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#95618 09/25/05 03:54 PM
Joined: Jan 2004
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DiverDan- There is no argument that temperature plays a part in voltage drop. Here's how - Since resistance increases with temperature, voltage drop increases. the amount of increase is not significant (IMHO) Would you perform your magic and see what voltage drop you come up with for the problem as presented in this thread? thank You in advance.


George Little
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#95619 09/25/05 11:01 PM
Joined: Jan 2004
Posts: 1,507
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Sorry Dan I missed one of your post where you had worked the problem using "Volts" Is the Volts program Mac compatible?

My conclusion:

DiverDan's "Volts" program came up with 2.93% VD or 7.032v.

Using NEC Table 8 and simple VD formula I came up with 2.62% VD or 6.3v.

So the difference is 0.732v.

I assume we both used 240v. as the voltage.

I think we can live with these differences.

Thanks Dan for your insight.

[This message has been edited by George Little (edited 09-25-2005).]


George Little
#95620 09/26/05 02:26 AM
Joined: Mar 2005
Posts: 38
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Member
George Little, my previous post's computed values DO NOT APPLY to this situation as they were based on I at 65A, 208VAC, 3-phase. As a degreed engineer of 30 years, I am liable for my computations and therefore protective that I am not miss quoted nor my computations miss used.

And, Quote:
There is no argument that temperature plays a part in voltage drop. Here's how - Since resistance increases with temperature, voltage drop increases. the amount of increase is not significant (IMHO)

"the amount of increase is not significant (IMHO)"
This is also just plain wrong!
Why not use the NEC the way that it is intended to be used with the temperature adjustment factors or try the IEEE temperature adjustment formulae with your VD computations. You'll be suprised how significant the role of ambient temperature can be in VD and conductor sizing.

As I stated earlier:
Your "beliefs" do not change physics.

For those who wish further discussion and information I can be reached at info@dolphins-software.com.

[This message has been edited by DiverDan (edited 09-26-2005).]


Dolphins Software
#95621 09/26/05 04:05 AM
Joined: Jan 2003
Posts: 4,391
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Moderator
OK Dan you want to step up to the plate and surprise us or do you just want to keep on saying the same thing over again?

Put some figures down to prove your point. [Linked Image]

Please demonstrate the 'major effect' ambient temp has on VD using an example.

A branch circuit, straight 240 volt, single phase, 100 amp load current (a unit heater), 200' one way length run in PVC.

What size copper conductor do I need to stay under 3% drop at 100 F ambient.

Now the same circuit in an ambient temp of 0 F.

Is there a significant difference?

Will we be surprised?

Bob


[This message has been edited by iwire (edited 09-26-2005).]


Bob Badger
Construction & Maintenance Electrician
Massachusetts
#95622 09/26/05 06:01 AM
Joined: Oct 2000
Posts: 2,749
Member
Here's a 1 hr. video by George Newton that discusses Voltage Drop, etc.
http://easylink.playstream.com/gnewton100/voltdrop/voltage_drop141.wmv
www.electrician.com


Joe Tedesco, NEC Consultant
#95623 09/26/05 08:17 AM
Joined: Jan 2004
Posts: 1,507
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Member
And when your done with Bob's (iwire) problem let me know what your voltage drop answer is for Tripp's problem. And I'll ask again is the "Volt's" software Mac compatible?


George Little
#95624 09/26/05 08:58 AM
Joined: Sep 2003
Posts: 650
W
Member
IMHO we are going to have to be very careful with wording and reading to make this a useful conversation; there are several subtle issues here. I've been working on this post for about an hour, running around in circles.

1) Voltage drop change with ambient temperature.

2) Wire size change with ambient temperature.

As I read DiverDan's posts, I believe that he is saying that the ampacity derating mandated for elevated temperature in table 310.16 is caused primarily by 1).

I disagree with this. When a conductor is operated at maximum ampacity, the internal self heating of the conductor raises the temperature to the maximum allowed by the insulation system. Once the wire is heated up to its operating temperature, it is at its operating temperature, _not_ at ambient temperature. The resistance of the copper wire is _not_ dependant upon the ambient temperature; it is dependant upon the _copper wire_ temperature. This means that for any given conductor insulation system, the resistance of the copper _at maximum ampacity_ can be taken as a constant.

Conductor ampacity is rated at 30C ambient with a given maximum allowed conductor temperature.

If we presume 90C conductors, then operation at 60C requires an ampacity correction of 0.71.

In both cases the conductor itself is presumed to be at 90C.

This is a reduction in allowed current capacity of 1/root(2) and a reduction in allowed internal self heating of 1/2.

