If 5 day BOD of waste water is 350mg/l what will be it's 7 day 25°C BOD? $K20=0.1/day$. Assume data if required:

given:-

5 day 20°C BOD =350mg/l

$K_{20}=0.1/d$

1)To find organic matter present(L) at the start of BOD:-

$Y_t=L[1-(10)^{-K_{20}*t}]$

Where $Y_t$:total amount of organic matter oxidised in t days =320mg/l

L= organic matter present at the start of BOD=?

$t=5 day K_{20}=0.1$

$350=L[1-(10)^{-0.1*5}]$

$350=L[1-(10)^{-0.5}]$

$350=L[1-0.316]$

350=L(0.684)

$L=\frac{350}{0.684}$

L=511.69mg/l

2)To workout $K_{D}$ at 25°C BOD:-

$K_t=K_{20}[1.047]^{T-20}$

$K_{25}=K_{20}[1.047]^{25-20}$

$=0.1[1.047]^5$

=0.1*1.258

$K_{25}=0.1258$

3)To calculate $Y_t$ for 7 days (i.e. $Y_7$ at 25°C)

We know

$Y_t=L[1-(10)^{-K_t*t}]$

=511.69[1-(10)^{-t_{25}*t}]

=511.69[1-(10)^{-0.1258*7}]

=511.69[1-(10)^{-0.8806}]

=511.69[1-0.131]

=511.69*0.869

=444.658

$Y_7(at \ 25°C)=444.66 \ mg/l$