And maybe I didn't.
"This means that turns ratio for my 240/120V tranny is not exactly 2:1."
Start from there and explain why that is a mistake.
No load losses - normal vs reverse fed
- Thread starter electrofelon
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"This means that turns ratio for my 240/120V tranny is not exactly 2:1."
Start from there and explain why that is a mistake.
- Location
- Massachusetts
The thread is here for all to see and make their own call.
I just prefer facts.
Electric-Light
Senior Member
Just checked out a 1:1 600VA isolation transformer. The no load is a boost of 1.03 on the secondary. Check the output voltage on an old school AC adapter if you have one around. The no load voltage on those can be high by as much as 50%. Load it down to the rated amp and the voltage drops down to something in line with the rated voltage. Step load regulation is more pronounced on tiny transformers. It's for this reason one neighbor's AC is much more likely to cause another neighbor's lights to dim noticeably every time their stat cycles when they're in a rural area that is fed from a tiny pole piglet.
In theory transformers conform exactly to the ratio but out in the real world, they're made of real world materials with resistance and needs compensation. We regularly ignore things like air friction, wiring resistance and such in the classroom for the sake of simplicity so the focus remains on the principle. If the wire you wind around a nail was an ideal coil, you could attach a battery, then short the wires together and it would stay on indefinitely like an MRI machine.
This surprised me a bit at first too but the worst inrush occurs when its switched on at zero crossing and the least when its turned on at the peak.
In theory transformers conform exactly to the ratio but out in the real world, they're made of real world materials with resistance and needs compensation. We regularly ignore things like air friction, wiring resistance and such in the classroom for the sake of simplicity so the focus remains on the principle. If the wire you wind around a nail was an ideal coil, you could attach a battery, then short the wires together and it would stay on indefinitely like an MRI machine.
This surprised me a bit at first too but the worst inrush occurs when its switched on at zero crossing and the least when its turned on at the peak.
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- Location
- Massachusetts
We are not talking about small transformers. It was established and not questioned that small transformers have compensated windings.
But larger NEMA transformers do not have this compensation.
Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
ElectroFelon…
Returning to you original query... "Impact of reverse-feeding on transformer losses!" Perhaps my answer using a single-phase xfmr was too elementary for some of the posters! So, following is additional detail.
While the Xfmr is considered a ‘simple’ electrical apparatus, the mfg’r has practical problems to consider. For example, consider a 3-core (or 3-limb or 3-leg) xfmr! Most assume it is symmetrical regarding development of magnetic flux density and excitation current! The simple fact is flux-density in each of the 3-cores is uneven!
It is not by accident the Xfmr is described as having a primary-winding and secondary-winding! Then design parameters are directed toward having the supply-source connected to the primary-winding! Why?
Unmatching magnetic flux-densities in the 3-cores result in uneven induced-voltages in the secondary-winding! Then, the mfg’r, ‘adjusts’ the effective turns-ratio of each phase so that secondary output-voltages are better matched! Usually, the ‘adjustment’ is made to the turns-ratio in the winding having a smaller wire, hence, more latitude in location and accessibility!
Now you can see why reversing-feed could affect the primary-winding’s voltage if used as the output, resulting in the Caveat to contact the Mfg’r!
Reur comment on ‘Inrush’… I suggest searching MHF files! There is a wealth of info! Here’s a hint… determining ’Inrush-current’ is not simple. The process is quite complicated! Any ‘Inrush-current’ problem must consider two components… the transient-period (the first few cycles, including the first-cycle-peak) and then the steady-state region (a few to many seconds)! Furthermore, the 3-phase currents never coincide together!
Gentle people! Please do not misconstrue the above as Xmrs-101!
Regards, Phil Corso
Returning to you original query... "Impact of reverse-feeding on transformer losses!" Perhaps my answer using a single-phase xfmr was too elementary for some of the posters! So, following is additional detail.
While the Xfmr is considered a ‘simple’ electrical apparatus, the mfg’r has practical problems to consider. For example, consider a 3-core (or 3-limb or 3-leg) xfmr! Most assume it is symmetrical regarding development of magnetic flux density and excitation current! The simple fact is flux-density in each of the 3-cores is uneven!
It is not by accident the Xfmr is described as having a primary-winding and secondary-winding! Then design parameters are directed toward having the supply-source connected to the primary-winding! Why?
Unmatching magnetic flux-densities in the 3-cores result in uneven induced-voltages in the secondary-winding! Then, the mfg’r, ‘adjusts’ the effective turns-ratio of each phase so that secondary output-voltages are better matched! Usually, the ‘adjustment’ is made to the turns-ratio in the winding having a smaller wire, hence, more latitude in location and accessibility!
Now you can see why reversing-feed could affect the primary-winding’s voltage if used as the output, resulting in the Caveat to contact the Mfg’r!
