THE PHYSICS OF... POWER

[FionaZuppa]

Your proposed definition of "power = work/time" fails for something as simple as an electric resistance heater. By that definition, the heater is using no power, as it is doing no work.

A definition in terms of energy transfer or conversion is broader and more appropriate for electrical engineering.

Cheers, Wayne

the heater does work, converts electrical potential into kinetic heat. write the math out in eV and you shall see the light.

a lossless inductor with a lossless cap in resonance does no work, energy just moves back and forth, no transformation being done.

 

[Ingenieur]

the heater does work, converts electrical potential into kinetic heat. write the math out in eV and you shall see the light.

a lossless inductor with a lossless cap in resonance does no work, energy just moves back and forth, no transformation being done.

that is a good posit
the molecules move faster/further causing heating
which could be constructed as 'work'

 

[GoldDigger]

1 I agree with you, ie, reactive power IS power, just not active power

2 yes, yes, yes, it exerts a pressure which induces stress in the walls
the walls will 'give' or move, I never said constant volume
in a constant vol all q no w

I beg to take issue with that.
You did not say "in the case of R heating raising the air temp will do some small incidental amount of work"
You said "in the case of R heating raising the air temp IS the work"
Of course that all depends on what the meaning of "is" is. :lol:

 

[Carultch]

1 I agree with you, ie, reactive power IS power, just not active power

2 yes, yes, yes, it exerts a pressure which induces stress in the walls
the walls will 'give' or move, I never said constant volume
in a constant vol all q no w

When the walls 'give', the gas is doing work on its surroundings. The amount of heat you add to the gas goes partly in to increasing the thermal energy of the gas, and partly in to pushing outward on its container.

The part that goes in to increasing its thermal energy, is not considered work in thermodynamics.
The part that goes in to pushing outward on its container, the cumulative sum of Pressure * change in Volume, is the part of that energy that is work.

 

[FionaZuppa]

that is a good posit
the molecules move faster/further causing heating
which could be constructed as 'work'

look at it from these two perspectives

1) a good heater has PF=1 and is rated in kW (real power), so it must be doing work :thumbsup:
2) look at it from the gen side, work must be done continuously to keep the house warm (eV potential), the heater then extracts eV into heat (work). it it were pur cap and inductor the the gen needs to do no work and no heat would ever come out, then me freeze in winter.

as for heat doing work to air, some of it yes, some eV of heat is then transferred to air. put the heater and your arm in a vacuum, you still get burned, but no work on any air

my real examples are almost complete, hopefully will have it posted shortly

 

[Ingenieur]

I beg to take issue with that.
You did not say "in the case of R heating raising the air temp will do some small incidental amount of work"
You said "in the case of R heating raising the air temp IS the work"
Of course that all depends on what the meaning of "is" is. :lol:

No need to beg,

Nor did I state that it was an unbounded system
quit distorting what people say to support your position
cherry picking and parsing
you have a bad habit of doing that along with snark and being arguementative
you fancy yourself a physicist but seem to lack basic knowledge
You consider it correcting people, setting the record straight, laying out the facts, look how big my brain is, etc...problem is you are usually wrong or using distortion to fit your 'answer':lol:

heating the air IS the work
burn fuel make power
heat the air
integrate P dt = W

my oriignal post stated I^2R does express work being performed
did not say how much, some q some w

 

[winnie]

Oy! Thermodynamics now

Clearly running current through a resistor to raise the temperature of a body involves energy moving from one place to another, and work being done. The work done is in the form of increasing the kinetic energy of the particles in the body.

Generally in thermodynamics, the basic calculations differentiate between thermal energy and other work because the whole point is to understand the relationship between randomly directed kinetic energy of the particles in a body (the thermal energy) and 'coherent' forms of energy such as a spinning mass.

So if you are intentionally differentiating between thermal energy and work, then you might say that the resistor is doing no work...but the equivalence of work energy and heat energy was demonstrated a couple of centuries ago...

-Jon

 

[wwhitney]

First Law of Thermodynamics:

Esys = q + w

delta E system change in energy
q heat added/removed
work performed/absorbed

That equation precisely shows that heat isn't work! There are two ways to put energy into the system: add heat, or do some work on it. They are mutually exclusive.

This argument about an electric heater heating the air, causing the pressure to rise and that doing work is a red herring. Take an electric water heater filled with 4 C water (so that the thermal coefficient of expansion of the water is 0). Where's the work?

Cheers, Wayne

 

[wwhitney]

The work done is in the form of increasing the kinetic energy of the particles in the body.

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.

Cheers, Wayne

 

[Ingenieur]

That equation precisely shows that heat isn't work! There are two ways to put energy into the system: add heat, or do some work on it. They are mutually exclusive.

This argument about an electric heater heating the air, causing the pressure to rise and that doing work is a red herring. Take an electric water heater filled with 4 C water (so that the thermal coefficient of expansion of the water is 0). Where's the work?

Cheers, Wayne

No
what it shows that the energy change of a system results in q or temperature rise AND work, in some proportion, 0-100% for either variable, usually a mix


Your red herring is physical fact

power = work/time
the work will happen over time as the water gets warmer you will have both q and W
if not no need for a pressure relief

 

[wwhitney]

what it shows that the energy change of a system results in q or temperature rise AND work, in some proportion, 0-100% for either variable, usually a mix

Sure, it's a mix, because heat is not work. They are different and accounted for differently. They are both energy, of course.

Cheers, Wayne

 

[Ingenieur]

Sure, it's a mix, because heat is not work. They are different and accounted for differently. They are both energy, of course.

Cheers, Wayne

But the added energy or power can produce both
heat or work
a large R with i flowing in a sealed vessel with water
constant power rate
first all q no W
over time shifts to W > q and the vessel ruptures, ie work expended

Energy is not work
energy is the capacity to perform work
big difference

 

[Ingenieur]

This thread is number 2 for replies in this sub-forum
lol

 

[Fulthrotl]

This thread is number 2 for replies in this sub-forum
lol

this one, overall, has been pretty civil. that "other one"
got a bit snarky in spots, iirc.

 

[FionaZuppa]

i can put the work and heat "issue" to an end.

W = delta KE

a quick study of a pure resistor
the mass of R gets hot as the R strips eV from amps, we can call this +KE
at the same time the heat energy in the mass begins to immediately travel out to the colder air, a heatsink if you will, AND, some heat energy emits out as RF, emission and transfer to air we can call -KE. the temp of the mass will rise into equilibrium (temp is constant) when +KE = -KE

so there's +delta KE, and -delta KE, so now we ask, if the net is zero at equilibrium is there any work being done. i say sure is, constant work being done, and the potential energy (eV) being ripped off of amps is being converted into another form of energy, just being done at the atom level. if you could stop the RF emission and you put R in a vacuum, it would simply get hotter and hotter until it turned into a liquid and then a gas and then plasma, hotter and hotter.

 

[FionaZuppa]

lets ask another way. the induction motor is 10kW, the output shaft has a perpendicular arm resting on it, a friction arm, the arm doesnt move or do anything but get hot, is the motor doing any work to the friction arm??

fundamentally, if you increase universe entropy then some work was done. with a lossless cap coupled to a lossless inductor, energy can move back and forth w/o increasing universe entropy :thumbsup:


what about a satellite in orbit, it flies at constant V_magnitude but the V and "a" vectors are continuously changing at a rate if 360 degrees per orbit. any work being done on the satellite ??

aint this physics stuff great!

 

[FionaZuppa]

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
real power of just inductor = 247.526watt

S = vector sum of P + Q = I²Z

S = I² x Z = VA = 351.809 watt
P = I² x 20.2 = 250.002 watt
Q = I² x 20 = 247.526 watt
vector sum of P+Q = 351.809 watt


PF = P/S = 0.71061 = cos(Φ)
P = PF x S (is this true?)
P = PF x S
P = 0.71061 x 351.809 = 249.999 (round err)


W_in = (I² x 50) + S
W_in = 618.816 + 351.809 = 970.625 watt

conclusion:
1) it takes 3.518amps to deliver 250.002watt at the load
2) the ratio of load R to load X_L is bad, poor PF
3) input work was 970.625watt

******************************************
Example-2

lets now attach a new load, less real power, but worse PF

what does "PF worse" mean, Φ is larger

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=10, X_L=18
Z = vector sum of 10.2 + 18 = 20.690

amps = 100/20.690 = 4.833amps
real power of just inductor = 233.579watt

S = vector sum of P + Q = I²Z

S = I² x Z = VA = 483.300 watt
P = I² x 10.2 = 238.250 watt
Q = I² x 18 = 420.442 watt
vector sum of P+Q = 483.254 watt


PF = P/S = 0.49297 = cos(Φ)

