Texas Waste Water Treatment Ch#15.3 Pond Problems

TLDR: 1 acre treats 35 lb BOD | MGD = gpd / 1,000,000 gal/MG = 600,000 gal / 1,000,000 gal/MG = 0.6 MGD | lb = MGD × 8.34 lb/gal × mg/L = 0.6 × 8.34 × 150 = 751 lb BOD | Acres = lb BOD / 35 lb BOD/acre = 751 lb/BOD / 35 lb BOD/acre = 21.5 acres! ||| EXPLANATION: This is yet another problem that the mnemonic in the top left can help you with but there is some preparation needed first. to get started we know from past explanations that the mnemonic only works with flow in units of millions of gallons per day(MGD) so we need to divide our gallons per day(gpd) flow by one million to put it in the correct scale. (MGD = gpd / 1,000,000 gallons per million gallons(gal/MG) = 600,000gpd / 1,000,000 gal/MG = 0.6MGD) Now we have all the values we need to determine how many pounds of biochemical oxygen demand(BOD) we are recieving per day requiring treatment. To find that using the mnemonic we will multiply our flow by the 8.34 pounds per million gallons(lbs/MG) conversion factor given in the mnemonic and possibly the reference sheet given with the exam. (pounds of BOD recieved at a rate of 1mg/L daily(lbs/day) = the number of millions of gallons recieved per day(MGD) X the number of pounds of BOD recieved if the concentration is 1mg/L/MG (lbs/MG) = 0.6MGD X 8.34lbs/MG = 5.004lbs/day) this is why we needed to convert our flow into MGD because the number of pounds of BOD recieved per gallon with one miligram per leiter BOD concentration(mg/L) is .00000834 (truncated) pounds. multiplying it by one million makes it 8.34 which happens to be the same number as how much one gallon of water weighs which makes it work nicely for the mnemonic. now that we know how many pounds of BOD we are receiving for each mg/L what comes in with the waste water we need to multiply that by the number of mg/L we are actually recieving to find out how much BOD we need to have space to process. (pounds of BOD recieved per day(lbs BOD/day) = number of pounds of BOD recieved per mg/L recieved in a day(lbs/day) X the number of mg/L we are recieving(mg/L) = 5.004lbs/day X 150mg/L = 751lbs BOD/day rounded up) now we know how much BOD is coming in to get treated daily we need to figure out how much space in acres is required to treat it. Per table 15.1 in the textbook and hopefully the reference sheet in your exam 1 acre of pond can accept 35 lbs of BOD per day(1acre = 35lbs BOD/day), so all we need to do now is divide the number of pounds of BOD we are recieving per day by how many pounds of BOD one acre can accept per day to find out how many acres we need. (Acres = lbs BOD/day / 35lbs BOD/acre/day = 751lbs BOD/day / 35lbs BOD/acre/day = 21.5acres!)

TLDR: capacity(gal) = L × W × D × 7.48 gal/ft3 = 850 ft × 250 ft × 4 ft × 7.48 gal/ft3 = 6,358,000 gal | DT(days) = Capacity(gal) / Flow(gpd) = 6,358,000 gal / 200,000 gpd = 3.18 days DT! ||| EXPLANATION: The question wants to know detention time(DT) in days of a pond. Detention time is the amount of time it takes for a given amount of flow to completely displace the volume already contained in a basin or pond. We were given the flow recieved in gallons per day(gpd) but not the number of gallons to be replaced which is the capacity of the pond so we will need to calculate that. Capacity is equal to the volume multiplied by the number of gallons that can fit in a cubic foot of volume. volume of a rectangular basin is length times width times depth. (Capacity(gal) = Volume(ft^3) X 7.48gal/ft^3 = L(ft) X W(ft) X D(ft) X 7.48gal/ft^3 = 850ft × 250ft × 4ft × 7.48gal/ft3 = 6,358,000gal) Now that we have the capacity of the pond and the flow all we need to do to determine the detention time is divide the capacity by the flow. The biochemical oxygen demand(BOD) given is extra information and not used. its just there to confuse you. (DT(days) = Capacity(gal) / Flow(gpd) = 6,358,000gal / 200,000gpd = 3.18 days!)