3 phase voltage drop

[ggunn]

Here's the scenario:

I have 6 inverters totaling 112kW, whose 480VAC 3 phase 4 wire outputs are gathered in an AC combiner. The combiner feeds a wye connected stepdown transformer (480 to 208) 205 feet away. What size single copper conductors do I need to run between the combiner and transformer to keep the voltage drop in that segment below 0.5%. Yes, I know that's very low, but I am under some unusual design constraints.

Please show your work (teach a man to fish) and don't just refer me to a calculator; I need to understand how this works.

 

[iceworm]

Here's the scenario: ... The combiner feeds a wye connected stepdown transformer (480 to 208) 205 feet away. ...

I don't know what "a wye connected ... transformer" means. Is the transformer 480D to 208Y?

ice

 

[ptonsparky]

What is the load?

 

[ggunn]

I don't know what "a wye connected ... transformer" means. Is the transformer 480D to 208Y?

ice

I am a neophyte when it comes to transformers, but I'm told by the inverter company that there has to be a neutral connection, so it has to be a 480Y/277 to 208Y/120 transformer.

 

[ggunn]

What is the load?

It is the output of a bank of PV inverters feeding a service through a step down transformer, not a load.

 

[ggunn]

I have gone on line and found three calculators for three phase voltage drop. Interestingly enough, when fed the same input parameters their answers are significantly different.

Aside from that, I am somewhat confused as to what value to use for the current. The three inverters total 112kW at 480V, and their rated output current totals 135A. 112kW divided by 480V is 233.3A, which is 135A times sqrt3. When I am using a calculator, which value for current should I use, 233.3A or 135A?

To confuse me even more, one site I found said that for a delta connection line current equals phase current times sqrt3, but for a wye connection, line current equals phase current. I had thought, possibly erroneously, that a balanced wye connection was basically the same as a delta connection since there would be no current on the neutral.

 

[ptonsparky]

It is the output of a bank of PV inverters feeding a service through a step down transformer, not a load.

You still need to.know what the load of the transformer and on the secondary side will be. How efficient is.the transformer? Transformers can have a pretty large inrush at startup. Does it need to be limited to.5% then? What type of loads?

 

[ggunn]

You still need to.know what the load of the transformer and on the secondary side will be. How efficient is.the transformer? Transformers can have a pretty large inrush at startup. Does it need to be limited to.5% then? What type of loads?

PV inverters are current sources; the current from them does not depend on loading. For the purpose of this calculation, assume the inverters connect to the transformer and feed directly into the service with no other loads.

At any rate, I am not concerned with transformer losses at the moment, just the voltage drop from the combiner to the transformer.

 

[iceworm]

gg -
Don't take any of this as making fun of you. I'm not. I'm just trying to keep it light.

I have gone on line and found three calculators. Interestingly enough, when fed the same input parameters their answers are significantly different. ....

Forget the on-line calculators. As you can see, they are worse than useless. If you don't know the algolrythm, you can't make any judgements as totheir validity

... I am somewhat confused as to what value to use for the current. ....

You need a lesson in basic three phase power. I'm guessing that never happened in school. Buy a copy of Electric Machinery Fundamentals by Chapman. Another is ieee 242-2001 (Buff book) Protection and Coordination of Industrial and Commercial Systems. I'd also add ieee 141-1993 (Red Book), Electric Power Distribution for Industrial Plants. Yes, it all cost money, but if this is what you are going to do for a living I suggest you get started accumulating references. Look at Amazon used. Half of mine are used - they work just as well.

Okay, 3ph power is a trig problem. Are you okay with trig? How about vectors? I'm thinking trig is okay, vectors maybe not so much. Let me know.

Draw out the vector diagram for a 480V (L-L), 3ph Y. Each leg is 277V, separated by 120 deg. The trig is all 30-60-90 right triangles. The ratio of the sides is 1: sqrt(3):2. sqrt(3) = 1.732. There will be a test on that value tomorrow morning. If you don't pass, put yourself in timeout Sat morning for two hours - NO CARTOONS

3ph power is P = VxIxsqrt(3)for 112kva, I = (112000/480)/sqrt(3) = 135A (Never, ever - not even one time say, "Amps per phase" or "Amps per leg", or anything that even remotely resembles them). Gee, Mr Worm, what if the three phases have differing currents? Then you list them out. A = xxAmps, B = yyAmps, C = zzAmps.

That 135A is for kva, so the pf doesn't matter.

I had thought, possibly erroneously, that a balanced wye connection was basically the same as a delta connection since there would be no current on the neutral.

