Volt drop from transfo to shed
Volt drop from transfo to shed
(OP)
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.
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.






RE: Volt drop from transfo to shed
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!
RE: Volt drop from transfo to shed
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!
RE: Volt drop from transfo to shed
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
RE: Volt drop from transfo to shed
If you have more questions, write it , perhaps i am missing something
RE: Volt drop from transfo to shed
RE: Volt drop from transfo to shed
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!
RE: Volt drop from transfo to shed
RE: Volt drop from transfo to shed
RE: Volt drop from transfo to shed
only 4.9 [two conductors live and neutral].
RE: Volt drop from transfo to shed
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.
RE: Volt drop from transfo to shed