# Chapter 5 : Flow Through Pipes¶

### Example 5.1 pageno : 131¶

In [5]:
import math
import numpy
#initialisation of variables
h = 4. 			#ft
h1 = 3. 			#ft
r = 3. 			#ft
h2 = 1.5 			#ft
#CALCULATIONS
m = (h*h1+(h1**2/2))/(h+(h/2)*math.sqrt(h1**2+(h1/2)**2))
a = 2*numpy.degrees(numpy.arccos(h2/r))
P = 2*math.pi*r*a/360.
H = A/(2*math.pi)
#RESULTS
print 'Hydraulic mean depth m = %.2f ft'%m
print  'hydraulic mean depth  = %.2f ft '%(H)

Hydraulic mean depth m = 1.54 ft
hydraulic mean depth  = 0.88 ft


### Example 5.2 pageno : 133¶

In [7]:
import math

#initialisation of variables
d = 3. 			#ft
l = 5280. 			#ft
v = 3. 			#ft/sec
f = 0.005
g = 32.2 			#ft/sec**2
C = 115.
#CALCULATIONS
hf = 4*f*l*v**2/(2*g*v)
m = d/4
hf1 = (v/C)**2*4*l/3
#RESULTS
print  'hf  = %.2f ft '%(hf)
print  'hf  = %.1f ft '%(hf1)

hf  = 4.92 ft
hf  = 4.8 ft


### Example 5.3 page no : 134¶

In [7]:
import math
#initialisation of variables
d = 6. 			#in
Q = 2. 			#cfs
l = 1000. 			#ft
f = 0.0055
w = 62.4 			#lb/ft**3
g = 32.2 			#ft/sec**2
h = 70. 			#ft
#CALCULATIONS
v = Q/(math.pi*(d/12)**2/4)
hf = 4*f*l*w*(Q/(math.pi*(d/12)**2/4))**2/((d/12)*2*144*g)
ft_water = round(hf*144/w + 70)
P = ft_water*w/144

#RESULTS
print  'pressure  = %.1f lb/in**2 '%(hf)
print  'presure difference  = %.0f lb/in**2 '%(P)

# Answers are slightly different because of rounding error.

pressure  = 30.7 lb/in**2
presure difference  = 61 lb/in**2


### Example 5.4 pageno : 135¶

In [12]:
import math
#initialisation of variables
d = 6. 			#in
hf = 7.7 			#ft
f = 0.005
l = 1000. 			#ft
g = 32.2 			#ft/sec**2
#CALCULATIONS
C = math.sqrt(2*g/f)
Q = math.pi*C*(d/12)**2.5*(hf/1000)**0.5 /8
#RESULTS
print  'Discharge  = %.2f cfs '%(Q)

Discharge  = 0.69 cfs


### Example 5.5 page no : 136¶

In [18]:
import math
#initialisation of variables
Q = 400000.
d = 4. 			#miles
h = 50. 			#ft
q = 40. 			#gallons of water
t = 8. 			#hr
f = 0.0075
w = 6.24 			#lb/ft**3
g = 32.2 			#ft/sec**2
#CALCULATIONS
Q1 = round(Q*q*0.5/(t*60*60*w),1)
d = (4*f*(d*5280)*Q1**2*16/(math.pi**2*h*2*g))**0.2*12

#RESULTS
print  'size of the supply  = %.3f in '%(d)

size of the supply  = 43.579 in


### Example 5.6 pageno : 138¶

In [4]:
import math

#initialisation of variables
Q = 0.7 			#cfs
d = 6. 			#in
v1 = 1.084*10**-5 			#ft**2/sec
v2 = 0.394*10**-5 			#ft**2/sec
R = 2320.
#CALCULATIONS
Re = (4*Q)/(math.pi*.5*v2)
v3 = R*v1/(d/12.)
v4 =R*v2/(d/12.)
v = Q*4/(math.pi*(d/12.)**2)

#RESULTS
print 'Re at 80 C %.0f'%Re
print  'crititcal velocity  = %.4f ft/sec '%(v4)
print  'actual velocity  = %.2f ft/sec '%(v)


Re at 80 C 452420
crititcal velocity  = 0.0183 ft/sec
actual velocity  = 3.57 ft/sec


### Example 5.7 pageno : 140¶

In [8]:
import math
#initialisation of variables
p = 0.91 			#units
u = 0.21 			#poise
q = 200. 			#gallons
h = 40. 			#ft
l = 200. 			#ft
w = 62.4 			#lb/ft**3
d = 3./4 			#in
g =32.2 			#ft/s**2
#CALCULATIONS
v = u/(p*(30.5)**2)
Q = q*10/(w*3600*p)
V = Q/(math.pi*(d/12)**2/4)
Re = V*(d/12)/v
F = 64/Re
Hf = F*l*V**2/(2*g*(d/12))
Ht = Hf+h
P = w*p*Ht/144
#RESULTS
print  'Pressure head   = %.1f lb/sq in '%(P)

