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Topic Title: Volt Drop 3 Phase unbalanced
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Created On: 09 December 2013 12:50 PM
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 09 December 2013 12:50 PM
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Chris123

Posts: 314
Joined: 08 February 2010

Hi all

If i have a 3 Phase distribution circuit which is not balanced, what is the correct way of calculating volt drop?

So if Ia = 100 at 0 degree P.f 1
Ib = 50 at -120 degree P.f 0.8
Ic =[ 40 at -240 degree P.f 0.7

I get a Neutral Current of 66.73 at 16.44 degree P.f 0.95

So should i add the voltage dropped in the neutral to the three individual lines and base it on a percentage of 230 V and not 400V as its now referenced to the neutral or is there another way of approaching it.

Many Thanks
Chris
 09 December 2013 03:56 PM
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Zuiko

Posts: 521
Joined: 14 September 2010

Forget about the power factor, it is a confusion.

Use

V = IZ

The drop in each phase will be the same as the magnitude of the source. So if the source is 230V; then the total drop is 230V.

So, you need to know the values of Z in the components in which you are calculating the volt drop. Is it the supply cables? Or the loads? Or both.





BTW
The currents you give are confusing?
I assume the angles are
Va = 0 degrees
Vb = 120 degrees
Vc = 240 degrees?

Are the current angles in reference to these? i.e. is Ib at -120 to Vb? If so the power factor is 0.5. Assuming the current and voltage waveforms are sinusodial, PF is the phase angle difference between the two.
 09 December 2013 04:47 PM
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Chris123

Posts: 314
Joined: 08 February 2010

Originally posted by: Zuiko

Forget about the power factor, it is a confusion.



Use



V = IZ



The drop in each phase will be the same as the magnitude of the source. So if the source is 230V; then the total drop is 230V.



So, you need to know the values of Z in the components in which you are calculating the volt drop. Is it the supply cables? Or the loads? Or both.











BTW

The currents you give are confusing?

I assume the angles are

Va = 0 degrees

Vb = 120 degrees

Vc = 240 degrees?



Are the current angles in reference to these? i.e. is Ib at -120 to Vb? If so the power factor is 0.5. Assuming the current and voltage waveforms are sinusodial, PF is the phase angle difference between the two.


Thanks for the reply

Yes i can see i the confusion caused


The Currents are

Ia = 100 at P.f 1
Ib = 50 at P.f 0.8
Ic =40 at P.f 0.7

Ia the reference phase at 0 Degrees, then -120, -240

I get a Neutral Current of 66.73 at 16.44 degree P.f 0.95

So my question is do i calculate using three phase mV/A/m or single phase and should i base it on % of 230V, so as you say just add the VD of the neutral to the volt drop of the individual Lines.

Thanks again
 09 December 2013 05:30 PM
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Zuiko

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I'm still not sure I understand what the question is?

If your supply voltage, Vsupply = 230V then your volt drop, Vdrop, assuming the neutral is at 0V, is also 230V.

What you need to know the is impedance of the various loads, and what share of 230V is dropped across each load. You haven't given the impedance of the loads, so there is no way of answering this question!


If you consider Ohm's Law

V = IZ

If you are calculating the volt drop across the suppy cable then because the loads are different per phase, the current is different per phase (as given), and so the volt drop per phase, over the cable is different. But you need to know the impedance of the cable!

If you assume a unit resistance (no reactance) for the cable, then the magnitude of the volt drop in ConductorB is half that of ConductorA; and the volt drop in ConductorC is 40% of conductorA.
 09 December 2013 05:44 PM
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Chris123

Posts: 314
Joined: 08 February 2010

Originally posted by: Zuiko

I'm still not sure I understand what the question is?



If your supply voltage, Vsupply = 230V then your volt drop, Vdrop, assuming the neutral is at 0V, is also 230V.



What you need to know the is impedance of the various loads, and what share of 230V is dropped across each load. You haven't given the impedance of the loads, so there is no way of answering this question!





If you consider Ohm's Law



V = IZ



If you are calculating the volt drop across the suppy cable then because the loads are different per phase, the current is different per phase (as given), and so the volt drop per phase, over the cable is different. But you need to know the impedance of the cable!



If you assume a unit resistance (no reactance) for the cable, then the magnitude of the volt drop in ConductorB is half that of ConductorA; and the volt drop in ConductorC is 40% of conductorA.


