CVG 2132 HW 4 with Solutions

.pdf

School

University of Ottawa *

*We aren’t endorsed by this school

Course

CVG 2132

Subject

Civil Engineering

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by DeaconMusic9132

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 + + → + + −

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

MW of NH 4 + as N = 14 mg/mmol Note: In the problem the ammonia concentration is given as nitrogen meaning that MW = 14 mg/mmol instead of 17 mg/mol. MW of O 2 = 2 x(16 mg/mmol) = 32 mg/mmol a) Determine ultimate CBOD - We know that, from lecture 12, L= L 0 (1 – e (-kt) ), here L = BOD 5 𝐿 0 = 𝐿 1 − ? −𝑘? = 165 ?? 𝐿 1 − ? −0.215×5 = ???. ? 𝒎𝒈/𝑳 b) Determine ultimate NBOD - 1 mol of ammonium (NH 4 + ) reacts with 2 mols of oxygen gas (O 2 ) ?𝐵?𝐷 = 27 ?? ?𝐻 4 + − ? 𝐿 × 1 ???? ?? ?𝐻 4 + − ? 14 ?? ?𝐻 4 + − ? × 2 ???? ? 2 1 ???? ?𝐻 4 + − ? × 32 ?? ? 2 1 ???? ? 2 = ???. ? 𝒎𝒈 𝑳 c) Determine remaining BOD - Remaining BOD = Ultimate BOD - BOD Utilized = (250.5 mg/L+123.4 mg/L) – 165 mg/L = 208.9 mg/L 3) A group of neighboring municipalities decide to combine their waste streams and treat them all at one existing wastewater treatment plant, initiating a need to upgrade the plant’s treatment capacity. The addition of the outside waste streams increases the UBOD loading on the plant to 45,000 kg/day which corresponds to a concentration of 238 g/m 3 with a flow rate of 202,000 m 3 /d. The wastewater discharged from the wastewater treatment plant is predicted to be at a temperature of 23˚C with no dissolved oxygen. The river upstream of the treatment plant has a flowrate of 5.5 m 3 /s, an average temperature of 16.5˚C, no UBOD and a dissolved oxygen concentration equal to the saturation concentration. After mixing with the wastewater the velocity and average depth of the river are estimated to be 0.38 m/s and 2.7 m, respectively. To avoid interfering with downstream fish habitat, a minimum DO of 5 mg O 2 /L must be attained at 43 km downstream from the wastewater plant. The carbonaceous reaction constant for the mixed river and wastewater is 0.14 1/d at 20˚C. Determine the following: a) the deoxygenation (k d ) and reaeration (k r ) constants at 20˚C and at the mixed temperature. b) the effluent UBOD (L W ) required to have a minimum DO concentration of 5 mg O 2 /L

or above. c) the degree of treatment the wastewater treatment plant is required to implement to meet the required effluent UBOD. Is primary treatment sufficient? 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: - There is no change in flow velocity from the addition of the two streams. - 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) reaeration is due to stream turbulence; and ii) deoxygenation is due to initial DO mixing and BOD effects. - The temperature of the combined streams is assumed to be constant. - Steady state, instantaneous mixing, and dilute conditions at the point of mixing. a) Determine the constants k r and k d at 20˚C and at the mixed temperature. Determine the constants at 20˚C – k d = 0.14 d -1 k r = 3.9u 1/2 /H 3/2 k r = 3.9 *(0.38) 1/2 /(2.7) 3/2 = 0.54 d -1 NOTE- The above equation determines the constant’s value at 20˚C Now we need to find the mixed temperature, 𝑇 0 = ? 𝑤 𝑇 𝑤 + ? ? 𝑇 ? ? 𝑤 +? ? = [(2.34 × 23) + (5.5 × 16.5)] × ? 3 ? × ℃ (2.34 + 5.5) ? 3 ? = ??. ?? ℃ At 18.44˚C, θ = 1.135 k d(T) = k d(20) θ (T-20) ; k d(18.4) = k d(20) 1.135 (18.4-20) = 0.14x1.135 (18.4-20) = 0.114 1/d k r(T) = k r(20) θ (T-20) ; k r(18.4) = k r(20) 1.135 (18.4-20) = 0.54x1.135 (18.4-20) = 0.441 1/d b) Determine the UBOD L W required to have a minimum DO concentration of 5 mg O 2 /L

