Comparison of this expression to Eq. 5.33 reveals that the second term within the braces is the oxygen requirement associated with soluble substrate removal. The first term is the requirement associated with the input of active biomass. When X|i HO is zero, Eq. 5.63 reduces to Eq. 5.33. The input of biomass debris has no effect on the oxygen requirement because the debris passes through the bioreactor without reaction. The oxygen requirement can also be calculated by performing a mass balance on COD across the bioreactor, as discussed previously for the soluble substrate case, with proper consideration of the input of active biomass on that balance.


Continue with the problem begun in Example The conditions are the same as those used previously (t = 8 hr, fc). = 160 hr, Ssi, = 200 mg/L as COD), except that 25 mg/L as COD of active biomass is added to the influent. No biomass debris is added.

a. What is the effluent m-cresol concentration?

To determine this we must use Eq. 5.52 with the kinetic parameters and stoichiometric coefficients given in Table E5.1:

+ (3.5 - 200)(1/160 + 0.01)Ss + (200)(3.5)( 1/160 + 0.01) = 0 Ss = 0.21 mg/L as COD

Thus, the presence of 25 mg/L as COD of active biomass in the influent to the CSTR decreased the effluent m-cresol concentration by 0.10 mg/L. This is approaching the minimum that could be obtained in such a bioreactor if no biomass was entering in the influent (as shown in Example

b. What is the minimum m-cresol concentration that could be obtained in this bioreactor?

This must be calculated with Eq. 5.55:

s""n 0.20{ 1 + [25/(0.34)(200)]} - 0.01 Ss„„„ = 0.13 mg/L as COD

If it was necessary lo decrease Ss below this value, either another bioreactor configuration would have to be used or more active biomass would have to be added to the influent, c. What is the MLSS concentration?

Insertion of the appropriate values into Eq. 5.54 gives: [1 + (0.20)(0.01)(160)]25

1 + (0.01)(160) 1 + (0.20)(0.01 )( 160)](0.34)(200 - 0.21)

Thus, the addition of 25 mg/L as COD of active biomass to the influent increased the suspended solids concentration in the bioreactor by 254 mg/L as COD. This is less than the impact of 25 mg/L of inert solids considered in Example because the active biomass underwent decay in the bioreactor, whereas the inert solids did not undergo any reaction.

d. What is the wastage rate of MLSS from the bioreactor?

This may be calculated with Eq. 5.62 or by multiplying the MLSS concentration by the wastage flow rate. Since we already know the MLSS concentration, the latter approach is easier:

The addition of 25 mg/hr as COD of active biomass to the bioreactor increased the mass of solids to be disposed of by 12.7 mg/hr as COD. The remainder of the influent biomass was destroyed by decay in the bioreactor.

e. How much oxygen must be supplied to the bioreactor?

This may be determined with Eq. 5.63:

Comparison of this value to the oxygen requirement calculated in Example shows that the oxygen requirement was increased by 12.4 mg/hr. Of this, 0.1 mg/hr was due to increased substrate removal and 12.3 mg/hr was due to decay of the added biomass. This latter fact could have been predicted by noting that 25 mg/hr as COD of biomass was added to the system, but only 12.7 mg/hr as COD of additional solids was wasted. The remainder (12.3 mg/hr) was oxidized and increased the oxygen requirement. This illustrates the advantage of expressing the concentrations of all organic constituents in COD units.

Frequently the waste solids stream from a bioreactor is directed to a CSTR to allow stabilization of the wasted biomass before ultimate disposal. This represents an extreme case of a CSTR receiving biomass in its influent because the influent MLSS concentration is generally high whereas the concentration of influent soluble substrate is very low. In that case, we are seldom concerned with the concentration of soluble substrate in the effluent and the term (Sso - Ss) can be considered to be negligible compared to the other terms in the performance equations, allowing their simplification. If this is done to Eqs. 5.51, 5.53, 5.54, 5.62, and 5.63, the resulting expressions are:

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