Texas Waste Water Treatment Ch#15.6 Chemical Problems

TLDR: Dosage in mg/L = lbs CL2 / (MGD X 8.34 lbs/MG) = 200lb / (1 X 8.34)lbs/day = 8.4mg/L/day! ||| EXPLANATION: This problem is a rather straight forward use of the mnemonic in the top left which tells us that the dosage is equal to the lbs of gas applied divided by the flow coming into the plant (MGD = gpd / 1,000,000 = 1,000,000 / 1,000,000 = 1MGD) times our 8.34 lbs/MG conversion factor. since this yeild the TLDR formula i dont feel it is necessary to copy it again here.... just go read it.......

TLDR: Dosage in mg/L = lbs CL2 / (MGD X 8.34 lbs/MG) = 200lb / (2.4 X 8.34)lbs/day = (200 / 20.016)mg/L = 810mg/L/day! ||| EXPLANATION: This problem is a rather straight forward use of the mnemonic in the top left and with the exception of the conversion of flow to MGD a carbon copy of the previous problem. i dont feel it is necessary to copy it again here.... just go read it....... again........

TLDR: Lbs 100% CL2 = lb hypochlorite X CL% = 11 X 0.65 = 7.15 lbs 100% CL2! ||| EXPLANATION: Pure chlorine gas, not to spoil the suprise, is 100% chlorine. As the name states however calcium hypochlorite is not. The up side to calcium hypochlorite is that at the cost of using more you get a more stable substance that is safer to handle. The percentage given in the problem is the amount of the calcium hypochlorite that is pure chlorine so all we need to do to determine how much pure chlorine it would take to do the same job is multiply the amount of calcium hypochlorite by the percentage to calculate what 65% of that value is. (see TLDR formula)

NOTE: imho this question has zero training value. the reading on your chlorinator would never be correctly read without including units. the only reason to leave out the units is to not give away the fact that all you have to do is divide the 120 POUNDS PER DAY you are given by 24 hrs/day. The day unit will cancel out and you will be left with 5 pounds per hour. i sincerely hope that all the math problems i see on this exam are this easy if not worded to be quite so blatantly useless.

TLDR: DTmin = Volume gal / Flow gpm = (Lft X Wft X Dft X 7.48gal/FT^3) / (gpd / 1440 min/day) = (30ft X 20ft X 10ft X 7.48 gal/ft^3)gal / (2,500,000gal/day / 1440min/day)gpm = (44880gal / 1736gpm)min = 26min! ||| EXPLANATION: This is a normal detention time problem with no added hurdles aside from conversion. First we calculate the capacity/volume of the basin in gallons. (Volume gal = Lft X Wft X Dft X 7.48gal/ft^3 = 30ft X 20ft X 10ft X 7.48gal/ft^3 = 44880gal) Then we convert the flow from gallons recieved per day to gallons per minute. (gpm = gpd / (24hrs/day X 60min/hr)min/day = (2,500,000gal/day / 1440min/day)gpm = 1736.1gpm) Finally we divide the capacity of one basin by the rate at which we will be filling it to determine how long it would take to completely fill the basin or displace the full volume of the basin. (DTmin = Capacitygal / Flowgpm = 44880gal / 1736gal/min = 26min!)

Yes, because the minimum is 20 minutes at peak flow. NOTE: The previous incomplete explanation was copied directly from the answer key that said it is incomplete. The following is taken from two different sections of the textbook. the first excerpt is from the section on design criteria and the second is from the section discussing disinfection. 1.) 30 TAC §217.281(b) Disinfection Contact Basins: (1) A chlorine contact basin must provide a minimum chlorine contact time of 20 minutes at the peak flow. (2) A chlorine contact basin must prevent short-circuiting to ensure that the wastewater is retained in a chlorine contact basin for at least 20 minutes at peak flow. 2.) Where chlorination is utilized, any combination of detention time and chlorine residual where the product of chlorine (Cl, mg/L) x Time (T, Minutes) equals or exceeds 20 is satisfactory provided that the minimum detention time is at least 20 minutes at peak flow and the minimum residual is at least 1.0 mg/L. The maximum chlorine residual in any discharge shall in no event be greater than 4.0 mg/L, or that necessary to protect aquatic life.