5.12 Advanced Applied Math: Mass Balance, Blending & Efficiency
Key Takeaways
- A mass balance states that mass in equals mass out plus mass accumulated, and it is the framework behind loading, removal efficiency, blending, and solids inventory calculations alike.
- A blending calculation is a mass balance on the constituent: the sum of each stream flow times its concentration equals the combined flow times the combined concentration.
- Percent removal equals the influent minus the effluent divided by the influent, multiplied by 100, and must be computed on the same basis, either concentration or mass, throughout.
- One population equivalent is conventionally 0.2 pounds of biochemical oxygen demand per person per day, which converts an industrial load into an equivalent number of people.
- On a timed exam, estimating the answer before calculating and checking that the units cancel correctly catches most arithmetic and setup errors faster than reworking the problem.
Advanced Applied Math: Mass Balance, Blending & Efficiency
The ABC Water Treatment Class IV exam contains 15 calculation items and the Wastewater Class IV exam contains 16, distributed across every content area. The basic dosage, detention, loading, and process-control formulas are covered earlier in this chapter. This section covers the calculation types that cut across those categories, and the discipline that keeps you from losing points to arithmetic.
1. The Mass Balance Framework
Almost every calculation in this field is one statement in different clothing:
Mass IN = Mass OUT + Mass ACCUMULATED (or destroyed, for a reacting system)
And mass, in US units, is almost always:
Pounds per day = Flow (MGD) x Concentration (mg/L) x 8.34
The 8.34 is pounds per gallon of water, and the MGD-times-mg/L product supplies the million that cancels the "per million" in mg/L. If you can state a problem as a mass balance, the arithmetic is bookkeeping.
Worked example - solids inventory
A plant's aeration basin holds 1.2 MG at 2,800 mg/L MLSS; the clarifiers hold 0.35 MG at 6,200 mg/L. What is the total system solids inventory?
- Aeration: 1.2 x 2,800 x 8.34 = 28,022 lb
- Clarifiers: 0.35 x 6,200 x 8.34 = 18,098 lb
- Total inventory = 46,120 lb
That number is the numerator of the MCRT calculation, and it is a mass balance on solids in storage.
Worked example - closing a balance to find an unknown
A plant receives 4,200 lb/day of BOD, discharges 210 lb/day in the effluent, and wastes solids containing an estimated 1,100 lb/day of BOD equivalent. How much was oxidized?
- IN = 4,200
- OUT = 210 (effluent) + 1,100 (waste solids) = 1,310
- Oxidized (destroyed) = 4,200 - 1,310 = 2,890 lb/day
That figure is the basis of the plant's oxygen demand, which is why the balance is worth closing.
2. Blending and Mixture Problems
When two or more streams combine, the constituent mass adds and the flows add:
(Q1 x C1) + (Q2 x C2) = (Q1 + Q2) x C(combined)
Worked example - two wells
Well A produces 400 gpm at 12 mg/L nitrate-N. Well B produces 250 gpm at 2 mg/L nitrate-N. What is the blended nitrate?
- Numerator: (400 x 12) + (250 x 2) = 4,800 + 500 = 5,300
- Denominator: 400 + 250 = 650 gpm
- Blended nitrate = 5,300 / 650 = 8.15 mg/L
Below the 10 mg/L MCL - which is exactly why blending is a recognized compliance strategy for nitrate.
Worked example - solving for the required blend
The same Well A at 12 mg/L must be blended with Well B at 2 mg/L to reach a target of 8.0 mg/L, with Well A running at 400 gpm. How much of Well B is needed?
- (400 x 12) + (Q2 x 2) = (400 + Q2) x 8
- 4,800 + 2Q2 = 3,200 + 8Q2
- 1,600 = 6Q2
- Q2 = 267 gpm
The same structure solves chemical dilution, sludge blending, recycle streams returning to the head of a plant, and combining two clarifier effluents.
Dilution shortcut
For a simple dilution of one stock into water, the mass balance collapses to:
V1 x C1 = V2 x C2
To make 500 gallons of 1.5 percent hypochlorite from 12.5 percent stock: V1 = (500 x 1.5) / 12.5 = 60 gallons of stock, diluted to 500 gallons.
3. Percent Removal and Efficiency
Percent removal = [(In - Out) / In] x 100
Two rules that generate wrong answers when broken:
- Use the same basis throughout. Compute removal on concentration or on mass, and never mix them. When flow changes between the two sampling points - which it does whenever a recycle stream, a plant water draw, or significant I&I is involved - the mass basis is correct and the concentration basis is misleading.
- Watch the direction. Percent removal uses influent in the denominator; a "percent increase" uses the starting value in the denominator. They are not interchangeable.
Worked example - the two bases disagree
Influent: 3.0 MGD at 210 mg/L BOD. Effluent: 3.4 MGD at 18 mg/L BOD (the extra 0.4 MGD is recycle and plant water returning).
- Concentration basis: (210 - 18) / 210 = 91.4 percent
- Mass basis: In = 3.0 x 210 x 8.34 = 5,254 lb/day; Out = 3.4 x 18 x 8.34 = 510 lb/day; removal = (5,254 - 510) / 5,254 = 90.3 percent
The VPDES secondary treatment standard requires 85 percent removal for BOD and TSS, and the permit specifies the basis. Know which one your permit uses.
4. Cost and Efficiency Calculations
The ABC Wastewater outline puts five calculation items in the Security, Safety, and Administrative Procedures area, which is where budget and unit-cost arithmetic lives.