Pressure head   = 31.6 lb/sq in


### Example 5.8 page no : 142¶

In [21]:
from math import log10
#initialisation of variables
h = 1.5         # H
v = 2.          # V

#CALCULATIONS
logh = round(log10(h),4)
logk = -0.415            # from fig.
logv = round(log10(v),4)
n = (logh-logk)/logv

#RESULTS
print  'n  = %.3f  '%(n)

n  = 1.964


### Example 5.10 page no : 146¶

In [5]:
import math
#initialisation of variables
pb = 20. 			#lb/in**2
w = 62.4 			#lb/ft**3
Q = 1.96 			#cfs
d1 = 0.5 			#ft
d2 = 1. 			#ft
f = 0.005
g = 32.2 			#ft/sec**2
l1 = 300. 			#ft
H = 14.015 			#ft of water
#CALCULATIONS
v1 = Q/(math.pi*d1**2/4.)
v2 = Q/(math.pi*d2**2./4.)
hf1 = 4*f*l1*v1**2./(2.*g*d1)
hf2 = 4*f*l1*v2**2/(2*g*d2)
h = (v1-v2)**2/(2*g)
h1 = v1**2/(2*g)
h2 = v2**2/(2*g)
P = H*w/144
#RESULTS
print  'Loss of head at C  = %.2f ft '%(h1)
print  'Loss of head at C  = %.3f ft '%(h2)
print  'Pressure differnece at discharge end  = %.2f lb/in**2 '%(P)

Loss of head at C  = 1.55 ft
Loss of head at C  = 0.097 ft
Pressure differnece at discharge end  = 6.07 lb/in**2


### Example 5.11 page no : 148¶

In [31]:
import math
#initialisation of variables
d = 8. 			#in
l = 6000. 			#ft
H = 100. 			#ft
H1 = 1000. 			#ft
f = 0.008
g = 32.2 			#ft/sec**2
h1 = 24. 			#ft
h2 = 34. 			#ft
h3 = 25. 			#ft
w = 6.24 			#lb/ft**3
#CALCULATIONS
v = math.sqrt(H*d*2*g/(4*f*l*12))
h = -h1+(v**2/(2*g))+h3+(4*f*H1*v**2/(2*g*(d/12)))
Q = round(math.pi*(d/12)**2*v*3600*w/4,-2)
#RESULTS
print  'minimum depth  = %.f ft '%(h)
print  'Discharge  = %.f gpm'%(Q)

minimum depth  = 18 ft
Discharge  = 37100 gpm


### Example 5.12 page no : 151¶

In [33]:
import math
#initialisation of variables
h = 25. 			#ft
l = 2000. 			#ft
d = 12. 			#in
g = 32.2 			#ft/sec**2
f = 0.005
dz = 16. 			#ft
zb = 25. 			#ft
zc = -16. 			#ft
#CALCULATIONS
v = math.sqrt(2*g*h/(1.5+(4*f*l/(d/12))))
Q = math.pi*(d/12)**2*v/4
l1 = (34-dz)*l/(zb-zc-dz)
#RESULTS
print  'Discharge  = %.1f cfs '%(Q)
print  'length of the inlet  = %.f ft of water '%(l1)

Discharge  = 4.9 cfs
length of the inlet  = 1440 ft of water


### Example 5.13 page no : 153¶

In [6]:
import math

#initialisation of variables
d1 = 2 			#in
l1 = 25 			#ft
d2 = 4 			#in
l2 = 140 			#ft
v = 4 			#ft/sec
g = 32.2 			#ft/sec**2
f = 0.0065
#CALCULATIONS
v1 = v*(d2/d1)**2
H = (0.5*v1**2/(2*g))+(4*f*l1*12*v1**2/(d1*2*g))+((v1-v)**2/(2*g))+(4*f*l2*12*v**2/(d2*2*g))+(v**2/(2*g))
#RESULTS
print  'necessaey height of water  = %.3f ft '%(H)

# Note : answer is slightly different because of rounding error.

necessaey height of water  = 22.688 ft


### Example 5.14 page no : 155¶

In [35]:
import math
#initialisation of variables
l1 = 3000. 			#ft
d1 = 18. 			#in
l2 = 1500. 			#ft
d2 = 15. 			#ft
l3 = 1000. 			#ft
d3 = 12. 			#in
#CALCULATIONS
d = ((l1+l2+l3)/((l1/d1**5)+(l2/d2**5)+(l3/d3**5)))**(1./5)
#RESULTS
print  'Diameter  = %.2f in '%(d)

