THE PHYSICS OF... POWER

[wwhitney]

FionaZuppa, here's an example like yours:

Say we have a voltage source of 100V with an internal resistance of 0.001 ohms, connected via transmission wires with a resistance of 0.2 ohms, to a load. Consider two different loads with (almost) the same active power:

a) A pure resistance of 9.8 ohms. The transmission plus load resistance is 10 ohms, so the current is 10 amps. The load power is I^2 R = 980 watts. Transmission losses are 20 watts, and the internal generator resistance dissipates 0.1 watts. The input power to our voltage source is 1000.1 watts. The electrical efficiency of the system is 980/1000.1 = 98.0%

b) A load with resistance 4.9 ohms and reactance 4.9 ohms, so the impedance is 4.9 sqrt(2) ohms = 6.93 ohms. The transmission plus load impedance is 7.07 ohms, so the current is 14.14 amps. The load power is I^2 R = 979.6 watts (and the load reactive "power" is 979.6 VAr). Transmission losses are now 40.0 watts, and the internal generator resistance dissipates 0.2 watts. The input power to our voltage source is 1019.8 watts. The electrical efficiency of the system is 96.1%.

So yes, when the load changes from (a) 980 watts + 0 VAr to (b) 979.6 watts + 979.6 VAr (apparent power 979.6 * sqrt(2) = 1385 VA), the input power goes up. However, it doesn't go up by a factor of 2, or a factor of sqrt(2). Rather the system losses basically double from 2% to 4%.

Cheers, Wayne

 

[Ingenieur]

P = R I^2

W = P/t rearranging P = W t

substituting
W t = R I^2
W = (R I^2)/t

Therefore:
a resistor with I flow will perform Work proportional to the power it dissapates (as heat q) over time

 

[GoldDigger]

P = R I^2

W = P/t rearranging P = W t

substituting
W t = R I^2
W = (R I^2)/t

Therefore:
a resistor with I flow will perform Work proportional to the power it dissapates (as heat q) over time

Try again please.

If P = Power, which it must to equal I²R, then W = Pt --> P = W/t, not P = Wt.

It looks like you may be confusing Power with Energy. Or else just some typos.

The English language phrase "power over time" as commonly used often implies power times elapsed time, not power divided by time.

No snarkiness, just fact.

 

[romex jockey]

Anyone brave enough to summarize the content of this thread???????:angel:

~RJ~

 

[jtinge]

~RJ~

Amen!

 

[Ingenieur]

Try again please.

If P = Power, which it must to equal I²R, then W = Pt --> P = W/t, not P = Wt.

It looks like you may be confusing Power with Energy. Or else just some typos.

The English language phrase "power over time" as commonly used often implies power times elapsed time, not power divided by time.

No snarkiness, just fact.

Snark is a fact in your case
your high opinion of yourself is not justified
I bet you have alot of friends :lol:

Just a typo
simmer down skippy, you've not cured cancer
this is an internet chat room

P = W/t = R I^2
W = R I^2 t
does not change the fact that work is performed

wrong
power over time = power/time
power times time = power x time

Every forum has one of your type
mouth bigger than brain
consider yourself smarter than the average bear
how's that working out for you? Rotflmao

 

[ggunn]

Snark is a fact in your case
your high opinion of yourself is not justified
I bet you have alot of friends :lol:

What is it that they used to say on Usenet? Oh, yeah: Pot. Kettle. Black.

 

[GoldDigger]

power over time = power/time
...
how's that working out for you? Rotflmao

So "the integral of power over time" always = integral((P/t)dt)?
Just mentioning that I have heard people who knew what they were talking about say "energy is power over time" when they meant power extended over a period of time" rather than power divided by time. Best to avoid that construction if you do not want to be ambiguous.
...
It's working out just fine for me, thank you. It seems that you have the problem.

A simple typo that extended so far through your post makes we want to double check everything you write.

 

[wwhitney]

P = W/t = R I^2
W = R I^2 t
does not change the fact that work is performed

Work is only performed by a resistor if you've expanded the definition of work to mean any change in energy (delta E). In which case, I'd rather just call it delta E.

Cheers, Wayne

 

[Carultch]

Work is only performed by a resistor if you've expanded the definition of work to mean any change in energy (delta E). In which case, I'd rather just call it delta E.

Cheers, Wayne

Work is performed ON a resistor by the voltage drop across it and the current through it. But the resistor doesn't keep this work in an ordered form of energy. It dissipates it as heat, and ceases to be available as work.

