System Volume from Turnover Time
Worked example: 1200 gpm with a 25 min turnover → 113,562 L (30,000 gal) — press Try an example to run it live, then adjust anything.
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System Volume from Turnover Time explained
Nine times out of ten the plant cannot tell you how much water is in the system, and the treatment program depends on knowing. If you have the pump curve or a flow meter, and you can time one complete circuit, the volume follows immediately: 1200 gpm circulating with a 25 minute turnover means 30,000 gallons in the loop. Timing the turnover is the practical trick — inject dye or a salt slug at the pump discharge and clock how long until it comes back around.
Design rules of thumb are worth carrying as a cross-check. Cooling towers are usually sized so the whole system turns over in about 10 minutes of recirculation, and hydronic heating loops in 15–20; if your measured turnover is far outside that, suspect the flow figure rather than the volume. And remember what "system volume" must include — sump, standpipes, riser, condenser bundles, and the full length of any buried run. Underestimating volume is the single most common reason a slug dose lands low.
System Volume from Turnover Time formula
- = System volume (L)
- = Recirculation rate (L/min)
- = Turnover time (s)
Missing one of these? Work it out first, then come back
- System volume — Dose Achieved from Chemical Added, Closed Loop Slug Dose Volume
- Recirculation rate — Cooling Tower Evaporation Rate, Cooling Tower Drift Loss
- Turnover time — Hydraulic Detention Time, Chick–Watson Inactivation