Reur comment on ‘Inrush’… I suggest searching MHF files! There is a wealth of info! Here’s a hint… determining ’Inrush-current’ is not simple. The process is quite complicated! Any ‘Inrush-current’ problem must consider two components… the transient-period (the first few cycles, including the first-cycle-peak) and then the steady-state region (a few to many seconds)! Furthermore, the 3-phase currents never coincide together!
Gentle people! Please do not misconstrue the above as Xmrs-101!
Regards, Phil Corso
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Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
Electric-Light
Senior Member
sooo, i wonder if those big ass transformers between the alternators and the grid at pump-up hydros are bi-directional
electrofelon
Senior Member
- Location
- Cherry Valley NY, Seattle, WA
Very interesting Phil. So does the above form of compensation commonly exist in the typical transformers used in electrical construction, say your typical 75KVA 480D to 120/208Y?
big john
Senior Member
- Location
- Portland, ME
They would have to be for those plants because the generators run as synchronous motors during off-peak hours to refill the reservoir. But that's also true for any powerplant GSU transformer I've ever seen: Even the plant isn't generating then the station service power is flowing "backwards" through that grid tie in order to keep the lights on and the ancillary equipment running.
jminer99er
Member
- Location
- Sacramento, CA
GE white paper on backfeeding dry type
Good find!
gar
Senior Member
- Location
- Ann Arbor, Michigan
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- EE
170116-2145 EST
Experimental results:
1. Transformer is a P-3197, probably Stancor, from later 1940s, but before 1950. Rubber insulated wire. Predates PVC. Input 117 V 60 CPS, and rated 80 W.
2. Primary resistance 4.7 ohms.
3. Secondary resistance 5.9 ohms.
4. Using no load to determine turns ratio.
Primary input 110.1 V secondary 116.2 V, ratio secondary to primary 1.055 .
Secondary input 116.1 V primary 110.1, ratio secondary to primary 1.054 .
Good correlation.
5. Using a Kill-A-Watt for source measurement, and a Fluke 27 for output voltage. An 0.2 V difference in calibration between the two instruments was corrected for.
6. First powering the primary with no secondary load. Then power input to secondary and no load on primary.
Vin = 110.1 V, Iin = 0.09 A, Pin = 4.5 W, VAin = 10.1 VA, PF = 0.41 --- Vout = 116.2 V.
Vin = 116.1 V, Iin = 0.09 A, Pin = 4.6 W, VAin = 11.0 VA, PF = 0.41 --- Vout = 110.1 V.
Using these values both tests had close to the same core flux density. To obtain a more accurate comparison more accurate instruments would be required and a more stable voltage source than my home line voltage.
The results, as should be expected, are essentially the same. If you do not work at the same flux density, then you should not expect core losses to be the same.
If the coils were wound side by side, then you would expect to see the resistance ratio about the square of the turns ratio. But that would make the secondary 5.2 ohms and not 5.9 ohms. Thus, the conclusion is that the secondary is wound on top of the primary.
With a 75 W incandescent as a secondary load the results are:
Vin = 110.1 V, Iin = 0.66 A, Pin = 72.7 W, VAin = 72.9 VA, PF = 0.99 --- Vout = 108.9 V.
.
Experimental results:
1. Transformer is a P-3197, probably Stancor, from later 1940s, but before 1950. Rubber insulated wire. Predates PVC. Input 117 V 60 CPS, and rated 80 W.
2. Primary resistance 4.7 ohms.
3. Secondary resistance 5.9 ohms.
4. Using no load to determine turns ratio.
Primary input 110.1 V secondary 116.2 V, ratio secondary to primary 1.055 .
Secondary input 116.1 V primary 110.1, ratio secondary to primary 1.054 .
Good correlation.
5. Using a Kill-A-Watt for source measurement, and a Fluke 27 for output voltage. An 0.2 V difference in calibration between the two instruments was corrected for.
6. First powering the primary with no secondary load. Then power input to secondary and no load on primary.
Vin = 110.1 V, Iin = 0.09 A, Pin = 4.5 W, VAin = 10.1 VA, PF = 0.41 --- Vout = 116.2 V.
Vin = 116.1 V, Iin = 0.09 A, Pin = 4.6 W, VAin = 11.0 VA, PF = 0.41 --- Vout = 110.1 V.
Using these values both tests had close to the same core flux density. To obtain a more accurate comparison more accurate instruments would be required and a more stable voltage source than my home line voltage.
The results, as should be expected, are essentially the same. If you do not work at the same flux density, then you should not expect core losses to be the same.
If the coils were wound side by side, then you would expect to see the resistance ratio about the square of the turns ratio. But that would make the secondary 5.2 ohms and not 5.9 ohms. Thus, the conclusion is that the secondary is wound on top of the primary.
