Answer:
[tex]t_c = 2.87\ DAYS[/tex]
Explanation:
Given data:
Initial BOD level is [tex]12 mg l^{-1}[/tex]
temperature of river is 10 degree C
Rate constant is [tex]kd = 0.30 day^{-1}[/tex]
Reaeration rate constant is kr [tex]0.40 day^{-1}[/tex]
As DO in river is at saturation hence initial Da is zero
critical temperature is calculated as
[tex]t_c = \frac{1}{K_r - k_d} ln[ \frac{k_r}{k_d}(1 -Da\frac{k_r -k_d}{k_d L_a})][/tex]
[tex]t_c = \frac{1}{K_r - k_d} ln[ \frac{k_r}{k_d}(1 -0)][/tex]
[tex]t_c = \frac{1}{0.40 - 0.30} ln[ \frac{0.40}{0.30}(1 -0)][/tex]
[tex]t_c = 2.87\ DAYS[/tex]