Chapter 9 : Reciprocating Pump

Example 9.1 Page No : 191

In [6]:
import math 

# Variables
D = 0.15   # cm
S = 0.25   # cm
N = 50.     # r.p.m
Hs = 5.  # m
Hd = 15     # m
ns = .6     
deltaD = 0.75
w = 9810.

# Calculations 
suction = w*math.pi*D**2*Hs/(4*ns) 
delivery = (w*math.pi*D**2*Hd)/(4*deltaD)
p_required = ((suction+delivery)*S*N)/(1000*60)


# Results 
print "Power required by the pump : %.4f kW"%p_required

# note : book answer is wrong.

Power required by the pump : 1.0233 kW

Example 9.2 page no : 192

In [8]:
import math

# Variables
D = 0.18
s = 0.36
Hs = 3.
Hd = 45.
N = 50.
n = 0.85

# Calculations 
a = math.pi*D*D/4
Q = (2*a*s*N)/60
g = 9.81
w = g*1000
P = w*Q*(Hs+Hd)/(n*1000)

# Results 
print "power in kw required to drive the pump : %.3f"%P

power in kw required to drive the pump : 8.458

Example 9.3 pageno : 192

In [10]:
import math

# Variables
D = 0.15
s = 0.3
Hs = 3.
Hd = 30.
n = 0.8
a = math.pi*D*D/4
N = 60./60
w = 9810.
Q = 0.62/60

# Calculations 
Qth = (2*a*s*N)
slip = (Qth-Q)/Qth
power = (w*Qth*(Hs+Hd))/(1000*n)

# Results 
print "power in Kw required to drive the pump : %.2f kW \
\npercentage slip : %.3f %%"%(round(power,2),round((slip*100),2))

power in Kw required to drive the pump : 4.29 kW 
percentage slip : 2.540 %

Example 9.4 Page No : 193

In [11]:
import math

# Variables
D = 0.15          #mm
s = 0.3           #mm 
N = 50./60        #r.p.m 
H = 25.           #m, height
Qact = 0.0042     #liter/s   
Ld = 22.          #m long  
d = 0.1           #mm  

# Calculations 
a = math.pi*D*D/4
Qth = a*s*N
w = 9810
power = w*Qth*H/1000
slip = (Qth-Qact)/Qth
W = 2*math.pi*N
a1 = math.pi*d*d/4
g = 9.81
Had = (Ld*a*W*W*s)/(g*a1*2)

# Results 
print "theoritical discharge : %.8f m**3/s \
\ntheoritical power : %.5f kW  \
\npercentage slip : %.2f %% \
\nacceleration head : %.2f m"%(Qth,round(power,5),round((slip*100),2),round(Had,2))

theoritical discharge : 0.00441786 m**3/s 
theoritical power : 1.08348 kW  
percentage slip : 4.93 % 
acceleration head : 20.75 m

Example 9.5 Page No : 194

In [13]:
import math

# Variables
s = 0.15        #m, strok length
Ls = 7.         #7 pipe
ds = 0.075      #cylinder    
N = 75./60       
Hs = 2.5        #m, water level
z = 16./9        
f = 0.01
W = 2*math.pi*N
g = 9.81

# Calculations 
Has = Ls*z*W*W*ds/g
H = Hs+Has
H1 = Has-Hs
Hfs = (4*f*Ls/(ds*2*g))*((z*W*ds)**2)
H2 = Hfs+Hs

# Results 
print "pressure head:beginning of suction stroke : %.4f m \
\nend of the suction stroke : %.4f m \
\nmiddle of the suction stroke : %.4f m"%(round(H,4),round(H1,4),round(H2,4))

pressure head:beginning of suction stroke : 8.3688 m 
end of the suction stroke : 3.3688 m 
middle of the suction stroke : 2.7087 m

Example 9.6 Page No : 196

In [16]:
import math

# Variables
D = 0.08                #m diameter
s = 0.15                #m stroke  
Hs = 3.                 # water for sump
ds = 0.03               #diameter  
g = 9.81                 
Ls = 4.5                
p = 78.86*(1000)       #kN/m**2  
w = 9810.                
W = 2*math.pi/60

# Calculations 
z = (D/ds)**2
Hsep = p/w
Habs = 10.3-Hsep
Has = Hsep-Hs
N = ((Has*g*2)/(z*W*W*s*Ls))**0.5

# Results 
print "maximium speed in rpm at which may run without separation",round(N,2),"r.p.m"

# note : rounding off error

maximium speed in rpm at which may run without separation 43.34 r.p.m

Example 9.7 Page No : 197

In [17]:
import math

# Variables
Hs = 5.
Ls = 10.
D = 0.15             #m diameter
d = 0.1              
N = 30./60
s = 0.15             #m delivery head
g = 9.81
W = 2*math.pi*N
w = 9810.
ha = 10.3

# Calculations 
z = (D/d)**2
H = (Ls*z*W*W*s/g)
Ph = Hs+H
Phabs = ha-Ph
f = 0.01
Hfs = (4*f*Ls/(d*2*g))*((z*W*s)**2)
H1 = Hs+Hfs
H1abs = ha-H1
H2 = Hs-H
H2abs = ha-H2
Hd = 15.
Ld = 25.
H11 = (Ld*z*W*W*s/g)
H12 = H11+Hd
H12abs = ha+H12
Hfd = (4*f*Ld/(d*2*g))*((z*W*s)**2)
H22 = Hd+Hfd
H22abs = ha+H22
H3 = Hd-H11
H3abs = ha+H3
a = 3.142*D*D/4
Q = a*s*2*N
power = (w*Q*(Hs+Hd+(0.6666*Hfs)+Hfd*0.6666))/1000

# Results 
print "pressure head at middle and end of suction stroke",round(H1abs,4),round(H2abs,4)
print "pressure head at beginning,middle,end of suction stroke",round(H12abs,4),round(H22abs,3),round(H3abs,4)
print "power in Kw required to drive the pump",round(power,3)

# note : rounding off error.

pressure head at middle and end of suction stroke 5.0708 8.6955
pressure head at beginning,middle,end of suction stroke 33.7888 25.873 16.8112
power in Kw required to drive the pump 0.534
In [ ]: