Question

A printed circuit board manufacturing plant is discharging the chemical 1,1,1-trichloroethane into a river that subsequently flows into a lake, the source of drinking water for a small town. the levels of TCA in this lake currently fail to meet drinking water objectives of 10ppb. by assuming that the lake is a well mixed system, that
concentrations are at steady state and the only processes acting on this chemical in the lake are volatization and biodegredation, how high can the influent concentration of TCA be at the point where the river enters?
given data:
TCA volatization rate= 3×10^-4 ug/(cm^2sec)
TCA biodeg. rate= 6×10^-3 uM/day
river inflow= 200gal/min
river outflow=200gal/min
lake volume= 1.5×10^6 gal
avg. lake depth=10ft

Answer 1

First, let's convert some of the data into a much more confortable units:

TCA volatilization rate = 3 mu g/(m2.s); TCA biodegradation rate = 9.2 Time 10-3 mu g/(m3.s); River inflow to lake = 12.6 L/s; River outflow from lake = 12.6 L/s; Lake volume = 5.7 Times 106 L; Average lake depth = 3 m.

Now solving the exercise:

rate of mass in =x* 12.6 ( where x is assumed concentration of TCA)

loss of TCA through Volatilzation = 3ug/m2s*{5.7*106/1000 m3}/3 =5700 ug/s

Loss due to biodegradation = 9.2X10^-3*5700 ug/s= 52.44 ug/s

outlet flow rate of TCA =10*12.6 ug/s =126

From conservation law

Rate of mass in = rate of mass out+ loss due to volatilization + loss due to biodegradation

12.6x =5700+52.44+126=5878.44

x= 5878.44/12= 489.87 ug/l

Hope this helps