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Topic Title: Direction of power flow for a single phase system
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Created On: 16 April 2012 07:55 PM
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 16 April 2012 07:55 PM
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AdrianWint

Posts: 265
Joined: 25 May 2006

A topic in another forum has got me thinking....

I know how to calculate the direction of power flow in a three phase system by using the quadrature voltage (ie. for red phase you polarise by using the yel-blue volts).

But.... how is direction determined with a single phase system? Its obviously possible but how does the instrument achieve this?

Adrian
 16 April 2012 08:48 PM
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ArthurHall

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Conventional electromechanical kwh meters would spin backwards if the current flow was reversed. A shading ring was used to provide quadrature. Most text books or google give a more detailed description.
 16 April 2012 09:08 PM
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AdrianWint

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Thanks for the insight, Arthur.

I guess, I'm thinking more about a modern digital multifunction power meter rather than a revenue meter, but the principles must still be the same, abiet in software.

Or even a reverse power relay....how does it reach the conclusion that the power flow is indeed reverse?

Adrian
 16 April 2012 10:16 PM
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HKafeman

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Adrian

Simply, it is determined by the phase angle between the voltage and current - i.e. lag or lead.

e.g. -90 to +90 deg = power consumed, +90 to -90 = power delivered

A modern digital meter samples current and voltage many times per cycle.

I hope that helps.

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 16 April 2012 10:47 PM
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AdrianWint

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Henry,

Thanks for that..... so, if it working on the angle between V & I, how does the instrument distinguish between import/export of real power & an inductive/capacitive load?
 17 April 2012 07:56 AM
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AdrianWint

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OK, I think I may have seen the light in the shower this morning......

Using the instantaneous values of voltage & current we can calculate the apparent power, S.

Using the zero crossing points of both waveforms we can determine the time difference & hence the phase angle between V & I....

We can now construct the power triangle & hence determine Q & P. Using the determined angle we can also place the triangle in the correct quadrant of the power circle & thus determine the direction of flow of both P & Q by convention.

Does this sound right?
 17 April 2012 08:48 AM
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ectophile

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I don't see why you should see any difference in phase if the load is resistive - shouldn't the voltage and current then be in phase?

I would have thought the answer was a lot simpler. A conventional DC ammeter puts a small resistance in series with the load and measures the voltage across it. If we also put a voltmeter across the supply, then we can work out the power flow as well. If the voltage is positive and the current positive then power is going in, and the same goes if the voltage is negative and the current negative. But if the voltage is positive and the current negative, or vice-versa, then power is going out.

In the AC case, you just need to sample several times per cycle and you can make the same calculation.

Edit: Or, to put it another way, if the phase difference between voltage and current is less than 90 degrees, positive or negative, then the power flow is positive (or inwards). If it's greater than 90 degrees (positive or negative) then the power flow is negative (or outwards).

-------------------------
S P Barker BSc PhD MIET

Edited: 17 April 2012 at 01:28 PM by ectophile
 31 May 2012 12:44 PM
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tonyfisher123

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This should be quite simple: Imagine you are working in software (as you will be for a digital solution).

Define a repetitive loop to be repeated endlessly at a specified rate (many times per cycle).

On each iteration:
1. Read the instantaneous voltage and current. Because this is AC, these can each be positive or negative.
2. Multiplty (voltage) * (current) to obtain the instantaneous power value. This will be positive or negative.
3. Multiply the instantaneous power by the time interval between readings to obtain instantaneous energy flow (in joules = watt-seconds or watt-hours, depending on the units of measurement)
4. Build a running total of energy flow (just add up successive results). This will show a steady progression according to average power, plus a fluctuating component due to harmonic effects. It will be positive or negative as circumstances demand (direction of average power flow).

