Question

	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	0.002

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

Answer 1

90% Biomass is Organic

=> Inorganic Mass in VSS = (1-.9)*500 = 50 mg/d

Weight of component require = Weight fraction * Inorganic mass weight

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

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