1. Consider the behavior of the chemical hexamethyldeath (HMD) in a lake. The la
ID: 886689 • Letter: 1
Question
1. Consider the behavior of the chemical hexamethyldeath (HMD) in a lake. The lake has a constant volume of 10^8 L, and it has one outflow and two inputs:
• a factory waste: flow = 7 L/s and HMD = 100 mg/L
• a river: flow = 50 L/s and HMD = 0 HMD decays with a rate k = 0.2/d. Assume well mixed conditions in the lake. Calculate the HMD concentration (mg/L) in the outlet of the lake.
2. Revisit the lake in Problem 1 above. Suppose the HMD concentration in the waste flow suddenly decreases to 0 mg/L.
a) Calculate the HMD concentration (mg/L) in the lake outlet 2 days later.
b) After the waste concentration decreases, is the change in HMD mainly due to reaction (decay) or dilution? Briefly explain.
ps: This is a fundamental Evironmental enginnering problem.
Explanation / Answer
1.Let'sconsider for 2 days
The final volume of water after 2 days including outflow = 108 L + 7L x60x60x24x2 + 50x60x60x24x2
= 108 L +1209600L +8640000L = 109849600 L
Mass of HMD in the totalvolume of water after 2 day = 100 mg/Lx(1209600L) = 120960000 mg.
Given Rate of decay of HMD = 0.2 mg/d
HMD decayed in 2 days = 0.2mg/d x 2 d = 0.4 mg
Total MHD left in the total volume of water = 120960000 mg. - 0.4 mg= 120959999.6 mg
Hence HMD concentration (mg/L) in the total volume of water = 120959999.6 mg / 109849600 L = 1.101142 mg/L
2(a): HMD concentration in the waste flow suddenly decreases to 0 mg/L.
t = 2 day
Total time = 2+2 = 4 d
Total volume of water entering in these 2 days = 7L x60x60x24x2 + 50x60x60x24x2
= 1209600L +8640000L = 9849600 L
Total volume of water including outflow in 4 days = 108 L + 2x9849600 = 119699200 L
Total HMD decayed in 4 days = 0.2mg/d x 4 d = 0.8 mg
HMD left in the total volume of water = 120960000 mg - 0.8 mg = 120959999.2 mg
HMD concentration (mg/L) in the lake outlet 2 days later(after a total of 4 days)
= 120959999.2 mg / 119699200 L = 1.01053 mg/L
(b) Hence the HMD concentration decreases when we consider decay in 2 days is from 1.101142 mg/L to 1.01053 mg/L
Hence the decrease in concentration = 0.090612 mg/L
Now if we conside neglect for this 4 days,
Total HMD in 4 days = 120960000 mg + 0 mg = 120960000 mg
Total volume of water including outflow in 4 days = 108 L + 2x9849600 = 119699200 L
Hence Concentration without decay = 120960000 mg / 119699200 L = 1.01053 mg/L, which is almost same as the concentration when we consider decay
Hence the decrease in concentration is mainly due to dilution
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