You'r right, a static, balanced Wye is indistinguishable from a Delta. Cut the neutral and you couldn't tell. (assuming no triplen harmonics) In fact there are equations (already worked out) for Y-D transformation and amazingly D-Y transformation as well.

ice

 

[iceworm]

Now for your quesion

... 112kW, 480VAC, 3 phase, 4 wire ... feeds a wye connected stepdown transformer (480 to 208) 205 feet away. What size single copper conductors do I need to run between the combiner and transformer to keep the voltage drop in that segment below 0.5%. ...

... I'm told by the inverter company that there has to be a neutral connection, so it has to be a 480Y/277 to 208Y/120 transformer.

It is the output of a bank of PV inverters feeding a service through a step down transformer, not a load.

I don't know what to do with the Y-Y xfm or how a Y-Y could (or would) afect the VD. I also would not know what to do with an unbalance load. I don't know what difference it matters if the secondary load is a service or customer load. Service just means the utility owns is - physics is still the same.

So - assumptions:
Load on inverters is balance 3ph, 480V to a xfm (112kva or bigger). Treating as a wye connected load. The basic physics is from NEC Table 9, READ Note 2. Read it again. It's really important - All the physics modeling is right there.

Calc is for copper in steel conduit. With no mor information that you have given, I would normally assume .85pf. See attached two sheets.

For non-paralleled conductors, .85pf - you damn near can't get there from here. 1000 is not quire big enough. For a unity pf, the size drops some. 500s are big enough. If you get stalled. tell where and I (we) will stretch out the stalled part.

ice
ice

 

[dkarst]

It's been a long day so apologies if I'm not thinking straight here. If the system line voltage is 480V, then a 0.5% voltage drop is 2.4V _L-L which implies a line to neutral voltage drop of 2.4/√3 = 1.38 V _L-N. I would prefer to use the method from IEEE red book and a spreadsheet but if I instead apply table 9 NEC as was done above with the line current given I get a conductor effective Z of 1.38/135 or 0.0102 Ω. Assuming 0.85 PF and steel conduit as nothing contrary given I would say a single set of 500 kmcil Cu would be sufficient with that conductor meeting 0.05 Ω to neutral per 1000' and since the one way run is ~ 200'. I guess it is close to the edge given run is 205' but we don't know power factor either so this is all just an engineering estimate until give better assumptions.

 

[dkarst]

With the clarity of some sleep, I ran an exact calculation given the assumptions listed above using IEEE red book method and got 0.498% vs. the desired 0.5% drop.

Getting back to the OP, I googled "voltage drop calculators" and picked the top one on list. It turned out to be Southwire here... http://www.southwire.com/support/voltage-drop-calculator.htm I agree with Ice that many of these provide answers at a furious pace but leave the assumptions behind them shrouded in secrecy. I ran the conditions you provided and it said:

1 conductors per phase utilizing a #400 Copper conductor will limit the voltage drop to 0.48% or less when supplying 135.0 amps for 205 feet on a 480 volt system.
For Engineering Information Only:
335.0 Amps Rated ampacity of selected conductor
0.033 Ohms Resistance (Ohms per 1000 feet)
0.04 Ohms Reactance (Ohms per 1000 feet)
2.4 volts maximum allowable voltage drop at 0.5%
2.26. Actual voltage drop loss at 0.48% for the circuit
0.9 Power Factor

So why the difference? The first thing you note is they used 0.9 power factor vs. our earlier assumption. I selected conduit on the menu pull down but if you compare their conductor properties to Table 9, you see they are using pvc or Al conduit. When I change my model above to match the power factor and conductor properties, the solutions match.

So I think if I was teaching someone voltage drop in today world of apps and spreadsheets, I would use Table 9 and Note 2 and make sure you know the assumptions. For example, Even table 9 equation in Note 2 is an approximation but in nearly all cases, you can ignore the difference between that and the "absolutely correct" solution.

Back to the OP, you asked to be "taught to fish" and maybe we missed that so please post back.

 

[ggunn]

Thanks for all the help, but here is a very basic question that goes to the heart of my understanding, or lack of it:

In a 240V split phase circuit, there is no difference between the current in the circuit and the current in the individual conductors, i.e., if a balanced 240V load pulls 10A, the current in the conductors is the same - 10A. Each conductor handles 10A@120V to neutral for 1200W, the two of them together make 2400W, and the neutral current cancels out. It's a 10A load on the 240V circuit and each conductor carries 10A. No matter how you look at it, it's 10A. This is clear to me.

Is 3 phase the same? If a three phase inverter spec says that its maximum output is 24A @ 480V 3 phase, does that mean that the current in each conductor is 24A and the overall current in the three phase line is also 24A?

 

[ggunn]

Thanks for all the help, but here is a very basic question that goes to the heart of my understanding, or lack of it:

In a 240V split phase circuit, there is no difference between the current in the circuit and the current in the individual conductors, i.e., if a balanced 240V load pulls 10A, the current in the conductors is the same - 10A. Each conductor handles 10A@120V to neutral for 1200W, the two of them together make 2400W, and the neutral current cancels out. It's a 10A load on the 240V circuit and each conductor carries 10A. No matter how you look at it, it's 10A. This is clear to me.

