# Chapter 5 : Notches and weirs¶

## Example 5.1 Page No : 98¶

In :
# Variables
q = 0.2
Cd = 0.62
g = 9.81

# Calculations
# using the relation
z = (3*q*(2**1.5))/(2*Cd*((2*g)**0.5))
b = z**0.4

# Results
print "the lenght of the notch in cm ",round((b*100),1)

the lenght of the notch in cm  62.5


## Example 5.2 Page No : 98¶

In :
import math

# Variables
b = 1.
H = 0.15
Cd1 = 0.62
x = 90.
g = 9.81
Cd2 = 0.58

# Calculations
Q1 = 2*Cd1*b*((2*g*H*H*H)**0.5)/3
z = (15*Q1)/(8*Cd2*((2*g)**0.5)*(math.tan(math.radians(x/2))))
H1 = z**0.4

# Results
print "the depth over the traingular veir in cm",round((H1*100),2)

the depth over the traingular veir in cm 35.98


## Example 5.3 Page No : 99¶

In :
import math

# Variables
x = 90.
Cd = 0.62            # co-efficient
H = 0.36
g = 9.81

# Calculations
Q = (8*Cd*math.tan(math.radians(x/2))*((2*g)**0.5)*(H**2.5))/15
q = Q*1000

# Results
print "the actual discharge in litres/sec",round(q,2)

the actual discharge in litres/sec 113.89


## Example 5.4 Page No : 100¶

In :
import math

# Variables
x = 90.
H = 0.2
b = 0.3
Cd = 0.62
g = 9.81

# Calculations
q1 = (8*Cd*math.tan(math.radians(x/2))*((2*g)**0.5)*(H**2.5))/15
q2 = 2*Cd*b*((2*g*H*H*H)**0.5)/3
q = q1+q2

# Results
print "discharge over the trapezoidal notch in m3/sec",round(q,6)

discharge over the trapezoidal notch in m3/sec 0.075327


## Example 5.5 Page No : 100¶

In :
# Variables
a = 20*(10**6)
x = 0.03
q = a*x
qf = q*0.4/3600
n = 2.
H = 0.6

# Calculations
# using Francis formula
L = (qf/(1.84*(H**1.5)))+(0.1*n*H)

# Results
print "the lenght of the weir in m",round(L,4)

the lenght of the weir in m 78.0786


## Example 5.6 Page No : 101¶

In :
# Variables
L = 36
v1 = 2
g = 9.81
H = 1.2
H1 = (v1*v1)/(2*g)
n = 2*12
w = 0.6
Nv = 11

# Calculations
Lf = L-(Nv*w)
Q = 1.84*(Lf-(0.1*n*(H+H1)))*((H+H1)**1.5-(H1**1.5))

# Results
print "dischsrge over the weir in m3/sec",round(Q,3)

dischsrge over the weir in m3/sec 75.261


## Example 5.7 Page No : 102¶

In :
# Variables
l = 0.77
H = 0.39
H1 = 0.6
Dp = H+H1
Cd = 0.623
g = 9.81

# Calculations
Q = (2*Cd*l*((2*g*H*H*H)**0.5))/3
v = Q/(l*Dp)
Ha = (v*v)/(2*g)
q = (2*Cd*l*((2*g)**0.5)*(((H+Ha)**1.5)-(Ha**1.5)))/3

# Results
print "discharge in m3/sec",round(q,6)

discharge in m3/sec 0.357447


## Example 5.8 Page No : 103¶

In :
# Variables
Q1 = 0.005
Cd = 0.62
g = 9.81
Q2 = 0.75
h = 0.07

# Calculations
z = (Q1*15)/(8*Cd*((2*g)**0.5)*(h**2.5))
H = h*((Q2/Q1)**0.4)
W = 2*H*z

print "width of the water surface in m",round(W,3)

width of the water surface in m 2.736


## Example 5.9 Page No : 104¶

In :
# Variables
b = 4.
H = 0.2
Cd = 0.62
g = 9.81

# Calculations
Q1 = 2*Cd*b*((2*g*H*H*H)**0.5)/3
Q2 = (2*Cd*((2*g)**0.5)*(H**1.5)*(b-(0.2*H)))/3
m = 0.405+(0.003/H)
Q3 = m*b*((2*g)**0.5)*(H**1.5)

# Results
print "discharge when end contraction are supressed in m3/sec",round(Q1,3)
print "discharge when end contraction are taken into account by francis formula in m3/sec",round(Q2,4)
print "discharge when end contraction are taken into account by bazin formula in m3/sec",round(Q3,5)

discharge when end contraction are supressed in m3/sec 0.655
discharge when end contraction are taken into account by francis formula in m3/sec 0.6485
discharge when end contraction are taken into account by bazin formula in m3/sec 0.66559


## Example 5.10 Page No : 104¶

In :
import math

# Variables
Cd = 0.6
x = 45.
H = 0.5
g = 9.81

# Calculations and Results
q1 = (8*Cd*math.tan(math.radians(x/2))*((2*g)**0.5)*(H**2.5))/15

print "rate of flow over the rectangular notch in m3/sec",round(q1,5)

dq1 = 0.025
dh = dq1*H/2.5
h1 = H+dh
h2 = H-dh
print "limiting values of head in centimeters",h2*100,h1*100

rate of flow over the rectangular notch in m3/sec 0.10379
limiting values of head in centimeters 49.5 50.5


## Example 5.11 Page No : 105¶

In :
import math

# Variables
Cd = 0.6
x = 90.
q = 0.05
g = 9.81

# Calculations
dh = 0.00025
z = (15*q)/(8*Cd*((2*g)**0.5)*(math.tan(math.radians(x/2))))
H = z**0.4
error = 2.5*(dh/H)

# Results
print "the percentage error in the discharge",round((error),5)

the percentage error in the discharge 0.00238

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