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Let's calculate an estimate of the solids retention time for a number of aerobic installations using the iterative formula - all data self-explanatory. Please be aware that these calculations are a simplification of the actual project data. For example, actual solids inventories could be smaller or larger and all other derived magnitudes adjusted accordingly. SRT is thus about ballpark.
| SRT Sludge Age Sizer | Dairy Biological Nutrient Removal | ||||||||||||
| T | 20 | °C | T | 20 | °C | T | 20 | °C | |||||
| Yoc | 0.5 | Yon | 0.31 | Ydn | 2.28 | ||||||||
| Kd | 0.200 | Kd | 0.200 | Kd | 0.200 | ||||||||
| flow in | 2000 | m3/d | 0.528 | mgd | |||||||||
| CODin | 483 | mg/L | 2126 | lbCOD/day | 964 | kgCOD/day | TKNin | 44 | mg/L | ||||
| TKNin | 44 | mg/L | (for nitrif!) | 194 | lbTKN/day | 88 | kgTKN/day | ||||||
| Nitrates | 44 | mg/L | 194 | lbCOD/day | 88 | kgNO3/day | 0 | used up for COD removal/growth | |||||
| nVSS | 0 | mg/L | 0 | lbSS/day | 0 | kgSS/day | TKNin used for growth | 0 | mg/L | ||||
| Percent | 100 | % Treated | 964 | kgCOD/day | TKNin avail for nitrification | 44 | mg/L | ||||||
| Percent | 95 | % COD removed | 916 | kgCOD/day | |||||||||
| Vreactor | 899300 | gallon | 3404 | m3 | (from order of magnitude sizing) | ||||||||
| 0.8993 | mg | 0.28 | kgCOD/day per m3 | ||||||||||
| Target MLVSS | ballpark bug mix | ||||||||||||
| 1086 | mg/L | heterotrophs | 3690 | kg heterotrophs | 1086 | mg/L | 68 | % of Target MLVSS | |||||
| 61.5 | mg/L | nitrifiers | 209 | kg nitrifiers | 62 | mg/L | 4 | % of Target MLVSS | |||||
| 450 | mg/L | denitrifiers | 1529 | kg denitrifiers | 450 | mg/L | 28 | % of Target MLVSS | |||||
| 0 | mg/L | other solids | 0 | kg other | 0 | mg/L | 0 | % other | (this entry's not used) | ||||
| 1598 | mg/L | ||||||||||||
| 1598 | mg/L | 11967 | lb bugs | 5428 | kg bugs | ||||||||
| Heterotrophic Calcs | |||||||||||||
| SRT | 30.04 | days | (iterates automatically) | ||||||||||
| Yn = | 0.134 | ||||||||||||
| biomass yield = | 123 | kg/day bugs | |||||||||||
| Global Sludge Age | |||||||||||||
| Nitrifiers | 30.01414 | 181 | kg/day | ||||||||||
| SRT | 30.16 | days | (iterates automatically) | synthesis | |||||||||
| Yn = | 0.083 | ||||||||||||
| biomass yield = | 7 | kg/day bugs | |||||||||||
| Denitrifiers | |||||||||||||
| SRT | 29.93 | days | (iterates automatically) | ||||||||||
| Yn = | 0.612 | ||||||||||||
| biomass yield = | 51 | kg/day bugs | |||||||||||
| HRT = V / Q = | 1.7 | days | |||||||||||
| 41 | hours | ||||||||||||
Task: select a mechanical and alternative aeration system with tubular diffusers to integrate a plant for the treatment of municipal wastewater, and the process is partially characterized as follows:
Flow rate to be treated: 1.43 mgd (5412 m3/day)
BOD5: 300 mg/L
TKN: 70 mg/L
The consulting firm proposed the use of 4 tanks of 25m x 14.2m each, with a depth of around 5 meters. It is desired to give dimension to the installation to be able to fully process/reduce the level of the two magnitudes present.
IMPORTANT: Just to illustrate the selection process of mechanical aerators and tubular diffusers, this presentation is outlined in a very simplified application context, with regard to treatability conditions, final calculations, drawing/guideline criteria, and which will be summarized by the specifications of the consulting firms that intervene.
PRACTICAL SCHEME: To compare the results obtained with existing tables in the classic literature of the specialization, English techniques will be used.
To estimate the need for oxygen, we must first calculate the daily pounds to be eliminated of each of the magnitudes using classical formulas. We will then have, for BOD5:
mgd's * BOD5 * 8.33 = 1.43 * 300 mg/L * 8.33 = 3574 lbs BOD/day
and corresponding to nitrification:
mgd's * TKN * 8.33 = 1.43 * 70 mg/L * 8.33 = 834 lbs TKN/day
We can now use the following "simplified suggestions" of design:
1.6 lbs O2 per lb BOD5 to be processed (BOD ultimate = 1.6 * BOD5)
4.6 lbs O2 per lb TKN processing ->
Adding the oxygen required for both processes:
3574 * 1.6 + 834 * 4.6 = 5719 + 3837 = 9556 lbs O2/day =9556 /24 = 399 lbs O2/hour = 6.637 lbs O2/minute
In the case of mechanical aerators, using AOR/SOR = 0.7 for the design we will have:
SOR = 399 lbO2/hour / 0.7 = 570 lbO2/hour.
