Quality And Characteristics Of Waste Water – GATE CE Environmental Engineering Practice Questions & PYQs | BattleExam | BattleExam

Q19GATE 2004MCQ

A portion of waste water sample was subjected to standard BOD test (5 days, 20 $^{\circ}$ C), yielding a value of 180 mg/L. The reaction rate constant (to the base'e') at 20 $^{\circ}$ C was taken as 0.18 per day. The reaction rate constant at other temperature may be estimated by $k_{T}=k_{20}(1.047)^{T-20}$ . The temperature at which the other portion of the sample shoudl be tested, to exert the same BOD in 2.5 days, is

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📖 Explanation

Given data:

\begin{aligned} k_{D1} &=0.434 \\ k_{e} &=0.434\times 0.18=0.0781\: d^{-} \\ t_{1} &=5\: \text{days} \\ Y_{1} &=180\: mg/l \\ Y_{2} &=180\: mg/l \\ L_{01} &= \text{ultimate BOD}=L_{02}\\ t_{2} &=2.5\: days \\ k_{D2} &=? \\ \because \; \; \frac{Y_{1}}{Y_{2}} &=\frac{1-10^{-k_{D1}t_{1}}}{1-10^{-k_{D2}t_{2}}} \\ \Rightarrow \; \; \frac{180}{180} &=\frac{1-10^{-0.0781\times 5}}{1-10^{-k_{D2}\times 2.5}} \\ \Rightarrow \; \; \;10^{2.5k_{D2}} &=10^{0.3906} \\ \Rightarrow \; \; k_{D2} &=\frac{0.3906}{2.5}=0.15624 \\ \text{Now}\; \; k_{D2} &=k_{D1}\left[ 1.047 \right]^{T-20} \\ \Rightarrow \; \; 0.15624 &=0.434\times 0.18\left[ 1.047 \right]^{T-20} \\ \Rightarrow \; \; 2 &=\left[ 1.047 \right]^{T-20} \\ \Rightarrow \; \; \log 2 &= \left ( T-20 \right )\log 1.047\\ \Rightarrow \; \; T &=35.09^{\circ}C \end{aligned}