How do you calculate percentage Vd for single phase? 4

[AusLee]

Hi,

Can someone please advise on which voltage the single phase voltage drop is calculated?

The application is simple: we have a cable 100m long. It is a 3 phase multi-core copper cable with cross section 25mm² per phase and neutral. The voltage is 230/400V 50Hz.

From page 87 in the following manual (Olex Handbook, the characteristic voltage drop is 1.61 mV/A/m

The loads at the end are 3 single phase loads, each consuming 10A and are a balanced load.

So the voltage drop in principle would be: 1.61 x 100 x 10 = 1.61V.

As the load is single phase, if you divide 1.61/230 = 0.7%. But if you divide by 400V, then it becomes only 0.4%. Which percentage value is correct?

Also, say for some reason two of the loads have been manually shut off, there remains only one load of 10 A on one phase in this 3 phase cable. As there are no loads on the other phases to balance it, the current in the neutral is then 10A also and not zero and consequently the voltage drop across the neutral should be accounted for. How do you calculate that in this case? 1.6 x 100 x 10 times 2?

Thanks.

 

[sibeen]

AusLee, going by your handle and the catalogue you have linked to, I perhaps recommend that you have a look at the relevant Australian/New Zealand standard for the installation of cables - AS/NZ 3008.1.1.2009. Perhaps look at tables 40 - 51 and pay special note to what it suggests way at the bottom of each of these tables. Perhaps also a thorough reading of Appendix A of the same standard may be a decent idea.

The standards are produced for a purpose, and a manufacturers handbook is not going to contain all the relevant information.

 

[AusLee]

Hi sibeen

Thanks for the advice, been giving it to myself for quite a while AS/NZS 3008.1.1.2009 has been amended in 2011. I read the note under the tables it says multiply by 2 and divide by sqrt(3). Appendix A6.2 talks about a single phase application but uses 3 phase values. There i've read it can you answer now?

Cheers!

 

[magoo2]

Since you're doing a per phase voltage drop, you can rule out 400 V because that's a phase to phase value. Does this help?

 

[AusLee]

Hi mahoo2,

yes it helps very much, because 400/230 = 1.73 so a Vd of 3.5% calculated based on 400V seems fine because it is less than 5%, but if you calculate it based on 230V then the 3.5% becomes 6% which means the cable is not enough!

Can we vote on this one?

 

[waross]

First throw out the darn percentages!! Mixing percentages from different bases leads to confusion.
Consider a three phase load; The phase to phase currents sum to a value that is in phase with the line to neutral voltages, so your voltage drops are in the right vector direction. For single phase cases the voltage drop on the neutral conductor will be the same as on the line conductor (for a 100% neutral conductor, for a reduced neutral you must consider the change in the resistance plus the reactive component).
So, single phase voltage drop in volts will be twice the three phase voltage drop in volts.
Now pick a voltage and calculate your percentage.
You didn't mention it but the condition where two phases are energized is interesting. The neutral current will still be equal to the current in either line current but not at the same phase angle.
A close approximation of the line to neutral voltage drop in this case will be the three phase voltage voltage drop times 1.5
Considering the voltage drop on A phase, the voltage drop on the neutral will be equal to the voltage drop on the phase conductor when only this phase is energized, but when B phase is energized a component of the B phase voltage drop on the neutral conductor will act as a voltage rise on the A phase neutral drop.
Summary;
Three phases energized, book value voltage drop (in Volts).
Two phases energized, book value voltage drop x 1.5 (in Volts).
One phase energized, book value voltage drop x 2 (in Volts).


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[healyx]

Hi Auslee,

The 1.61 quoted is for 3 phase voltage drop, so it applies to the 400V reference. Because your system is balanced, you don't need to include the return path drop, so no x2 required.

The correct % drop is 1.61V/400V = 0.4%. You can do that at 230v if you want to, but you need to divide the 1.61 by sqrt(3) first - it works out the same either way.

