Therefore, t„„„ = 5.36 hrs and the maximum permissible flow is F = 8.0 L/5.36 hrs = 1.49 L/hr b. The flow through the bioreactor is 1.0 L/hr and the wastage flow is 0.05 L/hr. What is the concentration of m-cresol (in COD units) in the effluent?
First, we must calculate the SRT using Eq. 5.2:
Then we must use Eq. 5.13 to find the m-cresol concentration:
_ 3.5(1/160 + 0.01) s ~ 0.20 - (1/160 + 0.01) Ss = 0.31 mg/L as COD
c. What is the minimum m-cresol concentration (in COD units) that can be obtained from a CSTR?
This may be determined by using Eq. 5.14:
_ (3.5)(0.01) Sm n 0.20 - 0.01 Ss„„„ = 0.18 mg/L as COD
If a lower concentration were desired, some other bioreactor configuration would have to be used.
The suspended solids in a bioreactor receiving only soluble substrate will contain two components, active biomass, X|)H, and biomass debris, X„. Thus, we must be able to calculate the concentration of each.
The only source of active biomass in the system is from growth due to substrate utilization and its concentration may be calculated from a mass balance on substrate. Substituting the appropriate terms into Eq. 5.10 gives:
This states that the active biomass concentration depends on both the SRT and the HRT. Furthermore, rearrangement of Eq. 5.19 shows that for a fixed SRT (which determines Ss), the product Xy jj * t is constant:
In other words, at a fixed SRT and a fixed flow rate, a fixed mass of substrate will be removed per unit time, generating a fixed mass of microorganisms. A bioreactor with a small volume (short HRT) will contain a higher concentration of active biomass than one with a large volume (large HRT), although they will both contain the same mass. Likewise, if the bioreactor volume is fixed and the flow is increased at constant SRT, the mass of microorganisms in the bioreactor must increase to keep it consistent with the mass of substrate being removed. As a consequence, the biomass concentration must increase proportionally. Equation 5.20 demonstrates why the steady-state performance of a CSTR equipped with a biomass separator is independent of the HRT. If the HRT is changed for any reason, the concentration of biomass will also change to maintain the mass of organisms sufficient to produce an effluent substrate concentration consistent with the SRT. Likewise, if the influent substrate concentration, Ss„, is changed, the concentration of active biomass will change until the effluent substrate concentration is consistent with the SRT, thereby making Ss independent of Sso.
The situation concerning the active biomass concentration is different when the bioreactor has no biomass separator so that all effluent contains biomass and the SRT is equal to the HRT. In that case, the biomass concentration depends solely on the HRT, as can be seen by substituting t for ©c in Eq. 5.19:
The major disadvantage of a CSTR without a biomass separator is that its performance depends on its HRT, and this follows directly from its inability to maintain a constant mass of microorganisms as the HRT is changed. However, Ss is still independent of Sso because a change in Sso will cause the active biomass concentration to change until Ss is consistent with the HRT (SRT).
The units used to express the biomass concentration will depend on the units used to define the true growth yield, YH. In this part of the book, we are using the quantity of biomass COD formed per unit of substrate COD utilized as the units for yield. Consequently, the biomass concentration will also be in COD units. It can be converted to a mass of solids basis by dividing by 1.20 g COD/g SS and to a mass of volatile solids basis by dividing by 1.42 g COD/g VSS as discussed in Section 2.4.1 and given in Table 3.1.
The concentration of biomass debris in the bioreactor can be obtained from a mass balance on the debris, recognizing that its concentration in the influent and the effluent from the biomass separator is zero:
The units of X,, will be the same as the units of X„M. Debris adds to the total suspended solids concentration in the bioreactor, but does not add to the degradative capability because it has no biological activity associated with it. If the bioreactor has no biomass separator, the SRT will equal the HRT and Eq. 5.23 should be modified appropriately in calculating X,,.
The total biomass concentration in a bioreactor, X,, is the sum of the active biomass and biomass debris concentrations. Adding Eqs. 5.19 and 5.23 gives:
/«A (1 + fp' bH ■ (")JY||(SS(i ~ Ss) ' W l+b„-(i
Examination of it reveals that like the X,mi-t product, the X, -t product is fixed if the SRT is fixed. As far as theory is concerned, once the SRT has been chosen to give a desired effluent substrate concentration, any combination of bioreactor size and total biomass concentration may be used as long as it gives the proper X, -t product. There are practical limits, of course, and these are discussed in Chapter 10.
