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The 5-day BOD of a waste water sample is obtained as 190 mg/l (with k = 0.01h–1). The ultimate oxygen demand (mg/l) of the sample will be
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- 3800
- 475
- 271
- 190
Correct Option: C
BOD5 = BOD4[1 – 10–kD.t]
| BOD4 = | ||
| [1 - 10–kD.t] |
kD = 0.434 k = .434 × 0.01 h–1
kDt = .434 × .01 × 5 × 24 = 0.5208
| BOD4 = | ||
| [1 - 10-0.5208] |
= 271 mg/l.
