# Chapter 6 : Elements of Hydrology¶

## Example 6.1 Page No : 6-6¶

In [1]:

#initialisation of variables
H = 1360	#ft
t = 60	#f
a = (10**3)*5.5*(10**-3)	#f
q = (1.36*10**3)*5.5*(10**-3)	#f
s = (4-1.36)*(10**3)*(3.2*10**-3)	#f

#CALCULATIONS
T = t-q-s	#F
T1 = T+3*a	#F

#RESULTS
print 'the temperature at the mountain top = %.0f F'%(T)
print 'the temperature on the plain beyond the mountain = %.1f F'%(T1)

the temperature at the mountain top = 44 F
the temperature on the plain beyond the mountain = 60.6 F


## Example 6.2 Page No : 6-12¶

In [3]:

#initialisation of variables
t = 60	#f
v = 0.52	#in
t1 = 80	#F
p = 40	#percent
v1 = 1.03*0.40	#in
w = 8	#mph
pa = 29.0	#in
p1 = 0.497	#ft
q = 1.32*10**-2	#ft
r = 0.268	#ft

#CALCULATIONS
E = p1*(1-q*pa)*(1+r*w)*(v-.41)	#in

#RESULTS
print 'the evaporation for the a day during = %.2e in'%(E)

# note : answer is slightly different because of rounding off error.

the evaporation for the a day during = 1.06e-01 in


## Example 6.3 Page No : 6-19¶

In [4]:

#initialisation of variables
t = 47	#f
q = 8000	#ft
a = 100	#ft
d = 0.10	#in
d1 = 7	#degree days
s1 = 14000	#ft
s2 = 7000	#ft
s = 1000	#ft
g = 32	#ft
h = 17.37	#ft
h1 = 1.547	#ft

#CALCULATIONS
T = q+s*(t-g)/3	#ft
T1 = t-3*1	#F
T2 = (T1+g)/2	#F
T3 = d1*d*a	#sq mile in
M = h*T3	#mgd
M1 = M*h1	#cfs

#RESULTS
print 'the upper boundary of the melting zone and temperature at the snow line = %.0f F'%(T1)
print 'The average temperature of  = %d cfs'%(M1)

the upper boundary of the melting zone and temperature at the snow line = 44 F
The average temperature of  = 1880 cfs