In a sludge digestion tank if the moisture content of sludge V1 litres is reduced from p1% to p2% the volume V2 is
A. $$\left( {\frac{{100 + {{\text{p}}_1}}}{{100 - {{\text{p}}_2}}}} \right){{\text{V}}_1}$$
B. $$\left( {\frac{{100 - {{\text{p}}_1}}}{{100 + {{\text{p}}_2}}}} \right){{\text{V}}_1}$$
C. $$\left( {\frac{{100 - {{\text{p}}_1}}}{{100 - {{\text{p}}_2}}}} \right){{\text{V}}_1}$$
D. $$\left( {\frac{{100 + {{\text{p}}_2}}}{{100 - {{\text{p}}_1}}}} \right){{\text{V}}_1}$$
Answer: Option C
This Question Belongs to Civil Engineering >> Waste Water Engineering
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