Sizing a hydraulic pump and motor starts with your load requirements: how much force (or torque) you need, how fast things need to move, and how long the system runs at a time. From there, a handful of straightforward formulas connect pressure, flow, displacement, and power into component specs you can actually shop for. The process is methodical, and getting it right means your system runs efficiently without oversized components wasting energy or undersized ones burning out early.
Start With Your Load Requirements
Before touching any catalog, define what your system needs to do. For a hydraulic cylinder application, that means the force required and the speed the cylinder must extend or retract. For a hydraulic motor application, it means the torque at the output shaft and the rotational speed (RPM). These are your non-negotiable starting points.
System pressure is determined by your load. A cylinder’s required pressure equals the force divided by the piston area. For a motor, the pressure is derived from the torque demand. Most industrial hydraulic systems operate between 1,500 and 5,000 psi, though mobile equipment can push higher. Once you know the pressure, you can begin selecting components rated for that range.
Flow rate determines speed. The volume of oil delivered per minute controls how fast a cylinder extends or how fast a motor shaft spins. So you need both numbers, pressure and flow, locked down before sizing anything.
Sizing the Hydraulic Motor
A hydraulic motor converts fluid flow and pressure into rotational force. To size one, you need to know the torque your application demands and the shaft speed required. Two core equations tie everything together:
- Torque: T = (P × D) / (2π), where T is torque, P is system pressure, and D is the motor’s displacement per revolution.
- Speed: N = Q / D, where N is shaft speed in RPM, Q is the flow rate delivered to the motor, and D is displacement.
In practice, you typically work these formulas in reverse. You know the torque and speed you need, so you solve for the required displacement. Rearranging the torque formula: D = (2π × T) / P. That gives you the minimum displacement per revolution to hit your torque target at a given pressure.
These are theoretical values. Real motors lose some energy to internal friction and leakage. Multiply theoretical torque by the motor’s mechanical efficiency (typically 0.85 to 0.95 for gear and piston motors) to get actual output torque. If you need 500 ft-lbs at the shaft and the motor runs at 90% mechanical efficiency, you need a theoretical torque capacity of about 556 ft-lbs. Similarly, volumetric efficiency (usually 0.90 to 0.97) means you’ll need slightly more flow than the theoretical calculation suggests, because some fluid slips past internal clearances without doing useful work.
Sizing the Hydraulic Pump
The pump needs to deliver enough flow at enough pressure to satisfy every actuator in the system. Its displacement and speed determine flow output:
Q = D × N, where Q is flow in gallons per minute (or liters per minute), D is the pump’s displacement per revolution, and N is the pump shaft speed in RPM. If you need 20 GPM and your electric motor or engine drives the pump at 1,800 RPM, the required pump displacement is 20 / 1,800 = 0.0111 gallons per revolution, or about 1.54 cubic inches per revolution.
Again, volumetric efficiency matters. Pumps don’t deliver 100% of their theoretical displacement due to internal leakage. A gear pump might run at 85 to 90% volumetric efficiency, while a piston pump can reach 95% or higher. Divide your required flow by the pump’s volumetric efficiency to find the actual displacement you need to specify.
For systems with multiple actuators running simultaneously, add their flow demands together. If actuators run sequentially, the pump only needs to supply the highest individual flow requirement, though you’ll need directional valves to route flow appropriately.
Calculating the Prime Mover Power
The pump doesn’t create energy. It needs an electric motor or engine to drive it. The standard formula for hydraulic horsepower is:
HP = (Q × P) / (1,714 × EM)
Where Q is flow in GPM, P is pressure in PSI, and EM is the pump’s mechanical efficiency. For a system running at 15 GPM and 3,000 PSI with a pump mechanical efficiency of 0.90, you’d need: (15 × 3,000) / (1,714 × 0.90) = about 29.2 horsepower.
This works perfectly when pressure and flow are constant. Many real systems aren’t that simple. A press might run at high pressure for 10 seconds, low pressure for 30 seconds, then sit idle for a minute. For these variable-duty applications, you calculate root mean square (RMS) horsepower, which accounts for the thermal load on the electric motor across the full duty cycle. The RMS method squares the horsepower at each stage, multiplies by the time spent at that stage, sums everything up, divides by total cycle time (with idle time adjusted by a cooling factor), and takes the square root. The cooling factor is 3 for open drip-proof motors and 2 for totally enclosed fan-cooled motors, reflecting how well each design dissipates heat during off periods.
