3.30%) (open system, unsteady state, change with time) Consider a lake having th

Question

3.30%) (open system, unsteady state, change with time) Consider a lake having the following properties: Lake volume 10m3 Initial concentration of contaminant C° = 1 mol/m3 Water outflow rate 0.5 m/s Water inflow rate = 0 m3/s (there is no inflow (advection) of the contaminant into the lake) 1st order reaction rate constant k = 10-2 h' (i.e., contaminant is lost by reaction) (a) Calculate the total loss of the contaminant from the lake in mol/s (it will be a number times C because C is unknown except at time 0 or Co-1 mol/m) (b) Calculate the concentration (mol/m3) of contaminant after 1 day and after 10 days. Hint: k for the 1st order loss reaction - 0.778 x 105 s1 (c) Comment on the ratio of the concentrations after 1d and 10d. Does the ratio 10? Explain. (d) Calculate the time for 90% of the contaminant to be lost from the lake.

Explanation / Answer

a) Total loss of contaminant is k[C0] = 10-2 /(60x60) x 1 = 0.27 x 10 -5 mol per second.

b) The concentration (C) of contaminant after 10 days can be calculated as

C = C0 e-k x (10x24x3600)

since d[C]/dt = -k[C]

Thus, C after 10 days = 1.e-0.778x10e-5 x (240x3600) = 0.0012 mol/m3

Similarly, C after 1 day = 1.e-0.778x10e-5 x (1x24x3600) = 0.51 mol/m3

c) The ratio of C after 1 day and 10 days indicate that the contaminant lost its concentration by 425 times (0.51 / 0.0012) over a period of 9 days. The ratio is thus not equal to 10. In other words, the concentration after 10 days decreased by 99.7% as compared to that after 1 day.

d) The time (t) for 90% of the contaminant to be lost from the lake can be calculated as:

e-(0.778x10e-5) x t = 0.1; since 0.1 of the initial concentration will remain.

t = 295962 seconds, i.e. it will take 3.42 days for 90% of the contaminant to be lost.

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