OT Electrical question

Hi jk

About how much difference is there in power loss with DC instead of AC? Again, I had considered RMS to be equalivant to DC in electrical circuits.

Jerry

Reply to
Jerry Martes
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In terms of heating value, RMS and DC are equivalent, by definition.

In high-voltage transmission lines, one can push more power through using DC than AC, because HV lines are breakover-voltage-limited, and the peak voltage of AC is Sqrt[2] larger than the rms voltage.

So, with DC, one can run Sqrt[2] larger voltages with the same breakover voltage. The current is limited by the wire diameter and material, not the voltage, so the transmitted power is Sqrt[2] larger for a given line.

The inductive effects, if important, are neutralized by adding HV capacitors as needed.

Joe Gwinn

Joe Gwinn.

Reply to
Joseph Gwinn

Joe sez:

" . . . > In terms of heating value, RMS and DC are equivalent, by definition.. . ."

True ! IMO, this statement negates all other points made re. losses in transmission. If one were to compare DC and RMS on the basis of collateral losses, then dielectric loss would have to be brought into play. Give it up ! DC and RMS are equivalents.

Bob (picking fly shit out of pepper) Swinney

Reply to
Robert Swinney

As a parting shot to this overlong thread, I would add: "All electrical current flows from a higher voltage gradient to a lower voltage gradient; how does the math expression go, [di / dt] ? When you get right down to it, all current is DC at any infinitesimal point in time -- only the voltage gradients vary"

Bob (Ben Frankl>>

Reply to
Robert Swinney

:)

The same DC and RMS voltage will cause equal dissipation of power in a fixed impedance load by definition. 110 volt RMS is defined to be the amount of AC voltage needed to deliver the same amount of power as 110 V DC will deliver to a given impedance load. However, the power loss in a given wire is not the same because the same wire will have slightly different impedance to DC than to 60 Hz AC.

In all typical power distribution applications most of us deal with day to day, the difference is so small that we can ignore it and just assume there is no difference. The major factor which is the resistance of the wire will have basically the same effect to DC and 60 Hz RMS AC.

Reply to
Curt Welch

Curt sez, again:

"However, the power loss in a given

Still not quite right ! See Don Foreman's posts in this thread where he tries to explain the reactive component(s) of an electrical load have no effect on power consumed. Only resistance can consume power. Granted, if one of the "Z" elements was inductive then the pure resistance of that conductor would draw power. The same might be said of the leakage R of capacitors. I believe this analogy holds true for any frequency, not just

60 Hz RMS.

Bob Swinney

Reply to
Robert Swinney

Joe - you a hi line guy ?

Skin effect is in use - the center is steel - the outside is Al.

In DC - no skin at all - goes through the steel - so ugly -so lossy.

In special runs, Copper or Al is used, but the support has to be increased big time!

Martin

Martin H. Eastburn @ home at Lions' Lair with our computer lionslair at consolidated dot net TSRA, Life; NRA LOH & Endowment Member, Golden Eagle, Patriot"s Medal. NRA Second Amendment Task Force Charter Founder IHMSA and NRA Metallic Silhouette maker & member.

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Joseph Gw> >

Reply to
Martin H. Eastburn

Equally, all currents are AC, given a long enough perspective. jk

Reply to
jk

Into a given Ideal resistance, the heat generated in that resistance is the same for a straight DC voltage across it, as for any other RMS equivalent voltage.

For a voltage applied at some wire terminals some distance away, that is not necessarily so. THe wires have a slightly higher AC resistance than DC, but if we assume that the wires are Ideal (0 ohms) It still will not be the same, because the inductance of the wires, will cause a voltage drop. THis voltage drop will cause the terminals of the resistor to see a lower voltage, so it will heat less. [THe inductance CAN be counteracted by added capacitance at the resistor.] jk

Reply to
jk

Only in terms of what happens to an IDEAL resistor

WHich turns out to have sand in it too.

jk

Reply to
jk

Well, as Ben Franklin said right after he "discovered" electricity, "SShhhiiiiiiittttt"

Bob Swinney

Reply to
Robert Swinney

No, although I do have an electronics background.

I have to wonder how big an effect this can be at 60 Hz. I'll have to run the computation, when I have a chance.

I recall reading that many HV lines are a set of parallel wires anyway, but I don't recall the reason.

