Hydraulic Motor Speed Control

Hydraulic motor and hydraulic motor speed control: Answers to 4 FAQs

Hydraulic motors are essential devices that are powered by hydraulic fluids. They convert the energy from pressurized fluid (generated by a hydraulic pump) into rotational energy (in the form of torque and angular displacement of a shaft).

However, given that hydraulic motors are among the most widely used motors in a broad range of industries today, many would assume that engineers don’t face challenges when specifying hydraulic motors for a particular application. But that assumption would be wrong. For instance, engineers are usually plagued with finding an optimal hydraulic motor size and achieving accurate motor speed control for a particular application.

This article answers four frequently asked questions about hydraulic motors and speed control methods. It will also cover some design calculations and considerations necessary for specifying hydraulics motors.

There are several frequently asked questions about hydraulic motor speed control
Source: [PhotoBetulo/Adobe Stock]

Question 1: What are the steps to size hydraulic motors?

The first step to take when specifying hydraulic motors is to know the application requirements. For instance, consider an application that calls for 4 hp at 2500 rpm with an available supply pressure of 2900 psi and return line pressure of 200 psi. The theoretical torque required to meet this application requirement can be calculated using:

Where:
    Power is measured in horsepower
    N = motor speed (in revolutions per minute)

Next, calculate the motor displacement using:

Where:
    T = torque (lb.in)
    D = motor displacement (in3/rev)
    ΔP = pressure differential between supply line and return line (psi)
    em = mechanical efficiency

Motor displacement describes the volume of hydraulic fluid required to turn the hydraulic motor shaft through one revolution. Suppose the mechanical efficiency is 80% in this case. By substituting the values of pressure, torque, and mechanical efficiency into the equation, the motor displacement is obtained as 0.293 in3/rev.

The final step is to calculate the hydraulic fluid flow required to achieve this motor displacement, and in turn, power and rotational speed requirement. The required flow can be estimated using:

Where:
    Q = Hydraulic fluid flow (gallons per minute or gpm)
    N = speed (rpm)
    ev = volumetric efficiency

So assume a volumetric efficiency of 90% is to be achieved, then the required flow will be obtained to be 2.87 gpm, as shown below:

Question 2: What are the methods of hydraulic motor speed control?

There are generally two methods of achieving speed control with hydraulic pumps. The first method involves using a variable displacement pump that controls the hydraulic fluid flow into the motor. In contrast, the second method involves the use of a servo valve (or control valve) powered by a constant pressure source to drive the motor.

Engineers should opt for the first method (also called hydrostatic transmission) if there is a need to reduce operating costs since it is energy efficient. While the second method is not as energy efficient as the hydrostatic transmission method, it offers engineers better responsiveness and accuracy of speed control.

Question 3 How to design a hydraulic motor control valve system for speed control?

Like every other component design process, the design process of a valve speed control system of hydraulic motors starts with determining the maximum application requirements. For example, engineers must know the maximum load torque (T) and the required speed at that torque (N).

Once this requirement has been determined, engineers can size the valve such that the supply pressure from the constant pressure source equals the sum of differential pressure losses for the valve lands and hydraulic motor. This can be obtained using the valve control of motor motion equation:

Valve control of motor motion equation (VCMM)

Where:
    D = motor displacement (in3/rev)
    N = motor speed (rpm)
    em = mechanical efficiency of the motor
    K = total valve coefficient for series combination of powered and return lands
    T = Load torque (lb-in)


Using this equation, engineers can obtain the motor displacement. However, it is recommended that engineers design for optimality by replacing the load torque with the stall torque in the VCMM equation. Stall torque is the maximum torque that can be applied to the shaft, causing it to stop rotating.

As a rule, the stall torque should be about one and half times the running torque. Also, approximations can be made for the motor efficiency to be around 70 to 90%.

[Learn more about hydraulic motors with Engineering360]

Hydraulic motor speed control: Engineers should reach out to manufacturers
While this article presents helpful tips about hydraulic motor sizing and speed control, there are several other things that engineers must consider when specifying hydraulic motors for an application. For instance, engineers need to choose between the different types (or designs) of hydraulic motors.

Therefore, engineers are advised to reach out to hydraulic motor manufacturers to discuss their application requirements.