Slim shady
Member
formula please
3 phase unbalance resistive load delta formation
3 phase unbalance resistive load delta formation
unbalanced delta resistive load
- Thread starter Slim shady
- Start date
- Status
ptonsparky
Tom
- Occupation
- EC - retired
All I know is that if you get it wrong, you can see the flash of the arc in those white fiberglass bodied 600v fuses.
Give a little more detail as to what you want and most likely someone here can help you.
Give a little more detail as to what you want and most likely someone here can help you.
Slim shady
Member
??
??
3 phase heater elements wired in delta 400v supply unbalanced load ( blown elements ) each element is 2.5 kw and from my current draw ratings i have an unbalanced load.
the formula is ??
i want 2 replace sum elements and pre order them .. stoppin the machine once and ordering once from overseas
L1 72amps
L2 51A
L3 66A
could be 390 volts not to concern
??
3 phase heater elements wired in delta 400v supply unbalanced load ( blown elements ) each element is 2.5 kw and from my current draw ratings i have an unbalanced load.
the formula is ??
i want 2 replace sum elements and pre order them .. stoppin the machine once and ordering once from overseas
L1 72amps
L2 51A
L3 66A
could be 390 volts not to concern
ptonsparky
Tom
- Occupation
- EC - retired
What is the total KW supposed to be?
How many elements do you have? For all we know you may have 73%, 51% or 98% of the original load still working.
How many elements do you have? For all we know you may have 73%, 51% or 98% of the original load still working.
In order to calculate the line currents you will need to figure out the phase currents inside the delta connection. Once you have all of the phase currents you then need to vectoraly add the respective phase currents for each line to calculate the line current. You will need both the phase currents magnitude and angle when adding these to determine line current.
If you are able to determine the resistances of each element left we may be able to help with this calculation.
If you are able to determine the resistances of each element left we may be able to help with this calculation.
Slim shady
Member
total kw
total kw
there should be a total of 39 elements that should be installed
39 x 2500watts = 97.5kw of total heating
total kw
there should be a total of 39 elements that should be installed
39 x 2500watts = 97.5kw of total heating
Slim shady
Member
why not?
why not?
could do. maybe i should replace all of them everywhere in the factory.
no so cost effective.
this is just one bank of 39
i have 2 more like it and 3 more halve size banks that just use this type element.
so maybe i change all 171 elements at 44 dollars each = 7.5 thousand dollars mmmmmmmmm just on elements no such a good move.
why not?
could do. maybe i should replace all of them everywhere in the factory.
no so cost effective.
this is just one bank of 39
i have 2 more like it and 3 more halve size banks that just use this type element.
so maybe i change all 171 elements at 44 dollars each = 7.5 thousand dollars mmmmmmmmm just on elements no such a good move.
You don't complex calculations,IMO. Count the good elements connected to each phase and balance your heater loads, if that's what you wanted. From your data above, you have thirteen elements per phase. Inspect and count the elements that are not busted and compare with other phase loads.
- Location
- Illinois
- Occupation
- retired electrician
I believe the issue is that he can't determine the number of non working heater elements without shutting the machine down and taking it apart. He is looking for a way to determine the number of non working elements from the current that the equipment is pulling. That will permit him to order the new heaters without shutting the machine down.
ptonsparky
Tom
- Occupation
- EC - retired
You need 6 units.
You also need some spares. At least 39.
Get the bean counters involved and make sure everone understands how production is hurt when equipment can only be run at half speed. $7500 is small change. Small industrial I work at figures production at +$2000 per hour. The only question they ask for $1000 items is "When?".
You also need some spares. At least 39.
Get the bean counters involved and make sure everone understands how production is hurt when equipment can only be run at half speed. $7500 is small change. Small industrial I work at figures production at +$2000 per hour. The only question they ask for $1000 items is "When?".
