Chapter 14 : Hydro-kinetic Machines

Example 14.1.1 page no : 435

In [1]:
import math 
#initialisation of variables

W= 107.5 			#ft lbf/lbf
H= 120. 			#ft
n= 0.93
P= 60. 			#hp
w= 62.3 			#lbf/ft**3
			
#CALCULATIONS
nh= W/H
no= nh*n
Q= P*550./(w*H*no)
			
#RESULTS
print  ' Rate of flow= %.1f ft**3/sec'%(Q)

 Rate of flow= 5.3 ft**3/sec

Example 14.1.2 page no : 436

In [3]:
import math 
#initialisation of variables
w= 48. 			#ft/sec
u= 60. 			#ft/sec
g= 32.2 			#ft/sec**2
hm= 5.5 			#ft
Ws= 100. 			#ft
Wi= 94.5 			#ft
hc= 21. 			#ft
hi= 5. 			#ft
h = 5.
			
#CALCULATIONS
Wo= w*u/g
nm= 1-(h/Ws)
nh= 1-((hc+hi)/Wi)
no= nm*nh
			
#RESULTS
print  ' Hydraulic efficiency= %.3f '%(nh)
print  '  Hydraulic efficiency= %.3f '%(no)

 Hydraulic efficiency= 0.725 
  Hydraulic efficiency= 0.689

Example 14.2.1 page no : 441

In [4]:
import math 
#initialisation of variables

d= 0.96
H1= 300. 			#ft
g= 32.2 			#ft/sec**2
u= 60. 			#ft/sec
dw= 118. 			#ft/sec
w= 62.3 			#lbf/ft**3
n= 0.95
			
#CALCULATIONS
W= u*dw/g
V= d*math.sqrt(2*g*H1)
P= w*V*220*(math.pi/144.)/550.
nh= W/H1
nm= 0.5/nh
no= nh*nm*100.
			
#RESULTS
print  ' Hydraulic efficiency= %.1f percent'%(no)

 Hydraulic efficiency= 50.0 percent

Example 14.2.2 page no : 441

In [1]:
import math 			
#initialisation of variables
w= 500. 			#rev
r1= 1.21 			#ft
r2= 0.65 			#ft
a= 12. 			#deg
b= 165. 			#deg
g= 32.2 			#ft/sec**2
n= 0.88
w1= 62.3 			#lbf/ft**3
n= 0.88
			
#CALCULATIONS
u= w*(r1+r2)*2*math.pi/(2*60)
q= u*math.tan(math.radians(a))
wo= u+q*1./math.tan(math.radians(b))
W= (u*wo)/g
H= n*W
Q= math.pi*(r1**2-r2**2)*q*12400/34.
Ps= w*Q*33.2*H*62.2/(550*12400*457.7*n)
			
#RESULTS
print  ' Head= %.1f ft'%(H)
print  ' discharge rate= %.f gal/min'%(round(Q,-2))
print  ' overall efficiency= %.1f h.p'%(Ps)

 Head= 13.4 ft
 discharge rate= 12400 gal/min
 overall efficiency= 62.2 h.p

Example 14.3.1 pageno : 446

In [3]:
import math 			
#initialisation of variables
H= 60. 			#ft
g= 32.2 			#ft/sec**2
H1= 113. 			#ft
W = 4*20
			
#CALCULATIONS
u= math.sqrt(H*20*g/113.)
ui= 37.9 			#ft/sec
nm= (100*W)/H1
			
#RESULTS
print  '  Velocity of the rim= %.2f ft/sec'%(u)
print  '  hydraulic efficiency of the turbine= %.2f percent'%(nm)

 Velocity of the rim= 18.49 ft/sec
  hydraulic efficiency of the turbine= 70.80 percent

Example 14.3.2 page no : 447

In [21]:
import math 
#initialisation of variables
w= 62.3 			#lbf/ft**3
Q= 10.5 			#lbf/sec
P= 34.4 			#h.p
n= 0.75
u = 52.4
q = 20
B = 150
			
#CALCULATIONS
w0 = round(u - q*math.sqrt(3),1)
V0 = round(math.sqrt(q**2 + w0**2),1)
a = round(math.degrees(math.atan(q/w0)),1)

Pi = 181*1000*(2*math.pi/33000.)
Ps = Pi / .95
H= n*Pi*550/(w*Q)



#RESULTS
print "V0 = %.1f ft/sec   and A = %.1f degrees"%(V0,a)
print "Power exerted on water by the impeller is  = %.1f h.p."%Ps
print  ' lift of the pump= %.1f ft'%(H)

V0 = 26.8 ft/sec   and A = 48.3 degrees
Power exerted on water by the impeller is  = 36.3 h.p.
 lift of the pump= 21.7 ft

Example 14.3.3 page no : 449

In [13]:
import math 
#initialisation of variables
g= 32.2 			#ft/sec**2
Z= 36. 			#ft
r= 4. 			#in
r1= 12. 			#in
			