The mV/A/m values in BS7671 are either single phase so current in the neutral or 3 phase assuming balanced.

The distribution is a 3 phase circuit though has neutral current due to being unbalanced.

If it were balanced i could use the 3 phase mV/A/m values and design to a 1 % of 400V .

So im looking as to what is the most accurate way of determining volt drop in an unbalanced 3 phase circuit.

Thanks again
 09 December 2013 05:53 PM
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Zuiko

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Originally posted by: Chris123
So im looking as to what is the most accurate way of determining volt drop in an unbalanced 3 phase circuit.


Volt drop across what? The cable? Or the load?
 09 December 2013 09:09 PM
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Chris123

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Originally posted by: Zuiko

Originally posted by: Chris123

So im looking as to what is the most accurate way of determining volt drop in an unbalanced 3 phase circuit.




Volt drop across what? The cable? Or the load?



Volt drop across the cable.

Thanks again
 10 December 2013 10:55 AM
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Fame

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Volt drop is relative to the phase in this case, so surely just calculate mV/A/m for each phase giving you the VD for each phase?
 10 December 2013 11:12 AM
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Chris123

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Hi

Loads:

Ia = 100 Amps PF 1 (2.3 + j0)
Ib = 50 Amps PF 0.8 ( 0.56 + j4.57)
Ic = 40 Amps PF 0.7 (3.99 + j4.14)

Va 0, Vb -120, Vc -240

I calculate a Neutral Current of 66.73Amps at 16.44 degree P.f 0.95

I haven't the mV/A/m for the Distribution cable, what im looking to determine is the correct and most accurate way of determining volt drop in this situation.

The mV/A/m values in BS7671 are either single phase so current in the neutral or 3 phase assuming balanced and no Neutral current.

The distribution is a 3 phase circuit though has neutral current due to being unbalanced.

If it were balanced i could use the 3 phase mV/A/m values and design to a 1 % of 400V .

So which mV/A/m value should be used and what approach is best suited to this scenario.

Many thanks
 10 December 2013 02:55 PM
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hpcompaq

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Joined: 27 September 2007

Hello Chris.

I think your 50A load has a power factor nearer to 0.121.

If you want to use the tables then you should use the single phase mV/A/m values because your loads are unbalanced.

It depends what you mean by "volt drop". Because the neutral will be at a potential above "0" and at a phase angle (relative to the A phase as you have calculated it) then it is probably best to calculate the load voltages, for example using J Millman, and then subtracting these individually to determine the "volt drop". This will be different from the "volt drop" that you would obtain by multiplying the conductor impedances by the load currents.
 10 December 2013 05:39 PM
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Zuiko

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Originally posted by: hpcompaq
I think your 50A load has a power factor nearer to 0.121.


I think that is because the original data is a little vague (just a little!)

The reference angle for the yellow phase is 120° so a power factor of 0.8 will give a angle leading or lagging that of 37°.

i.e. the yellow phase current is at 157° or 83° as referenced to the red phase voltage.
 10 December 2013 05:55 PM
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Zuiko

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Originally posted by: Chris123
I haven't the mV/A/m for the Distribution cable, what im looking to determine is the correct and most accurate way of determining volt drop in this situation.


Can you not see the contradiction here?

You cannot accurately determine, or even guess, the volt drop unless you know the impedance of the cable!


Yo have so little data it is impossible to model anything without making assumptions.

Let's make some reasonable assumptions, so you can at least get started.

1) The currents quoted are the currents through the phase windings of the distribution transformer.
2) The voltage at the head of the cable is 240V per phase
3) The load is star connected to a star wound TX
4) The neutral is at 0V
5) The load is simply connected in series with the cable - i.e. you have a three phase cable, and at the end of each conductor is an impedance. These three impedances are different giving the different phase currents.
6) The impedance of the cable does not vary with load nor temperature. It is fixed at all times.


We have to make one further assumption - which is the main one - and that is the impedance of the cable! Let's assume it is purely resistive and is equal to 1ohm.


Start with the Red Phase
The total volt drop, from the output of the TX through the cable, through the load to the neutral is 240V. Using Kirchoffs Law and working through the calculation in steps

Total Volt Drop = Volt Drop Across Cable + Volt Drop Across Load

Vdrop = (Ia * Zcable) + (Ia * Zaload)

240V = (100A * 1 ohm) + (100A * Zaload)

240V = 100 V + (100A * Zaload)

Therefore the volt drop across the cable is 100V and the volt drop across the load is 140V.