I think your amp readings indicate your heaters have lots of blown-out elements! With 94.5 kW of heating at 390V, your line amps should be around 144.34 amps (balanced). Assuming your amp readings are correct:
Ia = 72 A; Ib = 51 A; Ic = 66 A
My computations tell me that your phase to phase heater loads are:
Lab = 12,944 (roughly 5 heaters working) - 8 elements busted
Lbc = 9,958 (4 heaters working) - 9 elements busted
Lca = 19,273 (8 heaters working) - 5 elements busted
Or a total of 42.17 kW (65% of rated heating power).
A total of 23 elements are for replacement.
I am using my own vectorial addition spreadsheet, so others can correct me.
Hope this helps.
Ia = 72 A; Ib = 51 A; Ic = 66 A
My computations tell me that your phase to phase heater loads are:
Lab = 12,944 (roughly 5 heaters working) - 8 elements busted
Lbc = 9,958 (4 heaters working) - 9 elements busted
Lca = 19,273 (8 heaters working) - 5 elements busted
Or a total of 42.17 kW (65% of rated heating power).
A total of 23 elements are for replacement.
I am using my own vectorial addition spreadsheet, so others can correct me.
Hope this helps.
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
To my eye, topgone's calculations are as correct as is possible given all of the possible variables (heater resistance tolerance and age variation, supply voltage variation and balance, etc).
I also concur with rattus and ptonsparky. It is almost certainly most cost effective to simply replace _all_ of the elements.
While it may seem a waste to replace a 'perfectly good working element', consider the reasons for the failures and the failure modes. If all of the elements have been aging together, then the ones still working are probably pretty close to the end of their life anyway. One huge variable that your measurements cannot address: what is the failure mode? You are assuming that each element works fine at full capacity until it fails with an open circuit. Well, what happens if the real failure mode is that element resistance increases. It may be that _all_ of the elements are working, just at partial capacity
Ask yourself: if you knew that an element would fail the day after you did this maintenance, would you replace it? If you _knew_ that it would fail a month from now, would you replace it? Remember that while the replacement element costs money and you have the labor costs of each replacement, you also have all of the labor costs of getting prepared to do the work, the cost of getting things like the external enclosure open, plus the costs of shutting down the process (downtime), the cost of reduced performance prior to getting to the replacement, etc.
If it is extremely easy to replace one element (system is designed so you don't have to shut down the process, safe connections so that you are not 'working live', easy to identify which element is failed, etc.) then it makes sense to simply wait for failures and fix them as they happen. But you would not even be asking this question if this were the case.
You have a machine that you don't want to shut down to do the diagnostics, which means that there is significant cost to the diagnostics. You only provide external current measurements, which means that you don't have easy access to the individual elements while operating, etc. All of this means that if you suspect that an element will fail soon, then you want to replace it now.
IMHO you have a system in which more than half of the elements have already failed, or one in which most of the elements are working at partial capacity. This means that you should suspect that the rest will go soon, and you should replace them all.
-Jon
I also concur with rattus and ptonsparky. It is almost certainly most cost effective to simply replace _all_ of the elements.
While it may seem a waste to replace a 'perfectly good working element', consider the reasons for the failures and the failure modes. If all of the elements have been aging together, then the ones still working are probably pretty close to the end of their life anyway. One huge variable that your measurements cannot address: what is the failure mode? You are assuming that each element works fine at full capacity until it fails with an open circuit. Well, what happens if the real failure mode is that element resistance increases. It may be that _all_ of the elements are working, just at partial capacity
Ask yourself: if you knew that an element would fail the day after you did this maintenance, would you replace it? If you _knew_ that it would fail a month from now, would you replace it? Remember that while the replacement element costs money and you have the labor costs of each replacement, you also have all of the labor costs of getting prepared to do the work, the cost of getting things like the external enclosure open, plus the costs of shutting down the process (downtime), the cost of reduced performance prior to getting to the replacement, etc.
If it is extremely easy to replace one element (system is designed so you don't have to shut down the process, safe connections so that you are not 'working live', easy to identify which element is failed, etc.) then it makes sense to simply wait for failures and fix them as they happen. But you would not even be asking this question if this were the case.
