Texas Waste Water Treatment Ch#15.5 Fixed Film

TLDR: MGAD = (MGD X 43,560 ft^2/acre) / Area of filter in ft^2 = (MGD X 43,560 ft^2/acre) / (Pi X R X R)ft^2 = (1MGD X 43,560 ft^2/acre) / (3.14 X 60 X 60)ft^2 = (43,560 / 11,304)MGAD = 3.85MGAD! ||| EXPLANATION: To calculate this value you divide flow by the area to which it is applied. Doing this directly however will yeild Millions of gallons(MG) per foot squared(ft^2) and therefore is half way between two different commonly used units. The units most commonly used to measure the rate of hydraulic loading are million gallons per acre per day (MGAD) and gallons per day per square foot (gpd/ft^2) the answer key converts MG/ft^2 into MGAD by converting the square foot unit into an acre unit by multiplying by 43,560ft^2/acre. (shown above in the TLDR section) For the explanation and comparison i will go the other way and convert the MG unit to gallons for the gallons applied to each square foot of filter area per day(gpd/ft^2). (gpd/ft^2 = (MGD X 1,000,000gal/MG) / Area of filter in ft^2 = (MGD X 1,000,000gal/MG) / (Pi X R X R)ft^2 = (1MGD X 1,000,000gal/MG) / (3.14 X 60 X 60)ft^2 = (1,000,000gpd / 11,304ft^2) = 11,304,000,000gpd/ft^2!)

TLDR: IAW Table 9.1 (design criteria) on page 9-9 in my textbook the expected hydraulic loading for a standard rate trickling filter is 1-4 million gallons per acre per day (MGAD) or 25-90 gallons per day per square foot (gpd/ft^2) so yes this is within design loading.

TLDR: % reduction = ((In-Out) / In) X 100 = ((150-20)mg/L / 150mg/L) X 100 = 86.7%! ||| EXPLANATION: What we want to know is if the influent BOD is 100% what percentage of BOD is removed by the treatment process. First, take the amount of BOD that came in and subtract the BOD left after the treatment process to arrive at the total amount of BOD that was removed durring the treatment process.(BOD removed = Influent BOD - Final effluent BOD = (150 - 20)mg/L = 130 mg/L) then divide the amount removed by how much came in to determine what fraction of the incoming BOD was removed. (Fraction BOD removed = BOD removed / Influent BOD = 130 mg/L / 150 mg/L = 0.8666666666666666...) Finally, multiply the result by 100 to convert to a percentage. (% reduction = Fraction BOD removed X 100 = 0.86667 X 100 = 86.667%!)

TLDR: BOD in lbs = MGD X 8.34lb/gal X mg/L = 4MGD X 8.34lb/gal X 132mg/L = 4404lb | Filter volume in ft^3 = Pi X R^2 X D = 3.14 X 80ft X 80ft X 6ft = 120,576ft^3 | lb BOD/acre ft/day = (lb BOD X acre ft) / Filter volume = (4404lb X 43560ft^3) / 120576ft^3 = 1591lbs/acre ft/day! ||| EXPLANATION: This question wants to know how many pounds of biochemical oxygen demanding material are applied to the filter, but before we get to that i want to explain a few of the numbers used. for starters theres the 8.34 used in the answer key for finding the pounds of BOD delivered by a full days flow. this is from the mnemonic you see in the top left corner and represents both the weight of one gallon of water AND ALSO the weight of matter in each million gallons if the incoming concentration is 1mg/L. The mnemonic can also be used as in the TLDR answer to determine how much BOD is recieved per day in pounds. The 43,560 value can be written in either ft^2 or ft^3 the difference is that the AREA of an ACRE is 43560ft^2 (should be listed in your formula table provided during your graded evolution) and the volume of an acre area that is one foot deep (an ACRE FOOT) is 43560ft^2 X 1ft = 43560ft^3. The TLDR section uses the ft^3 number because it is saying one acre foot is equal to 43560ft^3 therefore converting the answer to be the number of pounds of oxygen demanding mater deposited in each acre foot of volume (instead of per ft^3 of volume) every day if flow remains constant. the answer key splits this into three separate problems but im going to do it in one long step to try to better show the process. (BODlb/acre foot/day = (BOD applied per day X 43560ft^3/acre ft) / volume of the filterft^3) = ((FlowMGD X 8.34lb/MG X mg/L)lbs/day X 43560ft^3/acre ft) / (Pi X Rft X Rft X Dft)ft^3 = ((4 X 8.34 X 132)lbs/day X 43560ft^3) / (3.14 X 80 X 80 X 6)ft^3 = (4403.52lbs/day X 43560ft^3) / ft^3 = (191817331.2 / 120576)lb/acre ft/day = 1590.841719745 = 1591 lbs/acre ft/day!)

No; an intermediate filter is designed for 700–1,400 lbs/acre ft/day IAW Table 9.1: Design Criteria in my textbook on page 9 - 9

TLDR: BOD in lbs/day = MGD X 8.34lb/gal X mg/L = 1.5MGD X 8.34lb/gal X 140mg/L = 1751.4lb/day | Filter volume in ft^3 = Pi X R X R X D = 3.14 X 40ft X 40ft X 6ft = 30144ft^3 | lb BOD/1000ft^3/day = lb BOD/day / (Filter volume / 1000) = 1751.4lbs/day / (30144ft^3 / 1000) = 58lbs/1000ft^3/day! ||| EXPLANATION: This question wants to know how many pounds of biochemical oxygen demanding material are applied to the filter, but unlike problem #4 we are solving for pounds of BOD applied to every thousand cubic feet of filter volume in a day instead of per acre foot of filter volume in a day. First thing we want to do is calculate the pounds of BOD delivered by a full days flow using the mnemonic you see in the top left corner of this quiz. (lbs of BOD recieved each day(#BOD) = MGD X 8.34 lb/gal X mg/L = 1.5 X 8.34 X 140 = 1751.4lbs) Next, we calculate the filter volume like we did in problem #4. (Volume of filter in ft^3 = Pi X R^2 X D = Pi X R X R X D = 3.14 X 40ft X 40ft X 6ft = 30144ft^3) Finally we divide the filter volume by 1000 to determine how many sets of 1000ft^3 there are and divide the BOD recieved by that number to determine how many pounds are deposited in each 1000ft^3. (lb BOD/1000ft^3/day = lb BOD/day / (Filter volume / 1000) = 1751.4lbs/day / (30144ft^3 / 1000) = 58lbs/1000ft^3/day) the answer key splits this into three separate problems but im going to do it in one long step to try to better show the process. (BODlb/1000ft^3/day = BOD applied per day / (filter volume / 1000)1000ft^3 = (FlowMGD X 8.34lb/MG X mg/L)lbs/day / ((Pi X Rft X Rft X Dft) / 1000)1000ft^3 = (1.5 X 8.34 X 140)lbs/day / ((3.14 X 40 X 40 X 6) / 1000)1000ft^3 = 1751.4lbs/day / (30144 / 1000)1000ft^3 = 1751.4lbs/day / 30.144 1000ft^3 = 58.1011146 lbs/1000ft^3/day = 58lbs/1000ft^3/day!)