W_in = (I² x 50) + S
W_in = 1167.894 + 483.300 = 1651.194 watt

conclusion:
1) it takes 4.833amps to deliver 233.579watt at the load
2) the ratio of load R to load X_L is bad, poor PF
3) input work was 1651.194watt

******************************************
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) 970.625/250.002 = 3.882
2) 3.518A/250.002W = 0.01407amps per watt

from example-2
PF=0.49297
1) 1651.194/233.579 = 7.069
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 ~3.8x ~250watts to deliver that ~250watts of useable power. in example-2 it takes ~7x ~233watts to deliver that ~233watts of useable power. on a per unit watt basis its 0.0152x per watt, and 0.03x per watt respectively. as you can see, the diff in PF (0.49297 - 0.71061) had a ~1.97x impact on the input work required to deliver same unit watt of useable load power.

now lets look at amps per watt ratios. as you can see between the two examples, the decrease in PF means it takes more amps to deliever same unit watt useable power to the load.

many like to ask "where does all that reactive power go?". simple answer is it is wasted by I²R losses through the system. it takes work on the gen side to create EMF to push amps, and reactance impedes the flow of amps, thus more reactance means i need to push harder to keep useable power watts constant on the load side. if load-A is 20kW PF=0.8 and load-B is 20kW PF=0.7, this just tells us the cost per unit watt is higher for the lower PF.

it is now trivial to determine the economics of it all in terms of business model. in general, low PF means less ROI than high PF. it is best interest of business owner to tune the characteristics of Z to reduce reactance.

 

[wwhitney]

But the added energy or power can produce both
heat or work

Let's apply the 1st law of thermodynamics (Delta E = q + w) to the electric resistance heater. Say we run 10 amps at 120V through an electric heater for 1 sec, so the electricity delivers 1200 joules of energy. Delta E = 1200 joules, yes?

OK, so maybe 1100 joules of that goes into heat, and we happen to get 100 joules of work out it incidentally due to some thermal expansion against a resisting force. Now, what power doing you want to assign to the electric heater during that 1 second?

Power = work/time, your proposed definition, gives us a power of 100 W.
Power = Delta E/time, my proposed definition, gives us a power of 1200W.

I think most people would call the heater a 1200 W heater, which requires using the expanded definition of power.

Cheers, Wayne

 

[wwhitney]

here's my examples, dig in.

Sorry I didn't have a chance to take a look tonight, I'll read it in the morning.

Cheers, Wayne

 

[GoldDigger]

lets ask another way. the induction motor is 10kW, the output shaft has a perpendicular arm resting on it, a friction arm, the arm doesnt move or do anything but get hot, is the motor doing any work to the friction arm??
Classical Mechanics answer: The shaft does no work on the arm since the arm does not move. But the arm does work on the shaft since the shaft is moving. The speed of the shaft does not change since there are additional forces (in the motor) doing work on the shaft to keep it turning. The frictional force at the interface converts relative motion against a force into heat.

fundamentally, if you increase universe entropy then some work was done. with a lossless cap coupled to a lossless inductor, energy can move back and forth w/o increasing universe entropy :thumbsup:
Classical Thermodynamics reply: You can increase entropy just by putting a hot mass in contact with a colder mass. No work is done in the process, but the total number of available states in the new system (side stepping for a moment into quantum physics) is greater than in the old system with the total energy remaining the same and the change is irreversible.


what about a satellite in orbit, it flies at constant V_magnitude but the V and "a" vectors are continuously changing at a rate if 360 degrees per orbit. any work being done on the satellite
Classical Mechanics answer: Gravitational force does no work on the satellite because the force is always at right angles to the direction of motion (assuming a circular orbit.) For an elliptical orbit work is done by gravity over part of the orbit since a component of the gravitational force is parallel to the velocity. Potential energy due to the satellite's position in the gravitational field is converted to kinetic energy. In the other half of the orbit, going back toward apogee, negative work is done on the satellite decreasing its velocity and storing the energy as increased potential energy of its mass in the gravitational field.??


aint this physics stuff great!
For something really interesting, consider the Moon orbiting the Earth as the earth orbits the sun. Approximate the Moon's earth orbit as circular. Now the Earth's gravity is doing work on the Moon over part of its orbit because the gravitational force is lined up with the Earth's motion in orbit. End result is that the Moon loses kinetic energy as it moves to the far side of the Earth from the Sun and gains it back again as it moves closer to the Sun.

See bold above.