As discussed above for the active biomass concentration, when a bioreactor has no biomass separator, thereby making the SRT equal to the HRT, the total biomass concentration depends solely on the HRT as can be seen by substituting t for (-), in Eq. 5.24:
The active fraction of the biomass, fA, is defined as the concentration of active biomass divided by the total biomass concentration. Division of Eq. 5.19 by Eq. 5.24 and rearrangement yields:
Examination of it shows that the active fraction declines as the SRT is increased due to the build up of biomass debris in the bioreactor.
As discussed in Sections 2.4.2 and 3.8, the observed yield in a biochemical operation is always less than the true growth yield because some of the energy in the substrate must go to meet the maintenance energy needs of the culture. The observed yield associated with a bioreactor is equal to the actual net mass of biomass formed per unit mass of substrate destroyed, taking into consideration the amount of biomass lost to decay. When engineers attempt to measure the concentration of biomass formed it is difficult to distinguish the active biomass from the biomass debris. The total biomass concentration is generally used for purposes of defining the observed yield. At steady-state the total biomass formed in the bioreactor must equal the mass wasted from it. Thus, the observed yield, YM„h„ is given by:
Substitution of Eq. 5.24 for X, and simplification gives:
It can be seen that the larger the SRT of the bioreactor, the smaller the observed yield will be. This is because longer SRTs provide greater opportunity for biomass decay and greater need for maintenance energy, leaving less energy for synthesis of new biomass.
Continue with the problem begun in Example 18.104.22.168.
a. What is the active biomass concentration in the bioreactor when the inlluent flow is 1.0 L/hr and the wastage flow is 0.05 L/hr?
From Example 22.214.171.124, for this condition the SRT is 160 hr and Ss is 0.31 mg/L as COD. Furthermore, the HRT is 8 hr. Thus, using Eq. 5.19:
_ 0.34(200 - 0.31) "'" ~ (1/160 + 0.01)8 Xn.n = 523 mg/L as COD = 436 mg/L as suspended solids b. What is the total biomass concentration under the same conditions?
Use Eq. 5.24 to calculate this:
x - / 16(A [l1 + (0-20)(0.01)(160)](0.34)(200 - 0.31)1 r ~ V s / t 1 + (0.01X160) J
X, = 690 mg/L as COD = 575 mg/L as suspended solids c. What is the active fraction of the biomass?
This may be calculated from its definition or from Eq. 5.26. Using the definition gives:
Using Eq. 5.26 gives:
d. What is the observed yield?
This may be calculated from Eq. 5.28:
_ [1 + (0.20)(0.01)(160)]0.34 ~ 1 + (0.01)(160) Y = 0.17 mg biomass COD produced/mg substrate COD destroyed
This is only 50% of the true growth yield, showing the impact of decay and maintenance energy requirements on the net production of biomass and debris.
5.1.4 Excess Biomass Production Rate, Oxygen Requirement, and Nutrient Requirements
The two major costs associated with the treatment of wastewaters in aerobic CSTRs are from the disposal of the excess biomass produced and the provision of ample oxygen. It is important to be able to determine the amount of excess biomass produced and the quantity of oxygen that must be supplied. In addition, because of the negative impacts of nutrient limitations it is important to be able to determine the nutrient requirements as well.
Excess biomass is removed from the bioreactor via the wastage stream and the mass that must be disposed of per unit time is simply the concentration in that stream times its flow rate. At steady-state, this must equal the net production rate. Letting W, represent the total biomass wastage rate gives:
Combining the equation relating F„ to the SRT (Eq. 5.2) with Eq. 5.24 for X, gives:
Since S.s depends only on the SRT, it can be seen that W, depends on the SRT, the flow rate of the wastewater, and the concentration of substrate in it. Furthermore, it can be seen that the excess biomass wastage rate will decrease as the SRT is increased. This is due to the increased importance of decay at long SRTs. One use of biochemical operations is the stabilization of insoluble organic matter (see Section 1.2.1). The decrease in the amount of excess biomass brought about by decay is one example of stabilization. As the SRT is increased, more and more of the active biomass is oxidized and converted to debris, meaning that less excess biomass must be disposed of.
