reduced neutral

[Davis9]

The reason for oversizing the conductors is voltage drop, which is caused by the resistance of the wire itself. If you don't increase the size of the egc, the same resitance (of the wire itself) will reduce the fault current. It's no longer a "low impedance path"
I think I understand this better than I explain it , so if someone can jump in and explain it better, please do!

I think we got that part , but VD is a FPN not a requirement. So if he wants to really save $$$ then he could just use #12.

Tom

 

[JohnJ0906]

Not going to save money doing it twice! That light 1000' away might not be bright enough for the customer if #12 is used!:grin:

 

[Bea]

I agree with iwire to slow typing

 

[winnie]

Design question: do you really need to keep to less than a 2% voltage drop? Could you use regulating ballasts in order to tolerate greater voltage drop?

Remember when doing your voltage drop calculations that this wire will be quite cool, not at the 75C temperature used in the resistance tables. Lower temperature means lower resistance, probably by 15-20%.

Also remember that the further you go into the run, the lower the current and the greater the allowed resistance for the same voltage drop. If you put the first 4 lamps on 2ga, the next 4 on 4ga, the next 4 on 6ga and the last 4 on 8ga, then I believe you will stay below 3% voltage drop. The voltage drop of each leg accumulates as you go down the chain, but the last leg is not carrying very much current.

-Jon

 

[ramsy]

Remember ..this wire will be quite cool, not at the 75C temp.. Lower temperature means lower resistance.. Also remember that the further you go into the run, the lower the current..

Nice one Jon, that makes perfect sense.
Split this run in two by putting the panel/feeder in the middle. Then feed each 120vac branch, 500 ft segment, as follows:

#14cu, last 250ft, 720VA, drops 4.06 volts @ 35.4?C
#12cu, first 250ft, 1440VA, drops 5.42 volts @ 45.0?C

The most sensitive nameplates I've seen are 108vac minimum (computers), check your ballasts, you would be running near 110vac.

We can thank "Smart $" for that slick R2 adjustment.

 

[Smart $]

Nice one Jon, that makes perfect sense.

It surely does!

Split this run in two by putting the panel/feeder in the middle.

That would be the ideal way to go, provided the middle of the run is closest to the source panel. Otherwise, it is wasting money, both on wire and I^2R loss. Whichever way the posts are geographically arranged, you should always "feed" the one having the shortest, physically-feasible route then branch out if determined advantageous.

 

[Smart $]

Design question: do you really need to keep to less than a 2% voltage drop? Could you use regulating ballasts in order to tolerate greater voltage drop?

If these light posts are ballast driven, the power factor involved could possibly reduce voltage drop.

...If you put the first 4 lamps on 2ga, the next 4 on 4ga, the next 4 on 6ga and the last 4 on 8ga, then I believe you will stay below 3% voltage drop.

I calculated it to be L1 @ 2.42% and L2 @ 3.03% using #3AL for first 4, #4AL for second 4, #6AL for next 4, and #8AL for last 4, alternating L1 and L2, with L1 obviously being the shorter-distance set. This includes voltage drop for the neutral on the last leg, as it is substantial enough to include. Providing all lamps are functioning [properly], there would be no other substantial voltage drop on the neutral [I believe]. Should one circuit not be operating while the other is, the VD would [nearly...] double [...in the case of L2].

 

[LarryFine]

Should one circuit not be operating while the other is, the VD would [nearly...] double [...in the case of L2].

I don't mean to seem argumentative, I just like discussing this stuff. :smile: Wouldn't de-energizing one leg of a MWBC merely displace the voltage drop from the de-energized line to the neutral?

It seems to me that you'd have to "convert" a 3-wire circuit into two 2-wire circuits for the VD to double. But we're discussing de-energizing one line of a 3-wire MWBC, halving the voltage across half the resistance: same current.

You could theoretically remove the neutral connection of a genuinely balanced MWBC, resulting in the equivalent of twice the voltage across twice the resistance, with the resultant total current equal on each line either way.

The line current is equal in both legs of any circuit, so the VD of either one 2-wire circuit or a balanced (zero neutral current) MWBC would be the same, i.e., that of two conductors with a given current.

