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A portion of wastewater sample was subjected to standard BOD test (5 days, 20°C), yielding a value of 180 mg/1. The reaction rate constant (to the base 'e') at 20°C was taken as 0.08 per day. The reaction rate constant at other temperature may be estimated by kr = (k 20 (1.047)T-20. The temperature at which the other portion of the sample should be tested, to exert the same BOD in 2.5 days, is
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- 4.9°C
- 24.9°C
- 31.7°C
- 35.0°C
Correct Option: D
BOD520 = 180 mg/L = (L520)
BOD520 = BOD2.5T = 180
L520 = L0 [1 – e–kt]
| L0 = | = | ||
| 1 – e–kt | 1 – e–kt |
180 = L0 (1 – e–k20×5) = L0(1 – e–kT×2.8)
Upon simplifying, we get,
–k20 × 5 = – kT × 2.5
kt = k20 (1.04T)T–20
∴ 0.18 × 5 = 0.18 × (1047)T-20 × 2.5
∴ T = 35°C
