Jpflex
Electrician big leagues
- Location
- Victorville
- Occupation
- Electrician commercial and residential
Don’t take me wrong, you have a lot of info and I’m sure there are times you’ll need more accuracy so I’ll have to review your post and I’m sure you got it down.In order to calculate the cable impedance, we need to know the conductor aluminium or copper, the number of cores 1 or 3 and if it is 1 the lay configuration flat or trefoil .In order to state the conductor cross section we need ampacity and then we need to know the run way: underground or overhead[in air, exposed to sun or not].
Let's say it is a 3 cores cable according to IEC 60502-2 of 7.2 kV, aluminum conductor, XLPE insulation, copper tape of 5 mils thick 12.5% overlap shielded, non-armored, underground in 1.5 RHO, 20oC Earth.
IEC 60502-2 Table B.6 16 mm^2 it is the minimum and ampacity is 67A at 90oC.
If secondary current is 400 A the primary will be 11.56 A[400*208/7200].
The resistance per 1 km at 20oC is 1.91 ohm and at 90oC 2.45 ohm/km.
The reactance is 0.12 ohm/km
Total resistance for 4000 ft R=2.45*4*.3048=2.987 ohm and the reactance X=0.12*4*.3048=0.1463 ohm
Let’s say the power factor [cosφ=0.8 and sinφ=0.6] VD=sqrt(3)*(2.987*0.8+0.1463*0.6)*11.56=49.6 V
However, the voltage drop through both transformers has to be added.
The apparent power required it is S=sqrt(3)*7.2*11.56=144.16 kVA[150 kVA]
IEEE C57.12.01 recommends 5-5.75% short-circuit impedance then the impedance seen from HV side
will be Z=7.2^2/0.15*5.75/100=19.872 ohm
The voltage drop per one transformer will be VDxfr=sqrt(3)*19.872*11.56=397.89 V
Total voltage drop will be 2*397.89+49.6=845.38 =11.74%
So,we have to increase the transformer size to 400 kVA Z=7.2^2/0.4*5.75/100=5.96 ohm
VDxfr=sqrt(3)*5.96*11.56=119.33 and total voltage will be 2*119.33+49.6=288.26 V[=288.26/7200=4%]
By chance are you familiar with calculating available fault current from a transformer to its entire run?