Understanding that the loads will be staggered along the run, look at the current on the neutral. An interesting thing happens: the greatest neutral current should be between each pair of loads. Follow me on this one.

The last pair of loads should have a net zero neutral current fromn the source; all of it would be in the length of run between the last two loads. Likewise, the next-to-last pair would have their neutral current only between them.

In theory (mine, anyway), the only neutral current there should be (with even staggering) is that caused by the resistance of the difference in wire length between the first load on each leg.

Everything else is pairs of 120-volt loads in series. The source neutral current is caused by the extra wire in the longer leg, which (were the neutral opened) would result in the voltage at the mid-point moving away from zero volts to earth.

Even staging wire size, the neutral current of each pair should be a net of zero, balancing at the point halfway between load pairs. The voltage at this midpoint should be halfway between that at the load hot points.

That these two voltages will be slightly different, relative to earth, is why the midpoint voltage will be slightly above zero to earth, which is where the neutral comes into play. As in any unbalanced MWBC, the neutral current is the result of pulling the neutral voltage back to zero.

Did you enjoy the journey into my universe? :wink:

 

[Bob NH]

Larry is totally correct on balanced load and neutral current.

I don't remember the type of lamps described when this was question was asked on an earlier thread a few weeks ago. However, the illumination available from compact fluorescents can significantly cut down the required current.

I replaced an outside floodlight with a compact flourescent that was sold by the local power company and that 65 Watt unit provides more light that I was getting from 300 Watts of incandescent.

There is probably more to be gained by getting more lumens per Watt than there is from reducing voltage drop.

There are 240 Volt compact fluorescents that will eliminate the need for the neutral.

 

[Smart $]

I don't mean to seem argumentative, I just like discussing this stuff. Wouldn't de-energizing one leg of a MWBC merely displace the voltage drop from the de-energized line to the neutral?

It seems to me that you'd have to "convert" a 3-wire circuit into two 2-wire circuits for the VD to double. But we're discussing de-energizing one line of a 3-wire MWBC, halving the voltage across half the resistance: same current.

That's true for the loads... but the voltage drop we discuss is from wire impedance. When both lines are energized, the circuits work like a resistor network.

De-energize one line and the network turns [essentially] into a ladder (er-r-r, somewhat sideways though in my image ).

There would probably still be current in the de-energized line conductor between loads, but I believe it would be a trivial amount, nowhere near enough to substantially reduce the amount of voltage drop in the energized circuit.

You could theoretically remove the neutral connection of a genuinely balanced MWBC, resulting in the equivalent of twice the voltage across twice the resistance, with the resultant total current equal on each line either way.

The line current is equal in both legs of any circuit, so the VD of either one 2-wire circuit or a balanced (zero neutral current) MWBC would be the same, i.e., that of two conductors with a given current.

Yep... if I follow you correctly... and that is, by 2-wire circuit you mean no neutral conductor to the first load.

Understanding that the loads will be staggered along the run, look at the current on the neutral. An interesting thing happens: the greatest neutral current should be between each pair of loads. Follow me on this one.

The last pair of loads should have a net zero neutral current fromn the source; all of it would be in the length of run between the last two loads. Likewise, the next-to-last pair would have their neutral current only between them.

In theory (mine, anyway), the only neutral current there should be (with even staggering) is that caused by the resistance of the difference in wire length between the first load on each leg.

Everything else is pairs of 120-volt loads in series. The source neutral current is caused by the extra wire in the longer leg, which (were the neutral opened) would result in the voltage at the mid-point moving away from zero volts to earth.

Even staging wire size, the neutral current of each pair should be a net of zero, balancing at the point halfway between load pairs. The voltage at this midpoint should be halfway between that at the load hot points.

That these two voltages will be slightly different, relative to earth, is why the midpoint voltage will be slightly above zero to earth, which is where the neutral comes into play. As in any unbalanced MWBC, the neutral current is the result of pulling the neutral voltage back to zero.

Well, IMO, that's quite close. The main difference I see is that the currents on the neutral wire between loads utilize the current from all loads rather than pairs. Yes, I see pairs as doing the major portion of the balancing, but we must remember, electricity does not choose exclusive paths, it utilizes all available paths. That is why, to evaluate the circuits to an ultra-accuracy level, it would have to be evaluated as a network (...which gives me a headache just thinking about it).