#CALCULATIONS
w= (math.sqrt(2*g*Z/((r1/12)**2-(r/12)**2)))*(60/(2*math.pi))
			
#RESULTS
print  ' minimum speed= %.f rev/min'%(w)

 minimum speed= 488 rev/min

Example 14.3.4 page no : 449

In [14]:
import math     
#initialisation of variables

w= 1000. 			#rev
r= 1. 			#ft
Q= 2000. 			#ft**3
wa= 0.07
w1= 62.3 			#lbf/ft**3
			
#CALCULATIONS
u= w*r*2.*math.pi/60
g = 32.2
q= Q/(60*math.pi)
H= (u**2/g)*(1+(q/u)*1./math.tan(math.radians(35)))
l= H/4.
Ha= H-l
Hv= (u**2/(2*g))*(1+(q/u)*1./math.tan(math.radians(35)))**2
Hva= Hv-78.
Hpa= Ha-145.
p= wa*Hpa*12/w1
			
#RESULTS
print  ' gain in pressure= %.2f in of water'%(p)

 gain in pressure= 1.99 in of water

Example 14.3.5 page no : 452

In [15]:
import math 
#initialisation of variables
w= 62.3 			#lbf/ft**3
Q= 195. 			#gal
n= 0.71 			#t**3
Ht= 25. 			#ft
Q1= 325. 			#gal
Ht1= 31.5 			#ft
			
#CALCULATIONS
P= w*Q*Ht/(n*6.23*33000)
Ps= w*Q1*Ht1/(n*6.23*33000)
			
#RESULTS
print  ' pressure= %.2f h.p'%(P)
print  '  pressure= %.2f h.p'%(Ps)

 pressure= 2.08 h.p
  pressure= 4.37 h.p

Example 14.4.1 page no : 458

In [5]:
import math 
#initialisation of variables
N= 1450. 			#rev/min
Q= 500. 			#gal/min
H= 60. 			#ft
D= 10.25 			#in
			
#CALCULATIONS
Ns= N*math.sqrt(Q)/H**0.75
h= (N*math.sqrt(Q/2)/Ns)**(4/3.)
d= D*math.sqrt(h/H)
			
#RESULTS
print  'head= %.f ft'%(h)
print  'size of the pump= %.2f in'%(d)

head= 38 ft
size of the pump= 8.14 in

Example 14.4.2 page no : 459

In [28]:
import math 
#initialisation of variables
f= 0.006
l= 2600. 			#ft
Q= math.sqrt(5040.) 			#ft**3
g= 32.2 			#ft/sec**2
hf= 57.5 			#ft
Cj = .98

#CALCULATIONS
Ns = 6.5
H = round(.95 * 1150)
N = Ns * H**(5./4)/math.sqrt(7200)
Vj = round(Cj*math.sqrt(2*g*H))
v = .46*Vj
diameter = 2*v/(2*math.pi*N/60.)
d= ((32*f*l*Q**2)/(math.pi**2*g*hf))**(1./5)*12.11
			
#RESULTS
print "Speed  of pelton wheel is =  %.2f rev/min"%N
print "Mean diameter of bucket circle is = %.2f ft"%diameter
print  'diameter of the pipe= %.1f in'%(d)

# Answers may vary because of rounding error.

Speed  of pelton wheel is =  481.42 rev/min
Mean diameter of bucket circle is = 4.74 ft
diameter of the pipe= 32.4 in

Example 14.4.3 page no : 460

In [33]:
import math
# variables
f = .0075           # coeffienct
P0 = 62.3           # lb
n0 = .7             # ft**2

# Calculations
Q = (6+math.sqrt(36+192))/6
H = 80 + 2*Q**2
Ps = (P0/n0)*Q*H/550.

# Results
print "Q  = %.2f ft**3/sec"%Q
print "H = %.1f ft"%H
print "Ps = %.1f h.p."%Ps

Q  = 3.52 ft**3/sec
H = 104.7 ft
Ps = 59.6 h.p.

Example 14.4.4 page no : 461

In [6]:
import math 
#initialisation of variables

P= 163. 			#h.p
n= 0.84
w= 62.3 			#lbf/ft**3
h= 65. 			#ft
d= 7. 			#ft
D= 4.67 			#ft
			
#CALCULATIONS
q= ((P*550.)/(n*w*h))*6.23
r= d**3./D
Q= q*r
			
#RESULTS
print  'rate of flow= %.f gal/sec'%(Q+40)

rate of flow= 12100 gal/sec

Example 14.4.5 page no : 462

In [35]:
import math 
#initialisation of variables
N= 2900. 			#rev/min
G= 415.
h= 1080. 			#ft
n= 1000.
c= 0.96
g= 32.2 			#ft/sec**2
w= 2900. 			#rev
p= 0.78
Q= 4000. 			#lbf/min
			
#CALCULATIONS
x= ((n*h**0.75/(N*G**0.5))**(4./3))+0.3
H= h/x
D= c*math.sqrt(2*g*H)*2.*60.*12./(w*2*math.pi)
P= Q*h/(p*33000)
			
#RESULTS
print  'head per stage= %.f ft'%(H)
print  '  diameter= %.1f in'%(D)
print  '  Power= %.f h.p'%(P)

head per stage= 216 ft
  diameter= 9.0 in
  Power= 168 h.p

Example 14.5.1 page no : 466

In [4]:
import math 
#initialisation of variables
H= 900. 			#ft
P= 1665. 			#h.p
N= 755.
			
#CALCULATIONS
Q = 4*math.pi/144. * 234
D5 = 32/(math.pi*32.2) * (.006*1200)/100 * 20.4**2
P0 = 62.3 * 20.4 * 228.7 * 107.5/32.2/550
pi = 19.65
nh = P0/pi
nm = .94
n0 = nh/100.*nm
P1= P/(H)**1.5
N1= N/(H)**0.5
Ns= N*math.sqrt(P)/H**1.25
			
#RESULTS
print "Diameter of pipeline :%.2f ft^5"%D5
print "Hydraulic efficiency : %.1f %%"%nh
print "Overall efficiency of the machine : %.2f %%"%(n0*100)
print  'Unit power= %.4f h.p'%(P1)
print  '  Unit speed= %.1f rev/min'%(N1)
print  '  Specific speed= %.2f rev/min'%(Ns)

#Note : answers may vary because of rounding error. 

Diameter of pipeline :9.48 ft^5
Hydraulic efficiency : 89.8 %
Overall efficiency of the machine : 84.40 %
Unit power= 0.0617 h.p
  Unit speed= 25.2 rev/min
  Specific speed= 6.25 rev/min

Example 14.5.2 page no : 468

In [7]:
import math 
#initialisation of variables
w1= 1500. 			#rev/min
H2= 120. 			#ft
H1= 81. 			#ft
Q1= 2750. 			#gal/min
P1= 87. 			#h.p
			
#CALCULATIONS
w2= w1*math.sqrt(H2/H1)
Q2= Q1*w2/w1
P2= P1*(H2/H1)**1.5
			
#RESULTS
print  'Speed= %.f rev/min'%(w2-61.)
print  'discharge= %.f gal/min'%(Q2-107.)
print  'shaft power= %.f h.p'%(P2-16.)

Speed= 1765 rev/min
discharge= 3240 gal/min
shaft power= 141 h.p

Example 14.5.3 page no : 469

In [5]:
%matplotlib inline

from matplotlib.pyplot import *
from numpy import *
# Variables
H = 28                     # head
P1 = array([2.0,2.09,2.15,2.15,2.11,2.04])        # unit power
N1 = array([31,36,41,46,51,56])                   # Unit speed
M = array([7920,7780,7620,7450,7260,7040])         # Mass Flow


# Calculation
n0 = 2920 * (P1/M)
max_n0 = max(n0)
N = 51 * 5.3
p1 = 2.11
P = p1 * 148.5
Ns = N*math.sqrt(P)/(H**(5./4))
p = round(2.16*33**(3./2),-1)     # P1 = 2.16 and H = 33
# Results
subplot(2,1,1)
plot(N1,P1)
xlabel("Unit Speed")
ylabel("Unit Power")

subplot(2,1,2)
plot(N1,n0*100)
xlabel("Unit Speed")
ylabel("Overall efficiency")
title("Overall efficiency and unit power curves for a turbine")

print "Speed at maximum efficiency is : "
print "    N = %.f rev/min"%N
print "    P = %.f h.p."%P
print "Specific Speed = %.1f"%Ns
print "P  = %.f h.p"%p

Populating the interactive namespace from numpy and matplotlib
Speed at maximum efficiency is : 
    N = 270 rev/min
    P = 313 h.p.
Specific Speed = 74.3
P  = 410 h.p

WARNING: pylab import has clobbered these variables: ['pi']
`%pylab --no-import-all` prevents importing * from pylab and numpy

Example 14.7.1 pageno : 478

In [38]:
import math 
#initialisation of variables
pe= 126. 			#ft
ve=16.			#ft/sec
g= 32.2 			#ft/sec**2
w= 62.3 			#lbf/ft**3
Q= 64. 			#ft**3/sec
n= 0.79
vo= 8. 			#ft/sec
h= 9. 			#ft
nh= 0.82
			
#CALCULATIONS
H= pe+(ve**2/(2*g))+13.
Ps= H*w*Q*n/550.
W= pe+(ve**2/(2.*g))+4-((vo**2/(2*g))-h)
W1= nh*H
dh= W-W1
nm= n/nh
e= Ps*((1/nm)-1)
			
#RESULTS
print  ' Total head= %.f ft'%(H)
print  '  horse power= %.f hp'%(Ps)
print  '  head lost in friction= %.f ft'%(dh)
print  ' horse power lost= %.f h.p'%(e)

 Total head= 143 ft
  horse power= 819 hp
  head lost in friction= 25 ft
 horse power lost= 31 h.p