Cvg 2132 Assignment Satisfaction

Homework 7 solutions M.J. Kirisits CE341 1. The waste contains 100 mg/L acetic acid and 50 mg/L of NH 3-N. a. Write the balanced redox reaction for the aerobic conversion of acetic acid to carbon dioxide. O H CO O COOH CH 2 2 2 3 2 2 2    b. Write the balanced redox reaction for the aerobic conversion of ammonia to nitrate. O H H NO O NH 2 3 2 3 2       c. Calculate the theoretical carbonaceous oxygen demand, the theoretical nitrogenous oxygen demand, and the total theoretical oxygen demand of the waste. Theoretical carbonaceous oxygen demand: L mg O mmole mg acid acetic mmole O mmole mg acid acetic mmole L acid acetic mg 107 32 2 60 1 100 2 2  Theoretical nitrogenous oxygen demand: L mg O mmole mg N mmole O mmole mg N mmole L N mgNH 228 32 2 14 1 50 2 2 3   L mg L mg L mg demand oxygen l theoretica Total 335 228 107    d. Would you expect the total BOD to be equal to the total theoretical oxygen demand? Briefly explain.

Unformatted text preview: RYERSON UNIVERSITY Department of Civil Engineering CVL 300: ENVIRONMENTAL SCIENCE AND IMPACT ANALYSIS Tutorial #4 1. How much (in kg/day) can a wastewater treatment plant of 5000 m3/day discharge to a river with a drought flow of 137.7 m3/min if the BOD5 in the river after initial dilution must not exceed 1 mg/L? What degree of treatment (% satisfied) does this represent, assuming the plant receives an influent wastewater of 250 mg/L BOD5? Answer: a. Assume steady state and completely mixed conditions. Let wastewater BOD be Cw mg/L Upstream river flow (Qr) = 137.7 m3/min Wastewater flow (Qw) = 5000 m3/d = 5000/(24*60) = 3.47 m3/min Upstream River BOD (Cr) = 0 mg/L Downstream River BOD after mixing (Cm) = 1 mg/L Apply volumetric balance, Downstream river flow (Qm) = Qr + Qw = 137.7 + 3.47 = 141.2 m3/min Apply mass balance after mixing, Qm Cm = Qr Cr + Qw Cw 141.2* 1 = 137 (0) + 3.47 Cw Cw = 40.7 mg/L Allowable BOD from wastewater= Qw Cw = 5000 * 40.7 * 1000/106 = 204 kg/day b. Degree of treatment = (250 – 40.7)/250 = 84% 2. For the Problem 1 above, the dissolved oxygen concentration (DO) and temperature (T) of the treated wastewater and the river are listed below: Treated wastewater: DOw = 2 mg/L; T = 20 C River upstream: DOr = 9 mg/L; T = 15 C Determine the DOm, Tm, and initial oxygen deficit (Do) after the complete mixing of the treated wastewater and the river. It is assumed that air pressure is 1 atm and it contains 20% oxygen and the saturated dissolved oxygen concentration (DOs) equals 9.75 mg/L. Answer: After mixing, Apply mass balance to DO gives Qw * DOw + Qr * DOr = (Qw + Qr) DOm 3.47 * 2 + 137.7 * 9 = 141.2 * DOm DOm = 8.8 mg/L Apply mass balance to temperature gives Qw * Tw + Qr * Tr = (Qw + Qr) Tm 3.47 * 20 + 137.7 * 15 = 141.2 * Tm Tm = 15.1 ˚C ≈ 15 ˚C Initial dissolved oxygen deficit after mixing (Do) = 9.75 – 8.8 = 0.95 mg/L 3. A municipal pollution control plant discharges continuously 21 MGD (1 MGD equals 0.0438 m3/s) of wastewater effluent into a stream of 8.7m3/s. The ultimate BOD and the DO of treated effluent is 50 mg/L and 2 mg/L while the stream background ultimate BOD and DO is 6 mg/L and 8.3 mg/L. The stream is flowing at approximately 0.3 m/s, an average depth of 3 m, and the stream temperature after mixing is 20 