ELEMENT DESIGN


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Copyright David Coggins BPQ KILNS 1999 - 2003


 

The design of kiln elements is very complex - many factors must be taken into account. Not all original kiln manufacturers' elements are designed to best specifications - sometimes, cost factors and convenience outweigh efficiency in element designs. Elements can be modified to give more heat if your supply voltage is low, or more power if you need to get to a higher temperature, or just to give a longer life than the original manufacturer's design.

 

Data and formulae to allow you to carry out element designs are given at the end of this article.

 

The element designer usually follows these steps.

 

  1. The power requirement of the kiln is decided. This is dictated by the volume of the kiln, the kiln construction material and thickness, the maximum temperature required, the speed of heating required and the mass of material being heated. There are multitudes of tables designed for this purpose but there are fairly well established precedents, and the power of elements in kilns manufactured by reputable makers can be used as examples. For example, a 4 cubic foot kiln for firing pottery to cone 10 with 4.5" thick K23 brick walls usually requires 7.2 Kilowatts, or 240 volt 30 amps single phase power.

 

  1. The next step is to decide how many elements will be used - this is dictated by the space available for grooves in the walls/door. Elements may be installed on a single level (once around the kiln) or multi-level (2,3 or more times around the kiln). Elements can be straight or hairpin (i.e. doubling back on another level) or even more complex shapes.

 

  1. From steps 1 and 2 above, the designer then decides how the elements will be connected together. This is the step that may need to be repeated many times to get a suitable result, depending on the results of the next steps. There are many options available - parallel, series, or series/parallel connections - one, two or three phase - if 3 phase, whether delta or star connection will be used.

 

  1. The designer now calculates the resistance of each element using the data from steps 1,2 and 3. At this time the power dissipated in each element is calculated. This is essential for later calculations.

 

  1. The gauge of the wire can now be selected. Previous experience is of great assistance at this point. Having selected a possible wire gauge, the required length of wire can be calculated from the element tables.

 

  1. Now the designer can calculate the most important result of all, which is power dissipation of the element in "watts per square inch" or "watts per square centimetre". All heat generated by electric current through a length of resistance wire must be dissipated into the surrounding air as soon as it is produced, or else the wire will keep getting hotter and hotter until it melts. Each type of wire is supplied with a dissipation table, which allows calculation of this important figure for the selected gauge. The power dissipation should not exceed a specified limit for the maximum required temperature in the kiln, as exceeding the limit would drastically shorten the life of the elements. The power dissipation calculated from the previous information is compared with the manufacturer's table, considering the maximum required kiln temperature. If the result is within the specified range, we can proceed to the next step. If the value is too high, we must select a thicker gauge of wire, and recalculate. If the dissipation is too low, we are wasting wire and money, and we can select a thinner gauge.

 

  1. When a satisfactory result is obtained from steps 5 and 6, which meets the dissipation requirements, the designer can then calculate the number of close wound turns of wire on the selected winding former. The length of the unstretched close wound element is then calculated.

 

  1. From steps 3 & 4 the designer knows the length of the groove for each element, which is the length that the element must be stretched to fit in the groove allocated in the kiln. This length must be at least twice the length of the unstretched element; so that when the element is stretched out to the required length there is a space between the coils at least equal to the wire diameter. Heat is dissipated from the inside of the coil as well as the outside, and unless there is a space between the coils to let the heat escape, the temperature inside the coil will build up and destroy the element. If there is less than one wire diameter between the coils of the element after it is stretched to length, the designer needs to return to step 2 and start all over again.

 

  1. If the stretched length of the element is acceptable, the designer then writes out the specifications of the element. This could include the gauge and type of wire, number of elements, resistance of each element, total length of wire, dissipation factor, former diameter, number of turns, unstretched length, stretched length, length and type of element tails and shape of stretched element. Also a calculation can be made of the cost of each element. Element tails are usually doubled and twisted to reduce the heat generated inside the walls of the kiln where the element passes through.

 

  1.  A prototype element is manufactured to the design specifications, and stretched to length. The element tails are formed, the element is stretched and formed, and the resistance and other details are checked against the requirements. If the prototype is acceptable, the remainder of the element batch are manufactured. If not, the specifications are modified accordingly. Each batch of element wire must be checked because there may be slight variations in resistance between batches, which can affect the results.

 

    

 

 

TABLE 1. DATA FOR KANTHAL A1 WIRE

 

GAUGE

B&S

DIAM

IN.

DIAM

 mm.

OHMS

/FT

IN2/FT

FT/LB

8

0.129

3.26

0.053

4.84

25.0

9

0.144

2.91

0.067

4.31

31.6

10

0.102

2.59

0.084

3.84

39.9

11

0.091

2.30

0.106

3.42

50.3

12

0.081

2.05

0.133

3.05

63.4

13

0.072

1.83

0.168

2.71

79.0

14

0.064

1.63

0.212

2.42

101.0

15

0.057

1.45

0.268

2.15

127.1

16

0.051

1.29

0.338

1.92

160.3

17

0.045

1.15

0.425

1.71

202.0

18

0.040

1.02

0.545

1.51

259.1

 

 

TABLE 2. INCHES PER TURN

 

(approximate only - depends on wire gauge)

 

FORMER SIZE

IN PER TURN

3/16"  4mm

0.8

¼"   6mm

1.0

3/8"  8mm

1.2

7/16" 10mm

1.5

½"  12.5mm

1.8

5/8"  16mm

2.5

¾"  19mm

3.0

1"   25mm

3.25

1 1/8"  27.5mm

3.75

 

 

TABLE 3. TURNS PER INCH

 

(approximate only - depends on tightness of winding)

 

WIRE GAUGE

B&S

TURNS PER INCH

18

24

17

22

16

20

15

18

14

15

13

13

12

12

11

10

10

9

9

8

8

7

 

CALCULATIONS

 

1.        From available information, we know the element resistance = R

 

2.      Calculate the power dissipated in each element = WATTS PER ELEMENT

 

3.      Select a possible wire gauge.

 

4.      Using Table 1, calculate the required length of wire

       LENGTH = R ¸ OHMS/FT

 

5.        Using Table 1, calculate the watts per foot

WATTS PER FOOT = WATTS PER ELEMENT ¸ LENGTH

 

6.        Using Table 1, calculate the dissipation in watts per square inch

WATTS PER IN2 = WATTS PER FOOT ¸ IN2 PER FOOT

 

7.        Dissipation should be within the range of 8 to 12 watts per square inch for kiln  

      temperatures up to 1280°C. Values up to 15 watts per square inch may be

      acceptable in kilns which will not exceed 1100°C. If the dissipation is too high or

      too low, recalculate with a different wire gauge.

 

8.        Select the winding former size.

 

9.        Calculate the number of turns from Table 2

      TURNS = LENGTH X 12 ¸ IN PER TURN

 

10.  Calculate the winding length from Table 3

WINDING LENGTH = TURNS ¸ TURNS PER INCH

 

11.  Check that the element groove length (from the kiln design) is greater than twice the winding length, so that the gap between the turns is greater than the wire diameter. If not, recalculation may be necessary.

 

12.  The weight of wire required for the kiln can now be calculated from Table 1

WEIGHT (in lbs.) = LENGTH ¸ FT/LB x number of elements

 

WEIGHT (in Kg) = WEIGHT in pounds ¸ 2.2

 

 


If you want to contact me, please send an

Copyright David Coggins BPQ KILNS 1999 - 2003