3 phase kiln questions?

[Darkwood]

If measuring phase-phase then you will see twice the resistance (of a single element of star-connected load) on a DC test.

?.. 4 elements between any 2 phases will equate to 4 X the resistance as we have 4 elements in series with each other per any 2 phase.

 

[HappyHippyDad]

P = 12kW = 12,000 W
V = 230V
I = 12,000 / 230 = 52.2A for single-phase (excluding control circuit, fans, etc)
I per phase = 52.2 / 3 = 17.4A
R = V / I = 230 / 17.4 = 13.2 ohm per set of elements
In practice elements R will increase as they heat, so cold measurmenets will give you the switch-on current and that ought to drop by a small amount as it heats up (off hand don't know by home much though).

A set of elements consisting of 4 elements, so 13.2 ohm /4 = 3.3 ohm p/element?

 

[pc1966]

?.. 4 elements between any 2 phases will equate to 4 X the resistance as we have 4 elements in series with each other per any 2 phase.

Treating the 4 series-connected elements as one "arm" of a star.

So if each "arm" is 13.2 ohm, then L-L would show 26.4 but as each arm has 4 elements, they would each be 3.3 ohms.

 

[westward10]

A set of elements consisting of 4 elements, so 13.2 ohm /4 = 3.3 ohm p/element?

Simple as that with series resistances. In your example all elements are identical so straight forward.
Stick 230v across it you get 17.42A using ohms law.

 

[Darkwood]

Treating the 4 series-connected elements as one "arm" of a star.

So if each "arm" is 13.2 ohm, then L-L would show 26.4 but as each arm has 4 elements, they would each be 3.3 ohms.

Did you edit the star bit into your post, when I quoted it it did not have that info but the quote showed that information as though you edited it while I was quoting it?..

 

[westward10]

Did you edit the star bit into your post, when I quoted it it did not have that info but the quote showed that information as though you edited it while I was quoting it?..

It was edited but "star" is in both versions.

 

[HappyHippyDad]

Group reading

L1 to L2
L1 to L3
L2 to L3

Each reading should be more or less the same.

Take this common value and divide by 4 then use basic ohms law.

I = V/R
Amps = 400/resistance.

Single element reading - resistance of single element (wiring disconnected)

Pick your voltage either 230 or 400, ideally 400 as that is the supply you are connecting.

Amps = 400/resistance of one element.

This is where I'm getting confused.

You say Amps = 400/resistance of one element

Therefore Amps = 400/3.3 = 121A.

The other examples show that an element with 3.3 ohm of resistance indicates 12kW of power which is 52.2A

i.e. 3.3 x 4 (elements) = 13.2 ohm p/set of elements.

230/13.2 = 17.4A per phase = 52.2A for all 3 phases.

 

[Darkwood]

My bad, @happyhippydad

You are been told a few differing methods here and I threw confusion into the mix, I edited that post to allow a star config of both single and 3 phase, you quoted it before the edit.
I actually posted wrong, I posted a job I am doing atm which is another config so apologies for the confusion here.

 

[westward10]

You don't have 400v across each group of elements you have 230v whether the supply is 230 or 400v.

 

[loz2754]

The other examples show that an element with 3.3 ohm of resistance indicates 12kW of power which is 52.2A

i.e. 3.3 x 4 (elements) = 13.2 ohm p/set of elements.

230/13.2 = 17.4A per phase = 52.2A for all 3 phases.

This part is correct.

 

[loz2754]

Maybe a bit over simplified, but this is the reading I would expect to see if connecting all three phases as if it were single phase.

 

[westward10]

This part is correct.

It is but would be better worded 52.2A for all three element groups on a 230v single phase supply.

 

[ipf]

My bad, @happyhippydad

You are been told a few differing methods here and I threw confusion into the mix, I edited that post to allow a star config of both single and 3 phase, you quoted it before the edit.
I actually posted wrong, I posted a job I am doing atm which is another config so apologies for the confusion here.

I should think so, too! ?

 

[Darkwood]

You don't have 400v across each group of elements you have 230v whether the supply is 230 or 400v.

Depends on where exactly you are testing from an what tests, if you are testing from the output of the contactor then you are measuring 8 elements between 2 phases (400v)
If you are testing from the output of the contactor to any PH / N then you have a group of 4 (230v)
If you are testing at the elements then it again depends on where your testing points are whether to apply 230v or 400v but crucially as I think you were saying - 1 group of 4 elements will not have 400v applied to it which I agree fully with.

 

[westward10]

Yes I was referring to each group of four.

 

[HappyHippyDad]

This has been very interesting. Thank you all for helping me understand it better.

Just to make sure I'm getting it...

Lets say I measure the resistance of 1 of the elements and it is 2 ohm (we'll assume this is correct for all of them). To find out the power rating of the kiln we would do the following:

2 ohm x 4 = 8 ohm p/set of elements.

230/8 = 28.75A p/set

x 3 = 86.25A for all 3 elements.

x 230v = 19.8 kW (to 1 dp)

Is this correct?

