11.5) Determine the methane gas production rate (m3/d) for the reactor from prob
ID: 1732523 • Letter: 1
Question
11.5) Determine the methane gas production rate (m3/d) for the reactor from problem 11.4. Assume COD equivalent of VSS equals 1.42 kg COD/kg VSS.
PLEASE ANSWER 11.5 thank you!
11.4) Design a UASB reactor to treat wastewater at 30°C from a food pro- cessing plant. The wastewater flow rate is 500 m3/d with a soluble COD concentration of 6000 mg/L. The design parameters are as follows:
Reactor effectiveness factor (E) 0.85 Upflow velocity -1.2 m/h Organic loading rate 12 kg sCOD/m3 d Y 0.08 kg VSS/kg COD kd 0.03 d-1 ?max = 0.25 d-l K. = 360 mg/L Height for gas collection 3 Using the given information, determine the following: a. The reactor area and diameter The reactor liquid volume c. Liquid depth and total height of the reactor d. The average SRT (assuming 97% degradation of sCOD)Explanation / Answer
Given data:
Q=500m3/day
soluble matter=6000 g/m3
efficiency=97%
Step 1:
Prepare a steady state mass balance for COD, to determine the amount of influent COD converted to methane
CODin=CODeff+CODVSS+CODmethane
Step 2:
CODin=COD concentration in influent x waste water inflow=500x6000=3000 Kg/day
CODeff=COD concentration in effluent x waste water inflow=500x6000x0.3=90 Kg/day
COD used=3000x0.97=2910 kg/day
Bio sythesis yeild=0.08 x2910x1.42=330.576 kg
CODmethane=?
step 3:
CODmethane=3000-(330.576+90)
CODmethane=2579.424 kg
Step 4:
As 64 gm comes from 1 mole, then
(1 mole/64)x2579.424=40.3 mole Methane gas/day
step 5:
From the STP conditions
V/T= constant
22.4/273=V2/(273+30)
V2=24.86Liters/mole
Step 6:
Amount of methane=24.86x40.3=1001.858 liters/day or 1.001858m3/day
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