8. An industrial park has 3 major businesses that empty a portion of their waste
ID: 153231 • Letter: 8
Question
8. An industrial park has 3 major businesses that empty a portion of their wastewater into a shared pond. The pond is well mixed and in steady state and has two outlet points. One outlet feeds into a nearby creek at 3001.75 ft3/day and the other outlet runs into the storm sewer. The volume of the pond is 155 m3.
Discharge for three
businesses: business Wastewater Flow rate (m^3/day) THM Concentration (mg/L)
1 30 2.5
2 40 4
3 25 1.7
a) What is the concentration of THM in the storm sewer stream?
b) If the THM from the industrial park were to suddenly cease flowing in the streams, how long would it take for the THM concentration to reach zero?
Explanation / Answer
Input of wastewater from the first business= 30 cubic meter=30000 L
Input of THM from the first business(per day)=(concentration of THM in the water X amount of water)=(30000X2.5) mg=75000 mg=75 gm
Similarly:
Input of THM from the second business=(40000X4) mg=160 gm
Input from the third business=(25000X1.7) mg=42.5 gm
Total input of THM in one day=(75+160+42.5) gm=277.5 gm
Total input of westwater from the businesses=(40+30+25) cubic meter=95000 L
Output of water into the creek=3001.75 cubic ft/day=85 cubic meter/day
So, output through the storm sewer=(95-85) cubic meter=10 cubic meter
So, concentration of THM in the sewer=concentration of THM after homogenization
=amount of THM/westwater amount=(277500/95000) mg/L=2.92105263 mg/L
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Volume of the pond=155 cubic meter
For the concentration of THM to reach 0; all of these water have to be ejected out of the pond.
Rate of output of water= 95 cubic meter/day
So, time required for this water to be expelled out of the pond=(155/95) days=1.631 days
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