Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The volume of each reactor is [tex]V= 691.53\ m^3[/tex]
b
The rate at which cells must be discharged in the unit is [tex]R = 1002.22 kg/day[/tex]
Explanation:
From the question we are told that
The water flow rate of the units is [tex]\r q = 0.5 m^3/s[/tex]
The [tex]BOD_5[/tex](Biochemical oxygen demand) of the effluent suspended solid is [tex]S = 70[/tex]% of allowable suspended solid concentration
Given that the flow rate for two identical parallel units then for a single unit the flow rate would be
[tex]Q = \frac{\r q}{2} = \frac{0.5}{2} = 0.25 m^3 /s[/tex]
Generally the required BOD in terms of EPA(Effluent Guidelines) is concerned is [tex]S_e = 30 mg/l[/tex]
Now Sludge age of this unit of reactor is mathematically evaluated from this expression
[tex]S_e = \frac{k_s (1 + k_d \theta_c)}{\theta_c (\mu_m - k_d) - 1}[/tex]
Where [tex]\theta_c[/tex] is the Sludge age
Substituting values
[tex]30 = \frac{30 (1 + 0.05 \theta_c)}{\theta_c (3 - 0.05) - 1}[/tex]
[tex]30 = \frac{ 30 + 1.5\theta_c)}{( 3 \theta_c - 0.05 \theta_c) - 1}[/tex]
=> [tex]\theta _ c = 0.69 \ days = 59616s[/tex]
Now the volume of the reactor is mathematically evaluated from this relation
[tex]VX = \frac{\r q Y (S_o - S) \theta_c} {1 - k_d \theta_c}[/tex]
Making V the subject of the formula
[tex]V = \frac{\r q Y (S_o - S) \theta_c} {1 - k_d \theta_c} * \frac{1}{X}[/tex]
Substituting values
[tex]V = \frac{0.25 * 4 (150 - 30 ) * 59616 }{1+ 0.05 * 0.69} * \frac{1}{1000}[/tex]
=> [tex]V= 691.53\ m^3[/tex]
We can obtain the cell mass been wasted per day using this expression
[tex]Q_w X_i = (\frac{VX}{\theta _c})kg /day[/tex]
Where [tex]Q_w[/tex] is given as [tex]Q_w = 10 L/s = 10 *10^{-3} \ m^3 /s[/tex]
[tex]Q_w X_i[/tex] is the sludge been wasted per day
[tex]X_i[/tex] is the discharge concentration for each unit
Substituting values
[tex]X_i Q_w = \frac{1002.22}{86400}[/tex]
[tex]X_i = \frac{1002.22}{86400} * \frac{1}{10*10^{-3}}[/tex]
Now the rate at which they discharge this concentration is mathematically evaluated as
[tex]R = X_i * Q_w * 86400[/tex]
substituting values
[tex]R = 1002.22 kg/day[/tex]
