voltage drop and steel conduit

[George Stolz]

(additionally, does anyone know of a better, non-image, free file host?)

I can't vouch for any of these, I just went looking and that's what I came up with.

 

[Rockyd]

In regard to voltage drop, at a reasonable PF (85%) we can use the following information -

Standard chart usage (tables 8 and 9) will get you the following formula -

VD=Q x K x I x D
CM​

Q as a Multiplier

For ac dircuits, the dc Resistance Constant (K) must be adjusted for the effects of eddy current and skin effect. The Q-Factor multiplier is calculated by dividing the ac resistance (chapter9 table 9) by the dc resistance (chapter9 table 8.) as listed in the NECode. Eddy currents and skin effect are insignificant for conductors No. 1/0 and smaller and their effects can be ignored.

Here is a q factor table -

Size COPPER ALUMINUM
AVG ... PVC.......AL.......STEEL... PVC.......AL..... STEEL
MCM - Conduit Conduit Conduit Conduit Conduit Conduit
2/0 -- 1.0341 1.0341 1.0341... 1.0062 1.0062 1.0062
3/0 -- 1.0052 1.0704 1.0313... 1.0317 1.0317 1.0317
4/0 -- 1.0197 1.1019 1.0362... 1.0000 1.0000 1.0000

250 -- 1.0097 1.1068 1.0485... 1.0035 1.0626 1.0153
300 -- 1.0256 1.1422 1.0489... 1.0042 1.0749 1.0184
350 -- 1.0354 1.1717 1.0627... 1.0083 1.0909 1.0413

400 -- 1.0280 1.1838 1.0903... 1.0201 1.1153 1.0397
500 -- 1.0465 1.2403 1.1240... 1.0142 1.1321 1.0613
600 -- 1.0748 1.3084 1.1682... 1.0198 1.1615 1.0765

 

[Smart $]

In regard to voltage drop, at a reasonable PF (85%) we can use the following information...

Where's the adjustment for inductance? Yes, it would be nice to pin all circuits down to a .85 PF. In the real world, it is only occasionally so.

 

[winnie]

I admit heat is not the issue as far as the OP is concerned. But while we're on the subject, I noticed Jon, too, mentioned heating in one of his posts. This leads me to the question I've often wondered about, never ran across any details, and never asked... Why are the resistance/impedance tables given at 75?C. Is this the temperature of the wire? ...ambient temperature inside the conduit? ...outside the conduit? ...the conduit, too? host?)

I am pretty certain that the 75C is the conductor temperature. See the note at the bottom of table 8 for temperature change; that is the standard formula used calculating resistance assuming a linear change in resistance versus temperature.

-Jon

 

[Rockyd]

Smart $,

That's my short answer, and I'm sticking to it. For more information in regard to formula and calculations, see the 2005 Handbook examples covering table 9. Other than that, I'd say it's up to the engineering department.

 

[Smart $]

I am pretty certain that the 75C is the conductor temperature.

I assumed that also.

See the note at the bottom of table 8 for temperature change; that is the standard formula used calculating resistance assuming a linear change in resistance versus temperature.

I was aware of that formula... but how does one determine T2. Say, for example, we know the following parameters of a circuit:

• 10AWG THHN/THWN
• 250' one-way circuit length
• 24A continuous load @ .8 PF
• 30?C ambient temperature
• L-N circuit in one of two 4-wire MWBC's in 1" EMT

At the very least, the line conductor is operating at a temperature somewhere above ambient.

 

[ramsy]

..how does one determine T2.

I believe T2 is strictly the conductor temperature rise that I'm permitted to work with: 60c, 75c, or 90c.

Ambients in NEC tables adjust ampacity; so do I, since not permitted to modify R2 with ambients unsupervised.

This would set your #10 temp. rise at 60c, with the following adjustments:
21-25c ambient L-N V-drop = 4.40%
26-30c ambient L-N V-drop = 4.76%
31-35c ambient L-N V-drop = 5.23%

 

[winnie]

I'm pretty sure that T2 is the actual conductor temperature at which you are trying to calculate voltage drop, not the allowed temperature rise. If you are looking at a conductor operating at rated ampacity, then the limiting temperature for that wire type is a reasonable guess at this temperature.

For most circuits, however, the '30C' ambient is probably a better guess.

