Question

A water treatment plant capacity, $1m^3/s$ has filter boxes of dimension $6\times 10m$. Loading rate to the filter is $120m^3/day/m^2$. When two of the filters are out of service for back washing, the loading rate (in $m^3/d/m^2)$ is

• A. 144

Answer 1

The correct option is A 144
Total water filtered in a day

= $24\times 3600\times 1$

$=86400m^3/day$

Total surface area of filter required

$=\frac{86400m^3/day}{120m^3/day/m^2}=720m^2$

Area of one filter = $6\times 10=60m^2$

Total no. of filter = $\frac{720}{60}=12filters$

If two filters are out of service no. of filters remain 12-2 = 10 filters

Total surface area of filter = $60\times 10=600m^2$

The new loading rate= $\frac{86400m^3/day}{600m^2}$
$=144m^3/day/m^2$