# Chapter 7 : Rainfall and Runoff¶

## Example 7.5 Page No : 7-15¶

In [1]:
import math

#initialisation of variables
n = 20.                 	#ft
s = math.sqrt(12676/19.)	#ft
c = 45.5	#ft
q = 551400	#ft
q1 = 12700	#ft
h = 8.5	#ft
w = s/c	#ft

#CALCULATIONS
D = q/(2*s*q1)	#cfs
D1 = D*(1+h/n)	#cfs

#RESULTS
print 'the record runoff of a stream draining = %.2f cfs'%(D1)

the record runoff of a stream draining = 1.20 cfs


## Example 7.6 Page No : 7-19¶

In [1]:

#initialisation of variables
i = 16./(62)**0.66	#in hr
q = (16*10**0.31)/(62)**0.66	#in hr
c = 1.0	#max
C1 = c*(0.01)**0.31	#in
C2 = c*(0.1)**0.31	#in
x1 = 640	#cfs

#CALCULATIONS
Y1 = C1*i*c*x1	#cfs
Y2 = C2*q*c*x1	#cfs

#RESULTS
print 'the time of concentration = %.f cfs'%(Y2)

# rounding off error.

the time of concentration = 672 cfs


## Example 7.8 Page No : 7-25¶

In [2]:
import math

#initialisation of variables
d = 163*48.5	#cfs
a = 48.5	#ft
q = 100	#cfs
Q = 45.5*a	#cfs
c = 0.57	#cfs
v = 1.8	#cfs
p = 0.45	#ft

#CALCULATIONS
P = d/(q*math.sqrt(a))	#percent
C = Q/(a**0.8*(1+2*a**-0.3))	#cfs
d1 = 2.6	#cfs
T = (1-p*c+v*c*2)	#cfs

#RESULTS
print 'the meyers rating  = %.1f percent'%(P)
print 'the magnitude of the maximum peak flood  = %.1f cfs'%(T)

the meyers rating  = 11.4 percent
the magnitude of the maximum peak flood  = 2.8 cfs