calculating impedance in feeder

[mshields]

I've got a medical CT which requires that it's incoming line have an impedance of less than or equal to 105.

The source is approximately 250 feet away and the circuit breaker is required to be 300. I'm not clear if I can use smaller conductors (I suspect I can) but lets assume I use conductors based on the CB and therefore use 350kcmil (rated for 310 under the 75 deg column)

To calculate impedance which table/column should I use in chapter 9 of the NEC.

Table 8 gives ohms per kFT for coated conductors under a subheading of Direct-Current Resistance at 75 deg C. Obviously I have AC current, is this the appropriate column for this application.

If I go to Table 9, here my only options are for "uncoated" copper. Obviously my conductors are insulated, how will that effect my results?

Thanks,

Mike

 

[bob]

Uncoated is bare wire, coated is tinned wire. The diameter of the conductor is the same after tinning so the diameter of the copper is reduced about .002" to make up for the addition of the tinning. This increases the resistance of the conductor. I would think you would use non coated.

 

[Smart $]

...an impedance of less than or equal to 105.

The source is approximately 250 feet away...

Shouldn't be a problem even if that's in milliohms.

As Bob stated, use uncoated copper, as long as you're using typical copper building wire.

 

[Haji]

Table 8 gives ohms per kFT for coated conductors under a subheading of Direct-Current Resistance at 75 deg C. Obviously I have AC current, is this the appropriate column for this application.

About above 2 AWG size, the inductive reactance of the cable to be taken into account in addition to DC resistance of the cable.

 

[mshields]

Effective Z or Resistance column in Table 9

Effective Z or Resistance column in Table 9

I found a Mike Holt article on determining the impedance on a particular feeder. He uses the resistance column rather than the effective Z column. This is of course a lower value giving much different results. Any idea why.

 

[ron]

an impedance of less than or equal to 105.

Impedance is R+jX. What is the units of 105?
Is it really to avoid voltage drop of a certain amount?

 

[Smart $]

I found a Mike Holt article on determining the impedance on a particular feeder. He uses the resistance column rather than the effective Z column. This is of course a lower value giving much different results. Any idea why.

Got a link to the article?

 

[Julius Right]

I agree with bob: you don't need tinned copper. See [for instance]:
http://www.southwire.com/commercial/StretchBudget.htm
Actually, the cable XHH or XHHW is EPR or XLPE insulated now and THHN and THWN cable are even PVC insulated still. No sulfur is involved here.
In my opinion, if the total impedance has to be 0.105 ohm [or 105 milliohm as Smart $ said] the voltage drop could be 48 V [10% of 480 V rated voltage at 0.85 pf and 270 A]. So, this has to be total impedance from the utility supplying point [if here will be not less than 480 V].
For 350 ft. and 0.105 ohm the impedance per 1000 ft. could be 0.105/350*1000=0.3 ohm/1000 ft.

 

[Phil Corso]

Julius...

Shouldn't it be (0.105 Ohm/k-ft) x (350 ft/1,000) = 0.0368 Ohm.

Regards, Phil Corso

 

[bob]

I found a Mike Holt article on determining the impedance on a particular feeder. He uses the resistance column rather than the effective Z column. This is of course a lower value giving much different results. Any idea why.

I believe he does not include the X component is because on short runs it does not make much difference in the resultant VD.

 

[Phil Corso]

Bob...

I used the Ohms/k-ft noted, 0.105, for the distance, 305 ft!

Phil

 

[steve66]

I found a Mike Holt article on determining the impedance on a particular feeder. He uses the resistance column rather than the effective Z column. This is of course a lower value giving much different results. Any idea why.

Using resistance gives the real power lost in the feeder or branch circuit. That's fine for finding efficiency, and for finding out how much you are going to pay the utility for power that is lost. I assume that might be what Mike was doing.

For power quality calculations, or for voltage drop calculations, I'd include the reactance portion.

I use a standard voltage drop calculator (one that includes reactance and the type of conduit). At 300 amps, your voltage drop should be less than 105 mOhms X 300 amps = 31.5 volts. So if you have less drop than that, you should be good to go. I'd include a safety factor, and cap the limit at 20 or 25 volts.

Also, make sure that isn't supposed to include the supply transformer voltage drop. If it does include that, use SKM or similar software to find the total voltage drop, and again, make sure its less than 25 volts or so.

 

[vicdog]

I'm curious too. 105 what? Has to be more to it. If it is some kind of impedance for the feeder, do they not care if it's 10 miles to the transformer? Do they have a required minimum system power requirement?

 

[Julius Right]

In my opinion, if current is 300 A the system has to be three phases.:roll: In this case a 1.73 factor has to be employed.
Phil Corso, what I meant is 0.105 ohm it would be "total impedance”. Then if we have only 250 ft. [my mistake I took 350 instead of] the resistance of the cable in ohm/1000ft -as it is shown in tables-has to be 0.105/250*1000=0.42[effective impedance as per NEC].

 

[mshields]

thank you

thank you

thank you all

By the way, your assumption that we my numbers were in milliohms were correct.

Unfortunately I can't find that Mike Holt article again.

Thanks again,

Mike