This is a philosophical discussion so I am going to start with a concrete example to focus where I'm trying to go. The proof is in the attachment.

You walk up to 3 phase component.
Ambient is 40C
Two phases are 50C, third phase is 70C.

We have a number of ways of evaluating it. To evaluate the deviation from normal we normally look at rise above the normal "sister" phases 70-50 = 20C.

I propose that when you sit and think about it, a more useful quantity to examine is
Rise-above-ambient-suspect-faulty / Rise-above-ambient-sister = (70-40) / (50-40) = 3.0

In an approximate way, this represents the ratio of the suspect faulty component electrical resistance to the normal sister component electrical resistance. As such, it is not greatly affected by load or wind conditions. In contrast, rise above sister is greatly affected by both wind and load.

Linked to this message is an attachment (RfaultOverRnormal.doc) which provides the math framework for these claims. I heartily recommend that when you view it you switch to View Menu and select "outline" view (available in both word and internet explorer when viewing this document). Then you can use the indentation of the outline will help show you the structuve of a discussion. Statements that support a previous statement are indented under that statement. Also will be helpful to turn off formatting (show formatting button on the outline toolbar). If you don't get the right format, it is going to look like a bunch of aimless rambling. (perhaps some people will conclude that anyway?.... hope not ;-).

This message has been edited. Last edited by: electricpete,

RfaultOverRNormal.doc (36 Kb, 54 downloads) Discussion

Here is a spreadsheet that provides examples of temperature rises computed using the model of the above word document. The punchline of course is that the ratio I propose depends only on equipment condition (resistance ratio) and not on load and wind. Also suitable for indirect heating situations.

The symboles are the same as the word document but I'll repeat them for info:
Tamb = ambient temp
I = current (load)
Rt = thermal resistance
Re_S = electrical resistance of the normal Sister connection
Re_F = electrical resistance of the suspect Faulty connection
T_S = Measured temperature of the Sister connection
T_F = Measureed temperature of the Faulty conneciton
ROA_S = Rise above ambient of the Sister connection
ROA_F = Rise above ambient of the Faulty connection
ROS = ROS = rise above sister = T_F-T_S=ROA_F-ROA_S
DI = Deviation Index = ROA_F/ROA_S = Proposed ratio to estimate Rfaulty/Rnormal

I would be interested to hear any comments on this approach.

RfaultOverRnormal2.xls (17 Kb, 39 downloads)

I agree, the existing IR “severity chart” for electrical applications makes no sense.

IR people do realize that absolute temperature of an electrically produced heat, at given electrical resistivity, depends on factors, such as current, wind, ambient temperature. What they don’t realize is the fact that the temperature difference (what they relying on now for severity assessment) between the Faulty and Sister objects is also a function of the same parameters (except ambient).

Usage of the proposed relative measure, such as DI, eliminates this dependency. Another variable that creeps in is the change of the specific resistivity ( 0.4% per deg for copper wire) that overstates the actual resistivity increase due to a poor contact. This can be taken into account.

In the expression:

DI=(Tfaulty – Tsister)/Tsister

(Where T is the absolute temperature reading from the IR instrument)

if DI is less then, say 5%, NO FAULT is present;
if DI = 5-10% ALERT severity level could be assigned, and so on.

Good idea!

Thx David, I appreciate the comments.