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## kiln wattage

### mo cain on wed 26 apr 00

------------------
Eventually we will understand this I hope. Our electric kiln is supposed to =
go
to cone 8 and doesn't. It barely reaches cone 4. The face plate indicates =
that
at 208 volts it should pull 49 amps for a wattage of 10.192 watts. In the =
real
world under load the voltage is 203 volts and the amperage pulled is 42 amps=
(
using one of those intruments that encircles the supply line) This equates =
to
8526 watts. I deduce that something has to be done to the elements to get to=
the
specified wattage. Do we have to inrease or decrease the resistance of the
elements and how do we do the math? The top two elements have a resistance =
of
13.4 ohms as do the bottom two while the middle two have a resistance of =
14.8
ohms

I hope one of you folks can solve this mystery for us
.
mocain xPat Brit ATLGA

### GSM_ENT on thu 27 apr 00

Hi!

IMPORTANT FORMULAS:

Watts=Amps x Amps x Ohms

also
Watts=Volts x Amps

and
Watts=Volts x Volts / Ohms

Ohms=Volts/Amps

Amps=Volts/Ohms

Volts=Amps x Ohms

The above Ohms Law and its derivate will assist you in determining the
efficiency of any kiln. The Amperage listed on the data plate of the kilns
is NOT the actual. After the kiln has gone into high the Amps will raise.
That is why the fuze rating recommended by the manufacturer varies with the
kiln model.

A kiln rated 45 Amps could either require a 50 or a 60 Amps breaker of fuse.
The kiln Operator's Manual will indicate it-normaly there is a chart with
the models and the elecrical requirements for each such as the wiring gauge
or size required (always copper) which needs to be increased for every 40
feet of length between the kiln and the fuse box, the kiln model Amps
rating, the required fuse Amperage, the NEMA configuration, and most also
provide you with several brands and part number for the required receptacle.

Each manufacturer kiln model will require different element resistance
(Ohms). what brand and model kiln do you have? If you let me know I may be
able to give you the factory readings.

Effective firing temperature decreases with use and abuse of the kiln. Any
contaminants in the groove where the elements rest also decreases and most
times cause the element to break. The type of firing done also has a great
bearing. Reduction firing on electric kilns tremendously reduces the half
live (number of good firings) of the elements by 50% or better.

Looking at the formulas above we see that the decreased Voltage also has
direct impact on the top temperature achievable, at a reasonable length of
time.

In you case:
Watts=203Volts x 42Amps or 8,525Watts if you /10,192 = 83.6%
efficient

If
Watts=208Volts x 42Amps or 8,736Watts if you /10,192 = 85.7%
efficient

or
WAtts=203Volts x 49Amps or 9,947Watts if you /10,192 = 97.6%
efficient

Note that the 5Volts only increases 2.1% in efficiency but on the other hand
a 7 units increase in the Amperage increases the efficiency 14%.

Since Amps are inversely affected by resistance (Ohms) we see that the lower
the resistance, the higher the Amperage. As elements wear out their
resistance (Ohms) increases therefore decreasing the Amperage and thus the
maximum possible Wattage of the kiln.

The answer is obvious - your kiln elements are worn out and need to be
replaced. The manufacturer of the kiln can tell you what is the factory
resistance for your kiln top/bottom and middle elements. The guideline is
that if the elements are 10% or more of the factory reading they need to be
replaced. For example: if the factory reading is 10 Ohms and yours read 11.8
Ohms they must be replaced because the original 10 Ohms plus the 10% (1 Ohm)
equals 11 Ohms as maximum allowance.

Manuel R A "Tony" Diaz Rodriguez
MAJ., US Army (Ret.)
Master Kiln Repair Technician/Senior Educator
Multi-Factory Qualified
Multi-Company Product Qualified Teacher

----- Original Message -----
From: mo cain
To:
Sent: Wednesday, April 26, 2000 10:00 AM
Subject: Kiln Wattage

----------------------------Original message----------------------------
------------------
Eventually we will understand this I hope. Our electric kiln is supposed to
go
to cone 8 and doesn't. It barely reaches cone 4. The face plate indicates
that
at 208 volts it should pull 49 amps for a wattage of 10.192 watts. In the
real
world under load the voltage is 203 volts and the amperage pulled is 42 amps
(
using one of those intruments that encircles the supply line) This equates
to
8526 watts. I deduce that something has to be done to the elements to get to
the
specified wattage. Do we have to inrease or decrease the resistance of the
elements and how do we do the math? The top two elements have a resistance
of
13.4 ohms as do the bottom two while the middle two have a resistance of
14.8
ohms

I hope one of you folks can solve this mystery for us
.
mocain xPat Brit ATLGA

### GSM_ENT on thu 27 apr 00

----- Original Message -----
From: mo cain
To:
Sent: Wednesday, April 26, 2000 10:00 AM
Subject: Kiln Wattage

----------------------------Original message----------------------------
------------------
Eventually we will understand this I hope. Our electric kiln is supposed to
go
to cone 8 and doesn't. It barely reaches cone 4. The face plate indicates
that
at 208 volts it should pull 49 amps for a wattage of 10.192 watts. In the
real
world under load the voltage is 203 volts and the amperage pulled is 42 amps
(
using one of those intruments that encircles the supply line) This equates
to
8526 watts. I deduce that something has to be done to the elements to get to
the
specified wattage. Do we have to inrease or decrease the resistance of the
elements and how do we do the math? The top two elements have a resistance
of
13.4 ohms as do the bottom two while the middle two have a resistance of
14.8
ohms

I hope one of you folks can solve this mystery for us
.
mocain xPat Brit ATLGA

### Louis H.. Katz on thu 27 apr 00

Hi Mo,
Watts(W) = Volts(E) X Amps(I) X a Power Factor which is 1 for electric kilns so
can ignore the Power Factor in this equation.
To raise the Watts you have to raise the Volts or Amps.

