By GARY C. SCHWARTZ, Sales and Product Manager and RICHARD 1. HEXEMER, Senior Application Engineer Cesiwid Inc., Globar Division Niagara Falls, NY 14302-0339

A

n easy approach to designing the proper size of the heating elements (Globar) and power requirements for a furnace is readily conducted in a series of steps based on simple calculations and known relationships. Before describing the use of these relationships, Table I, it is logical to briefly cover some of the operating characteristics and advantages of these elements that have made them commonly applicable for a myriad of furnace processes (see Table II).

1

ADVANTAGES Manufactured from high purity, high density closely controlled silicon carbide crystals, in a special recrystallization process, these elements (Globar) undergo an extremely slow change in resistance with age, which greatly reduces operational problems. Electric heating with these elements is economical, dependable, clean, quiet and safe. The heating units can be used for a wide range of temperatures to 3000°F and at a relatively high watt density to allow the required power for high temperature and high production work. Life of the units, varying with operating temperature, is very g o o d 4 months to 2 years. They can operate in air and under proper conditions in artificial atmospheres as described in this article.

Metallurgical Industry

Steatites

Electrical Porcelain Refractories

Grinding Wheels Dinnerware Whiteware Pottery Tile Spark Plugs Calcining: powders, ore

Hardening Forging Annealing Deoxidizing Sintering Brazing Melting and Holding

Sa- Ft) (Sa Ft Area of F u m a d

3)

Heat Storage &W)= (PeriodicFunace Only)

4)

Total Power Require~nents=ROCem Heat+Heat Loss+HeatStorage (Mini”)

5)

Select Globar Element Size to Fit PhysicalDimensionsof Furnace

6)

Watts/Element = (Sq In,of Heating Section)(WattslSqIn.)

7) Quantity of Elements=TotalPower Reauired (In Watts) Wattsmement (From Step 6 ) 8)

Volk/Element = dWatts/Element X ResistanceElement

9)

Volts Total = Volts/Element X Quality of Elements in Series

10) Maximum Amperage/Element = VoltsElement ResistanceElement

X 1.56

11) Amperage Total = AmpdElement X Quality of Parallel Circuits

3 Phase Circuits: 12) Delta circuik-

3 Phase Volts=Single Phase Volts 3 Phase Amps-1.73 X Single Phase Amps

13) Wye Circuits-

3 Phase Volts=1.73 X Single Phase Volts 3 Phase AmpsrSingle Phase Amps

Transformem With Taps

Electronics industry Crystal Growing Diffusion

Glass Industry Melting Holding Refining

14) Voltage (From Step 9) X .7=Lowest Tap Voltage X l.O=Nominal Tap Voltage X Z.O=High Tap

INDUSTRIAL HFATINC~MARCH 1995

69

time of furnace. OPERATING INFLUENCES Electrical Characteristics These elements have a negative resistance characteristic at temperatures to approximately 1200°F (650°C). Above that temperature, the characteristic changes to positive and remains positive throughout the normal operating temperature range as shown in Fig. 1. The solid curve as shown from room temperature to 1200°F (650°C) is an average curve, since traces of impurities have a noticeable effect on the cold resistance. These impurities, however, are not a factor in the higher temperature range. Therefore, nominal resistance (the value used when making calculations) is measured at 1960'F

(1071OC). Watt loading The watt loading for an element is established by dividing the watt input by the square area of its radiating surface. The recommended watt loadings of elements operated at various control temperatures are given in Fig. 2. The upper m e is used to select the number and size of elements for a given maximum input in atmospheres of air and inert atmospheres such as argon and helium. The lower curve applies to reducing atmosphere of hydrogen and nitrogen. Atmospheres Some atmospheres at certain temperatures shorten the element's life. The dew point of the atmosphere and the watt loading on the element also are factors which combine to affect element life. The elements can be operated at up to 2370°F (1300°C) in hydrogen and dissociated ammonia atmospheres. In atmospheres containing nitrogen, silicon nitride will form if a temperature of 2500°F (1370°C) is exceeded. Exothermic gas can be used up to a carbon monoxide content of 18%. Methane

