A lake of area 10^5 m^2 and depth 10 m receives an input of 400 mol/d of chemica

Question

A lake of area 10^5 m^2 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 10^4 m^3/d at a concentration of 0.01 mol/m^3. The chemical reacts with a first order rate constant of 10^-3 h^-1. The outflow is 8000 m^3/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/m^3), the output by flow and loss by reaction in units mol/d.

Explanation / Answer

Solution:

Volume (V) = 106 m3

Discharge = 400 mol/day

Inflow = 104 m3/day * 0.01 mol/m3 = 100 mol/day

Total Input = 500 mol/day

Reaction rate = VCk = 106 m3 * C mol/m3 * 10-3 h-1 * 24h/day = 24 * 103C mol/day (C = water concentration)

Outflow = 8000 m3 /day * C mol/m3 = 0.8 * 104 C mol/day

So,

500 =  24 * 103C mol/day + 0.8 * 104 C mol/day =  32 * 103C mol/day

C = 500 /  (32 * 103 mol/day)

C(concentration in lake flow)= 1.5625 * 10-2 mol/m3

Output by flow =  0.8 * 104 * 1.5625 * 10-2  mol/day = 125 mol/day

loss by reaction = 24 * 103 * 1.5625 * 10-2 mol/day = 375 mol/day

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.