Physics Help Forum Current Carrying Capacity

Aug 26th 2016, 05:17 AM   #1
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Current Carrying Capacity

I am trying to solve for the current carrying capacity of a wire in a glass-to-metal seal. I have been trying to use the neher-mcgrath formula to do so but have been unsuccessful. This may or may not be the appropriate formula for what I am trying to do. My end goal is to have a working understanding of a formula that will enable me to make a chart similar to the one titled "Current Capacities of Standard Kemlon Glass Seals (52 alloy) in Amperes with 50 Degree [Fahrenheit] Temperature Rise" on this page:

Kemtite High Pressure Connectors Technical Information

The main difference is I want to be able to make a chart in excel that will let me fill in the required properties and excel will do the math to give me current carrying capacity for that setup.

What I have done so far:
I have found the neher-mcgrath formula for current carrying capacity (ampacity).
This formula reads I = square root((T sub c - (T sub a - delta T))/(R sub dc * (1+Y sub c) * R sub ca)) Sorry if that is hard to understand, here is a link to where I found it:

Understanding the Neher-McGrath Calculation and the Ampacity of Conductors

Where I = ampacity, T sub c = conductor temperature, T sub a = ambient temperature, delta T = conductor temperature due to dielectric loss, R sub dc = conductor dc resistance, Y sub c = loss increment due to conductor skin, R sub ca = thermal resistance between conductor and ambient temperature

As I mentioned at the beginning of this post, the wire would be within a glass to metal seal. There is no conductor skin (unless air counts?) and I read that Y sub c was negligible for wire sizes smaller than AWG 2. I read that if the voltage will be less than 2000V delta T is negligible. That said I used the shortened version labeled "Equation 1" which omitted those values. For conductor temp I was trying to match the chart given as an example to verify I was performing the calculations correctly, so I used 35 degrees Celsius. I used 25 degrees Celsius as my ambient temperature. The conductor (52 alloy) dc resistance is .00000429 ohm-mm which I multiplied by the length of my pin then dived that by the cross sectional area to get .001517277 ohms. The thermal resistance I was using (which is most likely in error) was that of the 52 alloy is .0132 watt/mm Celsius which I divided by the length to get 481.060606 watts Celsius. When i run the calculation for a .030 dimater pin as shown in the chart i get 3.7 ampacity which is close to their 3.8. At this point i was feeling pretty good but then as I tried the other pin diameters and temperatures listed in the chart the numbers I got were way off mark. Any help in understanding where I went wrong would be greatly appreciated.

I will attach the excel file I was working with as well.
Attached Files
 Ampacity.xls (23.0 KB, 7 views)

 Tags capacity, carrying, current

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