dynamic limitation in the application of hydraulic motor

http://www.industrial101.com/equipment/hydraulic-motors.aspx

The acceleration of a hydraulic motor is usually not discussed as the load on the motor has a greater effect on acceleration than the motor itself. If maximum acceleration is what you are looking for then oversizing the motor allows greater "punch" to be applied to the load; but this will require a larger pump (if not on an accumulator system) which requires a larger drive (ie: electric motor) and on and on.

What is it you are actually after? If it is a number for a brochure, then tests must done under all of the various conditions of load, pressure, flow etc.

I AM LOOKING FOR CALCULATING TOTAL REVOLUTIONS HYDRAULIC MOTOR WILL ROTATE BEFORE REACHING ITS NORMAL SPEED FROM REST . THIS WILL TAKE ME FOR CALCULATING ACCELERATING TORQUE FOR A PARTICULAR SPEED .

delta T = (120 * delta REVOLUTION ) / delta N

delta T = time in second reaching speed change over

SO THERE IS TWO UNKNOWN FOR ACCELERATING TORQUE CALCULATION FOR A PATRICULAR SPEED OF A PARTICULAR HYDRAULIC MOTOR , THAT IS delta T & delta REVOLUTION .

SO I AM LOOKING FOR FORMULA FOR QUICK CALCULATING MINIMUM ACCELERATING TIME delta T .

IT IS VITAL FOR DYNAMIC RESPONSE OF THE SYSTEM . IT COULD BE CALCULATED FROM ITS NATURAL FREQUENCY . BUT ???????????????????????????????????????

Some of this confuses me? What is the load (if no load, with close coupling of the directional valve and high enough pressure, a change in rotation can be done in less than 1 rev of the motor). What is the pressure, is the pressure a constant value (ie accumulator pressure) or is the pressure the resultant load pressure. With no load pressure is quite low then when change over occures pressure can spike quite high (if pressure spike exceeds system limiting pressure reversal will be delayed) During reversal what is the motor speed (some internal inertia will delay the reversal) the only other delay is the compressiblity of oil during reversal.

THE PRESSURE IS CONSTANT , FROM ACCUMULATOR . LOAD IS POSITIVE . SYSTEM PRESSURE IS 210 BAR .MOTOR DISPLACEMENT IS 82 CU. CM / REVOLUTION . MOTOR FINAL SPEED 12 RADIAN / SPEED , DRAIN PRESSURE 1 BAR . MOTOR IS ACTUATED BY PROPOTIONAL VALVE . MAXIMUM PRESSURE AT MOTOR INLET IS 138 BAR / MOTOR OUTLET PRESSURE IS 82 BAR DURING STARTING . TOTAL TORQUE AT STARTING IS 55 N.m , when calculated using above data .

FROM ABOVE I CAN CALCULATE ACCELERATING TORQUE BY SUBTRACTING LOAD TORQUE .my 2nd question - WITHOUT NOTING ABOVE READINGS , HOW I CAN CALCULATE ACCELERATING TORQUE AS I COULD NOT GET J VALUE FROM MANUFACTURER , T = J x alpha .

1st question - how to calculate minium accelerating time or maximum accelerating torque for a particular motor from manufacturers motor performance data .

2nd question - how to calculate accelerating torque for a particular motor at a given speed from manufacturers motor performance data .

OK lets talk Hydraulic Motors. Pressure on inlet - pressure on outlet (differential pressure) x torque/bar (from graphs) gives outlet torque of motor.

TORQUE is usaully give either as Inch-lbs per PSI or inch-lbs per 100 PSI or Ft-lbs per 100 PSI etc. (your numbers in metric are the same N.m per bar, N.m per 10 bar etc.?)

You indicate a starting pressure of 82 bar on the outlet? does this change... with speed or is this system dependant from other valves etc.

Why, if the system pressure is 210 bar do you not "see" this at the initial start-up. ie. stalled motor should see instant system pressure if proportional is opened enough. This depends on load (large inertia load will show high pressure until load goes to speed) (low inertia load-car wash brush- will accelerate to speed quickly and only "load" will be what is required to keep brushes spinning).

