Chapter 4 : Flow Over Weirs Notches

Example 4.1 page no : 103

In [3]:
import math
#initialisation of variables
p = 70. 			#per cent
Cd = 0.6
Q = 50. 			#million gallons
H = 2. 			#ft
w = 62.4 			#lb/ft**3
g = 32.2 			#ft/sec**2
#CALCULATIONS
Q1 = p*Q*10**6*10/(100*w*24*3600)
L = Q1*3/(2*Cd*math.sqrt(2*g)*H**1.5)
#RESULTS
print  'length of the weir  = %.2f ft '%(L)

length of the weir  = 7.15 ft

Example 4.2 page no : 104

In [5]:
import math 

#initialisation of variables
L = 15. 			#ft
H = 1. 			#ft
Cd = 0.6
v = 80. 			#ft/min
g = 32.2 			#ft/sec62
w = 62.4 			#lb/ft**3
#CALCULATIONS
vo = v/60
Q = 2*Cd*math.sqrt(2*g)*L*((1+(vo**2/(2*g)))**1.5-(vo**2/(2*g))**1.5)*w*100/(3*550)
#RESULTS
print  'HP  = %.f HP '%(Q)

# This is accurate answer. Please calcualte manually.

HP  = 567 HP

Example 4.3 pageno : 104

In [23]:
import math 
#initialisation of variables
L = 11. 			#ft
H = 0.7 			#ft
Cd = 0.6
g = 32.2 			#ft/sec**2
h = 1.95 			#ft
Q = 20.65 			#cuses
Q1 = 21.2 			#cfs

#CALCULATIONS
Q = 2*Cd*math.sqrt(2*g)*L*H**1.5/3
vo = Q/(h*L)
h1 = vo**2/(2*g)
Q1 = 2*Cd*math.sqrt(2*g)*L*((H+(vo**2/(2*g)))**1.5-(vo**2/(2*g))**1.5)/3
v1 = Q1/(L*h)
Q2 = 2*Cd*math.sqrt(2*g)*L*((H+(v1**2/(2*g)))**1.5-(v1**2/(2*g))**1.5)/3
p = (Q2-Q1)*100/Q1

#RESULTS
print "Head to velocity approach = %.1f cu ft/sec"%Q1
print "Q2 = %.2f cu ft/sec"%Q2
print  'discharge percent  = %.3f per cent '%(p)

# Note : answers may vary because of rounding error. Please calculate manually.

Head to velocity approach = 21.3 cu ft/sec
Q2 = 21.29 cu ft/sec
discharge percent  = 0.148 per cent

Example 4.4 page no : 106

In [4]:
import math 
#initialisation of variables
b = 3. 			#ft
H = 1 			#ft
Q = 9 			#cfs
Q1 = 1.105      # log Q from fig.
h = 0.1 		# log H from fig. ft
#CALCULATIONS
K = Q/b
n = (Q1-math.log10(3*K))/h
#RESULTS
print  'K  = %.f  '%(K)
print  'n  = %.1f  '%(n)

K  = 3  
n  = 1.5

Example 4.5 page no : 108

In [28]:
import math 
#initialisation of variables
g = 32.2 			#ft/sec**2
Cd = 0.62
L = 7.573 			#ft
H = 1.2 			#ft
S = 2.85 			#ft
#CALCULATIONS
Q1 = 2*Cd*math.sqrt(2*g)*L*H**1.5/3
Q2 = 3.33*L*H**1.5
Q3 = math.sqrt(2*g)*L*H**1.5*(0.405+(0.00984/H))
He = H+0.004
Q4 = (3.227+0.435*(He/S))*L*He**1.5
#RESULTS
print  'Q  = %.2f cuses '%(Q1)
print  'Q  = %.2f cuses '%(Q2)
print  'Q  = %.2f cuses '%(Q3)
print  'Q  = %.2f cuses '%(Q4)

