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Topic Title: Where to apply power factor to a voltage drop calculation?
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Created On: 03 December 2013 05:20 PM
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 03 December 2013 05:20 PM
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earthing

Posts: 56
Joined: 17 December 2008

Just wondering if anyone can shed some light on this problem.

Page 60 of the electricians design guide states that power factor be applied to the tabulated value of (mV/A/M) so its on the top line of a voltage drop calculation. ie mV/A/m x L x cos Ø x Ib / 1000

I am working my way through all of the calculations and writing them down but i'm a bit confused now.

I am trying to work out how you calculate voltage drop in a three phase circuit with power factor correction.

However when I apply the power factor in the above way it seems to reduce the voltage drop.

I thought power factor increases voltage drop?

I have found another part of the design guide which gives the following equation for design current in a three phase system:

KVA or KW / ?3 x U x cos Ø

Now if I apply the power factor at this stage and then use the obtained Ib value already corrected for power factor the voltage drop increases.

Example:

Length = 60metres
Load = 3.5KW
mV/A/m for selected cable = 25
cos Ø = 0.89

P/1.732 x U = 3500/1.732 x 400 = 5.05 amps

Voltage drop = 25x0.89x60x5.05 / 1000 = 6.75 volts (as per page 60)

If you don't apply the power factor = 25x60x5.05 / 1000 = 7.59 volts

That seemed wrong so I tried another way.

P/1.732 x U x cos Ø = 3500/ 1.732 x 400 x 0.89 = 5.67 amps

Now using Ib of 5.67 instead of 5.05 gives voltage drop of:

25x60x5.67 / 1000 = 8.5 volts


I am trying to find the answer as I am preparing for my design exam on Thursday.

Which one is true?

Edited: 03 December 2013 at 10:12 PM by earthing
 03 December 2013 08:03 PM
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dlane

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Joined: 28 September 2007

I don't have a copy of the electricians design guide so I can't comment on that aspect of it but I would say that the power factor can only be applied to find the current from the power equation. It cannot be applied directly to the volt drop equation.

For single phase True Power (kW) = V x I x cos Ø. For a 3 phase circuit this value would then be multiplied by 1.732.

For a single phase Apparent Power (kVA) = V x I. Again multiplied by 1.732 if it is 3 phase.

If you think about it, a low power factor is equivalent to electrical inefficiency and gives a higher line current due to the higher kVA but the kW of the load will still be the same.

If you increase the power factor closer to 1, the line current drops and hence so will the kVA, but again the kW of the load will remain unchanged.

A higher line current give a higher voltage drop and overall will have a slightly higher kW due to the I^2 x R losses in the cable.

Kind regards

Donald Lane
 03 December 2013 08:27 PM
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earthing

Posts: 56
Joined: 17 December 2008

Hello Donald thanks for your reply.

What you say is the way I understand it too.

I notice you didn't apply power factor correction to the apparent power equation. Why is that?

Thanks, this gives me some confidence that the design guide has an error there and i'll stick with the theory you described.
 03 December 2013 09:18 PM
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Parsley

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See 7671 appendix 4 section 6.2 from memory.

The vd is lower because the cable and load are out of phase and less true power is actually consumed in the cable.

Amps =p/v x pf if your using va you have already included the pf to get va in the first place.

Regards
 03 December 2013 10:15 PM
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earthing

Posts: 56
Joined: 17 December 2008

Hi Parsley. I just read the section you referred to and it says exactly the same as the design guide but doesn't give an example.

So does Power factor improve voltage drop? Getting terribly confusing!

Thanks,

Dan
 03 December 2013 10:52 PM
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dlane

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Apparent power is the is the trigonometric sum of the true power and the reactive power in a circuit. It comes from what we refer to as the power triangle.

True power is what we turn into real work, heat, light or movement etc. Rective power is what is consumed by an electrical circuit to function.