Comparison of Eq. 5.30 with Eq. 5.28, and substitution of the latter into the first reveals that the excess biomass wastage rate is just the observed yield times the substrate removed, which is consistent with the fact that the mass of biomass wasted must equal the mass produced at steady-state:
It can be seen that knowledge of the observed yield makes it easy to estimate the amount of excess biomass that must be disposed of.
The units on W, as given by Eqs. 5.29, 5.30, or 5.31 are mass of total biomass COD per time. Often, however, it is desirable to know the actual mass of dry solids that must be disposed of. This may be obtained by dividing by 1.20 g COD/g SS, just as was done to convert the biomass COD concentration to mass units.
The rate at which the microorganisms utilize oxygen in the bioreactor is equal to the overall rate expression for oxygen as developed from Table 5.1 and expressed in Eq. 5.9. Consequently, oxygen must be supplied at the same rate. If we multiply that rate by the bioreactor volume to give the mass per unit time required (RO), the result is:
Substitution of Eq. 5.12 for p.H, Eq. 5.19 for X|1M and Eq. 4.15 for the HRT, and simplification gives:
Because the stoichiometric coefficients in Table 5.1 are from a COD balance for each reaction, Eq. 5.33 represents a COD balance across the bioreactor. This can be seen in the following way. A COD balance states that the amount of oxygen which must be supplied to a bioreactor must equal the total COD in minus the total COD out, including the COD of the biomass and the biomass debris:
Examination of Eq. 5.34 reveals that the last term is just W,, the excess biomass wastage rate. Substitution of Eq. 5.30 for W, yields an equation identical to Eq. 5.33 after rearrangement. Furthermore, substitution of Eq. 5.31 for W, gives an equation for the oxygen requirement in terms of only the observed yield and the amount of substrate removed:
Substitution of Eq. 5.28 for Y,,ohs also gives an expression identical to Eq. 5.33. Thus, if the observed yield is known, the oxygen requirement is also known.
All of the above follows directly from the COD-based stoichiomctrv discussed in Sections 3.1.1 and 3.8.1. Equation 3.90 stated that the COD removed equals the oxygen equivalents of terminal electron acceptor used plus the COD of the biomass formed, which is the same as Eq. 5.34. Thus, all of the above equations have their roots in the basic stoichiometry discussed earlier. The ability to calculate the steady-state oxygen requirement directly from such simple equations is the key advantage to expressing yields and biomass concentrations in COD units.
Equation 5.33 clearly shows that the oxygen requirement in a CSTR increases as the SRT is increased. This too, is indicative of the increased stabilization that occurs as the SRT is increased. Increased stabilization implies that more of the electrons in a material end up being transferred to the terminal electron acceptor. Thus, as less excess biomass is produced, more oxygen must be used.
The amount of nutrient required can also be determined directly from the stoichiometry of biomass growth as discussed in Section 3.8.2. It was seen that the amount of nitrogen required to form biomass that can be represented by the empirical formula CsH,0:N is 0.087 mg N/mg biomass COD. If we assume that the nitrogen content of biomass debris is the same as that of active biomass, then the amount of nitrogen required per unit of substrate COD removed, NR, is just 0.087 times the observed yield, or:
Furthermore, as seen earlier, the phosphorus requirement is about one-fifth of the nitrogen requirement on a mass basis and thus it may be calculated by replacing 0.087 in Eq. 5.36 with 0.017. The requirements for micronutrients may be determined in a similar manner by using appropriate factors from Table 3.3. Nutrients should be added in slight excess of the theoretical amounts to ensure that the organic substrate is rate limiting, as discussed in Section 3.2.9.
Continue with the problem begun in Example 126.96.36.199.
a. How many mg/hr of dry solids would have to be disposed of when the How through the bioreactor is 1.0 L/hr and the wastage rale is 0.05 L/hr?
w = , „ J (1 + (0.20X0.01)(160)j(0.34)(200 - 0.31) I
W, = 34.5 mg/hr as COD To convert this to a dry solids basis, divide by 1.20 g COD/g dry solids: W, = 28.7 mg/hr as dry solids b. How many mg/hr of oxygen must be supplied to the bioreactor?
This can be obtained from use of Eq. 5.34:
RO = 1.0(200 - 0.31) - 34.5 = 165.2 mg/hr It can also be obtained from the fundamental system parameters by using Eq
Finally, it can be obtained directly from the observed yield through use of Eq.
