3 phase kiln questions?

[HappyHippyDad]

A customer has asked me to install a supply for a Wenger Webcot kiln.

The faceplate is very worn and I cannot see the kW. They state the kiln it type 3670w and is rated at 12kW.

The paperwork does have 3670w on it but when I look at the spec sheet (see picture 2, SpecSheet) ) there is no sign of a kiln which is 12kW. This model has 12 elements therefore it must be one of the last 6 models on the spec sheet, but again no sign of 3670w. It looks like there is one called 3060w though but this is not 12kW. Does anyone have any ideas how I could find the kW, or does anyone recognise the kiln details and know the kW?

The wiring diagram (see picture 4) shows that it can be wired as single phase or 3 phase. This will be wired as single phase. My questions...

1. Do I just join the 3 phases together as per the diagram?
2. Do I need to increase cable size compared to the existing flex on the Kiln as I am joining 3 x 10mm cables into one conductor? Although the neutral carries the same current and there is only 1 neutral conductor! Why is this?
3. On the spec sheet (pic 2) it has 2 columns for electricity supply required. One for single phase and one for 3 phase. The last 4 columns for single phase have a dot. looking below it says treat as nominal rating 14kW for single phase, yet the actual column labelled nominal electrical .loading states different, higher values, why is this?

Pictures below show the almost unreadable faceplate, spec sheet, cable and part of the wiring diagram.

Picture 1, faceplate.


Picture 2, Spec sheet.


Picture 3, Cable


Picture 4

 

[HappyHippyDad]

Would it possible to do an example now? Its quite interesting!

If we assume the kiln is 12kW.....

Still cant see how its as simple as v = I x R though as V could be 230 or 400, giving us varying results of R if we assume 12kW.

 

[loz2754]

Would it possible to do an example now? Its quite interesting!

If we assume the kiln is 12kW.....

Still cant see how its as simple as v = I x R though as V could be 230 or 400, giving us varying results of R if we assume 12kW.

V=230 as the elements are wired line to neutral.

 

[HappyHippyDad]

I need numbers!

V = 230

I = 12000

R = ? single element. (Is this a reading literally of one element, if so what will the equation look like?)
R = ? group reading Ph to Ph (what is a group reading? L1 + L2 + L3 to N?)

 

[Darkwood]

V=230 as the elements are wired line to neutral.

Be careful we have a duel voltage system here, we are discussing testing the 400v approach if testing multiple elements at once, he is using 3ph here so lets not confuse matters.

 

[loz2754]

I need numbers!

V = 230

I = 12000

R = ? single element. (Is this a reading literally of one element, if so what will the equation look like?)
R = ? group reading Ph to Ph (what is a group reading? L1 + L2 + L3 to N?)

I is not 12000

 

[pc1966]

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).

 

[loz2754]

Be careful we have a duel voltage system here, we are discussing testing the 400v approach if testing multiple elements at once, he is using 3ph here so lets not confuse matters.

Thanks. I understand that. But the aim is to connect to the available single phase supply. I had that in mind.
So we have a series/parallel resistance to calculate, 3 parallel groups of 4 series resistances.

 

[Darkwood]

Group reading

L1 to L2
L1 to L3
L2 to L3

Each reading should be more or less the same.


I = V/R
Amps = 400/resistance.



L - N of 1phase

Amps= 230/resistance of element group

multiply by 3 banks to get single phase load.

 

[pc1966]

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

 

[HappyHippyDad]

I is not 12000

Of course not! I'm rushing and coffee overloading. I = 52.2A

 

[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?..