A local fishing village is located on the bank of a small lake with a volume of

Question

A local fishing village is located on the bank of a small lake with a volume of approximately 30 million liters. Upstream from the lake is a chemical waste plant that has been slowly polluting the waters for years now. As a solution to the pollution, the chemical company installed a water treatment system in the lake that removes a certain amount of the pollution each day. The current pollution level in the lake is 0.015 mol/L of toxic waste. The chemical waste plant deposits approximately 3500 kg of waste each day, and the water treatment system can remove 120 kg of waste per hour. a) Draw a system with boundaries and show all inlets and outlets. b) Use a mass balance to determine the accumulation or reduction of total waste in the lake in kg/day. c) If the fish in the lake can only survive at a pollution level at or below 0.4 mol/L, will the fish ever die? If so, how long will it take to reach a lethal pollution point? The molar mass of the toxic waste is 142.6 g/mol. Assume that the volume of the waste added to the lake is negligible. d) How many treatment systems would be required to completely prevent any waste accumulation? BE CAREFUL WITH MASS AND MOL IN THIS PROBLEM....WATCH YOUR UNITS!

Explanation / Answer

part(b)

toxic waste in = 3500 kg/hr

toxic waste out = 120 kg/hour

1 day = 24 hours

hence, in 1 hour = 120 kg waste is removed

in 24 hours, waste removed = 24*120 = 2880 kg/day

part (c)

molar mass of waste = 142.6 g/mol

current pollution level = 0.015 mol/L

moles already present = 0.015mol/L*30*106 = 450000 moles of toxic waste

maximum pollution level at which fishes will die = 0.4mol/L*30*106 = 12*106 moles

Hence, fishes will not die with current pollution level

taking basis of 1 day

3500 kg is added and 2880 kg is removed. So effective accumulation = 3500-2880 = 620kg

level of waste at which fishes will die = 12*106*142.6 = 1.7112*109 g = 1.7112*106 kg toxic waste

So, 620 kg waste is accumulates in = 1 day

1.7112*106 kg waste will get accumulates in = x

x = 2760 days = 7.56 years

part (d)

for 1 wastewater treatment system, 2880 kg waste is removed per day

to remove 3500 kg of toxic waste, for the same size of wastewater system, 2 will be required. With 2 systems in place, we can remove = 2880*2 = 5760 kg/day of waste.

2 systems will be more than enough to prevent any accumulation of waste

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