That's all

This method does not concern itself with phase angles, and even works when the voltage or current waveform is non-sinusoidal (or even when both are). In the non-sinusoidal case, it is impossible to define an unequivocal phase-displacement angle, anyway.
 06 June 2012 02:55 PM
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aroscoe

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Thats pretty much spot on. All I would add is that for a single-phase system the power flow actually fluctuates from 0 to twice the average, 100 times a second (for a 50Hz frequency), since P=VI, where V is proportional to sin(wt) and also I is proportional to sin(wt) for a perfect ohmic load. Therefore you get an average DC component of power flow but also an equal quantity of 2nd harmonic on your measurement. The sum of these components results in a power flow cycling from 0 to twice the average and back, at twice nominal frequency. So, for example, a 3kW kettle actually has a power flow which varies from 0W to 6kW and back to 0W, 100 times a second. To get a steady power flow readout you need to digitally filter the output a bit - averaging over 1/f is a great way to do this, and it gets rid of all the higher-order ripples due to harmonics and non-linear loads too.

Also, in a meter system, sometimes you need to account for sensor bandwidth, especially when metering modern lighting, computer, audio-visual equipment etc, since these can draw high order harmonics and the sensor frequency response sometimes needs to be accounted for - then you need to measure convert the measurements into the frequency domain (spectrum analyser approach) and correct for the sensor behaviour.







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 19 June 2012 12:28 PM
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pulse90

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Hi,
you don't need to measure things at two different positions along the wire to determine the power flow. You can do this with measurements taken at a single position. You must measure both current and voltage. The measuring device is built so that it rotates based on the relative phase of the current and voltage; and the direction of rotation corresponds to the direction of power flow. That's what happens in the power meter connected to your house. If you couldn't measure power flow this way it would be very hard for the electric utility to charge you for power.

-------------------------
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 29 September 2012 12:36 PM
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loyolacyriac

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the direction of power flow in a three phase system by using the quadrature voltage (ie. for red phase you polarise by using the yel-blue volts).


Hi
I just like to know about the method u said above .
How does thy work...for determining the direction of power flow in three phase

Pls explain
Thnx in advance

Rgds
Cyriac
 09 October 2012 04:34 PM
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timothyboler

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Originally posted by: tonyfisher123

3. Multiply the instantaneous power by the time interval between readings to obtain instantaneous energy flow (in joules = watt-seconds or watt-hours, depending on the units of measurement)

4. Build a running total of energy flow (just add up successive results). This will show a steady progression according to average power... It will be positive or negative as circumstances demand (direction of average power flow).


I don't understand how you could ever get a net-negative energy flow (i.e. power flowing backwards) with this method. Current and Voltage would have to be out of phase by greater than 90 degrees to get a net negative instantaneous power over a cycle. This is impossible with a conventional electrical system.

Put it another way imagine I had two single phase circuits with pure resistive loads but with the load connected at different ends so the direction of power flow was opposite in each circuit. How would my power meter know that the energy was flowing in different directions?

Regards, Tim

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 10 October 2012 08:53 AM
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ectophile

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It's down to how you connect the meters. The meters must be able to read both positive and negative currents/voltages. That needs to be instantaneous voltage or current for the AC case.

Suppose we arrange the circuit so that a load on the right gives positive voltage and positive current - and negative voltage and negative current on the other half-cycle of AC.

Now, if the load and source are swapped, a positive voltage will give a negative current, and vice-versa. To our meters, the voltage and current appear to be 180 degrees out of phase. So we know the power is going the other way.

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S P Barker BSc PhD MIET
 11 October 2012 08:53 AM
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timothyboler

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Ok, that makes sense. I think my example was probably misleading though because by switching the load/source around all we are doing is effectively reversing the current phase relative to the voltage.

What if instead I had two single phase induction motors connected together; one running as a generator and one running as a motor. If I then connected the prime mover to the other motor and ran that as the generator (therefore reversing the flow of power) how does the meter respond?

Regards, Tim

EDIT: Ok my small brain has worked it out now. The current with respect to the original reference voltage would indeed appear anti-phase in my motor example. What's interesting though is that unless you knew the original direction of power flow (and arranged your CT accordingly) you wouldn't be able to tell which way the energy was flowing via the meter - just that it was flowing in a particular direction. The meter will only tell you the power quadrant relative to the reference voltage which is arbitrarily chosen by the direction of the CT is placed.

-------------------------
Everyone loves a fireman - but hates the fire inspector.

Edited: 16 October 2012 at 04:07 PM by timothyboler
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