## Info

9.5.3 Calculation of Actual Oxygen Requirement, the AOR

Consider the contents inside an activated sludge reactor laden with dissolved oxygen and pollutants subjected to aeration. Also, consider the sedimentation basin that follows the reactor and all the associated pipings. For the purpose of performing the material balance, let the boundaries of these items encompass the control volume. According to the Reynolds transport theorem, the total rate of increase of the concentration of dissolved oxygen is equal to its partial (or local) rate of increase plus its convective rate of increase.

The convective rate of increase is equal to -Qo[Co] + (Qo - Qw)[Ce] + Qw[Cw], where Qo is the inflow to the reactor; [ Co ] is the concentration of dissolved oxygen in Qo; Qw is the outflow of wasted sludge; [ C e ] is the concentration of dissolved oxygen in the effluent of the secondary sedimentation basin; and [ Cw ] is the concentration of dissolved oxygen in the wasted sludge. [ Cw ] is equal to zero. The local rate of increase is simply given by (d[C ]/dt)V.

The total rate of increase is (d[C]/dt)V, where V is the volume of the system which, as mentioned, is composed of the reactor, secondary basin, and the associated pipings. From Equation (9.22), d[C]/dt is given by the rate of aeration, (KLa)w([Cosw] -[C]) = AOR, and the rate of respiration by the organisms, f. Thus, the total rate of increase is ddCV = [(KLa)w([Cos,w] - [C]) - f] V = [AOR- f] V (9.28)

The total rate of increase is equal to the local rate of increase plus the convective rate of increase, so the following equation is obtained:

ddCV = V[AOR- f] V = dV- Qo[Co] + (Qo - Qw)C] (9.29)

Note that [ Cw ] is equal to zero; thus, it is not appearing in Equation (9.29). It will be recalled that the left-hand side of the equation, (d[C ]/dt) V, is called the Lagfangian derivative and the right-hand side exp-ressions, (d[C]/dt) V - Qo[Co] + (Qo - Qw)[Ce], are collectively called the Eulefian derivative.*

* As mentioned in the chapter, "Background Chemistry and Fluid Mechanics," the Reynolds transport theorem distinguishes the difference between the full derivative and the partial derivative. As stated in that chapter, the environmental engineering literature is very confusing with respect to the use of these derivatives. Some authors use the full derivative and some use the partial derivative to express the same meaning.

At steady state, the local derivative, (d[C ]/dt) V, of the Eulerian derivative is zero. Also, as the wastewater enters the reactor, its dissolved oxygen content is practically zero; thus, [ Co ] is equal to zero. [ Ce ], as it goes out of the secondary basin must also be equal to zero. Thus,

The respiration rate r is due to the consumption of substrate BOD which is composed of CBOD and NBOD. Let (r)s be the respiration due to CBOD and (r)n be the respiration due to NBOD. r is then r = ( r )s + (r) n (9.31)

Now, apply the Reynolds transport theorem to the fate of the CBOD. As a counterpart to [C ] in the case of dissolved oxygen, let [5] represent CBOD. The Lagrangian rate of decrease (opposite to increase, thus, will have a negative sign) of CBOD is -(d [5 ]/dt) V. This decrease represents the consumption of CBOD substrates to produce energy by respiration, (r )s, and the consumption of the substrates for synthesis or to replace dead cells, (syn)s. Thus,

For the Eulerian derivative, the local derivative is -(d[ 5] Idt) V and the convec-tive derivative is -(- Qo[5o] + Qo[5]), where [5o ] is the influent CBOD concentration and [5] is the outgoing CBOD concentration from the control volume. The outgoing concentrations are those coming out from the effluent of the secondary basin and the wasted sludge. Note that the convective derivative is preceded by a negative sign. The negative sign is used to precede it, since this derivative is a convective rate of decrease, as distinguished from the convective rate of increase which has a positive sign preceding it.