Based on the requirement of minimum DO concentration at a location of 43 km downstream, we assume the critical location in the river occur at 43 km downstream. 𝑋 ? = 43 ?? × 1000 ? 1 ?? = 43000 ? t c = x c /u ? ? = 43000? 0.38 ? ? × 3600? 1ℎ? × 24ℎ? 1 ?𝑎? = ?. ?? 𝒅𝒂𝒚𝒔 At the temperature of 18.44˚C, the saturation concentration is 9.54 mg O 2 /L DO c concentration = 5 mg O 2 /L, we can find D c D c = DO S – DO c = 9.54 mg O 2 /L – 5 mg O 2 /L = 4.54 mg O 2 /L Now we can solve for L 0 using the following equation. 𝐷 ? = ? ? 𝐿 0 ? ? − ? ? (? −𝑘 ? ? ? − ? −𝑘 𝑟 ? ? ) + 𝐷 0 (? −𝑘 𝑟 ? ? ) But D 0 is unknown. D 0 = DO S – DO 0 𝐷? 0 = ? 𝑤 𝐷? 𝑤 + ? ? 𝐷? ? ? 𝑤 +? ? We need to find DO R before we can continue. We know that DO R is equal to the saturation concentration of the river before mixing. DO S,R = DO R = 9.74 mg O 2 /L at 16.5˚C 𝐷? 0 = [(2.34 × 0) + (5.5 × 9.74)] × ? 3 ? × ?? 𝐿 (2.34 + 5.5) ? 3 ? = ?. ?? 𝒎𝒈 𝑳 D 0 = DO S – DO 0 = 9.54 mg O 2 /L – 6.83 mg O 2 /L = 2.71 mg O 2 /L Rearrange the D c equation to obtain the following: 𝐿 0 = (𝐷 ? − 𝐷 0 (? −𝑘 𝑟 ? ? ) ) (? −𝑘 ? ? ? − ? −𝑘 𝑟 ? ? )? ? (? ? − ? ? ) 𝐿 0 = (4.54 − 2.71(? −0.441×1.31 )) (? −0.114×1.31 − ? −0.441×1.31 ) × 0.114 (0.441 − 0.114) = ??. ?? 𝒎𝒈 𝑳 Now that we have L 0 , so we can now determine L W.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

𝐿 0 = ? 𝑤 𝐿 𝑤 + ? ? 𝐿 ? ? 𝑤 +? ? As L R = 0 we can rearrange the equation to isolate L W 𝐿 𝑤 = 𝐿 0 × (? 𝑤 +? ? ) ? 𝑤 = 21.09 ?? 𝐿 × (2.34 + 5.5) ? 3 ? 2.34 ? 3 ? = ??. ? 𝒎𝒈 𝑳 Therefore, the effluent UBOD needs to be a maximum of 70.7 mg/L to obtain a DO c of 5 mg O 2 /L c) Determine the extent of treatment required for the wastewater treatment plant - The initial UBOD was given as 238 g/m 3 or 238 mg/L. The percent of UBOD that needs to be removed is: ??????? = 238 ?? 𝐿 − 70.7 ?? 𝐿 238 ?? 𝐿 × 100% = ??% Therefore, 70% needs to be removed. Primary treatment only removes 25- 40% of the BOD so that would be insufficient, therefore secondary treatment is required.

Recommended textbooks for you

Solid Waste Engineering

Civil Engineering

ISBN:9781305635203

Author:Worrell, William A.

Publisher:Cengage Learning,

Residential Construction Academy: House Wiring (M...

Civil Engineering

ISBN:9781285852225

Author:Gregory W Fletcher

Publisher:Cengage Learning

Engineering Fundamentals: An Introduction to Engi...

Civil Engineering

ISBN:9781305084766

Author:Saeed Moaveni

Publisher:Cengage Learning

Sustainable Energy

Civil Engineering

ISBN:9781337551663

Author:DUNLAP, Richard A.

Publisher:Cengage,

Construction Materials, Methods and Techniques (M...

Civil Engineering

ISBN:9781305086272

Author:William P. Spence, Eva Kultermann

Publisher:Cengage Learning

SEE MORE TEXTBOOKS

Recommended textbooks for you

Solid Waste Engineering
Civil Engineering
ISBN:9781305635203
Author:Worrell, William A.
Publisher:Cengage Learning,
Residential Construction Academy: House Wiring (M...
Civil Engineering
ISBN:9781285852225
Author:Gregory W Fletcher
Publisher:Cengage Learning
Engineering Fundamentals: An Introduction to Engi...
Civil Engineering
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Cengage Learning
Sustainable Energy
Civil Engineering
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,
Construction Materials, Methods and Techniques (M...
Civil Engineering
ISBN:9781305086272
Author:William P. Spence, Eva Kultermann
Publisher:Cengage Learning