BIZOMAT MAKEOS · Hydraulics
Hydraulic Motor Drive Sizing
What motor displacement, flow and pump power does this hydraulic drive need for the load torque and speed?
Sizing a hydraulic motor drive through an optional gear reduction: motor torque and speed, the displacement needed at the available pressure difference with mechanical efficiency, the flow with volumetric efficiency, hydraulic and pump input power, and, for a chosen motor size, the actual pressure needed (using the same relations as the hydraulic motor tool).
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Load torque | N·m | Torque needed at the load. |
| Load speed | rpm | Speed of the load. |
| Reduction between motor and load (1 = direct) | Gear reduction between the motor and the load. | |
| Gear efficiency | % | Efficiency of the gear reduction. |
| Pressure difference available | bar | Pressure difference across the motor available from the system (relief setting minus line losses and back pressure). |
| Motor mechanical efficiency | % | Mechanical (torque) efficiency of the motor. |
| Motor volumetric efficiency | % | Volumetric efficiency of the motor. |
| Pump overall efficiency | % | Overall efficiency of the pump. |
| Chosen motor displacement (0 = required) | cm³/rev | Displacement of a catalogue motor you are considering. |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Load torque
- What it is
- Torque the driven load needs at its own shaft at the working speed, including its friction.
- Units and format
- Enter a number in N·m: newton-metres. The form accepts 0.000001 N·m to 100000000 N·m. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 2000 N·m as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
What this means: Load speed
- What it is
- Rotational speed the driven load shaft must run at, on the load side of any gear reduction.
- Units and format
- Enter a number in rpm: revolutions per minute. The form accepts 0.000001 rpm to 100000 rpm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 40 rpm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
What this means: Reduction between motor and load (1 = direct)
- What it is
- Speed ratio of any gearbox or chain drive between the hydraulic motor and the load, motor speed ÷ load speed. 1 means the motor drives the load directly.
- Units and format
- Enter a plain number with no unit. The form accepts 1 to 500. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 4 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Motor speed rises and motor torque falls in proportion to the ratio.
What this means: Gear efficiency
- What it is
- Efficiency of the gear reduction between the motor and the load, from the gearbox maker.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 30% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 95% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- Use 100 % for a direct drive with no reduction.
What this means: Pressure difference available
- What it is
- Pressure difference across the motor ports that the system can supply: relief setting minus line losses and return back pressure.
- Units and format
- Enter a number in bar: bar (1 bar = 100 kPa). The form accepts 0.000001 bar to 700 bar. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 180 bar as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A higher pressure difference needs proportionally less displacement.
- Get it right
- A difference across the motor, not the pump pressure; back pressure in the return line reduces it.
What this means: Motor mechanical efficiency
- What it is
- Mechanical (torque) efficiency of the hydraulic motor at running speed, from the motor data sheet.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 30% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 92% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Lower efficiency needs a larger displacement or more pressure for the same torque.
What this means: Motor volumetric efficiency
- What it is
- Volumetric efficiency of the motor: the share of the supplied flow that turns the shaft rather than leaking past internally.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 30% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 95% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Flow needed = displacement × motor speed ÷ volumetric efficiency.
What this means: Pump overall efficiency
- What it is
- Overall efficiency of the pump supplying the motor, used to turn hydraulic power into pump input power.
- Units and format
- Enter a number in %: percent: the number is read as a percentage, so type 8 for 8 %. The form accepts 20% to 100%. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 85% as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
What this means: Chosen motor displacement (0 = required)
- What it is
- Displacement per revolution of a catalogue motor you are considering, used to check the pressure that motor would need.
- Units and format
- Enter a number in cm³/rev: cm³/rev. The form accepts 0 cm³/rev to 100000 cm³/rev. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0 cm³/rev as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- 0 = use the calculated required displacement instead of a catalogue motor.
Method and formulas
- Motor torque = T_L ÷ (i η_G); motor speed = n_L × i.
- Displacement needed V = 20π T ÷ (Δp η_m) cm³/rev (from T = V Δp η_m ÷ 20π).
- Flow Q = V n ÷ (1000 η_v) L/min; hydraulic power = Q Δp ÷ 600 kW; pump input = hydraulic ÷ η_pump.
- Example: 2000 N·m at 40 rpm through 4:1 needs 199.69 cm³/rev and 33.633 L/min.
Limitations
- Running efficiencies; starting torque efficiency is lower (check breakaway).
- Speed must lie within the motor’s rated range.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Size a hydraulic motor for a winch, wheel or conveyor drive
- Check whether a chosen motor has enough torque at the pressure available
- Size the pump for the drive
Who it is for
- Hydraulic engineers
- Machine designers
- Mobile equipment designers
What you get
- Motor torque and speed
- Displacement and flow
- Pressure needed and pump power
Interpreting the result
- A larger reduction allows a smaller, faster motor; check the motor’s minimum smooth speed for slow drives.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 3418–3425. Topic reference; no source tables or recommended settings are reproduced.
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