← Course contents · Part 6

· Visual lesson · Session 48 of 56


Coolant is flowing but is it enough

A running pump does not prove that every branch receives enough coolant. Read flow, pressure, temperature, and workload together.

Picture the idea

Half the flow changes the heat balance

0.5 kg/s

10 kW ÷ (0.5 × 4) = 5°C rise

0.25 kg/s

10 kW ÷ (0.25 × 4) = 10°C rise

Illustrative steady 10 kW load and heat capacity of 4 kJ/(kg·°C). Not an operating range or sizing recommendation.

Understand the measurements

Flow is how much fluid passes over time. Differential pressure is the pressure difference between two points.

Supply and return temperatures show another part of the picture. Label the measurement locations: a CDU reading is not a measurement at every GPU.

Trapped air, a restriction, or a changed valve position are possible causes to investigate, not conclusions from one chart.

Use a simple heat balance

Assume a steady 10 kW load, mass flow of 0.5 kg/s, and heat capacity of 4 kJ/(kg·°C).

The idealized temperature rise is 10 ÷ (0.5 × 4) = 5°C. Halving flow would double that rise under the same assumptions.

These are teaching values, not operating limits.

Try it yourself

Two racks have the same temperature rise, but one carries twice the heat. Must their flows match?

Check your answer

No. With the same fluid properties and steady-state assumptions, the higher-load rack needs twice the mass flow. Leave valve changes, venting, and fluid handling to qualified staff.

Sources and further reading

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