A lake of area 10 5 m2 and depth 10 m receives an input of 400 mol/d of chemical

Question

A lake of area 10 5 m2 and depth 10 m receives an input of 400 mol/d of chemical in an effluent discharge. Chemical is also present in the inflow water of 104 m3/d at a concentration of 0.01 mol/m3. The chemical reacts with a first order rate constant of 10-3 h -1. The outflow is 8000 m3/d, there being some loss of water by evaporation (assume none of the chemical volatilizes). Assuming that the lake water is well mixed, calculate the concentration in the lake outflow (moles/m3 ), the output by flow and loss by reaction in units mol/d.

Explanation / Answer

Volume of lake, V = area* depth= 10.*10= 1050 m3

Water entering =104 m3/day

Loss of chemical due to first order chemical reaction= KCA*V

Where K = rate constant

CA= concentration of chemical in the out flow

Moles balance of chemical in a continuous mixed CSTR can be written as

Moles in = Moles out+ loss due to chemical reaction

104 (m3/d)*0.01mol/m3+ 400 (mol/day)= 8000 (m3/d)*CA+ (10-3/hr)*24hr/day*1050

1.04 +400=8000*CA+CA*10-3*24*105 =CA*(8000+25.2)

CA= 401.04/8025.2 = 0.049 mol/m3

Loss of Chemical due to Chemical Reaction= (10-3*24)*0.049*1050=1.2348 mol/day

Output by flow =8000*m3/day* 0.049 mol/m3= 392 mol/day

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