Question

1. The East Hartford WWTP discharges 150 MGD of treated wastewater into the CT river. The treated was tested in the lab and found to have aBOD₅ of 17.5 mg/L and an ultimate BOD (Lo) of 92 mg/L at 20^oC.

1. Find the reaction rate K, for the treated wastewater.
2. The CT river has a flow of 250 m²/s and an ultimate BOD (Lo) of 6 mg/L. the DO of the river is 4.5 mg/L and the DO of the wastewater is0.2 mg/L. find the DO and the initial ultimate BOD (Lo) of the downstream mixture if the temperature of the water is 5^oC.

3. Find the initial deficit (D_o) in the DO after mixing. Saturated dissolved oxygen values can be found in the textbook appendix.

4. The deoxygenation rate in the river (after mixing) is k_d =0.3d^-1 . the reaction rate in the river is k_s = 0.5 d^-1. If the river is traveling at 0.25 m/s, what will the deficit, D, be 4.32 km from the point where the treated wastewater enters the river?

Answer 1

Answers:

a) BOD5 = UBOD * ( 1 - e^(-k*t))

17.5 = 92*( 1 - e^(-k*5)

k = 0.042 day^-1

b) UBOD = (Q1*UBOD1 + Q2*UBOD2) / (Q1 + Q2)

1 MGD = 0.043813 m3/s

150 MGD = 0.043813 * 150 m3/s

UBOD =( 0.043813 * 150 * 92 + 250*6 ) / (0.043813 * 150 + 250)

= 8.2 mg/L

DO = ( 0.043813 * 150 * 0.2 + 250*4.5 ) / (0.043813 * 150 + 250)

= 4.39 mg/L

deficit = DOsat - DO

critical travel time = Tc = 4.32 * 1000 / 0.25

= 17280 s

= 0.2

critical deficit = Dt =( kd * UBOD / (ks - kd)) * [ e^(-kd*t) - e^(ks*t) ] + [ Da*e^-ks*t]

assuming initial deficit Da = 0

Dt = ( 0.3 * 8.2 / (0.5 - 0.3)) * [ e^(-0.3*0.2) - e^(0.5*0.2) ]

=- 2 mg/L + Da