Power question

[gbw]

From a neophyte, on behalf of all neophytes, this question:

For a 'normal' tractor if there is such a thing, does the engine have to be running at full power in order to have the hydraulic system at deliver full force and speed?

Asked another way, if the engine is rated full power at 2700rpm, and delivers half power at 1900rpm:

How does the hyraulic system behave at at 1900? Would the FEL deliver the same lifting force at the same speed (seems unlikely), or the same force but at a slower speed, or at reduced max force and at a slower speed?

Or, to keep things from being simple, does the answer depend on the relative matching of the hydraulic systems and engine power? (For example, a tiny hydraulic system mated to a monster engine would, it seems, deliver all it can at almost any engine setting.)

If this is the case then my questions don't make sense in general, but can only be asked for a specific make/model of tractor - in my case I'd be interested in a JD 4320 CUT. What is the minimum rpm needed to allow full use of the hydraulic power and speed available on this tractor? Does it matter if the tractor is moving or not (eHydro drive)?

 

[kennyd]

Here is my short answer:

Pressure=Force or "power"
Flow=Speed

The pumps used in our CUT's are constant pressure, but variable flow. So theoretically, you have the same lifting power at low RPM vs. high RPM but at a reduced speed. But as the load increases on the pump like from lifting a heavy load, it demands more power input from the engine so you need to raise the RPM's to compensate...

 

[J_J]

Another answer. The pump requires a certain HP to run efficiently. If the pump is rated to flow 10 GPM at 2500 rpm, then to achieve the 10 GPM flow, the engine will have to run at 2500 rpm. At any speed lower that that, the pump is putting out less, and that is translated to cylinder speed, motor speed, etc. Pressure provides the power, only if the valves are activated, and resistance is realized.

The max work force is only generated when there is opposition to the flow of fluid. The cylinders will only provide that resistance to the flow when lifting heavy loads, or the relief pressure has been met.

The tractor HP is calculated by the amount of work that is required. The hydraulic system will only require a small portion of that HP. There may be several hydraulic pumps in use.

If the tractor is just idling, you are using very little HP, and if you try and lift max load on the loader, the engine may bog down and shut off. You have to at least meet the HP requirements of the pump to lift your max load.