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

[mbrooke]

Ok, I apologize if this is not the right place; as the question is purely a theoretical one- and not job related. However, I was asked this question by an EE student and I don't know how to answer it. Apologies ahead of time if its not well worded.


Picture a DC circuit first. 12 gauge THHN, 100 volts, 15 amperes.

At 50 feet the voltage drop is 2.38% resulting in 97.62 volts at the load.

Then consider 2000 feet where the voltage drop is 95.3% at 4.7 volts at the load.


Same wire, same 15 amp, same 2000 feet now at 10,000 volts:

Voltage drop is 0.95% and 9904.7 volts at the load.


So we can conclude a huge voltage discrepancy in voltage drop based on the voltage. The % voltage at the source remains the same for both 100 volts and 10,000 volts, but the % voltage at the load varies tremendously based on the source magnitude. I can understand this and it makes sense for a DC circuit. I can also conclude that for any given voltage, the longer the run the more % drop.


Now picture an AC circuit. At any point in time the "DC" voltage (snap shot frozen in time) varies in magnitude relative to other points in time of the sine wave. In essence a 2000 foot circuit should behave like the 100 volt DC version near the bottom of the sine wave giving 95.3% drop, yet perform like a 10,000 volt circuit at the 10,000 volt peak giving a 0.95% voltage drop. The AC RMS voltage would be 7000 volts at the source.

Now, wouldn't the difference in % drop cause cause a different sine wave to appear at the load when compared to the source? :blink::?

 

[Barbqranch]

The misconception is that the current doesn't stay at 15 amps throughout the sine wave, 15 amps is just the RMS current.

 

[mbrooke]

The misconception is that the current doesn't stay at 15 amps throughout the sine wave, 15 amps is just the RMS current.

:slaphead::slaphead: Good point.

But what grantees that current remains exactly at the same magnitude as the voltage in all portions of the sine wave? Don't some loads draw a bit more current at lower voltages then ohms law would dictate?

 

[GoldDigger]

Your error in math is in the assumption that 15A is flowing at each moment in the sine wave.
It is not.
For a linear (let's even say resistive) load the current is proportional to the voltage at the load. Thus the % drop ends up being the same at each time.
And a constant times a sine wave is a sine wave
If you had a non linear load, the load voltage sine wave would indeed be distorted.

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[Carultch]

Ok, I apologize if this is not the right place; as the question is purely a theoretical one- and not job related. However, I was asked this question by an EE student and I don't know how to answer it. Apologies ahead of time if its not well worded.


Picture a DC circuit first. 12 gauge THHN, 100 volts, 15 amperes.

At 50 feet the voltage drop is 2.38% resulting in 97.62 volts at the load.

Then consider 2000 feet where the voltage drop is 95.3% at 4.7 volts at the load.


Same wire, same 15 amp, same 2000 feet now at 10,000 volts:

Voltage drop is 0.95% and 9904.7 volts at the load.


So we can conclude a huge voltage discrepancy in voltage drop based on the voltage. The % voltage at the source remains the same for both 100 volts and 10,000 volts, but the % voltage at the load varies tremendously based on the source magnitude. I can understand this and it makes sense for a DC circuit. I can also conclude that for any given voltage, the longer the run the more % drop.


Now picture an AC circuit. At any point in time the "DC" voltage (snap shot frozen in time) varies in magnitude relative to other points in time of the sine wave. In essence a 2000 foot circuit should behave like the 100 volt DC version near the bottom of the sine wave giving 95.3% drop, yet perform like a 10,000 volt circuit at the 10,000 volt peak giving a 0.95% voltage drop. The AC RMS voltage would be 7000 volts at the source.

Now, wouldn't the difference in % drop cause cause a different sine wave to appear at the load when compared to the source? :blink::?

Interesting question.

Consider three functions of time:
V0(t), the voltage at the source
VL(t), the voltage at the load
I(t), the current in the circuit

And constant values:
RT, the transmission resistance
RL, the load resistance, considering a simple resistive load

Resistance does somewhat depend on current, due to thermal effects, but the termperature variation settles on a steady state value, so we'll assume the resistance is at a steady value too. If there were any distortion in a simple two-resistor circuit like this, this would be the source of it.

Voltage drop:
V0(t) - VL(t) = I(t)*RT

Current relation to voltage:
I(t) = V0(t)/(RL + RT)

Combine the above:
V0(t) - VL(t) = (V0(t)/(RL+RT))*RT


Solve for VL(t):
VL(t) = V0(t) - (V0(t)/(RL+RT))*RT

Factor:
VL(t) = V0(t)*(1 - RT/(RL + RT))

Observe that we have VL(t) = V0(t) times a constant. We also have I(t) equal to V0(t) divided by a constant. All functions of time, are proportional to V0(t), so there is no distortion.

 

[mbrooke]

Interesting question.

Consider three functions of time:
V0(t), the voltage at the source
VL(t), the voltage at the load
I(t), the current in the circuit

And constant values:
RT, the transmission resistance
RL, the load resistance, considering a simple resistive load

Resistance does somewhat depend on current, due to thermal effects, but the termperature variation settles on a steady state value, so we'll assume the resistance is at a steady value too.
Voltage drop:
V0(t) - VL(t) = I(t)*RT

Current relation to voltage:
I(t) = V0(t)/(RL + RT)

Combine the above:
V0(t) - VL(t) = (V0(t)/(RL+RT))*RT


Solve for VL(t):
VL(t) = V0(t) - (V0(t)/(RL+RT))*RT

Factor:
VL(t) = V0(t)*(1 - RT/(RL + RT))

Observe that we have VL(t) = V0(t) times a constant. We also have I(t) equal to V0(t) divided by a constant. All functions of time, are proportional to V0(t), so there is no distortion.

