# Question 9.3: Design an indirect resistance furnace as specified below. Wo......

Design an indirect resistance furnace as specified below.

Working Space      60 × 45 × 90 cm

Continuous temperature      1550°C

Occasional maximum temperature      1600°C

Estimated power      52 kW

Electric supply      380 V, 3 ph

Now determine:

Heater material

No. of heaters and connections

Voltage and current

Transformer specification

Heater dimension

Approximate inner and outer dimensions

Step-by-Step
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Working volume
= 45 (W) × 60 (H) × 90 (D)
2.5 × $10^5$ cm³ = 0.25 m³
As this is the working volume, actual volume will be a little larger
$\simeq$ 0.27 m³
Maximum continuous temperature = 1550°C
Occasional temperature = 1600°C
For these temperatures we select Kanthal Super ST grade which has a maximum element temperature 1700°C.
The power requirement is 52 kW. The muffle height is 600 mm. The heating portion Le of the element should be about 600/10 = 60 mm from bottom hearth, i.e.,
Le = 540 mm
Nearest to shank heater is 560 mm (Le) and has terminals 60 mm apart.*
To keep proper clearance with the bottom, the height will be increased to about 616 mm.
Assume that the distance between adjacent elements is about 100 mm (more exact calculations can be done from the handbook). Heaters are to be located on the two vertical sides, hence, the number of heaters should be divisible by 2.
The supply available is three-phase. To distribute the load equally, the number of heaters should be divisible by 3.

Hence, the number of heaters to be used must be 6 (3 × 2) or 12, or 18, and so on.
The required depth is 900 mm (each side) and one element will occupy 60 +100 mm. If the number of heaters chosen is 12, there will be 6 heaters on each side. We will have to increase the depth by about 100 mm. Final depth is 900 + 100 = 1000 mm

Note : If suitable heaters are available we can design for six heaters (two per phase) and keep the depth at 900 mm.

At 52 kW power each heater will be rated at 52/12 = 4.33 kW.

Assume an insulation thickness W/2

= 45/2 = 22.5 cm (225 mm).

The element shanks will have to be about 225 (in insulation) + 150 (terminals) which equals to 375 mm.

The catalog extract (See Fig 9.8) shows that the nearest elements are

1. Heating length 500 mm Shank 355 mm 4.57 kW

2. Heating length 560 mm Shank 355 mm 5.06 kW

3. Heating length 500 mm Shank 400 mm 4.60 kW

We will have to choose an element from the three sizes above.

The following element is chosen (refer to Table 9.8) is

Kanthal Super ST 9/18 mm
Heating zone Le = 500 mm
Tapered zone g = 30 mm
Shank Lu = 400 mm
Distance between shanks a = 60 mm
Diameter between elements b = 100 mm
Diameter of heating zone = 9 mm
Power per element = $\frac{52}{12}$ = 4.33 kW.
Distance of bend = a = 60 mm
Heating length 2Le + ($\pi$ /2 − 1) α = 2 × 500 + 0.57 × 60
= 1034 mm (103.4 cm)

Heated surface
= 103.4 × $\pi$ × 0.9 = 292 cm²
Surface load, (assuming 10% power lost in heating terminals)

= $\frac{0.9 \times 4.33}{292}$ = 13.3 W/cm²

The surface load diagram (Figure 9.10) shows that for a furnace temperature 1550°C and surface load 13.3 W/cm² current of 9 mm element is 250 A and the element temperature is 1650°C.

Electrical resistivity
Room temperature (20°C) = 25 × $10^{-6}$ ohm.cm
at 1650°C = 375 × $10^{-6}$ ohm.cm
at 900°C = 180 × $10^{-6}$ ohm.cm
Length of hot zone = 103.4 cm
Resistance of hot zone = 103.4 × 375 × $10^{-6}$
= 0.0387 ~ 0.04 ohm.

Assume average terminal (shank) zone temperature = 900°C
Terminal length = 2 × Lu = 2 × 400 = 800 mm (80 cm)
Resistance = 80 × 180 × $10^{-6}$
= 0.0144 ohm.
Total Resistance = 0.04 + 0.0144 = 0.0544 ohm.
Voltage required = current × resistance
= 250 × 0.0544 = 13.6 volts/element.
There are four elements in series in each phase
Phase voltage (secondary) = 4 × 3 × 13.6
= 94.2 volts ~100 V
This is the highest voltage for starting and control; suggested taps are 40 and 60 V.

As a higher current will be drawn when starting (because of low resistance) the current rating of the transformer should be higher than 250 A. Let it be 350 amp.
The power rating of the transformer

= $\sqrt{3} \times 350 \times 100 = 60 \ kVA$
The above calculations should be repeated for other element sizes and insulating materials to arrive at an optimum design. The outer dimensions and exact interior size can then be calculated.

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