Unfortunately there is no easy way to calculate T2 in normal situations, since it requires knowledge of the thermal conductivity between conductor and ambient; for a given current flowing through the wire in a given ambient temperature, the wire sitting in 'free air' will be much cooler than the wire buried in thermal insulation. Table 310.16 makes certain assumptions about the thermal resistance to ambient, IMHO pretty conservative estimates.

If you want to calculate T2 using the same thermal resistance assumptions as table 310.16, then you can work via 'successive approximations'.

1) Temperature rise is proportional to heat dissipated. You know the 90C ampacity, you know the ambient, and you know the resistance. So you can calculate the thermal resistance in K/W (Kelvin temperature rise per watt) For #10 conductors this is roughly 30K/W per foot of conductor.
2) Heat dissipated is proportional to resistance and the _square_ of current flow. Make a _guess_ at the conductor temperature, use this to calculate the conductor resistance, and multiply by the square of the current flow to get the estimated power dissipation.
3) Use the thermal resistance to calculate the temperature rise, see if your guess was good. If not, use the newly calculated temperature, and feed it back into the equation.

The above is based upon so many guesses and approximations that I doubt that it would get you much closer to reality than simply saying 'ambient'. For the circuit that Smart$ posted, my starting guess would be a conductor temperature of 45C.

-Jon

Gaak. I've had 8 days of 'The Festival of Lights' this past week. Makes today a normal work day. Up early with a head cold. Yuk. Enjoy your holidays!

 

[ramsy]

My figures also change if I select an ampacity table other than 310.16.

Santa was good to use this morning. Merry Xmass.

 

[Smart $]

I'm pretty sure that T2 is the actual conductor temperature at which you are trying to calculate voltage drop, not the allowed temperature rise...

Unfortunately there is no easy way to calculate T2 in normal situations, since it requires knowledge of the thermal conductivity between conductor and ambient; for a given current flowing through the wire in a given ambient temperature, the wire sitting in 'free air' will be much cooler than the wire buried in thermal insulation. Table 310.16 makes certain assumptions about the thermal resistance to ambient, IMHO pretty conservative estimates.

If you want to calculate T2 using the same thermal resistance assumptions as table 310.16...

You have pretty much confirmed my suspicions.

The formula R2 = R1 [1 + α (T2-75)] indicates temperature and resistance are directly proportional. Would it be appropriate to assume both temperature rise and resistance are directly proportional to I^2? If so, I could:

T2 = 30?C ambient temperature + (24A nominal current)^2 ? (32 maximum current; see below)^2 x (60?C operating temperature range) = 30?C + 33.75?C = 63.75?C operating temperature

Continuing with the example I have set forth, 10 AWG THHN/THWN (when used as THHN) is rated at 90?C and an ampacity of 40. To me this means when 40 amperes of current is drawn through this conductor continuously, the conductor's temperature rise will be +60?C over the ambient temperature of 30?C. However, since this is only one of six ccc's in the conduit (assuming all loads are linear), the ampacity has to be derated to 80% of it's table value. That results in an assumed +60?C temperature rise at a derated ampacity of 32.

Therefore:

VD = d/1000*(I*R*(1+α*(T2 - 75))*PF+I*X*SIN(ACOS(PF)))
VD = 250*2/1000*(24*1.2*(1+0.00323*(63.75-75))*0.8+24*0.063*SIN(ACOS(0.8)))
VD = 11.554992​

For comparison, without adjusting for conductor temperature:

VD = d/1000*(I*R*PF+I*X*SIN(ACOS(PF)))
VD = 250*2/1000*(24*1.2*0.8+24*0.063*SIN(ACOS(0.8)))
VD = 11.9736​

And for comparison, without adjusting for conductor temperature and inductance:

VD = d/1000*I*R
VD = 250*2/1000*24*1.2
VD = 14.4​

 

[ramsy]

I agree checking the rise with a calc beats assumptions.

11.42 = Temp. rise 60c, Steel conduit, PF=0.8 (Error if over 60c)
11.97 = Default 75c, Steel conduit, PF=0.8
14.40 = Default 75c, PVC conduit, PF=1

If derating #10 permits the 24A load, can we do it this way?

T2 = Ambient + (Load / I@MaxC * 60c )
T2 = 30 + 24 / 40 * 60 = 66c

Or, is it this way:

T2 = Ambient + (Load^2 / I@MaxC^2 * 60c )
T2 = 30 + 24^2 / 40^2 * 60 = 52c

 

[Smart $]

I agree checking the rise with a calc beats assumptions.