If you drop two phases, so it now becomes unbalanced, you will need to multiply the voltage drop % by 2 to allow for return path drop - which will go to 0.8%.

I would second sibeen's comment. Use the standard, not the catalogue. I don't have the standard on me at the moment, but I'll check when I get back to the office.

I'm pretty sure the 1.61 mV/A/m will be too high. If you are only sending 10A down a 25mm2 cable, the cable temp will only barely exceed ambient.

How is the cable installed and what is the load characteristic? I'll calculate the exact value for you.

 

[healyx]

BTW, I think Magoo might be incorrect in saying you can rule out 400V. You can't in this instance because the 1.61 mV/A/m quoted has been calculated for the 3 phase case (400V).

 

[AusLee]

Hi Waross,

Isn't what healyx is saying totally opposite what you're saying?

If I use your method and calculate the voltage drop in volts, then divide by whatever I want, that's one. healys is saying I can't divide but by 400V?

 

[waross]

Find the voltage drop in Volts. Then for a 230 Volt circuit use 230 Volts as the base to find the percentage voltage drop.
Cable reactance is a large component of cable impedance. Codes and standards publish typical values of voltage drop.
BUT
The reactive component of a cable's impedance is influenced by the conductor spacing. A thicker layer of insulation will affect the voltage drop.
The cable manufacturer may be able to supply a more accurate voltage drop value for a particular cable construction than either codes or anonymous internet advisers, but I could be wrong.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[healyx]

Auslee,

I checked the standard. 1.61 mV/A/m quoted by the manufacturer is straight from the standard. That is the 3 phase voltage drop at 90 degrees. You will probably only be slightly more than 40 degrees if installed in air (less if installed in ground). Using the 1.39 mV/A/m quoted at 45 degrees is a better estimate.

So for the balanced case voltage drop is 1.39*100*10 = 1.39 volts. % is 1.39/400 = 0.35%. For 1 phase or 2 phases, time that figure by 2 to account for return path current (which will be the same in either case).

If you want to work at 230v, volt drop is 1.39/sqrt(3) = 0.8025 mV/A/m. If balance you only include the active cable drop, so 0.8025*100*10 = 0.8025v. For %, 0.8025/230 = 0.35%, which is the same as the 3 phase case. To count return current in the neutral (which is same as active current in 1 or 2 phase case) it is the same drop again, so 2x.

FYI the mV/A/m quoted in the standard includes resistance and reactance for the given cable (including installation condition). It assumes that cable x/r ratio matches the load x/r - which in this case holds if the load is almost all resistance (pf = 0.995) because x/r of this cable is about 0.1. If the load pf is not purely resistive, voltage drop will be even less.

A more interesting question is: Why are you using a 25mm2 cable to carry only 10 amps. Sounds like overkill to me.

 

[AusLee]

Healyx,

Thanks for the clarification mate. I'll answer the question you point out as more interesting in a while but first regarding your calcs. Point #1 is for clarification only but my real question is Point #2 which waross is saying I can do it the way I think and I need your different point of view

Point #1 - Actual Calculation
I already got the answer to the second part of the question from sibeen.

Let's stick with one value 1.61 mV/A/m for now I agree 100% with what you're saying in this regard that is not the point. The reason is that this makes it easier just referencing the catalog which as you say, the standard is a copy of the catalog, even though it may seem the catalog to be a copy of the standard, again beside the point.

You're saying that for 1 phase or 2 phase, I need to multiply by 2, so your calculated value of 1.39V, or make that 1.61V, will become 3.22V.

Let's do this two separate not necessarily different ways:
Method #1: from AS 3008 note under the table as pointed out by sibeen, if you want to get the single phase/imbalanced Vd from the 3 phase value you multiply by 2 and divide by sqrt(3), totaling a multiplication by 1.15, not multiply by 2.