Diameter  = 14.86 in


### Example 5.15 page no : 156¶

In [36]:
import math
#initialisation of variables
g = 32.2 			#ft/sec**2
D = 9. 			#in
#CALCULATIONS
d = D/(2**0.4)
#RESULTS
print  'diameter of paralle mains  = %.2f in '%(d)

diameter of paralle mains  = 6.82 in


### Example 5.16 page no : 157¶

In [38]:
import math
#initialisation of variables
d = 2. 			#ft
l = 5280. 			#ft
f = 0.01
H = 100. 			#ft
g = 32.2 			#ft/sec**2
#CALCLATIONS
v = math.sqrt(H*2*d*g/(4*f*l))
Q = math.pi*d**2*v/4.
r = d
v2 = math.sqrt(H/((r**2+1)*(4*f*l/(2*2*2*g))))
Q1 = 2*math.pi*d**2*v2/4.
dQ = Q1-Q
p = dQ*100./Q
#RESULTS
print  'percentage increase in discharge  = %.1f %% '%(p)

percentage increase in discharge  = 26.5 %


### Example 5.17 Page no : 160¶

In [19]:
 import math

# Variables
d1 = 2          # ft
d2 = 1.5        # ft
d3 = 1          # ft
l1 = 2000       # ft
l2 = 3000       # ft
l3 = 1500       # ft
za = 100        # ft
zb = 70         # ft
zc = 50         # ft
zd = 80         # ft
f = 0.007       # ft
v3 = 7.93

# Calculations

v1 = round(math.sqrt(111-1.412*(v3**2)),1)
v2 = round(math.sqrt(-23.4+.745*(v3**2)),2)
pd_w = round((za - zd) - ( 29 * v1**2/64.4),2)
Q1 = (math.pi/4)*d1**2*v1
Q2 = (math.pi/4)*d2**2*v2
Q3 = Q1 - Q2

# Results
print "V1 = %.1f ft/sec"%v1
print "V2 = %.2f ft/sec"%v2
print "V3 = %.2f ft/sec"%v3
print "Pressure at the junction point : %.2f ft of water"%pd_w
print "Discharge in section : %.2f cfs"%Q3

# Note : Answers may vary because of rounding error. Please check it manually.

V1 = 4.7 ft/sec
V2 = 4.84 ft/sec
V3 = 7.93 ft/sec
Pressure at the junction point : 10.05 ft of water
Discharge in section : 6.21 cfs


### Example 5.18 page no : 164¶

In [43]:
import math
#initialisation of variables
A = 10000. 			#ft**2
H1 = 50. 			#ft
H2 = 40. 			#ft
l = 1500. 			#ft
d = 6. 			#in
f = 0.0075
g = 32.2 			#f/sec**2
#CALCULATIONS
t = 2.*A*math.sqrt((1.5+(4*f*l/(d/12)))/(2*g))*(math.sqrt(H1)-math.sqrt(H2))/(math.pi*(d/12)**2/4)
#RESULTS
print  'Time taken to lower the level of water  = %.f hours '%(t/3600)

Time taken to lower the level of water  = 25 hours


### Example 5.19 page no : 164¶

In [49]:
import math
#initialisation of variables
l = 24. 			#ft
b = 12. 			#ft
f = 0.006
d = 4. 			#in
l1 = 25. 			#ft
H1 = 6. 			#ft
H = 20. 			#ft
g = 32.2 			#ft/sec**2
Cd = 0.6
#CALCULATIONS
a = math.pi*(d/12)**2/4
A = l*b
H2 = H1+H
t = round(2*A*math.sqrt((1.5+(4*f*l1/(d/12)))/(2*g))*(math.sqrt(H2)-math.sqrt(H))/a,-1)
t1 = 2*A*math.sqrt((1.5+(4*f*l1/(d/12)))/(2*g))*math.sqrt(H1)/a
t2 = 2*A*math.sqrt(H1)/(Cd*a*math.sqrt(2*g))
#RESULTS
print  'Time taken to lower the pipe  = %.f sec '%(t)
print  'Time taken to lower the pipe  = %.f sec '%(t1)
print  'Time taken to lower the pipe  = %.f sec '%(t2)

# note : answers may vary becasue of ronding error.

Time taken to lower the pipe  = 940 sec
Time taken to lower the pipe  = 3660 sec
Time taken to lower the pipe  = 3358 sec


### Example 5.20 page no : 165¶

In [50]:
import math
#initialisation of variables
d = 2 			#ft
l = 1000 			#ft
f = 0.0075
H1 = 20 			#ft
A1 = 100000 			#ft**2
A2 = 50000 			#ft**2
g = 32.2 			#ft/sec**2
#CALCULATIONS
h = 2*A1/A2
H2 = H1-h
t = 2*A1*A2*math.sqrt(1.5+(4*f*l/2))*0.47/((A1+A2)*(math.pi*d**2/4)*math.sqrt(2*g))/60
#RESULTS
print  'Time taken to lower the level of water  = %.f min '%(t)

