Chapter 11 : Pumping Machinery¶

Example 11.1 Page No : 223¶

In [2]:
#initialisation of variables

import math
h= 75. 	#ft
e= 0.75
k= 0.01
Q= 3000. 	#gal/min
k1= 1.2
N= 1500.
g= 32.2 	#ft/sec**2
D= 0.836 	#ft

#CALCULATIONS
W= h/e
v1= math.sqrt((W-h)/k)
Q1= Q/374.06
f1= Q1/(k1*D**2)
u1= math.pi*D*N/60
w1= W*g/u1
B= math.degrees(math.atan((f1/(u1-w1))))

#RESULTS
print  'Diameter of impeller = %.3f ft '%(D)
print  ' Blade angle at outlet edge of impeller = %.f degrees '%(B)

Diameter of impeller = 0.836 ft
Blade angle at outlet edge of impeller = 30 degrees


Example 11.3 Page No : 226¶

In [4]:
import math

#initialisation of variables
V= 150. 	#ft**3/sec
A1= 750. 	#r.p.m
di= 21. 	#in
do= 30. 	#in
v= 50. 	#ft/sec
A= 70. 	#degrees
w= 4.	#in
p= 30. 	#per cent
p1= 25. 	#per cent
sv= 12.8 	#ft**3/lb
g= 32.2 	#ft/sec**2

#CALCULATIONS
u= A1*2*math.pi*di/(24*60)
u1= A1*2*math.pi*do/(24*60)
f1= V/(math.pi*(do/12)*(1./3))
v1= math.sqrt(f1**2+w1**2)
h= 30*v1**2/(100*2*g)
Nh= v1**2/(20*2*g)
Prt= P+Nh
W= u1*w1/g
e= Prt*100/W
Power= Prt*V/(sv*550)

#RESULTS
print 'Total pressure rise = %.1f ft of air'%(Prt)
print ' manometric efficiency = %.1f percent'%(e)
print  ' Power = %.2f hp '%(Power)

#The answer is a bit different due to rounding off error in textbook

Total pressure rise = 137.9 ft of air
manometric efficiency = 58.5 percent
Power = 2.94 hp


Example 11.4 Page No : 228¶

In [5]:
import math

#initialisation of variables
g= 32.2 	#ft/sec**2
u1= 90. 	#ft/sec
w1= 70. 	#ft
e= 0.8
h1= 10. 	#ft
h2= 16. 	#ft
h3= 5. 	#ft
k= 2./5
f1= 20. 	#ft/sec
f= 18.   	#ft/sec
a= 45. 	    #degrees
x1= 164.4 	#ft

#CALCULATIONS
Hm= u1*w1/g
Hm1= e*Hm
lh= Hm-Hm1-h1-h2-h3
vg= k*math.sqrt(f1**2+w1**2)
pr1= x1-pr
ge= pr1*g*2*100/(vg/k)**2

#RESULTS
print  'manometer Head = %.1f ft '%(Hm1)
print  ' outlet velocity from guides = %.1f ft/sec '%(vg)
print  ' Pressure rise through impeller only = %.1f ft '%(pr)
print  ' Guide balde efficiency = %.f per cent '%(ge)

manometer Head = 156.5 ft
outlet velocity from guides = 29.1 ft/sec
Pressure rise through impeller only = 102.4 ft
Guide balde efficiency = 75 per cent


Example 11.6 Page No : 231¶

In [6]:
import math

#initialisation of variables
D1= 7.5 	#in
Q1= 850. 	#gal/min
p1= 62.4 	#lb/ft**3
N1= 1800.
D2= 15. 	#in
Q2= 12000. 	#gal/min
p2= 64. 	#lb/ft**3
N1= 1800. 	#r.p.m
H1= 14. 	#lb/ft**2

#CALCULATIONS
N2= Q2*N1*(D1)**3/(Q1*D2**3)
P1= p1*H1/144
P2= P1*N2**2*D2**2*p2/(N1**2*p1*D1**2)

#RESULTS
print  'N2 = %.f r.p.m '%(N2+4)
print  ' P2 = %.f lb/in**2 '%(P2)

N2 = 3180 r.p.m
P2 = 78 lb/in**2


Example 11.8 Page No : 234¶

In [7]:
#initialisation of variables
r= 5.

