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GATE CE 2014 Official Paper: Shift 1

Option 1 : 60.50

__Concept:__

From Chick’s Watson’s Law:

\({{\rm{C}}^{\rm{n}}} \times {\rm{t\;}} = {\rm{constant}}\)

Where C is the concentration of disinfectant

We know Concentration C = W/Q

Where W is the weight of disinfectant per day

Q is the discharge per day

‘t’ is the detention time or contact time

‘n’ is the dilution factor

\({\rm{Contact\;time\;}}\left( {\rm{t}} \right) = \frac{{{\rm{Volume\;of\;contact\;unit}}}}{{{\rm{Flow\;rate}}}}\)

\({\rm{t}} = \frac{{\rm{V}}}{{\rm{Q}}}\)

As \({{\rm{C}}^{\rm{n}}} \times {\rm{t}} = {\rm{constant}}\)

\({\rm{C}}_1^{\rm{n}} \times {{\rm{t}}_1} = {\rm{C}}_2^{\rm{n}} \times {{\rm{t}}_2} = {\rm{constant}}\)

\({\rm{C}}_1^{\rm{n}} \times \frac{{\rm{V}}}{{{{\rm{Q}}_1}}} = {\rm{C}}_2^{\rm{n}} \times \frac{{\rm{V}}}{{{{\rm{Q}}_2}}}\)

As volume being constant and n = 1

Equation becomes

\(\frac{{{W_1}}}{{{Q_1}}} \times \frac{1}{{{Q_1}}} = \frac{{{W_2}}}{{{Q_2}}} \times \frac{1}{{{Q_2}}}\)

\(\frac{{{W_1}}}{{Q_1^2}} = \frac{{{W_2}}}{{Q_2^2}}\)

__Calculation:__

Given,

The weight of chlorine dose initially (W1) is 32 kg/day

Flow rate initially (Q1) = 16 MLD

Dilution factor = 1

(i) When flow rate changes from 16 MLD to 22 MLD

\(\frac{{{W_1}}}{{Q_1^2}} = \frac{{{W_2}}}{{Q_2^2}}\)

\(\frac{{32\frac{{{\rm{kg}}}}{{{\rm{day}}}}}}{{16^2{\rm{\;MLD}}}} = {\rm{\;}}\frac{{{{\rm{W}}_2}}}{{22^2{\rm{\;MLD}}}}\)

W2 = 60.5 kg/day

∴ The amount of chlorine dosage added should be 60.5 kg/day.