Prepare a recipe for an inorganic medium to be used in a laboratory CSTR to grow
ID: 701805 • Letter: P
Question
Prepare a recipe for an inorganic medium to be used in a laboratory CSTR to grow 500 mg VSS/d of bacteria biomass, assuming that the composition of the biomass is as listed below. Determine the concentration of essential inorganic compounds required for a feed rate of 1 L/d. Assume that phosphorus is added as KH2PO4, sulfur as Na2SO4, nitrogen as NH4Cl, and other cations associated with chloride. Note: assume that 90% of the biomass (sludge) is organic, i.e., VSS = 0.90·TSS
Fraction dry weight N 0.12 P 0.02 S 0.01 K 0.01 Na 0.01 Ca 0.005 Mg 0.005 Cl 0.005 Fe 0.002Explanation / Answer
90% Biomass is Organic
=> Inorganic Mass in VSS = (1-.9)*500 = 50 mg/d
Weight of component require = Weight fraction * Inorganic mass weight
1 mole KH2PO4 gves 1 moles P and 1 moles K
Using Molecular weight
=> 136 gm KH2PO4 gives 30 gm P and 39 gram K
=> 4,53 mg KH2PO4 gives 1 mg P and 1.3 mg K
Both required K and P qty are satisfied using 4.53 mg KH2PO4
1 moles Na2SO4 gives 2 moles Na and 1 mole S
Therefore, 142 mg Na2SO4 gives 46 mg Na and 32 mg S
=> 2.2 mg Na2SO4 gives 0.71 mg Na and 0.5 mg S
Both required Na and S qty are satisfied using 2.2 mg Na2SO4
1 moles NH4Cl gives 1 moles N
=> 53.5 gm NH4Cl gives 14 gm Nitrogen
=>22.98 gm NH4Cl gives 6 gm Nitrogen
N qty is satisfied using 22.98 mg NH4Cl
1 mole CaCl2 gives 1 mole Ca
=> 110 gm CaCl2 gives 40 gm Ca
=>0.6875 mg CaCl2 gives 0.25 mg Ca
1 mole MgCl2 gives 1 mole Mg
=> 95 gm MgCl2 gives 24 gm Mg
=>0.98 mg MgCl2 gives 0.25 mg Mg
1 mole FeCl3 gives 1 mole Fe
=> 162 gm FeCl3 gives 55 gm Fe
=>0.29 mg FeCl3 gives 0.1 mg Fe
Component Mass fraction component weight (mg) N .12 6 P .02 1 S .01 .5 K .01 .5 Na .01 .5 Ca .005 .25 Mg .005 .25 Cl .005 .25 Fe .002 .1