#CALCULATIONS
sr= r**2
sr1= r**2/r

#RESULTS
print  'Corresponding ratio = %.f '%(sr)
print  ' Corresponding ratio = %.f '%(sr1)

Corresponding ratio = 25
Corresponding ratio = 5


Example 11.9 Page No : 236¶

In [3]:
import math

#initialisation of variables
e= 0.88
w= 1.25 	#in
d= 10.   	#in
q= 630.  	#gal/min
a= 40.   	#degrees
g= 32.2 	#ft/sec**2
e1= 0.83

#CALCULATIONS
Q= q/(6.24*60)
f1= Q/(e*math.pi*(d/12)*(w/12))
u1= 1000*(w*4/12)*2*math.pi/60
W= u1*w1/g
mh= e1*W
p= mh-lr
v1= math.sqrt(f1**2+w1**2)
ke= v1**2/(2*g)
pke= p*100/ke
me= 100*lr/W

#RESULTS
print  'Velocity of flow = %.f ft/sec'%(f1)
print  ' Work done = %.1f ft-lb/lb'%(W)
print  ' manometric efficiency = %.1f ft'%(mh)
print  ' Pressure recovered = %.1f ft head'%(p)
print  ' Kinetic energy discharge = %.f ft-lb/lb'%(ke)
print  ' Percentage of kinetic energy recovered = %.1f per cent'%(pke)
print  ' manometric efficiency = %d percent'%(me)

#The answer is a bit different due to rounding off error in textbook

Velocity of flow = 7 ft/sec
Work done = 47.8 ft-lb/lb
manometric efficiency = 39.7 ft
Pressure recovered = 11.2 ft head
Kinetic energy discharge = 20 ft-lb/lb
Percentage of kinetic energy recovered = 55.7 per cent
manometric efficiency = 59 percent


Example 11.10 Page No : 239¶

In [11]:
import math

#initialisation of variables
W1= 7640. 	#gal/min
W2= 11400. 	#gal/min
Hm= 63. 	#ft
Hm1= 80. 	#ft
ep1= 72. 	#per cent
ep2= 76. 	#per cent

#CALCULATIONS
whp1= W1*Hm/(60*550)
whp2= W2*Hm1/(60*550)
bhp1= whp1*100/ep1
bhp2= whp2*100/ep2
w1= W2/10

#RESULTS
print  'For both pumps discharge = %.f gal/min against an 80-ft head'%(W2)
print  ' delivery from one pump = %.1f h.p '%(bhp1)
print  ' delivery from two pumps = %.1f h.p '%(bhp2)

For both pumps discharge = 11400 gal/min against an 80-ft head
delivery from one pump = 20.3 h.p
delivery from two pumps = 36.4 h.p


Example 11.11 Page No : 241¶

In [12]:
import math

#initialisation of variables
h= 94.   	#ft
w= 62.4 	#lb/ft**3
e= 0.58
p= 73.5 	#per cent

#CALCULATIONS
WHP= h*e*w/550
BHP= WHP/(p/100)

#RESULTS
print 'W.H.P= %.2f h.p'%(WHP)
print ' Brake horse power= %.1f'%(BHP)

W.H.P= 6.19 h.p
Brake horse power= 8.4


Example 11.12 Page No : 243¶

In [13]:
import math

#initialisation of variables
sl= 12. 	#ft
l= 20.  	#ft
d= 4.   	#in
dp= 6.  	#in
lst= 18. 	#in
k= 0.025
H= 32.   	#ft
g= 32.2 	#ft/sec**2
pf= 6.   	#ft
a= 33.83
a1= 9.53

#CALCULATIONS
A= math.sqrt((H-sl-d)*g/a)*a1
Q= 2*math.pi*(dp/12)**2*lst/(12*4*60)
v= Q/(math.pi*(d/12)**2/4)
kh= v**2/(2*g)
fh= k*l*v**2*12/(2*g*d)
N= math.sqrt((H-sl-pf)/(kh+fh))

#RESULTS
print  'premissible speed = %.1f r.p.m'%(A)
print  ' maximum premissible speed = %.1f r.p.m'%(N)

#The answer is a bit different due to rounding off error in textbook

premissible speed = 37.2 r.p.m
maximum premissible speed = 168.8 r.p.m


Example 11.13 Page No : 245¶

In [1]:
import math

#initialisation of variables
b= 6.    	#in
s= 12. 	    #in
d= 4. 	    #in
a1= 30. 	#degrees
a2= 90. 	#degrees
a3= 120. 	#degrees
N= 120. 	#r.p.m
n= 4.
#calculations
A= 2*math.pi*N/60
V= math.pi*(b/12)**2*n/4
v= (b/12)**2*A*(b/12)/(d/12)**2
Q4= V-Q1
Q5= Q2-V
Q6= Q3-V
a4= math.degrees(math.asin(V/(v*math.pi*(d/12)**2)))+a1
A= 180-a4

