Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A printed circuit board manufacturing plant is discharging the chemical 1,1,1-tr

ID: 914304 • Letter: A

Question

A printed circuit board manufacturing plant is discharging the chemical 1,1,1-trichloroethane (TCA) 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 10 ppb. As an environmental specialist with the state department of environmental protection, you have been asked to help determine discharge requirements for the manufacturing plant. By assuming that the lake is a well-mixed system, that concentrations are at steady state, and that the only processes acting on this chemical in the lake are volatilization and biodegradation, how high can the influent concentration of TCA (C_in) be at the point where the river enters the lake Some pertinent data are

Explanation / Answer

Let the maximum qunatity of 1,1,1- tricholoroethane be X g/L.

Lake volume = 1.5 x 106 gal = (1.5 x 106 x 0.133681) ft3 =  200521.5 ft3          [ Since 1 gal = 0.133681 ft3]

Area of the lake = 200521.5 ft3 / 10 ft = 20052.15ft2 = ( 20052.15 x 929.03) cm2 = 18629048.91 cm2 [ 1 ft2 = 929.03 cm2]

Lake volume in L = 5670000 L [1 gal = 3.7851 L ]

Volatilization rate = (3 x 10-4 x10-6 ) ( 86400) g/ cm2 . day = 0.00002592 g/ cm2 . day

= (0.00002592 g/ cm2 . day ) ( 18629048.91 cm2 ) = 482.864 g/day

Biodegradation rate = 6 x 10 -3 X 10 -6 M /day = 6 x10-9 M /day = 0.03402 mol / day = 4.538 g/ day

River inflow into the lake = 200 gal / min = ( 200 x 3.78541) ( 1440 ) L/day = 1090198.08 L/day

Mass of chemical inflow = ( 1090198.08 X ) g/ day

Mass of chemical outflow = 0 [ Since concentration is constant]

Maximum mass of the chemical in the lake / L = { ( 1090198.08 X) - 0 + 0 - 482.864 - 4.538} / 5670000

By the problem the maximum mass should be = 0.01 g/L

0.01 = 0.19227x - 8.59 x 10 ^ -15

x = 0.052

C(in) = 0.052g / L = 3.93 x 10 ^ -4 M

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote