THE PHYSICS OF... POWER

[mike_kilroy]

here's my examples, dig in.

lets look at two "realistic" cases of AC power gen deliver energy to two loads that have different impedance characteristics. we'll fix the gen and transmission. then we will compare some important ratios to understand how the reactance affects power transmission
*************************************************
Example-1

input work to create the voltage needed to push amps = W_in
R_gen = 50ohms
transmission line = 0.1 x 2 (no X_C and no X_L) = 0.2
V_gen-output = 100_vac-rms (60hz)
endpoint Load = inductor k, R_k=20, X_L=20
Z = vector sum of 20.2 + 20 = 28.426

amps = 100/28.426 = 3.518amps


...

Can a voltage source of 100v with 50 ohms R generate more than 2 amps?

 

[winnie]

Hi Jon,
So what is your definition of work? The one I'm familiar with is strictly mechanical work, F dot d. I don't see how raising the temperature of an object is work in that sense.

As I said, classical thermodynamics is all about differentiating between mechanical work and thermal energy. In this situation you define work to be strictly mechanical work, and work out the relations with other forms of energy.

But you have the equivalence of different forms of energy including thermal energy, electrical energy, kinetic energy, etc.

And the kinetic theory ties thermal energy to strict mechanical energy, in that thermal energy is simply the kinetic energy of the mass of individual particles which make up a mass.

So I will stick with a very general (but rough) definition of 'work' as any operation that moves energy from one place or form to another. Specific situations will require more precise definitions and equations that match.

If I supply 100J of electrical energy to a resistor I can heat up a mass of material by some amount.

If I supply that same 100J of electrical energy and use it to stir that material with a motor, then I can heat it up the same amount, but via the intermediate of force * distance.

For the purpose of discussing power factor, any delivery of electrical power to a load is 'work'.

-Jon

 

[Carultch]

Can a voltage source of 100v with 50 ohms R generate more than 2 amps?

In that example, by definition, the output voltage of the source is 100V. This equals the EMF minus the I*R of the source resistance. So the actual EMF is considerably higher that 100V.

 

[Besoeker]

Can a voltage source of 100v with 50 ohms R generate more than 2 amps?

According to Fiona, it can. Considerably more.

 

[mike_kilroy]

In that example, by definition, the output voltage of the source is 100V. This equals the EMF minus the I*R of the source resistance. So the actual EMF is considerably higher that 100V.

correct, so supply side watts and kva calcs need to use THAT voltage (50ohm*3.518=175.9v) for case 1, and 50*4.833amp=241.7v for case 2) , not 100v

 

[wwhitney]

So I will stick with a very general (but rough) definition of 'work' as any operation that moves energy from one place or form to another.

So I get the classical thermodynamics idea of distinguishing different forms of energy, work versus heat. But if you expand the term 'work' to mean any delta E, it seem to me it is no longer a useful name. We can just use the name 'delta E' directly.

Anyway, it's just a terminology question, I think we agree on the math/physics.

Cheers, Wayne

 

[wwhitney]

lets look at two "realistic" cases of AC power gen . . .
R_gen = 50ohms

First question, is R_gen = 50 ohms "realistic"? That seems like a pretty poor generator.

Cheers, Wayne

 

[FionaZuppa]

Can a voltage source of 100v with 50 ohms R generate more than 2 amps?

that term is rather irrelevant in the examples. you can drop it completely if you want. if it makes you feel better, make it 0.001 ohms.

 

[FionaZuppa]

Can a voltage source of 100v with 50 ohms R generate more than 2 amps?

First question, is R_gen = 50 ohms "realistic"? That seems like a pretty poor generator.

Cheers, Wayne

that term is rather irrelevant in the examples. you can drop it completely if you want. if it makes you feel better, make it 0.001 ohms.
i could also say that the gen with load attached had a output voltage of 100v, R_gen doesnt really matter, R_gen is not calculated into S, etc.

 

[wwhitney]

W_in = (I² x 50) + S

No, W_in = (I² x 50) + P

Cheers, Wayne

 

[Besoeker]

that term is rather irrelevant in the examples.