Two things must be true for a proper match: the RMS horsepower must be equal to or less than the electric motor’s rated horsepower, and the motor’s pull-up torque (its lowest torque during acceleration) must exceed the maximum torque the pump demands. That peak torque occurs at the highest system pressure, since a fixed-displacement pump draws more torque as pressure rises. Calculate it with: T = (D × P) / (12 × 6.28 × EM), where D is pump displacement in cubic inches and T comes out in ft-lbs.
Continuous vs. Intermittent Duty
Pump ratings are often published for intermittent duty, meaning the pump isn’t expected to run at full pressure and speed nonstop. If your application runs continuously, you need to derate. A common starting point is reducing operating speed by 15 to 30% below the intermittent rating. The exact derating depends on which component is most vulnerable to sustained load: the bearings, seals, or fluid-end components. Manufacturers can usually provide continuous-duty data sheets if you ask, but if you’re working from intermittent specs alone, that 15 to 30% speed reduction is a reasonable safety margin.
For the electric motor or engine, continuous duty is simpler. Electric motors have well-defined service factors (typically 1.0 to 1.15 for standard motors). A 1.15 service factor means the motor can handle 15% above its rated horsepower continuously without overheating. Match this to your RMS horsepower calculation and you’re covered.
How Fluid Viscosity Affects Your Sizing
Hydraulic fluid viscosity plays a bigger role in pump and motor performance than many people expect. Fluid that’s too thin reduces volumetric efficiency because more oil slips past internal clearances instead of doing useful work. This internal leakage generates heat, which thins the fluid further, creating a feedback loop that can overheat the system. Fluid that’s too thick causes poor mechanical efficiency, hard starts in cold weather, and cavitation damage as the pump struggles to pull thick oil from the reservoir.
The sweet spot for most hydraulic components falls between 13 and 860 centistokes (cSt), a measure of fluid thickness. Within that range, there’s an optimum zone where both volumetric and mechanical efficiency peak. The practical way to find it: determine the lowest temperature at startup and the highest fluid temperature during operation, then use a temperature operating window (TOW) chart to select the right ISO viscosity grade. For example, a machine shop with startups around 45°F and peak system temperatures around 150°F would perform well with an ISO VG 68 fluid, which tends to keep the system operating on the more efficient side of the viscosity curve.
Outdoor and mobile equipment that sees wide temperature swings generally requires multigrade hydraulic fluids to maintain acceptable viscosity across the full range. If you’re sizing a system for extreme environments, factor this in early, because a pump that’s perfectly sized for warm-weather viscosity may be undersized for the volumetric losses that occur when cold fluid thins out at operating temperature, or it may cavitate on cold morning startups.
Accounting for Heat Generation
Every percentage point of inefficiency in your pump and motor becomes heat in the hydraulic fluid. A system running at 85% overall efficiency converts 15% of input power into heat. That heat has to go somewhere, or fluid temperatures climb until seals degrade, viscosity drops, and components wear prematurely.
Two formulas help estimate heat load. For heat generated by system inefficiency: BTU/hr = Q × 210 × ΔT, where Q is flow rate in GPM and ΔT is the temperature rise in degrees Fahrenheit. For heat from pressure drops across valves or relief valves where flow isn’t doing useful work: BTU/hr = Q × 1.485 × pressure drop in PSI. That second formula is especially relevant if your system dumps flow over a relief valve during idle portions of the cycle, because all that energy becomes heat.
These numbers tell you whether your reservoir can handle the heat through natural convection (a common rule of thumb is 1 BTU/hr per square foot of reservoir surface per degree Fahrenheit of temperature difference with ambient air) or whether you need a heat exchanger. Undersizing the cooling system doesn’t change your pump and motor specs, but it will shorten their lifespan and degrade performance as fluid temperatures creep up.
Putting the Sizing Process Together
The sequence, start to finish, looks like this:
- Define force/torque and speed at every actuator in the system.
- Calculate pressure and flow from those load requirements, including simultaneous demands.
- Size the motor by solving for displacement using your torque and pressure targets, then adjusting for mechanical and volumetric efficiency.
- Size the pump by solving for displacement using your total flow requirement and drive speed, then adjusting for volumetric efficiency.
- Calculate prime mover power using the hydraulic horsepower formula for constant loads, or the RMS method for variable duty cycles.
- Select hydraulic fluid based on your operating temperature range to keep viscosity in the efficient zone.
- Estimate heat load to determine whether you need auxiliary cooling.
At each step, efficiency losses compound. A pump at 90% volumetric efficiency feeding a motor at 90% mechanical efficiency means your overall system efficiency is closer to 81% before accounting for line losses and valve drops. Build those losses into your calculations from the start rather than discovering them after installation.