A continuous process is set up for treatment of wastewater. Each day, 103 kg cellulose and 105 kg bacteria enter in the feed stream, while 102 kg cellulose and 1.5 x 102 kg bacteria leave in the effluent. The rate of cellulose digestion by the bacteria is 8 x 102 kg d-1. The rate of bacterial growth is 4x 102 kg d-l; the rate of cell death by lysis is 6 x 102 kg d-1. Write balances for cellulose and bacteria in the system.

Correct Answer: 1× 102 kg, 9.965 × 104 kg
Cellulose is not generated by the process, only consumed. Using a basis of 1 day, the cellulose balance in kg is : (103 – 102 + 0 – 8 x 102) = accumulation Therefore, 1× 102 kg cellulose accumulates in the system each day. Performing the same balance for bacteria: (105 – 1.5 x 102 + 4 x 102 – 6 x 102) = accumulation Therefore, 9.965 × 104 kg bacterial cells accumulate in the system each day.