3

OC

-20

200

480

650

870

1090

1315

1540

In a carbonaceous atmosphere, the elements will tend to pick up carbon. This is a slow process, and the resistance drop can be noted by a gradual increase in amperage. The carbon has no detrimental effect on the element. However, it can short-circuit sections of the element, thus increasing the electrical load on the remaining section and reducing the life. The carbon can be burned out by shutting off the atmosphere and introducing air into the furnace chamber at periodic intervals. Excessive moisture, methane, and hydrocarbon vapors should be kept out of the high temperature zone of the furnace where the elements are located. The combination of element watt loading, temperature and atmosphere should be kept in balance at all times and recommended limits should not be exceeded (see Table III). EXAMPLE OF ELECTRIC ELEMENT FURNACE DESIGN An example in designing the heating elements (Globar) and power requirements for a fumace follows in simplified manner, using round numbers and a few approximationsand assumptions. Consider a small heat treating fumace with interior dimensions of 4' in width, 4' high and 5' deep. The M a c e will be used to heat 1500 lb of steel to 2100°F within 2 hours. The insulating firebrick refractory is 9"thick. The elements are going to be placed horizontally along the 2 sidewalls. Thus, the heating element will have a 6 0 effective heating length; this information will be used later in this step by step fumace element design in accord with the formulas in Table I. Step 1: Process Heat The process heat is the amount of power required to heat just the steel during the furnace cycle. Referring to the formula in Table I, plug in the appropriate numbers from the above information as follows:

"F "C

1800 960

2000 1090

2200 1200

2400 1315

2600 1425

2800 1540

3000 1650

Surface Temperature of Heating Element Chamber Temperature

Fig. 1 Typical resistance temperature characteristics of a heating element (Globar) at a standard calibration of 7960°F (1077 "0.

Fig. 2 Recommended watt loading for heating elements

(Globar).

I

--

"F Heat Up/Hour

Continuous furnaces are usually brought up to temperature without a load so the power is used just to heat the refractory on start-up. After the fumace is at equilibrium temperature, HS is no longer a factor and power is used for the process and heat losses.

2100"F = 1050"F/Hr. 2 Hr.

The numerator product is the amount of heat required in Btu and the denominator converts Btu into kilowatts. Step 2: Heat Loss The surface area of this furnace is approximately 112 sq ft. Assume 275 Btu/sq ft/hr for the heat loss. Although the heat loss will not reach this amount until after the furnace is at operating temperature, use this number for the calculations. 275x112 = 9kW 3412

Step 3: Heat Storage The heat storage of 9" insulating firebrick at 2100°F is approximately 6100 Btu/sq ft. The HS times the area divided by 2 hr (heat up time) requires the largest percentage of total power. filOOXll2 = 200kW 3412 2 Hr.

=

100kW/Hr.

76 kW +9 kW + 100 kW =185 kW

Normally, from 10% to 50% additional power would be added in order to compensate for extra heat losses through door openings and cracks, larger load sizes or efficiency losses through power controllers. However, in this case, for simplicity, design for 185 kW. Step 5: Select Element Size Life Line (LL) type elements are the choice for this application because the temperature is moderate and the atmosphere is not harmful enough to require the use of CR elements. SG elements would not be used for the same reasons, plus the 60" span is too long for the SG element. The same factors apply to the SGR element and being more expensive it offers no real advantage over the conventional LL elements.

Attacks silicon__carbide Attacks silicon carbide

.__-_____

Carbon monoxide

Step 4: Total Power Requirements Merely add up the totals from Steps 1,2 and 3 to get the minimum power necessary to heat the loaded furnace within 2 hr.

-

I _

2800°F

25

Endothermic: 18% carbon monoxide Max. Max. No effect 20Y0carbon monoxide 2500°F 25 Carbon pick-up . Exothermic Max. Max. No effect Halogens 1300°F 25 Attacks silicon carbide and reduces silica . . Helium . . . . . . . Max. Max. No effect . Hydrocarbons ........... 2400°F 20 Hot spotting from carbonpick-up..... Hydrogen 2370°F 25-30 Reduces silica film;forms methane from silicon carbide 20 Hot spotting ........... .Methane ....................................................... 2400°F ..............from carbonpick-up.......... Nitrogen 2500°F 20-30 Forms insulating silicon nitrides ........................................................ ............................... ............................................... Oxygen 2400°F 25 Oxidizes silicon carbide . sodium ............................... ....... 2400°F .... 25 Attacks silicon carbide .................................. ................ ............. ..................................................... Sulfur dioxide 2400°F 25 Attacks silicon carbide ............................................................................................................................................................. ...... Vacuum __ 2200°F 25 Below 7 microns, vgorizes silicon carbide . . . . . . . . Water: Dew Point 60OF 2000°F 20-30 Reacts with silicon carbide to form silicon hydrates 50°F 2200°F 25-35 0°F 2500°F 30-40 -50°F 2800°F 25-45

_ _ _

~

INDUSTRIALH U T I N G / ~ R C 1995 H

71

would be desirable to use the largest diameter available, which is 2l/;. A 9" refractory wall will necessitate cold ends; therefore, the element size will be 84" X 60" X 2l/{. This element has 401 sq in. of radiating surface area and a resistance of .86 Ohms. Step 6: WattdElement This consideration requires a little trial and error work. It is necessary to pick a number for watt loading; start with 35 watts/sq in. In this case; 35 X 401 sq in.=14,035 watts. Step 7: Quantity of Elements Step 6 and 7 must be used interchangeably at this point. We need 185,000 watts (185 kW) so the formula in Table I and the following calculation suggests that 13 elements are needed.

