Someone just had to ask.

[Alaric]

Originally Posted by ndzied1

I think it was either a chemical or nuclear reaction when I traaced it back. Further than that I'm at a loss...

My plan to solve the world's energy problems with hydraulics.


If that doesn't work, we'll try beer.

[ndzied1]

Funny you should say that. UPS is poised to start testing a way to save energy with hydraulic hybrid trucks:


http://blog.wired.com/cars/2008/10/ups-hydraulic-h.html

Sometimes the truth is stranger that friction.

[fluidpower1]

Originally Posted by ndzied1

Funny you should say that. UPS is poised to start testing a way to save energy with hydraulic hybrid trucks:


http://blog.wired.com/cars/2008/10/ups-hydraulic-h.html

Sometimes the truth is stranger that friction.

However he left out the Hydraulic Battery (Acumulator) that the UPS trucks are using.

In 1966 I rode on a Mercedes Tour Bus in Germany that was using the UPS system to save fuel. Worked well in the mountains and reduced brake shoe replacement intervals.

[fluidpower1]

Alaric quipped a short time back:

"I'll say it again:the pump doesn't make anything, it transfers power."

I would like to see you after you jumped in front of a running 50 GPM pump Outlet port.

I call that profuse amount of Liquid, FLOW.

It would also might be Powerful enough to knock you down if you were too close to the Outlet port. Your resistance would probably be overcome by that profuse amount of FLOW.

[Tom Jenkins]

Originally Posted by ndzied1

You just proved the opposite of what I think you intended. You closed the valve all the way so there is no flow! The pump is making pressure until one of your afformentioned situations (tube break, motor o.l. etc) occures.

Wrong. The flow from the pump continues, which is what causes the tubing to swell like a balloon until it bursts. The flow through the valve, however, is zero. The pump is creating flow. The flow has to go someplace, which is why the pressure rises until something blows when you completely block the discharge.

I don't know why you guys are fixated on F = ma. This applies to acceleration, like a drag racer or the acceleration of gravity. It has nothing to do with our problem. It just means that if the tubing and valve aren't fastened tightly enough to the pump body to resist the force trying to pull them apart they will accelerate and fly off the pump. It has nothing to do with pump flow or orifice (control valve) pressure drop. Ohms law (V = I x R) is also correct, but it has nothing to do with our hydraulic problem either. Laws of physics don't have universal applicability to the solution of all problems.

And I think it is fairly obvious that an electric motor or a gas engine or a windmill or dog powered treadmill is required as an input power source to the pump. It doesn't matter for the purposes of the hydraulic system performance as long as the dog or whatever can keep the pump input shaft turning. You are right - the pump is an energy transfer device, but that isn't significant to the "pumps make flow" discussion. Semanticly it would be more correct to state "The pump takes energy from rotational motion of a motor and transfers it to movement of a fluid from the inlet port through the pump to the discharge port" but really, who cares about semantics in this case.

And yes, Alaric, some pressure and losses occur inside the pump - remember I asked you to ignore them as not significant. However, at the point the fluid leaves the system and starts flying through the air the static pressure (the kind measured by a pressure guage) is zero. Bernoulli's law sticklers will argue correctly that the velocity pressure is not zero, but really, in a hydraulic system, who cares?

[Peter Nachtwey]

Originally Posted by Tom Jenkins

Peter and Alaric, I am a goin' to stick to my guns on this one. Ya ain't lookin at the whole picture in an exactly keerect way, fellers.

First, do an experiment, or imagine one if you don't have a pump and reservoir handy. Put a pressure guage on the pump at the discharge flange. Run the pump with a free discharge - no tubing or valves or anything else. (Except the bits needed to attach the pressure guage, of course.)

Result, lots of flow, but zero pressure. Power output of the motor will be close to zero - just enough to overcome the internal friction losses in the pump. For the future in this discussion lets assume that this is negligible and ignore these friction effects. Let's also agree for the purposes of discussion to ignore internal leakage in the pump. These aren't important to the concept under discussion.

OK, but the pump is still transfering energy to the oil in the form of kinentic energy. At the piston/oil interface there is still a pressure increase but this energy is converted to kinetic energy when there is no restriction.

[quote]

Now add a valve to the discharge of the pump. Close the valve to say 90% open. What happens is a function of the kind of pump you are using. Let's keep it simple and say that our pump is a gear pump running at constant rpm. This will serve to illustrate the concept I'm after.

With a gear pump or other fixed displacement pump there will be measurable pressure at the discharge of the pump, and flow will be unchanged. The pressure will be proportional to the square of the velocity through the valve and the Cv of the valve at its current position. This is based on the formula Q=Kv*sqrt(ΔP) in Peter's first post. Motor output power will be equal to gpm x psi / 1714.

