Why doesn't long distance power tranmission cause sine wave distortion?

[mbrooke]

That takes care of the cutoff. Now look at when the load turns on again: when the source voltage reaches -5 volts?
At that point the load voltage suddenly goes positive for awhile!

If you wanted symmetry you would have to turn the load current back on at whatever time it would cause the new load voltage to become -5V and proceed more negative from there.


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I get it now... At -5 volts as soon as the load cut in it would drop back out, then cut back in, then out and until those 15 amps did not cause the voltage to plummet below 5 volts at the load. The device would have to be smart enough to know when to cut in.

 

[Carultch]

Any way to graph this? If I get my hands on a scientific calculator Im willing to try it out.

Here's a graph in Excel. The time axis represents fractions of a cycle. I have 49 data points from 0 to 48, and then I make time t equal that over 48. Then V0 = 170V * sin(2*pi()*t). 170V is the peak voltage for nominal 120Vac.


The red shows V0. The blue shows VL(t) = V0(t) - I(t)*R, where I(t) is 15A in the same sign as V0(t). I(t) is a square wave that snaps between +15A and -15A. I set the line resistance at 5 Ohms.

 

[mbrooke]

Here's a graph in Excel. The time axis represents fractions of a cycle. I have 49 data points from 0 to 48, and then I make time t equal that over 48. Then V0 = 170V * sin(2*pi()*t). 170V is the peak voltage for nominal 120Vac.


The red shows V0. The blue shows VL(t) = V0(t) - I(t)*R, where I(t) is 15A in the same sign as V0(t). I(t) is a square wave that snaps between +15A and -15A. I set the line resistance at 5 Ohms.

Ok, its sinking in- still a tad confused why VL(t) is going into the negative while the source is still in the positive... but I am sure to understand... unless this is not physically possible in the real world? And won't the load end voltage be heavily distorted if we tried say 5000 ohms Vt?

Maybe the graph is throwing me off... V0 seems to be in blue :?

The red shows V0.

 

[GoldDigger]

The reason that the load voltage seems to start going down before V0 reaches zero is that the graph is just connecting points and the value of the function at the zero crossing is theoretically undefined but the spreadsheet sets it to zero instead.
The smooth function has a vertical line at the zero crossings.
If we tried Rt = 5000 ohms and I = 15 the voltage would just look like a 75000V square wave..

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[Carultch]

Ok, its sinking in- still a tad confused why VL(t) is going into the negative while the source is still in the positive... but I am sure to understand... unless this is not physically possible in the real world? And won't the load end voltage be heavily distorted if we tried say 5000 ohms Vt?

Maybe the graph is throwing me off... V0 seems to be in blue :?

I labeled the graph wrong. My bad. Red is V0, blue is VL.

 

[Carultch]

Ok, its sinking in- still a tad confused why VL(t) is going into the negative while the source is still in the positive... but I am sure to understand... unless this is not physically possible in the real world? And won't the load end voltage be heavily distorted if we tried say 5000 ohms Vt?

Maybe the graph is throwing me off... V0 seems to be in blue :?

The reason why it goes negative, is that we defined current to always be in the direction of V0, fixed at 15A. This would imply that your special load is actually releasing power back to the grid, during that brief period. Because the load would be pushing 15A against the voltage that is across it itself, so the load would be supplying power to the line resistance.

To make a pure consumptive load, define current to diminish to zero the instant voltage across the load does so as well. Don't let current resume at -15A, until the voltage across the load is enough to provide -15A to the load.


Attached is such a graph. This sort of looks like the output waveform of a diode bridge rectifier output, when you include diode losses, that has been inverted after rectifying when main supply voltage is negative.

 

[mbrooke]

The reason why it goes negative, is that we defined current to always be in the direction of V0, fixed at 15A. This would imply that your special load is actually releasing power back to the grid, during that brief period. Because the load would be pushing 15A against the voltage that is across it itself, so the load would be supplying power to the line resistance.

To make a pure consumptive load, define current to diminish to zero the instant voltage across the load does so as well. Don't let current resume at -15A, until the voltage across the load is enough to provide -15A to the load.