I do know that some inter-ties are DC, partly to eliminate phasing issues between nets, and sometimes to stuff more power through the plant.

Joe Gwinn

Reply to
Joseph Gwinn

"jk" wrote: Equally, all currents are AC, given a long enough perspective. ^^^^^^^^^^^^^^^ And all are DC, given a short enough perspective. DC could be considered the degenerate case of AC at zero frequency.

Reply to
Leo Lichtman

"Robert Swinney" wrote: Well, as Ben Franklin said right after he "discovered" electricity, "SShhhiiiiiiittttt" ^^^^^^^^^^^^^^^^^ I thought he discovered kite string.

Reply to
Leo Lichtman

He pointed out that the inductive load in itself doesn't consume power. What he didn't point out, but I'm sure understands, is that an inductive load changes the current flowing in the wires delivering the power, and in doing so, changes the power dissipated in the wire delivering the power.

Right. The entire conductor. We have been talking about power delivery systems here, not just a small circuit with an inductor in it. If the load at the end of the power line has any non resistive Z component then it causes the load to have a non unity power factor which means the current will be out of phase with the voltage. As such, the current ends up being larger than it would be to deliver the same power to the resistive component of the load. As such, the current flowing in the transmission line is higher and the voltage drop in the transmission line is higher and the power dissipated in the transmission line is higher.

In other words, if you have a pure resistive wire delivering to a pure resistive load, then then DC voltage and RMS voltage of the same value applied to the system will cause it to dissipate the exact same power.

But add to that a small inductor at the load and it will cause extra current to flow for the AC power which is out of phase with the voltage. The fact that it's out of phase is what makes the power consumed in the inductor net out to zero. But that extra current flow is also flowing through the resistive wire delivering power to the load, and as such, you get extra power consumed in the wire delivering the current to the inductor.

Any none resistive impedance in the load will cause the DC voltage and RMS voltage of the same value to dissipate different amounts of power in the resistive transmission line.

Yes, true. The impedance will change as a function of frequency so the effect will change as a function of frequency. In a circuit tuned for 60 Hz you can make the capacitive impedance cancel out the inductive impedance but at any other frequency, they will not cancel each other out. The current and the power consumption will change as a function of frequency. At 0 Hz (DC) the power consumption will be different than at 60 Hz.

Reply to
Curt Welch

Hi jk

I'm sure you are a really sharp engineer, but, this reference to third order effects is BS when we are discussing Malibu Lights on a guys lawn. I realize that you have a need to express the fact that you really know alot about electricity. That was evident when you wrote "Uh, no it doesnt" in response to don Foreman's accurate and correct reply to the OP. I'm sure that you know (by now if you didnt before) that Don is so well informed on things electrical that he is equal to any expert in the general electrical field. Don sure doesnt deserve to be responded to with a curt response that implies that he doesnt know about RMS to DC comparison.

Jerry

Reply to
Jerry Martes

Which is why I shut up when the conversation took a left turn from Practical Electricity, off into applied trig and other voodoo. Malibu lights work just fine with AC, and rectifying it just to be DC adds parts to the system that can and will fail.

The KISS Principle far and away trumps any slight gains.

There my be a tiny advantage in the real world, but it isn't going to be notable until you get out to REALLY long runs, several hundred feet. Which is why they make some of the more expensive transformers are 24V Center Tapped secondaries, and they sell 3-wire cable.

Then again, if you're going to that length might as well just ship

120V or 240V out to the Back 40, put the transformer at the far end.

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Reply to
Bruce L. Bergman

But whenever your car's electrical system 'turns into AC Current' and passes the zero point to full opposite polarity (usually due to a friend "Helping" you with a jump start) very exciting, messy, and expensive things happen.

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Reply to
Bruce L. Bergman

Sooooo...

How many engineers DOES it take to change a lightbulb??

On the bright side though, topics like this are both educational and entertaining.

Aaron

Reply to
Aaron

Voltage losses due to wiring resistance rise with the square of the load current (I^2 x R, or current squared, times resistance), so the smaller the current, the lower the losses. For a given power consumption (ie; Watts or VA), you can lower the current by raising the voltage down the wires, then lower it again at the other end. (ie; in your gizmo). This is expensive & difficult to do with DC, but can be done cheaply & easily on AC with just a transformer.

Reply to
Lionel

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