Did you enjoy the journey into my universe? :wink:

Sure!

 

[Smart $]

...

There is probably more to be gained by getting more lumens per Watt than there is from reducing voltage drop.

There are 240 Volt compact fluorescents that will eliminate the need for the neutral.

Two great points, Bob.

What kind of base do the 240V CFL's have?

 

[Bob NH]

Two great points, Bob.

What kind of base do the 240V CFL's have?

This is a result of a quick web search.

http://www.rabweb.com/product_detail.php?product=WPTF26W

http://www.asltg.com/DiStarCompactFluorescent.htm

Some Edison base. Some special base as part of fixtures. It will take some shopping to find the best fit. Since the line power is operating the ballast, it probably makes little difference except in the ballast.

Some foreign sources use 240 Volts to start with.

Some ballasts are tolerant of low voltage, so it might be possible to accept more voltage drop in the circuit.

 

[ramsy]

There are 240 Volt compact fluorescents that will eliminate the need for the neutral.

In the game of spades, I believe that's called a trump card.

I fold.

 

[tallgirl]

Well, IMO, that's quite close. The main difference I see is that the currents on the neutral wire between loads utilize the current from all loads rather than pairs. Yes, I see pairs as doing the major portion of the balancing, but we must remember, electricity does not choose exclusive paths, it utilizes all available paths. That is why, to evaluate the circuits to an ultra-accuracy level, it would have to be evaluated as a network (...which gives me a headache just thinking about it).

While it's true that all paths are used both in theory and in practice, in practice the amount of neutral current flowing between loads other than the pairs of lamps is nil compared to the uncertainty in the lamps themselves.

 

[winnie]

The original poster did not say, but I ASS-U-ME-d that at 180W per fixture that this wasn't incandescent.

I wouldn't use a standard screw base at 240V. That screw shell is supposed to be connected to a _grounded_ conductor. Though using 240V fixtures would eliminate the neutral issue entirely Any type of ballast based lighting should be available in multiple voltages, and ballasts with good tolerance for voltage drop should also be available.

-Jon

 

[Smart $]

The original poster did not say, but I ASS-U-ME-d that at 180W per fixture that this wasn't incandescent.

It's a stretch, but not much of one. However, IMO, 180W is an uncommon wattage for a ballast-driven fixture. Being far from an expert in lighting, I very well could be wrong on that. On the other hand, I know there are 3-lamp incandescent light posts where the lamps are 60W each.

I wouldn't use a standard screw base at 240V. That screw shell is supposed to be connected to a _grounded_ conductor.

You'd have to find such a screw base CFL before you could use it... and I don't believe any to be readily available

Though using 240V fixtures would eliminate the neutral issue entirely Any type of ballast based lighting should be available in multiple voltages, and ballasts with good tolerance for voltage drop should also be available.

As I said before, the power factor of ballast-driven lighting coupled with the reactance of AC wiring could reduce voltage drop on a circuit (...it could also increase voltage drop, as either way depends on circuit and load parameters).

 

[Smart $]

...uncertainty in the lamps themselves.

Can't say I've ever heard of electrical uncertainty being applied to lamps. Care to elaborate?

 

[LarryFine]

Can't say I've ever heard of electrical uncertainty being applied to lamps. Care to elaborate?

My guess is that she meant 'inconsistency.'

 

[tallgirl]

Can't say I've ever heard of electrical uncertainty being applied to lamps. Care to elaborate?

Lamps aren't high-precision loads.

The variation in voltage drop across a bulb in a network like what you've drawn -- based on plus or minus however many watts a 180w bulb actually consumes under operating conditions -- exceeds the amount of voltage drop needed to change the direction of the current flowing between pairs of pairs of bulbs on the neutral conductor.

 

[ramsy]

I wouldn't use a standard screw base at 240V. That screw shell is supposed to be connected to a _grounded_ conductor.

Excellent point, which prohibits 240V CFL screw shells on a Wye Xfmr. Perhaps a grounded 480/240 Delta.