Estimate of typical resisdential feed utility transformer secondary inductance

[ELA]

Can anyone provide an estimate or approximate range of secondary inductance for a typical size residential area transformer (7.2kV/240V)?

I am interested in the inductance value. Understood it may vary over a range based on other factors. Thus a range would also be good.
Or the secondary impedance ( @60HZ across full secondary) if I can then use this value to back calculate the inductance, so this value can then be used to estimate the impedance at other higher frequencies?

In particular I would like to estimate the impedance that a frequency of 131Khz would "see" when attempting to flow through the secondary.
The 131Khz originating inside of a customers home and traveling out to the transformer via one leg. I want to estimate how much might be expected to return on the other leg back into the home ( ignoring the center tap initially as well as service conductors impedance). If a value is known, can secondary to primary capacitance be ignored in this calculation as insignificant or not?

Hope this is not too vague, I am just trying to get a ball park figure.

 

[mivey]

Here are some sample sets. The size will depend on the type & # of homes on the transformer. 15-25 kVA is probably average for a rough calc.

10 kVA: %R = 1.9, %Z= 2.3
15 kVA: %R = 1.7, %Z= 2.1
25 kVA: %R = 1.6, %Z= 2.2
50 kVA: %R = 1.3, %Z= 2.2


10 kVA: %R = 1.4, %Z= 1.6
15 kVA: %R = 1.3, %Z= 1.7
25 kVA: %R = 1.2, %Z= 1.8
37.5 kVA: %R = 1.1, %Z= 1.9
50 kVA: %R = 1.1, %Z= 1.7


10 kVA: %R = 2.5, %Z= 2.6
15 kVA: %R = 2.1, %Z= 2.6
25 kVA: %R = 1.6, %Z= 2.6
37.5 kVA: %R = 1.6, %Z= 3.0
50 kVA: %R = 1.2, %Z= 2.3

 

[ELA]

Thank you for your input Mivey,

From a quick search I see that %Z is typically used for short circuit calculations. I searched a bit more and came up with this site:

http://www.jcmiras.net/jcm/item/117/

I do not have time right now but I will see if I can use that information to work out some values. Or if you would be so kind as to elaborate/calculate that would be much welcomed.

 

[mivey]

And x = sqrt(z^2 - r^2)

 

[mivey]

Thank you for your input Mivey,

From a quick search I see that %Z is typically used for short circuit calculations. I searched a bit more and came up with this site:

http://www.jcmiras.net/jcm/item/117/

I do not have time right now but I will see if I can use that information to work out some values. Or if you would be so kind as to elaborate/calculate that would be much welcomed.

You will need to convert to ohms for your frequency difference calc. I am mobile on my cell so can't help you right now. Post again if you don't get it worked out and I will calculate it for you when I get back to my office.

 

[gar]

120513-1327 EDT

ELA:

It looks like you are looking at PLC (power line communication). I am doing some work directly looking at the PLC signal from a TED 1000 MTU (measurement and transmitting unit). Their carrier frequency is about 130 kHz. X10 is slightly lower.

Presently I use a parallel resonant circuit of about 2.5 mH and 600 pfd. Obviously it is fed as a series resonant circuit. Actually I am using the output from a secondary coil on the toroid with a primary of 2.5 mH. The turns ratio is 180 to 36.

On the phase with the TED MTU and at the same socket the peak voltage is

600 pfd 2.90 V about 130 kHz
500 pfd 0.33 V about 142 kHz
700 pfd 0.32 V about 120 kHz

On the opposite phase with no coupling between the phases, except the pole transformer, and maybe, but doubtful a neighbor load, the signal was

600 pfd 0.030 V
and with TED disconnected
600 pfd 0.004 V --- this is the noise floor.

This is why a capacitive bridge is needed between the phases for PLC.

In the TED 5000 MTU they provide output to both phases, but this may reduce the signal level on each phase. The various different loads on the phases should have a substantial effect on the PLC signal level.

PLC was a very poor communication choice for the TED system. Many users have substantial problems.

Since mid summer of last year I have had a 50 kVA pole transformer. Also the transformer location was moved so now there is an added wire length of about 80 feet.

.

 

[ELA]

Hello and thank you for your input Gar,

Yes this would be powerline carrier and all of what you said is a given.

In the reference I pointed to, they mention the calc they do is primary and I desire secondary?

Can you help with my unknowns please? When you try to explain to people that very little returns on the opposite leg they sometimes don't believe it. The biggest reason being the relative low impedance path back to the source via the center tap (neutral), without intentional coupling of course. I had wanted to model it so that a calculation can be presented to help with that.

 

[gar]

120513-1626 EDT

ELA:

I do not believe the source impedance of the pole transformer has much to do with 130 kHz coupling from 1/2 of the secondary to the other secondary half.

For power frequencies, 60 Hz, a fairly simple equivalent circuit from input to output would be a series inductance and resistance associated to the primary, and another series resistance and inductance for the secondary. A shunt inductor would be located from the mid-point to neutral. Capacitances are neglected. The input or output impedances can be reflected to the output or input by the turns ratio squared.

The series resistances have to be associated with the copper resistance of the coils. The series inductances result from leakage flux between the primary and secondary, and likely would be split equally between the primary and secondary after adjustment for the turns ratio.

The coupling between the two secondary halves is very close to unity. Same is true primary to secondary. The deviation from unity is the result of leakage flux. So we are talking in the high 90% range.

But a power transformer at 130 kHz is a very lossy device and therefore coupling between coils is very poor.

A 10 to 1 transformer when viewed from the secondary --- rated at 1 kVA, 100 V, and a series resistance of 1% would have a voltage drop of 1 V at 10 A, or a resistance of 0.1 ohms. Reflected to the primary this is 100 * 0.1 or 10 ohms.

Generally the resistive component of the primary would be less than N² times the resistance of 1/2 of the secondary because you would want about equal power dissipation in equal volumes of copper in the coils under proper full load loading.

.

 

[gar]

120514-2152 EDT

ELA:

I did some tests on a Signal Transformer 241-6-20. This is designed for rectifier type circuits. This is a 115 V input 60 Hz, split bobbin, with a 20 V @ 1.5 A center tapped secondary. The secondary is all on one side of the bobbin.

At 60 Hz the unloaded coupling between the two halves of the secondary was about 0.995 . At 120 kHz this dropped to about 0.83 . The performance was not nearly as bad as I had expected. There was a resonance somewhat above 200 kHz, but this did not change the coupling ratio much. How the core material compares to a pole transformer I do not know. The equivalent series inductance from leakage flux between the the two halves of the pole transformer and the resistive loads on each half of the secondary may be more dominate than the core losses in the transformer.

It may be necessary for you to run tests on an actual pole transformer to see what its characteristics are at the 120 to 130 kHz frequency range.

.

 

[gar]

120515-0836 EDT

ELA:

More to consider.

First, I think PLC is a very poor way to get rock solid communication with no problems. Thus, I would not design a system around this method as TED has done. If you look at many of the problems TED users have, then it is seen that PLC is a significant one.

To create a reliable communication path for TED put an X10 in-line filter at the main panel. On the output side of this filter create a branch circuit that has nothing but TED devices on it. The filter does two things.

1. The filter creates a high impedance between the isolated branch circuit and all other loads on the main panel at the 120 to 130 kHz frequency range. This reduces loading on the TED transmitter, and increases the signal level to all devices on the isolated branch circuit. Also reduces the TED signal on all the non-isolated circuits. This way neighbors don't receive your measurements, and X10 or other similar systems on the non-isolated side have interference from TED minimized.

2. The filter reduces noise on the isolated circuit that originate from the non-isolated circuits. Fluorescent lights and dimmers seem to be a source of noise.


A single original TED 1000 sends a continuous packet of data for about 100 milliseconds, 0.1 second, about 6 cycles at 60 Hz. This consists of eleven data bytes at 1200 baud. Any claim they only send on zero-crossings is nonsense.

I believe if this continuous signal is transferred thru a power transformer that the signal amplitude will be modulated thru the period of the 60 Hz cycle. This would result from the change in leakage flux as the core is driven into saturation each half cycle.

.

 

[ELA]

Hi gar,
You stated an awful lot without answering my question :huh: But thanks anyways.... :happyyes:

I calculated ~234 microhenries for a 15KVA transformer, %Z =2.3 This represents approx. 193 ohms at 131Khz.

I have tons of first hand experience with power line communications if you want to exchange information on that we can PM if you like.

 

[mivey]

Looks like you two are off to the races. But for grins I made the following estimates this morning for a 7200-240V 25 kVA transformer:

R1 = 18.06 Ω
R2 = 17.7 mΩ
X1 = 21.57 Ω
X2 = 24.0 mΩ
Xm = 120 kΩ
Rm = 202 kΩ

 

[gar]

120515-1144 EDT

ELA:

I did not answer your question because I do not know that you can extrapolate 60 Hz data to what happens at 130 kHz. Also I am not inclined to think that a simple equivalent circuit exists for a 60 Hz power transformer at 120 kHz.

Relative to Insteon I have no experience. In the TED discussions there are a lot of reports on cross interaction between X10 and TED. Apparently fewer problems with Insteon.

TED is a whole house power/energy measuring system.

Note in my previous post that I indicated a resonance in the Signal transformer around 200 kHz. I don't know where this might be in a large power transformer. Also there are likely multiple different resonances.

With your tester what range of attenuation of the Insteon signal do you see from one half of a secondary to the other half?

.

 

[ELA]

Thanks Mivey,
but I do not understand what those values are supposed to represent?

Gar,
As I told you I would be happy to exchange information on power line communications via PMs if you like.
Or if you want to start a thread on it I would be happy to participate there.

 

[Speedskater]

Here is a Bill Whitlock paper on audio transformers. (Mr. Whitlock makes very nice audio transformers)
You might find some of the math sections helpful.

http://www.jensen-transformers.com/an/Audio Transformers Chapter.pdf

 

[mivey]

Thanks Mivey,
but I do not understand what those values are supposed to represent?

Values for the transformer equivalent circuit:
R1 & X1 are for primary series impedance followed by the primary Rm & Xm parallel impedances for the no-load losses and magnetizing reactance. These are followed by an ideal transformer. This is followed by R2 & X2 for secondary series impedance.

[FONT=Courier New]                           N1:N2
-----R1---X1------|--------)||(-[FONT=Courier New]---R2---X2----|[/FONT]
|                 |        )||(               |
S              -------     )[FONT=Courier New]||(               |[/FONT]
O              |     |     )[FONT=Courier New]||(               L[/FONT]
[FONT=Courier New]U              |     |     )[FONT=Courier New]||(               O[/FONT][/FONT]
R             Rm     Xm    )[FONT=Courier New]||(               A
[/FONT][FONT=Courier New][FONT=Courier New]C              |     |     )||(               D[/FONT]
E              |     |     )[FONT=Courier New]||(               |[/FONT]
[FONT=Courier New]|              |     |     )[FONT=Courier New]||(               |[/FONT][/FONT]
---------------|-----|-----)||(---------------|[/FONT]
[/FONT]

 

[gar]

120515-2153 EDT

ELA:

I do not want to start a PLC thread. The only reason I am working with PLC is because the TED people use it as the means of getting data out of the main panel.

The Jensen discussion on audio transformers looks quite good.

.

 

[ELA]

Thank you very much for your efforts Mivey,

Your X2 value of 24 mohms is roughly 1/2 of what I got using your 25KVA example with %Z =2.1.

I am by no means an expert at these types of calculations. The discrepancy is not really all that important to me as I was just looking for a ball park value.
I will divide the difference and call it good.

Thanks,
Ela

 

[mivey]

Thank you very much for your efforts Mivey,

Your X2 value of 24 mohms is roughly 1/2 of what I got using your 25KVA example with %Z =2.1.

I am by no means an expert at these types of calculations. The discrepancy is not really all that important to me as I was just looking for a ball park value.
I will divide the difference and call it good.

Thanks,
Ela

Did you account for the fact that the primary and secondary impedances are normally combined by transferring one set to the other side?

 

[ELA]

Hi Mivey,
Since I did not know what to do I followed the example at the site I posted a link to early on. Being aware of the dangers in relying on arbitrary internet sites ( coupled with my interpretation of same) I was looking for some help here.

The site referenced found a "Zbase" calculated from MVA and KV. From there they used the Zbase and %Z to find the impedance. I was not familiar at all with using "per unit" values that are common in the power distribution realm.

Would you please detail the method you use so that I may learn from it?

Thanks,
Ela