Then why state it then ignore it in your calculations which appears to be what you did?

 

[FionaZuppa]

No, W_in = (I² x 50) + P

Cheers, Wayne

so the amps associated with Q you just discard? math please.

 

[Carultch]

So I get the classical thermodynamics idea of distinguishing different forms of energy, work versus heat. But if you expand the term 'work' to mean any delta E, it seem to me it is no longer a useful name. We can just use the name 'delta E' directly.

Anyway, it's just a terminology question, I think we agree on the math/physics.

Cheers, Wayne

It is usually the definition of power that is generalized to account for any rate of change, conversion, or transfer of energy. While work is kept with a specific defintion to distinguish it from heat and thermal energy. In thermodynamics, work can be both mechanical work as is the case in the common definition, or electrical energy transmission. Any kind of "ordered" energy transmission.

 

[wwhitney]

many like to ask "where does all that reactive power go?". simple answer is it is wasted by I²R losses through the system.

Nope, the reactive power is a mathematical contrivance, and its nominal value in VA is grossly different from the power involved over any time period longer than a quarter cycle. It contributes to some additional I^2 R losses, but those show up as additional real power.

So the answer to "where does all that reactive power go?" is "back and forth".

Cheers, Wayne

 

[wwhitney]

that term is rather irrelevant in the examples. you can drop it completely if you want. if it makes you feel better, make it 0.001 ohms.

If you repeat your computations with R_gen = 0.001 ohms, then you will get grossly different answers (for any of the computations involving W_in).

Cheers, Wayne

 

[FionaZuppa]

If you repeat your computations with R_gen = 0.001 ohms, then you will get grossly different answers (for any of the computations involving W_in).

Cheers, Wayne

ok, maybe you dont see that it doesnt really matter because the power associated with R_gen is linear, and R_gen is not in S, the impact of PF diff yields same results
the discrepancy of 1.97x and 1.54x is due to rounding err.

R_gen = 0.001
now lets look at some ratios
what ratios to look at
1) W_in:Watts(load)
2) Amps:Watts(load)

from example-1
PF=0.71061
1) 351.821/250.002 = 1.407
2) 3.518A/250.002W = 0.01407amps per watt

from example-2
PF=0.49297
1) 483.323/233.579 = 2.069
2) 4.833A/233.579W = 0.02069amps per watt

Conclusions:
Regardless of R value of load, PF plays major role on the effort needed to take useable power from gen to load. in example-1 it takes ~1.4x ~250watts to deliver that ~250watts of useable power. in example-2 it takes ~2x ~233watts to deliver that ~233watts of useable power. on a per unit watt basis its 0.0056x per watt, and 0.0086x per watt respectively. as you can see, the diff in PF (0.49297 - 0.71061) had a ~1.54x impact on the input work required to deliver same unit watt of useable load power.

 

[Carultch]

so the amps associated with Q you just discard? math please.

By definition, you gave the source a pure resistive impedance.


The power produced by the source equals the power supplied by the source, plus the power loss in the source resistance. I^2*R is the power loss in the source resistance, and is based on the I that is supplied to the rest of the circuit, and already accounts for any reactive power.

 

[wwhitney]

the discrepancy of 1.97x and 1.54x is due to rounding err.

I haven't had a chance to look at your new numbers (still catching up), but a change from 1.97x to 1.54x is not due to rounding errors. It's too big a change for that.

Cheers, Wayne

 

[FionaZuppa]

By definition, you gave the source a pure resistive impedance.


The power produced by the source equals the power supplied by the source, plus the power loss in the source resistance. I^2*R is the power loss in the source resistance, and is based on the I that is supplied to the rest of the circuit, and already accounts for any reactive power.

ok, i somewhat follow you. I(t) is same everywhere, Q is not. Q is a load characteristic, associated with X portion of Z_load.

 

[wwhitney]

so the amps associated with Q you just discard? math please.

No, they aren't just discarded, they contribute to the input power required via R_gen and via R_transmission.

Cheers, Wayne