Exam 1 Questions | Water and Wastewater Treatment | ENV 4514C, Exams of Engineering

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ENV 4514C

Exam # Name:_____________________________________________________________ 1 gal = 3.8 L 1 lb = 454 g t = V/Q G = √(P/μV) First‐order reaction: C = C 0 e‐kt^ Hazen‐Williams: V = 1.318CR0.633S0.

1. (a) Compare the contaminants commonly removed in surface water treatment plants versus ground water treatment plants in two lists. Please provide four (4) contaminants for each. (b) Which set of contaminants are more difficult to treat and why? Groundwater - Minerals (Ca2+, Mg2+)  hardness - Natural occurring elements (As, F, Mn, Fe)  health risk - H 2 S - Lack of oxygen - Little bacteria Surface water - turbidity - color - taste and odor compounds - pathogens - nutrients - Fe, Mn, H 2 S Groundwater has a constant composition so is easier to treat (Alum or soda to remove hardness, purging to remove H 2 S and to add oxygen). Often time, surface waters require more treatment to treat to the same standards and require operation flexibility (due to seasonal variations).
2. (a) What is the primary difference between primary drinking water standards and secondary drinking water standards? (b) How are disinfection by‐products produced? (a) Primary drinking water standards are regulated while secondary drinking water standards are suggested for aesthetic qualities (b) Disinfection by‐products are produced by the interaction of Cl 2 with TOC/NOM

3. An alum dose of 1.3 mg/L is applied to a rapid mix basin with a flow rate of 200 gallons per minute. How many pounds of alum are added per hour of operation? 200 gal/min * 3.785 L/gal * 60 min/hr * 1.3 mg/L * 1 g/1000 mL * 1 lb/453.6 g = 0.13 lb/hr
4. Derive the first‐order rate equation for a CSTR (i.e., rapid mix basin). Note: if you studied by analyzing previous exams, this question is different (i.e., in the past I asked for the derivation of a batch reactor and here I want the derivation for a CSTR that has flow). Please be sure to begin with the mass balance on the rapid mix basin. Accumulation = In‐Out+Generation dN/dt = QinC 0 ‐QoutC+V(dC/dt) For completely stirred tank reactors, the generation term is zero and flow rate and volume is assumed to be constant 0 = QC 0 ‐QC+ V dC/dt First order reaction: dC/dt=r=kC^1 0 = QC 0 ‐QC+ V (kC) 0/C = QC 0 /C‐QC/C+ V (kC)/C 0 = QC 0 /C‐Q+ kV ‐kV = Q(C 0 /C‐ 1 ) ‐kV/Q = (C 0 /C‐ 1 ) V/Q = ‐(C 0 /C‐1)/k t=V/Q t=( 1 ‐C 0 /C)/k

5. Chemical addition in a 25‐MGD water treatment plant will be performed in two parallel mixing basins with mechanical mixers. The mixing intensity as measured by G should be at least 750 s‐^1 and the residence time should be 15 s. Find the volume of the tanks and the power required for the mixers. Assume the mixers are 70% efficient. The dynamic viscosity of water is 1.139 x 10‐^3 kg/m‐s. (If you have performed Example problem 10.3 in your text, this is the same problem.) G should be between 600 – 1000 s‐^1 for rapid mix Assuming the flow is split equally t=V/Q  V=tQ=(15 s)(25 10^6 gal/day /2 day/24 hrs * hr/60 min * min/60 s)= 2170 gal per tank * 1 m^3 /264.172 gal = 8.2 m^3 G = (P / μ V) ½  P=G^2 μ V = (75 0 s‐^1 )^2 (1.139*10^‐ 3 kg/m‐s)(8.2 m^3 )= 5253.6 W = 5.3 kW/0.7= 7.5 kW
6. Two basins have the same volume, flow rate, and water viscosity. The first basin has a G of 1500 s‐^1 , whereas the second basin has a G of 150 s‐^1. a) What is the ratio of power input to the first basin, as compared to the second? b) Which G would you employ for a rapid mix basin and why? P 1 = G 12 μ V= (1500 s‐^1 ) 2 μ V P 2 = G 22 μ V= ( 150 s‐^1 ) 2 μ V  μ V = P 2 /( 150 s‐^1 ) 2 P 1 =(1500 s‐^1 ) 2 (P 2 /( 150 s‐^1 ) 2 ) P 1 / P 2 =(1500 s‐^1 ) 2 /( 150 s‐^1 ) 2 = 100 I would employ the first G because generally rapid mix requires a G above 6 00 , so although a G of 1500 is very large, a G of 150 would not sufficiently mix.

7. A chemical reaction that follows first‐order kinetics has a measured reaction constant of 50 at 20 deg. C. What is the time for the concentration of species A to decrease in concentration from 100 to 10 mg/L? t = 1/k (ln(C 0 /C)) = 1/( 50 (1/day)) (ln (100 (mg/L)/10 (mg/L)) = 0.046 days (or any given units)
8. Given a V‐shaped pipe with a slope of 0.001, a top width of 8 ft. and a depth of 3 ft, determine the velocity of flow. Please use a C of 130. V = 1.318CR0.633S0. R = A/P = 0.5 wh / (2L+w) = 0.583/( 2 (3^2 + 42 )1/2+8) = 12/18 = 0. V = 1.318130(0.67)0.633(0.001)0.54= 3.18 ft/s