Since the oxygen demand associated with the inlluent substrate is 2(H) mg/hr, these calculations show that it is necessary to supply sufficient oxygen to meet 83ci of that demand. The remainder is associated with the excess biomass formed and wasted from the system, c. How many mg/L of nitrogen and phosphorus should the influent contain'.'
The nitrogen requirement can be calculated with Eq. 5.36
nr = (0.087)[ 1 + (0.20X0.01 X 160)](0.34) 1 + (0.01)(160) NR = 0.015 mg N/mg substrate COD removed
Since the substrate COD removed was 200 -- 0.31 = 199.69 mg/L, the biomass will require 2.95 mg/L of nitrogen. If we allow an extra 0.5 mg/L to prevent
Using Eq. 5.30
nitrogen from being rate limiting, the influent should contain approximately 3.5 mg/L as N.
The phosphorus requirement will be about one-fifth of the nitrogen requirement, and the biomass will use about 0.59 mg/L. If we allow an extra 0.25 mg/L to prevent phosphorus from being rate limiting, the influent should contain approximately 0.85 mg/L as P.
Before the widespread use of SRT as the basic independent variable for design and control of a CSTR, most designers used the process loading factor, also called the food to microorganism (or F/M) ratio. The process loading factor, U, is defined as the mass of substrate applied per unit time divided by the mass of microorganisms contained in the bioreactor."' Because it is difficult to distinguish active biomass from biomass debris, the mass of microorganisms has generally been defined in terms of the total biomass concentration, X, :
The SRT is an important design parameter because at steady-state it is related in a simple way to |xH, the specific growth rate coefficient of the biomass, which in turn controls the substrate concentration in the bioreactor and its effluent. The relationship of the process loading factor to |xH can be obtained by rearranging the mass balance on substrate (Eq. 5.17) and substituting into it the fact that the active biomass concentration is equal to the total biomass concentration times the active fraction, giving:
For the conditions generally found in bioreactors used in wastewater treatment, Ss « Sso. Therefore:
This shows that the active fraction must be known before the specific growth rate or the specific substrate removal rate of the biomass can be determined exactly from the process loading factor. Conversely, Eq. 5.12 showed that at steady-state the specific growth rate can be determined directly from the SRT without requiring such knowledge. Since the active fraction depends on the characteristics of the bioreactor and requires knowledge of the SRT (Eq. 5.26), it is simpler to work directly with SRT as the fundamental design and operational parameter for a CSTR at steady-state. Furthermore, when a wastewater contains particulate substrates, the active fraction cannot be calculated easily from an expression like Eq. 5.26 and thus its determination becomes a major problem. Consequently, the relationship of the process loading factor to the specific growth rate of the biomass is more difficult to determine than it is for the simplified situation under consideration here. This has led to the use of the SRT instead of the process loading factor by many designers. Nevertheless, because the process loading factor is related to the specific growth and substrate fx' M-M _ F(Ss<> - Ss) Y„ " V-X,
removal rates, there are situations in which it provides valuable information, particularly in tanks-in-series systems where the applied substrate varies from tank to tank but the active fraction does not. In Chapters 10 and 11 we consider such situations.
As discussed in Section 3.2.2, occasions arise in which the steady-state substrate concentration in a CSTR is much less than the half-saturation coefficient so that the Monod equation may be simplified into a first-order equation:
This situation often arises when the SRT is long, thereby making it difficult to evaluate p.,, and Ks independently from steady-state data. In that case, the mean reaction rate coefficient, k,., is often used. Recalling Eqs. 3.44 and 3.45 and substituting them into Eq. 3.38 shows that:
The use of Eq. 5.40 does not alter any of the mass balances nor does it alter the fact that p,„ is controlled by the SRT as expressed in Eq. 5.12. All it does is simplify the relationship between Ss and the SRT. Substitution of Eq. 5.40 into Eq. 5.12 and rearrangement gives:
Comparison of Eq. 5.41 with Eq. 5.13 shows the effect that the first-order approximation has. None of the other equations are affected, except through the effect on Ss. It should be emphasized that care should be exercised in the application of the first-order approximation of the Monod equation to ensure that the basic assumption (i.e., Ss « Ks) is valid.
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