convert shaft-torque to heatflow?

I have the shaft torque... what I need is a half horsepower of heatflow at 300 F.
I'm having visions of paddles inside of an insulated tank full of gear
oil. Bet it would work just fine. But it smells a bit klugey.
Any more *elegant* way to do it?
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alanh snipped-for-privacy@yahoo.com wrote:

Torque is in units of energy, heat flow in units of energy/second. You can't convert them.
hvacrmedic
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"RP" wrote: Torque is in units of energy, heat flow in units of energy/second. You can't convert them. ^^^^^^^^^^^^^^^ If you're going to adopt a superior attitude, and criticize the OP, at least try to get it right. Torque is NOT in units of energy. Lb-ft and ft-lbs are not the same. I'll leave it to you to figure out what I am talking about.
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What about Newton Meters?? or Meter Newtons?? I think I will have another Fig newton :-)
On the serious side, the paddle wheel thing... Seems to me that when dealing with fluid dynamics, there is no heat unless there is friction.......but I could be wrong
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unless
As long as you are able to do work on the system, heat will be generated. It's not reversible though, That is, if he were to apply heat, the paddle wheel would not turn...
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Dear Rick:

*only* apply heat.
Because we apply heat to a turbine, allowing it to escape (along with the heated gas), and the turbine turns. You can apply heat to one "side" of this "dashpot", and cool the other, and convective currents in the fluid can be made to turn the shaft...
David A. Smith
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message

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...
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True. But "lb-ft-radians" *is* a unit of energy and the radians, being unitless, could have been neglected, as in most rotary calculations. In other words, turning a shaft with a torque of 1 lb-ft, through an angle of one radian, does 1 ft-lb of work.
But I know what you meant, Leo. I'm just being pedantic.
Don Kansas City
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Leo Lichtman wrote:

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

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
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RP wrote:

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
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dave.harper wrote:

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 <sic> 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
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RP wrote:

<SNIP>
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." http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.intro.html

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
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dave.harper wrote:

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
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RP wrote:

<SNIP>
Incorrect. Torque is a vector, not a scalar. Let me repeat: TORQUE IS NOT A SCALAR, IT IS A VECTOR. As you stated above, you are treating it as a scalar. Your oversimplification of F*d eliminates all vector components, and thus treats torque as a scalar (which is incorrect).
See the following quote: "Vectors are not ordinary numbers so normal multiplication has no direct analog" http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/VectorProducts.html
What you are trying to do is multiply two vectors (R and F) and a cross product is not the same as scalar multiplication.
Dave
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dave.harper wrote:

Lots of engineering quantities are vectors. Force is a vector. That doesn't change its units at all. A vector has magnitude and direction. The magnitude is indicated with units, like newtons or lbs. The direction is never indicated in the magnitude units; this would be absurd.
Vector math does not explain why torque appears to have similar units to energy. Your cross-product will maintain proper vector direction for the torque, if dealing with vector equations for force and radius, but it does not reconcile the concept of energy, which is what we are talking about.
As I explained in a previous post, the answer is simple. Angular energy is torque multiplied by angle traveled. Since angles are measured in radians, which technically have no units, the units for the angular work look deceivingly the same as the torque alone.
The scalar product does not take any angle of rotation into account...but then neither does your cross-product. It will only return a equation of torque which has magnitude and direction. When you multiply it by an angle of rotation, it *will* be energy and it is no longer a vector. Work is a scalar quantity because it doesn't matter in what direction the work takes place, you have still expended energy.
Don Kansas City
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eromlignod wrote:

<SNIP>
Vector math DOES expain the reason if you look carefully at the formula for torque and don't oversimplify things.
Question: How come torque, which is a VECTOR, has the units of a SCALAR? Answer: because it's difficult to express a cross-product in the units, so we simplify.
Example:
Volume=l * w * h, so the units are in * in * in (in^3 or equivalent). The units of each component (inches) are multiplied together to get a SCALAR.
Torque=r X F (that's r CROSS F). It's NOT using the same mathematical operation (multiplication) as for volume. Yet the units N-m imply that we DID use the same mathematical operation. So why do we want to relate units that imply we multiplied force and distance? As mentioned before, it's easier to say ft-lbs or N-m instead of worrying about how to express a cross-product in the units.
Dave
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dave.harper wrote:

The units *ARE* the scalar portion of a vector equation. When coupled with a real number they are literally the magnitude. Pounds are not "force"; pounds are a "magnitude" that describes force, irrespective of direction. A "force" has a directional component *added* to it. Pounds have no direction. For something to be treated as a vector, it must *be* a vector and have an innate directional component with i's, j's and k's. This is freshman physics.
All ordinary units are scalar. That's the whole point of having them--They are pure magnitudes. Which way is a ft/s? Which way is a newton?
Don Kansas City
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eromlignod wrote:

That's really what I'm getting at. The units are simplified down to a scalar quantity. If we wanted to be exactly right with the units, we'd have a dimensional formula beside all answers: force-dot-meter (aka Joule), newton-cross-meter, magnetism, etc. But that'd be a hassle and would scare away little kids from ever learning physics.
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Dear eromlignod:
...

Just as a sidelight... Note that voltage has units of work, but you don't actually get power until current flows, and you don't get work until you integrate that power. A similar argument to yours...
David A. Smith
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