Is 3 phase the same? If a three phase inverter spec says that its maximum output is 24A @ 480V 3 phase, does that mean that the current in each conductor is 24A and the overall current in the three phase line is also 24A?

It looks to me like the overall current should be 41.6A.

 

[ggunn]

Thanks for all the help, but here is a very basic question that goes to the heart of my understanding, or lack of it:

In a 240V split phase circuit, there is no difference between the current in the circuit and the current in the individual conductors, i.e., if a balanced 240V load pulls 10A, the current in the conductors is the same - 10A. Each conductor handles 10A@120V to neutral for 1200W, the two of them together make 2400W, and the neutral current cancels out. It's a 10A load on the 240V circuit and each conductor carries 10A. No matter how you look at it, it's 10A. This is clear to me.

Is 3 phase the same? If a three phase inverter spec says that its maximum output is 24A @ 480V 3 phase, does that mean that the current in each conductor is 24A and the overall current in the three phase line is also 24A?

It looks to me like the overall current should be 41.7A.

Sorry to reply to myself, but the edit function timed out. Here's why I think that:
The inverter spec says that it produces 20kW of power at 480V 3 phase and its maximum current is 24.1A.

(24.1A)(277V) = 6675.7W, which is 1/3 of 20kW. 1/3 the power on each leg, that sounds right.

20kW/480V = 41.7A = (24.1A)(sqrt3)

So it looks to me like the overall current is the current in one of the conductors times sqrt3.

 

[dkarst]

This is a bit different than the voltage drop earlier but you need the line current to get that so let's start here. It is easy to get confused when you start trying to figure out "per-phase" power so try this. Say your inverter is connected to a load in a black box with 3 current carrying conductors and a neutral if you like. I'm not telling you what is in the black box or how it is connected. The power assuming balanced load is ALWAYS 3Φ Power = √3 V _line I_line cosθ. Doesn't make any difference wye or delta. So in your example, P = √3 x 480 x 24 = 20kW (I've assumed unity PF for simplicity). The line current is simply the current you would measure if I gave you a clamp on meter and you snapped it on one of the wires. Remember the voltage is the LINE voltage not the line to neutral voltage you were using. I'll maybe add to this later but does this make sense?

 

[ggunn]

This is a bit different than the voltage drop earlier but you need the line current to get that so let's start here. It is easy to get confused when you start trying to figure out "per-phase" power so try this. Say your inverter is connected to a load in a black box with 3 current carrying conductors and a neutral if you like. I'm not telling you what is in the black box or how it is connected. The power assuming balanced load is ALWAYS 3Φ Power = √3 V _line I_line cosθ. Doesn't make any difference wye or delta. So in your example, P = √3 x 480 x 24 = 20kW (I've assumed unity PF for simplicity). The line current is simply the current you would measure if I gave you a clamp on meter and you snapped it on one of the wires. Remember the voltage is the LINE voltage not the line to neutral voltage you were using. I'll maybe add to this later but does this make sense?

Yes, except that the load is not relevant. A PV inverter acts as a current source with its output voltage clamped by the grid.

 

[dicklaxt]

Do you know the size of the transformer in KVA ,if so solve for the primary current and size the conductors for your VD limitation you have to work with..

If you waqnt to work with the 112KW then solve that for max current at 480 ditto on calcs.

Or did I miss the train?

dick

 

[Smart $]

Yes, except that the load is not relevant. A PV inverter acts as a current source with its output voltage clamped by the grid.

Do you know the size of the transformer in KVA ,if so solve for the primary current and size the conductors for your VD limitation you have to work with..

If you waqnt to work with the 112KW then solve that for max current at 480 ditto on calcs.

Or did I miss the train?

dick

The xfmr size would only make a difference if it was rated less than the total rated output of the inverters (which would make for a bad design).

Is 3 phase the same? If a three phase inverter spec says that its maximum output is 24A @ 480V 3 phase, does that mean that the current in each conductor is 24A and the overall current in the three phase line is also 24A?

As stated, the 24A output rating would be line current capability. 24A ? 480V ? √3 = 20kVA

Having a xfmr rated greater than 20kVA you will not get any more than 20kVA. Having a xfmr rated less than 20kVA will limit the output potential.

I'd like to give more specific answers but the numbers seem to be jumping all over the place. Do you have an exact make and model number of the inverter(s) so we can look the spec's up on the web. Also the number of inverters... Your OP said six... further down you said three.... :slaphead:

 

[dicklaxt]

The xfmr size would only make a difference if it was rated less than the total rated output of the inverters (which would make for a bad design).



That may very well be but the OP question was asking for VD and how to get there,I cited two methods as starting points.

The Old Man