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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 |
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Answer» 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. |
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