convert shaft-torque to heatflow?

Well, yeah...

Again, I have no idea what the original topic is-this thread was crossposted to RCM where I found it. It looked like he's trying to just heat up some oil from here...

Torque and Energy are dimensionally the same. Convention doesn't change that fact. Blow me.

hvacrmedic

Well hell, as long as you brought it up, why not turn this useless thread into something interesting. Question: What situation can you come up with in which allowing radians as a dimensional quantity will lead to mathematical contradiction. After studying dimensional analysis, and formal logic, I put the two together and concluded that there really was no logical justification in prescribing dimensionless factors, period. Now, let's allow for the moment the radian as a dimension, immediately the anomaly that led to your out of line reply to me would no longer exist. What are your thoughts on this. BTW, I wasn't attempting to be superior, it was just a bit of dry humor, there's too little of that on usenet, so no wonder it's almost always misinterpreted. It wasn't directed toward the OP, since he most likely didn't even understand what I said.

hvacrmedic

OK . You've impressed us all with your academic brilliance and absolute adherence to the accuracy of the language but it hasn't added much to the solution of the problem.

Tom

LOL! I'm glad that you could make enough sense of the post to extract a problem to solve ;) What exactly was the question? :)

hvacrmedic

I don't know what you mean by "heat flow"; but if you just want to turn torque (and RPM I presume) into heat, you might just use the shaft to turn a generator, and run the electrical output through a big resistor. That's the way dynamic braking systems work on the servos and spindle motors in many machine tools. Once you have a nice hot resistor, then you can collect, store, or transfer the heat any way your heart desires.

This method might be more "elegant" just because it doesn't need tanks of oil, or plumbing, hydraulics, etc. It might also be more efficient, because the heat all happens in one specific place (the resistor), instead of all over the volume of a tank. And, depending on how you collect/store/transfer the heat, it doesn't need to be lost in the plumbing.

With any method you choose, keep in mind that as you increase the load on the shaft, you'll also increase the heat generated by the source of your power. The real, total heat created by the system will be the sum of heat dissipated at the source, and at the "target". If the source-end heat is a substantial percentage of the total, then you might want to think about collecting heat on that end, too.

KG

-- I'm sick of spam. The 2 in my address doesn't belong there.

r.c.m added back on, because those guys actually make real metal things. You're always welcome to edit the target groups of your own Original Posts. Respectfully request that you don't edit mine?

i don't think those things were designed to shed 1272.5 BTU/hour on a continuous basis. I think they are optimized for converting bolii of shaftpower, into heat for radiation into the ambient air, intermittently. Pretty much the same difference between a golf-cart battery, and a engine-starting battery.

Anyway, I'm not looking to "shed" heat, I'm looking to "pump" it, so I can bake real things.

I think the pro's use super-heated steam, but I think my entire mooted power capacity is smaller than their calculation error-budgets....

Maybe your project requires rapid heat-up. If mine did, I would have said so. Mine wants to minimize capital expense & maintenance expenses, and puts a premium on KISS.

great answer. Too bad it doesn't answer the question. Dr Pine, you're a professor... if a student turned in such a answer in one of your exams, how many points would you give him?

Sure. You didn't provide enough details for specifics but I'll offer these generalities up.

Eddy current. Spin a magnet in close vicinity of a conductive metal like copper or aluminum. It will be heated by the high induced eddy currents. If you're heating a vessel then if a well can be provided on the outer surface, the rotor from a permanent magnet stepper or brushless DC motor could be spun inside the well. Regulate the heat by varying the insertion depth.

Hydraulic pump. Drive the pump from your shaft. Supply a closed loop of hydraulic oil to the pump. Provide an orifice suitable to build enough back pressure to properly load the shaft. Pressure ~~ torque input for a positive displacement pump. The oil leaving the orifice is hot. Circulate this hot oil through your process.

Permanent magnet generator and resistive heater. A small PM stepper, BLDC or DC servo motor would do. Select the immersion heater to properly load the generator.

Shear type absorber. This absorber was invented by Lockheed in the

50s to load jet engines and is the basis of most automotive engine dynamometers such as the SuperFlow brand. This absorber consists of a series of tightly fitting discs, alternately fixed and driven. The discs have holes on the same radii but in an even/odd combination so that the working fluid is accelerated by the rotating disc, slowed by the stationary disc and sheared at the interface. The working fluid is usually oil or water, although high pressure gas will work. A dense, high MW gas like SF6 would work best. Fluorocarbon refrigerants should also work well, as would air.

The load is controlled over a very wide range by varying the level of the working fluid in the absorber. In this instance, the heat would be transferred to the process via the heated fluid.

This type of absorber is very compact for the power absorbed. One capable of absorbing 400 watts might be as small as a 35mm film can. One capable of absorbing 10,000 HP is small enough for a man to hold in his arms. A rule of thumb is that with 6" discs, figure 1000 hp per disc, stacking as many as needed.

