Hydraulic Reservoir Size and Cooler Heat Rejection
How much heat a hydraulic system makes, how big its reservoir should be, and what is left for a cooler.
Example
You enter
- Pump flow (gpm) 20
- System pressure (psi) 2000
- Pump and drive efficiency (0-1) 0.85
- Fraction of input that becomes heat (0.25 typical) 0.25
- Reservoir multiplier (3 industrial, 1-2 mobile) 3
- Reservoir dissipation at design oil temp (BTU/hr) 4000
You get
- Hydraulic (hp) 23.3372
- Input (hp) 27.4556
- Heat (Btu/hr) 17468.6
- Reservoir volume 60 gal
- Required cooler duty 13469 BTU/hr
Details, formula, and sources
Everything the system does that is not useful work becomes heat in the oil: pressure drop across valves and lines, relief-valve flow, and pump and motor inefficiency. A quarter of input power is a common figure for a system with ordinary metering losses, and on a system that spends much of its cycle over relief it is far more. The reservoir does three jobs -- de-aerate, settle contamination, and shed heat -- and the classic industrial rule of three times the pump's per-minute flow is really a DWELL TIME rule: it gives the oil about three minutes in the tank to release entrained air before it goes around again. Mobile equipment cannot carry that much oil and runs one to two times instead, which is exactly why mobile systems need coolers and industrial power units often do not. A 20 gpm pump at 2,000 psi is 23.3 hydraulic horsepower and 27.5 at the input at 85% pump efficiency; a quarter of that to heat is 6.9 hp, which is 17,469 BTU/hr, and a 60 gal tank shedding 4,000 leaves 13,469 for a cooler -- a real heat exchanger and a real fan, not an afterthought. Note what the heat fraction does: a system designed so only 15% of input becomes heat needs barely half that rejection. Circuit design, not cooler selection, is where hydraulic heat is actually controlled. A sizing estimate; the component manufacturers' published efficiencies, the duty cycle, and a measured oil temperature govern.
hydraulic_hp = pump_gpm x pressure_psi / 1714; input_hp = hydraulic_hp / pump_efficiency; heat_hp = input_hp x heat_fraction; heat_btu_hr = heat_hp x 2545; reservoir_gal = pump_gpm x reservoir_multiplier; cooler_duty = max(0, heat_btu_hr - reservoir_dissipation).
Hydraulic power from gpm x psi / 1,714, the heat fraction of input power converted at 2,545 BTU/hr per horsepower, and the classic reservoir dwell-time rule of three times pump flow for industrial units (one to two for mobile), by name -- public fluid-power practice. The component manufacturers' published efficiencies, the duty cycle, and a measured oil temperature govern.
The power and heat conversions are public physics; the efficiencies, heat fraction, and tank dissipation are the designer's own values.
Estimate. AHJ and licensed professional govern.
Field names used by the API: pump_gpm, pressure_psi, pump_efficiency, heat_fraction, reservoir_multiplier, reservoir_dissipation_btu_hr, hydraulic_hp, input_hp, heat_btu_hr, reservoir_gal, cooler_duty_btu_hr
- Heat fraction about 25% of input power for a system with ordinary metering losses; far more on one that lives over relieffluid power practice
- Reservoir rule 3x pump flow industrial, 1 to 2x mobile -- a dwell-time rule for de-aeration, not a heat rulefluid power practice
- If nothing rejects it oil temperature climbs until viscosity drops far enough that leakage balances input -- stable and destructivefluid power practice