3 phase heating element connected only to two phase.

Hello Teams!

I have got 3 phase/9kW heating element like on below picture,
Im wonder what will happend if during the working i disconnect one phase?
I suposse heating power will drop down....
But what else can happend?
If the heater will be damage?
1594119433009.png
1594119466085.png
Regards,
Andy
 
Hello Teams!

I have got 3 phase/9kW heating element like on below picture,
Im wonder what will happend if during the working i disconnect one phase?
I suposse heating power will drop down....
But what else can happend?
If the heater will be damage?
View attachment 362
View attachment 363
Regards,
Andy
Hi Andy,

My thinking is that there unbalancing load or current occuring in the case that you described .

You should know how this would affect power supply ( transformer ...) protection relay ....

Heating power will surely drop down as you mentionned.


What is the wiring type & rated voltage of the tranformer.

You can submit a diagram ( old vs new installation) , then we may have better overview of what can be done or not, for a safe and smooth operation of the Heater & installation.

What kind of heater is it ?? i use to work on space heater and other heater tubes like boiler heater .

Hope this can help,
James
 
Hi James,
Glad to hear you again!

Its a simple heater using for heat up the water(or oil),
Assembly inside boilers or water batch,

The circuits its very simple,, unfortunately i dont have i diagram, im sorry:(
Please consider it like a heater insiade the boiler,
Heater heat up the water to set-up temperature then switch off,
When water temperature drop down the heater start heating agian...

Im dont want to connect this heater into two phases,
Im just wonder what will happend if one phase will be disconnect,
Do you think that current will rise-up and will damage the heater?
regards,
Andy
 
Hi James,
Glad to hear you again!

Its a simple heater using for heat up the water(or oil),
Assembly inside boilers or water batch,

The circuits its very simple,, unfortunately i dont have i diagram, im sorry:(
Please consider it like a heater insiade the boiler,
Heater heat up the water to set-up temperature then switch off,
When water temperature drop down the heater start heating agian...

Im dont want to connect this heater into two phases,
Im just wonder what will happend if one phase will be disconnect,
Do you think that current will rise-up and will damage the heater?
regards,
Andy
Hi Andy ,

For sure there will be current/load unbalancing in the case you described.
The thing is that this can make damage heater .
The case you are describing is like Overload operation and the associated protection ( thermal relay ) would trip in my opinion.
Ifthe power supply/protection relay... and wiring, are not sized correctly indeed then you can surely get some troubles, if you have one phase disconnected by any reason!

That what i can tell you from my own experience on power plant commissioning and other outages around the world .

Hope that this can help!
Regards,
James.
 
Hi Andy ,

For sure there will be current/load unbalancing in the case you described.
The thing is that this can make damage heater .
The case you are describing is like Overload operation and the associated protection ( thermal relay ) would trip in my opinion.
Ifthe power supply/protection relay... and wiring, are not sized correctly indeed then you can surely get some troubles, if you have one phase disconnected by any reason!

That what i can tell you from my own experience on power plant commissioning and other outages around the world .

Hope that this can help!
Regards,
James.
Thanks a lot James :)
Its clear and understood!
Regards,
Andy
 
Andy...
Is the source 3-phase, 3-wire, 400-V, or 3-phase, 4-wire 400/231 ?
BTW, if one element is removed neither source above will endanger the remaining elements ! Nor will it trip a protective device!
Full explanation will be illustrated when you answer the question above!
Phil Corso
 
Andy (if I may)...
I believe the fellas that alluded to destructive or disruptive events were referring to a phenomenon called "Displacement Neutral Voltage" ! Yes, it can occur ! But, only if the 3 'Y' loads are complex, meaning containing inductive or capacitive components! It isn't possible in your situation because your elements are all "resistive" !
Phil
 
Hi Phil,
Good day!
Sorry for not reply sooner,

Answering to your question:
"Is the source 3-phase, 3-wire, 400-V, or 3-phase, 4-wire 400/231 ? "
Its 3 -phase, 4-wire(3 live + PE), Voltage between 2 phases is 400VAC,

Im asking about this heating element because im just curious:
In case of 3-phase/4wire/400VAC water pump (or any other 3 phase engine) when engine lost one live wire = its bad,
Due to fast coil heating you can easy damage enginee( in case if i you dont not have thermisor or overload relay for avoid destruction),
Im wonder if the same situation will happend when one live wire will be disconnect from my heating element...
Regards,
Andy
 
Andy...
The requirement you just proposed is interesting. Essentially you want to know if there is any situation which by accident could cause harm to the Heater Elements. I have to admit there is one, but to date no one has recognized it. Furthermore, it would have to be done deliberately by someone stupid enough to do it!
I realize you are not an EE... may have a slight interest in it... but would rather be done with it. I'll return in a little while!
Phil
 
Hello Phil,

Yes you have absolutely right im not electric engineer,
Im just a guy who looking a knowledge,
I couldn't find the answer for my question(or doubts) in other forums so i just tried to ask people who have more knowledge and practice,
thats story,
You wrote:
"Furthermore, it would have to be done deliberately by someone stupid enough to do it! "
Im just consider situation when one wire(phase) was not good jointed with heater and disconnected itself ,
i think such thing could happend..
And i wonder what "bad" can happend with the heater when this wire is disconnect?
Hope this make sense,
Regards,
Andy,
 
Andy, here is my answer using my Gi.Fi.E.S. method (pronounced Jiffy’s), where:
o Gi means ‘Given’
o Fi “ ‘Find’
o E “ ‘Equation’
o S “ ‘Solution’ The procedure follows:
Gi.ven: System Facts
o Supply = 3-phase, 4-wire (3- current carrying Line conductors, and 1- current carrying GroundED conductor (Neutral) for unbalanced operation.
o Note: there is a 5th GroundING conductor for safety (PE). Italics refer to USA NEC practice.
o Power supply's phase-to-phase voltage, Vpp = 400Vac.
o Load consists of 3, Wye-connected Heating Elements, each rated, Pe = 3,000 Watts
F.ind: System Parameters and Unusual Operation
o Element Voltage, Vpn !
o Element Amperes, Ie !
o Element Resistance, Re !
o Any situation that could be dangerous !
E.quations, To use.
o Element Voltage, Vpn = Vpp/Sqrt(3) = 400/Sqrt(3) = 400/1.7332 = 230.9 Volts.
o Element Current, Ie = Pe/Vpn = 3,000/230.9 = 13.0 Amperes.
o Element Resistance, Re = Vpn/Ie = 230.9/13.0 = 17.76 Ohms.
S.olution, Case 1, Operation With 1 Element Removed
o Because the supply has 4 current carrying conductors (3-line, 1-neutral), the remaining element currents are each 13.0 A. And total heater element power available is reduced to 6,000 W. Thus, no danger.
S.olution, Case 2, 1 Operation with 1 Element Removed, And Neutral Supply conductor ‘Opens’
o Because the Neutral-Supply conductor is unavailable, the 2 remaining heater elements are series connected across the two remaining Line conductors. The current through the 2 remaining heater elements is Vpp/(2*Re) = 400/35.36 = 11.2 A. And total heater power available is reduced to 2*(Ie * Ie * Re) = 4,460 Watts. Thus, no danger because output of each element is only 2,230 Watts !
S.olution, Case 3, Operation with 1 Element Removed, And a 2nd Line Conductor ‘Opens’.
o Because both the Neutral-conductor and Line-conductor are available only 1 heater element is still connected. Total heater power available is reduced to 3,000 Watts. Thus no danger
S.olution, Case 4, The Only Dangerous Situation.
o A heater element is inadvertently or deliberately connected across 400 Vac. Then, the current, Ie = Vpp/Re = 400/17.76 = 22.5 A. Total heater element power available is increased to 8,986 Watts.

Andy, since you already have overload protection, I suggest installation of an Under-Load Detection unit, like a Power Meter !

Regarding similar events with other loads. Yes, the calculations can be made, but no other situation is as simple as one having just a resistive load ! Furthermore, except for Under-Load every situation discussed above is detectable !

If anything else is needed let me know!
Regards, Phil
 
Thank you Phil for your fully explanation and for patience :)
Now is totally clear,
(Gi.Fi.E.S method its very interesting and i think could be usefull not only for such "hypothetical considerations " but in daily trouble too)
Wish i have at least a small piece of your knowledge in the future,
have a good day!
regards,
Andy
 
Andy...
This forum's old-timers are always on the ball. They never forget. I was asked, "What about "PE" ? Once more, I edited Jiffy's. I also changed the Heating Element's rated-current label to Ie, to make it more "visible !
Phil
 
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