240v kiln

Status
Not open for further replies.

ddkeith

Member
i have a kiln thhat operates on 240v 30amp 7200w single phase at a max temperture of 2300 degrees.the equipment isnt dual rated but is operating on 208v. my question is at 208v will the heating element ever reach the desired temperture? i know the best thing to do is buy a kiln that operates on 208v but due to government red tape i dont have that option. thanks in advance
keith
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: 240v kiln

The temperature that the unit will reach will depend on three things:
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">The heat input from the heating elements.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">The heat loss to the surrounding area.</font>
<font size="2" face="Verdana, Helvetica, sans-serif"></font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">The temperature that you set into the unit?s control system.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">If you operate the unit at 208 volts, instead of 240 volts, then the first item on this list is a significantly lower value (about 75 %). Therefore, it will take more time to get up to temperature.

But will it ever reach the desired temperature? That depends more on the second item on the list. If the unit loses much of its heat to the surrounding room, then the heat input might not overcome the heat loss, and you might not get it up to temperature.

The amount of time the unit needs to reach temperature may be a critical factor. That depends on the requirements of whatever pottery (or other stuff) you intend to cook in this unit.
 

physis

Senior Member
Re: 240v kiln

I'm just curious.

208 is 86⅔% of 240.

Because of that, the operating temperature, and therefore the resistance, will be lower.

Lower resistance means higher current.

So I would expect the power to come out a little above the 86⅔% as opposed to the 25% less generalization that you guys are posting.

:confused:
 

tshea

Senior Member
Location
Wisconsin
Re: 240v kiln

The real constant is the resistance of the heating elements. It doesn't change significantly.

It will take much longer for the kiln to get to 2300 degrees.

Your options are:
1. Live with it.
2. Transformer 208-240 about 7.5kva
3. Buck boost transformer about 1kva SqD 1S46F
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: 240v kiln

First, power varies with the square of the voltage (V^2 / R). So I came up with 75% by taking (208 / 240), and squaring it. Peter's numbers match this method, since 5400 / 7200 is also 75%.

Secondly, the operating temperature of the unit does not vary with voltage. The heating elements will stay on until the unit reaches the temperature you set on the control panel. If the heaters are running on lower power, then it will take longer to get to temperature. If the heat loss is too high (in comparison to the heat input), then you might not reach the desired temperature.

But I suspect that you meant operating temperature of the wires, or perhaps of the heating elements. Peter and I have disregarded the effect of temperature on the resistance of these items. It will change the percentages, as you correctly point out. But it is a second order effect. I think the original question had to do with more general concepts.
 

physis

Senior Member
Re: 240v kiln

Actually Charlie, I'm only talking about the element. I neglected to incorperate the square and if I do our math matches pretty well. :)
 

ddkeith

Member
Re: 240v kiln

first of all thanks for all the help. i still dont understand how you guys come up with 5400w can someone inlighten me on this. i have a feeling that ill have to explain this and i would like to be able to back it up. i know that e*i=p but wouldnt amperage change in a drop in voltage?
 

petersonra

Senior Member
Location
Northern illinois
Occupation
engineer
Re: 240v kiln

For practical purposes the resistance of the heating element does not change when you apply a different voltage to it (it does because the element heats to a different final temperature, but it is not usually significant enough to worry about).

P= V^2/R


at 240V P1=240*240/R
at 208V P2=208*208/R

P2/P1=(208/240)^2= 0.75

so the power dissipated (in the form of heat) at 208V is 75% of what it would be at 240V.

<edited to correct spelling errors>

[ May 09, 2005, 12:54 PM: Message edited by: petersonra ]
 

rattus

Senior Member
Re: 240v kiln

This is rough, but if 2300 degrees is the temperature at 100% duty cycle, then the thermal resistance to 100 degrees ambient is 0.3 degrees/Watt.

Then with 5400 Watts, the temp rise would be:

0.3deg/W x 5400W = 1620 degrees.

Add the 100 degrees ambient to obtain 1720 degrees max.

If the unit has a heat control, then this calculation does not apply.
 

john m. caloggero

Senior Member
Re: 240v kiln

Operating a 240V rated heating unit at 208V will reduce the wattage developed down to 75% of rated wattage. .75 x 7200W = 5400W. However, due to the reduced voltage, the heating element might not reach the desired red hot glow that is necessary to arrive at a temperature of 2300 degrees F no matter how long it is energized. I suggest a 1 KVA buck-boost transformer with 208 primary, 32 volt secondary, connected to boost the voltage. 208V + 32V = 240V
 
Status
Not open for further replies.
Top