#RESULTS
print  'rate of flow at a1 = %.3f cuses'%(Q4)
print  ' rate of flow at a2 = %.3f cuses'%(Q5)
print  ' rate of flow at a3 = %.3f cuses'%(Q6)
print  ' crak angle = %.1f degrees'%(a4)
print  ' crak angle = %.1f degrees'%(A)

#The answer is a bit different due to rounding off error in textbook

rate of flow at a1 = 0.169 cuses
rate of flow at a2 = 0.448 cuses
rate of flow at a3 = 0.283 cuses
crak angle = 39.2 degrees
crak angle = 140.8 degrees


Example 11.14 Page No : 247¶

In [2]:
import math

#initialisation of variables
n= 2. 	#strokes/sec
dp= 6. 	#in
ds= 18. 	#in
ds1=4. 	#in
l= 20. 	#ft
l1= 20. 	#ft
f= 0.008
la= 5. 	#ft
A= 60. 	#r.p.m
f= 0.008
w= 62.4 	#lb/ft**3
g=32.2

#CALCULATIONS
V= math.pi*(ds/12)*n*(dp/12)**2/4
vmp= 2*math.pi*A*(ds/24)/60
vmp1= vmp*(dp**2/ds1**2)
hfmax= 4*f*(l-la)*vmp1**2/(2*g*ds1/12)
H1= round(2*hfmax/3,1)
H2= H1*13
Wls= (H1+H2)*w*math.pi/16*1.5*2
mv= V/(math.pi*(ds1/12)**2/4)
lh= round(4*f*(l-la)*mv**2/(2*g*(ds1/12)),2)
lhf= 12*lh
Wls1= (lh+13.21)*w*math.pi*1.5/16 *2
WS= Wls-Wls1

#RESULTS
print 'Work lost per second= %.f ft lb/sec'%(Wls)
print  ' Work saved per second = %.f ft-lb/sec'%(WS)

#The answer is a bit different due to rounding off error in textbook

Work lost per second= 875 ft lb/sec
Work saved per second = 352 ft-lb/sec


Example 11.15 Page No : 248¶

In [16]:
import math

#initialisation of variables
d= 7.5  	#in
s= 15.   	#in
l= 36.  	#ft
h1= 34. 	#ft
h2= 12. 	#ft
L= 10. 	    #ft
g= 32.2 	#ft/sec**2
f= 0.008
l1= 20. 	#ft
d1= 4.  	#in
h3= 110. 	#ft
w= 62.4 	#lb/ft**3
l2= 180. 	#ft

#CALCULATIONS
Q= (math.pi/4)*(d)**2*(s/12)*2*(l/60)/144
v= Q/((math.pi/4)*(d1/12)**2)
a= (d/4)**2*(d/12)*(l*2*math.pi/60)**2
H= h1-h2-(L*a/g)-(v**2*0.5/g)-(4*f*l1*v**2/(2*g*(d1/12)))
H1= h1+h3+(L*a/g)+(v**2*0.5/g)+(4*f*l2*v**2/(2*g*(d1/12)))
dh= (H1-H)*w/144
NP= dh*(math.pi/4)*d**2

#RESULTS
print  'Head at piston = %.2f ft of water absolute'%(H)
print  ' Head at piston = %.2f ft of water absolute'%(H1)
print  ' Difference on head of piston = %.f lb/in**2'%(dh)
print  ' Net load on piston = %.f lb'%(NP)

#The answer is a bit different due to rounding off error in textbook

Head at piston = 11.04 ft of water absolute
Head at piston = 161.59 ft of water absolute
Difference on head of piston = 65 lb/in**2
Net load on piston = 2882 lb


Example 11.16 Page No : 250¶

In [15]:
import math
from numpy import *
from numpy.linalg import *

#initialisation of variables
f= 0.009
dc= 3.5 	#in
ds= 6. 	#in
r= 0.25
sl= 8. 	#ft
d= 2.5 	#in
l= 14. 	#ft
el= 8. 	#ft
ed= 22.5 	#in
ph= 4. 	#ft
g= 32.2 	#ft/sec**2
f= 0.009

#CALCULATIONS
BC= el+l
v= math.sqrt(BC*g/(l*(d/2)*(r)*(dc/d)**2))*9.55
vec=roots([2,1/r,-1])
H1= 77

pump speed = 86.8 r.p.m