Question

the bod of a sample is 120mg/l.if k for the sample is 0.28 day.calculate how much bod is exerted after 10 days

Answer 1

Explanation:

Given:

Ultimate of BOD $(L)=120$ $mg/l$

Time in days $(t)=0.28$

Deoxygenationrate rate constant $(k)$ in $days^{-1} =10$.

To find:

$(y_{t} )$- Amount of BOD exerted at the time $(t)$ in $mg/l$ .

Solution:

Biochemical Oxygen Demand (BOD) exertion formula,

⇒ $y_{t} =L(1-e^{-kt} )$

• $y_{t}$ is the amount of BOD exerted at the time $t$ in $mg/l$.

• $L$ is the ultimate of BOD in $mg/l$.

• $t$ is the time in days.

• $k$ is the deoxygenation rate constant in $days^{-1} .$

Now apply the given values in the formula,

⇒ $y_t =120(1-e^{-(0.28)(10)} )$

⇒ $y_{t} =120(1-e^{-2.80000000000000003})$

⇒ $y_{t} =120$ × $0.9391899373747821$

⇒ $y_{t} =112.70$ $mg/l$

∴ $y_{t} =113$ $mg/l$

Answer:

Hence the value of $y_{t}$ is $113$ $mg/l.$