Biocide slug dose and when to re-dose

Cooling tower · biocide dosing and holding time

A routine dip slide on the same 750 gpm tower comes back at 10⁵ CFU/mL and the decision is a manual slug of non-oxidising biocide before the next service visit. A dye trace puts the full system turnover — basin, risers, condenser bundles and the buried return — at 12 minutes, the label calls for 25 ppm of active ingredient, and the drum in the truck is 20 % active. The tower is holding 5 cycles on a 10 °F range. Find the evaporation, the bleed, the system volume, the product dose to aim at, the mass of product to weigh out, and the holding time index that says when the slug is half gone.

Step 1 of 6 · solve for Evaporation rate

The bleed rate is what washes the biocide back out, and the bleed follows from evaporation, so the water balance has to be done before the dosing question can be answered at all.

Rearranged for E
E=0.001RΔTE = 0.001 \, R \, \Delta T
Your values, in your units
E=0.001(750 gpm)(10 F)E = 0.001 \, \left( 750\ \text{gpm} \right) \, \left( 10\ \text{F}^{\circ} \right)
Converted to base units
E=0.001(2,839.06 L/min)(5.55556 C)E = 0.001 \, \left( 2{,}839.06\ \text{L/min} \right) \, \left( 5.55556\ \text{C}^{\circ} \right)
Answer
E=28.4E = 28.4

Carried onward at full precision, not this rounded figure.

Open the Cooling Tower Evaporation Rate solver →

Step 2 of 6 · solve for Blowdown rate

At five cycles the controller bleeds 1.875 gpm. Every gallon of it leaves at whatever biocide concentration is in the basin at that moment — this is the drain the slug is fighting.

Rearranged for B
B=ECOC1B = \frac{E}{\text{COC} - 1}
28.4 L/mincarried from step 1
Your values, in your units
B=(0.000473176 m3/s)(5)1B = \frac{\left( 0.000473176\ \text{m}^{3}\text{/s} \right)}{\left( 5 \right) - 1}
Converted to base units
B=(28.3906 L/min)(5)1B = \frac{\left( 28.3906\ \text{L/min} \right)}{\left( 5 \right) - 1}
Answer
B=7.1B = 7.1

Carried onward at full precision, not this rounded figure.

Open the Blowdown Rate from Cycles solver →

Step 3 of 6 · solve for System volume

A slug dose is a mass divided by a volume, so the volume has to be real. 750 gpm circulating with a 12 minute measured turnover is 9000 gallons — trust the trace, not the drawings, which never include the buried run.

Rearranged for V
V=RtV = R \, t
Your values, in your units
V=(750 gpm)(12 min)V = \left( 750\ \text{gpm} \right) \, \left( 12\ \text{min} \right)
Converted to base units
V=(2,839.06 L/min)(720 s)V = \left( 2{,}839.06\ \text{L/min} \right) \, \left( 720\ \text{s} \right)
Answer
V=34.1V = 34.1

Carried onward at full precision, not this rounded figure.

Open the System Volume from Turnover Time solver →

Step 4 of 6 · solve for Product dose

Nobody doses neat active ingredient. The label's 25 ppm is active, the drum is 20 % active, so the water has to reach 125 ppm of product to carry 25 ppm of the thing that actually kills.

Rearranged for D_p
Dp=100DaAD_p = \frac{100 \, D_a}{A}
Your values, in your units
Dp=100(25 ppm)(20 %)D_p = \frac{100 \, \left( 25\ \text{ppm} \right)}{\left( 20\ \text{\%} \right)}
Converted to base units
Dp=100(0.0025 %)(20 %)D_p = \frac{100 \, \left( 0.0025\ \text{\%} \right)}{\left( 20\ \text{\%} \right)}
Answer
Dp=125D_p = 125

Carried onward at full precision, not this rounded figure.

Open the Product Dose from Active Strength solver →

Step 5 of 6 · solve for Mass of chemical added

Now the one-shot: the product dose from step 4 applied to the volume from step 3. About 4.26 kg — 9.4 lb, or a shade over a gallon of a typical 9 lb/gal product — poured into the sump at the pump suction where it will mix in one turnover.

