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A manufacture in Ontario has been fined because it has been releasing 5 L/s effl

ID: 986837 • Letter: A

Question

A manufacture in Ontario has been fined because it has been releasing 5 L/s effluent having zinc concentration of 0.1 mg/L into a river. Upstream of the factory, the stream water rate is 100 L/s with zinc concentration of approximately zero. The factory has been forced to reduce the zinc concentration of the effluent to below 20 pg/L. The engineer that is an employee of the factory recommends that they divert parts of the stream into the site and dilute the effluent to bring its zinc concentration to the required level. a) Calculate the present concentration of zinc several miles downstream of the plant where the zinc is diluted in stream completely. b) Compute the amount of water needed to be diverted to the site to achieve the required zinc concentration of effluent. c) Determine the concentration of zinc downstream of the plant where the zinc is diluted in stream completely if the engineer's plan is put into operation. 3) The following two-pond system receives water with a flow rate of 1.25 MGD (million gallon per day) and a chloride concentration of C0= 45 mg/L. Due to reaction with a decay rate of k = 0.24 /day, the chloride concentration reduces downstream of each pond. The volume of the left and the right ponds are, respectively, 6.4 and 4.2 million gallons. Determine Ci and C2. Q= 1.25 MGD 4) Consider the air over a city (60x25 km). Wind with 18 km/hr speed brings the fresh air into the city from the west side of the region (25 km wide). Assume that 15 kg/s carbon monoxide (CO) is emitted from different sources in the city to the air. Assume, also, that the CO is conservative and uniformly spread all over the area up to an altitude of one kilometer. Modeling the air over the city as a steady-state system, find the CO concentration of the air over the city in ppm if the temperature is -20 degree C and pressure is 1 atm.

Explanation / Answer

4) use PV=nRT

n/V =P/RT

C =P/RT = 0.000475 moles

3) k = C/C0

C1 = 45*0.24 = 10.8

C2 = 2.592

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