unbalanced delta resistive load

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mull982

Senior Member
topgone said:
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.
Click to expand...

Yes I understand your method now. So baically you plug in the equations plug in the line currents that were given and let excell calculate and spit out the L-L loads? Is this all done by the solver function, I'm not familiar with it?

Isn't it possible however that there is more than 1 L-L loading combination that can result in these line currents?
 
S

Slim shady

Member
supplied by

supplied by

the heater bank has been switched around from machine to machine so its prop been blown from another machine but yeah supplied by 160-200 amp mcb with individual fuses down stream. controlled with SSR.
 
mbeatty

mbeatty

Senior Member
Location
Illinois
topgone said:
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.
Click to expand...

Sounds reasonable to me.
Regards,
Mark
 
S

Slim shady

Member
investigating

investigating

further investigating has shown me that elements have been disconnected due to i'm guessing a blow on that circuit. i pretty much need down time on this machine to isolate each line and test each element individually... long process but got to be done
 
R

rattus

Senior Member
topgone said:
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.
Click to expand...

My graphical solution agrees with your results within reason. Now, tell us how you solved for the phase currents--just the main points, no need for fine detail.
 
Smart $

Smart $

Esteemed Member
Location
Ohio
rattus said:
My graphical solution agrees with your results within reason. Now, tell us how you solved for the phase currents--just the main points, no need for fine detail.
Click to expand...
He used Excel's solver add-in.

The solver add-in allows the user to input an answer and the add-in back solves for the input variables.
 
R

rattus

Senior Member
For what it's worth:

For what it's worth:

Maybe someone has already said so, but it is a simple matter to compute the angles from the line current magnitudes in a delta load. Just solve a set of 3 equations--simple but a little messy.
 
S

Slim shady

Member
answer

answer

Element box was removed on friday elements all look in great condition.
what was overlooked by me was that the load was feed with 16mm2 cu cables to a junction box in which these cables would never have the grunt to hold 97 kw of heating . inside such junction box legs from the elements had been removed to derate this aspect and in turn elements not being used. possible 23. not going to much further noting is needing to be ordered and having over halve the elements sitting there spare .
thank you all for such help with vector diagrams
most appreciated
 
mbeatty

mbeatty

Senior Member
Location
Illinois
% of rated heating power?

% of rated heating power?

topgone said:
With 94.5 kW of heating at 390V, your line amps should be around 144.34 amps (balanced).

Or a total of 42.17 kW (65% of rated heating power).
Click to expand...

Topgone,
Wouldn't the total percent of rated power be about 44.7%? IMHO.
Mark
 
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