convert shaft-torque to heatflow?

N:dlzc D:aol T:com (dlzc) wrote:


It is similar, but only when corrected :) V=E/Q But note that Torque=E/theta
Even though you missed it a bit, thanx for the suggestion, because this example does serve very well to illustrate my point in much simpler fashion than my previous attempts.
Now, do any of you see the logical inconsistency between the forms of these equations?
hvacrmedic BTW David, you've known me all along as Richard Perry over there in sci.physics land. How goes it lately in the nuthouse? SOS no doubt :)
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Dear RP:

..
I suspected.

SSDD (same sh*t, different day).
I have a problem with both the coulomb and the mole as well. Dimensional or non-dimensioned... They seem a whole lot like "a dozen" to me.
But radians turn radius into area, then into volume. Let's see if any "bright sparks" have been lit...
David A. Smith
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-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1
wrote:

Let's not get out on a tangent, now.
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iQA/AwUBQtQaagIk7T39FC4ZEQJ6lgCgnRfG+XQkmOb+lEvkRr36UBrnR0cAn1ei ww2HHXkpH/y72gWUqikGHOd6 ox -----END PGP SIGNATURE-----
--
-john
wide-open at throttle dot info
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That's exactly what I wanted to suggest. Mechanical energy can easily be (directly) converted to thermal by friction. The paddle solution is what is used in brakes for engine test stands, just with the intent to put some load onto the motor.
Nick
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Nick Mller wrote:

Yes, but torque can't be converted to thermal energy. Nor is there such a thing as heat flow at 300F. That's gibberish. I assume he means heat flow from a fluid held at an equilibrium temperature of 300F. In this case, the Btuh generated at a given torque from a given motor will depend upon the angular velocity of the shaft. This will in turn correspond to a specific resistance to rotation. Thus for measurable temperature rise of the fluid in the shortest time period, it's best to use a small arm radius and higher angular velocity-- less fluid will be required to provide the same back pressure and thus the same torque. The sensible heat requirement will be lower with lower fluid mass. How he is going to provide 300F is another matter. He'll have to do some heavy calculations or by trial and error adjust the flow of heat removing media and/or insulation value of the tank.
hvacrmedic

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even if the termology is wrong (and it is), what he wanted was clear.
Nick
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Nick Mller wrote:

It isn't clear how he's going to do it given that he doesn't even know what the hell he's saying, so it's moot anyway :) OTOH, yes, I could gather that he wanted to heat some fluid by friction.
hvacrmedic
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If you only accept stricly scientific asked questions, you also might have used some inductive logic to restucture and resolve the question - by the one solution the OP has given - as a challenge.
Nick
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Nick Mller wrote:

What challenge? This effect dates way back to the very discovery of heat/energy equivalence. It is the very experiment that was used to demonstrate it. Neither was deciphering his message a challenge, the real challenge is on his part, to learn basic thermodynamics which is covered in most every seventh grade physical science class. Now if he's *in* seventh grade then he doesn't need to be spamming the sci. groups.
hvacrmedic
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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
wrote:

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Tom Miller wrote:

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

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On Sun, 10 Jul 2005 12:53:49 +0200, in sci.engr.mech snipped-for-privacy@gmx.de (Nick Mller) wrote:

It's so absolutely clear that in the three posts I so far see you've made, exactly how many answer the OP's question? ---------- Ed Ruf Lifetime AMA# 344007 ( snipped-for-privacy@EdwardG.Ruf.com) http://EdwardGRuf.com
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Exactly the first one, if you take the efford to read. My other exactly two were about RP's useless bashing. Exactly this posting is about your challenge in reading answers and remembering names over exactly 4 postings.
exact enough?
Anyhow, EOT Nick
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RP wrote:

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.
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Nick Mller wrote:

Some MODERN dynos are water pumps. But the original Prony brake was just a friction brake. For a few horsepower or less, a drum brake would work fine. Modern brakes, either disk or drum, can dissipate hundreds of hp for a short while.
While some modern dynos use water pumps, others use generators. A generator takes approximately as much power to turn it as the electrical power delivered. One can turn a generator with the shaft, run the heat into power resistors. Since power resistors can get pricy, light bulbs are a cheap substitute. A modern generator should be able to take a horsepower indefinitely. So load up the generator with a dozen headlight bulbs or so- that will generate oddles of heat. This has the advantage of moving the heat away from the generating location by a few feet of (very heavy) wire.
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You mean axial/radial pumps. Not piston pumps. They exist for a long time. At least before WW II. Schenck is one of the producers.

ACK
This is a good solution, if the OP is willing to buy a generator (scrap yard).
The elegant thing with the "paddle pump"*) is, that it is the most direct way. Also, they can be easily regulated by controlling the amount of liquid in the pump body.
*) we (Germany) call them water-whirl-brakes, I don't know your terminus technicus.
Nick
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Say Tau, in inch-pounds.

rpm = 63000HP/Tau. For instance, 0.5 HP and Tau = 105 in-lb makes 300 rpm.

Use a pump?
Nick
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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?
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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
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On 9 Jul 2005 22:07:08 -0700, alanh snipped-for-privacy@yahoo.com wrote:

[hobby group knocked off]
Hook up a disk brake from a compact car.
Brian Whatcott Altus, OK
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