Now the Yellow Phase:
The yellow phase current is 40+30J in reference to the yellow phase voltage
240V = (40+30j * 1 ohm) + (40+30j A * Zbload)
240V = 40+30j V + (40+30j A * Zbload)

Therefore the volt drop across the cable is 40+30j V (50V)

Rearranging for Zbload we get the volt drop across the load as 200-30JV (202V)


Blue Phase
You can work that one out.

You should get a cable drop of 40V and a load drop of 214V.


cheers
W

Edited: 10 December 2013 at 06:01 PM by Zuiko
 10 December 2013 06:34 PM
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Chris123

Posts: 314
Joined: 08 February 2010

Thankyou for Your replies

Ill go over check my calcs, had numbers everywhere

Hpvompac, thanks ill have a nosey at millmans therom and see if I can fathom it out, Thankyou

Zuikio, sorry for the confusion, I realise I needed the cable mV/A/m, it was more the method to use more so. Thanks for your in depth reply ill have read and digest it all.

Thanks again
Chris
 10 December 2013 09:07 PM
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hpcompaq

Posts: 78
Joined: 27 September 2007

I have read the OP as a 4 wire circuit and therefore the voltage across the load must be the difference between the source voltage and the common neutral voltage - this is a vector relationship.
Power factor of the load is R/Z - the B load of is therefore 0.121.
The various load voltages are best worked out using Millman or Maxwell. A bit long winded but you could always use superposition.
Best regards.
 10 December 2013 09:10 PM
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hpcompaq

Posts: 78
Joined: 27 September 2007

I have read the OP as a 4 wire circuit and therefore the voltage across the load must be the difference between the source voltage and the common neutral voltage - this is a vector relationship.
Power factor of the load is R/Z - the B load of is therefore 0.121.
The various load voltages are best worked out using Millman or Maxwell. A bit long winded but you could always use superposition.
Best regards.
 10 December 2013 09:16 PM
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hpcompaq

Posts: 78
Joined: 27 September 2007

I have read the OP as a 4 wire circuit and therefore the voltage across the load must be the difference between the source voltage and the common neutral voltage - this is a vector relationship.
Power factor of the load is R/Z - the B load of is therefore 0.121.
The various load voltages are best worked out using Millman or Maxwell. A bit long winded but you could always use superposition.
Best regards.
 11 December 2013 06:51 AM
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Zuiko

Posts: 521
Joined: 14 September 2010

yes, the original question was vague, but it was cleared up later...the three references were 0, -120, -240 and the power factors 1, 0.8 and 0.7 respectively.

You don't really need to use superposition unless you have loads across different voltages, it adds needless complications.

I think using plain old Ohms Law, V = IZ, as above, is the easiest. It only looks long winded because every stage of the calc was broken down. You can do it on a calculator in 30 seconds.

cheers
W

Edited: 11 December 2013 at 10:27 AM by Zuiko
 11 December 2013 03:16 PM
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hpcompaq

Posts: 78
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Hello W.

I think that there may be an error in your method to answering this question.

You posted

"Start with the Red Phase
The total volt drop, from the output of the TX through the cable, through the load to the neutral is 240V. Using Kirchoffs Law and working through the calculation in steps

Total Volt Drop = Volt Drop Across Cable + Volt Drop Across Load

Vdrop = (Ia * Zcable) + (Ia * Zaload)

240V = (100A * 1 ohm) + (100A * Zaload)

240V = 100 V + (100A * Zaload)

Therefore the volt drop across the cable is 100V and the volt drop across the load is 140V"

The load Z is 2.3 ohms. This would therefore give a load voltage of 230V as opposed to 140V.

Best regards.
 11 December 2013 05:00 PM
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Zuiko

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Originally posted by: hpcompaq

Hello W.

I think that there may be an error in your method to answering this question..


The numbers used were only for illustration.

I assumed a voltage of 240 V at the transformer, which is reasonable,(point 2 above) and a cable impedance of 1 ohm (which is an easy number to multiply!)

Whilst the parameters used may not be realistic, there is no error in the method!

cheers
W
 11 December 2013 05:11 PM
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hpcompaq

Posts: 78
Joined: 27 September 2007

Hello W.
Again, if using your method, the voltage across the yellow phase load would be 230V not 202V. The correct method needs to allow for the magnitude and the vector relationship of the displaced load neutral.
Best regards.
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