Question

2. A tannery wastewater treatment plant has 2250m/d of effluent having de-oxygenation rate of 0.23/d at 20°C. Determine the ultimate Biochemical Oxygen Demand (BOD) (L.) and the remaining O, after 5 days (L.). Then, compute de-oxygenation rate constant (kd) and determine the rate of de-oxygenation in mg/L/d at 25°C. Note that the stream has 5 days BOD of the waste (BOD) OF 450 mg/L. The avg speed, u, of the stream flow is 0.03 m/d. The depth.h, is 1.2m and the bed activity coefficient, n is equal to 0.32 and 0 = 1.056 (10 marks)

Answer 1

Ans) We know, BOD at any time can be given by BOD_t ,

BOD_t = L_o (1 - 10^{-Kd. t} )

where, t = time in days

L_o = Ultimate BOD

Kd = de-oxygenation rate constant

=> BOD₅ = L_o ( 1 - 10^{-Kd. t} )

Putting values,

450 = L_o ( 1 - 10^-0.23(5))

L_o = 484.28 mg/L

Now, remaining O₂ after 5 days , L_t

L_t = L_o 10^{-Kd .t}

=> L₅ = 484.28 10^-0.23(5) = 34.28 mg/L

Now, de-oxygenation rate constant at any temperature greater than 20 deg C (K_{d, T}) can be given by,  

K_{d, T} = K_d,20$\theta$^{(T - 20)}

Given , $\theta$ = 1.056 ,

=>   K_{d, T} = K_d,20 1.056^{(T - 20)}

=> K_d,25 = 0.23 (1.056)^{(25 - 20)} = 0.302 /day

Now, use relation K_d = K + ( u/H)$\eta$

where, K = BOD rate constant

u = stream velocity

H = depth of stream

=> 0.302 = K + (0.03/1.2)0.32

=> 0.302 = K + 0.008

=> K = 0.294 /day

According to Kinetics of organic matter,

dL_t/dt = - K L_t

= - 0.294(34.28)

= 10.1 mg/L/day