Determine the biomass cell yield Y for the biological oxidat

Determine the biomass (cell) yield (Y) for the biological oxidation of hydrogen (as H2) by CO2 for microbes that can utilize H2 and CO2. Express your answer as Grams of biomass (cells) per gram of H2.

Solution

Answer:
For the better understanding a detail calculation is given below. Some data of the calculation are hypothetical. Please consider with kindness reply me if any error you got. Partly this problem is pasted from a different source.
Problem: An anaerobic reactor, operated at 35oC, treats wastewater with a flow of 2000m3
/d and a
biological soluble COD (bsCOD) concentration of 500g/m3
. At 90% bsCOD removal and a biomass
synthesis yield of 0.04 g Volatile Suspended Solids/ g bsCOD used, estimate the amount of methane
produced in m3
/d.
Step 1: Prepare a steady state mass balance for COD to determine the amount of influent COD
converted to methane.
CODin = CODeff + CODVSS(biomass) + CODmethane
CODin = COD concentration in influent x Wastewater Inflow
CODeff = COD concentration in effluent x Wastewater Inflow
CODVSS = 1.42 g COD/g VSS x Yield coeff g VSS/g COD x Efficiency of system
CODmethane= ?
Step 2: Determine the volume of gas occupied by 1 mole of gas at 35oC.
Step 3: The CH4 equivalent of COD converted under anaerobic conditions =
(L/ mole)/ (64 g COD/ mole CH4)
Step 4: CH4 production = CODmethanex CH4 equivalent of COD converted
Solution:
CODin = COD concentration in influent x Wastewater Inflow = ( 500g/m3
) *(2000m3
/d)=1000 Kg/d
Biological soluble COD removal = 90%
CODeff = 0.10* 1000 Kg/d = 100 Kg/d
CODused = 0.90* 1000 Kg/d = 900 Kg/d
Biomass synthesis yield = 0.04 g Volatile Suspended Solids/ g bsCOD used
Biomass synthesis yield = (900*1000g/day)*(0.04 g VSS)=36000 g/day
COD corresponding to biomass = (36000 g VSS/day)*(1.42 g COD/g VSS)=51.120 Kgday
CODin = CODeff + CODVSS(biomass) + CODmethane
CODmethane = CODin -{CODeff+CODVSS(biomass)}
CODmethane= (1000 Kg/d)- {100 Kg/d+ 51.120} =848.88 Kg/d (answer)
CODmethane=848.88 Kg/d (answer)
As 64 g COD comes from 1mole of CH4. So 848.88 Kg/d COD would result from
= (1 mole CH4 /64)*(848.88 Kg/d) = 13.26 mole CH4/day
Now we know that for STP conditions (i.e., 273 K and 1 atm pressure, 1 mole gas occupies 22.4 Liter
volume). Now we need to calculate volume of methane gas at 35°C (i.e., 308K) and 1 atm pressure:
V/T=constant (as P is constant here)
[22.4/273] = [V2/308] ==> V2=25.27 Liters/mole (this is true for methane gas also)
Now as biological process is producing 13.26 moles methane gas/day, it indicates that 335.1 liters/day (or
0.335 m3/day).