If they were low speed aerators, e.g. 3.25 lbO2/h per HP:
570 / 3.25 = c. 175 - 180 HP
If they were high speed aerators, e.g. 2.5 lbO2/h per HP:
570 / 2.5 = c. 228 HP
Being an activated sludge process, with power density requirements ca. 100 HP per mg results:
HP mixing = 1.876 mg * 100 HP/mg = c. 190 HP
In the case of aeration through diffusers, we just have to have an idea of the amount of air we need to move forward with our design. We will repeat calculations similar to those exposed in Metcalf (p.525) for two slightly different operating contexts, which we present below:
oxygen content in the air 23.2 % 23.2 %
transfer efficiency 8.0 % 12.65 %
air density 0.075 lb./cu.ft. 0.075 lb./cu.ft.
Using contemporary equipment from the famous academic text:
CFM's = 6.637 / ( 0.08 * 0.075 * .232 ) = 4768 CFM = 4768 * 60 / 35.3145 = 8100 m3/h
Using slightly more sophisticated equipment:
CFM's = 6.637 / ( 0.1265 * 0.075 * .232 ) = 3016 CFM = 3016 * 60 / 35.3145 = 5123 m3/h
Let us remember that we must consider the EPA standard (0.12 cfm/sq.ft.) regarding mixing, i.e. keeping particles in the tanks suspended and preventing them from depositing there.
When being four tanks:
4*25*14.2=1420 m2=1420 m2*10.76=15280 sq.ft.->
15280 sq.ft. * 0.12 cfm/sq.ft. = 1834 CFM = 1834 * 60 / 35.3145 = 3116 m3/h
In any case, the value of the air flow/flow rate given by the oxygen requirement is important. However, we can already see it clearly now, and even in spite of all the voluntary and involuntary simplifications we have made in order to expand the inexorable, the proposals are beginning to appear that are quite difficult to reconcile with each other. If even this small lack of connection, we introduce variants of constructive design (e.g. mechanical) of the diffuser in relation to maintaining transfer efficiency over time, the elaboration of the comparative synthesis of the various offers can become unmanageable.
In both cases, let us calculate the power of the blowers in the two installations. Since the depth of the tanks will be around 5 m (5m * 3.28 = 16.4') we can think that the diffusers act at a depth of 15'. Therefore, regarding the size of the motors we will have:
15' = 15 * 2.9861 kPa = 15 * 2.9861 / 6.8948 psi = 6.5 psi
4' = 4 * 2.9861 kPa = 4 * 2.9861 / 6.8948 psi = 1.8 psi
Totaling 6.5 + 1.8 = 8.3 psi ("pressure loss")
Using 8.5 - 8.6 psi for the drawing we can arrive at the estimate:
4768 CFM * 8.5 psi
bhp = ----------------------- = ca. 290 hp (BOD + nitrification)
20 CFM/hp * 7 psi
3016 CFM * 8.5 psi
bhp = ---------------------- = ca. 184 hp (BOD + nitrification)
20 CFM/hp * 7 psi
Even though we are all "telling the truth", as a quantity or as a difference, 100 hp is difficult to conceal. As a final exercise, we will reiterate the calculation logic assuming that the public notice only indicated the BOD5 process.
Using the indication mentioned above:
1.6 lbs O2 per lb BOD5 processing ->
Oxygen required:
3574 * 1.6 = 5719 = 5719 lbs O2/day = 5719 /24 = 239 lbs O2/hour = 3.972 lbs O2/minute
In one case (transfer efficiency 8%):
CFM's = 3.972 / ( 0.08 * 0.075 * .232 ) = 2854 CFM = 2854 * 60 / 35.3145 = 4848 m3/h
In another case (transfer efficiency 12.65%):
CFM's = 3.972 / ( 0.1265 * 0.075 * .232 ) = 1805 CFM = 1805 * 60 / 35.3145 = 3066 m3/h
Maintaining all other conditions, it follows that:
2854 CFM * 8.5 psi
bhp = ----------------------- = ca. 174 hp (BOD only)
20 CFM/hp * 7 psi
1805 CFM * 8.5 psi
bhp = ----------------------- = ca. 110 hp (BOD only)
20 CFM/hp * 7 psi
Note: Comparing with the values calculated in the previous part, we have the following proportions:
174 hp / 290 hp = ca. 60 %
110 hp / 184 hp = ac. 60 %
Other calculations/auxiliary factors of interest
F/M (food-to-mass ratio)
where V is the total volume of tanks:
V = 4 * 25 * 14.2 = 1420 m2 * 5 = 7100 m3 = 7:100000 l = 7:100000 / 3.785 gallon = 1:875826 gallon = 1.876 mg
mgd's * BOD5 * 8.33 mgd's * BOD5
F/M = ---------------------- ---------------
MLSS design * V * 8.33 MLSS*V
design Using a nominal MLSS design value = 3500 mg/L in the
tanks, then 1.43*300*8.33 1.43 * 300
F/M = ------------------- = ------------- = 0.065
3500 * 1.876* 8.33 3500 * 1.876
A classification (Hammer.413) of processes suggests the following dimensions:
"high rate" 0.5 < f/m < 1
"conventional" 0.2 < f/m < 0.5
"extended" 0.05 < f/m < 0.2 (extended aeration)
Retention Time
H.R.T. (hours) = 24 * total tank volume / flow rate = 24 * 1.876 mg / 1.43 mgd = ca. 32 hours
Volumetric Organic Load (lbsBOD day per 1000 cu.ft.)
