LESSON 25. Hydraulic motors


Hydraulic motors  are the devices which  receives pressure energy and converts it to mechanical output in the form of  motion or rotation. These are  rotary actuators. A rotary actuator produces torque and rotating motion  is called  a hydraulic motor or simply  motor. Hydraulic actuators  are the muscles of many machines. They are produced in varying sizes from miniature units to massive actuators. A rotary actuator is a device that  produces a rotary  motion or torque. The motion developed by an actuator may be either continuous rotation or an angular rotation..It basically transforms the hydraulic energy into rotary energy in the form of mechanical rotations which can be applied to an object with the help of a shaft.

Hydraulic Motors

A hydraulic motor is a device which  converts the flow of a liquid under pressure into rotational motion. It is similar in construction to a hydraulic pump. Instead of mechanical parts are moved and  fluid is pushed ,  high-pressure fluid pushes the mechanical parts, causing them to move. The rotary motors are provided in the hydraulic systems to convert the fluid power into shaft power by forcing the shaft to rotate .When fluid pushed is at faster speed , the shaft speed is more and vise versa.  Basically  there are three types  of hydraulic motors, like- gear, vane and piston.

Gear Motor

The gear motor is  consist of two meshed gears in a housing, with inlet and outlet pipes on opposite sides. Both gears are driven gears, but only one is connected to the output shaft. Its working  is reverse of a gear pump. Fluid at high pressure is injected into the housing, where it flows , between the  gear teeth and the housing wall. Similar to the gear pump gears in gear motor are closely fitted into the housing. This forces  the gears  to rotate. The meshing teeth of gears prevent the fluid from flowing back, and results in continuous rotations. This rotary motion is then available in the form of rotary work at the output shaft. The direction of motion can be reversed by changing the direction of fluid through the motor.

L 25 fig.25.1 

Vane Type motor

This type of motor is a positive displacement motor. Vane motors is having  a single rotor with vanes sticking at the  housing wall of motor. The inlet and outlet pipes are  located on opposite sides of the motor. The fluid flow from the pump enters the inlet of the motor and  forces the rotor and vanes to rotate, and passes out through the outlet pipe. As motor rotate it  causes the output shaft to rotate . As the vane is near  to inlet pipe, it get rotated due to fluid and a  tight seal is created .The seal is  loosening  and letting the fluid  flow from the outlet pipe of the housing of the motor. Since the fluid  is always pushing against the vanes tangentially, the rotor is rotated by the fluid pressure.

L 25 fig.25.2

Piston Motors

This type of motor is a positive displacement motor which can develop torque at shaft with the help of pressure of the fluid. It may be of following three types

  • In line piston

  • Axial piston

  • Radial piston

Axial  piston

This type of motor consists of a cylinder fitted with a number of pistons. As the fluid is forced into the cylinder barrel with pressure, the pistons forced outward against an angled plate. By pushing on the plate at an angle, the pistons rotate the plate and generate torque. Y In this manner fluid energy is converted into rotary output .

Radial piston

Radial piston motors are best suited for low speed high torque applications, like in winches. Radial piston motors are designed in in varying capacities ranging from 20 to 10,000 cm3/rev at higher pressures like 450x105N/m2 or so.  In radial piston motor the pistons are connected to a shaft with the help of crank . The pump flow into the cylinder forces the piston to reciprocate and rotate the shaft. In this way a high power and torque is produced.

L 25 fig.25.3 Points to be considered while choosing a hydraulic motor

  • Torque needed

  • Working pressure of the hydraulic system

  • Speed required

  • Size of fly wheel required

  • Heat generated if any

Calculations in Motor

Motor Power

It is the output power of motor shaft which can be defined as the product of shaft torque and speed .

                                      Ps     =       \[\frac{{2\pi N}}{{60}}\]

where                       N      =         speed (rpm)

                                     \[\tau \]       =          torque

Fluid Power Calculations

It can be defined as the energy per second by the fluid in the form of pressure and the quantity of fluid. It is product of fluid quantity (flow rate)  and the differential pressure. Hence

                                       Pf         =          Q x Δp

Where                       Pf            =          Fluid power

                                       Q          =          flow rate , m3/s

                                       Dp        =          differential pressure, N/m2

Overall Efficiency

It can be defined as the ratio of output power to the input power or shaft power to the fluid power, which can be given as-

\[\eta \]ova        =     L 25 EQ.1

\[\eta \]ova        =     \[\frac{{{\text{Shaftpower}}}}{{{\text{Fluidpower}}}}x100\]

\[\eta \]ova        =     \[\frac{{{\text{Ps}}}}{{{\text{Pf}}}}x100\]


\[\eta \]ova        =     overall efficiency

Pf            =          Fluid power

Ps           =          Shaft power

Speed and Flow Rate            

The flow rate and speed can be given as-

Flow rate                 Q         =          Kq x speed

Where                     Kq       =          Nominal displacement of motor, cm3/rev           

Torque and Pressure

Let motor be 100% efficient, then the shaft power will be equals to fluid power.


                                      Shaft power    =          Fluid power

                                                Ps                    =          Pf

                                                \[\frac{{2\pi N}}{{60}}\] =       Q x Dp

\[\tau\]    =          \[\frac{{{\text{Qxpx}}60}}{{2\pi N}}\]

Volumetric Efficiency

It can be defined as the ratio of theoretical flow rate to the actual flow rate of fluid.


                                    Qa       =          actual flow rate

                                    Qt        =          theoretical flow rate

                                    \[\eta \]vol        =          volumetric efficiency

                                    \[\eta \]vol        =    \[\frac{{{\text{Qt}}}}{{{\text{Qa}}}}\]       x 100


A hydraulic motor is running at a speed of 395 rpm with the differential pressure of 8.2x106 N/m2. The nominal displacement is 5 cm3/rev and the overall efficiency and volumetric efficiency of 86 and 89% respectively. Calculate the followings-

a) Theoretical flow rate

b) Actual flow rate

c) Fluid power

d) Shaft power

e) Torque of shaft


Given              N         =          395 rpm

                                  Δp        =          8.2x106 N/m2

                                  \[\eta \]ova        =              86%

                                  \[\eta \]v           =              89%

Kq        =          5cm3/rev

=          5x10-6m3/rev

Theoretical flow rate      Qt            =          KqxN

                                                                =          5x10-6x395/60

                                                                =          3.29x10-5 m3/s

                                                    \[\eta \]vol        =   \[\frac{{{\text{Qt}}}}{{{\text{Qa}}}}\]        x 100

          \[{\text{Qa}}\]        =   \[\frac{{{\text{Qt}}}}{{\eta {\text{vol}}}}\]        x100

                                                       =    \[\frac{{3.29{\mathbf{x}}10 - 5}}{{89}}\]      x100

                                                       =    3.69x10-5 m3/s

Fluid power                    Pf        =          Q x Δp

                                                            =          3.69x10-5 x 8.2x106   

                                                            =             303.27 Nm/s

                                                            =             303.27 watt

            Shaft Power                 \[\eta \]ova        =       \[\frac{{{\text{Ps}}}}{{{\text{Pf}}}}x100\]  

                                               Ps                   =       \[86x303.27{\text{}}/{\text{}}100\]                

                                                                     =        260.8 watt

           Shaft Torque                               

                                                  Ps        =     \[\frac{{2\pi N}}{{60}}\]    

                                                  \[\tau \]    =  \[\frac{{Psx60}}{{2\pi N}}\]

     =  \[\frac{{260.8x60}}{{2x\pi x395}}\] 

                                                           =     6.3 Nm

Last modified: Tuesday, 18 March 2014, 7:23 AM