With a 75 W incandescent as a secondary load the results are:
Vin = 110.1 V, Iin = 0.66 A, Pin = 72.7 W, VAin = 72.9 VA, PF = 0.99 --- Vout = 108.9 V.
.
electrofelon
Senior Member
- Location
- Cherry Valley NY, Seattle, WA
Thanks, good info. Strange transformer. Now repeat with 75 kva 480-120/208 
Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
Gar...
Can you perform a Short-Circuit Test? Also what are xfmr's physical dimensions?
Phil
Can you perform a Short-Circuit Test? Also what are xfmr's physical dimensions?
Phil
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170117-1147 EST
electrofelon:
It is not a strange transformer. It is what you should expect. This is a single phase two winding transformer.
This is a small transformer 80 W on nameplate, thus 80 VA is possibly a better descriptor.
I don't have a 75 kVA to test or an appropriate load. But, transformers whether large or small follow pretty much the same theory.
.
electrofelon:
It is not a strange transformer. It is what you should expect. This is a single phase two winding transformer.
This is a small transformer 80 W on nameplate, thus 80 VA is possibly a better descriptor.
I don't have a 75 kVA to test or an appropriate load. But, transformers whether large or small follow pretty much the same theory.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170117-1155 EST
Phil Corso:
I have not run a short circuit test. I shall try to do that.
However, after my post last night I did some calculations from the available data and did not get a good correlation on power. But there is a reason that does not change the core loss measurement which is basically correct.
I made a determination of the transformer turns ratio from the open circuit voltage measurements that resulted in a ratio of 1.055 . This is incorrect relative to the actual turns ratio because of the percentage of flux lines that do not couple the two coils, leakage flux.
If I use as the turns ratio the value calculated from the open circuit voltage ratio, which is low because of the leakage flux, to predict the secondary current under load from the 75 W bulb, then I predict a higher secondary current than actually occurs, calculated load power from V*I is 108.9*0.66/1.055 = 68.1 W. The measured input power is 72.7 W. The difference 72.7 - 68.1 = 4.6 W, just the ballpark for core loss, nothing for I*R loss. This results from using a turns ratio that is low from ignoring the leakage flux.
Today I did an additional experiment to determine the actual secondary currrent. With the same 75 W bulb load and 110.1 V into the transformer the primary was 0.66 A, as before. and the measured secondary current was 0.58 A. Turns ratio calculated based on the current ratio is 0.66/0.58 = 1.14 . To be expected based on there being leakage flux.
Now the load power calculates to 0.58*108.9 = 63.1 W. The sum of the losses is now 0.66*0.66*4.7 = 2.04 W (primary I*I*R), core 4.5 W, 0.58*0.58*5.9 = 1.98 W, and summed = 8.5 W. The difference between input and load is 72.7 - 63.1 = 9.6 W. Now we have a better correlation, and the results make sense. More accurate measurements should provide a better correlation.
If I used a high peramability material tape wound, close to no air gap, then the input to output voltage ratio would be much closer to the actual turns ratio.
.
Phil Corso:
I have not run a short circuit test. I shall try to do that.
However, after my post last night I did some calculations from the available data and did not get a good correlation on power. But there is a reason that does not change the core loss measurement which is basically correct.
I made a determination of the transformer turns ratio from the open circuit voltage measurements that resulted in a ratio of 1.055 . This is incorrect relative to the actual turns ratio because of the percentage of flux lines that do not couple the two coils, leakage flux.
If I use as the turns ratio the value calculated from the open circuit voltage ratio, which is low because of the leakage flux, to predict the secondary current under load from the 75 W bulb, then I predict a higher secondary current than actually occurs, calculated load power from V*I is 108.9*0.66/1.055 = 68.1 W. The measured input power is 72.7 W. The difference 72.7 - 68.1 = 4.6 W, just the ballpark for core loss, nothing for I*R loss. This results from using a turns ratio that is low from ignoring the leakage flux.
Today I did an additional experiment to determine the actual secondary currrent. With the same 75 W bulb load and 110.1 V into the transformer the primary was 0.66 A, as before. and the measured secondary current was 0.58 A. Turns ratio calculated based on the current ratio is 0.66/0.58 = 1.14 . To be expected based on there being leakage flux.
Now the load power calculates to 0.58*108.9 = 63.1 W. The sum of the losses is now 0.66*0.66*4.7 = 2.04 W (primary I*I*R), core 4.5 W, 0.58*0.58*5.9 = 1.98 W, and summed = 8.5 W. The difference between input and load is 72.7 - 63.1 = 9.6 W. Now we have a better correlation, and the results make sense. More accurate measurements should provide a better correlation.
If I used a high peramability material tape wound, close to no air gap, then the input to output voltage ratio would be much closer to the actual turns ratio.
.
Electric-Light
Senior Member
////////.m,m,m,.
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Electric-Light
Senior Member
Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
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