In any circuit Volts (E) = Amps (I) X Resistance (R)
Stated another way Amps = Volts/Resistance or I=E/R
Since your voltage is given to you by the electric company, the easiest way to
raise the amps is to lower the resistance.

Ten feet of wire has twice the resitance of five feet. So if you want to decreas
the resistance by ten percent, decrease the length by ten percent. But, if you
this your element life will decrease. It is probably a better bet to use a sligh
thicker gauge of wire.

If your elements are old they have gotten thinner from oxidation. They will have
more resistance, reducing the amps, reducing the wattage. Maybe you just need ne
elements?

Call the manufacturer and ask them what the element resistance should be, or cal
Euclids. They make replacement and custom elements.

By the way your kiln uses approximately ten KILOwatts not ten watts.

Louis
--

Louis Katz
NCECA Director of Electronic Communication and Webmaster(Ad-Hoc)
Texas A&M-CC Division of Visual and Performing Arts
Visit the NCECA World Ceramics Image Database Online
Looking for a school or a class? Visit NCECA Ceramics Educational Programs Datab
Online

### chris@euclids.com on thu 27 apr 00

>Do we have to inrease or decrease the resistance of the
>elements and how do we do the math?
Hi,
It depends on how the elements are wired. In a parallel connection, to
increase the power of an element, you would decrease the resistance. In a
series connection, a change to one element effects the output of the others.
Alot of care must be taken when altering elements in a series connection.
If you give us a call, one of our technicians can go over it with you & do
the math for you. It would be better to know the make, model, size, ect...
chris

chris@euclids.com
www.euclids.com
800-296-5456

### WHew536674@cs.com on thu 27 apr 00

As you have indicated the wattage is calculated by the formula P (Power in
Watts) = I (current in Amps) X E (voltage in Volts). Increasing E or I will
increase Power. By Ohm's law E = I X R (Resistance in Ohms). If R is
decreased, I must increase to maintain the equality of the expression. This
increase when translated to the power formula produces increased power.
Now that all that is said the values reported indicate a possible wiring
problem. The resistance of the elements is constant so a 2% decrease in the
applied voltage should translate to a 2% decrease in the current to maintain
the equality of the expression E = IR. You report a 14% reduction in current
flow. It may be that an element is not connected or a switch is bad.

The reported values for voltage and current imply a resistance of 4.8 ohms
but the resistance value calculated from the measurement of each element
produces a value of 3.51 ohms through the formula for parallel resistances:

R total = ____________1______________

_1_ + _1_ + _1_ + _1_
13.4 13.4 14.8 14.8

If you perform the calculation with only three elements the answer is closer

Bill Hewlett for Joyce Hewlett
whew536674@cs.com

### Michael on thu 27 apr 00

I have had to adjust my kiln also for line voltage variations...

watts are volts times amps (to get kilowatts they are divided by 1000)

or amps squared divided by resistance

lower element resistance = higher wattage

lower voltage = lower wattage

age=increased resistance of elements=lower wattage

some steps:

1. Find out from the manufacturer the limits on the element resistance

they might recommend changing elements

2. measure your voltage and current under full load at top temperature for worst

3. monitor your line voltage - varies with time of day, other loads etc

4. Check the refractory insulation and drafts around the kiln area for excessive

If the amperage draw is below the manufacturers specification you can change to

a transformer called a buck/boost autotransformer can be installed in the line t

My kiln is 1000 ft from the mains supply and a long underground feed to my isola

I also added additional insulation to the top and sides of the kiln, some are no

Michael

At 01:00 PM 04/26/2000 EDT, you wrote:

>----------------------------Original message----------------------------

>------------------

>Eventually we will understand this I hope. Our electric kiln is supposed to go

>to cone 8 and doesn't. It barely reaches cone 4. The face plate indicates that

>at 208 volts it should pull 49 amps for a wattage of 10.192 watts. In the real

>world under load the voltage is 203 volts and the amperage pulled is 42 amps (

>using one of those intruments that encircles the supply line) This equates to

>8526 watts. I deduce that something has to be done to the elements to get to th

>specified wattage. Do we have to inrease or decrease the resistance of the

>elements and how do we do the math? The top two elements have a resistance of

>13.4 ohms as do the bottom two while the middle two have a resistance of 14.8

>ohms

>

>I hope one of you folks can solve this mystery for us

>.

>mocain xPat Brit ATLGA

>

..........................

Michael Clark

FAX 804-985-4985 Phone 804-985-3570

<<
email@planetearthdiversified.com>

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