relays

On Sat, 22 Nov 2008 08:12:04 -0800, John Larkin wrote:

Leaving aside the fact that the above is not true (you divided joules by joules *per second* (watts), not joules by joules, and the word "average" was added later in an attempt to hide your original error) your math is wrong.

Dividing joules by joules results in energy gain, not power gain. Joules are a unit of energy, not power. To get power gain, you need to divide watts by watts. Watts are a unit of power, not energy. This is basic engineering math that you should have learned in high school.

Inventing a new term which you call "averaged power" with a definition that is exactly identical to the standard definition for energy and then using it to calculate another new term which you call "average power gain" with a definition that is exactly identical to the standard definition for energy gain does not change the fundamental fact that energy is not power, power is not energy, energy gain is not power gain, and power gain is not energy gain.

The correct answer consists of using the standard engineering unit for power to describe power gain and the standard engineering unit for energy to describe energy gain. Your math error consists of using the the standard engineering unit for energy to describe power gain instead of energy gain.

This basic math error, when applied to a system (a latching relay) that has a fixed power gain and an arbitrarily large energy gain, led you to make the basic error if saying that a latching relay has an arbitrarily large (actually, you claimed infinite, but that's another math error of yours) power gain. It does not.

The formula that you used to make your original error is: Wrong Units / Right Units = Wrong Answer.

The formula that you used to make the above, new error is: Wrong Units / Wrong Units = Wrong Answer.

The correct formula is: Right Units / Right Units = Right Answer.

This has been explained to you several times.

How else would you compute gain?

Ditto.

Show us some math. This one's really easy.

Yet it appears to be over JKK's head.

John

Hint: if you scale both the numerator and the denominator of a fraction by the same amount, the quotient doesn't change. High school math back at'ya.

What? Power is watts. Average power is watts. Energy is joules. I gave an example, the 7-watt night light, recently somewhere nearby.

and then using it to calculate another new term which

Over a fixed time interval, they are the same. So whatever way is the handiest works.

Your

Of course it does. Try a real case. With numbers.

Quit being a dork. Stop ranting nonsense, find a relay data sheet, and do some real math. I've already posted two or three actual, real-life examples.

John

Nothing is incumbent on me but breakfast (coffee and cherry cobbler.) If neither you nor JKK can read a datasheet and work some real numbers, well, nobody will be really shocked. Engineering is not a high-school debating team, much less a gaggle of little girls calling names. Only the numbers matter.

Some are in fact rated in VA. But many are rated for maximum working voltage, and maximum current, and there is a mathematical operation that will convert those into VA. See if you can discover that.

Well, I suppose that forbids you from even thinking in terms of VA. You are NOT ALLOWED TO MULTIPLY.

John

My God, you don't know how to multiply.

Do a little

Why would NEC rate a part for 400 mA continuous, when it can only handle 250?

The answer is obvious: your math is once again nonsense.

(0.4)^2 * 1.2 = 0.192, comfortably below 0.3 watts.

WHY can't you check your work? (You forgot to actually take the root.)

So, the rest is all wrong too...

More nonsense! The fet doesn't drop 1.2 volts with a 250 mA load. At

Some people really shouldn't do math in public.

NEC specs 400 mA, in multiple places on their datasheet. You have chosed to ignore their datasheet and re-define what is "legal" for their part. This disconnect didn't suggest that your math ought to be checked? Wouldn't a usenet dispute over mathematical accuracy also suggest checking?

Based on what I've seen so far, I think not.

John

Cite?

The cite must be a rating of the relay contacts, as you claimed above. Many relay datasheets list acceptable motor loads in HP, acceptable transformer loads in VA, acceptable incandescent lamp loads in watts, etc. This is a convenience for the electrician who is choosing a relay out of the catalog. The actual contact ratings are never in VA units.

There exists no such mathematical operation. It would require the contacts to be open and closed at the same time.

Maximum working voltage is a rating for the contact while it is open (it's worst when just slightly open, but open nonetheless), and is usually different for AC and DC because DC does not have a zero-crossing which self-extinguishes the arc. You cannot achieve maximum working voltage across a closed contact.

Maximum current is a rating for the contact while it is closed (it's worst right as it closes/opens and the contacts are wiping[1], but closed nonetheless) and is a measure of heating and thus is influenced by ambient temperature. You cannot achieve maximum current across a closed contact. There are several current ratings such as continuous load current, fault withstand current, breaking current, etc, but none of them are specified in units of VA.