# Note : answers may vari because of rounding error. Please check manually.

Q  = 33.02 cuses 
Q  = 33.15 cuses 
Q  = 33.01 cuses 
Q  = 34.12 cuses

Example 4.6 pageno : 109

In [29]:
import math 
#initialisation of variables
H = 2.5 			#ft
L = 10 			#ft
A = 10 			#miles
p = 30 			#per cent
a = 2 			#in/hr
w = 2 			#ft
#CALCULATIONS
Q = L*1760**2*3**2*a*p/(60*60*12*100)
n = ((Q/(3.33*H**1.5))-(L-0.1*w*H))/(L-0.1*w*H)
#RESULTS
print  'n  = %.f  '%(n)

n  = 30

Example 4.7 page no : 109

In [30]:
import math 
#initialisation of variables
L = 2.5 			#ft
H = 1 			#ft
g = 32.2 			#ft/sec**2
Cd = 0.61
L1 = 1.75 			#ft
L2 = 2.25 			#ft
#CALCULATIONS
Q1 = 2*Cd*math.sqrt(2*g)*L*H/3
Q2 = 2*Cd*math.sqrt(2*g)*L1*(L1**1.5-1)/3
Q3 = 2*Cd*math.sqrt(2*g)*H*(L2**1.5-L1**1.5)/3
Q = Q1+Q2+Q3
#RESULTS
print  'Total discharge  = %.1f cfs '%(Q)

Total discharge  = 19.1 cfs

Example 4.8 page no : 110

In [32]:
import math 
#initialisation of variables
g = 32.2 			#ft/sec**2
h1 = 16.63 			#cm
h2 = 10.18 			#cm
h3 = 16.53 			#cm
#CALCULATIONS
H1 = h1-h2
H2 = h3-h2
p = (H1**1.5-H2**1.5)*100/H1**1.5
#RESULTS
print  'Percent decrease in discharge  = %.2f %% '%(p)

Percent decrease in discharge  = 2.32 %

Example 4.9 pageno : 111

In [33]:
import math 
#initialisation of variables
Cd = 0.6
a = 20000 			#yd**2
H2 = 12 			#in
L = 5 			#ft
H1 = 2 			#ft
g =32.2 			#ft/s**2
#CALCULATIONS
t = 2*a*9*(L-H1)*((1/math.sqrt(H2/12))-(1/math.sqrt(H1)))/(2*60*Cd*math.sqrt(2*g)*L)
#RESULTS
print  'time required to lower level of reservoir  = %.2f min '%(t)

time required to lower level of reservoir  = 109.49 min

Example 4.10 pageno : 113

In [34]:
import math 
#initialisation of variables
L = 3. 			#ft
H = 6. 			#in
Cd = 0.62
Cd1 = 0.59
a = 45. 			#degrees
g = 32.2 			#ft/sec**2
#CALCULATIONS
H = ((2./3)*Cd*math.sqrt(2*g)*L*(H/12)**1.5/((8./15)*Cd1*math.sqrt(2*g)))**0.4
#RESULTS
print  'depth of water  = %.3f ft '%(H)

depth of water  = 1.142 ft

Example 4.11 page no : 114

In [36]:
import math 
#initialisation of variables
V = 20. 			#litres
g = 981. 			#cm/sec**2
Cd = 0.593
r = 2.5
r1 = 1.5
e = 2. 			#mm
Cd1 = 0.623
L = 30. 			#cm
#CALCULATIONS
H = (V*1000*15/(8*Cd*math.sqrt(2*g)))**0.4
dH1 = e/10.
p = r*dH1*100/H
H1 = (V*3*1000/(2*Cd1*math.sqrt(2*g)*L))**(2./3)
p1 = r1*dH1*100/H1
#RESULTS
print  'percentage error of discharge over the weir  = %.2f %% '%(p)
print  'percentage error of discharge over the weir  = %.2f %% '%(p1)

percentage error of discharge over the weir  = 2.74 % 
percentage error of discharge over the weir  = 2.74 %

