# Chapter 14 : Wastewater Flows¶

## Example 14.1 Page No : 14-4¶

In [1]:

#initialisation of variables
n = 0.013	#ft
s = 4.90	#ft
v = 0.590	#ft
d = 0.463	#ft
w = 3.9*10**-2	#ft
p = 1.696	#ft

#CALCULATIONS
V = s*v	#fps
Q = s*d	#cfs
N = (w*p)**2*1000	#percent

#RESULTS
print 'the velocity of flow and rate of discharge = %.2f percent'%(N)

the velocity of flow and rate of discharge = 4.38 percent


## Example 14.2 Page No : 14-7¶

In [5]:
import math

#initialisation of variables
v = 1.34	#fps
s = 3.7*10**-3	#fps
k = 0.8	#ft
r = 20	#ft
k1 = 0.04	#ft
v = 3.0	#fps
v1 = 5.0	#fps
d = 10**-1	#ft
d1 = 1.34	#ft

#CALCULATIONS
V = round((1.49/1.3*10**-2)*(1./4)**(1./6)*(k1*d/30.48*(2.65-1)/1)**(1./2) * 10000,2)
# for v = 3
v1 = 3.
D1 = d*(v1/V)**2

# for v = 5
v2 = 5
D2 = d*(v2/V)**2

#RESULTS
print 'For velocity = %d, the gradient at the which coarse quartz = %.1f cm'%(v1,D1)  #incorrect answer in the textbook
print "For velocity = %d, the gradient at the which coarse quartz = %.1f cm"%(v2,D2)

For velocity = 3, the gradient at the which coarse quartz = 0.5 cm
For velocity = 5, the gradient at the which coarse quartz = 1.4 cm


## Example 14.3 Page No : 14-10¶

In [3]:

#initialisation of variables
v = 2.5	#fps
q = 0.873	#cfs
s = 5.20	#percent
a = 0.252	#ft
r = 0.684	#ft
r1 = 1.46	#ft
v1 = 0.776	#ft
q1 = 0.196	#ft
n = 0.78	#ft
R = 0.939	#ft

#CALCULATIONS
V = v1*v	#fps
Q = q1*q	#cfs
R1 = r1*s	#percent
Vs = R*v	#ft
N = n*Vs	#fps
Qs = a*R*q	#cfs
N1 = n*Qs	#cfs

#RESULTS
print 'the required grades and associated velocity and rates = %.3f cfs'%(V)
print 'the depth and a grade = %.3f cfs'%(Q)
print 'the self cleaning flow = %.3f cfs'%(N1)

the required grades and associated velocity and rates = 1.940 cfs
the depth and a grade = 0.171 cfs
the self cleaning flow = 0.161 cfs


## Example 14.4 Page No : 14-11¶

In [4]:

#initialisation of variables
Q = 0.873	#cfs
s = 5.20	#percent
d = 0.161	#cfs
q1 = 0.185	#ft
d2 = 2.5	#ft
v = 0.91	#ft
s1 = 1.70	#ft
s3 = 1.46	#ft
w = 0.185	#ft
d1 = 0.30	#ft
v1 = 0.732	#ft

#CALCULATIONS
q = d/Q	#cfs
Vs = v*d2	#fps
Ss = s1*s	#percent
Va = v1*d2	#fps
Ss1 = s3*s	#percent

#RESULTS
print 'the depth and velocity of flow and the required slop = %.1f percent'%(Ss1)

the depth and velocity of flow and the required slop = 7.6 percent


## Example 14.5 Page No : 14-13¶

In [5]:

#initialisation of variables
d1 = 0.67	#ft
h1 = 2.00	#ft
h2 = 4.04	#ft
hv1 = 0.062	#ft
hv2 = 0.254	#ft
d = 0.19	#ft
h = 0.2	#ft
h1 = 0.04	#ft
q = 0.644	#ft
q1 = 0.65	#ft
v = 0.92	#ft
d2 = 6.5	#ft
v1 = 3.69	#ft
d3 = 0.542	#ft
hv3 = 0.21	#ft
delv = 0.15	#ft
d4 = 0.02	#ft

#CALCULATIONS
H = d1+hv1	#ft
H1 = d1+hv2	#ft
he = h*d	#ft
hi = d+h1	#ft
H2 = d3+hv3	#ft
he1 = h*delv	#ft
S = d4+h1	#ft

