Suppose a coil is to be designed to fit a ceramic hot plate spiral:

We know that:

1. We want an 800-watt element

2. The groove is 30" long and J" wide.

3. The element is to operate on 115 volts.

4. To insure the best possible life, we will use NICHROME 80.

To find total cold coil resistance: Table I shows that for NICHROME 80 we need 15.44 ohms in

the coil and the recommended wire gauges are #18 to #22.

To select coil diameter and wire gauge:

Selecting heavier wire will lower the element temperature, but if the voltage stays the same the current will increase and the watts output will increase. There is no ultimate way to determine the temperature so usually you have to experiment to determine the element temperature.

For the present element let us select #20 gauge wire. Looking in Table IIa we find the resistance per inch of close would coil for 7/32" dia. (thus allowing 1/32" clearance in the 1 inch groove) #20 gauge. This resistance is .967 ohms per inch.

Length of coil and stretch factor:

We now know that we need a total coil resistance of 15.44 ohms and a resistance per inch of .967 ohms thus 15.44/.967 = 16" of closed coil. This would mean a stretch factor of 30/15 or 2:1. A stretched coil should be 1.5 to 4 times its closed length for good design.

Other calculations on this coil might be:

Estimated wire temperature: Can be found in Table V.

We are using #20 gauge wire and can calculate that the current will be:

I = W = 800 = 6.95 amps.

E = 115

Looking across the table we find the coil temperature in air will be about 1300 F. When surrounded by the ceramic this temperature would probably be about 200 F. higher or about 1500 F.

Watts per square inch of element surface:

One method used to test the soundness of an element design is a check on the power density per square inch of radiant surface. This is done in the following manner:

Our element is to be constructed of #20 gauge wire (.032" dia.) with a total resistance of 15.44 ohms. From Table V we find that .032" NICHROME 80 has a resistance of .6347 ohms per foot. Thus the total length of wire = (Total ohms) / (Ohms per foot of Wire).

L = 15.44 / .6347 = 24.4 feet of wire

The surface area of the element is:

Surface area = Pi*(wire dia.)(length of wire in inches)

or = (3.14)*(.032) *(24.4)*(12)

= 29.4 square inches

The power density is thus = Watts / Area = 800 / 29.4 = 27.2 watts per square inch

For safe design the watts density on an open helical coil should not exceed 35 watts per square inch when the element is operated in still air. Naturally, if the coil is to be used to heat a rapidly moving air stream, the power density may be safely increased. Conversely, if the coil is to be imbedded in a refractory material, the power density must be decreased to prevent overheating of the element.

Number of coils per pound of wire:

Using #20 gauge wire with 15.44 ohms per coil, from Table V, we find that there are .6347 ohms per foot of wire, thus

Ohms per Pound = Ohms/FT X 1000 = .6347 X 1000 = 216.4 LBS. per 1000 FT.

1 pound of wire = 216.4 / 15.44 = 14 coils

RIBBON ELEMENT DESIGN

Suppose a ribbon wound mica element is to be designed for a single piece toaster.

We know that:

1. We want 500 watts.

2. The element will operate on 115 volts.

3. The mica sheet has space for 10 feet of ribbon. Again we will use NICHROME 80.

To find the cold ribbon resistance:

Table I shows that we need 24.704 ohms using Nichrome 80.

We want 10 ft. of ribbon. This is a resistance of 24.704 / 10 = 2.470.

To estimate ribbon size:

From the ribbon resistance for NICHROME 80 in Table IV we find that 1/16 X .004 ribbon has a resistance of 2.458 ohms per foot which should work very well and wrap easily on the mica form.

To make an approximate check on the soundness of our element design we should calculate the watts density per square inch of element surface.

Our ribbon is 120 inches long, 1/16 inch wide and .004" thick. Therefore, the total surface area is:

120 X .062 X 2 = 15.0

120 X .004 X 2 = .96

Total surface area = 15.96 square inches

The watts density is thus: 500 / 15.96 = 31.4 watts per square inch.

For approximate design calculations the following table gives the range of watts per square inch for various types of elements:

Element Type Range of Watts Per Square Inch

Furnace Strip 6 to 20

Toaster Strip 20 to 40

FlatIrons(Ribbon Elements) 30 to 60

It should be pointed out that the matter of watts per square inch is a factor which varies considerably due to design characteristics of the unit. In general, the better a means is provided for heat conduction from the element the higher is the safe limit of power density. This is especially true for electric irons where, due to close contact, the sole plate can be considered the radiating surface.