Engineering Technician Supplemental Questionnaire * 1. Community Water Systems (
ID: 295730 • Letter: E
Question
Engineering Technician Supplemental Questionnaire * 1. Community Water Systems (CWSs) routinely test their water for the presence of total coliform and Escherichia coli bacteria. In 200 words or less, explain why these two groups of bacteria provide important information to CWSs 2. Three wells are pumping their effluent into a common reservoir. The water in Well A has 10mg/L of nitrate and the well runs at 1,200 gallons per minute (gpm), the water in Well B has 15 mg/L of nitrate and the well runs at 800 gpm, and the water in Well C has 8 mg/L and the well runs at 2,000 gpm. Assuming perfect mixing, what is the concentration of nitrate in the effluent to the reservoir. Please explain in 200 words or ess 3. A CWS system must sample one of its wells (Well A) for perchiorate on a daily basis, a second well (Well B) twice per week, a surface water source once a week, a treatment plant each Monday and Friday, a reservoir every two weeks, and four points in its distribution system once per month. The same CWS must also test all of wells and surface water sources for total coliform bacteria once per week, all distribution sample points every two weeks, and all of its reservoirs on Mondays. In 200 words or less, explain how you would organize the sampling schedule.Explanation / Answer
1. If we are studying the quality of community water standard then we go for Total coliforms counts and Escherichia coli (E. coli) counts as these are "indicator" organisms of the source pathogenic organisms and since source pathogenic organisms are present in a very small number which makes it impractical to test for pathogens in every water sample collected and hence we go for the test for Total coliforms counts and Escherichia coli (E. coli) counts which will indicate the presence of pathogenic bacteria.
2. well A is pumping 10mg/L and runs at 1,200 gpm
We know 1 gallon = 3.78541 liters
1,200 gallons = 1,200 x 3.78541 = 4542.494 liters
4542.494 liters will have a nitrate concentration of = 4542.494 x 10 = 45424.94 mg = 45.42494 grams per minute
Similarly for well B 15mg/L and runs at 800 gpm
800 gallons = 800 x 3.78541 = 3028.33 liters
3028.33 liters will have a nitrate concentration of = 3028.33 x 15 = 45,424.95 mg = 45.42495 grams per minute
Similarly for well C 8mg/L and runs at 2,000 gpm
2,000 gallons = 2,000 x 3.78541 = 7570.824 liters
7570.824 liters will have a nitrate concentration of = 7570.824 x 8 = 60,566.592 mg = 60.566592 grams per minute
Nitrate concentration in reservoir = 60.566592 + 45.42495 + 45.42494 = 151.416482 grams per minute
3. Test for perchiorate will be done for well A daily, then for well B it can be done on Tuesday and saturday, Surface water source would be checked on sunday every week, treatment plant will be is sampled on monday and friday every week, reservoir would be sampled on wednesday in alternate weeks, four points will be sampled on wednesday, thursday, friday and saturday. Total coliform test will be scheduled on tuesday. Distribution sample points will be sampled on sundays and on tuesday and reservoirs on mondays.
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