Delta wired calculations for heating elements in a kiln

[OSLaser]

Hi! I'm working on an electric kiln project in Europe and can't figure out why the B16 circuit breaker breaks the circuit after about 30 seconds of heating the elements. I've wired the elements in delta connection like in diagram below. Please take a look at my calculations and tell me where I went wrong. I have underlined the possible places where I went wrong in my calculations.


- 3 phases
- 380 volts
- 34 ohms resistance wire total (11.3 ohms between each two phases)
- 11.18 amperes of current flow - is this correct or is it 380/11.3= 33.6 amperes?
- 4248watts - or is it 380vX33.6A=12 768watts
- Kanthal wire 1.4mm diameter with resistance 0.94 ohms/meter
- used coiled resistance wire total length 36 meters
- watt loss is around 3 watts/mm2, which is acceptable for this kind of wire to heat up to about 1200celsius.


Thanks for your help!

 

[Besoeker]

Hi! I'm working on an electric kiln project in Europe and can't figure out why the B16 circuit breaker breaks the circuit after about 30 seconds of heating the elements. I've wired the elements in delta connection like in diagram below. Please take a look at my calculations and tell me where I went wrong. I have underlined the possible places where I went wrong in my calculations.
View attachment 9998

- 3 phases
- 380 volts
- 34 ohms resistance wire total (11.3 ohms between each two phases)
- 11.18 amperes of current flow - is this correct or is it 380/11.3= 33.6 amperes?
- 4248watts - or is it 380vX33.6A=12 768watts
- Kanthal wire 1.4mm diameter with resistance 0.94 ohms/meter
- used coiled resistance wire total length 36 meters
- watt loss is around 3 watts/mm2, which is acceptable for this kind of wire to heat up to about 1200celsius.


Thanks for your help!

Are each of the three elements 11.3 ohms?

 

[iceworm]

...can't figure out why the B16 circuit breaker breaks the circuit after about 30 seconds of heating the elements. I've wired the elements in delta connection like in diagram below. Please take a look at my calculations and tell me where I went wrong. I have underlined the possible places where I went wrong in my calculations.
View attachment 9998

- Kanthal wire 1.4mm diameter with resistance 0.94 ohms/meter
- used coiled resistance wire total length 36 meters
- watt loss is around 3 watts/mm2, which is acceptable for this kind of wire to heat up to about 1200celsius.

From this data, I infer: There is 12 meters of heating element wire per each delta leg. Each heating element is 1/3 x .94 x 36 = 11.3 ohms
Which matches with "- 34 ohms resistance wire total (11.3 ohms between each two phases)"
if the above is true, then:

- 11.18 amperes of current flow
No
- is this correct or is it 380/11.3= 33.6 amperes?
No
- 4248watts - or is it 380vX33.6A=12 768watts
and No

Here's my recomendation:
Power for each delta leg = (E^2)/R, where R = ohms per leg
P = (380^2)/11.3 = 12,800 watts

Power total for the three delta legs = 3 x P
Pt = 3 x 12800 = 38,400 watts

Current = (Pt/E)/sqrt(3)
I = (38400/380)/1.732 = 58.3 Amps

Tripping a 16A CB is 30 seconds? That sounds about right.

ice

 

[iceworm]

Hi! I'm working on an electric kiln project in Europe and can't figure out why the B16 circuit breaker breaks the circuit after about 30 seconds of heating the elements. I've wired the elements in delta connection like in diagram below. ...

View attachment 9998

If you connected the elements Wye (probably called "star"), the power would drop to 12.800 watts (1/3 of the delta connection power). The current would still be 19.4 amps. It would still trip a 16A CB.

ice

 

[OSLaser]

Are each of the three elements 11.3 ohms?

Yes it's 11.3 for each element

 

[Besoeker]

Yes it's 11.3 for each element

OK.
Each element dissipates 12.8kW (V²/R)
Total is 38.3kW
The line current is 58.2A
The current in each leg is 33.6A.

 

[OSLaser]

If you connected the elements Wye (probably called "star"), the power would drop to 12.800 watts (1/3 of the delta connection power). The current would still be 19.4 amps. It would still trip a 16A CB.

ice

Thank you very much for the suggestions. I think I'll have the main circuit breaker upgrade to 32A or more.

I tried connecting one leg separately- one phase to neutral

my calculations are as follows for one leg 12 meters of 0.94ohm/m resistance wire:

force (E)= 230volts
resistance (R) = 11.5ohm
Power (P)=(230^2)/11.5ohm=4600watt
Current (I)= here is where I don't really understand. I thought there are several formulas, but I don't understand where 1.732 number comes from? Here are the formulas that I usually use: I=E/R, I=P/E, Sqrt(P/R) Aren't these correct?
If so, then current for this is I=230/11.5= 20Amps, which means it should trip the B16 breaker, but it didn't. should I devide 20Amps/1.732, which is 11.54Amps? Is this correct and why should I devide by 1.732?

 

[OSLaser]

OK.
Each element dissipates 12.8kW (V²/R)
Total is 38.3kW
The line current is 58.2A
The current in each leg is 33.6A.

If the current in each leg is 33.6A, then why does the the breaker doesn't trip immediately, but after about 30 seconds?

 

[Sahib]

If the current in each leg is 33.6A, then why does the the breaker doesn't trip immediately, but after about 30 seconds?

Check with the current-time characteristic curve of the B16 breaker available with the breaker manufacturer. Locate the current value and time in that curve to confirm the tripping time of 30 seconds in your case.
Moreover your breaker is equipped with thermal element to initiate tripping action. Naturally it takes some time for the element to heat up to trip the breaker.

 

[Smart $]

...
If so, then current for this is I=230/11.5= 20Amps, which means it should trip the B16 breaker, but it didn't. ...