Close the valve further. The pressure will go up, the power will go up, the flow from our ideal pump will be unchanged.

The pump is producing flow, the system resistance to flow is producing more pressure.

NO NO NO!!! This is looking at it backwards. The pump produces the pressure/energy to move the oil through the resistance or to push the pistons. If the pump reaches its pressure limit then I don't care how much more resistance to flow you add there will be no more pressure. Resistance does not produce energy.

Something else produces energy like a diesel engine or electric generator on a dam, coal, or nuke plant etc.

This is sort of like telling an electrician that if he adds more resistance to the output of a 24 volt supply he will some how get more voltage. No electrician is going to buy that. Unfortunately too many mechanical and hydraulic people do.

I don't see how the rest of your post applies. What part of the VCCM equation do you disagree with? The VCCM equation assumes that the piston will accelerate until the net force is 0. How can you disagree with that?

[Tom Jenkins]

Yes, the pump is producing kinetic energy. That is why the Bernouli's law guys are right in saying that the velocity head (velocity presure) is not equal to zero. But a pressure guage measures static pressure, and that is zero if there is no resistance to flow. In most hydraulic systems the static pressure is much much greater than velocity pressure, which is why we can neglect it in most hydraulic systems analysis.

There wasn't a piston in my system, but it doesn't matter. If you are accelerating a load with a constant volume pump the initial pressure will equal the sum of dead load plus force required to accelerate the load divided by the area of the cylinder. The load determines the pressure, not the pump. The ability of system components to withstand the pressure determines the max pressure, unless a relief valve opens. Then the relief valve establishes the max presure limit at the pump. As long as the loads don't produce a pressure above the relief valve setting the speed of cylinder travel is determined by the pump flow rate. Flow makes it go. The pump produces flow, the system resistance to that flow establishes the pressure needed to move that flow.

This is like a 4-20 mA transmitter. The voltage produced by the transmitter rises until it transmits the current through the loop. The voltage at the transmitter is determined by the load according to Ohm's law. The limiting factor is the voltage available in the system. That's why a transmitter can handle more total resistance in a 24 VDC powered loop than it can in a 12 VDC powered loop. Just like a pump can handle more "load" in a system with a 1,000 psi relief setting than it can in a system with a 500 psi relief setting.

Peter you said "This is sort of like telling an electrician that if he adds more resistance to the output of a 24 volt supply he will some how get more voltage. No electrician is going to buy that. Unfortunately too many mechanical and hydraulic people do."

This analogy doesn't mimic a constant flow pump. This is analagous to a constant pressure variable flow system like an accumulator. If you put a cylinder on this system the flow rate and cylinder speed will increase until the total pressure from the load PLUS the friction loss through valves and fittings exactly matches the prssure at the accumulator. If you increase the valve resistance, the total pressure at the accumulator won't change, but the flow rate will drop because a higher percentage of that total pressure is used up as pressure drop at the valve.

This is similar to a circuit with a constant voltage source like a 24 VDC power supply. If you have low resistance the current rises until it matches V/R. If you have a low resistance (like a short circuit) the current rises until the power supply maxes out or a fuse blows, or the smoke gets out of something.

And of course energy is required to rotate the pump. The rate of energy required (power) increases in direct proportion to the pressure in the system, but for our ideal gear pump the flow is still the same.

And you are absolutely correct: the piston will accelerate and the flow rate increase until the net force is zero. And the force available to accelerate the load (call it F1) is the pressure at the accumulator minus the pressure across the valve minus the static pressure required to lift the load. The acceleration is = F1/m. That is in no way contradictory to anything I said.

If you open a perfect valve between my ideal constant flow rate gear pump and your cylinder, the pressure available to accelerate the load is equal to the relief setting. As the load accelerates and its velocity increases the flow rate over the relief valve drops and the flow rate to the cylinder increases. When all of the flow is going to the cylinder and none over the relief, the max velocity of the cylinder matches the pump flow rate and the pressure at the cylinder equals the load divided by the area.

The pressure drop over valves and fittings is propotional to the flow rate squared. This presure is added to the load pressure, and the pressure at the pump equals the sum of the two. The pump flow makes it go, and the cylinder's load and system losses determines the pressure.

[fluidpower1]

Tom Jenkins wrote:
"If you open a perfect valve between my ideal constant flow rate gear pump and your cylinder the pressure available to accelerate the load is equal to the relief setting. As the load accelerates and its velocity increases the flow rate over the relief valve drops and the flow rate to the cylinder increases. When all of the flow is going to the cylinder and none over the relief the max velocity of the cylinder matches the pump flow rate and the pressure at the cylinder equals the load divided by the area. The pressure drop over valves and fittings is propotional to the flow rate squared. This presure is added to the load pressure, and the pressure at the pump equals the sum of the two. The pump flow makes it go, and the cylinder's load and system losses determines the pressure."

