A well-mixed pond has a volume of 2135 m^3 with flow through the pond of 175 m^3

Question

A well-mixed pond has a volume of 2135 m^3 with flow through the pond of 175 m^3 per day. Typically the POND WATER has a BOD level of about 2.4 mg/L. A sewer begins leaking into the INFLUENT STREAM. Without an appreciable change in the inflow volume, the incoming concentration is increased to 150 mg/L of BOD. If the first order degradation rate for BOD in the pond is 0.12/day: a) What will be the outflow concentration three days after the leak begins? b) What will be the concentration in the pond three days after the leak begins? a) 29 mg/L b) 29 mg/L

Explanation / Answer

Unsteady state mass balance of BOD can be written as

Rate of mass in-Rate of mass out- rate of loss due to chemical reaction= Rate of accumulation

Rate of mass in =175 m3*1000 l/day*150 mg/L

Rate of mass out = 175*1000*CA mg/L ( CA is concentration of leaving stram of BOD)

Rate of mass lost due to chemical reaction= KCA*V= 0.12*CA*2135*1000

175*1000*150- 175*1000*CA-0.12CA*2135*1000= d/dt( CA*2135*1000)

=175*150- 175CA- 0.12*2135CA= d/dt(2135CA)

=26250-413.2CA= 2135*dCA/dt

dCA/dt= (26250-413.2 CA)/2135

dCA/dt= 12.3-0.19CA

dCA/(12.3-0.19CA)= dt

when integrated, -ln(12.3-0.19CA)/0.19= t+C, C is integration constant

at t=0, CA= 2.4 mg/L, -ln(12.3-0.19*2.4)/0.19= C

C= 13

Hence –ln(12.3-0.19CA)/0.19= t-13

When t= 3 days, -ln(12.3-0.19CA)/0.19= 3-13= -10

Or ln(12.3-0.19CA)/0.19= 10

Ln(12.3-0.19CA)= 1.9

12.3-0.19CA= exp(1.9)= 6.7

0.19CA= 12.3-6.7= 5.6

CA= 5.6/0.19 =29.47 mg/L

Since the tank is well mixed, the outlet concentration is same as inlet concentration of 29.47 mg/L

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.