Attached is such a graph. This sort of looks like the output waveform of a diode bridge rectifier output, when you include diode losses, that has been inverted after rectifying when main supply voltage is negative.

Ok, thats what I was thinking! Yes, that would be it. :thumbsup:


Now... applying the same constant 15 amps between 5 through 170s and -5 through -170, what would the wave look like at the load vs at the source assuming a long transmission line?

 

[GoldDigger]

Ok, thats what I was thinking! Yes, that would be it.


Now... applying the same constant 15 amps between 5 through 170s and -5 through -170, what would the wave look like at the load vs at the source assuming a long transmission line?

Since the offset voltage (15x5=75) is greater than 5 there will still be two jumps per half cycle. And you might as well relate to the source voltage, not the load voltage to avoid chattering.

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[mbrooke]

Since the offset voltage (15x5=75) is greater than 5 there will still be two jumps per half cycle. And you might as well relate to the source voltage, not the load voltage to avoid chattering.

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Id agree, but even then would the load not chatter? I think when I think about the chattering I answered my own question. you know its moments like this that you realize whom ever created the universe really thought of everything :happyyes::happyyes::happyyes::happyyes:

But going back to linear loads; assuming 6 awg 5000 foot transmission line, at 5000 volt AC with a total fixed resistance of 100 ohms (load plus 6 awg conductor):




5000 volts, 50 amps with ohms: 3.95% drop with a load voltage of 4802.49

2500 volts, 25 amps with ohms law: 3.95% drop with a load voltage of 2401

1000 volts, 10amps: 3.95% with 960.5 load volts

500 volts, 5 amps: 3.95% with 480.25 load volts

250 volts, 2.5amps: 3.95% with 240.12 load

100 volts, 1 amp: 3.95% with 96.05 volts

50 volts, .5 amps: 3.96% with 48.02 volts

25 volts 0.25: 3.96% with 24.01 volts

10 volts 4.00% with 9.6 volts

5 volts 4% at 4.8 volts


Math looks good? my VD calculator for some reason gives more % drop at lower voltages, but this is just rounding?

So, the conclusion by simple math we can gather this: When current is in phase with the voltage as with any pure resistance, the voltage drop will remain the same (proportional) in all portions of the sine wave. So even a 765,000 volt transmission line will still run at 10% voltage drop when the sine wave is at 5 volts because the load is also pulling 1/4 an amp instead of the 4000amps it does at the peak.

Now, how does this hold up to the real world? Dare I ask what motors, HID lamps and power factor do when in theory these concepts are not linear and do not pull current in phase with the voltage?

 

[GoldDigger]

Yes, it's just rounding.
And non linear loads get complicated. Reactive loads, OTOH, just change the phase of the load voltage sine wave slightly in addition to decreasing the amplitude.

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[romex jockey]

Everytime i read the EE threads here my head hurts ,and i end up with more Q's than A's....yet i am compelled to risk a relevant (i hope) Q....

Just what is a 'sine wave'?

Way back i was told the electrons in any given atom (CU) don't necessarily 'flow' , they charge each other via some big back/forth 'bump' to any given load...

The magnitude ,or peak, was simply the flux @ certain degree of generator rotation.

Am i close....??

How the physics of VD are further altered here escapes me , but i have to start somewhere


~RJ~

 

[mbrooke]

Yes, it's just rounding.
And non linear loads get complicated. Reactive loads, OTOH, just change the phase of the load voltage sine wave slightly in addition to decreasing the amplitude.

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But that phase and amplitude does not behave in a "none linear" fashion? And a bit off topic... but does the reactance and capacitance in a transmission line have anything to do with maintaining or distorting a sine wave?

 

[mbrooke]

Everytime i read the EE threads here my head hurts ,and i end up with more Q's than A's....yet i am compelled to risk a relevant (i hope) Q....

Just what is a 'sine wave'?

Way back i was told the electrons in any given atom (CU) don't necessarily 'flow' , they charge each other via some big back/forth 'bump' to any given load...

The magnitude ,or peak, was simply the flux @ certain degree of generator rotation.

