METHOD NOTE 01 · REPRODUCIBLE EXAMPLE

A 120 kW liquid-cooling loop: getting the heat balance right

bupt.ai · AI-assisted method note · Illustrative, not measured · Revised2026-11-01

A cooling-capacity question need not start with a complex model. Define the control volume and units, and a steady-state energy balance can reveal many obvious errors. This teaching example uses 120 kW of heat absorbed by a fluid, a mass flow rate of 4 kg/s, an inlet temperature of 30 °C and a constant specific heat of 4.18 kJ/(kg·K). These are example inputs, not project measurements.

Define where the 120 kW enters

Let the control volume contain only the liquid heat-transfer section. Neglect heat loss to the surroundings and changes in kinetic and potential energy, exclude pump work from this section, and assume single-phase flow with approximately constant specific heat. Then use Q = m × cp × (Tout − Tin). If 120 kW is the server’s total electrical input rather than the heat entering this liquid loop, it cannot be substituted without further justification.

Quantity	Example value	What to check
Heat absorption rate Q	120 kW	Not cumulative meter energy in kWh
Mass flow rate m	4 kg/s	Not volumetric flow in m³/h
Specific heat at constant pressure cp	4.18 kJ/(kg·K)	An example constant, not valid for every coolant
Inlet temperature Tin	30 °C	Temperature differences have the same numeric value in K and °C

Recalculate in one step

ΔT = 120 / (4 × 4.18) = 7.17703349 K, giving Tout = 37.17703349 °C. Dimensionally, kW / [(kg/s) × kJ/(kg·K)] = K. A common error is to treat 4.18 as J/(kg·K), producing a temperature rise that is 1,000 times too large.

Copy the model below into ThermoForge’s EES text input. It does not call a fluid-property library. This reproduces an algebraic heat balance under the stated specific-heat assumption; it does not validate property accuracy.

Q=120 [kW]
m=4 [kg/s]
cp=4.18 [kJ/kg-K]
Tin=30 [C]
Q=m*cp*(Tout-Tin)

What follows when the flow rate changes?

At the same heat load and example specific heat, the temperature rise is 14.35406699 K at 2 kg/s and 4.78468900 K at 6 kg/s. This shows the effect of heat-capacity flow rate; it does not establish that higher flow makes the system more energy-efficient. Pump energy, pressure drop, heat-exchanger temperature differences and control constraints require separate analysis.

This example also cannot directly determine chip junction temperature. Packaging, interface materials, the cold plate and convective resistance lie between the chip and fluid. Coolant outlet temperature is neither the junction temperature nor the highest local temperature.

From a first calculation to engineering judgment

For large temperature changes, changing coolant composition or phase change, use properties or enthalpy differences appropriate to the conditions. Balance the heat-transfer section and whole facility separately, include pump work, bypasses and branch flows, and compare against independent measurements. First check units and order of magnitude with this example, then proceed to system energy-efficiency comparisons; this provides a stronger basis than comparing two dashboards alone.

Method: a sensible-heat approximation of the steady-state energy balance for an open control volume. Values follow directly from the listed inputs. For property inputs and units, see CoolProp High-Level Interface. This example fixes cp; that documentation is not cited as precise property evidence for cp = 4.18.