C (Saturated DO concentration equals 9.1 mg/L). The deoxygenation constant is 0.2/day. Determine the DO sag curve. Answer: Qw = 21 MGD = 21 * 0.0438 m3/s = 0.92 m3/s Qr = 8.7 m3/s Lw = 50 mg/L DOw = 2 mg/L Lr = 6 mg/L DOr = 8.3 mg/L Vr = 0.3 m/s Dr = 3 m Tm = 20 C Kd = 0.2/day Apply mass balance to determine the characteristics after mixing, Qw * Lw + Qr * Lr = (Qw + Qr) Lo Lo = (0.92*50+8.7*6)/(0.92+8.7) = 10.2 mg/L Qw * DOw + Qr*DOr = (Qw + Qr) DOm DOm = (0.92*2+8.7*8.3)/(0.92+8.7) = 7.7 mg/L DO saturation at 20C = 9.10 mg/L Initial oxygen Deficit after mixing (Do) = 9.10-7.70 = 1.4 mg/L kr = 3.9U0.5 / H1.5 = 3.9*(0.3)0.5 / 31.5 = 0.41/day tc = ln ((kr/kd)*(1-Do*(kr-kd)/(kd*Lo)))/(kr-kd) = ln((0.41/0.2)*(1-1.4*(0.41-0.2)/(0.2*10.2)))/(0.41-0.2) = 2.68 days Xc = Utc = 0.3 m/s * 2.68 days * 24 hr/day * 3600 s/hr = 69,466 m Dc = (kd*Lo)*(exp(-kdtc) – exp(-krtc))/(kr-kd) + Do* exp(-krtc) = (0.2*10.2)*(exp(-0.2*2.68) – exp(-0.41*2.68))/(0.41-0.2) + 1.4*exp(-0.41*2.68) = 2.45 + 0.47 = 2.91 mg/L Minimum DO at Xc = DO saturation – Dc = 9.1 – 2.91 = 6.2 mg/L 4. There are three pollution sources to a river: (1) a municipal wastewater treatment plant; (2) a fish processing plant; and (3) a creek which drains agricultural land. The effluent flow rate from the treatment plant is 12.0 m3/s and the organic concentration is 30.0 mg/L. The organic loading rate from the processing plant is 5000.0 kg/day. The summer low flow in the small creek is 10 m3/s with the organic concentration of 5 mg/L. The summer low flow in the river is 120.0 m3/s and the background organic concentration upstream of the plants is 0.5 mg/L. If the discharge rate of the processing plant is negligible in the river, determine: (a) the concentration of organic in the river downstream of the plant; (b) the required treatment level (in terms of percent of mass controlled) at the processing plant in order to have the concentration in (a) not exceeding 1.0 mg/L. Comment on the treatment level. State all your assumptions. (answer a) 3.7 mg/L; b)???) Since we don’t know the relative locations of all these pollution sources, we have to consider a few possibilities. Case 1: Answer: a) Volumetric balance gives Qd=Qr + Qc + Qw Mass balance gives QrCr + QcCc + QwCw + Lf = QdCd Each term is calculated as follows: QrCr = 120 m3/s*0.5 mg/L* 1000 L/m3*1/106 kg/mg * 3600s/hr* 24 hr/day = 5184 kg/day QcCc =10 m3/s*5 mg/L * 1000 L/m3*1/106 kg/mg * 3600s/hr* 24 hr/day = 4320 kg/day QwCw = 12 m3/s*30 mg/L* 1000 L/m3*1/106 kg/mg * 3600s/hr* 24 hr/day = 31,104 kg/day QdCd = 142 m3/s*Cd* 1000 L/m3*1/106 kg/mg * 3600s/hr* 24 hr/day Thus, 5184+4320+31104+5000 = 12,268.8Cd Cd = 3.7 mg/L b) Assume X is the required treatment efficiency at the fish processing plant, 5184+4320+31,104+(1-X)5000 = 12,268.8*1 X = 6.6 which is impossible. By inspection, if all the discharge from the fish plant is eliminates, 5184+4320+31,104 = 12,268.8*Cd Cd= 3.3 mg/L greater than 1 mg/L The treatment efficiency of the fish plant is irrelevant. Concentration downstream of the plant is always greater than 1mg/L. Case 2 a) b) Answer: Volume balance gives Qd=Qr+Qw = 120+12 = 132 m3/s Mass balance gives QrCr + QwCw + Lf = QdCd 5184+31,104+5000=132*Cd*1000*10-6*3600*24 Cd = 3.6 mg/L 5184+31,104 + (1-X)5000 = 132*1*1000*10-6*3600*24 X=5.97 mg/L which is impossible. If all discharges from the fish plant is eliminated, Cd= 3.2 mg/L The treatment efficiency is irrelevant. 