Voltage drop two methods inconsistent

[Jpflex]

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%]

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.

By chance are you familiar with calculating available fault current from a transformer to its entire run?

 

[tortuga]

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

Excellent post but I am wondering why I get 33.9 not 49.6V?

So,we have to increase the transformer size to 400 kVA

And here is why I calculate voltage drop to the load and not the breaker, to avoid purchasing larger wire conduit and transformers than necessary. Keep in mind unless its a Art. 647 [Sensitive Electronic Equipment] VD calc (which are insane) if your in CA you only have to comply with:

California title 24 energy code 130.5 (c) Voltage Drop. The maximum combined voltage drop on both installed feeder conductors and branch circuit conductors to the farthest connected load or outlet shall not exceed 5 percent.

If you do not have load figures or know everything that will be added later,

I would not be allowed to size a feeder without any load figures from a client, what if they ended up with a 420A load?
We make the customer supply nameplate data over email or in writing, its a CYA, often its not what they end up having (sometimes more or less) but if the service/feeder is too small then its on them and it makes them think thru what they are doing ahead of time.
If you think too large is not a problem think again, one time a customers corporate VP and their new finance consultant grilled me on why we had put in a service that was 'too large' aka too expensive, and it was really handy to have all those emails from the plant manager detailing all the equipment they had planned. There was a 150k difference due to transformer size.

 

[wwhitney]

Excellent post but I am wondering why I get 33.9 not 49.6V?

Calculator error? I concur that "sqrt(3)*(2.987*0.8+0.1463*0.6)*11.56=49.6".

Cheers, Wayne

 

[tortuga]

Z=7.2^2/0.15*5.75/100=19.872 ohm

for that one I get
7.2^2 / 0.15*5.75 / 100

51.84 / 0.8625 / 100

60.10 / 100

=0.601
what did I miss there?

 

[Julius Right]

for that one I get
7.2^2 / 0.15*5.75 / 100

51.84 / 0.8625 / 100

60.10 / 100

=0.601
what did I miss there?

Try this way:
=(7.2^2/0.15)*5.75/100

 

[tortuga]

increase the transformer size to 400 kVA Z=7.2^2/0.4*5.75/100=5.96 ohm

So should that one be
(7.2^2/0.4) * 5.75 / 100 = 5.96

( 51.84 / 0.4 ) * 5.75 / 100

129.6 * 5.75 / 100

745.2 / 100

=7.452 ?

 

[wwhitney]

So should that one be

You replaced 0.15 by 0.4.

Cheers, Wayne

 

[tortuga]

You replaced 0.15 by 0.4.

Cheers, Wayne

Yeah for a 400 kVA instead of a 150 right?

 

[Julius Right]

So, Julius forgot to change 0.15 with 0.5 in order to get 5.96 and I am sorry!

 

[Jpflex]

Excellent post but I am wondering why I get 33.9 not 49.6V?

And here is why I calculate voltage drop to the load and not the breaker, to avoid purchasing larger wire conduit and transformers than necessary. Keep in mind unless its a Art. 647 [Sensitive Electronic Equipment] VD calc (which are insane) if your in CA you only have to comply with:



I would not be allowed to size a feeder without any load figures from a client, what if they ended up with a 420A load?
We make the customer supply nameplate data over email or in writing, its a CYA, often its not what they end up having (sometimes more or less) but if the service/feeder is too small then its on them and it makes them think thru what they are doing ahead of time.
If you think too large is not a problem think again, one time a customers corporate VP and their new finance consultant grilled me on why we had put in a service that was 'too large' aka too expensive, and it was really handy to have all those emails from the plant manager detailing all the equipment they had planned. There was a 150k difference due to transformer size.

You would be surprised on how work was requested at my last job as an industrial and residential electrician at a mine site. Before they lost investor funds and closed down, the owners, not being electricians, would NEVER know all the loads that were to be supplied by a 3 phase 4800/480 delta delta transformer.

When I wired up an entire laboratory including an expensive plasma particle analyzer computer, the owner selected a pad mounted transformer based on his assumption on what the KVA size should be. He kept adding buildings and loads but could not do a load calculation because “they” did not know everything that would need to be powered. After they wasted money on two transformers, they ended up with a 300 kva transformer for an MCC trailer, 200 ampere rock crushing station, 3 room laboratory, 100 ampere rock crushing pilot station, black smith shop with welding, and another dryer building.

These people had no clue what they were doing. They would often go to junk yards or auctions for used electrical equipment with SOOW cord having wires torn out of insulation, breaker panels with bent covers and sides, covers that would not hold, 1940 switch disconnects. Their former electricians would use any type of undersized wire for feeders etc

 

[tortuga]

So, Julius forgot to change 0.15 with 0.5 in order to get 5.96 and I am sorry!

I was not trying to nit pick just wanted to make sure I was following along and that number was the transformer kva/1000