IEEE 242-2001 - Intermediate Cable Overload Capacity - Equation 9.5.2.4 inconsistent with Table 9-6? 2

[crjohnson]

Hi Folks,

I think I may have found an inconsistency in IEEE 242-2001 between the equations Section 9.5.2.4 and Table 9-6.
I hope that I am making a mistake somewhere, and y'all will be able to help point out my error.

Edit: I believe if you follow just the equation for any value of K other than 1, the equation results are way off. For values of k<1, you could overestimate the amount of current the cables could handle by a factor of 2 or more.

Based on the standard, values in table 9-6 are calculated based on the equation shown in 9.5.2.4:

I[sub]e[/sub]/I[sub]n[/sub] % = SQRT(((T[sub]e[/sub]-T[sub]o[/sub])/(T[sub]n[/sub]-T[sub]o[/sub])-e[sup](-Θ*K)[/sup])/(1-e[sup](-Θ*K)[/sup])*((230+T[sub]n[/sub])/(230+T[sub]e[/sub])))*100​

Where:

I[sub]e[/sub] is emergency operating current rating,
I[sub]n[/sub] is normal current rating,
T[sub]e[/sub] is conductor emergency operating temperature,
T[sub]n[/sub] is conductor normal operating temperature,
T[sub]o[/sub] is ambient temperature,
K is a constant, dependent on cable size and installation type (see Table 9-5 in IEEE 242-2001),
230 is zero-resistance temperature value (234 for copper, 228 for aluminum),
e is base for natural logarithms.
​

However, if you plug in the values as indicated the results returned do not equal that of the table.

For example, take first line of table 9-6 calculated for EPR-XLP with T[sub]e[/sub]=130, T[sub]n[/sub]=90,and T[sub]o[/sub]=40
From the table:

k=0.5, %=1136
k=1.0, %=1602
k=1.5, %=1963
k=2.5, %=2533​

From Equation as written:

k=0.5, %=2265.49
k=1.0, %=1603.885
k=1.5, %=1311.5
k=2.5, %=1018.061​

But, if you change the equation to divide by k instead of multiply by k then it appears to work.

I[sub]e[/sub]/I[sub]n[/sub] % = SQRT(((T[sub]e[/sub]-T[sub]o[/sub])/(T[sub]n[/sub]-T[sub]o[/sub])-e[sup](-Θ/K)[/sup])/(1-e[sup](-Θ/K)[/sup])*((230+T[sub]n[/sub])/(230+T[sub]e[/sub])))*100​

From Equation as corrected:

k=0.5, %=1136.859
k=1.0, %=1603.885
k=1.5, %=1962.765
k=2.5, %=2532.281​

Does the corrected equation work for y'all?
Did I miss somewhere in the section where is says the inverse of K should be used?
I attached my spreadsheet where I ran my tests.

Thanks,

Chris

 

[7anoter4]

In my opinion, it could be better. However, there are still errors, in any way. If Io will be 88% of In then the error is only 0.1-0.5% for 10-100 sec but 3-7% for 1000 sec and more.

 

[crjohnson]

I was using the assumption that the cable was already loaded to full load current, I.e. Io=In.

There maybe other approximation errors,but in case it wasn't clear, the results of the equation as listed cause much higher errors than 10%.
For k=0.5, the equation as written results in overestimating the allowable current by a factor of 2. So 200% error.
I.e. for a cable with a rating of 100 amps, the table estimates it could handle 1136 amps, but the equation says it could handle 2266.
So if you followed only the equation, you would would be overloading the hypothitcal 100 amp cable by over 1000 amps.

 

[cuky2000]

crjohnson, good catch, the revised formula is in agreement with the previous version of IEEE Std 242-1986 (see excerpt below). As a suggestion, Fill free to notify the IEEE committee to correct the error of your finding.

 

[crjohnson]

Thanks for the reply cuky2000, and for helping me confirm this.
I was so confused trying to get my spreadsheet to work

 

[7anoter4]

Sorry. I put To=20oC instead of 40.However,k=0.5 to 2.5 it is for cable in air and could be 40oC ambient, K=4 and 6 it is for underground cables then To has to be less-in my opinion.