5. A single source of BOD causes an oxygen sag curve with a minimum downstream DO equal to 6.0 mg/L. If the BOD of the waste is doubled (without increasing the waste flow rate), what would be the new minimum downstream DO? In both cases, assume that the initial oxygen deficit just below the source is zero, and the saturated value of DO is 10.0 mg/L. (Note that when the initial deficit is zero, the deficit at any point is proportional to the initial BOD). (Answer: New DO minimum equals 2.0 mg/L) D = k d L0 k r - k d e -k d t - e - k r t + D 0 e -k rt Answer: If Do is zero, D is directly proportional to Lo. Given the current minimum downstream DO (DOc) is 6.0 mg/L and the saturated DO (DOs) is 10 mg/L, the critical dissolved oxygen deficit (Dc) = 10 – 6 = 4 mg/L. Since doubling Lo will double Dc, the new Dc will be 8 mg/Lm. The new minimum downstream DO (DOc) will be = 10 – 8 = 2 mg/L. 6. Two point sources of BOD along a river (A and B) cause the oxygen sag curve shown in Fig. 1. Sketch the rate of reaeration vs. distance down river and Lt (BOD remaining) vs. distance down river. Fig. 1 Answer: 7. Suppose the only source of BOD in the river is untreated wastes that are being discharged from a food processing plant. The resulting oxygen sag curve has a minimum value of DO, somewhere downstream, equal to 3.0 mg/L. Just below the discharge point, the DO of the stream is equal to the saturation value of 10.0 mg/L. (a) (b) (c) (d) By what percent should the BOD of the wastes be reduced to assure a healthy stream with at least 5.0 mg/L of DO everywhere? Would a primary treatment plant (typically removes about 35% of the BOD) be sufficient to achieve this reduction? Answer: 29% If the stream flows 60 mi/d, has a reaeration coefficient (kr) equal to 0.8/d, and has a deoxygenation coefficient kd of 0.2/d, how far downstream (miles) would the lowest DO occur? Answer: 138.6 mi What ultimate BOD (Lo mg/L) of the mixture of river and wastes just downstream from the discharge point would cause the minimum DO to be 5 mg/L? Answer: 31.7 mg/L Sketch the oxygen sag curve before and after treatment recommended in (a), labeling the critical points (DOc location and value). Answer: (a) The minimum DO of 3 mg/L means the maximum DO deficit (before fixing it) is Dc = 10-3= 7 mg/L. For healthy conditions, we want DOc to be 5 mg/L. So, it means we want Dc = 10-5 = 5 mg/L Dc ( preferred ) 5 0.71 Dc (now) 10 3 We need to remove 29% of the BOD. Since a primary treatment plant removes about 35% fo the BOD, it should do the job. (b) Using the following equation, the critical time (tc) and the critical distance (Xc) downsteam are: k 1 tc ln r 2.31 _ days if initial Do is zero. kr kd kd Xc = 60 mi/d * 2.31 days = 138.6 mi. (c) What ultimate BOD (Lo) to assure 5 mg/L of DO any where in the river? k d Lo e k d tc e kr tc 5 kr kd Lo = 31.7 mg/L Dc (d) Note: All of the above questions except Question 4 are from Introduction to Environmental Engineering and Science by Gilbert M. Masters and Wendell P. Ela, Pearson Canada, 3rd edition, 2008. ...
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