d = 0.4 # diameter of the pump in m
b = 0.03 # width in m
theta = pi/3 # blade angle
N = 1500 # speed in RPM
Q = 0.4 # flow rate in m**3/s
g = 9.81 # acceleration due to gravity in m/s**2
# Solution
w = 2*pi*N/60 # anggular velocity in rad/s
u2 = (d/2)*w # blade velocity in m/s
V2r = Q/(2*pi*(d/2)*b) # relative velocity in m/s
print "(a)"
print "Radial velocity at exit =",round(V2r,1),"m/s"
V2t = u2 - V2r*(cos(theta)/sin(theta))
print "Whirl velocity = ",round(V2t,1),"m/s"
v2 = V2r/sin(theta)
print "Relative velocity = ",round(v2,2),"m/s"
V2 = sqrt(V2t**2+V2r**2)
print "Actual velocity =",round(V2,2),"m/s"
print "(b)"
H = u2*V2t/g
print"Head added for no inlet whirl =",round(H,0),"m"
print "(c)"
P = g*Q*H
print "Power required =",round(P,1),"kW"
```

In [4]:

```
# Example 12.3
from math import *
from __future__ import division
# Given
d = 0.36 # diameter of the impeller of pump
N = 1500 # Speed of impeller in RPM
# Solution
# For best efficiency
Q1 = 82 # discharge in l/s
H1 = 17.5 # Head in m
Eta = 0.8 # efficiency
Q2 = 100 # discharge in l/s
H2 = 20 # head in m
# Solving the simulataneous equation we get
D2 = 38.45
print "Impeller size =",round(D2,2),"cm"
N2 = 1500
print "Speed of the pump =",round(N2,0),"RPM"
```

In [5]:

```
# Example 12.4
from math import *
from __future__ import division
# Given
q = 500 # discharge in cgm
Q = 500/449 # discharge in ft**3/s
D = 0.667 # diameter in ft
A = pi*D**2/4
V = Q/A # velocity in ft/s
g = 32.2 # acceleration due to gravity in ft/s**2
N = 1800 # speed in RPM
# Solution
# for water at 65 deg C
nu = 1.134*10**-5 # viscosity in ft**2/s
e = 0.00085 # epssilon in ft
r = 0.001275
R = V*D/nu # reynolds no
f = 0.022 # from moody's diagram
Hl = V**2*(12.1+(f*224.9))/64.4
hs = 119.4 + Hl
print "Discharge, Q = ",round(Q,2),"ft**3/s"
print "Dynamic head of the pump, H =",round(hs,1),"m"
Ns = N*sqrt(q)/(hs)**(3/4)
print "Specific speed of the pump, Ns =",round(Ns,0),"RPM"
```

In [6]:

```
# Example 12.5
from math import *
from __future__ import division
# Given
H = 60 # height in m
Pb = 98*10**3 # barometric pressure in N/m**2
Hl = 1 # head in m
Pv = 1707 # vapour pressure
sigma = 0.08
w = 9810 # specific weight
# Solution
Npsh_m = sigma*60 # minimum NPSH
Hsm = (Pb/w)-(Pv/w)-Npsh_m-Hl
print "Minimum value of static suction lift = ",round(Hsm,2),"m"
```

In [ ]:

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