# Chapter 10 : Centrifugal Pump¶

## Example 10.1 Page No : 210¶

In [11]:
import math

# Variables
N = 900./60
x1 = 90.
D1 = 0.2
D2 = 0.4
n = 0.7
g = 9.81
u1 = math.pi*D1*N
u2 = 2*u1 			# as D2 = 2D1
y1 = 20.

# Calculations
Vf2 = Vf1
Vr2 = Vr1
x = (Vr2*Vr2-Vf1*Vf1)**0.5
Vw2 = u2-x
B1 = 0.02
Q = math.pi*D1*B1*Vf1
H = Vw2*u2/g
w = 9810
P = (w*Q*Vw2*u2)/(g*1000)
inputpower = (w*Q*H)/(1000*n)
print "discharge through the pump %.4f litre/s \
\nheat developed %f m \
\npower in Kw at outlet %.3f \
\ninput power if overall efficiency is 70%% : %.4f kW" \
%(Q*1000,H,P,inputpower)

# note : rounding off error

discharge through the pump 43.1069 litre/s
heat developed 18.109366 m
power in Kw at outlet 7.658
input power if overall efficiency is 70% : 10.9401 kW


## Example 10.2 Page No : 212¶

In [2]:
# Variables
Hs = 2.
Hd = 20.
Hfs = 1.
Hfd = 5.
Q = 1./60
N = 1450./60
ds = 0.1
dd = ds
n = 0.75
g = 9.81
w = 9810.

# Calculations
a = 3.142*ds*ds/4
Vs = Q/a
Vd = Vs
Ht = Hs+Hd+Hfs+Hfd+(Vs*Vs/(2*g))+(Vd*Vd/(2*g))
Pi = (w*Q*Ht)/(n*1000)
Ns = ((N*(Q**0.5))/(Ht**0.75))*60

# Results
print "total head developed by the pump,power input to the pump,specific speed of pump in r.p.m",round(Ht,4),round(Pi,5),round(Ns,3)

total head developed by the pump,power input to the pump,specific speed of pump in r.p.m 28.4589 6.20404 15.192


## Example 10.3 Page No : 213¶

In [3]:
import math

# Variables
d2 = 0.6
Q = 20./60
N = 1400./60
V1 = 2.8
g = 9.81
y2 = 30.
w = 9810.
Vf1 = V1
Vf2 = V1

# Calculations
u2 = 3.142*d2*N
Vw2 = u2-x
Hm = Vw2*u2/g
P = (w*Q*Hm)/1000

# Results

head developed, pump power 309.5484 1012.2231


## Example 10.4 Page No : 214¶

In [4]:
# Variables
N = 1450./60
N1 = 1650./60
H = 12.
P = 6.

# Calculations
H1 = H*((N1/N)**2)
P1 = P*((N1/N)**3)

# Results
print "head developed and power required if pump runs at 1650 r.p.m",round(H1,4),round(P1,4)

head developed and power required if pump runs at 1650 r.p.m 15.5386 8.841


## Example 10.5 Page No : 215¶

In [6]:
# Variables
Q = 0.03
Hs = 18.
d = 0.1
l = 90.
n = 0.8
w = 9810.
a = 3.142*d*d/4
f = 0.04
g = 9.81

# Calculations
Vd = Q/a
H1 = (4*f*l*Vd*Vd)/(d*2*g)+(Vd*Vd/(2*g))
Hm = Hs+H1
P = (w*Q*Hm)/(n*1000)

# Results
print "power required to drive the pump",round(P,3),"kW"

power required to drive the pump 46.279 kW


## Example 10.6 Page No : 216¶

In [6]:
# Variables
Q = 0.04
Hm = 30.
n = 0.75
w = 9810.

# Calculations
p = w*Q*Hm/1000
P = p/n

# Results
print "output power of the pump,power required to drive the motor",p,P

output power of the pump,power required to drive the motor 11.772 15.696


## Example 10.7 Page No : 216¶

In [7]:
# Variables
Q = 1.8/60
d = 0.1
n = 0.72
Hs = 20.
w = 9810.
Hl = 8.

# Calculations
Hm = Hs+Hl
p = (w*Hm*Q)/1000
P = p/n
print "water power required to the pump,power required to run the pump",p,P

water power required to the pump,power required to run the pump 8.2404 11.445


## Example 10.8 Page No : 217¶

In [12]:
import math

# Variables
d2 = 0.6
Q = 15./60
N = 1450./60
V1 = 2.6
g = 9.81
y2 = 30.
w = 9810.
Vf1 = V1
Vf2 = V1

# Calculations
u2 = math.pi*d2*N
Vw2 = u2-x
Hm = Vw2*u2/g
P = (w*Q*Hm)/1000

# Results

head developed, pump power 190.6161 467.4859


## Example 10.9 Page No : 217¶

In [8]:
# Variables
Q = 0.05
p = 392.4*1000
n = 0.65
s = 0.8
w1 = 9810.