That's right. You are not allowed to multiply the voltage that only occurs while the contact is open with the current that only occurs while the contact is closed.

Note [1]: The physics are a bit different for contactors with no wiping compared to relays with wiping.

The obvious conclusion follows that no relay contact can switch any power, since the E*I product is always zero. That sort of ends the discussion, no?

Contactors aren't relays?

Well, JF sure bit the bag on computing power gain for this one...

Wanna have a go?

John

False.

Best regards, Spehro Pefhany

Where does the above imply that contactors aren't relays?

Looks like I have to explain basic logic to you...

What I wrote above makes two claims:

From those claims, it is invalid logic to conclude that the logic of the two claims above lead to a claim that

Given your ongoing pattern of making silly mistakes and then digging in your heels rather than admitting that you were wrong, I fully expect you to reply that you did not come to a false conclusion in your logic, but you have not (so far) made any error concerning what a relay is and what a contactor is, so I am hoping that you are still educable on that topic. Perhaps this will help:

There are no good definitions in the standards that we use to define electrical components. Normally I would simply go to IEEE Std 100-1992 (Standard Dictionary of Electrical and Electronic Terms), but it has multiple definitions for Relay and an inadequate definition for Contactor.

The AIEE (which became the IEEE) had a slightly better definition:

Relay: A device by which contacts in one circuit are operated by a change in conditions in the same circuit or in one or more associated circuits.

Contactor: A device for repeatedly establishing and interrupting an electric power circuit.

The Struthers-Dunn Relay Engineering book from the 1940s gives these definitions:

Relay: An electrically controlled device that opens and closes electrical contacts to effect the operation of other devices in the same or another circuit.

Contactor: A magnetically-operated device, for repeatedly establishing and interrupting an electrical power circuit. It is usually applied to devices controlling power above 5kW, whereas the term 'relay' is ordinarily employed below 5kW. The terms are often used interchangeably.

In my experience, common usage is that a "Relay" is small and usually has contacts that wipe, while a "Contactor" (sometimes called a "Contactor Relay") is big, has a crossbar that bridges dual contacts without wiping, usually has a way to actuate it mechanically as well as a low current signal contact (often used to drive an indicator lamp), and at times contains a thermal overcurrent protection device on each contact. That being said, I certainly wouldn't correct someone who uses the two terms interchangeably. That's why I added the qualifiers in what I wrote above.

And, of course, from the context I am assuming electrical relays and not pneumatic or hydraulic relays and I am ignoring such things as motor-driven contactors or ballistic contactors...

Please tell me more. I am always eager to learn something new, and having someone correct me when I am wrong is a great way to learn, but the single word "False" is not helpful.

It is my understanding that the voltage rating of a contact assumes an open contact -- you cannot achieve maximum voltage across a closed contact. Am I wrong?

It is my understanding that the current rating of a contact assumes a closed contact -- you cannot achieve maximum current through an open contact. Am I wrong?

It is my understanding that VA (volt-amperes) is a measure of apparent power -- the vector sum of real power and reactive power, that when the contacts are closed the voltage across them is near-zero, and thus the VA is near zero. In like manner, when the contacts are open the current through them is zero, and thus the VA is zero. Am I wrong?

I would be very interested in seeing a data sheet that gives a VA rating for a contact. Is that measured with the contact open or closed?

Not trying to give you a hard time, just trying to understand your comment.

Gosh, are you an engineer or a lawyer?

John

I am an Engineer. You?

"We trained hard but it seemed that every time we were beginning to form into teams, we would be reorganized. I was to learn later in life that we tend to meet any new situation by reorganizing. And what a wonderful method it can be for creating the illusion of progress while producing inefficiency, confusion and demoralization." -Gaius Petronius Arbiter, 1st century CE

Counter-example to "The actual contact ratings are never in VA units":

See the "Contact Data" table and compare with the "Maximum Switching Capacity" curve. The operating voltage must not exceed 250VAC AND the carrying current must not exceed 30A AND the switched VA must not exceed 6600VA (resistive) or 5500VA (cos(theta) = 0.4 inductive).

That's why the curve is rounded off at the top corner- there really is a limit on the VA that should be switched-- like the SOA on a transistor.

Best regards, Spehro Pefhany

Worse and worse. And in the public record for, maybe, decades. Maybe forever.

John