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Small voltage drop calc
2

Small voltage drop calc

Small voltage drop calc

(OP)
Hi,

Can I ask if somone could please do this calculation:

Cable length from transformer: 900 meters, 3 phase + neutral all 185 mm2 XLPE insultation, 230V phase-neutral, 400V phase-phase.
Connected single phase load on phase A: 60A
Connected single phase load on phase B: 50A
Phase C: zero

What is the voltage drop at the end of this cable?

Thanks.

RE: Small voltage drop calc

Neglible, would be the short answer.
For better precision, I think that the cos(phi) of the loads are needed. Also, are you asking for voltage drop in each single phase? I don't think there is a representative single voltage drop. Don't forget to take the voltage drop in the Neutral into account. That will produce a (perhaps) unexpected voltage difference between N and C.
I have some spare time and will run a simulation for you.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

RE: Small voltage drop calc

(OP)
Thanks Skogsgurra. The single phase loads are pure resistive and connected very very close to the end of the cable. That's right, the current in the neutral is not zero. Looking forward to your simulation results.

RE: Small voltage drop calc

Here it is:
http://www.gke.org/pub/files/EngTips%20Low%20volta...

I was surprised to get such a high voltage drop. It stems from the cable inductance. I assumed 1 uH/m (standard value if you don't have the correct data).
Actually, a jw calculation would have been almost as simple to do. But...

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

RE: Small voltage drop calc

(OP)
The value you put for the resitance is 0.085. That is correct but only for 1 meter length; the cable is 900m long.

For the inductance, the value is Lw = 0.112 Ohm/m

Can you please check again?

RE: Small voltage drop calc

R=17.5*.9/185=.085 ohms, you probaly use milliohms?

As for inductance, .112 ohms/m would correspond to 360 uH/m -> .324 H for the 900 m cable. That is a very high value. I then get voltage drops almost equal to the grid voltage. Are you sure about those values?

Time From 1.0000000e-005
Time To 1.0000000e-001
VP4 2.3032000e+002
VP5 2.3014786e+002

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

RE: Small voltage drop calc

Sorry, forgot to change L in Netral. With new inductance in Neutral you get:

Time From 1.0000000e-005
Time To 1.0000000e-001
VP4 2.0215624e+002
VP5 2.0425296e+002

A little less voltage drop. But still huge.

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

RE: Small voltage drop calc

Hi Auslee,

Assuming 40 degree ambient in air. 90 degree XLPE Cu SDI cables arranged in trefoil touching, with the neutral also touching and equal spaced from each current carrying phase cable. The temperature of each cable will be about 41 deg C (assuming 468 A current rating in each cable - from AS3008 table 8 spaced from surface). This gives an AC resistance of 0.0988 ohms for each 900m cable. Coming off a 1000kVA 5% transformer, 400V delta/wye.

It's probably overkill but I used the impedance matrix method to model this, which takes into account the dependencies of mutual impedance - thought it might be important in such a long length with reasonably large cables.

Anyway - I get 6.65V drop across the neutral cable and the loads are 4.4 and 3.74 ohms resistive each. Therefore you drop 220v across the 4.4 ohm load and 224.4v across the 3.74 ohm load. Given you started with Phase to Neutral voltages of 230.94v at the transformer, you are effectively losing 4.7% voltage on the 4.4 ohm load (50A) phase and 2.8% voltage on the 3.74 ohm load (60A) phase. Neutral current is 53.5A.

These results will probably vary with different phase voltage sequencing.




RE: Small voltage drop calc

(OP)
Hi Skogsgurra,

Not sure about those values, but the manufacturer's data are on the following link if you could please have a look:

Olex 185mm

RE: Small voltage drop calc

Auslee,

The AC Resistance quoted in the catalogue is at 90 degrees. As I said, the cable will operating at about 41 degrees, probably more like 45 degrees now that I notice it is multicore. That will give you a resistance of 0.1008 ohms over 900 meters. Gunnar has used the basic DC resistance at 20 degrees. It doens't include proximity or skin effect.

The inductance will be about 0.0648 ohms for the 900 metre length. That is based on table data, but because the currents aren't exactly equal and it is such a long length I modelled it with an impedance matrix - which is more accurate.

RE: Small voltage drop calc

(OP)
Hi Healyx,

Ok thanks, can you please let me know why this is wrong then:
  1. Open the small Olex catalogue page 89 3x1 core XLPE and read Vd for trefoil for 185 = 0.269 mV/A.m
  2. Multiply this value by 2/sqrt(3) to get single phase: 0.269 x 1.155 = 0.3106
    [li]Calculate Vd for the worst loaded phase as: 60 x 900 x 0.3106 /1000 = 16.77 volts
  3. Calculate the current in the neutral approx 54A.
  4. Voltage drop across neutral: 54 x 900 x 0.3106 / 1000 = 15.1
  5. Total Vd = 16.77+15.1 = 31.9V, or over 230V = 13.8%
? :)

RE: Small voltage drop calc

Auslee,

I don't have that catalogue but 0.269mv/Am is lifted straight out of the standard.
(1) That figure is at 90 degrees temp with single core trefoil cables - you should be looking at multicore figure at 45 degrees - which is 0.234.
(2) Those published figures assume the worst case where cable pf equals load pf. Your case doesn't doesn't because it is purely resistive. where as cable pf has almost as much X as R.
(3)The x 2 term in (2/sqrt(3)) is to allow for the path out the phase cable and back on the neutral, you don't use it if you are only considering the neutral path. The divide by sqrt 3 is to convert 400V to 231V, which you do need to do.

Therefore using impedances from tables, you have R = 0.112*0.9 = 0.1008 ohm/900 metres at 45 degrees. X = 0.072*.9 = 0.0648 ohm/900 metres.

Z = sqrt(r^2 + x^2) = 0.21 ohm. V = IZ. V = 54*0.12 = 6.5v drop on the neutral conductor. THis is very close to my exact method above. All these figures were taken from AS3008.

By the way I re-did the calc for multicore cable.

R goes to 0.0993 at 41 degrees. A load = 4.41 ohm, B load = 3.73 ohm. It all works out about the same. Neutral voltage drop goes to
6.32 volts.

Out of interest - does this cable exist? Obviously putting some of the load onto the C phase will improve voltage drop again, but based on my calcs, you don't have a voltage drop problem as it stands.

RE: Small voltage drop calc

Auslee,

Just rechecked your calc. You are effectively counting the neutral voltage drop twice and using values that are too conservative for the case you describe. That is the short answer to your question smile

RE: Small voltage drop calc

Auslee - sorry, I should hasten to add a disclaimer, when I said you don't have a voltage drop problem with 4.5% (with multicore cable) - that of course would depend on what other drops you have upstream and downstream from this segment.

RE: Small voltage drop calc

(OP)
Thanks mate, that's what I saw other people used to do. Cheers!

RE: Small voltage drop calc

It is a very good idea to publish the cable data!

Now, we get this:

VD1 6.64 V
VD2 5.63 V
VD3 0.00 V
VDN 5.98 V



Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

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