# Calculations
Hw = p/w1
Hoil = p/(w1*s)
Pw = (w1*Q*Hw)/(n*1000)
Poil = (w1*s*Q*Hoil)/(n*1000)

# Results
print "power in Kw to drive the pump with water and oil of s,p = 0.8",round(Poil,6),round(Pw,6)

power in Kw to drive the pump with water and oil of s,p = 0.8 30.184615 30.184615


## Example 10.10 Page No : 218¶

In [15]:
import math

# Variables
Q = 0.118
N = 1450./60
Hm = 25.
d2 = 0.25
B2 = 0.05
n = 0.75
g = 9.81

# Calculations
u2 = math.pi*d2*N
Vf2 = Q/(math.pi*d2*B2)
Vw2 = g*Hm/(n*u2)
y2 = math.degrees(math.atan(Vf2/(u2-Vw2)))

# Results
print "vane angle in degree at the outer nperiphery of the impeller",round(y2,2)

# note : rounding off error

vane angle in degree at the outer nperiphery of the impeller 59.75


## Example 10.11 Page No : 219¶

In [18]:
import math

# Variables
Hm = 14.5
N = 1000./60
y2 = 30.
d2 = 0.3
B2 = 0.05
g = 9.81
n = 0.95

# Calculations
u2 = math.pi*d2*N
Vw2 = g*Hm/(n*u2)
Q = math.pi*d2*B2*Vf2

# Results
print "discharge of pump in m3/sec if manometric efficiency if 95%% : %.3f litre/s"%(Q*1000)

discharge of pump in m3/sec if manometric efficiency if 95% : 168.024 litre/s


## Example 10.12 Page No : 220¶

In [21]:
import math

# Variables
d2 = 1.2
N = 200./60
Q = 1.88
Hm = 6.
y2 = 26.
g = 9.81
Vf2 = 2.5
d1 = 0.6
u2 = math.pi*d2*N

# Calculations
n = g*Hm/(Vw2*u2)
z1 = (math.pi*d2/60)**2
z2 = (math.pi*d1/60)**2
N1 = (Hm*2*g/(z1-z2))**0.5

# Results
print "least speed to start pump : %.3f r.p.m \
\nmanometric efficiency : %.2f %%"%(N1,(n*100))

least speed to start pump : 199.395 r.p.m
manometric efficiency : 62.95 %


## Example 10.13 Page No : 222¶

In [26]:
import math

# Variables
Q = 0.125
Hm = 25.
N = 660./60
d2 = 0.6
d1 = d2*0.5
a = 0.06
y2 = 45.
g = 9.81

# Calculations
u2 = math.pi*d2*N
u1 = u2*0.5
Vf2 = Q/a
n = g*Hm/(Vw2*u2)
Vf1 = Q/(a)
y1 = math.degrees(math.atan(Vf1/u1))

# Results
print "manometric efficiency %.2f %% \
\nvane angle at inlet : %.2f degrees"%((n*100),y1)

# note : rounding off error.

manometric efficiency 63.42 %
vane angle at inlet : 11.36 degrees


## Example 10.14 Page No : 223¶

In [28]:
import math

# Variables
n = 3.
d2 = 0.4
B2 = 0.02
y2 = 45.
da = 0.1
nm = 0.9
w = 9810.
no = 0.8
g = 9.81
N = 1000./60
Q = 0.05

# Calculations
Vf2 = Q/(math.pi*d2*nm*B2)
u2 = math.pi*d2*N
Hm = nm*Vw2*u2/g
Ht = n*Hm
P = w*Q*Ht/1000
Ps = P/no

# Results
print "shaft power in Kw %.2f"%Ps

shaft power in Kw 66.21


## Example 10.15 Page No : 225¶

In [12]:
# Variables
n = 6.
Q = 0.12
p = 5003.1*1000
N = 1450./60
w = 9810.

# Calculations
Ht = p/w
h = Ht/n
Ns = (N*(Q**0.5)/(h**0.75))*60

# Results
print "radial impeller would be selected",round(Ns,2)

radial impeller would be selected 17.94


## Example 10.16 Page No : 225¶

In [13]:
import math

# Variables
sg = 1.08
w = 9810.*sg
Q = 0.3
H = 12.
no = 0.75

# Calculations
P = w*Q*H/(no*1000)
p = w*H

# Results
print "power in Kw required by the pump,pressure developed by the pump in  N/m2",round(P,3),p

power in Kw required by the pump,pressure developed by the pump in  N/m2 50.855 127137.6


## Example 10.17 Page No : 226¶

In [14]:
# Variables
d1 = 0.3
N1 = 2000./60
Q1 = 3.
Hm1 = 30.
Q2 = 5.
N2 = 1500./60
Ht = 200.

# Calculations
Hm2 = ((N2/N1)*((Q2/Q1)**0.5)*(Hm1**0.75))**1.3333
n = Ht/Hm2
d2 = ((Hm2/Hm1)**0.5)*(N1/N2)*d1

# Results
print "number of stages and diameter of each impeller in cm",round(n,3),round((d2*100),2)

number of stages and diameter of each impeller in cm 6.96 39.15

In [ ]: