Chapter 3 : Flow of Water

Example 3.1 Page No : 67

In [1]:
		
#initialisation of variables
d1 = 1. 		#ft
d2 = 6. 		#in
h1 = 5. 		#ft
h2 = 15. 		#ft
Pa = 15. 		#lbs
v1 = 10. 		#ft/sec
w = 62.4 		#lbs/ft**3
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
v2 = v1/(d2/12)**2
Pb = (w*((Pa+(Pa*144/w)+(v1**2/(2*g)))-h1-(v2**2/(2*g))))/144
		
#RESULTS
print  'Pb = %.2f lbs/in**2 '%(Pb)  

Pb = 9.24 lbs/in**2

Example 3.2 Page No : 69

In [3]:
import math 
		
#initialisation of variables
d1 = 4. 		#ft
d2 = 2. 		#ft
h1 = 50. 		#ft
h2 = 45. 		#ft
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
r = (d1**2/d2**2)
v1 = round(math.sqrt((h1-h2)*2*g/(r**2-1)),1)
Q = v1*math.pi*d1**2/4

#RESULTS
print  'discharge through pipe = %.2f cubic feet per second '%(Q)

# rounding off error

discharge through pipe = 57.81 cubic feet per second

Example 3.3 Page No : 70

In [3]:
import math 
		
#initialisation of variables
z1 = 10. 		#/m
h1 = 10. 		#m
v1 = 12. 		#ft/sec
v2 = 4. 		#m/sec
k = 0.6
w = 62.4 		#lb/in**2
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
p = (w/144)*(z1+h1+(v1**2/(2*g))-(v2**2/(2*g))-(k*(v1-v2)**2/(2*g)))
		
#RESULTS
print  'pressure at bottom end  = %.2f lb/in**2'%(p)

pressure at bottom end  = 9.27 lb/in**2

Example 3.4 Page No : 73

In [4]:
import math 
		
#initialisation of variables
d = 4.   		#ft
d1 = 5./4 		#ft
g = 32.2 		#ft/sec**2
h = 3. 	    	#ft
K = 1.
		
#CALCULATIONS
C = (math.pi/4)*d**2*math.sqrt(2*g)/(math.sqrt((d**2/d1**2)**2-1))
Q = K*math.sqrt(h)*C
V = Q/(math.pi*d1**2/4)
		
#RESULTS
print  'Velocity at the throat = %.2f ft/sec '%(V)

# ronding off error

Velocity at the throat = 13.97 ft/sec

Example 3.5 Page No : 74

In [5]:
import math 
		
#initialisation of variables
d = 9. 		#in
d1 = 4. 		#in
g = 32.2 		#ft/sec**2
dh = 10. 		#in
sg = 13.6 
K = 1.
		
#CALCULATIONS
C = (((math.pi/4)**2*(d*d1)**2*math.sqrt(2*g)/144**2)/(math.sqrt((math.pi*d**2/12**2)**2-(math.pi*d1**2/12**2)**2)))+0.52
h = (sg-1)*dh/12
Q = K*C*math.sqrt(h)
		
#RESULTS
print  'Discharge passing through the pipe = %.2f cuses '%(Q)

Discharge passing through the pipe = 2.26 cuses

Example 3.6 Page No : 76

In [1]:
import math 
		
#initialisation of variables
sm = 13.6
so = 0.8
di = 8. 		#in
dt = 4. 		#in
K = 0.98
v = 1.  		#ft
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
s = sm/so
dp = v*12*(s-1)/12
A = math.pi*(di/12)**2/4
At = math.pi*(dt/12)**2/4
C = A*math.sqrt(2*g)/(math.sqrt((A/At)**2-1))
Q = C*math.sqrt(v*12+dt)*K
		
#RESULTS
print  'Discharge passing through the pipe = %.2f cuses '%(Q)

Discharge passing through the pipe = 2.84 cuses

Example 3.7 Page No : 77

In [7]:
import math 
		
#initialisation of variables
s = 1./10
d1 = 6. 		#in
d2 = 2. 		#in
l = 20. 		#in
p = 15. 		#lbs/in**2
p1 = 6. 		#lbs/in**2
K = 0.95
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
H = (l*s/12)-(p1*144/(2*g))+(p*144/(2*g))
C = math.sqrt(2*g)*(math.pi*(d1/12)**2)/(4*(math.sqrt((d1**2/d2**2)**2-1)))
Q = C*K*math.sqrt(H)*374.7

#RESULTS
print  'Discharge passing through the pipe = %.f Gallons/minute '%(Q)

# note : rounding off error at value of H in textbook. so answer is slightly different

Discharge passing through the pipe = 282 Gallons/minute

Example 3.8 Page No : 78

In [8]:
import math 
		
#initialisation of variables
d1 = 12. 		#in
Q = 4.25 		#ft**3/sec
h = 18. 		#ft
K = 0.98
g = 32.2 		#ft/sec**2
sm = 13.6
		
#CALCULATIONS
R = math.sqrt((K*math.sqrt(2*g)*math.sqrt(h)*(math.pi*(d1/12)**2/4)/Q)+1)
d2 = math.sqrt(d1**2/(144*R))
dh = (sm-1)*(h/(12*2))
d3 = Q*math.sqrt(dh/h)
		
#RESULTS
print  'Diameter of the throat = %.2f ft '%(d3)

Diameter of the throat = 3.08 ft

Example 3.9 Page No : 81

In [3]:
import math 
		
#initialisation of variables
R = 4. 		#in
r = 0.5 		#in
c = 0.007
K = 33.96
w = 62.4 		#lb/ft**3
pa = 12.13 		#lb/in**2
pb = 14.7 		#lb/in**2
w1 = 2.5 		#lbs
Q = 40. 		#gals/min
h = 1.86
		
#CALCULATIONS
va = Q*4*(2*r*12)**2/(6*w*math.pi)
vb = Q*(2*r*12)**2/(6*w*2*R*math.pi*0.32)
vx = vb*R/2
pu = 2*math.pi*w*h
pd = pb*math.pi*R**2
RP = pb*math.pi*R**2-2*math.pi*w*(0.5*K*((R/12)**2-(r/12)**2)-c*math.log(R/r))-pa*math.pi*r**2+w1
		
#RESULTS
print  'velocity va = %.1f ft/sec'%(va)
print  'velocity vb = %.2f ft/sec'%(vb)
print  'velocity vx = %.2f ft/sec'%(vx)
print  'pressure px = %.1f lbs/in**2'%(pb)
print  'upward pressure  = %.1f lbs'%(pu)
print  'downward pressure  = %.1f lbs'%(pd)
print  'Resultant pressure  = %.1f lbs'%(RP)

velocity va = 19.6 ft/sec
velocity vb = 1.91 ft/sec
velocity vx = 3.83 ft/sec
pressure px = 14.7 lbs/in**2
upward pressure  = 729.3 lbs
downward pressure  = 738.9 lbs
Resultant pressure  = 9.4 lbs

Example 3.10 Page No : 86

In [10]:
import math 
		
#initialisation of variables
d = 1.  		#ft
h = 4. 	    	#ft
h1 = 3. 		#ft
p = 25. 		#percent
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
h2 = ((h/4)-(h1/4))*h*2
w = math.sqrt(h2*2*g/(d/2)**2)
N = w*60/(2*math.pi)
h3 = (h-h1**2/4)*2
w1 = math.sqrt(h3*2*g/(d/2)**2)
N1 = w1*60/(2*math.pi)
		
#RESULTS
print  'original volume = %.1f R.P.M '%(N1)

original volume = 286.7 R.P.M

Example 3.12 Page No : 89

In [11]:
import math 
		
#initialisation of variables
R2 = 2. 		#ft
R1 = 1. 		#ft
w = 200. 		#r.p.m
g = 32.2 		#ft/sec**2
		
#CALCULATIONS
v2 = R2*math.pi*w*R2/60
v1 = R2*math.pi*w*R1/60
H = (v2**2-v1**2)/(2*g)
		
#RESULTS
print  'centrifugal head = %.1f ft of watrer '%(H)

centrifugal head = 20.4 ft of watrer