CVG 2132 HW 4 with Solutions

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CVG 2132

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Apr 3, 2024

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Department of Civil Engineering November 20 th , 2023 CVG2132 – FUNDAMENTALS OF ENVIRONMENTAL ENGINEERING Homework 4 Professor: Chris Kinsley Due Date: November 29 th , 2023 – Location: Online submission on Brightspace 1) 26 million gallons per day of secondary effluent is discharged by a municipal wastewater treatment plant. This effluent contains 36 mg/L of ultimate BOD at 25.7°C and 1.7 mg/L of DO. The receiving stream before mixing with the effluent has a flow of 5.1 m 3 /s, a temperature of 22°C, is 80% saturated with oxygen and has an ultimate BOD of 2.1 mg/L. The reaeration (k r ) and deoxygenation (k d ) rate coefficients of the stream at 20°C are 0.35 d -1 and 0.2 d -1 , respectively. The velocity of the stream after mixing is 0.5 m/s. Determine the following: a) The wastewater flow rate in m 3 /s. b) The temperature of the combined wastewater and stream. c) The DO concentration of the mixture at the location of initial mixing. d) The DO deficit of the mixture at the location of initial mixing. e) The ultimate BOD concentration of the mixture at the location of initial mixing. f) The distance downstream (x c ) to the point of minimum DO concentration. g) The minimum DO concentration in the stream. Solution – NOTE: The subscript 0 refers to the mixing point, the subscript W refers to the wastewater treatment plant, the subscript R refers to the receiving waters, the subscript S refers to saturation and the subscript c refers to the critical point. Assumptions: - The river is completely mixed in the lateral direction but not in the longitudinal direction. - 1-dimensional model in the direction of flow. - Two reaction processes are assumed to be essential i) atmospheric reaeration and ii) deoxygenation due to BOD effects. - Steady state, instantaneous mixing, and dilute conditions at the point of mixing. a) Unit conversion –

𝑉???????𝑖? ?????𝑎?? ?? ?𝑖??ℎ𝑎???, ? 𝑤 = 26 × 10 6 ?𝑎????? ?𝑎? × 0.00378541 ? 3 1 ?𝑎???? × 1 ?𝑎? 24 ℎ?? × 1 ℎ? 3600 ? = ?. ?? 𝒎 ? 𝒔 b) Temperature of combined discharge and rive or stream, T 0 𝑇 0 = ? 𝑤 𝑇 𝑤 + ? ? 𝑇 ? ? 𝑤 +? ? = [(1.14 × 25.7) + (5.1 × 22)] × ? 3 ? × ℃ (1.14 + 5.1) ? 3 ? = ??. ? ℃ c) Dissolved oxygen concentration of the mixture at the location of initial mixing, DO 0 Dissolved oxygen of river, DO r = 80% of saturation at 22°C DO at 22°C is 8.83 mg /L (from table in lecture 13) = DO s DO r = 0.80 x 8.83 = 7.06 mg/L 𝐷? 0 = ? 𝑤 𝐷? 𝑤 + ? ? 𝐷? ? ? 𝑤 +? ? = (1.14 × 1.7) + (5.1 × 7.06) × ? 3 ? × ?? 𝐿 (1.14 + 5.1) ? 3 ? = ?. ?? 𝒎𝒈 𝑳 d) DO deficit of the mixture at the location of initial mixing, D 0 𝐷 0 = 𝐷? ? − ? 𝑤 𝐷? 𝑤 + ? ? 𝐷? ? ? 𝑤 +? ? = 8.83 ?? 𝐿 − 6.08 ?? 𝐿 = ?. ?? 𝒎𝒈 𝑳 e) Ultimate BOD concentration of the mixture at the location of initial mixing, L 0 𝐿 0 = ? 𝑤 𝐿 𝑤 + ? ? 𝐿 ? ? 𝑤 +? ? = (1.14 × 36) + (5.1 × 2.1) × ? 3 ? × ?? 𝐿 (1.14 + 5.1) ? 3 ? = ?. ?? 𝒎𝒈 𝑳 f) Distance downstream (x c ) to the point of minimum DO concentration – Adjust the rate coefficients for the new temperature of the river after mixing: k d(T) = k d(20) θ (T-20) ; k d(22) = k d(20) 1.056 (22-20) = 0.2x1.056 (22-20) = 0.22 d -1 L 0 Q w + Q r L w , Qw L r, Q r

k r(T) = k r(20) θ (T-20) ; k r(22) = k r(20) 1.056 (22-20) = 0.35x1.056 (22-20) = 0.39 d -1 Determine the travel time (t c ) to the point of minimum DO: Substituting all the values obtained in previous steps, ? ? = 1 0.39 − 0.22 × ?? [ 0.39 0.22 × (1 − 2.75 × 0.39 − 0.22 0.22 × 8.29 )] = 1.63 ?𝑎?? ? = ? ? × ? = 1.63 ? × 24 ℎ 1 ?𝑎? × 3600? 1ℎ? × 0.5? ? × 1?? 1000? = ??. ? 𝒌𝒎 g) 𝐷 ? = 0.22 × 8.29 0.39 − 0.22 × (? −0.22×1.63 − ? −0.39×1.63 ) + 2.75 × (? −0.39×1.63 ) = 3.269 ?? 𝐿 DO c = DO S – D c DO c = 8.83 mg/L – 3.269 mg/L = 5.561 mg/L 2) A wastewater sample collected has a BOD 5 equal to 165 mg/L and a carbonaceous reaction rate constant (k) equal to 0.215/day. The total ammoniacal nitrogen (TAN) was reported as 27 mg/L NH 3 as N. a) Find the ultimate carbonaceous BOD (CBOD). b) Find the ultimate nitrogenous BOD (NBOD). c) Find the remaining BOD (nitrogenous + carbonaceous) after five days. Solution – BOD 5 = 165 mg/L k = 0.215 d -1 Concentration of ammonia as N = 27 mg/L O H H NO 2O NH 2 3 2 3 + + → + + −

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