Message Board Thread - "Cost for Loose Connection"

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Cost for Loose Connection Hal 9/30/2005
I would like to put a dollar amount saved for identifying a hot connection.
I welcome your input.

480 Volt 3 phase connection with one leg having 110 deg temperature rise carring 157 amps current.

Heat ~ Resistance

For Copper: ohms per degree = .0068
Increased Resistance with 110 degree temperature rise:

R=.0068 * 110= 0.748 ohms

Additional Watts needed with 157 amps of current:

Watts=I^2R =157^2 * 0.748 = 18,437 watts

Additional kilowatt-hours (KWh) operating one year:

8736 hours/year * 18,467 watts / 1000 = 161,069 KWh/year

Wasted Energy Cost at $.035 per KWh
not including demand rate & taxes (residential rates $.07+)

161,069 kWh * $0.035/KWh = $5,637 per year
 
Re:Cost for Loose Connection Hal 10/1/2005
I see that the equation (delta R=alpha * delta T) is in error

delta R = R0 * alpha * delta T

where: alpha the temperature coefficinet of resistance= .0068 ohms/deg C for copper

I found the following equation at this web site:

http://230nsc1.phy-astr.gsu.edu/hbase/electric/restmp.html#c3

depending on the guage of wire R0 could be 1.74 ohms/cm to 0.00165 ohms/cm for 40 to 10 guage wire respectively (CRC Handbook of Chemitry and Physics)

 
Re:Cost for Loose Connection Manuel 10/2/2005
Interesting topic Hal.

would like to ask to re-evaluate your case.

i see you have 3 wires per fase, maybe each one seems 1/0 ??

please develop your case in order i can take this info like a very good example of thermography survey 'return of investment'

my very best regards.
roberto.cruz@thermoimagen.com
 
Re:Cost for Loose Connection Doctir bob 10/11/2005
Heat transfer is a complex science. At the risk of oversimplifing, let me give you some numbers from our experience at ITC. We simulate problem connections with significant temperature rises using a 30 Watt soldering iron tip. So, my gut feeling is the watt dissipation level for this problem is in this range or lower.

We have developed watt loss calculation software that calculates the radiant and convective losses from a surface that in steady state conditions enable reasonably accurate calculations of watt loss. It does not consider conductive losses through the wires or cables, as it was originally developed for equipment such as oil filled circuit breakers. But the calulation could be run for this case with knowledge of the average temperature over a known surface area, emissivity of the surface, ambient air temperature and reflected apparent temperature (background temperature).

Frankly, I think the larger questions are: what is using 157 amps that will quit using it when the connection fails, who may be injured, how much parts and labor cost will it take to replace it? Our experience is personnel safety, lost production and equipment loss are the key issues when it comes to these types of problems.
 
Re:Cost for Loose Connection Top Gun 11/8/2005
Perhaps another approach is needed to figure the cost of a loose connection.

Your image is showing the line-side connections of a large circuit breaker. If this issue were to progress to critical failure, your costs would be far higher than any energy losses based on the cost of electricity. Your organization would be pulling in new wires from the illustrated terminals to the other terminals at the far end of those wires, probably replacing the circuit breaker with a new one, and down time to the equipment to which this circuit breaker supplies power. If this CB is in a hospital, perhaps someone could lose their life; in a major computer center, some data loss; in an industrial setting, perhaps an automotive assembly line, over a million dollars.

These things are additional costs to whatever loss you can calculate on the track followed before this comment.
 
Re:Cost for Loose Connection OPG1 11/9/2005
You need to consider what a catastphic failure would cost. You will find that energy cost is minimal compared to this. If you are going to calculate energy costs, remeber that as the connection heats up the R value is going to change. This in turn will cause the current to change and circle continues.
I hope with the delta T you have in the image the breaker was isolated and not allowed to continue to stay online. The power station I work at would consider this atype a anomaly, which requires immediate repair
 


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