KEVOS® MAKEOS · Hydraulics

Coolant Flow for a Heat Load

What flow of water or oil removes this heat load for the allowed temperature rise?

Liquid flow needed to carry away a heat load at a given temperature rise, or the temperature rise for a given flow, for water, glycol mixtures or oil.

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How this tool works

Inputs, units and where to find them

InputUnitTypical source
CalculatechoiceFind the flow, or the temperature rise.
Heat loadkWHeat to be removed (machine losses, hydraulic power loss, process heat).
Allowed temperature riseKOutlet minus inlet temperature you can accept.
Flow (rise mode)L/minAvailable flow (rise mode).
Specific heat of fluidkJ/(kg·K)Specific heat (water 4.18, 30% glycol about 3.7, mineral oil about 1.9 kJ/kg·K).
Fluid densitykg/m³Fluid density (water 998, 30% glycol about 1040, oil about 870 kg/m³).

Method and formulas

  1. Heat Q = ṁ cp ΔT.
  2. Volume flow = heat ÷ (ρ cp ΔT), converted to L/min.
  3. Temperature rise = heat ÷ (ρ cp × flow).
  4. Example: 15 kW with a 5 K rise needs 43.148 L/min of water.

Limitations

  • Steady state; fluid properties taken at the mean temperature.
  • Heat exchanger size is a separate calculation (see the LMTD tool).
  • Example values are illustrations. Replace them with values for your actual task.

Typical uses

  • Size cooling water for a machine, mould or oil cooler
  • Check temperature rise in a hydraulic return line
  • Size a chiller circuit

Who it is for

  • Process engineers
  • Maintenance engineers
  • Design engineers

What you get

  • Flow or temperature rise
  • Mass flow

Interpreting the result

  • A smaller allowed temperature rise needs proportionally more flow.

References

  • Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 371–374. Topic reference; no source tables or recommended settings are reproduced.

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Open the Coolant Flow for a Heat Load calculator