Awesome

I could be wrong, but doesn't that steady state for RL and RT not vary in real time? IE, any heating element cools off as the voltage drops leading to a lower impedance, especially where immersed in a liquid like water instead of air. Also, in any system, can't the resistance of RL change more than the resistance of RT for changes in current? I know I am throwing a heavy curve ball in the variables department.

 

[GoldDigger]

Awesome

I could be wrong, but doesn't that steady state for RL and RT not vary in real time? IE, any heating element cools off as the voltage drops leading to a lower impedance, especially where immersed in a liquid like water instead of air. Also, in any system, can't the resistance of RL change more than the resistance of RT for changes in current? I know I am throwing a heavy curve ball in the variables department.

We can usually assume that any large chunk of metal that is conducting a sine wave electric current stays at a relatively constant temperature over the period of a 60Hz sine wave.
The DC stable temperature model would be more complicated.

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[mbrooke]

Your error in math is in the assumption that 15A is flowing at each moment in the sine wave.
It is not.
For a linear (let's even say resistive) load the current is proportional to the voltage at the load. Thus the % drop ends up being the same at each time.
And a constant times a sine wave is a sine wave
If you had a non linear load, the load voltage sine wave would indeed be distorted.

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Ok, going down the road of science fiction to have a theoretical reference point... :angel: Lets say I have a none linear load that pulled exactly 15 amps at all points in a sine wave... Would it be possible to know what the wave would look? Im guessing the bell like curve at the source would look like a sharp, narrow cone at the load?

 

[mbrooke]

We can usually assume that any large chunk of metal that is conducting a sine wave electric current stays at a relatively constant temperature over the period of a 60Hz sine wave.
The DC stable temperature model would be more complicated.

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Why would the DC model be more complicated? Id imagine at 60Hz the temperature would vary far more than with a constant DC current.

 

[GoldDigger]

Ok, going down the road of science fiction to have a theoretical reference point... :angel: Lets say I have a none linear load that pulled exactly 15 amps at all points in a sine wave... Would it be possible to know what the wave would look? Im guessing the bell like curve at the source would look like a sharp, narrow cone at the load?

That theoretical load would have an interesting behavior in the vicinity of zero volts!
Imagine a DC offset near zero volts that suddenly reverses direction. Sort of a snap switch behavior or over centering.

Behavior near the peaks will be far less interesting!

Taking a second look, it becomes obvious that for your hypothetical load the voltage waveform is just the original source sine wave with a square wave of average voltage zero added to it.

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[Carultch]

Ok, going down the road of science fiction to have a theoretical reference point... :angel: Lets say I have a none linear load that pulled exactly 15 amps at all points in a sine wave... Would it be possible to know what the wave would look? Im guessing the bell like curve at the source would look like a sharp, narrow cone at the load?

V0(t) - VL(t) = I(t)*RT

Set I(t) = 15A, with an sign that matches V0(t), if you desire that the output current switch directions. Solve for VL(t)

VL(t) = V0(t) - 15A*RT*sign(V0(t))

 

[mbrooke]

That theoretical load would have an interesting behavior in the vicinity of zero volts!
Imagine a DC offset near zero volts that suddenly reverses direction. Sort of a snap switch behavior or over centering.

Behavior near the peaks will be far less interesting!

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Good one :lol: Id imagine it would cut out at say 4 volts dropping its current to zero.

 

[mbrooke]

V0(t) - VL(t) = I(t)*RT

Set I(t) = 15A, with an sign that matches V0(t), if you desire that the output current switch directions. Solve for VL(t)

VL(t) = V0(t) - 15A*RT*sign(V0(t))

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

 

[GoldDigger]

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

See the verbal description at the end of post #10.

Note that we punningly see that a *sign* wave is really a square wave.

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[mbrooke]

Taking a second look, it becomes obvious that for your hypothetical load the voltage waveform is just the original source sine wave with a square wave of average voltage zero added to it.

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Dumb this down for me a bit :dunce:

 

[GoldDigger]

Dumb this down for me a bit :dunce:

Take the positive half of the sine wave and lower it by a fixed voltage. That means that near each end it will actually cross the zero line and be negative.
Take the negative half of the sine wave and raise it by the same voltage.
The ends of the two will NOT meet, corresponding to the instant rising and falling edges of the square wave.

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[mbrooke]

Take the positive half of the sine wave and lower it by a fixed voltage. That means that near each end it will actually cross the zero line and be negative.
Take the negative half of the sine wave and raise it by the same voltage.
The ends of the two will NOT meet, corresponding to the instant rising and falling edges of the square wave.

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Even cutting out at 5 volts?

 

[GoldDigger]

Even cutting out at 5 volts?

Depends on whether you mean 5 volts at the source or 5V at the load!
In either case the simple square wave becomes a modified square wave with an interval of zero between + and - stages.
The actual load voltage waveform would look different depending on which end you apply the 5V reference to, but in either case there will now be four discontinuities (jumps) in the load voltage in each cycle instead of two.

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[mbrooke]

Depends on whether you mean 5 volts at the source or 5V at the load!
In either case the simple square wave becomes a modified square wave with an interval of zero between + and - stages.
The actual load voltage waveform would look different depending on which end you apply the 5V reference to, but in either case there will now be four discontinuities (jumps) in the load voltage in each cycle instead of two.

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At the load. And now that I think about it, as soon as those 5 volts cutoff at the load the voltage would shoot back up to 100 volts, possibly re-triggering the current draw- but for the argument lets say it does not until the polarity reverses.

 

[GoldDigger]

At the load. And now that I think about it, as soon as those 5 volts cutoff at the load the voltage would shoot back up to 100 volts, possibly re-triggering the current draw- but for the argument lets say it does not until the polarity reverses.

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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