Voltage drop two methods inconsistent

[Jpflex]

Another member posted a question on another post regarding what size wire to use for a 4,000’ foot run. The question was worded as:

Just wondering what size wire do i need if my run is 4000 feet at 480 from my switch wanting to step it up to 7200 an then from 7200 down again to 208/120 what size wire will i need. [to a 3 phase 400i ampere service]

When trying to help this guy out on figuring the proper size wire for the 4,000 foot span I first determined the size of the downstream stream transformer at

Secondary feeding the 400 ampere service
208E x 400i x 1.732 = 144.1 kVA transformer

Primary side amperes which would be seen on the 4,000 ft run = 144,102 VA / 7,200 VA x 1.732 = 11.555 i amperes

Now that the full load amperes on the 4,000 foot run is known between transformers we can calculate voltage drop and minimum wire size.

However when I use either one of two equations permitted with one using the resistance values for wires in NEC chapter 9 Table 8, they never seem to result in consistent answers? Why

For example, first determine max 2% voltage drop at feeder with 7,200 volts. 7,200 volts x 0.02 = 144 volts dropped max

Calculation method #1

Vd = 1.732 x K x I x D / CM

144 = 1.732 x 12.9 x 11.555i x 4,000’ / CM

Solve for CM (wire size in circular mill) = 7,171.42 cm

Match 7,171 circular mills to wire size chapter 9 Table 8 =

A #10 AWG wire with 144 volts dropped max



Calculation method # 2
I X R

VD = dc resistance (Ch 9 Table 8) x i x 1.732 x distance / 1,000

1.21 x 11.55i x 1.732 x 4000’ / 1,000 = 96 volts dropped

144 volts dropped in first equation does not equal 96 volts dropped in second equation but both are NEC acceptable methods. Both equations never result in the same results so how would you handle this during an NEC journeyman’s test?

 

[david luchini]

Calculation method #1

Vd = 1.732 x K x I x D / CM

144 = 1.732 x 12.9 x 11.555i x 4,000’ / CM

Solve for CM (wire size in circular mill) = 7,171.42 cm

Match 7,171 circular mills to wire size chapter 9 Table 8 =

A #10 AWG wire with 144 volts dropped max



Calculation method # 2
I X R

VD = dc resistance (Ch 9 Table 8) x i x 1.732 x distance / 1,000

1.21 x 11.55i x 1.732 x 4000’ / 1,000 = 96 volts dropped

144 volts dropped in first equation does not equal 96 volts dropped in second equation but both are NEC acceptable methods. Both equations never result in the same results so how would you handle this during an NEC journeyman’s test?

calculation method 1:
vd = 1.732xKxD/cm

vd = 1.732 x 12.9 x 11.555 x 4000' / 10380cm = 99 volts dropped.

Pretty close to the 97 volts dropped per calculation method 2.

 

[wwhitney]

144 volts dropped in first equation does not equal 96 volts dropped in second equation but both are NEC acceptable methods. Both equations never result in the same results so how would you handle this during an NEC journeyman’s test?

They are much closer than you realize. The first equation told you that a conductor with 7171 cm would drop 144V. Then you went to Chapter 9 Table 8 and saw that #12 is 6530 cm, and #10 is 10380 cm, so you chose #10.

In your second equation, you aren't using a conductor with 7171 cm, so there's no reason you'd get 144V. Since you're using a conductor with 10380 cm, you'd expect to get 144 V * (7171/103800) = 99V. Instead you got 96V, which is pretty close.

The remaining discrepancy is your having chosen in Table 8 1.21 ohms/kft for #10 solid, rather than 1.24 ohms/kft for #10 stranded. However, I have no idea why the Table 8 DC resistance values are slightly different for solid and stranded, when they are both listed as 10380 cm.

Cheers, Wayne

 

[tortuga]

FYI voltage drop only applies to the calculated load not the breaker size.

 

[Jpflex]

FYI voltage drop only applies to the calculated load not the breaker size.

But if you don’t know the calculated load but only know the maximum amperes available at a service or sub pannel (limited by main breaker and feeder size) then wouldn’t these methods apply?

 

[Hv&Lv]

I have no idea why the Table 8 DC resistance values are slightly different for solid and stranded, when they are both listed as 10380 cm.

Cheers, Wayne

the cross section of a stranded cable isn't all copper.
the insertion loss of smaller gauge wires used to make stranded cables have higher losses, by about 30% over solid conductors

 

[Elect117]

AC uses the surface area for conductivity and stranded increases the area of travel for electrons.
DC uses the interior cross sectional area which is why solid is better for DC.

 

[Jpflex]

Yea i get it. The first equation provides the smallest circular mill size wire but if it’s not a standard size you go up in wire size

The second equation relays on the table and a standard size cm wire size so the circular mills are close but not the same

 

[wwhitney]

the cross section of a stranded cable isn't all copper.

Correct, but my understanding is that the copper portion of the cross section is the same for #10 solid and #10 stranded. I.e. the solid cross section is one circle of area 10380 cmil, while the stranded cross section is 7 circles, each circle with an area of 10380/7 cmils.

So I would expect the DC resistance to be identical, not differ by 2%.

Cheers, Wayne

 

[Jpflex]

The remaining discrepancy is your having chosen in Table 8 1.21 ohms/kft for #10 solid, rather than 1.24 ohms/kft for #10 stranded. However, I have no idea why the Table 8 DC resistance values are slightly different for solid and stranded, when they are both listed as 10380 cm.

Cheers, Wayne

Actually i did read in a book a while back to explain the different resistance values for coated wire vs uncoated wire for the same AWG conductor size

If i remember correctly the coated wire referred to wire that was coated with lead to mitigate the chemical reaction between copper and older style rubber insulation so this isn’t used much anymore since pure rubber insulation apparently no longer is manufactured or sold

 

[Jpflex]

Correct, but my understanding is that the copper portion of the cross section is the same for #10 solid and #10 stranded. I.e. the solid cross section is one circle of area 10380 cmil, while the stranded cross section is 7 circles, each circle with an area of 10380/7 cmils.

So I would expect the DC resistance to be identical, not differ by 2%.

Cheers, Wayne

See post below but I don’t know either why resistance would change with the same overall diameter of copper

 

[Jpflex]

AC uses the surface area for conductivity and stranded increases the area of travel for electrons.
DC uses the interior cross sectional area which is why solid is better for DC.

I have read about skin effect for AC but haven’t heard that dc travels within the middle of a conductor and solid was better on dc but I’ll have to look into this some time

 

[Jpflex]

FYI voltage drop only applies to the calculated load not the breaker size.

If you do not have load figures or know everything that will be added later, why not size the feeder to its maximum and account for maximum voltage drop + eliminate having to redo with larger wires in the future if cost is not a problem?

 

[Jpflex]

AC uses the surface area for conductivity and stranded increases the area of travel for electrons.
DC uses the interior cross sectional area which is why solid is better for DC.

Isn’t stranded required on conductors over 6 AWG or in a raceway?

 

[wwhitney]

DC uses the interior cross sectional area which is why solid is better for DC.

??

Stranded and solid conductors have the same cross-sectional area (the bundle diameter on stranded conductors is larger than the diameter of a solid conductor), so that doesn't explain the DC resistance discrepancy.

Cheers, Wayne

 

[Elect117]

Isn’t stranded required on conductors over 6 AWG or in a raceway?

It has to do with the field created when electrons move back and forth vs in one direction.

AC crosses the 0V point and so the field pushes and pulls accordingly creating a skin effect where the outer diameter of a bounded size is travelable.

In DC the field is only in one direction, and there is no alternating force pushing or pulling. This is also why DC has no reactive power. As such, using the formula for Z=R+jX you can see that AC would have a different impedance.

And yes, i believe stranded is used because you cant really bend solid copper like you can stranded.

 

[Jpflex]

It has to do with the field created when electrons move back and forth vs in one direction.

AC crosses the 0V point and so the field pushes and pulls accordingly creating a skin effect where the outer diameter of a bounded size is travelable.

In DC the field is only in one direction, and there is no alternating force pushing or pulling. This is also why DC has no reactive power. As such, using the formula for Z=R+jX you can see that AC would have a different impedance.

And yes, i believe stranded is used because you cant really bend solid copper like you can stranded.

But this is a rule that is often overlooked, since grounding electrode conductors from a pannel may be 6 AWG or larger and are installed in conduit for mandatory protection, but are installed as solid copper instead of stranded?

 

[wwhitney]

However, I have no idea why the Table 8 DC resistance values are slightly different for solid and stranded, when they are both listed as 10380 cm.

A plausible answer that google turned up is that it's due to the twist of the strands. So each strand is longer than the cable as a whole (except maybe the center strand?) and the effect adds up to 2-3% of the cable length.

Cheers, Wayne

 

[Julius Right]

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%]

 

[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%]

I know there are a lot of other variables such as a conductor within ambient temperature but to simplify calculations, certain assumptions are made in NEC such as wire being within an ambient temperature of 30 degrees Celsius, otherwise you can spend all day trying to account for every variable but never be exact since variables change at any given instant and fractions being dropped