Method #2: If we take the value from the book for 25 mm2 2 Core XLPE, it is 1.86 mV/A/m (Page 82 of the catalog). According to the manufacturer, that factor already accounts for the double length (Vd in phase and in neutral).

So doing the calcs: 1.86 x 100 x 10 = 1.86V. Note that 1.86 = 1.61 x 1.15 which makes perfect sense of calculating the voltage drop for a single phase application in either single phase or 3 phase cable.

Point #2 - Percentage
I'm not really seeing why if I use 1.61 x 1.15 = 1.86 mV/A/m I have to divide by 400V.

As per item #1 above, if I have a single phase cable (2 core), the published value is 1.86 mV/A/m and in this case I can - or am supposed to - divide by 230V. Why can't I just calculate the Vd and divide by whatever I want?

Let's put this differently. I know the published x mV/A/m value includes resistance and inductance at 50Hz (which makes it an upper limit when using the cable for a DC application). Say I want to calculate the Vd in percentage to a 24V CCTV camera at a gate 100m away from the board. The camera takes 24V AC to make things easy and it comsumes 24W = 1A at 24V. Using the same 25mm2 cable:

1.86 x 100 x 1 = 0.18V; that is 0.75% of 24V. Why do I have to divide by 230V in this case?

To cut the chase, please say yes that a single phase application of 10A at 100m supplied from a 3 phase cable of 25 mm2 of 1.61 mV/A/m will have a voltage drop of 1.86 V equivalent to 1.86/230 = 0.8% at 230V or advise why not

 

[healyx]

Hi Auslee,

Perhaps I could have been clearer. I might be wrong but when a three phase voltage drop is quoted in mV/A/m I think they are referring to the line to line voltage drop, so you need to use the 400V base to work out the percentage voltage drop when this figure is quoted. To get it back to the line to neutral value you divide by root 3.

If you are quoted 1.61 mV/A/m for three phase voltage drop, that is the line to line voltage drop, you can only compare this to the line to line supply voltage (400V).

For line to ground, divide by root 3, which gives 0.93 mV/A/m. Now this would be the actual volts dropped along the active cable in a line to neutral connected circuit. So it is actually less BUT you need to also include the neutral return conductor, which is where the x2 comes in. Now if you have 3 single phase circuits all sharing the neutral (e.g a wye connected load) and they have equal current (each from a different phase), you don't need to include the return path. Because you are now using line to neutral, you compare to the line to neutral voltage (230v). Whether you compare 1.61 mV line to line at 400V or 0.93 mV line to ground at 230V gives the same % drop - if percentage drop is what you are interested in. Of course if you need to consider the neutral return path, the line to ground reference becomes 1.86 mV.

"I'm not really seeing why if I use 1.61 x 1.15 = 1.86 mV/A/m I have to divide by 400V."

In this case you don't. If you want % vd, divide the resulting voltage by 230v or 24v, it doesn't matter what the voltage is as long as it is the line to neutral voltage.

"As per item #1 above, if I have a single phase cable (2 core), the published value is 1.86 mV/A/m and in this case I can - or am supposed to - divide by 230V. Why can't I just calculate the Vd and divide by whatever I want?"

You can. See above.

"1.86 x 100 x 1 = 0.18V; that is 0.75% of 24V. Why do I have to divide by 230V in this case?"

You don't. Your result looks right.

"To cut the chase, please say yes that a single phase application of 10A at 100m supplied from a 3 phase cable of 25 mm2 of 1.61 mV/A/m will have a voltage drop of 1.86 V equivalent to 1.86/230 = 0.8% at 230V or advise why not "

If the 3 phase cable is only carrying a single phase (line to neutral) circuit, that is correct (YES), remembering the 1.86v represents the line to neutral voltage. If it is carrying 3 balanced single phase circuits it will only drop 0.93v line to ground and 1.61v line to line.

