[Help] Delta connection resistance computation

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Enbi

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Location
Philippines
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Maintenance Engineer
Hello, everyone.
I've been searching on what is the solution to my problem and I saw this forum may give me some ideas.
So this is the situation, we are supposed to measure the resistance of heater in delta connection and the heater(3pcs) specs are below:

Voltage = 220V
Power = 1680W

All 3 heaters are the same specs.

Its terminal connection is as follows:
R1S1 (for heater 1) = 19.6 ohms
S1T1 (for heater 2) = 19.8 ohms
R1T1 (for heater 3) = 19.6 ohms

When measuring resistance using Fluke, it gives the resistance given above (all heater is in good condition)

Here's the thought way(in our section) on getting the resistance.

R = ( ( ( V x V ) / P) x 2) / 3
= ( ( ( 220 x 220 ) / 1680 ) x 2) / 3
= ( ( ( 48,400 ) / 1680) x 2) / 3
= ( ( ( 28.81) x 2) / 3
= ( (57.62) / 3
R = 19.21 Ohms

and we derive to a standard reading of 19.2 ohms for R1S1, S1T1 and R1T1 heater.

My question is what is the proper computation to explain that the heater resistance are still in acceptable condition? or is my computation correct?

Thank you so much,
 
Not sure I follow the question fully, but if you have 3 identical resistors of resistance R connected in a delta (triangle) configuration, and you measure the resistance between any two corners of the triangle, the measurement should be 2/3 R.

That's because the direct path between the two measurement point (the resistor R directly between them) is in parallel with the path around the other two sides of the triangle (the two resistors R in series). So you are measuring a resistance of R in parallel with a resistance of 2R. And the parallel resistance is 1/(1/R + 1/2R) = 2/3 R.

That fits fairly well with the numbers you provides. The resistance of one heater is R = V2/P = (220*220)/1680 = 28.8 ohms, nominally. And 2/3 of 28.8 ohms is 19.2 ohms.

Cheers, Wayne
 
To simplify, the formula is ( VxV/P ) x 2/3. Summing it up, that formula we are using is quite right.
I will take note of your explanation and relay to my team mates.

Thank you very much,
 
wwhitney said:
Yes, I noticed on rereading the OP your computation ends up the same as mine. You just used the 2/3 factor without explaining where it came from.

Cheers, Wayne
Click to expand...
Yes, the explanation is what I'm looking for. I can't present the computation without stating why this is the formula we use.
Thank you,
Enbi
 
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