Lighting voltage drop

[sevlander]

Can someone tell me if there is any other way to calculate voltage drop for lighting other than 3% for branch circuit and 5% for BC and feeder combined? We are involved in the design of a building renovation in which existing 100 watt MH soffit recessed fixtures are to be fed and added to. The building is approximately 300' long with around 60 fixtures. We are calculating # 8 wire and larger depending on circuit wattage and length. The contractor contests that the existing lights work fine with #12 wire and that as long as the VD is less than the manufacturer's minimum voltage requirements it is fine. I don't know of any allowance of this kind in the NEC, am I correct? Thanks, Scott

 

[steve66]

As far as the NEC is concerned, the 3% and 5% voltage drops are fine print notes, which means they are a suggestion, not a requirement.

However, some of the energy codes do limit voltage drops.

The fixtures may work fine with more voltage drop, but that also means the owner is going to pay a little more every month to run those lights. If you have 10% voltage drop, then that's about 10% of the power that is lost in heat, and the owner is going to pay about 10% more every month to run those lights.

So just making it work isn't the only issue here.

 

[brian john]

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So just making it work isn't the only issue.

A finer statment is seldom made.

 

[gar]

090727-1126 EST

steve66:

Suppose the light fixtures are constant current loads, then independent of the added series resistance the power supplied from the source remains constant up to the point where the added resistance has a voltage drop across it equal to the source voltage. After this point increasing the added series resistance reduces the power required from the source and the constant current load is no more a constant current load.

Suppose the lamps are constant resistance devices independent of voltage and you add series resistance, then the input power drops as the series resistance is increased.

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[steve66]

090727-1126 EST

steve66:

Suppose the light fixtures are constant current loads, then independent of the added series resistance the power supplied from the source remains constant up to the point where the added resistance has a voltage drop across it equal to the source voltage. After this point increasing the added series resistance reduces the power required from the source and the constant current load is no more a constant current load.

Suppose the lamps are constant resistance devices independent of voltage and you add series resistance, then the input power drops as the series resistance is increased.

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I'm not sure what your point is, but I think you are making it more complicated than it needs to be.

I think ballasts are normally "constant power" devices. Any power lost in the wiring is just that - lost. But that lost power still has to be paid for.

 

[mivey]

gar,

Bring it home. You stated some facts, but what was the point you were making?

Which of steve66 points were you addressing?

I think you may have been saying the total bill might not change. In that case, the value the customer received is what changed.

 

[gar]

090727-1230 EST

steve66:

If a ballast is non-electronic, then it probably looks like an approximately constant current device. A tungsten incandescent is lamp is between a pure resistance and a constant current device.

A switching power supply type of ballast is probably an approximately constant power device.

My point was that if you have a constant current or resistive load increased line resistance does not increase your electric bill. However, you would get less power to the load with increased line resistance.

.

 

[steve66]

090727-1230 EST

steve66:

If a ballast is non-electronic, then it probably looks like an approximately constant current device. A tungsten incandescent is lamp is between a pure resistance and a constant current device.

A switching power supply type of ballast is probably an approximately constant power device.

My point was that if you have a constant current or resistive load increased line resistance does not increase your electric bill. However, you would get less power to the load with increased line resistance.

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So with some ballasts, instead of paying a little more money, they are getting a little less light.

To me, it seems like six one way, and a half dozen the other way. No matter what, the owner of the building still winds up paying for more power than the light fixtures get.

Steve

 

[jghrist]

090727-1230 EST

My point was that if you have a constant current or resistive load increased line resistance does not increase your electric bill. However, you would get less power to the load with increased line resistance.

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You may be correct if the lighting is resistive. I remain to be convinced on the constant current load. You will, however, get less lumen output if the voltage is lower, so the owner still does not get the value paid for.

 

[mivey]

You may be correct if the lighting is resistive. I remain to be convinced on the constant current load.

That should be easy enough: Supply voltage is essentially constant. Current is constant. VA is constant.

So (if you let pf be constant), then the bill will be the same: Energy = Watts * time = Vsupply * Isupply * pf * time, regardless if the watts are dissipated in the lamp or wire.

 

[gar]

090727-1346 EST

mivey:

My point was based on the statement:

The fixtures may work fine with more voltage drop, but that also means the owner is going to pay a little more every month to run those lights. If you have 10% voltage drop, then that's about 10% of the power that is lost in heat, and the owner is going to pay about 10% more every month to run those lights.

The owner does not pay about 10% more per month unless additional lights are added to compensate for the lower light output of the initial number of lights. The owner possibly pays less per month, but not much.

The statement is correct if the load is a constant power load, and in that case there will be no loss of light output.

.

 

[mivey]

090727-1346 EST

mivey:

My point was based on the statement:

The owner does not pay about 10% more per month unless additional lights are added to compensate for the lower light output of the initial number of lights. The owner possibly pays less per month, but not much.

The statement is correct if the load is a constant power load, and in that case there will be no loss of light output.

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That's what I thought you meant. It would be hard to know which would apply as the OP said existing plus added MH.

FWIW, it might be worth replacing the existing MH with something more energy efficient and economical if you were doing a renovation.