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#11
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OK, I'm a little confused. Well, maybe more than a little.
Starting with energy as the "ability to do work" and power as the rate at which energy is "transformed" into work, things quickly get muddy. Energy passing through an imaginary surface (or point or plane) would not actually do any work in passing through, and in fact would retain its full potential to do work after having passed through. What then is power density? Is it the amount of work that the energy passing through a unit area of the surface "could have done" had it been actually and fully "captured" at that surface? There is no real power at that surface, is there? While power (and work) absolutely require energy, it strikes me as metaphysical whether all energy ultimately does do work and produce power. I don't think physics requires that, and it seems that lot of radiated energy is not obviously being transformed into work. So is energy without power really impossible, Cecil? Been away from this for a longer bit than I'm comfortable mentioning. Chuck. NT3G Cecil Moore wrote: chuck wrote: Technically, is it not energy that leaves the transmitter and is received by the receiver? Technically, RF energy passing a point/plane during a unit of time is RF power (joules/sec). We can't have one without the other. |
#12
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chuck wrote:
OK, I'm a little confused. Well, maybe more than a little. Starting with energy as the "ability to do work" and power as the rate at which energy is "transformed" into work, things quickly get muddy. Power is the rate at which energy is transferred or used. Period. Energy passing through an imaginary surface (or point or plane) would not actually do any work in passing through, and in fact would retain its full potential to do work after having passed through. Yes. What then is power density? Is it the amount of work that the energy passing through a unit area of the surface "could have done" had it been actually and fully "captured" at that surface? No. Power is not an "amount of work", nor is power density. Power is the rate at which power is being transferred, so it can tell you only the rate at which work can be done, not the amount. The rate at which energy is being passed through a given cross sectional area of a surface is the power density. If you integrate the power going across the boundary for some period of time, you then know the amount of energy which has passed, and therefore the amount of work which can be done. There is no real power at that surface, is there? There is real energy passing that surface, and the rate at which it's passing is the power at that surface. So yes, there is. While power (and work) absolutely require energy, it strikes me as metaphysical whether all energy ultimately does do work and produce power. I don't think physics requires that, and it seems that lot of radiated energy is not obviously being transformed into work. So is energy without power really impossible, Cecil? Cecil loves metaphysical arguments, so this is an ideal question for him. Been away from this for a longer bit than I'm comfortable mentioning. Try finding a basic physics textbook at your local library. Most high school level texts should cover the topic adequately. For a more mathematical and quantitative treatment, a freshman level college text would be fine. A couple of the more popular ones are Resnick & Halliday, and Weidner & Sells. In the older editions of both, at least, electrical phenomena are covered in the second volume. However, power, work, and energy aren't restricted to electricity so are covered in general terms in the first volume. Roy Lewallen, W7EL |
#13
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Correction:
Roy Lewallen wrote: . . . No. Power is not an "amount of work", nor is power density. Power is the rate at which power is being transferred, so it can tell you only the rate at which work can be done, not the amount. . . I of course meant ". . .Power is the rate at which *energy* is being transferred. . ." Thanks, Owen! Roy Lewallen, W7EL |
#14
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chuck wrote:
So is energy without power really impossible, Cecil? Roy has already answered the basic question from an engineering viewpoint. Physicists often consider "power" to have a different definition than the engineering definition. From the IEEE Dictionary: "power - The rate of generating, transferring, or using energy." (agrees with Roy) From "University Physics" by Young and Freedman: "power is the time rate at which work is done." i.e. only the "using energy" portion of the engineering definition. A certain physicist I know will argue that no work is being done in a lossless transmission line so there is no power there. He will say the existence of 100 watts at 1000 points along the line means there must be 100,000 watts in the line. He will say that a Bird wattmeter doesn't measure watts. He will say that a power generating plant doesn't generate power and a transmission line doesn't transfer power. He will also say that reflected power doesn't exist because it is not doing any work. He will say that for an EM wave in free space, ExH has the dimensions of watts but it isn't power because no work is being done. As Roy indicated, engineers have a wider definition of "power". Energy without power is certainly possible, e.g. a DC battery with zero current. However, for a constant steady-state power level associated with an EM wave, energy and power are inseparable. I like to use a one-second long lossless transmission line in some of my examples because it is impossible to hide the joules. A one-second long line with 200 watts forward power and 100 watts reflected power contains 300 joules that have been generated but have not reached the load. Since EM wave energy cannot stand still (or slosh around side to side) it is only logical to assume that 200 of those joules are in the forward wave and 100 of those joules are in the reflected wave both traveling at the speed of light. -- 73, Cecil http://www.qsl.net/w5dxp |
#15
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chuck wrote:
Energy passing through an imaginary surface (or point or plane) would not actually do any work in passing through, and in fact would retain its full potential to do work after having passed through. What then is power density? The full name is power *flux* density, implying the rate at which energy *flows through* unit area of a defined reference plane. SI units are watts per square metre. Is it the amount of work that the energy passing through a unit area of the surface "could have done" had it been actually and fully "captured" at that surface? Yes, that is the implication - except that it's the *rate* of energy capture, ie the amount that could be captured from unit area in unit time. This is only a concept, because it isn't physically possible to intercept the power flux through unit area of an EM wavefront - your wave-catcher would disturb the wavefront around its edges, and the shadow behind it would be filled in by diffraction. However, very similar concepts apply to power flux in a transmission line - and in that case you really *could* capture the exact steady-state power flux at any point, by cutting the line and substituting a dummy load of the correct impedance. There is no real power at that surface, is there? That rather depends on your personal definitions of the words "real", "at" and possibly "is" :-) I think you'd caught it correctly in the previous paragraph... but if you squeeze too hard, the waves will slip through your fingers. -- 73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
#16
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Thank you all for your assistance. It will take a little time (and a
little work) for the difference in physicist/engineer definitions of power to sink in. Chuck, NT3G Ian White GM3SEK wrote: chuck wrote: Energy passing through an imaginary surface (or point or plane) would not actually do any work in passing through, and in fact would retain its full potential to do work after having passed through. What then is power density? The full name is power *flux* density, implying the rate at which energy *flows through* unit area of a defined reference plane. SI units are watts per square metre. Is it the amount of work that the energy passing through a unit area of the surface "could have done" had it been actually and fully "captured" at that surface? Yes, that is the implication - except that it's the *rate* of energy capture, ie the amount that could be captured from unit area in unit time. This is only a concept, because it isn't physically possible to intercept the power flux through unit area of an EM wavefront - your wave-catcher would disturb the wavefront around its edges, and the shadow behind it would be filled in by diffraction. However, very similar concepts apply to power flux in a transmission line - and in that case you really *could* capture the exact steady-state power flux at any point, by cutting the line and substituting a dummy load of the correct impedance. There is no real power at that surface, is there? That rather depends on your personal definitions of the words "real", "at" and possibly "is" :-) I think you'd caught it correctly in the previous paragraph... but if you squeeze too hard, the waves will slip through your fingers. |
#17
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I think the differences in this discussion go back to very fundamental
definitions of power and energy (and power flux, which might not be the same as power). One set of terminology came from classical Thermodynamics and the other set from more general principles. More fun may be had dealing with the Second Law (which arrived from two very different viewpoints) and, for example, quantum effects in the same discussion. The discussion can quickly hinge on precise meanings of common words, such as "power." Mixing terminology produces such wonderful concepts as measuring 100 watts flowing at 100 points in a transmission line and concluding that we have found 10,000 watts. Bill W2WO |
#18
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Why unnecessarily complicate matters with words.
Watts = Amps times Volts. .. . . . and that's all there is to it. ---- Reg. |
#19
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Reg Edwards wrote:
Watts = Amps times Volts. Only for DC or in-phase AC/RF. Volt-Amps = Amps times Volts = SQRT(watts^2 + vars^2) Watts = Amps times Volts times cos(A) = real power Vars = Amps times Volts times sin(A) = reactive power Reference: "Alternating Current Circuits", Kerchner/Corcoran, 3rd edition, (C) 1938, 1943, 1951. -- 73, Cecil http://www.qsl.net/w5dxp |
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