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Compensate for Load Voltage Drop with a Transformer 1

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brandon05

Electrical
Jan 10, 2005
5
I have a situation where I have a 277 Vac, 20 A load about 750 ft from the source. Due to the voltage drop I would like to us a 480/277 transformer at the source and bump the secondary voltage to approximately 285 Vac to compensate for the voltage drop.

Two problems I have is; that I will need to purchase components (Lighting Panel and Circuit Breakers) rated for the higher voltage and finding an appropriate transformer. If I adjust the primary taps on the transformer I will exceed the rating of the secondary because the source is 480 V. If I use a Buck-Boost Transformer it will have to be at the load, which is not acceptable. Please correct me if I’m wrong.

I would appreciate any information or assistance.

Brandon
 
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Tap the secondary ~4% above nominal to account for the drop based on full load. I'm assuming the secondary (at source) will be 480/277. Size the feeder such that you get about a 4% drop at the panel at full load. Worst case is a positive 4% swing in voltage at the panel from full load to no load. Of course the larger the feeder, the closer to nominal you can tap the transformer.


Hopes this helps,

Sense
 
Thank you sense. You mentioned using the secondary tap however I have not been able loacte a transformer with a secondary tap. Is this a special order item or standard on most industrial transformers.

Sometime I get confused when denoting the voltage. For 480/277 source the primary voltage is 480 and the secondary 277. Please adive.

Thanks again

Brandon
 
Better and more economical option would be to use a buck-boost transformer, just sized for the load, at the load end.
 
I agree rbulsara but it not an option for this application. Thank you.
 
sorry, i should have read the whole question..

as for the rating of breaker, cables etc. goes, most devices are rated up to 600V, if you check its compelete listing.

But above all 285 is barely 2.9% change is well within tolerance and average deviation of utitly voltages of -15% to +10%..for which all electrical devices are expected to fucntion..so I would not worry about device ratings. In fact this range would be covered by tranformer taps.

If you already have 480Y/277 supply, you can use buck-boost tranfromer at the "Supply" end too.
 
You could also have an open circuit voltage of 550 volts phase to phase. I have seen some electrical systems where the power quality problem is voltage swells.

Also, a 2 winding dry transformer has an impedance of around 5.4% which means that an extra transformer will be murder on inductive loads, particularly when motors start. Also, each transformer introduces no load and full load resistive losses. You would also need power factor correction capacitors that are hooked up to both current and voltage sensors so that the power factor can be made capacitive on the secondary of a 2 winding transformer so that the inductance of the transformer acts a a voltage booster; however, the capacitors will cause a severe voltage swell if a load suddenly drops out.

What would be more economical is to go to and buy some aluminum alloy weldment lugs that you can tungsten-inert-gas weld to the ends of aluminum conductors. For what copper costs you can put in twice as much aluminum cross section and get better voltage drop. For this distance fatter wires would be the best bet for controlling voltage drop.

If First Energy had used welded lugs for all wire terminations and splices on the Chamberlin-Harding 345 KV line the 14 August 2003 blackout would never have happened. The only way that line could have sagged into trees at 44% of rated current ( 20% of maximum theoretiucal heating ) is if 1/3 or 1/2 of the wire strands were conducting.
 
I suppose I'm confused. When referring to the "source", I'm assuming something like a 13.8 kV utility. Is this application outdoors or inside a building? For some reason I was assuming you were wanting to use a pole mounted transformer with a 480/277 secondary and then route alluminum (overhead??) the 750 ft to the panel (which has a load of 20 amps).
 
Senselessticker

I believe I'm providing the confusion. I don't think I used the proper notions to describe the situation and I didn't understand the complete system.

To clarify I have a 277 Light Panel in an electrical room. Due to the voltage drop the lighting panel would not be an option without using either extremely large copper wire (aluminum is not an option) or a buck-boost transformer per circuit either at the load 750ft away or in the electrical room near the panel.

I thought that there was a 480 to 277V transformer supplying the light panel, which is stupid. A 277 Vac panel is powered from a 480V 3phase Y supply. 277 line to neutral and 480 line to line. The correct notion should have been 4160 - 480/277. Correct?

One Alternative would be to use a 3 phase 480V buck-boost transformer to supply a 480/277V Light Panel at a voltage of 493/285V. I will still need to use large cable but not quite as large as without the buck-boost.

I apologize for the earlier confusion. I am new to this arena of engineering and still learning. Again thank you for your assistance and patience.

Brandon
 
If you are only dealing with one 20 amp load (or one circuit for that matter) that is 750 ft from a lighting panel, just size the conductor approprietly. Also, is the load actually 20 amps or is that the breaker setting? If you are powering luminaires and the breaker is set at 20, do your calcs for VD based on actual load not the breaker.

Sense
 
Most medium voltage transformers come standard with 2 ±2.5% taps. You would need to make sure that the 4 kV voltage is not high at light load in order to use the transformer taps (or to use a buck-boost transformer). The utility might provide as much as 5% above the nominal voltage at light load.

I would agree with Sense that if your lighting load is only 20A, the best solution is to use larger conductor. You will also lower losses this way which would partially offset the extra conductor cost.

If you have multiple fixtures, you could run 3-phase and serve about 1/3 of the fixtures from each phase. This would greatly reduce the voltage drop. The current would be about 1/3 and the impedance about 1/2 for the same size wire.
 
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