Am i close....??

How the physics of VD are further altered here escapes me , but i have to start somewhere


~RJ~

X2. I hope to broaden the discussion now that I understand the basics.



Anyone know how do you define a sine wave? Or know the equation?

 

[GoldDigger]

But that phase and amplitude does not behave in a "none linear" fashion? And a bit off topic... but does the reactance and capacitance in a transmission line have anything to do with maintaining or distorting a sine wave?

Keep those cards and letters coming in....
Yes. A constant reactive load, X + jY ohms, is linear because the current is proportional to the voltage over a very wide range. A real motor is approximated by a network of impedances whose resultant current varies with speed and slip, but is still linear unless you let the motor react t a change in applied voltage by changing speed.
And the sum of two linear functions of voltage is still a linear function.

The inductance and capacitance of a transmission line join together to make a characteristic impedance which is purely resistive. But if the line is not terminated at that impedance (and it never is) the result is standing waves that appear in the form of voltage and phase changes from one end to the other, but still linear even though frequency dependent.
No major effect on the sinusoidal nature of the waveform.

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[mbrooke]

Keep those cards and letters coming in....
Yes. A constant reactive load, X + jY ohms, is linear because the current is proportional to the voltage over a very wide range. A real motor is approximated by a network of impedances whose resultant current varies with speed and slip, but is still linear unless you let the motor react t a change in applied voltage by changing speed.
And the sum of two linear functions of voltage is still a linear function.

The inductance and capacitance of a transmission line join together to make a characteristic impedance which is purely resistive. But if the line is not terminated at that impedance (and it never is) the result is standing waves that appear in the form of voltage and phase changes from one end to the other, but still linear even though frequency dependent.
No major effect on the sinusoidal nature of the waveform.

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But- what about the case of the 3000 mile transmission line where its said to be impractical because all the energy would radiate off into space?

 

[GoldDigger]

Everytime i read the EE threads here my head hurts ,and i end up with more Q's than A's....yet i am compelled to risk a relevant (i hope) Q....

Just what is a 'sine wave'?

Way back i was told the electrons in any given atom (CU) don't necessarily 'flow' , they charge each other via some big back/forth 'bump' to any given load...

The magnitude ,or peak, was simply the flux @ certain degree of generator rotation.

Am i close....??

How the physics of VD are further altered here escapes me , but i have to start somewhere


~RJ~

The conduction electrons in a metal are not bound to a particular atom, but float around freely through the crystal structure of the metal.
Solid state physics works really hard to explain that, but you should just take it as given.

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[GoldDigger]

But- what about the case of the 3000 mile transmission line where its said to be impractical because all the energy would radiate off into space?

If you use a coaxial line (or an approximation to one) the losses are more related to resistance than to any radiation, particularly at power line frequencies. But the losses are lower with DC.

AFAIK the practical problems have more to do with phase control than with energy losses.

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[mbrooke]

If you use a coaxial line (or an approximation to one) the losses are more related to resistance than to any radiation, particularly at power line frequencies. But the losses are lower with DC.

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I may dispute that... Its been argued that long lines become impractical even at 60Hz due to radiation- at least what I have heard- either that HVDC proponents got me good.


FWIW, I am not sure about its accuracy- but this is rather interesting in relation to the discussion:

https://www.google.com/?gws_rd=ssl#q=alex+mceachern+power+quality+teaching+toy

 

[mbrooke]

The conduction electrons in a metal are not bound to a particular atom, but float around freely through the crystal structure of the metal.
Solid state physics works really hard to explain that, but you should just take it as given.

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Only in the presence of a magnetic field correct? I am all ears though about conduction electrons :happyyes::thumbsup:

 

[GoldDigger]

Only in the presence of a magnetic field correct? I am all ears though about conduction electrons :happyyes:

Nope. Magnetic field has nothing to do with it (except for niceties like Hall Effect and skin effect.)

The outermost electrons in each atom spread out (share) across neighboring atoms, causing the force that holds the metal together.
An electron with enough energy to get into those top quantum shells can move freely between the locations of different atoms in the crystal structure.

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