Monday, December 29, 2008

Change in Load Question

The system consists of a fixed RPM electric motor, fixed displacement pump, system relief, cylinder, directional valve, and reservoir. If the cylinder is extending at a constant speed and the load opposing cylinder travel increases what happens to the speed of the cylinder?

16 comments:

FluidPowerWorks said...

Speed will reduce by whatever flow reduction is caused due to pump efficiency and any circuit Bypass the higher pressure required to move the increased load.

Hardly noticeable on short stroke resilient sealed cylinders using a Vane Pump.

A Spool type Directional Control Valve will bypass more oil to Tank would be another Bypass path.

If the pump was a Gear on Gear design the speed reduction would be more.

All the above is based on the increased pressure not going at or higher than the relief valve setting on a Fixed Volume Pump or withinn 100-300 PSI of the Compebsator set pressure.

Spring Loaded Pressure Compensated Vane Pumps would see reduced flow sooner than Pilot Operated Compensators found on most Piston and some Vane pumps.

It's in the Book.

pnachtwey said...

I agree if the increased load does not exceed the pump's power capability The pump does slow down a bit because of the extra load causes more slippage. The increased difference in pressure across the pump will increase leakage.

A fixed displacement pumps has almost a vertical curve until it is power limited. Normally the pump is flow is RPM limited.

Adding a load moves the system curve up but because the pump curve is almost vertical the flow doesn't change much.

If moving the load requires more power than the pump can generated it will overheat and fault.

JDK said...

My HYDRAULIC ANSWER will stick with the "givens" in the post.

The fixed displacement pump is driven by a fixed RPM motor. Fixed means relatively fixed.

As FluidPowerWorks points out the flow generated by the pump can go out to the system or back to the suction port through internal leakage paths in the pump. There are also leakage paths in the valve if it is a spool valve. There could also be leakage paths in the cylinder depending on the type and condition of the seals. The system relief, depending on the type, could also have leakage paths.

As FluidPowerWorks points out as long as the presure created by the system at the pump and system relief is below the cracking pressure of the relief, an increase in load will slow the cylinder down proportional to the increased leakage(a relatively small decrease). Once the relief cracks open the velocity of the cylinder will go from the velocity at that point (I'll call this cracking point velocity) down to zero as the load increases. How much of an increase in load it takes to reduce the cylinder velocity from cracking point velocity down to zero depends on the pressure gain vs. increase in flow across the system relief.

The pressure at the system relief is created by the sum of the resistance to flow to and from the work and the pressure required to do the work. In this simple system, if the load continues to increase, the relief will eventually pass all of the flow(exclude leakage flow) and the cylinder will stop. Any increase in load past this point will fully retract the cylinder.

FluidPowerWorks said...

JDK wrote:
"Any increase in load past this point will fully retract the cylinder."

Only if the load is capable of moving, as in a vertical Rod Up condition and the Load is weight added to the vertical cylinder.

If the Load was Horizontal and the added weigt caused enough Friction Drag to cause a pressure increase greater than the actuators maxmum force, all pump flow would pass through the Relief Valve and the Load would remain in its last position.

JDK said...

"Only if the load is capable of moving" Thanks for the correction; I agree.

"If the Load was Horizontal and the added weigt caused enough Friction Drag to cause a pressure increase greater than the actuators maxmum force, all pump flow would pass through the Relief Valve and the Load would remain in its last position." I agree with one qualification: The only way to acheive a "pressure" increase after stopping is for the cylinder to start retracting. The flow across the relief would increase to pump + return flow increasing th pressure. Did you mean "Force" increase or am I missing something?

FluidPowerWorks said...

Jdk countered with:
"I agree with one qualification: The only way to acheive a "pressure" increase after stopping is for the cylinder to start retracting. The flow across the relief would increase to pump + return flow increasing th pressure. Did you mean "Force" increase or am I missing something?"

Evidently I was not fully clear.

Scenario:
10 Sq.In. Area Cylinder, Relief Valve set at 1,000 PSI.

Horizontal Load requires 8,000# of Force/Thrust to move it. Cylinder Extends.

Next, Place another load requiring an added 2,200# of Force/Thrust on top of the first Load so the Force required increases to 10,200#. At that point the cylinder would stall and all pump flow would retirn to tnk through the Relief Valve at 1,000 PSI.

