Does Torque Drop?

We were talking about the same shaft. My theory was based on that if the pump requires X amount of torque to do its work, by putting the pulley there the torque at the pump will drop. And that theb engine will need to compensate in torque so that the pump can do its work at the new RPM. Do we increase the RPM at the engine to do this or this is Simeon they the engine controller does automatically?

The torque at the pump shaft will not change unless the mechanical load changes.

HP = (Torque x Speed)/63025 inlbs

(55x63025)/1000=Torque

If speed goes up, Torque goes down

Is the 55 HP engine electric or IC?

Your question brings to my mind another point here. The power output rating of most motors is their peak power output not a constant power output. The relationship accurately cited by others here of power, torque and angular velocity is the power transferred from the motor to the mechanical load.

I had a choice of asking two questions here. I chose that one.

The other one was to be, what color is the engine?

Then we could always discuss whether it is really an engine at all, or rather a motor.

Ultimately we might even have gotten into a discussion about how much power is really needed to achieve the desired pump output.

You've got to let go of the assumption you've got 55hp come what may. It may well be capable of producing 55hp at 1000rpm. If there's nothing attached to the shaft it will produce 0hp at 1000rpm.

Now, if you introduce some speed-changing mechanism between the engine and the load there will be a loss in that mechanism before some work is done by the output shaft of that mechanism. The actual work done by that shaft is the product of the angular velocity and the torque presented by the load, in this case the hydraulic pump. It too has losses, so you'd better expect something rather less than 55hp on the fluid output!

So your colleague is mistaken.

All clear now?

All clear! Thank you very much!

Hi

Your torque drops but don't worry about it. Use Power required in Kws = Pressure (bars) x Flow (lpm) over 600 then factor this by the inefficieny of the pump ie 85% for a piston unit say 80% for a gear unit. That is increase the power requirement. You might like to add a little more for the losses of the reduction unit depending on what you use / how you do it. It is worth remembering that with IC engines pull down if you try and raise the pressure beyond the power available. So either be generous on your allowance for inefficieny's or be confident that you know what they are.

Oliver Dunthorne - Hydraulic engineer

If it's an IC engine, that sort of size will be capable of running much faster than 1000rpm, which is not much above idle speed. Open the throttle to give required speed, you will have higher available power into the bargain.

If it's an electric motor and you need higher speed for the pump, provided the pump shaft power is less than 55HP (minus say 5% for losses) you will be OK. The torque at the pump is lower than at the engine, as others have said, but that's allowed for in the power calculation.

Developed torque is a function of engine design that can only be altered by changing components in or on the engine.

The torque demand on the engine will be increased because the pulley ratio increases.

There are two ways to identify if the engine will handle the increased loading;

1. Contact the engine manufacturer and discuss all of the parameters involved.

2. Physically do the alteration and observe the engine and pump performance.

I change the torque developed by my ICE all the time by adjusting the throttle position. If I don't I can quickly lose control on the highway.

Quite, that's what I said in #12

Here's a twist, normally a hydraulic analogy is used for electricity but I'm going to flip that. Think of a transformer, it's a fixed power device, MVA in equals MVA out. When you raise the output voltage the output current goes down, when you lower the voltage the output current goes up. That's because the voltage times the current always equals the power.

Same thing with any type of speed changing mechanism, the shaft power is the same but as you raise the speed the torque goes down and as you lower the speed the torque goes up (all this ignores the prime mover/load design, control system, speed torque curve, etc.). Otherwise you'd have the magical ability to say raise the speed while holding the torque constant with the resulting HP going up. Ain't gonna happen any more than having the voltage and the current simultaneously go up on one side of our theoretical transformer with no change in the input power.

As an aside I'm a bit curious to know if this has anything to do with the previous 55HP RPJ engine thread. Way too many physical laws violated there too.