At steady state, the local derivative is equal to zero. Thus, from the Reynolds transport theorem, (Lagrangian derivative = to the Eulerian derivative):

-(dif)V = {(r)s + (syn)s} V = -(- Qo[5o] + Qo[5]) (9.33)

V (syn )s is the equivalent amount of CBOD in the body of organisms in the waste sludge. If [Xu] is the concentration of organisms in the waste sludge, V(syn)s = fs Qw [Xu], where fs is the factor that converts the mass of microorganisms in the waste sludge to the equivalent oxygen concentration V (syn )s. Therefore,

The factor fs converts Qw[Xu] to its equivalent oxygen value. The value of fs is normally taken as 1.42; however, to do an accurate job for a given specific waste, it should be determined experimentally.

By analogy with Equation (9.35), the respiration rate due to NBOD, (r )n, may be obtained from v(r)n = fN Qo([No] - [N]) - fs Qw [Xu] (9.36)

where fN is the factor for converting nitrogen concentrations to oxygen equivalent, [No] and [N ] are the nitrogen concentrations in the influent and effluent, respectively, and fn is the factor for converting the nitrogen in the wasted sludge (Qw [ Xu ]) to the oxygen equivalent.

Having determined the expressions for (r)s and (r)n, the equation for AOR is finally

V(AOR) = Qo([So] - [S]) - fsQw[Xu] + fNQo([No] - [N]) - fnQw[Xu] (9.37)

In the derivations of the equations above, V was the volume of the control volume which is composed of the volume of the reactor, secondary clarifier, and the associated pipings. As the effluent from the reactor is introduced into the secondary clarifier, it is true that microorganisms continue to respire. In the absence of aeration in the basin, however, this respiration is but for a few moments and consumption of substrates ceases. It is also true that there will be no respiration and consumption of substrates in the associated pipings. Thus, the only volume of the control volume applicable to the material balance is the volume of the reactor. Therefore, V may be considered simply as the volume of the reactor.

Determination offs, fN, and fn. The formula for microorganisms has been given as C5H7NO2 (Mandt and Bell, 1982). To find the oxygen equivalent of the mass synthesized, fs, react this "molecule" with oxygen as follows:

From this equation, the oxygen equivalent of the mass synthesized is 1.42 mg O2 per mg C5H7NO2. Therefore, fs = 1.42.

The reduction in the concentration of nitrogen is also brought about by reaction with oxygen for energy and for the requirement for synthesis. NBOD is actually in the form of NH3. Reacting with O2,

From this reaction, the oxygen equivalent per mg NH3 - N is 4.57 mg. Thus, fN is equal to 4.57.

The nitrogen for synthesis goes with the sludge wasted, which can be expressed in terms of the total Kjeldahl nitrogen (TKN). Bacteria (volatile solids, VS) contain approximately 14 nitrogen. (Protein contains approximately 16% nitrogen.) Thus, the equivalent oxygen of the nitrogen in the sludge wasted is 4.57(0.14)Qw[Xu] = 0.64 Qw[Xu] and the value of fn is 0.64.

The total actual oxygen requirement, AOR, is therefore

AOR = 1 {Öo([5o] - [5]) -1.42Qw[Xu] + 4.57Qo([N0] - [N]) -0.64Qw[Xu]} V

This AOR is used to find SOR in order to size the aerator needed.

Example 9.9 The influent of 10,000 m /day to a secondary reactor has a BOD5 of 150 mg/L. It is desired to have an effluent BOD5 of 5 mg/L, an MLVSS (mixed liquor volatile suspended solids) of 3000 mg/L, and an underflow concentration of 10,000 mg/L. The effluent suspended solids concentration is 7 mg/L at 71% volatile suspended solids content. The volume of the reactor is 1611 m and the sludge is wasted at the rate of 43.3 m /day. Calculate the SOR. Assume the aerator to be of the fine-bubble diffuser type with an a = 0.55; depth of submergence equals 2.44 m. Assume ß of liquor is 0.90. The influent TKN is 25 mg/L and the desired effluent NH3 - N concentration is 5.0 mg/L • f = 1.43. The average temperature of the reactor is 25°C and is operated at an average of 1.0 mg/L of dissolved oxygen.

