Power Question

[colinsk]

I have been searching for a clear answer to this question for a few weeks and I am not convinced I understand the limitations of P=IV. If I make a circuit that clips the sine wave at zero crossing I can create a circuit that breaks the P=IV equation. This leads me to a question: If i use a boost transformer and boost a 208V 120phase voltage (from a three phase source) to 240V 120phase and attach it to a resistive load, will it draw the same power as a 240V 180phase (from a split phase transformer) power supply? Sorry for the basic question but it is driving me crazy.

 

[gar]

181015-1952 EDT

colinsk:

On an instantaneous basis p = i*v is valid for any circuit and waveform.

P = I*V implies an average over some time period, and is valid only for certain conditions.

First, the statistics must be stationary. This means that the statistical properties remain constant with time. So the average value remains constant, the standard deviation is constant, the probability distribution is constant, etc.

Second, if the waveform is periodic, then averaging is performed over an integral number of cycles.

Third, the integral of the instantaneous product of i*v with respect to time when averaged is a constant. Averaging needs to be over an integral number of cycles, or over a large enough number of cycles with random start and end times that the error is less than some desired value.

Fourth, if the load is resistive, then the values of I and V must be RMS.

Other conditions or modifiers can exist for various waveforms and loads.

.

 

[Jraef]

I have been searching for a clear answer to this question for a few weeks and I am not convinced I understand the limitations of P=IV. If I make a circuit that clips the sine wave at zero crossing I can create a circuit that breaks the P=IV equation. This leads me to a question: If i use a boost transformer and boost a 208V 120phase voltage (from a three phase source) to 240V 120phase and attach it to a resistive load, will it draw the same power as a 240V 180phase (from a split phase transformer) power supply? Sorry for the basic question but it is driving me crazy.

Your circuit doesn't "break" the P=IV equation. Other parts of your statements are confusing, or indicative of confusion on your part. You are seeming to conflate phase angle with voltage / power. Voltage and power doesn't give a rat's butt what the phase angle is, 120, 180 etc., irrelevant.

What IS relevant is the nature of this LOAD. Not all loads are created equal. The P=IV equation is only true for purely resistive loads. P (power, as in watts) is actually I x V x PF (Power Factor), but with a purely resistive load, the PF = 1, so it becomes irrelevant and is often left out. But if the load is in any way inductive, then the PF is no longer irrelevant. But that doesn't address your question.

Looking at just the purely resistive load in order to fit with your question, there is another mitigating factor, the Resistance, R. Yes, P = IV, but I and V are influenced by R, in that I = V/R, and V = I x R. If you change the V on your load that is purely R, and assuming the load remains at the the SAME value of R, then P = V² x R, meaning the P actually increases, just not as a direct 1:1 ratio of the increase in V. I will also increase, because I = V/R.

This by the way is why Ohm's law is often depicted as a "wheel". You really can't change one thing without considering the effects of the others.

 

[GoldDigger]

181015-1952 EDT

colinsk:

On an instantaneous basis p = i*v is valid for any circuit and waveform.

P = I*V implies an average over some time period, and is valid only for certain conditions.

First, the statistics must be stationary. This means that the statistical properties remain constant with time. So the average value remains constant, the standard deviation is constant, the probability distribution is constant, etc.

Second, if the waveform is periodic, then averaging is performed over an integral number of cycles.

Third, the integral of the instantaneous product of i*v with respect to time when averaged is a constant. Averaging needs to be over an integral number of cycles, or over a large enough number of cycles with random start and end times that the error is less than some desired value.

Fourth, if the load is resistive, then the values of I and V must be RMS.

Other conditions or modifiers can exist for various waveforms and loads.

.

Another way of looking at it is that P=IV is representative of the instantaneous equation P(t)=I(t)V(t) which is valid at any instant in time, t. That means that the integral of that P(t) over any period of time will be the energy transferred over that period of time. It is a fundamental property of the integration operation that the integral of the product of two functions is not necessarily equal to the product of the individual integrals over the same time period.
For periodic functions, particularly sine waves, you can make specific useful generalizations, among them those described by gar.

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[jumper]

All that is way too complicated. kW is kW.

 

[LarryFine]

If i use a boost transformer and boost a 208V 120phase voltage (from a three phase source) to 240V 120phase and attach it to a resistive load, will it draw the same power as a 240V 180phase (from a split phase transformer) power supply? Sorry for the basic question but it is driving me crazy.

In theory, yes, the same overall power used would equal the power delivered.

In practice, transformers are not 100% efficient. Some power goes toward heating a buck-boost.

 

[Besoeker]

I have been searching for a clear answer to this question for a few weeks and I am not convinced I understand the limitations of P=IV. If I make a circuit that clips the sine wave at zero crossing I can create a circuit that breaks the P=IV equation. This leads me to a question: If i use a boost transformer and boost a 208V 120phase voltage (from a three phase source) to 240V 120phase and attach it to a resistive load, will it draw the same power as a 240V 180phase (from a split phase transformer) power supply? Sorry for the basic question but it is driving me crazy.

For non-sinusoidal waveforms, I normally plot the instantaneous values of voltage and current on a spread sheet at one degree intervals. multiply (V*I), and average over a cycle. I've routinely done that at the bid stage where I have had to provide guaranteed NNT figures for system efficiencies etc.
Getting those wrong would have incurred stinging financial penalties.

Just my take. Others may have different suggestions.

 

[LMAO]

Your circuit doesn't "break" the P=IV equation. Other parts of your statements are confusing, or indicative of confusion on your part. You are seeming to conflate phase angle with voltage / power. Voltage and power doesn't give a rat's butt what the phase angle is, 120, 180 etc., irrelevant.

What IS relevant is the nature of this LOAD. Not all loads are created equal. The P=IV equation is only true for purely resistive loads. P (power, as in watts) is actually I x V x PF (Power Factor), but with a purely resistive load, the PF = 1, so it becomes irrelevant and is often left out. But if the load is in any way inductive, then the PF is no longer irrelevant. But that doesn't address your question.

Looking at just the purely resistive load in order to fit with your question, there is another mitigating factor, the Resistance, R. Yes, P = IV, but I and V are influenced by R, in that I = V/R, and V = I x R. If you change the V on your load that is purely R, and assuming the load remains at the the SAME value of R, then P = V² x R, meaning the P actually increases, just not as a direct 1:1 ratio of the increase in V. I will also increase, because I = V/R.

This by the way is why Ohm's law is often depicted as a "wheel". You really can't change one thing without considering the effects of the others.
View attachment 21383

you have a typo. P=V² / R

 

[ptonsparky]

you have a typo. P=V² / R

it was a test to see if you read the whole thing. You passed.

 

[Jraef]

you have a typo. P=V² / R

Brain fart... It was right there in front of me and I still wrote it backward...