Dosage & Detention Time Calculations

13.1 Dosage & Detention Time Calculations

Quick Answer: Chemical dose in pounds uses lbs = mg/L × MG × 8.34. Detention time is volume ÷ flow after units match. Always adjust feed rate for percent purity, and convert gallons, ft³, MGD, and gpm before dividing.

Texas TCEQ operator exams expect you to set up water-treatment math the way a plant operator does on shift: pick the right formula, convert units first, then calculate. This section covers the pounds formula, feed-rate and purity adjustments, detention (retention) time, and an introduction to weir overflow rate—all with fully worked examples.

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Why 8.34 Appears Everywhere

One gallon of water weighs about 8.34 pounds. One milligram per liter (mg/L) is the same concentration as one part per million (ppm) in dilute water solutions. Multiplying concentration (mg/L) by volume in million gallons (MG) by 8.34 converts that concentration into a mass of chemical in pounds:

[ \text{lbs} = \text{mg/L} \times \text{MG} \times 8.34 ]

For a daily dose when flow is given in million gallons per day (MGD), the same relationship becomes a feed rate:

[ \text{lbs/day} = \text{mg/L} \times \text{MGD} \times 8.34 ]

If the problem gives volume in gallons instead of MG, convert first: (\text{MG} = \text{gallons} \div 1{,}000{,}000).

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Worked Example 1 — Basic Pounds Formula

A clearwell holds 0.75 MG. Operators dose chlorine to 2.0 mg/L. How many pounds of chlorine are needed for one full volume change?

Step 1: Identify knowns: mg/L = 2.0, MG = 0.75.

Step 2: Apply the formula:

[ \text{lbs} = 2.0 \times 0.75 \times 8.34 = 12.51\ \text{lbs} ]

Answer: 12.51 pounds of chlorine (as pure Cl₂ equivalent) are required.

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Worked Example 2 — Daily Feed Rate from Plant Flow

A plant treats 2.4 MGD and maintains a finished-water chlorine residual dose of 1.5 mg/L. What is the chlorine feed rate in lbs/day?

[ \text{lbs/day} = 1.5 \times 2.4 \times 8.34 = 30.024\ \text{lbs/day} ]

Rounded for operations paperwork, that is about 30.0 lbs/day. On the exam, keep at least two decimal places unless the choices force rounding.

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Percent Purity Adjustments

Commercial chemicals are rarely 100% active ingredient. If a bag is 65% available chlorine, you must feed more product to deliver the same pure chlorine mass:

[ \text{lbs product to feed} = \frac{\text{lbs of pure chemical needed}}{\text{decimal purity}} ]

Worked Example 3 — Purity Correction

From Example 2, the plant needs 30.024 lbs/day of pure chlorine. The hypochlorite on hand is 12.5% available chlorine. How many pounds of product must be fed per day?

[ \text{lbs product/day} = \frac{30.024}{0.125} = 240.19\ \text{lbs/day} ]

Answer: Feed about 240 lbs/day of the 12.5% product. A common trap is multiplying by 0.125 instead of dividing—that would underfeed by a factor of 64.

Worked Example 4 — Alum Dose with Purity

Jar tests set an alum dose of 25 mg/L. Flow is 1.8 MGD. Dry alum assays at 92% active. Find pure alum lbs/day, then product lbs/day.

Pure chemical:

[ 25 \times 1.8 \times 8.34 = 375.3\ \text{lbs/day pure} ]

Product feed:

[ \frac{375.3}{0.92} = 407.9\ \text{lbs/day product} ]

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Detention Time (Retention Time)

Detention time is how long water stays in a tank, basin, or clearwell:

[ \text{Detention time} = \frac{\text{Volume}}{\text{Flow}} ]

Units must cancel. Common exam setups:

Useful conversions:

• (1\ \text{ft}^3 = 7.48\ \text{gal})
• (1\ \text{MGD} = 1{,}000{,}000\ \text{gal/day} = 694.4\ \text{gpm}) (because (1{,}000{,}000 \div 1{,}440 \approx 694.4))
• Hours = minutes ÷ 60; days = hours ÷ 24

Worked Example 5 — Detention Time in Minutes

A rectangular flocculation basin holds 45,000 gallons. Flow through the basin is 750 gpm. What is the detention time in minutes?

[ \text{DT} = \frac{45{,}000\ \text{gal}}{750\ \text{gpm}} = 60\ \text{minutes} ]

Worked Example 6 — Convert Volume from Cubic Feet

A circular clearwell has volume 8,500 ft³. Plant flow is 1.2 MGD. Find detention time in hours.

Step 1 — gallons:

[ 8{,}500 \times 7.48 = 63{,}580\ \text{gal} ]

Step 2 — flow in gpm:

[ 1.2\ \text{MGD} \times 694.4 = 833.28\ \text{gpm} ]

Step 3 — minutes, then hours:

[ \text{DT} = \frac{63{,}580}{833.28} \approx 76.3\ \text{minutes} \approx 1.27\ \text{hours} ]

Alternatively in MG and MGD:

[ \text{MG} = 63{,}580 \div 1{,}000{,}000 = 0.06358\ \text{MG} ]

[ \text{DT (days)} = \frac{0.06358}{1.2} = 0.0530\ \text{days} \times 24 = 1.27\ \text{hours} ]

Same result either path—choose the unit path that matches the answer choices.

Worked Example 7 — Contact Time Check

A chlorine contact basin volume is 0.35 MG. Nighttime flow drops to 0.50 MGD. Daytime peak is 1.40 MGD. Compare detention times.

Night:

[ \frac{0.35}{0.50} = 0.70\ \text{days} = 16.8\ \text{hours} ]

Day peak:

[ \frac{0.35}{1.40} = 0.25\ \text{days} = 6.0\ \text{hours} ]

Detention time shrinks as flow rises. CT compliance and baffling factor questions build on this idea: shorter hydraulic residence at peak flow is the critical case.

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Weir Overflow Rate (Introduction)

Clarifier and sedimentation weirs are rated by how much flow passes per length of weir:

[ \text{Weir overflow rate} = \frac{\text{Flow}}{\text{Weir length}} ]

Typical units are gpd/ft or gpm/ft. High overflow rates can pull floc over the weir and raise turbidity.

Worked Example 8 — Weir Loading

A rectangular clarifier has 120 ft of effluent weir length. Influent flow is 1.5 MGD. What is the weir overflow rate in gpd/ft?

[ 1.5\ \text{MGD} = 1{,}500{,}000\ \text{gpd} ]

[ \text{WOR} = \frac{1{,}500{,}000\ \text{gpd}}{120\ \text{ft}} = 12{,}500\ \text{gpd/ft} ]

If the exam asks for gpm/ft:

[ \frac{1{,}500{,}000}{1{,}440} = 1{,}041.7\ \text{gpm},\quad \frac{1{,}041.7}{120} \approx 8.68\ \text{gpm/ft} ]

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Exam Setup Habits That Save Points

1. Convert before you divide. Mixing MG with gpm without converting is the most common detention-time miss.
2. Purity always divides the pure requirement (unless the problem already gives product strength in a way that is pre-adjusted).
3. 8.34 is for water mass conversion, not a random fudge factor—know why it is there.
4. Peak flow gives the shortest detention time; use that case when a question asks whether contact time is adequate.

Master these setups and the same arithmetic pattern repeats across disinfection, coagulation, and basin design questions on the TCEQ exams.