I could'nt have said it better Tom. However, I don't believe it will convince Peter. It's just too simple for him. Heck, everybody can understand your writeup!!!

[ndzied1]

How about this for simple:


Can you have pressure without flow? Yes

Can you have flow without pressure? No

Q.E.D.

"What make the Flow Go?"

[m_turk]

Originally Posted by ndzied1

How about this for simple:


Can you have pressure without flow? Yes

Can you have flow without pressure? No

Q.E.D.

"What make the Flow Go?"

I 'm still undecided on this one.. many good arguments..

Does a free river flow have pressure?

[fluidpower1]

Originally Posted by ndzied1

How about this for simple:


Can you have pressure without flow? Yes

Only if there is flow potential. At the bottom of a Water Tower with 100' of water column a pressure gauge will read 43.3 PSI even when there is no flow. However the potential for flow is there as can be seen anytime a valve at the bottom of the 100' tank is opened to atmosphere is opened.

Or a Pressure Compensated pump feeding a Dead End Circuit. A Flow Meter at the Punp Outlet reads Zero but the Potential for flow is equal to the Pump's volume capability.

Can you have flow without pressure? No

Q.E.D.

"What make the Flow Go?"

Pressure diffference.

Flow is always from a Higher Pressure source to a Lower Pressure state. No Pessure Difference, No Flow.

At least that is the way I understand it.

[fluidpower1]

This Fluid Power person certainly thinks "Flow Makes it Go" is a correct statement. He even put it in writing in the first paragraph of this writeup.

http://www.noria.com/learning_center...up=Lubrication

[Tom Jenkins]

Originally Posted by ndzied1

How about this for simple:

Can you have pressure without flow? Yes
Can you have flow without pressure? No

Q.E.D.
"What make the Flow Go?"

Very good, Norm, and correct as far as it goes, but really beside the point. My discussion was directed to the factors that determine pressure at the discharge of a pump and what determines velocity of a system.

It is absolutely correct that the gears of a pump exert some miniscule force and therefore some pressure in moving the fluid from the pump inlet to the pump discharge. This is part of the frictional losses that I said I was going to ignore for purposes of discussion. Again, I'm talking about what happens at the discharge.

The pressure exerted to move the pump from inlet to discharge is used up in overcoming friction. When the fluid gets to the discharge of the pump that pressure is gone - turned to heat in overcoming internal friction. For free discharge on a pump the static pressure at the discharge point is zero.

If there is a load or restriction on the pump discharge, that is what determines the pressure at that point. The pump tries to move a fixed volume of flow for every rotation, regardless of this pressure.

And yes, a free flowing river has a pressure difference, created by the static pressure caused by a difference in elevation from one point to the next. Without an elevation and pressure difference you have what is commonly called a lake.

And you can't have "flow" without pressure difference, just like you can't have current without voltage difference.

And before anybody goes off on another tangent, please note that with my free discharge pump there is flow until the point of discharge. Once the fluid is moving through the air it is simply a flying object with kinetic energy just like a thrown stone. There isn't any pressure, just movement, just like a thrown stone.

[m_turk]

All that would mean that in nature, Peter's point of view is correct and the pressure difference is the cause and flow is the result, but when a machine, a pump is the case, it creates directly flow (potential), and that is the cause , while the pressure (difference) is the result?

Kinda weird...

[ndzied1]

Originally Posted by Tom Jenkins

And before anybody goes off on another tangent, please note that with my free discharge pump there is flow until the point of discharge. Once the fluid is moving through the air it is simply a flying object with kinetic energy just like a thrown stone. There isn't any pressure, just movement, just like a thrown stone.

Maybe this is where the disconnect lies. True that oil sprayed into the air is like throwing a stone. I impart some force on the stone to accelerate it. It flies through the air with a certain kinetic energy. Some time later it hits something and turns its kinetic energy into something else (depending on what it hits).

But that's not how a pump, lines and a cylinder work. The gear tooth in your pump doesn't smack a slug of oil and then wait for that slug of oil to get to a cylinder to make it move. Imagine a closed system like this:


You don't have to wait for the oil molecules on the right side to get to the left side before the piston on the left goes up. They don't have to flow there to make the left piston move up. They push on the molecules next to them and so on and so on... This pushing on each other is pressure. It's almost instantaneous. If we had to wait for the oil to flow there it would take time. It would be like hitting the cue ball and waiting for it to go across the table to drop the 3-ball in the corner pocket.

And again, nothing moves until I press on the cylinder on the right with a force. That's why I keep coming back to f=m*a.