IEEE 142 I2t Calc Error?

[mityeltu]

I am trying to learn as much as possible for a project I will be working on soon to ground my first substation.

I am reading through IEEE 142 to get my feet wet. and came to section 2.7.4.4 which gives an example of calculating the thermal energy (I2t) based on available informtion.

I followed the equation given in a previous section and it made perfect sense, but now in the application.... I must be doing something really wrong.

Here's the given information:
400 A feeder with a No. 3 AWG equipment-grounding conductor sized from the reference Table 250-95. Using an initial temperature of 60 deg C and a final temperature of 250 deg C for cross-linked polyethylene insulation, I2t = 17.8 x 10^6.

And here's the applicable equation:
I2t = A*[0.0297 log ((Tm + 234) / (Ti + 234))]

Where:

t = time of fault-seconds
I= fault current through conductor-amperes
A= conductor cross-sectional area-circular mils
Ti= initial operating temperature- degrees Celsius
Tm = maximum temperature for no damage - degrees Celsius

From tables I find A = 52620 cmils
and solving I2t = 338.35

This is only off by about 3 orders of magnitude. So, I'm certain I'm doing this wrong, but I don't see where.

Can anyone clarify this mess for me?

 

[Mayimbe]

Here's the given information:
400 A feeder with a No. 3 AWG equipment-grounding conductor sized from the reference Table 250-95. Using an initial temperature of 60 deg C and a final temperature of 250 deg C for cross-linked polyethylene insulation, I2t = 17.8 x 10^6.

Hello, I dont understand from where did you get the I2t data, isnt that what do you want to find? I mean, i believe the purpose of that expression is to find either the current or the time in which the equipment or the feeder in this case, will be damaged.

And here's the applicable equation:
I2t = A*[0.0297 log ((Tm + 234) / (Ti + 234))]

Where:

t = time of fault-seconds
I= fault current through conductor-amperes
A= conductor cross-sectional area-circular mils
Ti= initial operating temperature- degrees Celsius
Tm = maximum temperature for no damage - degrees Celsius

From tables I find A = 52620 cmils
and solving I2t = 338.35

I got the same values, so the calc is right. But that result would not be for much use if you dont have the I value or the t value.

 

[mityeltu]

The values come straigt from IEE 142. It is an example they are using. You're right that the purpose is to find the time or the current, but the example is used to show how the size of the gounding conductor and the SC magnitude affects the time for the OCD to operate. I don't see how you could possibly get the same values. What I'm trying to do is repeat their calculation sop that it makes more sense to me. As it is, it doesn't make any sense because my calculation isn't anywhere near the one in 142. When you say "same values" are you talknig about the ones I posted or the ones from the example in IEEE 142?

log[(250 + 234) / (60 + 234)] = 0.2165
0.0297*0.2165 = 6.43x10^-3

if A = 52620,

I2t = 338.35

If you managed to get the same I2t value they did (17.8x10^6), can you please explain what I did wrong above?

 

[Mayimbe]

No no, I got the same values that you got...

I have the IEEE 142 2007; It didnt have the example that you mentioned. this is what it says, after showing the expressions:

The maximum temperature, Tm, is given as 150 ?C for thermoplastic insulation and
250 ?C for cross-linked polyethylene and ethylene propylene rubber insulation. If the
EGC is undersized for the fault current and the clearing time, insulation damage to phase
conductors in a conduit may occur due to the proximity of the EGC to the phase
conductor. If fusing is a criterion, then a final temperature of 1000 ?C (1832 ?F) for copper
and 630 ?C (1166 ?F) for aluminum may be used (see Kaufmann, “Application
Limitations”).

sorry, im reading the example...

let me see

 

[Mayimbe]

well, I saw the example, its quite confussing, they are saying that if you design the ground fault to be a 500% of 400 A, which is 2000 A, at 4.5 sec you will get your feeder at 250 degrees. right? that means that the protective device will have 4.5 seconds to act, or you will lose your feeder. then they say that at 800% that will be reduce at 1.7 sec. :S

Well, 2000A*2000A*4.5sec=I2t=18*10^6 right? something must be wrong,

My expressions doesnt have the log on the division of the temperature, it says:
(I2t)/A = 0.0297 in ((Tm + 234) / (Ti + 234))]

it says IN, not LOG as you said.

I believe the expression is an integration

 

[mityeltu]

I don't believe an integration would be appropriate as there is no variable. Initial and final temps are constants as far as this equation is concerned. The log term is nothing more than a temperature conversion (if I read it right).

I have tried both natural and common logs and still have not managed to find how they got the value for I2t.

i don't know where the other constants came from, so I can't even verify them as correct.

I understand that small changes in time can have drastic effects on current when speaking from a protective device TCC, but this is escaping me.

 

[mityeltu]

I see that the newer version (2007) does not even contain the example I have shown here -- my copy of Ieee 142 is 1991 version.

I cannot account for the "in" in the equation... That really makes no sense. I find this to be somewhat disappointing given the repuation IEEE is supposed to have.

I guess it's possible that the 1991 version contained an error in that example, hence its removal, but I tend still to believe that there is something I'm missing.

 

[mityeltu]

I FOUND THE ERROR!!!!!!!

The equation SHOULD read [I/A]^2 x t = 0.0297 log ((Tm + 234) / (Ti + 234))

Working to find I2t looks like this:

I2t = A^2 x 0.0297 x log ((250 + 234) / (60 + 234))

assuming A is 52620 cmil for #3 AWG

I2t = 17.8038 x 10^6

I wish these folks at IEEE would check and verify their own material.

 

[Mayimbe]

congrats!!

I found out that the expression comes from a Joule equation of the specific energy produced by the current at any amount of time.

Eei=S i.dt

S is the integration symbol.

If you want to dig more on that.