## Q. 9.1

A resistance-heated furnace is to be designed with following requirements:

Working space = 30 × 40 × 100 cm

Occasional maximum temperature = 1250°C

Continuous working temperature = 1150°C

Atmosphere = Air

Estimated power = 24 kW

Available voltage = 440 V–3ph

Specific resistivity (average) = 150 × $10^{-6}$ ohm.cm

Now determine:

1. Material for heating resistors
3. Suggest method and material for construction
4. Wire or strip size
5. Element distribution, shape, and construction
6. Heating rate achievable

## Verified Solution

The furnace atmosphere is oxidizing. Working temperature 1150°C and occasional maximum temperature 1250°C suggest Fe-Cr-Al alloy (Kanthal A1 or Kanthal AF or other similar alloy) having a maximum permissible temperature of 1400°C and a probability of long service life (>1000 h).
We can use nonmetallic elements such as SiC or MoS$i_2$ (Kanthal Super). They have a much higher temperature capability and are costly. Higher temperatures (> 1250°C) are not specified.
The recommended surface load (q) in the range 1150–1250°C is 1–1.8 W/cm² . For trial, a preliminary load of 1.0 W/cm² is assumed. The chosen alloy is available in both wire and strip form.
For the first trial choose a round wire of diameter d cm.
The power requirement given is 24 kW with a 380 V 3 ph supply. Assume a star connection so that each phase has 24/3 = 8 kW power at 220 V.

Diameter of Wire
Voltage per phase (V) = 220 V
Power dissipated in each phase (P) = 8 kW
Specific resistance = 150 × $10^{-6}$ ohm.cm

\begin{aligned}d & =0.74 \sqrt{\left[\frac{8000}{220}\right]^2 \times \frac{150 \times 10^{-6}}{1.0}} \\& =0.43 \ \mathrm{~cm} \ ( 4.3 \mathrm{~mm})\end{aligned}

Alternatively,

Resistance $R= \frac{V^2}{P} \\ = \frac{220^2}{8000} = 60.5 \ ohm$

Surface area = $\frac{P}{q} = 8000 \ cm^2 \\ \frac{ohm}{cm^2} = \frac {8000}{6.05} = 1.323 \ cm^2 / ohm$

From Table 9.5, nearest wires are

Kanthal A1 – 1550 cm²/ohm – 4.5 mm
Kanthal AF – 1363 cm²/ohm – 4.25 mm
As the diameters determined by both methods are of the same order, we choose Kanthal A1, 4.5 mm diameter wire. This wire has following data:
0.0912 ohm/m at 20°C 113 g/m
Length required for 6.05 ohms is

$\frac{6.05}{0.0912} = 66.33 \ m \ (663 \ cm)$

At 113 gm/m the weight of wire is
66.33 × 113 × $10^{-3}$ = 7.5 kg

Add 10% for terminals, breakage etc.
Wt = 8.25 kg

Table 9.5 Design Data for Nikrothal (80Ni-20Cr) Heating Wires (Courtesy: Kanthal AB)

Wire mm 6.5–0.02 mm ∅

Resistivity Ω mm² $m^{−1}$ 1.09

Density, $gcm^{-3}$ 8.30

$\mathrm{cm}^2 / \Omega \frac{I^2 C_t}{P}$

I = Current

$C_t$ = Temperature factor

To obtain resistivity at working temperature multiply by the factor $C_t$ in the following table.
 °C 20 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 $C_t$ 1 1.01 1.02 1.03 1.04 1.05 1.04 1.04 1.04 1.04 1.05 1.06 1.07