Question

5. A well-mixed lagoon is used to pre-treat an agricultural waste stream before it is sent to the local municipal wastewater treatment plant. The waste stream has a flow rate of 400 m3/d and contains 300 mg/L of biodegradable waste. The waste decomposes via a first- order biodegradation process with a rate constant of 0.35 d-1.

   1. How large must the lagoon be in order to drive the concentration of waste down to 20 mg/L (as required by the municipal wastewater treatment plant)? Give your answer in m3.

   2. A colleague suggests you instead design a pre-treatment system of two equally- sized, well-mixed lagoons in series. How large must each lagoon be to achieve the same effluent concentration? Hint: You will need to write a separate mass balance for each lagoon.

Answer 1

Waste stream flow rate vo = 400 m3/day

Waste stream concentration of biodegradable Cao = 300 mg/L

Cao = 300 g/m3

Biodegradable is decomposed by first order reaction with rate constant k = 0.35 day-1

Final concentration of biodegradable Ca = 20 mg/L = 20 g/m3

Well mixed lagoon it is treated as cstr ,

Material balance for well mixed lagoon,

inlet flow rate of biodegradable - outlet flow rate of biodegradable - loss of biodegradable due to reaction = Accumulation of biodegradable into the lagoon

voCao - voCa - kCa * V = 0

V =( voCao - voCa) /kCa

V = vo(Cao - Ca) /kCa =( 400m3/d)(300 - 20)(g/m3)/(0.35d-1)*(20 g/m3) = 16000 m3

Volume of mixed lagoon V = 16000 m3

If we are using mixed lagoon in series with equal volume V for same outlet concentration Ca2 = 20 g/m3

Applying material balance :

For tank1:

voCao - voCa1- kCa1*V = 0

400*300 - Ca1(400 +0.35*V) = 0

Ca1 = 120000/(400 + 0.35V)

For tank2:

voCa1 - voCa2 - kCa2*V = 0

400*(120000)/(400+0.35V) - 400*20 -0.35*20*V = 0

120000/(400+0.35V) =20 + 0.0175V

120000 = 8000 + 7V + 7V +0.006125V²

0.006125V² +14V -112000 = 0

V = -14 +√(196 + 4*0.006125*112000) /2*0.006125 = 3283.4 m3

Volume of each lagoon V = 3283.4 m3