Fredrick Paget on sat 11 jan 03
W=EI. That is: watts equals the voltage, in this case the voltage drop in
volts, times the current in amps.
So if there were 3 volts drop from the meter to the kiln and it draws 30
amps there is 90 watts lost in the wiring causing it to heat up a little
(spread out over a number of feet of wire).
Fred
>I know that there is a formula for figuring what a voltage drop to the kiln
>"costs" in wattage loss. Something divided by something....The formula is
>in a notebook in my studio at the Farm and I cannot get to it..SNOW!
>
>Would someone post it, so that people with firing problems can check it
>out..?
>
>Lili Krakowski
From Fred Paget, Marin County, California, USA
Lily Krakowski on sat 11 jan 03
I know that there is a formula for figuring what a voltage drop to the kiln
"costs" in wattage loss. Something divided by something....The formula is
in a notebook in my studio at the Farm and I cannot get to it..SNOW!
Would someone post it, so that people with firing problems can check it
out..?
Lili Krakowski
P.O. Box #1
Constableville, N.Y.
(315) 942-5916/ 397-2389
Be of good courage....
Mike on sat 11 jan 03
Lili,
The American Electricians Handbook has an 18 plus page discussion on voltage
drop. There are many factors to consider depending on how "expert" you want
to be.
That said: Here is a basic formula that can get you close.
VD = 2R x L x I
where VD is voltage drop in volts
R is the resistance of wire in ohms per foot, for this you need to know the
type and size of wire and must have a chart to reference, the NEC (National
electric Code) has the chart or Table.
L is the one way length of circuit in feet, and
I is the current in amperes
There are other formulas, but the easiest way (with Web access) is to go to
http://electrician.com/
Roll Over on Calculators and then click on voltage drop calculator and enter
the needed info, no need for the chart here.
"It is 10 degrees here, snow covered and I am putting off my morning jog as
long as possible"
Mike
>I know that there is a formula for figuring what a voltage drop to the kiln
>"costs" in wattage loss. Something divided by something....The formula is
>in a notebook in my studio at the Farm and I cannot get to it..SNOW!
>Would someone post it, so that people with firing problems can check it
>out..?
>Lili Krakowski
Jim Kasper on sat 11 jan 03
Hi Lili,
THe simple answer is the cost in Watts will be :
Voltage drop times the amps your kiln is drawing.
IF your kiln is on High, and it draws 48 Amps:
1 volt drop = 48 watts
2 volts 96 watts
...
You can check this experimentally by reading the voltage at the panel and then read the voltage at your kiln when it is running.
( YOu probably don't want to do this.)
As Mark mentioned, you can calculate the resistance of a given length of wire from a chart:
http://electrician.com/calculators/vd_calculator.html
I checked the chart for a 50 foot run with #6 copper and # 4 copper
#6 had a 2.3 volt drop : 2.3volts x 48 amps = 110 Watts
#4 had a 1.3 volt drop 1.4 volts x 48 amps = 67 Watts
At 100 feet :
#6 had a 2.3 volt drop : 4.6 volts x 48 amps = 220 Watts
#4 had a 1.3 volt drop 1.4 volts x 48 amps = 134 Watts
Below is the ramblings of someone who read the wrong question the first time through:)
In the following formulas V= Volts, R is resistance in Ohms,
I is ampereage in Amps, and P is Power in Watts.
A simple analogy of how electrity travels though wires is
to compare it to water through a pipe.
Voltage would be the pressure, Amperage would be the flow rate, and resistance
would be how constricted the pipe was.
THe higher the resistance, the less current will flow for a given voltage.
Voltage = Amps x Resistance (V = I x R )
Amps = Voltage/resistance (I = V/R )
Resistance = Voltage/Amps (R= V/I)
Watts = Amps x voltage (P=E X x I )
Power = the square of the amperage times resistance
(P=I^2 x R )
I have not done measurements on my kilns yet.
I can pretty much guess though. My kiln is rated at 48 Amps 240 volts, and there are 12 elements.
This means that 48/12 each element will use 4 amps at full load.
Now , since the kiln runs at 240 volts: 240 volts / 4 amps = 60 Ohms of resistance per element.
Okay, i'm done now, but i had fun remembering at the expense of those who read this far :)
Regards,
Jim Kasper
Zafka Studios
Jensen Beach, FL.
http://zafka.com
Roger Korn on sat 11 jan 03
Hi Lili,
Power is proportional to the square of voltage. Let's say that instead
of 240 volts, your kiln line voltage only measures 230 volts when the
kiln is fully on. The loss in power, in percent, would be 100 x (240^2 -
230^2) / (240^2), or
100 * 4700/57600 = 8.16%. Here's a table for 240 volt kilns:
Actual Line Voltage 240 238 236 234 232
230 228 226 224 222 220
Power loss (percent) 0.00 1.66 3.31 4.94 6.56
8.16 9.75 11.33 12.89 14.44 15.97
In terms of peak temperature reached, the temperature (in degrees above
absolute zero = -459 F or 273 C) is approximately proportional to the
cube root of the power (assuming that heat loss is due entirely to
radiation). Think: "Eight times (2x2x2) the power doubles the
temperature." So if the peak temperature is 2421 F (= 2421 + 459, or
2880 Absolute) when the line voltage is 240 V, the peak temperatures and
cones for reduced line voltages would be:
Actual Line Voltage 240 238
236 234 232 230 228 226 224 222 220
Aprox. Peak Temperature (degrees F) 2421 2373 2326 2279
2232 2186 2140 2095 2050 2005 1961
Approx. Cone (300 C/Hr heating rate, 10 8 7
6 5 3 1 02 03 04
05
rounded to nearest cone)
Pretty dramatic! 2 volts of line drop means dropping from cone 10 to
cone 8. The moral probably is: check your line voltage with the kiln on
'Hi' befor assuming the elements are shot.
Hope this helps,
Roger
Lily Krakowski wrote:
> I know that there is a formula for figuring what a voltage drop to the
> kiln
> "costs" in wattage loss. Something divided by something....The
> formula is
> in a notebook in my studio at the Farm and I cannot get to it..SNOW!
>
> Would someone post it, so that people with firing problems can check it
> out..?
>
--
Roger Korn
McKay Creek Ceramics
In AZ: PO Box 463
4215 Culpepper Ranch Rd
Rimrock, AZ 86335
928-567-5699
In OR: PO Box 436
31330 NW Pacific Ave.
North Plains, OR 97133
503-647-5464 <-
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