 

[westward10]

Yes but as @darkwood suggested earlier measure two elements in case you pick on a duff one.

 

[HappyHippyDad]

Yes but as @darkwood suggested earlier measure two elements in case you pick on a duff one.

Hence me popping the brackets in saying we'll 'assume' it's correct ?

 

[Lucien Nunes]

Is this correct?

Yes.

One thing that deserves explaining here is the actual connection configuration of the element groups. We can ignore that each group is actually made of four physically separate sections and just consider them as single elements.

Each element group is connected from one line terminal to neutral (via the main switch and contactor.) In other words the kiln contains three identical, single-phase 230V heaters, just like three fanheaters sat on the floor with separate flexes and plugs. It wouldn't matter whether you plugged those fanheaters into three sockets on the same phase, or three sockets on different phases, because they are independent. Same with the kiln - the three line terminals can be connected to the same phase or different phases, so long as each terminal is 230V from neutral.

This is a different situation to (e.g.) a 3-phase motor, which is not like three independent single-phase motors because the magnetic fields of the three windings interact with each other. The windings must be connected to alternating voltages that vary in time with the correct phase relationship to each other. You cannot simply connect all three to one phase.

Likewise, a different situation obtains in a heater consisting of three identical elements connected in star (i.e. each between a line terminal and a star point) but where the star point is not connected to the supply neutral. This is a handy configuration for industrial supplies where a neutral might not be available When connected to a 3-phase 400V supply, the three elements form a balanced load in which each element receives 230V between the line terminal and the star point. Through symmetry, the star point will lie at a voltage close to that of the supply neutral, hence the element currents and voltages would be just like those of the klln (although note that the 230V control ccircuit of the kiln absolutely requires the neutral to be present). But clearly, if the star point of our heater is not connected to neutral, wiring all three line terminals to the same phase won't work because now there is no complete circuit. Like the motor, this needs a 3-phase supply.

Returning to the kiln and considering the currents in the circuit conductors; on 3-phase the currents from the three elements sum to zero at the neutral terminal (except for the small control circuit load) unless one element fails O/C in which case the neutral current will equal the line current. On single phase the line and neutral currents are inherently equal. The line and neutral currents can be taken as equal both on single-phase and 3-phase supplies, with the current on single phase supply three times that on 3-phase.

What I hope to have shown in the above explanation is that the kiln elements, by virtue of being connected L-N, are less like a 3-phase load and more like three single-phase ones, for which ordinary single-phase calculations can be applied.

You can supply your own sketches to accompany the above

Oh, BTW

230/12000 = 0.0192ohms

R=V/P ???
For this, you are to beat yourself with a big lump of SWA until it really hurts.

 

[loz2754]

Yes.

One thing that deserves explaining here is the actual connection configuration of the element groups. We can ignore that each group is actually made of four physically separate sections and just consider them as single elements.

Each element group is connected from one line terminal to neutral (via the main switch and contactor.) In other words the kiln contains three identical, single-phase 230V heaters, just like three fanheaters sat on the floor with separate flexes and plugs. It wouldn't matter whether you plugged those fanheaters into three sockets on the same phase, or three sockets on different phases, because they are independent. Same with the kiln - the three line terminals can be connected to the same phase or different phases, so long as each terminal is 230V from neutral.

This is a different situation to (e.g.) a 3-phase motor, which is not like three independent single-phase motors because the magnetic fields of the three windings interact with each other. The windings must be connected to alternating voltages that vary in time with the correct phase relationship to each other. You cannot simply connect all three to one phase.

Likewise, a different situation obtains in a heater consisting of three identical elements connected in star (i.e. each between a line terminal and a star point) but where the star point is not connected to the supply neutral. This is a handy configuration for industrial supplies where a neutral might not be available When connected to a 3-phase 400V supply, the three elements form a balanced load in which each element receives 230V between the line terminal and the star point. Through symmetry, the star point will lie at a voltage close to that of the supply neutral, hence the element currents and voltages would be just like those of the klln (although note that the 230V control ccircuit of the kiln absolutely requires the neutral to be present). But clearly, if the star point of our heater is not connected to neutral, wiring all three line terminals to the same phase won't work because now there is no complete circuit. Like the motor, this needs a 3-phase supply.

Returning to the kiln and considering the currents in the circuit conductors; on 3-phase the currents from the three elements sum to zero at the neutral terminal (except for the small control circuit load) unless one element fails O/C in which case the neutral current will equal the line current. On single phase the line and neutral currents are inherently equal. The line and neutral currents can be taken as equal both on single-phase and 3-phase supplies, with the current on single phase supply three times that on 3-phase.

What I hope to have shown in the above explanation is that the kiln elements, by virtue of being connected L-N, are less like a 3-phase load and more like three single-phase ones, for which ordinary single-phase calculations can be applied.

You can supply your own sketches to accompany the above

Oh, BTW


R=V/P ???
For this, you are to beat yourself with a big lump of SWA until it really hurts.

Interesting point about connecting as star with no neutral.