Now let us look at the temperature change of resistance.
Let us presume a conductor that is not subject to significant self heating. This, for example, would be a long conductor sized for voltage drop rather than for maximum ampacity. The change in resistance of this conductor between 30C and 60C is not insignificant:
The temperature correction for resistance is
R2 = R1 [ 1 + K(T2 - 75) ] which is an approximate linear correction for resistances tabulated at 75C
At 30C the resistance of the conductor is 85% of its 75C value
At 60C the resistance of the conductor is 95% of its 75C value

In other words, comparing the same conductor at 30C versus 60C, the voltage drop at 60C will be about 12% greater. (meaning that if you had a 3% VD at 30C, you would expect a 3.36% VD at 60C.) The power dissipated per unit length at the same current would be 12% greater.

But the thermal ampacity correction shows that the heat dissipation capability of the conductor is far more significant.

Now jumping over to Bob's example. The answer to Bob's question would incorporate a number of factors which would totally hide the point of the discussion, with wire size being set by OCPD requirements (125% requirements), and factors of derating for wire temperature rating versus terminal temperature rating, maximum ampacity of heater circuits, etc. I'm not going to answer the question as stated. Instead I will focus entirely on the NEC 310.16 ampacities and use Bob's question as the basis.

Say we have 75C wire and 75C terminations, and the conductor must safely carry 100A. Without considering ambient thermal issues, table 310.16 says that a #3 conductor can be used.

At 0F, the temperature correction factor for 75C wire is 1.05, meaning that the ampacity of this #3 wire is 105A; we are just fine. In fact, the table doesn't go down to 0F; at 0F the temperature correction factor is probably on the order of 1.4!

At 100F, the temperature correction factor for 75C wire is 0.88. We now must use a #2 conductor with a derated ampacity of 102A.

Now let us consider voltage drop.

At 100F, using the #2 conductors, we are quite close to the thermal ampacity of the wire, so we can assume that the wire is nearly at 75C. The resistive component of the voltage drop will be:
(100A * 0.194 Ohm/kFT * 0.4kFT) / 240V = 3.23%

At 0F, using the #3 conductors, we are not even close to the thermal ampacity of the wire. Since we would be at the thermal ampacity at 30C, we can estimate that the conductor temperature will be 75C - (30C - 0F) = 75C - (30C - -17C) = 28C
The resistive component of the voltage drop will be:
(100A * 0.245 Ohm/kFT * 0.4kFT * [1 + 0.00323 * (28 - 75)] ) / 240V = 3.46%

So using Bob's example numbers, a #2 conductor would be sufficient for ampacity at 100F, but not sufficient for voltage drop. A #3 conductor would be sufficient for ampacity at 0F, but not sufficient for voltage drop.

But now let us figure the voltage drop of the #2 conductor in the 0F environment:
A #2 conductor (75C rating) carrying 100A is at its thermal ampacity at an ambient of 40C. This means that the conductor, in a 40C ambient, will be at 75C when carrying 100A. So in a -17C environment, we can expect a conductor temperature of 18C. The voltage drop is
(100A * 0.194 Ohm/kFt * 0.4kFt * [1 + 0.00323 * (18 - 75)]) / 240V = 2.64%

In this example, the temperature coefficient of resistance means a _significant_ difference in voltage drop, enough to mean the difference between needing to use a #1 conductor and a #2 conductor. However it seems pretty clear to me that the ampacity limits set by 310.16 are not changed by the temperature coefficient resistance of copper.

-Jon

#95625 09/26/05 03:17 PM
Joined: Jan 2003
Posts: 4,391
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Moderator
Jon, first I thank you for explaining what I believed but did not know how to express. [Linked Image]

Quote
Now jumping over to Bob's example. The answer to Bob's question would incorporate a number of factors which would totally hide the point of the discussion,

Yes, that came to me as I was driving to work this morning. [Linked Image]

I think my question would have been better if I simply said what is the voltage drop of a certain conductor at 0 F and 86 F.

Thanks again for spending your time on this, during my ride in I was realizing how complicated this little problem was.

Bob


Bob Badger
Construction & Maintenance Electrician
Massachusetts
#95626 09/26/05 04:41 PM
Joined: Mar 2005
Posts: 38
D
Member
I'll jump back into this after work this evening.


Dolphins Software
#95627 09/27/05 12:47 AM
Joined: Jan 2004
Posts: 1,507
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Member
Winnie- I follow your approach and that means you did an excellent job of explaining. Tough to do, I've taught code classes for a number of years and so I appreciate your detail. There is however a "Fly in the ointment" as they say. Based on Bob's problem our conductor sizing also needs to comply with NEC Article 424.3 Section 423.3(B). This IMHO would require us to upsize the conductors by 25%;

I came up with a 1/0 conductor and a voltage drop of 4.88v. and would be within the recommended 3% max.


George Little
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