CEE 330 Homework 6 Problem 1 A waste has an ultimate CBOD of 1 000 mg L and a kL of 0 1 day What is its 5 day CBOD Problem 2 Calculate the ultimate BOD of a waste that has a measured 5 day BOD of 20 mg L assuming a BOD rate coefficient of 0 15 day Problem 3 7 32 textbook Calculate the dissolved oxygen deficit for a river at 30 C and a measured dissolved oxygen concentration of 3 mg L The Henry s law constant at that temperature is 1 125 10 3 mole L atm and the partial pressure of oxygen is 0 21 atm Problem 4 A waste having an ultimate BOD of 1 000 mg L is discharged to a river at a rate of 2 m3 s The river has an ultimate BOD of 10 mg L and is flowing at a rate of 8 m3 s Assuming a reaction rate coefficient of 0 1 day calculate the ultimate and 5 day BOD of the waste at the point of discharge 0 km and 20 km downstream The river is flowing at a velocity of 10 km day Problem 5 7 35 Textbook A river traveling at a velocity of 10 km day has a dissolved oxygen content of 5 mg L and an ultimate CBOD of 25 mg L at distance x 0 km that is immediately downstream of a waste discharge The waste has a CBOD decay coefficient k1 of 0 2 day The stream has a reaeration rate coefficient k2 of 0 4 day and a saturation dissolved oxygen concentration of 9 mg L a What is the initial dissolved oxygen deficit b What is the location of the critical point in time and distance c What is the dissolved oxygen deficit at the critical point d What is the dissolved oxygen concentration at the critical point Problem 6 A problem on river water quality is posted here http techalive mtu edu envengtext ch08 riverquality htm Use the River Dissolved Oxygen Simulator to complete 5 scenarios In your answer include the calculation results Solution 1 y5 Lo 1 e kL t y5 1000 mg mg 1 e 0 1 day 5 day 393 L L Solution 2 1 y5 Lo 1 e kL t mg Lo 1 e 0 15 day 5 day L mg Lo 38 L 20 Solution 3 Determine DOsat from the appropriate temperature dependent Henry s Law constant and the oxygen partial pressure DOsat 2 36 10 4 moles O2 1 125 10 3 moles 0 21 atm L atm L and converting to mg O2 L DOsat 2 36 10 4 moles O2 32 g O2 1 000 mg O2 7 6 mg O2 L mole O2 g O2 L Now calculate the oxygen deficit D DOsat DOact D 7 6 mg mg mg 3 4 6 L L L Problem 4 A waste having an ultimate BOD of 1 000 mg L is discharged to a river at a rate of 2 m3 s The river has an ultimate BOD of 10 mg L and is flowing at a rate of 8 m3 s Assuming a reaction rate coefficient of 0 1 day calculate the ultimate and 5 day BOD of the waste at the point of discharge 0 km and 20 km downstream The river is flowing at a velocity of 10 km day Solution Determine ultimate BOD at the point of discharge Lo 2 Lo Qriver BOD river Qwaste BOD Qriver Qwaste waste m3 mg m3 mg 10 2 1000 L s L 208 mg Lo s 3 m L 10 s 8 Determine BOD5 at the point of discharge 5 day BOD Ultimate BOD 1 e kt mg 5 day BOD 208 1 e 0 1 day 5days L mg 82 L Determine ultimate BOD downstream t x 20 km 2 days U 10 km day Lt L0 e kL t Lt 208 mg 0 1 day 2 days mg e 170 L L Determine BOD5 downstream 5 day BOD Ultimate BOD 1 e kt mg 5 day BOD 170 1 e 0 1 day 5days L mg 67 L Problem 5 a D0 DOsat DOact 9 5 4 mg L b Use Equation 8 7 to determine the critical time and knowledge of the river s velocity to determine the critical distance tcrit k D k2 k1 1 ln 2 1 0 k1 k2 k1 k L 1 0 3 tcrit mg 4 0 4 day 0 2 day 1 0 4 day ln 1 L mg 0 4 day 0 2 day 0 2 day 0 2 day 25 L tcrit 2 6 day xcrit 2 6 day 10 km day 26 km c Use Equation 8 5 to determine the oxygen deficit and Dt Dt k1 L0 e k1 t e k2 t D0 e k2 t k2 k1 0 2 day 25 mg L 0 4 day 0 2 day Dt 7 4 e 0 2 day 2 6 day e 0 4 day 2 6 day 4 mg e 0 4 day 2 6 day L mg L d Use Equation 8 2 to determine the actual dissolved oxygen concentration for the critical time as just calculated DO 9 7 4 1 6 mg L Problem 6 SCENARIO 1 Annual Average versus Critical Conditions Conditions Model inputs for annual average conditions are pre populated on the spreadsheet as AA Under critical conditions the river flow falls to 1 5 m3 s 1 and the temperature rises to 30 C All other model inputs are as for average annual conditions Critical conditions are represented as CC Use Trial as an adjustable working trial Analysis Questions Observe the simulated DO sag curves and data for AA and CC with a treatment efficiency of 80 4 1 Is the dissolved oxygen standard violated for annual average conditions AA with the present level of wastewater treatment No DO 7 3 mg L which is higher than the standard 5 mg L 2 Is the dissolved oxygen standard violated for critical conditions CC with the present level of wastewater treatment Yes DO 0 4 mg L which is lower than the standard 5 mg L 3 RESET Trial to Critical Conditions Adjust the BOD Removal efficiency until the water quality standard for DO is met 5 mg L 1 Record the BOD Removal efficiency If BOD removal efficiency is 98 DO 5 mg L SCENARIO 2 Addition of Post Aeration Conditions Unless special provisions are made the effluent may leave the WWTP with very low oxygen levels e g the 2 mg L 1 of the critical conditions scenario Oxygen can be added to the effluent post aeration to increase DO levels in the mixing basin and reduce adverse effects on the receiving water Analysis Questions RESET Trial to Critical Conditions Apply post aeration treatment WWTP Effluent DO 8 mg L 1 Adjust the BOD Removal efficiency until the water quality standard for DO is met Record the BOD Removal efficiency If WWTP effluent DO 8 mg L BOD removal efficiency 96 DO 5 mg L SCENARIO 3 Addition …