23.6 Percent Removal, Process Efficiency & Solution Blending

23.6 Percent Removal, Process Efficiency & Solution Blending

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Percent Removal

$\% \text{ Removal} = \frac{\text{In} - \text{Out}}{\text{In}} \times 100$

The denominator is always the influent value. Dividing by the effluent is the single most common error and, on a multiple-choice exam, it lands on an answer choice that was placed there for exactly that reason.

Worked example. Primary influent BOD is 260 mg/L; primary effluent is 165 mg/L.

$\frac{260 - 165}{260} \times 100 = \frac{95}{260} \times 100 = 36.5\%$

That figure is typical for primary clarification, which normally removes about 25 to 40 percent of BOD and 50 to 70 percent of suspended solids.

Removal in Pounds

A percentage says nothing about the actual load a process handled. Pounds are what matter for design and for troubleshooting.

$\text{lb/day removed} = (\text{In} - \text{Out}) \times \text{Flow (MGD)} \times 8.34$

Worked example. Using the values above at 3.8 MGD:

$95 \times 3.8 \times 8.34 = 3{,}011\text{ lb BOD/day removed}$

[!IMPORTANT] Percent removal and pounds removed can move in opposite directions. If flow doubles while percent removal holds constant, the process is handling twice the load. Conversely a plant whose percent removal drops during a low-flow, low-strength period may be removing the same pounds as always. Always ask which one the question is about.

Overall Plant Removal

Overall removal is calculated from the plant influent and final effluent, not by adding the removals of the individual processes.

Worked example. Influent 240 mg/L BOD, primary effluent 155 mg/L, final effluent 12 mg/L.

• Primary removal: (240 − 155) ÷ 240 × 100 = 35.4%
• Secondary removal: (155 − 12) ÷ 155 × 100 = 92.3%
• Overall: (240 − 12) ÷ 240 × 100 = 95.0%

Note that 35.4 + 92.3 = 127.7, which is obviously wrong. Removals in series do not add, because each stage acts on what remains after the previous one.

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Mass Balance

The principle: what goes in must come out or accumulate.

$\text{In} = \text{Out} + \text{Accumulation}$

Worked example — solids balance. A plant receives 2,900 lb/day of solids in the influent and discharges 180 lb/day in the effluent. How much must be removed as sludge?

$2{,}900 - 180 = 2{,}720\text{ lb/day}$

If the plant is actually wasting only 2,100 lb/day, the difference of 620 lb/day is accumulating in the system — building inventory in the aeration basin or the clarifier blanket — and the operator has a developing problem that a single day's data would not reveal.

Solids in a sludge stream: $\text{lb/day} = \text{Flow (MGD)} \times 8.34 \times \% \text{ Solids} \times 10{,}000$

Because 1 percent equals 10,000 mg/L, this is the pounds formula with the percentage converted to a concentration.

Worked example. 0.028 MGD of sludge at 3.4 percent solids:

$0.028 \times 8.34 \times 3.4 \times 10{,}000 = 7{,}940\text{ lb/day}$

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Dilution and Blending

The Dilution Relationship

$C_1 V_1 = C_2 V_2$

Worked example. How much 12.5 percent sodium hypochlorite is needed to make 400 gallons of 1.0 percent solution?

$12.5 \times V_1 = 1.0 \times 400$ $V_1 = \frac{400}{12.5} = 32\text{ gallons}$

Add 32 gallons of the 12.5 percent solution and dilute to 400 gallons total — not add 400 gallons of water to it. The distinction between "dilute to" and "add to" is a standard trap.

Blending Two Streams

$C_{\text{blend}} = \frac{(C_1 \times Q_1) + (C_2 \times Q_2)}{Q_1 + Q_2}$

[!WARNING] This is a flow-weighted average, not a simple average. Averaging the two concentrations directly is correct only when the two flows happen to be equal, and the exam reliably includes the simple average as a distractor.

Worked example. A utility blends 1.8 MGD of well water at 1,450 mg/L TDS with 4.2 MGD of surface water at 380 mg/L TDS.

$C = \frac{(1{,}450 \times 1.8) + (380 \times 4.2)}{1.8 + 4.2} = \frac{2{,}610 + 1{,}596}{6.0} = \frac{4{,}206}{6.0} = 701\text{ mg/L}$

The simple average of 1,450 and 380 is 915 mg/L — meaningfully different, and wrong, because the lower-concentration stream carries more than twice the flow.

Blending is a real Arizona treatment strategy. Where a well exceeds a standard for arsenic, nitrate, or fluoride, blending it with a compliant source can bring the combined water into compliance, and the blend calculation is how the allowable ratio is determined.

Worked example — allowable blend. A well contains 0.018 mg/L arsenic against a maximum contaminant level of 0.010 mg/L, and the alternate source contains 0.002 mg/L. What fraction of the blend may come from the well?

Let x be the well fraction: $0.018x + 0.002(1-x) = 0.010$ $0.018x + 0.002 - 0.002x = 0.010$ $0.016x = 0.008 \quad \Rightarrow \quad x = 0.50$

No more than 50 percent of the blend may come from the well, and prudent operation targets a margin below that rather than the exact limit.

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Efficiency

Pump, Motor, and Wire-to-Water

$\text{Water Horsepower} = \frac{\text{Flow (gpm)} \times \text{Head (ft)}}{3{,}960}$

$\text{Brake Horsepower} = \frac{\text{WHP}}{\text{Pump Efficiency}}$

$\text{Motor Horsepower} = \frac{\text{BHP}}{\text{Motor Efficiency}}$

$\text{Wire-to-Water Efficiency} = \text{Pump Efficiency} \times \text{Motor Efficiency}$

Worked example. 1,100 gpm against 190 ft of total dynamic head, with a pump efficiency of 76 percent and a motor efficiency of 91 percent.

$\text{WHP} = \frac{1{,}100 \times 190}{3{,}960} = \frac{209{,}000}{3{,}960} = 52.8\text{ hp}$ $\text{BHP} = \frac{52.8}{0.76} = 69.5\text{ hp}$ $\text{MHP} = \frac{69.5}{0.91} = 76.4\text{ hp}$ $\text{Wire-to-water} = 0.76 \times 0.91 = 0.692 = 69.2\%$

[!IMPORTANT] Wire-to-water efficiency is always lower than either component efficiency, because the losses multiply rather than average. A pump at 76 percent driven by a motor at 91 percent delivers 69 percent of the electrical energy as useful water horsepower — nearly a third is lost. This is why pump efficiency testing is a genuine energy conservation measure and why it appears explicitly in the Need-to-Know Criteria as a scored task.

A Reasonableness Habit

Every efficiency must be less than 100 percent. Every percent removal must be between 0 and 100. Any blended concentration must fall between the two input concentrations. Results outside those bounds mean the setup was wrong, and catching that takes two seconds.