Substrate concentration
Volume of aerated tank =V=
V= 500 m^3 = 500*1000 =500000 liters
Waste water flow rate =Q= 4*10 ^5 = 400000 liter /day
Retention time = ?= V / Q =500000 /400000 =1.25 Day = 1.25 *24 =30 hrs
According to monod expression
Μ= μ= μm *S / Ks +S
Where S = Concentration of substrate ( mg/L )
Μ= μ= SPECIFIC GROWTH RATE
FINAL DISCHARGE WASTE = 10mg BOD /L
INITIAL INFLUENT WASTE CONTAINS 1500 mg BOD /L
Ks = 30 mg /L AS BOD, μm = 3 mg VSS /mg VSS d
Y = 0.6 mg VSS / mg BOD
Ks = = half velocity constant or substrate concentration at utilization rate of μm /2 in mg /L
Μ= μ= μm * S / ( Ks +S ) == 3 *S / ( 30 + S)
=== > μ= μm *1500 / ( 30 +1500 ) =3 *1500 / 1530 =2.94
S =1500 mg /L
= d X /d? = ( μ - kd ) *X
==== > X = ( 2.94 – kd ) * ?==== > 1500 = (2.94 -kd ) *30 === > kd = 1500 /30 -2.94 =47.06 mg / L
NOW THE RATE OF GROWTH
Y * d S/ d?= dx /d? = μm*Xav *S /Ks
Integrating ,we have
Y *Se = μm * Xav *S *?/Ks
0.6 *Se = 3 *Xav *1500 *30 /30
CHANGE IN REACTOR = V * d S /d?= 0
THE Material balance
V*d S /d? = 0 = Q*Si - Q*Se - V*Se *k
Where Si = influent concentration
Se / SI = 1/ ( 1 +K*?)
Se / Si = 1 / (1 +2.94 *30 )
10 / SI = 1 / ( 89.2 )
Si = 10 *89.2 =892 mg /L
MAXIMUM INFLUENT CONCENTRATION = Si =892 mg /L
MAXIMUM FLOW
V*d S /d? = 0 = Q*Si - Q* Se - V*Se *K === > Q ( Si – Se ) = V*Se *K
Q max = V*Se *K / ( Si – Se ) = 500*1000 * 10 *2.94 / ( 892 - 10 ) = 16666.67 liter /hr
Daily production of solid biomass = Q max *Y * ( SI –se ) / [1 + k*?] = 16666.67 *o.6 * (1500 -10 ) / (1 + 2.94 *30 ) =14900002.98 / 89.2 = 167040.4 kg /day