Some other notes:
(1) The catalogue data comes from the standard, not the other way round. You should really be using the standard.
(2) The 24V CCTV camera was just an example right? You're not really using a 25mm2 cable to feed a 24W camera are you?

 

[sibeen]

healyx, nice and succinct explanation.

I wonder why the note under the tables, in the standard, says to multiply by 2 and divide by sqrt(3)?

 

[AusLee]

Hi healyx / sibeen

I was trying to compare the software output with manual calcs for a street lighting application. One last question please:

Is there anything in the regulations that does not allow you to use all cores of a 3 phase cable for a single phase application? Say the above calculated Vd was marginally close to 5%, or you want lower than 5% just because you can, and you have 2 spare cores in the cable that are just sitting there. If you bring one core in parallel with the phase, (say A // B) and the second core parallel with the neutral ( C // N), you can have half the voltage drop. I know there is a requirement to identify the phases, so how about permanent signs and tags and what have you?

This still meets the requirements that the cross section of the neutral should be equal to the cross section of the phase :]

I know it's not ideal but is it specifically not permitted?

 

[healyx]

Hi Auslee,

Good question, I don't know the answer but I'll have a think about it...

Out of interest, what software are you using?

Some things you might want to consider if you have a voltage drop problem.

(1) Are you calculating voltage drop with the correct temperatures applied? If you are using the figures you have quoted, those temperatures are probably too high.
(2) When calculating upper segments (larger cables) have you used the maximum voltage drop figure from the tables or have you applied a load power factor that doesn't match the cables? Large cables have higher X/R ratios, which don't match the load, so the VD is actually lower than shown for the maximum case.
(3) What is the power factor of the load? If street lighting, if the actual pf is more like 0.9 this will buy you some wiggle room, but not much.
(4) Remember that for distributed loads, AS3000 allows you to divide the current by 2, but you need to use the breaker rating for the current, not the actual load current. Street lighting will likely be a distributed load.
(5) Have you conducted voltage drop calcs to the breaker rating or to the actual load for upstream segments?

Don't forget to consider the effect of long runs on minimum fault current and disconnect time requirements. An RCD might fix disconnect time for an earth fault, but a phase to neutral fault could miss the instantaneous setting on the breaker, which isn't really addressed anywhere in the standards but I don't like it happening.

 

[sibeen]

AusLee, I'd probably be loathe to do it. 4 core cables are normally colour coded and even with tagging / identification there is a chance for mistakes to be made. Whilst AS3000 won't specifically prohibit it, an electrical inspector would probably have conniptions upon seeing an installation set out that way.

The other question would be, if you only have a single phase circuit why are you running a three phase cable?

 

[AusLee]

Thanks for the replies, I have PowerCAD which does all that sort of stuff you're talking about ... (i guess)

The cable is available in stock and must be used, extra length from another job.

The inspector will put up a defects notice (CCEW) and will put a Defect Code, Details (Location & Comments) and Rule Reference. If there is no rule or reference that says otherwise then he can't defect it can he?

 

[sibeen]

If there is no rule or reference that says otherwise then he can't defect it can he?

I'd imagine an inspector could just cite the colour code reference in AS3000. Of course, if he's had a fight with his wife the night before, he could defect the job because he doesn't like the colour of your yes.

 

[waross]

In Canada you may run smaller cores in parallel to reduce the voltage drop when the breaker setting or size will adequately protect any single core.
eg. #14 AWG cable normally requires a maximum 15 Amp breaker. So, #14 AWG conductors may be run in parallel to reduce voltage drop as long as the breaker is a maximum of 15 Amps.
The issue would be identifying the second neutral conductor in a way acceptable to the AHJ.
Do you have anything in your code re identifying neutral conductors?
Even if you are not allowed to parallel the neutral, paralleling the hot conductors may gain a percent or so.
This is probably worth a chat with the AHJ.

Bill
--------------------
"Why not the best?"
Jimmy Carter