Question

A continuous process is set up for treatment of wastewater. Each day, 10^3 kg cellulose and 10^5 kg bacteria enter in the feed stream, while 10^2 kg cellulose and 1.5 x 10^2 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 8 x 10^2 kg d^-1. The rate of bacterial growth is 4x 10^2 kg d^-l; the rate of cell death by lysis is 6 x 10^2 kg d^-1. Write balances for cellulose and bacteria in the system.

(a) 1× 10^3 kg, 9 × 10^3 kg

(b) 1× 10^2 kg, 9.965 × 10^4 kg

(c) 1× 10^3 kg, 9.964 × 10^4 kg

(d) 1× 10^2 kg, 9 × 10^3 kg

I had been asked this question in class test.

The above asked question is from Law of Conservation of Mass topic in division Material Balance of Bioprocess Engineering

Answer 1

Right answer is (b) 1× 10^2 kg, 9.965 × 10^4 kg

Easiest explanation: Cellulose is not generated by the process, only consumed. Using a basis of 1 day, the cellulose balance in kg is :

(10^3 – 10^2 + 0 – 8 x 10^2) = accumulation

Therefore, 1× 10^2 kg cellulose accumulates in the system each day.

Performing the same balance for bacteria:

(10^5 – 1.5 x 10^2 + 4 x 10^2 – 6 x 10^2) = accumulation

Therefore, 9.965 × 10^4 kg bacterial cells accumulate in the system each day.