However, the same Cylinder Mounted Vertically, Rod Up, would lift the the 8,000# Load easily. However if a second 2,200# Load was placed on the first 8,000# Load the Force required would increase to 10,200#. At that time the Cylinder would start Retracting and Pump Flow plus the Return Flow from the Cylinder, being forced back by the added weight, would both be Bypassing through the Relief Valve sat at 1,000 PSI.

If the Releief Valve was a Direct Acting type pressure would increase slightly due to the added flow. Probably less than 25 PSI due to the small amount of flow from the small amoount of Force over the Cylinders capable thrust of 10,000#.

If the Releief Valve was a Pilot Operated type pressure would increase much less until flow exceded its Rated Flow and thrrelieving Orifice size starts adding Flow Resistance like a Flow Control valve.

You can see the difference in Direct Acting and Pilot Operated Relief Valves in Ch. 9 of the basic book. It explains the operating diferences between Direct Acting and Pilot Operated Relief valves.

At least that is how I see the situation.

JDK said...

FluidPowerWorks:

I agree with everything you said in your last comment.

In the prior comment you said ""If the Load was Horizontal and the added weigt caused enough Friction Drag to cause a pressure increase greater than the actuators maxmum force, all pump flow would pass through the Relief Valve and the Load would remain in its last position."

In my comment I was asking if "pressure increase" in the above statement should have been more correctly called "force increase"? Pressure is in the oil; Force is from the load.

pnachtwey said...

Those that believe "flow makes it go" should read this.

Anonymous said...

Here is the situation, assuming the relief valve is set infinitely higher, the electric motor/prime mover is very big that it won't overload, the pump is fixed displacement/no internal leakage, the directional valve has no leakage and the cylinder has no leakage. The assumption is that all the hydraulic components are 100% and the cylinder has very very long stroke. The lines are fixed and external friction disregarded.

What is going to happen if at constant speed and load the cylinder encounter an additional load. I can say it will slow down a little then the speed will be restored back in a certain time. Why?

Note that if the relief valve is adjusted slightly higher than required to run the system at constant speed and at initial load, the speed will slow down until the relief valve is readjusted to the new total load(initial+additional).

JDK said...

I agree. If you read all of the comments, especially from Fluid Power Works, there is agreement there too.

Anonymous said...

The reason why the speed of the actuator decreases when there is an increase in load(from initial constant velocity) is due to pressure drop. If you consider all the conductors as orifice, there will always have a pressure drop. That is from one point to another.

This can be illustrated by using a flow control valve. In most hydraulic systems, there is a metering of flow. At a constant load and speed of an actuator using a flow control valve(say meter-in), the load pressure is equal to the flow control valve input pressure minus the pressure drop across the valve. If we increase the load without changing the valve input pressure, the load pressure will increase then the pressure drop decreases(Pd=Pin-Pl).

Low valve pressure drop means low flow across and vise versa.

JDK said...

Thank you for your comment.

I agree partially with your comment. Please read all of the comments on this topic. The point of this discussion is that, given the posted situation, the cylinder speed will not significantly slow down until the pressure at the relief equals the cracking pressure of the relief valve. This pressure is made up of load induced pressure and as you pointed out pressure drops due to flow restrictions.

Anonymous said...

I read all the post and we just need to agree to disagree.

If I have my hydraulic system, I will always see to it that the relief valve setting is for maximum load. And if your example has an infinite relief valve setting, how did the speed slow down when the load is added?

Try this example:
A simple pump, relief, flow control(meter-in)and a hydraulic motor. If you are running at half load and the other port of hydraulic motor is connected to return at ATM pressure and assuming that the hyd motor and pump has no internal leakage, at steady flow across the flow control valve derived from Q=k* sqrt(Pin-Pl) will result. Now if you add load, a different value of Q will result. This example also assume that the relief setting is way way above the maximum load pressure.

JDK said...

"I read all the post and we just need to agree to disagree........
And if your example has an infinite relief valve setting, how did the speed slow down when the load is added?"

Not sure where you read about the infinite relief valve setting. The speed slow down is discussed at length:
1. Small slow down due to internal leakage paths
2. Larger slow down when relief valve cracks open

Anonymous said...

Assuming there is no internal leakage and the relief valve will not crack open at half load or below the maximum load, why will it slow down if the load is increased from half to below maximum load? Forget about acceleration/deceleration sequence, why would the velocity change after acceleration/deceleration with changing loads?

JDK said...

With no external leakage and PSI generated below the relief cracking point, the speed won't change. Who said it would and in which comment? Thank you.