A circuit that has a load that draws 32 amps when 240V is applied. The circuit is supplied with a 2 conductor #8 R90 XLPE and has a maximum voltage drop of 3%. What is the maximum distance to the load?

We're on code right now, so I'm supposed to figure these questions out from the tables provided in my 2009 Code Book.

I just don't know the steps I have to take.

thanks to anyone who can help me!

 

Hi,

What country are you in?
Is that an aluminum, or copper wire?

The run length is limited because the wire is resistive and causes a voltage drop which increases with the length of run. 3% is an amount generally agreed to in the industry to help limit the minimum working voltages for loads and to limit temperature rise in the conductor. This ties in with maximum current allowed versus conductor size.

In Canada Section 8 deals with voltage drops.

Regards,
Ifixit

 

A circuit that has a load that draws 32 amps when 240V is applied. The circuit is supplied with a 2 conductor #8 R90 XLPE and has a maximum voltage drop of 3%. What is the maximum distance to the load?

We're on code right now, so I'm supposed to figure these questions out from the tables provided in my 2009 Code Book.

I just don't know the steps I have to take.

thanks to anyone who can help me!

The allowable voltage drop is 3% at 32A.

3% of 240V is 7.2V.

For the given cable type there will be a tabulated voltage drop per ampere current per standard length (1000 meters, 1000 feet ... etc.)

Suppose the voltage drop per ampere for a particular cable type is given as 2.5V per ampere per 1000 feet. At 32A that means the given cable will have 2.5*32V or 80V drop per 1000 feet. You are allowed only 7.2V drop which means the maximum allowable length of the given cable would be (7.2/80)*1000 feet or 90 feet.

So look up in your conductor data tables the volatge drop factor for your particular cable type [#8 R90 XLPE] and apply the same ideas.