#RESULTS
print 'the required slope = %.2f ft'%(hi)
print 'the lower sewer and the invert drop in the transition = %.2f ft'%(S)

the required slope = 0.23 ft
the lower sewer and the invert drop in the transition = 0.06 ft


## Example 14.6 Page No : 14-17¶

In [6]:

#initialisation of variables
q = 60	#cfs
D = 4	#ft
w = 0.177	#ft
s = 0.59	#ft
h = 4.0	#ft
d1 = 1.0	#ft
v = 0.90	#ft
d1 = 0.42	#ft
h1 = 6.0	#ft
h2 = 1.5	#ft
dl = 1.3	#ft
p = 0.41	#ft
u = 0.8	#ft
u1 = 3.2	#ft
y = 0.45	#ft

#CALCULATIONS
H = s*D	#ft
d2 = d1*D	#ft
V = v*D	#ft
P = p*D	#ft
D1 = y*D	#ft

#RESULTS
print 'the critical depth = %.1f ft'%(H)
print 'the alternate stages for an energy  = %.1f ft'%(V)
print 'the alternate stages for an energy head = %.1f ft'%(P)
print 'the lower alternate stage with upper alternate stage = %.1f ft'%(D1)

the critical depth = 2.4 ft
the alternate stages for an energy  = 3.6 ft
the alternate stages for an energy head = 1.6 ft
the lower alternate stage with upper alternate stage = 1.8 ft


## Example 14.7 Page No : 14-19¶

In [1]:

#initialisation of variables
d = 106	    #cfs
q = 400.	#cfs
d1 = 0.40	#cfs
w = 10	#ft

#CALCULATIONS
D = d/q	#cfs
D1 = d1*w	#cfs

#RESULTS
print 'the water level in this well rises = %.f ft'%(D1)

the water level in this well rises = 4 ft


## Example 14.8 Page No : 14-19¶

In [13]:
import math
#initialisation of variables
Q = (400/78.5*math.sqrt(10*32.2))**2	#ft
N = 0.012	#ft
d = 0.47	#ft
q = 10	#ft

#CALCULATIONS
D = d*q	#ft

#RESULTS
print 'the water surface in the sewer when it is flowing at maximum capacity = %.1f ft'%(D)

the water surface in the sewer when it is flowing at maximum capacity = 4.7 ft


## Example 14.9 Page No : 14-23¶

In [8]:
import math

#initialisation of variable
g = math.sqrt(3)	#ft
d = 5.67	#ft

#CALCULATIONS
C = g*d	#ft

#RESULTS
print 'The rate of propagation of a discontinuous surge = %.1f ft'%(C)

The rate of propagation of a discontinuous surge = 9.8 ft


## Example 14.10 Page No : 14-24¶

In [2]:
#initialisation of variables
Q1 = 30	#cfs
Q2 = 16	#cfs
a = 32	#sq ft
r = 1.6	#ft
i = 10**-4	#ft
n = 1.25*10**-2	#ft
h2 = 0.50	#ft
c = 3.33	#ft
h1 = 5.20	#ft
l = 72	#ft
s = 12320	#ft

#CALCULATIONS
L = s-l	#ft
h1 = 0.49022 - 8.04*10**-5*l
#RESULTS
print 'the forchheimer s methos  = %.0f ft'%(L)
print (Q1**2-Q2**2)/(2*32.2*a**2)

the forchheimer s methos  = 12248 ft


## Example 14.11 Page No : 14-27¶

In [14]:
import math

#initialisation of variables
q = 1.0	#cfs
g = 2.0	#percent
g1 = 5.6	#percent
r = 0.015	#cfs
w = 90.	#percent
Q = 10*0.9*q	#ft
p = 0.10	#ft
h = 3.48*g1**(1./3)	#ft
i = 5.6*10**-2
Q0 = 1.
s = 2*10**-2
n = 1.5*10**-2

#CALCULATIONS
q_l = round(1.87*i**0.569*(Q0/(math.sqrt(s)/n))**0.563,2)
l = q_l * 100
Q = 10*0.9*Q0
d = (3.48*32.2**(1./3)) * (q_l)**(2./3)

#RESULTS
print 'The maximum depth of flow in the gutter = %.1f in'%(d)

The maximum depth of flow in the gutter = 2.4 in