Chemical cost per million gallons
Cost per MG = Dose (mg/L) x 8.34 x Unit price ($/lb)
At 28 mg/L alum costing $0.24 per pound: 28 x 8.34 x 0.24 = $56.05 per million gallons treated. At 5.0 MGD that is $280 per day, or about $102,000 per year - which is how a 3 mg/L dose optimization becomes a real budget line.
Energy cost of pumping
kWh/day = (Q gpm x TDH ft x 24 hr) / (3,960 x wire-to-water efficiency) x 0.746
A 900 gpm pump at 180 ft TDH and 70 percent wire-to-water efficiency running continuously:
- WHP = (900 x 180) / 3,960 = 40.9 hp
- Input hp = 40.9 / 0.70 = 58.4 hp
- kW = 58.4 x 0.746 = 43.6 kW
- kWh/day = 43.6 x 24 = 1,046 kWh, and at $0.11/kWh that is $115 per day, about $42,000 per year
Raising wire-to-water efficiency from 70 to 78 percent saves roughly 10 percent of that - about $4,200 a year from one pump.
Population equivalent
1 population equivalent (PE) = 0.2 lb BOD5 per day (also 0.2 lb TSS per day; approximately 100 gallons per day)
An industry discharging 900 lb/day of BOD has a population equivalent of 900 / 0.2 = 4,500 people, which is the number that goes into a capacity and cost-allocation discussion.
5. CT and Log Inactivation
CT (mg-min/L) = Disinfectant residual (mg/L) x T10 contact time (min) T10 = Theoretical detention time x Baffling factor CT ratio = CT achieved / CT required, and the ratio must be at least 1.0
Log inactivation achieved is estimated as CT ratio x the log credit the required CT value represents, and inactivation from multiple disinfection segments in series is additive.
Worked example
A clearwell holds 480,000 gallons. Peak flow is 2,400 gpm. Baffling factor is 0.5. Free chlorine residual at the clearwell outlet is 1.1 mg/L, and the required CT for 0.5-log Giardia inactivation at that pH and temperature is 27 mg-min/L.
- Theoretical detention = 480,000 / 2,400 = 200 min
- T10 = 200 x 0.5 = 100 min
- CT achieved = 1.1 x 100 = 110 mg-min/L
- CT ratio = 110 / 27 = 4.07
- Log inactivation = 4.07 x 0.5 = 2.04-log Giardia
Combined with a conventional filtration credit of 2.5-log, the plant delivers about 4.5-log, comfortably above the 3.0-log requirement.
6. Recirculation and Recycle Ratios
Recirculation ratio (R/Q) = Recycle flow / Forward flow
Where it appears:
- Trickling filter recirculation, typically 0.5:1 to 4:1, which dilutes influent strength and maintains a wetting rate on the media.
- RAS ratio, typically 0.3:1 to 1:1.
- Internal nitrate recycle (IMLR) in an MLE process, typically 2:1 to 4:1.
- Filter backwash recycle, capped at roughly 10 percent of plant flow.
The dilution effect of recirculation
A trickling filter receives 1.0 MGD of primary effluent at 140 mg/L BOD, with 2.0 MGD of recirculated filter effluent at 30 mg/L. What is the applied strength?
- (1.0 x 140) + (2.0 x 30) = 140 + 60 = 200
- Total flow = 3.0 MGD
- Applied BOD = 200 / 3.0 = 66.7 mg/L
The organic mass applied has not changed, but the concentration has been cut by more than half, which is precisely the point of recirculation.
Denitrification limit from recycle
In an MLE process the theoretical maximum nitrogen removal is set by how much nitrate the recycle can deliver to the anoxic zone:
Maximum removal (%) = [(IMLR + RAS) / (1 + IMLR + RAS)] x 100
At IMLR = 3.0 and RAS = 0.6: (3.6 / 4.6) x 100 = 78 percent. Pushing beyond that requires a second anoxic stage with supplemental carbon, not more recycle - a fact that saves a great deal of futile pumping.
7. Exam Arithmetic Discipline
- Estimate first. Round every input to one significant figure and get a ballpark. If your calculated answer is ten times the estimate, you have a decimal or unit error, and you will find it faster than by re-solving.
- Check the units. Write them and cancel them. If gallons per minute divided by square feet is supposed to give gallons per day per square foot, the missing 1,440 will announce itself.
- Convert once, early. Get everything into MGD, mg/L, feet, and minutes at the start.
- Know the conversion constants cold: 7.48 gal/cu ft, 8.34 lb/gal, 1,440 min/day, 2.31 ft of head per psi, 0.433 psi per ft, 3,960 for horsepower, 1.604 for gpm/sq ft to in/min, 43,560 sq ft per acre, 0.785 for pi over four.
- Read what is asked. A question may give five inputs and ask for a value that needs three. Extra data is a distractor.
- Use the provided formula sheet. Both exams supply the ABC formula and conversion table at check-in. Practice with it in front of you so that on exam day you are selecting formulas, not recalling them.
- Answer every question. There is no penalty for a wrong answer, and eliminating two implausible options makes a guess a coin flip.
Well A produces 500 gpm at 14 mg/L nitrate as nitrogen, and Well B produces 300 gpm at 3 mg/L. What is the nitrate concentration of the blended water?
A plant receives 2.8 MGD at 195 mg/L BOD and discharges 3.1 MGD at 16 mg/L BOD, the extra flow being recycle and plant water. What is the BOD removal on a mass basis?
A modified Ludzack-Ettinger process operates with an internal nitrate recycle ratio of 2.5 and a return activated sludge ratio of 0.5. What is the theoretical maximum total nitrogen removal from recycle alone?