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Texas Waste Water Treatment Ch#15.1 Water Composition

Take the test..... duh....... Study questions for the class C waste water license. Info source: TEEX infrastructure training & safety institute waste water treatment resource book Module #15.1 Water Composition v2 2023.
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Last updated: January 17, 2024
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First submittedAugust 24, 2023
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1. What should be the flow capacity of a wastewater treatment plant (flow rate in MGD and GPM) needed handle the discharge from 300,000 people?
30MGD & 20,833GPM
3MGD & 2,083GPM
300MGD & 208,333GPM
60MGD & 41,666GPM
TLDR: # people × 100 gal/person/day = GPD => 300,000 people × 100 gal/person/day = 30,000,000GPD! ||| GPD / 1,000,000 = MGD => 30,000,000GPD / 1,000,000 = 30MGD! ||| GPD / 1440 min/day = GPM => 30,000,000GPD / 1440 min/day = 20,833GPM! ||| EXPLANATION: It is assumed that on average 100 gallons of waste water will be collected per person in the collection area each day therefore to find the number of gallons flowing into the plant you would multiply the number of people served by the amount each person is assumed to contribute (300K people X 100GAL/DAY = 30,000,000GPD) ||| The next section simply converts gallons per day to millions of gallons per day to make the number generally easier to work with for things like flow monitoring and more advanced calculations. This figure answers part one of the question but will not be used again in this problem. it is found by dividing the calculated number of gallons recieved by one million to find out how many millions of gallons are recieved. (30,000,000GPD / 1,000,000 = 30MGD) ||| now the question wants to know how many gallons per minute that amount is. luckily we already know how many gallons per day are recieved and we know there are 1440 minutes in each day. (24 hours in each day times 60 minutes in each hour is 1440 minutes) therefore if we divide the number of gallons recieved in a day by the number of minutes in a day it will tell us how many of those gallons we recieve in each minute. (30,000,000GPD / 1440 min/day = 20,833GPM) this gives us a figure that would help us choose pumps, figure out required pipe diameters, and much more.
2. A Publicly Operated Treatment Works (POTW) has a flow of 3.0 MGD and an influent Biochemical Oxygen Demand (BOD) of 200 mg/L. Find the pounds of BOD in the raw wastewater.
25.02 lbs/day
5,004 lbs/day
1,668 lbs/day
5,400 lbs/day
TLDR: MGD × 8.34 lb/million gal × mg/L = lbs/day => 3.0 × 8.34 lb/million gal × 200 mg/L = 5,004 lbs/day! ||| EXPLANATION: This problem is made much easier by remembering the mnemonic provided in the top left corner. The problem tells you how many millions of gallons are recieved each day, how much BOD is suspended in each of the liters received, and wants to know how much all the BOD received in a day weighs in pounds. To explain why the mnemonic works you have to understand that the "8.34" conversion factor is BOTH what one gallon of water weighs AND how much one milligram of a substance in 1 liter of water weighs in pounds per million gallons. (1mg/L = 0.000008345... lbs/gal = 8.34 lbs/million gallons) For this problem it is that second reason that is why we are using it in this problem. To find the answer you multiply the millions of gallons recieved in each day by how much one miligram per leiter in each of one million gallons of water would weigh and then multiplying it by the number of miligrams per leiter you are actually recieving. (MGD × 8.34 lb/million gal (which results in the weight in pounds if you were recieving 1mg/L) × # of mg/L recieve = LBS/day => 3.0 × 8.34 lbs/million gal × 200 mg/L = 5,004 lbs/day!)
3. The raw BOD of a wastewater treatment plant is 220 mg/L and the effluent BOD is 20 mg/L. Find the percent reduction of BOD.
95%
80%
85.3%
90.9%
TLDR: ((In - Out) / In) X 100 = % Reduction => ((220 mg/L - 20 mg/L) / 220 mg/L) X 100 = 90.9% ||| EXPLANATION: First you need to figure out how much BOD you removed. To do that you subtract the amount that left the plant from the amount that entered the plant. (In - Out) the difference is the amount that was removed in processing. then we divide that number by the amount we started with which will yeild the decimal form of our percentage. ((In - Out) / In) when we multiply that by 100 the decimal moves two places to the right yeilding the final answer. ((In - Out) / In) X 100 = % Reduction => ((220 mg/L - 20 mg/L) / 220 mg/L) X 100 = 90.9%
4. If the BOD population equivalent is 0.17 pounds of BOD per person, about how many people are contributing their wastewater to the Sewage Treatment Plant (STP) in problem #2?
30,000 people
20,000 people
29,435 people
27,342 people
TLDR: lb BOD / 0.17 lb/BOD/person = # people => 5,004 lb / 0.17 lb/person = 29,435 people ||| EXPLANATION: First go collect the answer you got from question two... dont worry, ill wait...... then divide that number by the conversion factor they gave in the question. the end..... the population equivelent it is referencing is the assumption that on average every person in the collection area will contribute 0.17 lbs of BOD every day. Assuming you got the correct answer for question #2 you have 5,004 lbs of BOD and it wants to know how many people it took to produce that amount of waste. (5,004 lbs of BOD / 0.17 lbs of BOD per person = 29,435 people) you can dummy check this by using the flow given in gallons per day divided by the assumed contribution of 100 gallons per day per person and if you have the right answer with the other equations itll be pretty close (3,000,000 GPD / 100 gallons per person = 30,000 people)
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