You have a machine that you don't want to shut down to do the diagnostics, which means that there is significant cost to the diagnostics. You only provide external current measurements, which means that you don't have easy access to the individual elements while operating, etc. All of this means that if you suspect that an element will fail soon, then you want to replace it now.
IMHO you have a system in which more than half of the elements have already failed, or one in which most of the elements are working at partial capacity. This means that you should suspect that the rest will go soon, and you should replace them all.
-Jon
ptonsparky
Tom
- Occupation
- EC - retired
Had to go back & see what I did wrong. I approached this from an entirely different view because I lack the skills of topgone. I just figured the average current imbalance (63) which turns out to be about 45% of what it should be. My error was taking that times 13 elements instead of the 39. Should be closer to 18 bad units. Just stating how I came up with my answer.
Let us know.
Let us know.
Cold Fusion
Senior Member
- Location
- way north
rattus, pt, winnie -
Your comments appear to be implying the elements are deteriorating with age. I'm curious on how you are determining the failure mode.
How is it you know the elements are not exhibiting popcorn failures - ie. random?
How is it you know that failures are not process related? Over temp, run dry, mechanical damage, chemical (corrosion) for a few examples.
Just curious - Do all heater elements burn out with age?
cf
Your comments appear to be implying the elements are deteriorating with age. I'm curious on how you are determining the failure mode.
How is it you know the elements are not exhibiting popcorn failures - ie. random?
How is it you know that failures are not process related? Over temp, run dry, mechanical damage, chemical (corrosion) for a few examples.
Just curious - Do all heater elements burn out with age?
cf
That has been my experience with strip heaters and open coil heaters. Especially with the latter, the nichrome wire corrodes and evaporates to the point that it finally opens. Even oven and range elements which are encased in a metal shell open up and start arcing.
Certainly a hostile environment would accelerate this aging process, and if that be the case, that is even more reason to replace all of them--perhaps on a regular basis.
Now it should be possible to build a heating element that does not age, but that might be too expensive. And, low temp heaters like you would find in a print dryer may last forever, or at least for one's lifetime.
Slim shady
Member
thank you all
thank you all
love this site so much experience and help from all
most appreciated
thank you all
love this site so much experience and help from all
most appreciated
Can you explain briefly how you went about this calculation. I understant that it needs to be done vectorally but dont know where to start considering there are probalby multiple combination with L-L heater loads that result in the given phase currents.
Sorry for the delayed response. Here are my spreadsheet assumptions:
Voltages: (on an ABC phase rotation)
Vab =390 @ an angle 30 degrees
Vbc =390 @ an angle -90 degrees
Vca =390 @ an angle 150 degrees
My initial spreadsheet use was solving line-to-neutral loads that were unbalanced, my reference voltage was Van (line-to-neutral) hence the 30-degree shift. I just modified the spreadsheet to cater to line-to-line loading.
With the cells intended for input of phase-to-phase loads empty (Lab, Lbc, and Lca), I forced iteration of values unto those cells using the Solver Add-in of Excel with the following constraints;
Ia = 72; Ib = 51, and Ic = 66
Note that:
Ia = Iab - Ica;
Ib = Ibc - Iab; and
Ic = Ica - Ibc (there are no line to neutral currents) and
Iab =Lab/Vab;
Ibc =Lbc/Vbc; and
Ica =Lca/Vca.
Just be careful in the formula you use to correctly compute the respective vector angles as this will get your results wrong if not well formulated.
Hope this is clear to you.
What's all this supplied from? A single 175A or 200A breaker?
With almost 100KW of electric heat, there should be some supplementary OCP somewhere. Are you sure you don't have some blown fuses somewhere? That's a lot of heaters to be out.
With almost 100KW of electric heat, there should be some supplementary OCP somewhere. Are you sure you don't have some blown fuses somewhere? That's a lot of heaters to be out.
- Status