For a power factor of 1, apparent power = true power and reactive power = 0 and therefore we can effectively ignore the power factor. Typically this would be a purely resistive circuit. At all other power factor values we multiply the apparent power by the power factor to get the true power.

If we assume the voltage is constant at a source altering the power factor is going to alter the current from that source.

The voltage drop on a cable is dependent upon the current flowing down that cable and the resistance of the cable, which is affected by the temperature, length, size etc of the cable. This is Ohms Law, V = I x R. All the power factor does is change the amount of current flowing down a cable, to my knowledge, it cannot be used directly in the voltage drop calculation.

So, improving the power factor, getting it closer to 1, would reduce the current and hence the voltage drop over the cable will be lower. If the power factor gets worse, i.e. decreases down to 0.8, then the current will increase to get the same amount of true power, due to the increase in reactive power, and the voltage drop will increase.

Kind regards

Donald Lane
 03 December 2013 11:33 PM
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earthing

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Joined: 17 December 2008

Feel like I just won the lottery!!! Although iv'e probably got it all wrong again. Been at this all day going around and around.........

Thanks for the input by the way.

Parsley and Donald you both mentioned KVA and that to get to KVA you must have accounted for power factor first. So I went back to my level 2 books and found the calculations and went through them.

The penny then dropped that the calculation I referred to - KVA or KW / 1.732 x U x cos Ø was incorrect.

The other reference from the electricians design guide I referred to was on page 50 of the Electricians design guide and it specifically states: KVA / 1.732 x U and KW / 1.732 x U x cos Ø

So when you calculate Ib you only take account of Pf when using KW.

But then as the regs and the design guide say, you must apply power factor correction to the tabulated values (mV/A/m) when calculating voltage drop with power factor correction.

So would I apply the power factor to obtain the design current and then again to obtain the voltage drop?

That would mean the following:

Line current (Ib) = KW/1.732 x U x cos Ø = 3500/1.732 x 400 x 0.89 = 5.67 amps

Voltage drop = (mV/A/m) x cos Ø x L x Ib / 1000 = 25 x 0.89 x 60 x 5.67 / 1000 = 7.56

Applying the power factor twice doesn't seem like a logical thing to do but thats what the indicated equations seem to suggest?

Thanks,

Dan

Edited: 04 December 2013 at 11:27 AM by earthing
 04 December 2013 12:05 AM
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Zuiko

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Originally posted by: earthing

I thought power factor increases voltage drop?


If you go back to very first principles, kirchoffs law tells us that the total voltage drop around a circuit equals the applied voltage.

This is true regardless of the power factor (ie you can't get something from nothing!)

The easiest way to work it out is to use

V = I Z

where I and Z are complex numbers.

You will see that as the real and imaginary part of the impedance Z (the resistance and the reactance respectively) varies; the current, I, will vary accordingly to ensure that Vdrop is constant. That is, if Vsource is 240V; then Vdrop is 240V.


edit: realised that 3-phase calc was not needed.

Edited: 06 December 2013 at 08:17 AM by Zuiko
 04 December 2013 11:07 AM
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Parsley

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Dan

As Donald stated poor PF results in more amps being drawn and more VA being required for the same KW of useful power and larger current carrying conductors, switchgear etc being required.

Where PF is not unity the VD is less and may allow a cable to be a bit longer but you might have already needed to select a larger cable to carry the extra current required due to the PF not being at unity.
Improving pf will be more beneficial as smaller conductors will be able to be selected as the Ib will be lower, which will also improve the VD and lower energy bills depending on metering tariffs as less reactive power is being supplied.

Also don't forget you can also correct VD for lower operating temperatures.

Regards
 04 December 2013 03:58 PM
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OMS

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OK chaps - we aren't answering the OP, and risk confusing him before an exam.

Page 60 of the electricians design guide states that power factor be applied to the tabulated value of (mV/A/M) so its on the top line of a voltage drop calculation. ie mV/A/m x L x cos Ø x Ib / 1000


It does and that is correct as it applying Cos theta to either mV/A/m or to mV/A/m z - using either of those values for cables above or below 16mm respectively will over estimate the voltage drop due to the resistance/reactance ratio (X/R)

If you apply the respective Cos theta and Sin theta to the R and X components of the tabulated voltage drops then with a range of power factors you should be getting a correct value and note that increasing current will increase voltage drop.

