Hydraulic Cooling Calculator

The calculator computes heat generated as input horsepower × (1 − overall efficiency) × 2545, so heat is literally defined as the fraction of hydraulic power that isn't delivered as useful work. At 90% efficiency only 10% of input HP turns into waste heat, while at 60% efficiency 40% does — so a less efficient system driven by the same PSI and GPM produces roughly four times the heat load and needs a correspondingly larger cooler.

A larger reservoir volume increases the estimated surface area (via the cube-root approximation used to turn gallons into a box shape), which raises natural dissipation and lowers the net heat load the cooler has to remove. Because surface area scales with volume to the two-thirds power rather than linearly, doubling reservoir size doesn't double natural cooling — it's a real but diminishing-returns effect, which is why undersized or enclosed reservoirs should get an oversized cooler rather than relying on tank cooling alone.

That figure assumes a water-to-oil shell-and-tube heat exchanger where the cooling water is allowed to rise 20°F as it absorbs the net heat load, using the simplified relationship GPM = BTU/hr ÷ (500 × ΔT). If your actual cooler uses air cooling instead, or a different water ΔT, this number doesn't apply — use the fan-HP estimate for air-cooled units instead.

Required cooler capacity (BTU/hr or kW) is how much heat the exchanger core needs to remove; fan HP is a rough estimate of the electrical power needed to drive air across an air-cooled exchanger of that capacity, using a rule of thumb of about 0.1 HP per 10,000 BTU/hr. They answer different questions — one sizes the heat exchanger core, the other sizes the motor driving its fan — and the fan figure is only relevant if you're specifying an air-cooled rather than water-cooled unit.