11.42 = Temp. rise 60c, Steel conduit, PF=0.8 (Error if over 60c)
11.97 = Default 75c, Steel conduit, PF=0.8
14.40 = Default 75c, PVC conduit, PF=1

Not sure if you are aware of it, but Table 9 lists circuits in PVC as having inductive reactance (#10Cu XL = 0.050Ω/1000'), same column as Aluminum Conduit. The use of PF = 1 does negate any effect it has on the circuit.

If derating #10 permits the 24A load, can we do it this way?

T2 = Ambient + (Load / I@MaxC * 60c )
T2 = 30 + 24 / 40 * 60 = 66c

Or, is it this way:

T2 = Ambient + (Load^2 / I@MaxC^2 * 60c )
T2 = 30 + 24^2 / 40^2 * 60 = 52c

I'm inclined to go with the latter. However, without empirical evidence to back that conclusion up, I'm afraid it is but mere speculation on my part. Did you notice I used a derated I@max?C of 32, because of the 6 ccc's in the conduit? Do you not consider this a logical assumption?

 

[ramsy]

Not sure if you are aware of it, but Table 9 lists circuits in PVC as having inductive reactance..

Yuk, didn't see that before. Perhaps 14.40 matches only for voltages below 600vac.

I'm inclined to go with the latter. However, without empirical evidence to back that conclusion up, I'm afraid it is but mere speculation on my part.

I don't have a white book either. However, Google searches showing IEE references related to Temp.rise are using load^2 / I@MaxC^2 in the equation. See near bottom page.
http://www.cda.org.uk/megab2/elecapps/pub116/appen32.htm

Did you notice I used a derated I@max?C of 32, because of the 6 ccc's in the conduit? Do you not consider this a logical assumption?

NEC 310.15 tables don't show your 12 conductor factor of 0.5 affecting the temperature, perhaps my brain isn't flexing enough to translate this for T2, but building a derating limit for ampacity into T2 is not a bad idea. I'm just used to doing that seperately.

 

[Smart $]

Yuk, didn't see that before. Perhaps 14.40 matches only for voltages below 600vac.

In reality, the 14.40 VD would only match a DC circuit, and only under a completely different and unique set of circumstances than the example. But there are many that consider such a valid voltage drop calculation result for an AC circuit!

I don't have a white book either. However, Google searches showing IEE references related to Temp.rise are using load^2 / I@MaxC^2 in the equation. See near bottom page.
http://www.cda.org.uk/megab2/elecapps/pub116/appen32.htm

So Load^2 / I@max?C^2 it is.

NEC 310.15 tables don't show your 12 conductor factor of 0.7 affecting the temperature, perhaps my brain isn't flexing enough to translate this for T2, but building a derating limit for ampacity into T2 is not a bad idea. I'm just used to doing that seperately.

First off, I only have (8) #10 AWG conductors in the 1" EMT. I said the example circuit is an L-N circuit in one of two 4-wire MWBC's. That's two "edisons", "full boats", or A,B,C,N sets. Assuming linear loads (i.e. not non-linear), only the line conductors count as current carrying conductors, of which there are six (6), and thus they are derated from 40 to an ampacity of 32.

NEC310.15(B)(2)(a) and its associated Table derate ampacity to prevent more than three conductors within a raceway from exceeding their temperature rating. That is to say, less heat per conductor is dissipated... or conversely, more heat is retained in the conductor, causing the conductor to reach max?C at a lower amperage. It only seems logical to me to use the derated ampacity as the current which corresponds to max?C. To use any other more precise method would require usage of the formula in 310.15(C).

 

[ramsy]

..more heat is retained in the conductor, causing the conductor to reach max?C at a lower amperage. It only seems logical to me to use the derated ampacity as the current which corresponds to max?C. To use any other more precise method would require usage of the formula in 310.15(C).

That makes perfect sense in conduit with > 3 ccc's. NEC table values must be adjusted.

I like your formula:
T2 = T1 + (TR-T1) * Load^2 / Imax^2

Where:
T1 = Ambient Temperature
T2 = Estimated Conductor Temperature
TR = Conductor Max Temperature Rating
Load = Circuit Current
Imax = Maximum Allowable Current at TR, appropriately derated for ccc's

 

[tilib]

Yes, only if the cables are single conductors only, otherwise not.