Question

1. A wastewater treatment plant (WWTP) releases effluent into a stream with mean depth 2 m and mean velocity 0.75 m/s. The BOD concentration at the WWTP is 15 mg/L, and the oxygen deficit is negligible. The deoxygenation rate in the stream is 0.8 d-1 and the reaeration rate is 1.2 d-1. a) Calculate the BOD concentration and DO deficit at a point 20 km downstream from the WWTP. (10 pts) b) What assumptions are inherent in these predictions (give at least two)

Answer 1

A) BOD = 6.51 mg/l , DO = 2.46 mg/l

B) BOD of stream is negligible and DO of stream is at saturation level

Step-by-step explanation:

Mean depth = 2 m

Mean velocity = 0.75 m/s

Bod concentration at WWTP = 15 mg/L

deoxygenation rate = 0.8 d-1

reaeration rate = 1.2 d-l

a) Calculate the BOD concentration and DO deficit

at 20 km

tc = (20 * 10^3) / (0.75 * 3600 * 24 )

= 0.309 days

`BOD_(t)` = lo ( 1 - 10^- 0.8 * 0.309 )

= 15 ( 1 - 10^ - 0.2472 )

= 15 ( 0.434 ) = 6.51 mg/l

DO = ( Kd * lo / Kr ) * 10^ -Kd*tc

= ( 0.8 * 6.51 / 1.2 ) * 10 ^ - 0.8 * 0.309

= 4.34 * 10^-0.2472 = 2.46 mg/l

B) The assumptions are : BOD of stream is negligible and DO of stream is at saturation level