# Chapter 4 Dispersion¶

## Example 4_2_1 pgno:100¶

In [2]:
#initialization of variables
z = 80. # metres
c1 = 410. #ppm
c = 860. # ppm
d = 2. #km
v = 0.6 #km/hr
r = 3600. #sec/hr
from math import log

#Calculations
t1 = (d/v)*r#sec
E = (-((z**2)/(4*t1))/(log(410./860.)))*10**4# cm**2/sec#answer in textbook is wrong
d2 = 15. #km
c2 = c*((d/d2)**0.5)#ppm
#Results
print"The value of dispersion coefficent is  cm**2/sec",round(E,3)
print"\n The value of maximum concentration  at 15 km downstream is  ppm",round(c2)

The value of dispersion coefficent is  cm**2/sec 1799.916

The value of maximum concentration  at 15 km downstream is  ppm 314.0


## Example 4_2_2 pgno:100¶

In [4]:
#initialization of variables
d = 10. #cm
s = 3. # km
v = 500. #cm/sec
nu = 0.15 # cm**2/sec
#Calculations
E = 0.5*d*v # cm**2/s
c1 = 1000. # m/km
c2 = 1./100. # m/cm
z = (4*E*c1*c2*s/v)**0.5
percent = z*100/(s*c1)
#Results
print"the concentration change is",round(z)
print" The percent of pipe containing mixed gases is  percent",round(percent,2)

the concentration change is 24.0
The percent of pipe containing mixed gases is  percent 0.82