How do you calculate the temperature rise of the coolant?

Via the heat balance ΔT = (Q · 60 · 1000) / (ρ · V̇ · c): the heat load Q divided by the product of density, flow rate, and specific heat capacity. The factor 60 · 1000 converts the flow rate from l/min to m³/s.

Which flow rate does my cooling system need?

The required flow rate follows from V̇ = (Q · 60 · 1000) / (ρ · ΔT · c) for a desired temperature rise ΔT. The smaller the permissible temperature rise, the higher the required flow rate.

How much may the coolant heat up?

For liquid coolers in electronics, the temperature rise is typically between 5 and 10 K. A small temperature rise ensures an even temperature distribution and a low risk of hotspots, but requires a higher flow rate and therefore more pumping power.

Does the calculator also apply to water-glycol mixtures?

Yes. Select the coolant or enter density and heat capacity directly. Glycol lowers the heat capacity, so the medium heats up more at the same flow rate.

Calculate cooling power and flow rate for liquid cooling

How much does the coolant heat up under a given heat load? And which flow rate keeps the temperature under control? The calculator provides both values for sizing liquid cooling circuits.

Why temperature rise and flow rate go together

The cooling power of a liquid cooling system depends on the flow rate and on the temperature rise of the coolant between inlet and outlet. A small temperature rise means an even temperature distribution across the heat sink and fewer hotspots. This calculator is for sizing liquid cooling circuits in electronics and power electronics, not for air-conditioning or room cooling.

Cooling Power Calculator

Heat load to dissipate Q

Flow rate V̇

l/min

CoolantDensity ρ

kg/m³

Heat capacity c

J/(kg·K)

Formula and derivation

The temperature rise of the coolant follows from the heat balance Q = ṁ · c · ΔT. With the mass flow rate ṁ = ρ · V̇ and the conversion of the flow rate from l/min, this gives:


ΔT = (Q · 60 · 1000) / (ρ · V̇ · c)

Q: heat load to dissipate [W]
ρ: density of the coolant [kg/m³]
V̇: flow rate [l/min]
c: specific heat capacity [J/(kg·K)]
The factor 60 · 1000 converts the flow rate from l/min to m³/s.


V̇ = (Q · 60 · 1000) / (ρ · ΔT · c)

Rearranged for flow rate, this determines the required V̇ for a desired temperature rise.

Worked example from practice

A cold plate dissipates the power loss of six chips: Q = 3,000 W. The coolant is water (ρ = 997 kg/m³, c = 4,180 J/(kg·K)) at a flow rate of 5 l/min.

ΔT = (3,000 · 60 · 1000) / (997 · 5 · 4,180)
ΔT ≈ 8.6 K

The cooling water heats up by about 8.6 K between inlet and outlet. That is a typical value for a properly sized liquid cooling system.

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