voltage drop on feeders 1

[deltawhy]

Hello, I have a pretty simple question regarding voltage drop on feeders in a building. I have a 100A aluminum feeder at 120V supplying a load of 10kW. If the drop is for example, 10% (not actual value), then will my delivered current be raised by nearly 10%? Does line loss account for this?

Thanks,
Daniel

 

[davidbeach]

Depends on the type of load. If a constant impedance load, the current goes down. If a constant power load, the current goes up. Real loads will be some mix of the two and can even result in a constant current situation.

 

[jghrist]

The effect of voltage drop on current depends on the nature of the load. If it is a resistive load, then a 10% drop in voltage will result in about 20% drop in current. If it is a constant power load like a motor, then a 10% drop if voltage will result in about 10% increase in current. Line loss accounts for the voltage drop, not the current change, although line loss is proportional to the square of the current.

 

[deltawhy]

Thanks for the quick responses. The load is mostly real (ie. resistive cooking devices). If I were to put a capacitor near the end of the line, would I be able to decrease the voltage drop? If not, are there any ways of decreasing the drop without increasing the size of the conductor (I am at maximum size for this application).

Thanks again

 

[jghrist]

Capacitors would reduce the voltage drop, but the reduction is proportional to the cable reactance. The reactance of 120 volt cables is relatively small, so capacitors would not be very effective.

Can you run a 3-wire 120/240 volt circuit and split the loads between the legs? This could cut the current in half?

Can you use copper instead of aluminum to reduce the resistance?

What is limiting the size of the conductor?

 

[deltawhy]

Well the cables, whether they are al or cu are only available up to 100A for this application. Sorry, I did not mention the cables are feeding a 120/208V panel that then branches to the resistive loads close by. The main issue is the drop is too large to meet code and the only solution apparent to me is very expensive (ie. installing another main panel, meter, and feeding it). I think I may be SOL.

Thanks for the time,

Daniel

 

[wareagle]

Quote jghrist

If it is a resistive load, then a 10% drop in voltage will result in about 20% drop in current.

You may want to check Ohms Law again.

 

[boshnaq84]

If it is possible you could ask the panel Manufacturer to allow for 2 incoming cables to be connected to the panel. Then you can use 2 smaller 100A copper cables instead of the 1 larger 100A Aluminum cable.

This applies to new installations of course and given a large enough installation/manufacturing quantity, price difference would be negligible to installing a new Meter, Panel, etc...

 

[jghrist]

wareagle,
Sorry, I was thinking about power.