Solution:

AOR = V{Qo([5o] - [5]) -1.42Qw[Xu] + 4.57Qo([No] - [N]) -0.64Qw[Xu]} Qw[ Xu ] = 43.3 [ 10,000 (0.001)] = 433.0kg/d

AOR = -—-—- {10,000(0.15-0.005)(1.43) - 1.42(433.3) + 4.57 (10,000) 1611

x (0.025 - 0.005) - 0.64(433.3)}= 2092.8 kg/d; AOR = 1.30-^--

P = Pb + -2 y; assume plant is located at sea level

Therefore,

F 113 257 3

[ Cos J = FF- [ Cos, sp J = \qi 3253 ( 8.4) = 9.39 mg/L

9.5.4 Time of Contact

Having determined the overall coefficient of mass transfer (KLa)w, the differential equation ddC = (KLa)w([Cos,w] - [C]) - r (9.41)

may be integrated. Take note that the only variables in this equation are t and C, all the others are constant. Integrating from t = 0 to t = t and from [C] = 0 to [C] = [C],

=__1_ (*~La)w([Cos,w]-[C])-r =__1_ (KLaUfl[C0S]-[C])-r

t is the time of contact for aeration or mass transfer. The apparatus must provide this time if the liquid phase is to have the concentration of [C ]. For example, for a spray tower, this time must be the time it takes for the droplets to fall from the top of the tower to the bottom. For a cascade aerator, this time is the time for the water to fall through the cascade from the top to the bottom.

9.5.5 Sizing of Aeration Basins and Relationship to Contact Time

The sizing of aeration basins (for the activated sludge reactor, for example) are determined using the parameters of hydraulic detention time and mean cell retention time. Hydraulic detention time is the average time that the particles of water in an inflow to a basin are retained in the basin before outflow. Mean cell retention time, on the other hand, is the average time that cells of organisms (not water) are retained in the basin.

A third parameter that is used not to size aeration basins but to design the aerators used in the basin is the contact time (as derived previously). The size of the basin must be such that it provides this time to effect the required length of time of contact between the gas phase and liquid phase. There are situations where contact time for aeration is equal to the hydraulic detention time. For example, in the case of the trickling filter, as the water flows down the bed, a volume of this water occupies the interstices of the bed. The total volume of these interstices multiplied by a factor to account for the partial filling of the interstices divided by the inflow Qo is the hydraulic detention time. Because contact between the water phase and the air also occurs during the time that the water occupies the interstices, however, detention time is equal to contact time.

In other situations, contact time is not equal to detention time. For example, in the case of the activated sludge reactor, contact time is the rise time of the bubbles to the surface, whereas detention time is the average time of travel of the water between the inlet to the outlet of the reactor. Contact time refers solely to the bubbles contacting the water during the time they were rising from the diffusers. Hydraulic detention time, in this case, has no relation to the contact time.

Example 9.10 An experiment was performed on a trickling filter for the purpose of determining the overall mass transfer coefficient. Raw sewage with zero dissolved oxygen was introduced at the top of the filter and allowed to flow down the bed. A tracer was also introduced at the top to determine how long it takes for the sewage to trickle down the bed. At the bottom, the resulting DO was then measured. From this experiment, the overall mass transfer coefficient (KLa)w was found to be 2.53 per hour. This information is then used to design another trickling filter. The ¡3 of the waste = 0.9 and the respiration rate r = 1.0 mg/L • hr. Assume the average temperature in the filter is 25 °C and the effluent should have a DO = 1.0 mg/L. What should be the detention time of the sewage in the filter to effect this DO at the effluent? What void volume should be provided in the interstices of the bed to effect this detention time? The inflow rate is 10,000 m /d. Assume the voids are 0.70 filled with water.

Solution:

, 1 , ( KLa)w (¡\ Cos ] - [C ]) -r Kt wrxr i a/■ n r^i

' = {KLa)3iC0s\-r At 25 C,[Cos,*] = 84 mg/L = [Cos]

Therefore, f = - 1 n 2.53(0.9x8.4-1.0)-1.0 = 0.059 hr Ans

Volume occupied by water = 10,000 (0|f9j = 24.58

Therefore,

Volume of interstices = = 35.1m3 Ans

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