20A is only a 1.25 multiple of 16A rating. Look at trip curve for the breaker. Don't forget to account for ambient temperature. May take some time for it to trip.

 

[OSLaser]

Thank you guys for clearing things up! What a great forum this is :thumbsup: I'll be back again. Hopefully I can contribute when I gain some more knowledge and understanding of electricity

 

[iceworm]

Thank you very much for the suggestions. I think I'll have the main circuit breaker upgrade to 32A or more.

I tried connecting one leg separately- one phase to neutral

my calculations are as follows for one leg 12 meters of 0.94ohm/m resistance wire:

force (E)= 230volts
resistance (R) = 11.5ohm
Power (P)=(230^2)/11.5ohm=4600watt

This is good. You connected one element single phase, to 230V. The power is:
P= E^2/R

... Current (I)= here is where I don't really understand. I thought there are several formulas, but I don't understand where 1.732 number comes from? Here are the formulas that I usually use: I=E/R, I=P/E, Sqrt(P/R) Aren't these correct? ...

I don't see what exactly is the purpose of Sqrt(P/R). I suspect this is a typo.

I=E/R, I=P/E
These are both good single phase formulas.

...Aren't these correct?
If so, then current for this is I=230/11.5= 20Amps, which means it should trip the B16 breaker, but it didn't.

Yes, one element connected single phase, line to neutral, will draw about 20A. As already noted by S$, look at the trip curve. It will either eventually trip (but itwill take a long time, maybe hours) - or may not trip. I don't know how long you tested, but not tripping is okay.

.... should I devide 20Amps/1.732, which is 11.54Amps? Is this correct and why should I devide by 1.732?

In this case, no. The element is connected single phase.

The 1.732 comes up in three phase calculations. Remember 1.732 = sqrt(3). For three phase power:
P = E x I x 1.732, where E = line-to-line voltage, I = line current.

Couple of things to remember:

E(line to line) = 1.732 x E(line to neutral)
For balanced 3phase, deta connected load, I(line) = I(in each delta leg) x 1.732

Thank you very much for the suggestions. I think I'll have the main circuit breaker upgrade to 32A or more. ...

No. As connected (11.3 ohms per element, connected delta), the load is 38.4 KW. This will draw 58.3A line current. You will need a 70A to 80A CB. (I don't know what the nominal sizes are for IEC CBs)

ice

 

[Besoeker]

No. As connected (11.3 ohms per element, connected delta), the load is 38.4 KW. This will draw 58.3A line current. You will need a 70A to 80A CB. (I don't know what the nominal sizes are for IEC CBs)

ice

63A, 80A, 100A.

 

[Smart $]

63A, 80A, 100A.

Just guessing here, 63A should be fine for a thermostatically-controlled heating application. If not thermostatically-controlled, may have to go to 80A.

Guessing because I also am not familiar with IEC nominal circuit ratings, breaker ratings vs. trip curves, etc.

 

[kwired]

This is good. You connected one element single phase, to 230V. The power is:
P= E^2/R


I don't see what exactly is the purpose of Sqrt(P/R). I suspect this is a typo.

I=E/R, I=P/E
These are both good single phase formulas.


Yes, one element connected single phase, line to neutral, will draw about 20A. As already noted by S$, look at the trip curve. It will either eventually trip (but itwill take a long time, maybe hours) - or may not trip. I don't know how long you tested, but not tripping is okay.


In this case, no. The element is connected single phase.

The 1.732 comes up in three phase calculations. Remember 1.732 = sqrt(3). For three phase power:
P = E x I x 1.732, where E = line-to-line voltage, I = line current.

Couple of things to remember:

E(line to line) = 1.732 x E(line to neutral)
For balanced 3phase, deta connected load, I(line) = I(in each delta leg) x 1.732


No. As connected (11.3 ohms per element, connected delta), the load is 38.4 KW. This will draw 58.3A line current. You will need a 70A to 80A CB. (I don't know what the nominal sizes are for IEC CBs)

ice

I will add a little to that, if you don't fully understand the vector math think of it this way:

Line A current is comprised of some of the single phase current of line A to B, and some of the single phase current from A to C, because of the phase angle it only increases current in A by a factor of 1.73 instead of by 2. If the phase angle were the same then the current would double, if the phase angle were 180 degrees then you have zero current on the common - which is exactly what we have in the neutral conductor with a typical 120/240 single phase supply and equal current on each side of the neutral.

 

[Besoeker]

Just guessing here, 63A should be fine for a thermostatically-controlled heating application. If not thermostatically-controlled, may have to go to 80A.

Guessing because I also am not familiar with IEC nominal circuit ratings, breaker ratings vs. trip curves, etc.

I don't know what the temperature coefficient of the resistance element is or the time constant for it to get to steady state conditions.
That could have a significant bearing on selecting the overload protection.

 

[iceworm]

I don't know what the temperature coefficient of the resistance element is or the time constant for it to get to steady state conditions.
That could have a significant bearing on selecting the overload protection.

Also guessing - but likely close:
.7 x 10^-3/degC
A few seconds for the element to reach SS temp
(You didn't ask, but) about 700C

Given that - what would be your selection?

ice

 

[Besoeker]

Also guessing - but likely close:
.7 x 10^-3/degC
A few seconds for the element to reach SS temp
(You didn't ask, but) about 700C

Given that - what would be your selection?

ice

63A C curve.

BTW the OP mentions 1200C.

 

[iceworm]

63A C curve.

BTW the OP mentions 1200C.

Yes I saw that.
But Red hot is 700 - 800C,
White hot is 1100C,
This stuff boils about 1400C

But, still maybe it is running at 1200C

However, if your selection works for 700C, it should work for 1200C

Zero importance here, just idle curiousity on my part about selecting an IEC CB

ice