Please give speed in RPM ---12Radian /speed does not compute with me, sorry I'm not a scientest, just a hydraulic geek.

Your total applied torque, is this from the torque-pressure curve from the motor specs.

Because of the low inlet pressure (210 system verses 138 inlet and the high outlet (0 verses 82) it appears that the proportional valve is not opening suffiently to "drive" the motor to its maximum capacity. Total delta P is only 56 bar about 30% of system capability.

You indicate a starting pressure of 82 bar on the outlet? does this change... with speed or is this system dependant from other valves etc.

82 BAR & 138 BAR IS CHANGING . THESE READINGS ARE AT THE INITIAL START OF MOTION . AS THE HYDRAULIC MOTOR STABILZES ITS ROTATION , BOTH READINGS ARE DECREASED AND REMAINED AT REDUCED READINGS . WE ARE GETTING OUR DESIRED SPEED 12 RADIAN / SECOND = 110 RPM

Your total applied torque, is this from the torque-pressure curve from the motor specs.

IT IS CALCULATED FROM DATA .

NOW I CAN CALCULATE ACCELERATING TORQUE .

BUT HOW TO CALCULATE ACCELERATING TORQUE IN ADVANCE ?

Sorry about not replying sooner I did't see your last reply until recently.

I am curious as to why you need to know the acceleration torque "in advance".

Questions: what presures do your system stabilize at; inlet, outlet.

what load is applied to the motor

if the required rpm is achieved why the concern

Other things that come to mind; if you use data for calculations could you put a pressure tranducer in the line (maybe one in the outlet also for differential pressure) and aquire the pressure therfore the torque thereby calculating the acceleration numbers?

Not to forget these units are only effient to about 85% (volumetric and mechanical eff.) so calculations I do normally include a fudge factor which will also include normal system wear and continuos lower and lower effiency until the unit requires replacement!

Again the load on the motor is the limiting factor in accelerating a hydraulic motor. Without the load the acceleration is limited to the moment of the rotating mass, the friction of the internal parts (this is a lot) and also the compression of the oil both inlet and outlet.

Mbsahoo I'm a little confused by your original question in combination with additional posts.

Do you want to calculate the minimum angular acceleration time of just the hydraulic motor OR the minimum angular acceleration time of the hydraulic motor AND driven mechanism?

Regards Woody

BUT HOW TO CALCULATE ACCELERATING TORQUE IN ADVANCE ?

Can you not calculate the moment of inertia of your driven mechanism?

Then... angular acceleration obeys the law.

a = t/I where t = torque, I = moment of inertia

Regards Woody

WoodyGB is right

http://www.engineersedge.com/fluid_flow/displacement_torque_hydraulic_motor.htm

On another forum I suggested that mbsahoo use the inertia from a similar size motor as an estimate for the inertia. Now we can calculate the acceleration time which is

t=ω/α

and α=Τ/J

and Τ=ΔP*Displacment/2*π

This simplified to

t=(2*π*ω)/(ΔP*MD)

Where:

MD is the motor displacement

The tricky part is knowing the pressure drop across the motor at all times

At this point the problem turns into a system of differential equations. If you can estimate what the average ΔP is while accelerating you can get a ball park estimate on the acceleration time. I would start with the ball park estimate because solving the system of non-linear differential equations requires a bit of work. However, if you are going to use this design over and over again then you should do the most accurate simulations you can.

A more expensive but easier way to do this is to buy a motion controller and a feed back device. A quadrature encoder or a tachometer will provide velocity/acceleration feedback. A closed loop system will provide precise control over the velocity and acceleration within the capabilities of the hydraulic system. The accelerations can be programmed to be very smooth and take any time you wish. A hydraulic motion controller with auto tuning will allow for precise control.

So mbsahoo must make a difficult choice. He must make decide whether he is going to pursue by making accurate calculations or let a motion controller do the closed loop control.

The natural frequency only plays a part when calculating the how fast the motor can be ramped up in closed loop mode but a good controller can ramp up the motor much faster than indicated by the 35/ω equation provide there is enough torque.