Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 05/10/03 08:53 GMT Daylight Time
Message-id:

On 05 Oct 2003 02:31:09 GMT, a (Dave Baker) wrote:
Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 05/10/03 01:22 GMT Daylight Time
Message-id:

On 04 Oct 2003 02:29:49 GMT, a (Dave Baker) wrote:
Torque bears no relation to any capacity to do work.

Well, it has a relationship. Power is defined as the capacity to do work,
and torque times RPM is power. But power is the independent variable
here. Torque and RPM can be traded off in any fashion as long as their
product equals the input power.

Although the two
quantities are expressed in the same units of force and distance, torque
is
a
vector quantity and work is scalar one.

Hmmmm. Force is a vector. In the definition of mechanical work given in
Machinery's Handbook, the distance used to calculate work is the total
displacement *in the direction the force is applied*. I think that makes it

a vector too. If I recall my elementary physics correctly, it doesn't
matter
what path is taken to the final position when calculating energy states
(and work is alternatively defined as the difference between the beginning
and ending energy states of a closed system). That's consonant with the
summation of an arbitrary number of vectors, so the distance should
indeed be considered a vector quantity.

Now if distance were considered scalar, there'd be no question, a vector
times a scalar is a vector, and so work would be a vector quantity. But if
the distance is considered a vector too, then there are two possibilities.
The cross product of a vector and another vector is a vector. But the dot
product of two vectors is a scalar.

In this particular case of aligned vectors, the *magnitude* of the answer
is
the same, but one resultant is vector and the other scalar. So we are left
to
wonder why we would apply the dot product instead of the cross product in
calculating work. I'd be interested in hearing others' thoughts on that.

The laws of conservation of energy determine that work is always a scalar
quantity. If we simultaneously apply two forces of magnitude 10 units each
to
an object we clearly need to know the directions to obtain the resultant net
force. This can be as high as 20 units if the forces are in the same
direction,
as low as zero if they oppose or anything in between. However, If we do 10
units of work to the object by moving it in a particular direction and then
another 10 moving it back again the net work is always 20 units and never
zero.
The equation has to be balanced by the total expenditure of energy consumed
in
doing the work.

That *sounds* good, but I'm not sure it holds up to analysis. For example,
if I raise a 1 kilogram mass 1 meter, I do work, requiring energy to
accomplish.
But if I then let the mass return to its original position, I regain the
energy.
So we have a situation where there has been a zero net change in energy.
Doesn't that also require that negative work was done so that net work is
also zero? Otherwise, we seem to have a violation of the conservation laws
because we'd have net work with no net energy expenditure.

I was anticipating that reply and if it hadn't been so late last night I was
going to mention it in advance. The reason being is you had alluded to it in
the bit about energy states at the top of your earlier post. Potential energy
states and work musn't be confused. You don't do work to an object when it
falls against gravity, the earth does. Whether you arrange to have that energy
recovered, hydroelectricity etc, doesn't affect the amount of work that got
done during the process. Work isn't something that gets conserved, only energy
is.


Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
I'm not at all sure why women like men. We're argumentative, childish,
unsociable and extremely unappealing naked. I'm quite grateful they do though.