There is an SAE paper documenting the design. If this catches your attention, I can probably dig up the number. I have the paper around here somewhere.

A non-obvious version of this could be a gearbox spun at sufficient speed that the internal losses amount to the power you need. Circulate the lube oil to the process.

Heat pump. Drive a small compressor with the shaft power and set up a refrigeration cycle. This would have the advantage of a COP >1. Arrange the condenser in the process to be heated.

John

Thanks :-)

It answered your stated need:

I'm afraid not.

If I gave an exam with this question, I might want to see where the

63000 came from. That's more than your need to know your shaft rpm.

If a 1' radius Tau ft-lb arm moves 2Pirpm/60 feet around a circle in 1 second, the power is 2PipmTau/60 ft-lb/s, and 1 HP is 550, so rpm = 12x60x550xHP/(2xPixTau) = 63025HP/Tau, with Tau in inch-pounds.

Happy now?

Nick

Now you're floating my boat. Might more than make up for the conversion-inefficiency of going through a conversion to electricity....

The wind driving the windmill gets hotter? Then why do they call it "chill factor"?

The river driving the micro-hydro, gets hotter?

Generator and electrical heater?

Not enough information given to reach a conclusion. But ... if you assume a given RPM on a brake shaft, say 1000 RPM, and 6.812 pounds of cooling water either agitated at 100% efficiency or used as a perfect cooling medium on a brake:

When the brake is doing 1/2 HP of work, torque would be 2.626 ft. pounds. There is 0.001285 BTU per ft/lb. or 2043.58 BTU from 2.626 ft./lb. A BTU is the amount of heat required to raise 1 lb. of water 1 degree F. Therefore (2043.58 / 6.812 = 300) deg F temperature rise in 6.812 lbs. of water (or of course, 6.812 F temp rise in 300 lbs. of water).

Bob Swinney

Torque is NOT energy. The units "newton-meter" or "ft-lb" is a simplification of the real vector formula for torque: r x f (that's radius CROSS PRODUCT force, not radius TIME force). Radius "times" force is an oversimplification assuming that the radius and force are perpendicular.

If you want to be dimensionally accurate with torque, call it "ft-cross-lbs" or "newton-cross-meters".

Dave

No, not so you'd notice. I was thinking about an electric motor that draws more amperage to run a load, or an engine that burns more fuel. Or a set of bicyle pedals pushed by someone who burns more calories.

I suppose there MUST be some increased pressure/friction/heat where wind or water come into contact with fan or turbine blades. There IS more work being done if the load is bigger. But you'd be hard pressed to measure or collect that kind of heat, I'm sure.

KG

The fundamental difference between rxf and f*d, even if we assume the convention of perpendicular force in the former case such that r*f (which is more consistent with our convention that the force is all ready defined to act along the line of motion in f*d), is that r and d are fundamentally different dimensions. d is displacement, which is of course relative to the FoR, while r is invariant (barring lorentz/fitzgerald contraction of course). There is a fundamental difference between length and displacement, and thus in reality these should be expressed with differing dimensions. IOW, you were sort of frame jumping between conventions in order to make your case. If we then adopt delta_s (or d) as a different dimensional quality than r, then we would be forced to set radians as a dimensional quality in order to complete the set of relationships between d and r in a rotating system. That's more or less the gist of my initial speculation :)

hvacrmedic

Bingo. That's just it. With work, force acts ALONG the line of motion. With torque, it acts perpendicular. Back to HS physics: if a block of ice is pushed along a flat plate of ice, how much work is gravity doing on the block? Answer: none, because the force is perpendicular to the direction of the gravity component. Same with torque.

See the following quote: "However, torque is not energy."

The point is that F and r are in perpendicular DIRECTIONS.

Why are you bringing frame jumping into this? There's no frame jumping needed to make my case. R CROSS F still doesn't equal R TIMES F. R TIMES F is a simplification, and the vectors lay in perpendicular directions.

Dave

If you have a force vector and motion along some line other than this, then you can either use vector math or you can do it longhand, e.g. diagonal of the rhomboid and replace the vector with the scalar of d*f. Thus there is nothing absolutely requiring that we define force as acting along the line of motion, it's just a convention, or IOW, another method of solution. In the former case the force doesn't act along the line, only a component of it does. How is the operator thus distinct from the cross product in your r x f ? You are changing conventions from one equation to the other and thus comparing apples and oranges. Your argument about the difference between torque and energy is invalid. If the entire problem were worked out in vector format then the only difference, dimensionally, between the two terms would be the inclusion of radians in the torque expression to make them equal, i.e. to convert them into each other in a rotational system, as pointed out by Don Gilmore. More specifically some angular velocity reference must be supplied. It isn't the cross product that separates these terms from each other, it is the non-equivalence of the dimensionality of r and d. The radian should not be dimensionless.

hvacrmedic