185.000 = 13.2 Elements 14,035 Since 13 elements is an odd number to work with, especially with 3 phase power, use 12 elements and increase watt loading a little. To save time, work the formula backwards. 185.000 W = 15.416 W/Element = 38.4 W/Sq In. 401 Sq.In.

12 Elements

The 38 watts/sq in. is still acceptable so proceed to Step 8 tu calculate the voltage required to power the elements. Step 8: Volts/Elements Multiply 15,416 (wattdelement) by .86 (Ohms per element). Take the square root of this product to get volts/element. In this case, it is 115 volts. Step 9: Volts Total The voltage which will be used in the plant is 230 volts. Therefore, 2 elements in series (115 X 2), will provide the needed 230 volts. Thus, no transformer is required; it is possible to operate off direct line voltage. Step 10: Amperage/Element The nominal current draw for one element (one parallel circuit) is: 115 VoltsElement = 134 Amps .86 OhmdElement

Due to the special resistance characteristics of elements, they draw more current during start-up than after they are at operating temperature. Multiplying the nominal Amps by 1.56, Table I, gives the maximum current draw for each parallel circuit. 134 X 1.56 = 209 Amps Maximum

This value is necessary to know in order to properly size the power circuits and controllers. Step 11: Amperage Maximum Now it is necessary to know if the available power is single phase or 3 phase. Consider that it is single phase. Since 2 elements are in series, there will be 6 parallel circuits.

Therefore, from the formula in Table I, total current for the fumace will be 209 Amps X 6=1254 Amps. That is alot of current; if 460 volts were available, current could be cut in half by putting 4 elements in series and using 3 parallel circuits. Three phase power is much more economical than single phase, as shown by the following calculations. In this case, there would be 4 elements on each phase as a separate single phase circuit. With 230 volts and 4 elements/phase, there are 2 elements in series with 2 parallel circuits. The current/phase then would be 209 X 2, or 418 Amps. Step 12: Delta Circuits In a Delta circuit, the volts per phase are the same as the 3 phase volts; that is, 230 volts 3 phase will provide 230 volts on each leg of the Delta. The current is what changes. The single phase Amps (418) must be multiplied by the square root of 3 (1.73) to leam how much 3 phase current is required.

418X1.73 = 723Amps Notice that this amperage is less than that required in the single phase example. With 3 phase circuits, less power is required to generate the same amount of work (watts) as with single phase. Step 13: Wye Circuits A Wye circuit is just the opposite of a Delta. The single phase Amps equal the 3 phase Amps. The voltage changes according to the formula in Table I, as shown:

230X1.73 = 398 Volts Since 398 volts are not available, Wye configurationcannot be used for this fumace. Step 14: Transformers Transformers are often required on the element-fired furnaces because there usually is not enough voltage needed for the bars to equal the line voltage. Furthermore, a variety of voltage settings is desired to compensate for varying production schedules and aging of the elements. To maintain constant power, the applied voltage must be increased to compensate for the gradual increase in resistance during element use. A typical transformer will have 6 taps. In the case of this example, if the incoming line voltage was 460, use would be made of a nominal tap of 230. There would be one tap lower than the nominal at .7 X 230, or 161 volts (see Table I). This will provide 1 / 2 power to the elements. The highest tap would be 2 X 230, or 460 volts. This will compensate for a 4 times increase in resistance to achieve maximum life from the elements. The other 3 taps would be spread between 230 and 460 to provide even steps when increasing voltage. SUMMARY There are many variables which can complicate an electric element furnace design, as indicated by the questionnaire (Table IV) and the previous discussion, but the preceding example will provide a good basis for proper designs.

J

Process Operation: Continuous Chamber Loading Space Wall Thickness Normal Operating Temp.

Batch "

"wide Composition

"high Maximum Design Temp.

Temperature Control Required Composition of Hearth Will Atmosphere be Contained in Muffle Composition, Shape, Size of Load Pounds of Material to be Heated Per Hour

Atmosphere Muffle

Belt

Firing Cycle: Heat-Up Soak Will Load be in Furnace When it is Brought Up to Temp. What Volatiles Will be Given Off from Load During Heating Power Input kW Power Supply: Method of Control: SCR Multiple Tap Transformer Voltages Full Load at What Voltage Other Information:

"deep

Cool

Volts A. C. Contactor

Phase:

Hz

Current Capacity Each Tap

INDUSTW HEATlNGhfARCH 1995

73