Rearranged for m
m=C100Vρwm = \frac{C}{100} \, V \, \rho_w
125 ppmcarried from step 4
34.1 carried from step 3
Your values, in your units
m=(0.0125 %)100(34.0687 m3)ρwm = \frac{\left( 0.0125\ \text{\%} \right)}{100} \, \left( 34.0687\ \text{m}^{3} \right) \, \rho_w
Converted to base units
m=(0.0125 %)100(34,068.7 L)ρwm = \frac{\left( 0.0125\ \text{\%} \right)}{100} \, \left( 34{,}068.7\ \text{L} \right) \, \rho_w
Answer
m=4.26m = 4.26

Carried onward at full precision, not this rounded figure.

Open the Dose Achieved from Chemical Added solver →

Step 6 of 6 · solve for Holding time index

The slug starts washing out the instant it goes in. The holding time index is its half-life in this system: about 199 600 s, which is 55 hours, so by Wednesday afternoon a Monday morning slug is at half strength and the re-dose is due.

Rearranged for HTI
HTI=0.693VB\text{HTI} = \frac{0.693 \, V}{B}
34.1 carried from step 3
7.1 L/mincarried from step 2
Your values, in your units
HTI=0.693(34.0687 m3)(0.000118294 m3/s)\text{HTI} = \frac{0.693 \, \left( 34.0687\ \text{m}^{3} \right)}{\left( 0.000118294\ \text{m}^{3}\text{/s} \right)}
Converted to base units
HTI=0.693(34,068.7 L)(7.09765 L/min)\text{HTI} = \frac{0.693 \, \left( 34{,}068.7\ \text{L} \right)}{\left( 7.09765\ \text{L/min} \right)}
Answer
HTI=55.5\text{HTI} = 55.5

Carried onward at full precision, not this rounded figure.

Open the Holding Time Index solver →

Why this order

This chain runs in the order it does because a slug dose is bracketed by two different things and both have to be found before any chemical is weighed. The front end is volume: a dose is a mass spread through a system, and 9000 gallons measured by turnover is the only defensible number — nameplate sump volumes ignore risers, condenser bundles and buried returns and routinely run 30–50 % light, which is why so many slugs land under target. The back end is the bleed, which is why steps 1 and 2 come first even though they look like a detour: the same blowdown that protects the tubes from scale is a drain pulling the biocide straight back out, and its size is fixed by evaporation and the cycles setpoint, not by anything to do with the biocide. Only between those two does the dosing arithmetic happen — active to product in step 4, then product to mass in step 5 — and the holding time index in step 6 converts the whole thing from a one-off number into a schedule. HTI is a true half-life; the 0.693 in it is ln 2, the same constant that governs radioactive decay, because a bled system is a stirred tank diluting itself exponentially.

The mistake that costs money — and, with a biocide, costs control of the system — is confusing product with active. Weigh out 25 ppm worth of product on a 20 % active drum and you have dosed 5 ppm of active, one fifth of label rate: not a kill, just a slow selection of the most resistant organisms in the tower, followed by a dip slide that looks worse a fortnight later and a customer who is paying for chemistry that cannot work. US EPA biocide labels are written around a use rate and a contact time, so under-dosing is also an off-label application. The second error is computing the holding time index from the bleed valve alone. Add drift, add a passing bleed solenoid, add the leaking gland on the standby pump, and the real half-life can be a third of the calculated one — which is exactly what is happening when residuals vanish overnight and nobody can find the reason. Best practice on a slug is to lock the bleed out for a few hours on the controller so the contact time is real. Keep an eye on the trade constants underneath as well: 8.34 lb per gallon is water near 60 °F and is wrong for a hot basin or a brine, and the 0.001 evaporation rule that fixed the bleed is a design-day figure that drifts between about 0.00085 in cold dry weather and 0.0011 on a humid afternoon — a reminder that a 55 hour half-life is a plan, not a guarantee, and the dip slide is still the referee.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.