 

[FionaZuppa]

Work is performed ON a resistor by the voltage drop across it and the current through it. But the resistor doesn't keep this work in an ordered form of energy. It dissipates it as heat, and ceases to be available as work.

darn entropy. maybe with specialized heat nets we can stop the never ending increase of universe entropy :huh:

 

[Sahib]

a resistor with I flow will perform Work proportional to the power it dissapates (as heat q) over time

It is not an accurate statement because of thermodynamic second law: If you try to convert all heat emitted by the resistor to work (even by theoretical Carnot cycle), it is impossible.

 

[FionaZuppa]

here's one for Wayne and GoldDigger

R is a term related to pure resistance. amps through R makes nothing but heat.

ok, if we say its pure R then 100watts being converted into heat, but not any useful output work at all, its all just heat that we cannot use for any meaningful purpose. do you agree (i suspect yes).

if R changes electrical potential into useless heat then why do you say R is the real power when heat cannot do work?

now, an induction motor has a R component, so we can say the amps running through that R of the motor does nothing more than create heat, a useless conversion of electrical potential energy. do you agree (i suspect yes).

so an ideal induction motor would have no R value. do you agree (i suspect yes).

so what you have is a motor that has nothing but Q. do you agree (i suspect yes).

this ideal induction motor can still do work? do you agree (i suspect yes).

but you say Q cant do any work, so how can this ideal motor be doing work?

if the ideal induction motor has PF=0 and can produce enough torque to do 1kW worth of work, how many amps are needed? how much input work to gen is required?

 

[Besoeker]

its all just heat that we cannot use for any meaningful purpose. do you agree (i suspect yes).

I don't agree. Motor and panel anti-condensation have a meaningful purpose.

 

[Sahib]

here's one for Wayne and GoldDigger

R is a term related to pure resistance. amps through R makes nothing but heat.

ok, if we say its pure R then 100watts being converted into heat, but not any useful output work at all, its all just heat that we cannot use for any meaningful purpose. do you agree (i suspect yes).

if R changes electrical potential into useless heat then why do you say R is the real power when heat cannot do work?

now, an induction motor has a R component, so we can say the amps running through that R of the motor does nothing more than create heat, a useless conversion of electrical potential energy. do you agree (i suspect yes).

so an ideal induction motor would have no R value. do you agree (i suspect yes).

so what you have is a motor that has nothing but Q. do you agree (i suspect yes).

this ideal induction motor can still do work? do you agree (i suspect yes).

but you say Q cant do any work, so how can this ideal motor be doing work?

if the ideal induction motor has PF=0 and can produce enough torque to do 1kW worth of work, how many amps are needed? how much input work to gen is required?

An induction motor even ideal one with PF=0 is possible provided it has no load, operates in vacuum and no internal losses. But PF=1 is an impossibility, because the motor requires a magnetic field and so its associated reactive power to transfer power to load.

 

[wwhitney]

R is a term related to pure resistance. amps through R makes nothing but heat.

Nope, stop right there. A resistor has power factor 1 and converts the incoming power into heat. But not all loads with power factor 1 will convert the incoming power into heat. Some may do useful work instead.

Cheers, Wayne

 

[wwhitney]

Work is performed ON a resistor by the voltage drop across it and the current through it. But the resistor doesn't keep this work in an ordered form of energy. It dissipates it as heat, and ceases to be available as work.

OK, how do you craft a formal definition that makes that distinction?

I guess my question boils down to this: in the context of electrical engineering, how and why is it useful to distinguish some ΔE as work as some as not? Or what is the issue with just defining power as P as dE/dt?

Cheers, Wayne

 

[Sahib]

OK, how do you craft a formal definition that makes that distinction?

I guess my question boils down to this: in the context of electrical engineering, how and why is it useful to distinguish some ΔE as work as some as not? Or what is the issue with just defining power as P as dE/dt?

Cheers, Wayne

Not more than around 35% of total heat can be converted to work.

 

[Besoeker]

An induction motor even ideal one with PF=0 is possible provided it has no load, operates in vacuum and no internal losses. But PF=1 is an impossibility, because the motor requires a magnetic field and so its associated reactive power to transfer power to load.

Permanent magnet brushless DC motors don't.

 

[Sahib]

Permanent magnet brushless DC motors don't.

We are talking about A.C.