Example 4.12 page no : 116

In [37]:
import math 
#initialisation of variables
L = 16. 			#in
H = 9. 			#in
h = 18. 			#in
g = 32.2 			#ft/sec**2
w = 2. 			#ft
Cd = 0.63
W = 62.4 			#lbs/ft**3
#CALCULATIONS
Q = 2*Cd*math.sqrt(2*g)*(L/12)*(H/12)**1.5/3
v = Q/(w*(h/12))
H1 = v**2/(2*g)
Q1 = 2*Cd*math.sqrt(2*g)*(L/12)*(((H/12)+H1)**1.5-H1**1.5)*W*6/3.
#RESULTS
print  'Discharge  = %.f gpm '%(Q1)

Discharge  = 1122 gpm

Example 4.13 pageno : 118

In [48]:
import math 
#initialisation of variables
L = 100 			#ft
H = 2.25 			#ft
Cd = 0.95
w = 120 			#ft
h = 2 			#ft
g = 32.2 			#ft/sec**2
#CALCULATIONS
Q = round(3.087*Cd*L*H**1.5)
v0 = round(Q/(w*(h+H)),2)
Q1 = 3.087*Cd*L*((H+(v0**2/(2*g)))**1.5-(v0**2/(2*g))**1.5)
#RESULTS
print  'Discharge  = %.0f cuses '%(Q1)

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

Discharge  = 1024 cuses

Example 4.14 pageno : 119

In [51]:
import math 
#initialisation of variables
L = 6 			#ft
H1 = 0.5 			#ft
H2 = 0.25 			#ft
g = 32.2 			#ft/sec**2
Cd1 = 0.58
Cd2 = 0.8
w = 6.24 			#lb/ft**3
#CALCULATIONS
Q1 = 2*Cd1*math.sqrt(2*g)*L*(H1-H2)**1.5/3
Q2 = Cd2*L*H2*math.sqrt(2*g*(H1-H2))
Q = round((Q1+Q2)*w*3600,-3)
#RESULTS
print  'Discharge  = %.f gph '%(Q)

Discharge  = 160000 gph

Example 4.15 pageno : 120

In [52]:
import math 
#initialisation of variables
W = 100 			#ft
h = 10 			#ft
v = 4 			#ft/sec
h1 = 3 			#ft
g = 32.2 			#ft/sec**2
H = 5.4 			#ft
Cd1 = 0.58
Cd2 = 0.8
#CALCULATIONS
v0 = (W*h*v)/(W*(h+h1))
h0  =v0**2/(2*g)
H2 = (W*h*v-(2*Cd1*W*math.sqrt(2*g)*((h1+h0)**1.5-h0**1.5)/3))/(Cd2*W*math.sqrt(2*g*(h1+h0)))
dh = h-H2
#RESULTS
print  'height of anicut which is drowned  = %.f ft '%(dh)

height of anicut which is drowned  = 8 ft

Example 4.16 page no : 123

In [53]:
import math 
#initialisation of variables
x = 6. 			#in
l = 200. 			#ft
d = 10. 			#ft
v = 4. 			#ft/sec
Ce = 0.95
g = 32.2 			#ft/sec**2
#CALCULATIONS
l1 = math.sqrt(l**2/(Ce**2*(((x/12)*2*g/v**2)+(d**2/(d+(x/12))**2))))
#RESULTS
print  'length  = %.f ft '%(l1)

length  = 123 ft

Example 4.17 page no : 124

In [57]:
import math 
#initialisation of variables
g = 32.2 			#ft/sec**2
H = 25. 			#ft
l = 2.5 			#ft
b = 5. 			#ft
Cd = 0.64
Q = 3200. 			#cuses
L =150. 			#ft
C =3.2
depth =0.5 			#ft
A1 =5000000. 			#sq yards
#CALCULATIONS
Q1 = Cd*l*b*math.sqrt(2*g*H)
n = Q/Q1
h = (Q/(3.2*L))**(2./3)
hr =h-depth
Area =A1*9
V = round(Area*hr,-6)
#RESULTS
print  'number of spilways  = %.f  '%(n)
print "Volume of extra water stored  = %d cu ft"%(V)

number of spilways  = 10  
Volume of extra water stored  = 137000000 cu ft
In [ ]: