KVA & VD

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I need a small transformer to supply misc 120v loads. Problem is the distance from xfmr. A 5 KVA is more than adequate but how do I, or do I need to factor transformer size? Distance is 800’. All other things being equal would a 7.5 be better at motor start? Motor is on an overhead door. 1 HP or less.
 

oldsparky52

Senior Member
If you are saying the 120V load is 800' from the transformer, then can't you get the transformer closer to the load?

As to your question, the VD is on the wire and not the transformer. I don't think a 7.5kVA transformer will offer any better performance than a 5kVA if the load is 5kVA or less.
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
Do you know the starting current required by the garage door opener motor? Or the letter code to indicate kVA/HP with locked rotor.
For example, PSC motor starting current would be significantly less than split-phase type.
Perhaps a smaller transformer could serve loads along the way, with another for the door opener motor closer to it? What works best depends on the detailed requirements and the total costs of wiring, transformers, etc.
 
I need a small transformer to supply misc 120v loads. Problem is the distance from xfmr. A 5 KVA is more than adequate but how do I, or do I need to factor transformer size? Distance is 800’. All other things being equal would a 7.5 be better at motor start? Motor is on an overhead door. 1 HP or less.

I think you are questioning the voltage drop from the transformer and questioning if a larger tranny will have less VD correct? I have wondered this too and how to weigh it out. I believe a larger tranny will have less VD and less load losses for the same current, but more no load losses and of course more capital cost.
 

paulengr

Senior Member
I think you are questioning the voltage drop from the transformer and questioning if a larger tranny will have less VD correct? I have wondered this too and how to weigh it out. I believe a larger tranny will have less VD and less load losses for the same current, but more no load losses and of course more capital cost.

Think of it this way. Voltage is electron pressure. You have 120 Volts of electron pressure. As it flows down through your undersized pipe it loses pressure. So at the end of the pipe you see a voltage (pressure) loss. That’s as long as the biggest loss is the wiring which it usually is. But if the transformer is running over name plate then it becomes a contributor too since it will start to drop in voltage. But you can easily measure that. Under normal circumstances a transformer voltage will dip with load but voltage regulation is generally pretty tight (under 1%). If you are arguing 5 kVA vs. 7.5 kVA don’t expect anything. 5 kVA vs 50 kVA you might see a small change. Or using our pressure analogy we start to notice that the transformer is a pump and it’s a big pump but it’s not infinite. So as long as we’re at name plate or less we shouldn’t expect to lose pressure.

For the truly nerdy though you might not believe anything without a calculator. So here we go...

Check %Z on the name plate. Tiny transformers like this are usually only 1-2%. I work with industrial equipment so to me transformers under around 1000 kVA are tiny. So let’s say it’s 2% Z. So max rated amps out at 5 kVA single phase 120 V is 5000 / 120 = 41.67
A full load. So if we want to calculate a dead short it’s 41.67 /.02 = 2083 and at that point we have 5000 / 2083 = 2.4 V. This is where the only resistance is the transformer secondary coil. The transformer is a 120 V ideal voltage source in series with a 120 / 41.67 = 2.88 ohm impedance. This assumes infinite available current on the primary side which is usually a good assumption for these types of issues. So if you know the load and the impedance of your cable you can easily estimate voltage drop including the transformer itself. Transformer engineers might use some much more complicated math but for our purposes that’s a waste of time.


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paulengr

Senior Member
I think you are questioning the voltage drop from the transformer and questioning if a larger tranny will have less VD correct? I have wondered this too and how to weigh it out. I believe a larger tranny will have less VD and less load losses for the same current, but more no load losses and of course more capital cost.

Think of it this way. Voltage is electron pressure. You have 120 Volts of electron pressure. As it flows down through your undersized pipe it loses pressure. So at the end of the pipe you see a voltage (pressure) loss. That’s as long as the biggest loss is the wiring which it usually is. But if the transformer is running over name plate then it becomes a contributor too since it will start to drop in voltage. But you can easily measure that. Under normal circumstances a transformer voltage will dip with load but voltage regulation is generally pretty tight (under 1%). If you are arguing 5 kVA vs. 7.5 kVA don’t expect anything. 5 kVA vs 50 kVA you might see a small change. Or using our pressure analogy we start to notice that the transformer is a pump and it’s a big pump but it’s not infinite. So as long as we’re at name plate or less we shouldn’t expect to lose pressure.

For the truly nerdy though you might not believe anything without a calculator. So here we go...

Check %Z on the name plate. Tiny transformers like this are usually only 1-2%. I work with industrial equipment so to me transformers under around 1000 kVA are tiny. So let’s say it’s 2% Z. So max rated amps out at 5 kVA single phase 120 V is 5000 / 120 = 41.67
A full load. So if we want to calculate a dead short it’s 41.67 /.02 = 2083 and at that point we have 5000 / 2083 = 2.4 V. This is where the only resistance is the transformer secondary coil. The transformer is a 120 V ideal voltage source in series with a 120 / 41.67 = 2.88 ohm impedance. This assumes infinite available current on the primary side which is usually a good assumption for these types of issues. So if you know the load and the impedance of your cable you can easily estimate voltage drop including the transformer itself. Transformer engineers might use some much more complicated math but for our purposes that’s a waste of time.
 
Think of it this way. Voltage is electron pressure. You have 120 Volts of electron pressure. As it flows down through your undersized pipe it loses pressure. So at the end of the pipe you see a voltage (pressure) loss. That’s as long as the biggest loss is the wiring which it usually is. But if the transformer is running over name plate then it becomes a contributor too since it will start to drop in voltage. But you can easily measure that. Under normal circumstances a transformer voltage will dip with load but voltage regulation is generally pretty tight (under 1%). If you are arguing 5 kVA vs. 7.5 kVA don’t expect anything. 5 kVA vs 50 kVA you might see a small change. Or using our pressure analogy we start to notice that the transformer is a pump and it’s a big pump but it’s not infinite. So as long as we’re at name plate or less we shouldn’t expect to lose pressure.

For the truly nerdy though you might not believe anything without a calculator. So here we go...

Check %Z on the name plate. Tiny transformers like this are usually only 1-2%. I work with industrial equipment so to me transformers under around 1000 kVA are tiny. So let’s say it’s 2% Z. So max rated amps out at 5 kVA single phase 120 V is 5000 / 120 = 41.67
A full load. So if we want to calculate a dead short it’s 41.67 /.02 = 2083 and at that point we have 5000 / 2083 = 2.4 V. This is where the only resistance is the transformer secondary coil. The transformer is a 120 V ideal voltage source in series with a 120 / 41.67 = 2.88 ohm impedance. This assumes infinite available current on the primary side which is usually a good assumption for these types of issues. So if you know the load and the impedance of your cable you can easily estimate voltage drop including the transformer itself. Transformer engineers might use some much more complicated math but for our purposes that’s a waste of time.

I am not seeing the part in red as being meaningful or even valid. You dont have nameplate KVA at secondary SCC :?

Here is a good past thread on transformer voltage regulation:
https://forums.mikeholt.com/forum/a...ering/69409-transformer-voltage-drop?t=119577

and a good online calculator:

https://voltage-disturbance.com/engineering-calculators/transformer-calculator/

Both indicate that regulation is highly dependent on load PF. At full loading, with a poor PF regulation seems to be about %Z. With close to unity PF, voltage regulation seems to be much lower, perhaps 25% of %Z. Of course both are dependent on the X/R ratio.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
FWIU, for practical purposes, the full load VD at the transformer would be the impedance of that transformer. I couldn't find the impedance of my transformer but let's guess it's at 2%. At full load the VD @ the transformer would be 120-2.4=117.6. I would use that as my starting voltage and figure VD from the conductors after that. Looks like it should be of very little concern.

I have not gone to the reference calculator yet.
 
FWIU, for practical purposes, the full load VD at the transformer would be the impedance of that transformer. I couldn't find the impedance of my transformer but let's guess it's at 2%. At full load the VD @ the transformer would be 120-2.4=117.6. I would use that as my starting voltage and figure VD from the conductors after that. Looks like it should be of very little concern.

I have not gone to the reference calculator yet.

I do not believe that is accurate. Try that calculator. With a relatively high PF, VD will be quite a bit less than %Z. You need a real low Pf to get VD to be close to %Z. Its kinda weird, but transformer losses dont all show up as VD on the secondary.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
I do not believe that is accurate. Try that calculator. With a relatively high PF, VD will be quite a bit less than %Z. You need a real low Pf to get VD to be close to %Z. Its kinda weird, but transformer losses dont all show up as VD on the secondary.

I had to guess at the X/R ratio. The nearest 1ph unit was 1.1, that I could find. Calculator jumps from 1 to 1.5.

Guessing again @ 2% transformer impedance. X/R of 1.5, 100% load @ 80% PF gave a Xfmr VD of 1.89%. Fifty per cent load was .99 Volt.

X/R of 1 gave 1.98%



Thank you. I will try to remember to take measurements when we are done.

eta: changing to high PF and lower xfmr impedance reduced the per cent VD by about half.
 
I had to guess at the X/R ratio. The nearest 1ph unit was 1.1, that I could find. Calculator jumps from 1 to 1.5.

Guessing again @ 2% transformer impedance. X/R of 1.5, 100% load @ 80% PF gave a Xfmr VD of 1.89%. Fifty per cent load was .99 Volt.

X/R of 1 gave 1.98%



Thank you. I will try to remember to take measurements when we are done.

eta: changing to high PF and lower xfmr impedance reduced the per cent VD by about half.

Yeah X/R is hard to find. There are some tables out there that give approx values and low ones seems about right for small single phase.

I found this which has a bunch of data, also losses. Looks dated however, I see something that says "1966 designs" in it:blink:
https://www.fs.fed.us/database/acad/elec/greenbook/10_shortcalc.pdf
Good thread. Before this I assumed %Z was essentially the VD.
 
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