Volt drop from transfo to shed 2

[elecquest]

Hi
Could you please correct me if i am wrong?

I have this situation: To supply a shed 1.3 km away from a distribution transformer (22 kV/415V - three phase)on 230V, load = 1kW. 415 kV cable used = 70 mm2 - three phase. V/A/km of 70 mm2 cable = 0.87.

Volt drop up to shed = 0.87 x 1.3 km x 4.4 A = 5 V ??

Am i doing this right?

Thanks for reading this and pointing out errors.

 

[burnt2x]

1 kW delivered at 415V is just 1.4 amperes!
If you have used a 70mm2 wire, it's a big wire! (equiv. to 2/0)
You mentioned 12kV/415V and you also mentioned 230V, which is which?
If you are supplying 1 kW on a 415V, three-phase line, your Vd at end-of-line will not even exceed 0.5 volt!

 

[ScottyUK]

Assuming the 22/0.415kV transformer is a normal star-wound type and you're taking a line and neutral supply from it then yes, your calcs look reasonable. You might want to consider load power factor, unless this is a heater or tungsten light at the far end. If it is a motor load then also consider the starting current, which will be about 6x the normal running current.

That cable you propose is quite generously sized in these days of high copper prices.


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If we learn from our mistakes I'm getting a great education!

 

[elecquest]

Ah yes thx

They are tungsten lights at the far end and the 70 mm2 cable is Aluminium ..

Mistake: 415 V instead of 415kV in original post ..ack

Phew .. just started working as an engineer :| Takes a lot of focus

Thx for the support! cheers

 

[elecquest]

Oops, Burnt2x .. i mean i have to supply the shed with 1 phase power supply only (230 V : PHASE-NEUTRAL) ..

If you have more questions, write it , perhaps i am missing something

 

[bacon4life]

Since this is just 1 phase, you will have a voltage drop in both the phase and the neutral conductor.

 

[7anoter4]

For the record only the 06/1 kV 3*70 sqr.mm AL conductor cable is R+j*X
where R=0.443 ohm/km [20oC,d.c.] X=0.12 ohm/km [PVC insulated 60hz] As skin effect and proximity effect for 70 sqr.mm AL conductor is about 1.008 may be neglected. Also if the cable is laid in underground the temperature rise will be not more than 1-2 degrees C so could be neglected. Then the resistance will remain 0.443 ohm/km[according
to IEC 60228].
The resistance of the tungsten lights is 230^2/1000=52.9 ohm.
The current flowing through the 2 conductor of 3*70 sqr.mm AL-live and neutral- and the resistance of the tungsten lights is :
I=415/SQRT(3)/(2*1.3*0.443+52.9)=4.433 A
The Net voltage possible drops to 400 V[ instead of 415 V] that means the source end voltage will be 400/sqrt(3)=231 V.
I=231/(2*1.3*0.443+52.9)=4.27A [if you neglect the cable reactance] and if you don't you'll get:
I=4.27-j*0.0247 A so you may neglect the reactive part.
As we know the tungsten light luminosity will support very well a drop voltage of 5% that means even at 230*.95=218.5 will be o.k. and usually 220 V is still good.
Now we can neglect the cable reactance and calculate the voltage drop [ as actually was done]:
DV=2*0.443*1.3*4.27=4.9 V So the worst case the tungsten light supply voltage will be:231-4.9 =226.1 V.
But if you'll take 3*50 you'll get 224 V still good!

 

[elecquest]

wow, thanks very much .. very informative!

 

[elecquest]

Another thing .. voltage drop in the neutral conductor stands too? as mentioned by bacon4life .. it would be 4.9*2 = 9.8V?

 

[7anoter4]

bacon4life is right.If you'll take the maximum resistance for a stranded class 2 conductor according to IEC 60228 is only 0.443 ohm/km [it is not 0.87 ohm/km as you noted in o.p.] you'll get
only 4.9 [two conductors live and neutral].

 

[7anoter4]

What is V/A/km noted in some of the cable catalogues it is the voltage drop for a three phase cable,
load pf=0.8 and conductor temperature maximum admissible[for PVC=70oC for XLPE =90oC]cosfi=pf=0.8 sinfi=sqrt(1-pf^2)=0.6
Vdrop=I*(R*cosfi+X*sinfi)*SQRT(3) [V/km] I[A] R[ohm/km] X[ohm/km]
or Vdrop/I=(R*cosfi+X*sinfi)*SQRT(3)
In your case R=0.443*(1+.00403*70)=0.568 ohm/km X=0.09 ohm/km [for XLPE insulation]
V/A/km=sqrt(3)*(0.568*.8+0.09*.6)=0.88 ohm/km.
But your pf [cosfi] =1 sinfi=0, you don't need sqrt (3) as you system is single phase, so you need only R for ambient [20oC] as 4.5 A even in underground cable is not able to rise the temperature more then 1-2 degrees then R=0.443 ohm/km.

 

[elecquest]

thanks everyone for prompt reply and explanation!