1.43*300 mg/L*8.33 3574 lbsBOD/day
cov= --------------------- = ------------------ = 14 lbsBODday p1000cu.ft.
7100m3*35.3145cu.ft. ca. 251*1000 cu.ft.
Selection and arrangement of diffusers
We calculated that we would need around 3016 CFM to be able to eliminate our problem. We have not made any statements about safety margins. It is common to use factors of Let's leave this in the hands of the moderator specialist and continue the development for selection of diffusers. The following logic may not be applicable to another type of broadcaster.
Without elaborating on the possible reasons, let us suppose that the construction firm unexpectedly built 4 tanks of 50m x 7.1m instead of the original 4 tanks of 25m x 14.2m proposed, although with a depth close to 5m. Despite what happened, the preliminary numerical developments remain valid and mayWe will think about finishing the calculations of our project before more errors are introduced., returning to the "English" units:
50m = 50 * 3.28 = 164'
7.1m = 7.1 * 3.28 = 23.3'
To determine the number of diffusers (in meters of diffuser tube) to be used, we must counterpose:
a. Usage patterns, i.e. air flow per linear meter of tubular diffuser, versus.
b. the available geometry.
a.We have 4 tanks, so each tank should receive 3016/4= ca. 754 CFMs
Using 8 CFM per meter of tubular diffuser we will have:
754 / 8 = ca. 95 meters of tubular diffuser per tank
when there are four tanks:
95 * 4 = 380 m of tube
Or 380 diffusers of 1-m
If we use diffusers assemblies (duplex assembly has two tubular diffusers, of 1-m each) we would then need 190 duplex assemblies.
b. Each tank let's say has two side ducts (164' long each), so the length of the side ducts is 2 * 164' = 328'
Placing the duplex assemblies approximately 5' (ca. 1.5m) to each other:
328 / 5 = ca. 66 assemblies.
If each assembly has two meters of pipe -> we will
need 66 * 2 = 132 1-m tubes per tank
If we have four tanks, then we will have:
132 * 4 = 528 1-m tubes,
or, using the language of duplexes, 264 duplex assemblies.
Comparing the results of both parts a. and b.
190 duplex vs. 264 duplex -> taking the higher value to meet/satisfy with the two conditions ->
Final selection: 264 duplex assemblies spaced 5' (ca. 1.50m), each meter of tubular diffuser will be operating at:
3016 CFM / 528 m = 5.7 CFM = ca. 6 CFM per 1-m tube.
Schematic diagram A-8128, included as a final illustration, shows an alternative configuration for this same design, which accommodates the entire aeration system in only three 25m x 14.2m tanks.
Check: 3 tanks * 88 assemblies * 2m = 528m diffuser tube.
Final grade and recognitions
Full knowledge of the techniques is absolutely essential for the right project. In this regard, it is necessary/advisable to consult technical literature. In practice, it is up to the designer to select the most appropriate method in the given case, and he must work in close cooperation with the various possible suppliers of equipment.
The preparation of this primer was largely achieved due to the technical contribution of mentors and engineers, Dr. James Young, P.E., Dr. Clifford W. Randall, P.E., Engineer Nils Young, P.E. of DBS Engineering, and Engineer Charles Tharp, P.E., from Environmental Dynamics Inc., Missouri.
The person in charge of the area must consult the innumerable specialized publications and specific monographs both regarding the type of industry, e.g. dairy, slaughterhouse, pharmaceutical, tannery, agribusiness, yeast, brewery, distillery, as well as the type of treatment process selected: conventional, intensive, extensive. These notes have been prepared simply as a guide to guide the task of facilitating the design of a preliminary design in the face of a concrete situation, in this case municipal. It is important to have access to engineering manuals from the supplier companies.
It was the object of this exposition simply to outline a logic of selection for a popular application, it is necessary to remark the fundamental directing role of the consultant, particularly for industrial applications, as the final determinant of the success of the installation.
Sample keywords: hydrogen sulfide EGSB tannery (no commas!)