Time taken to lower the level of water  = 84 min


### Example 5.21 page no : 167¶

In [52]:
import math
#initialisation of variables
H = 1000. 			#lb/in**2
Hf = 100. 			#lb/in**2
l = 10. 			#miles
HP = 100.
g = 32.2 			#ft/sec**2
w = 64.4 			#lb/ft**3
f = 0.006
#CALCULATIONS
n = (H-Hf)*100/H
v = Hf*550/((math.pi/4)*n*10*144)
r = Hf*144*2*g/(w*4*f*l*5280)
d = (v**2/r)**(1./5)
#RESULTS
print  'Diameter  = %.4f ft '%(d)

# answer may vary because of rounding error.

Diameter  = 0.4808 ft


### Example 5.22 page no : 170¶

In [58]:
import math
#initialisation of variables
h1 = 1640. 			#ft
h2 = 40. 			#ft
d = 8. 			#in
l = 2. 			#miles
D = 3. 			#ft
f = 0.006
Cv = 0.98
g = 32.2 			#ft/sec**2
w = 62.4 			#lb/ft**3
#CALCULATIONS
r = (d/12)/D
vact = Cv*math.sqrt(2*g*(h1-h2)/(1+(4*f*l*5280*r**4/D)))
HP = round(round(w*vact**3*(math.pi*(d/12)**2/4)/(550*2*g),-2),-3)
#RESULTS
print  'Horse Power of Jet  = %.f HP '%(HP)

Horse Power of Jet  = 15000 HP


### Example 5.23 page no : 171¶

In [59]:
import math
#initialisation of variables
p = 60. 			#lb/in**2
l = 300. 			#ft
D = 2.5 			#in
d = 7./8 			#in
f = 0.018
g = 32.2 			#ft/sec**2
w = 62.4 			#lb/ft**3
#CALCULATIONS
r = (D/d)**4
V = math.sqrt(2*g*144*p/(w*(r+0.5+(4*f*l/(D/12)))))
Q = V*(math.pi*(D/12)**2)/4
#RESULTS
print  'Volume of flow  = %.3f cu ft/sec '%(Q)

Volume of flow  = 0.246 cu ft/sec


### Example 5.24 page no : 171¶

In [60]:
import math
#initialisation of variables
D = 3. 			#in
l = 800. 			#ft
H = 120. 			#ft
f = 0.01
g = 32.2 			#ft/sec**2
w = 62.4 			#lb/ft**3
#CALCULATIONS
d = ((D/12)**5/(8*f*l))**0.25
hf = H/3
dh = H-hf
v = math.sqrt(hf*(D/12)*2*g/(4*f*l))
HPmax = w*math.pi*((D/48)**2/4)*v*dh/550
#RESULTS
print  'HPmax  = %.3f HP '%(HPmax)

HPmax  = 0.125 HP


### Example 5.25 page no : 177¶

In [61]:
import math
#initialisation of variables
l = 2. 			#miles
Q = 2.*10**6 			#gal/day
d = 12. 			#in
t = 16. 			#sec
w = 62.4 			#lb/ft**3
g = 32.2 			#ft/sec**2
#CALCULATIO
Q1 =Q*10/(w*24*60*60)
hi = l*5280*Q1/((math.pi*(d/12)**2./4)*(g*t))
#RESULTS
print  'height  = %.1f ft '%(hi)

height  = 96.8 ft


### Example 5.26 pageno : 178¶

In [62]:
import math
#initialisation of variables
d = 6. 			#in
Q = 0.7854 			#cfs
E = 30*10**6 			#lb/in**2
t = 0.25 			#in
g = 32.2 			#ft/sec**2
w = 62.4 			#lb/ft**3
K = 300000. 			#lb/in**2
#CALCULATIONS
v = Q/(math.pi*(d/12)**2/4)
p = v/(math.sqrt(144*(g/w)*((1/K)+(d/(t*E)))))
#RESULTS
print  'rise of presure in the pipe  = %.f lb/in**2 '%(p)

rise of presure in the pipe  = 228 lb/in**2


### Example 5.27 page no : 183¶

In [63]:
import math
#initialisation of variables
w = 62.4 			#lb/ft**3
f = 0.005
Q = 100. 			#cuses
m = 40. 			#Rs
n = 0.75
n1 = 0.065
K = 15. 			#Rs
#CALCULATIONS
d = ((5*w/(1.5*550*10))*n*f*Q**3*m/(K*n1))**(1/6.5)
#RESULTS
print  'economical diameter of pipe line  = %.3f ft '%(d)

economical diameter of pipe line  = 3.795 ft

In [ ]: