Formula · 2 min read

Coolant flow rate

Coolant flow rate is the volume of coolant per minute needed to carry a given heat load at a given temperature drop: P kW at ΔT °C needs P × 60 / (4.186 × ΔT) L/min — 50 kW of liquid-captured heat at ΔT 8 °C needs 89.6 L/min.

Formula

4.186 kJ/(kg·K) is the specific heat of water and a liter of water-glycol weighs about a kilogram, so kilowatts convert directly to liters per minute. P_liquid is the heat the loop actually captures — in direct-to-chip systems typically 60-80% of total IT load, not all of it.

Worked example

A rack sending 50 kW to liquid with a supply/return ΔT of 8 °C circulates roughly 90 L/min — about a bathtub a minute. The same 50 kW at ΔT 15 °C needs only 47.8 L/min: wider ΔT trades flow for temperature.

Try it

Common mistakes

• Feeding total IT load instead of the liquid share. Direct-to-chip captures 60-80%; the rest cools by air. Using 100% oversizes the loop by a quarter to two-thirds.
• Wrong ΔT. ΔT is the supply-return coolant temperature difference at the rack — not room air vs coolant, and not the facility water loop's ΔT.
• Ignoring glycol. PG25's specific heat runs roughly 10% below pure water, so real systems move a few percent more flow than the water formula suggests. First-pass math with 4.186 is standard practice.

Related

Rack builder computes flow for a whole assembly · Approach temperature · Room AC load

Sources

First principles: q = ṁ × c × ΔT with c(water) = 4.186 kJ/(kg·K), ρ ≈ 1 kg/L. Same model as this site's rack builder.