However when I apply the power factor in the above way it seems to reduce the voltage drop


The book is absolutely correct, it does, but you need to recognise the fact that it's being applied to a value of mV/A/m z that would have over estimated voltage drop. Be careful of drawing the wrong conclusion from what you are reading - terminology and the basis of data is as important to understand as the manipulation of that data.

I thought power factor increases voltage drop?


It does - but see above as to why it appears to reduce it when applied to values of mV/A/m z

So does Power factor improve voltage drop? Getting terribly confusing


Repeat the exercise for any random circuit using Cos theta applied to mV/A/m z and then using mV/A/m r and mV/A/m x and use Cos theta and Sin theta respectively and then vary the power factor - you'll soon see what is really happening.

Good luck

OMS

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Failure is always an option
 04 December 2013 07:34 PM
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earthing

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Hi OMS thanks for your reply. Having to prioritise what I'm studying as my exam is tomorrow so not going to investigate any further down the sin - cos routes just yet. Thanks tho I will refer back to this post at some point and practice what you said. I'll be starting to study maths more intensively shortly in preparation for either uni or an HNC so i'm sure i'll have the opportunity to open all these barrels of worms yet!
 04 December 2013 07:43 PM
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OMS

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OK mate, no drama - I was just trying to get the idea the book was wrong out of your mind before the exam

Just concentrate on the bits you need for the exam, come back anytime if you need a view or a bit of a hand with something - there 's plenty on here who'll help

What exam are you sitting out of interest - and good luck with it of course

Regards

OMS

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Failure is always an option
 04 December 2013 08:08 PM
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earthing

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Hi OMS its the City & Guilds 2396. Formerly 2391-20 and 2400 before that. I also have a project to pass.

Yes not good battling the book! As you said earlier its understanding the terminology and the basis of that data that will really help understanding but the course is an intensive 1 week crash course and it really only covers the basics with very little three phase or electrical theory for larger cables or things like harmonics, unbalanced loads and so forth.

I thought i'd be a fully fledged design engineer after this but it feels like this course hasn't even scratched the surface!
 04 December 2013 08:40 PM
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OMS

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Yes, I know what you mean - but you are also not likley to have to contend with power factor as part of volt drop considerations in the exam nor with harmonics, disturbing loads or the like.

Suffice that you know there is an effect

Do you have access to past papers ? - from what i know of it (which isn't much) I didn't think the written exam was overly technical, just that there is loads of it and you will struggle for time - so keep an eye on the watch

The exam and project won't make you a fully fledged designer, but it will give you a good introduction into the process of design - remember that cable calcs aren't design - they are just the calculations at the end that serve to validate the previous 90% of effort that is the design.

If you can't select Ib, decide on cable type, containment and route etc, then you can't design any circuit, remember

The course will have scratched the surface, in the sense that it has started off your thought process, - the rest is up to you

As I said, good luck with it - it's going to be a good first step into design for you, the result being perhaps a secondary thing

Regards

OMS

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 04 December 2013 09:14 PM
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Zs

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Good luck earthing. I have been wondering if this was 2396.

You won't need the complexities of three phase power factor correction.

You will need the adiabatic off pat, absolutely off pat, so make sure you are comfortable with the curves in the back of the regs. The quicker you can do a cable calc from first principles to confirming the CPC size and show your workings the better.

And you will need to know your way around that regs book like a demon.

It's a regs exam with some sums thrown in. Three hours needs to be four hours for all the writing.

The project is a massive bore, very time consuming (150 hours went into mine) but it is do-able. Start that project on Friday or you will forget everything. Really, Start it on Friday and start with circuiting up your project and doing the calcs.

Very best of luck,

Zs
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