Question

7. You work for the Ministry of Environment to evaluate the maximum ultimate BOD concentration in the effluent of a new wastewater treatment plant such that the oxygen deficit never exceeds 1.5 mg/L in the receiving river. During dry weather (i.e., lowest river flow), the wastewater will be diluted 10 times in the river water and the river water contains 3.5 mg/L of ultimate BOD. Also during dry weather, you know that the river velocity after the wastewater discharge would be 0.045 m/s and that the river depth is 1.5 m. Assume that the mixture of river 0 1 2 3 4 5 6 7 8 9 10 water and wastewater at the point of mixing is Time (day) saturated in oxygen, and that the temperature of the mixture remains constant at 30 °C. In order to determine ka, you perform a lab experiment at 20 °C by monitoring BOD concentration in river water mixed with wastewater. The results of this experiment are presented below. Remaining BOD (mg-02/L) Based on this information, answer the following questions. a. What is ka to be used for river modeling? b. What is kr to be used for river modeling? c. What is the maximum ultimate BOD concentration in the wastewater effluent?

Answer 1

• from the lab data you can calculate the kd at 20oc by using BOD rmaining formula for 5 days and then convert the value for 30oc as following

10 Solution :- (a). Dota given - Critical deficit of oxygen = (1.5 mg/l) Dilution factor at the O ze. I part of waste water in'g' part of river water. > Ultimate BoD of river = 3.5mg le > velocity = 0.045 mis → Depth of river =1.5m - Temperature of mix=30°C (as. from performed lab experiment at 20°C Linie Wl timate BoD of mixe é 4o my II Bod remaining after 5 days= 15 my 1 I = Lt formula Lt = Lomine 10 Here ka ei deoxigenation constant at w°c Lomin 10 kat -K4 X5 -1540X10 450) = 10-K4X5 2) (0.375) = 10-kol x5 + loglo 10.375) = -KaX5 -) Kd at 20°C = 0.085 d-'.

→ But temperature of river i 30°C ka at 30°c = Ka 20cm 11.048)' ** = 0.0858/1048) 30-20 = 0.085 x 11.098)'0 = 10.1360" / (6) » Formula for kr = 2.148* kr = 2.148 XV 0.878 wot:48 i x 1.5-1.48 / Varela Br=2148X0.0460.875 481 va velocity driver TH= depth of river = 0.0774 at 20° c wer, T-20 Krat 30°c =Kri'at 20°C X(.024) = 0.0774x (4:024) Krat 30°c = 0.098 which is less thanked It meane data is abrupt. že velocity should be 0.45 mis / TAAX (1.024 130-20 (C2 Lmin = 40 myll [mix = Loxo + Lux Ver Kotvo =) Vo= volume (discharge) of waste water in river. Vra discharge of river fra ultimate BoD of river = 3.5 myll lor ultimate BoD of effluent a given dilution factor in 10

It means - af - Vo= 1 part Dr=g part apply 1. Lox1 + 3.589 40 = jo |L0 = 368.5gle
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