Question 9.4: Design a heating element for heating a metal specimen in a h......

The heating element required has to be very compact. Considering the high temperature and oxidizing environment, precious metals appear to be the only choice.

A glance at Table 9.3 shows that platinum, 10  or 40% rhodium alloy, is suitable.
Pt – 10% Rh has higher resistivity, hence we will try this alloy.
The heating element will be spiral wound, free coil, with pure Al_{2}O_{3} pieces to separate turns.
Considering a specimen size (25 diameter × 15 ht), let the element size be 40 mm diameter and 50 mm ht.
The recommended surface loading is 2.5 W/cm². As the service expected is intermittent, 5 W/cm² is chosen as the permissible load

Surface area required = \frac{800}{5} = 160 cm²

As the heater is essentially free for rigidity considerations, let the wire diameter be 1.0 mm and the coil pitch 3 mm.

Wire cross-section area = \frac{\pi \times 0.1^2}{4} = 7.8 \times 10^{-3} \ cm^2
The specific resistivity of Pt–40% Rh alloy is 18.5 × 10^{-6} ohm.cm. The resistance of 1 cm length of wire will be

=\rho \frac{\ell}{a}=\frac{18.5 \times 10^{-6} \times 1}{7.83 \times 10^{-3}}=2.37 \times 10^{-3} \ \mathrm{ohm}
The coil diameter is 40 mm, hence, the length of wire in one turn is \pi × 4.0 = 12.6 cm.
The number of turns in 50 mm height at 3.0 mm turns is 50/3 = 16.7 turns.
Total length of wire in coil = 12.6 × 16.7 = 210 cm
Resistance R = 210 × 2.37 × 10^{-3} = 0.497 ~ 0.5 ohm
Assume a voltage 20 V, the current is
I = 20/0.495 = 40 A
Power dissipated = I^2 R = 40^2 \times 0.5 = 800 W
Current carrying capacity of Pt–40% Rh alloy at 1600°C is 2240 Amp/cm² , hence, the calculated current is much lower.
Surface area for 0.1 cm diameter and 210 cm length = \pi × 0.1 × 210 = 66 cm²

Surface load is = 800/66 = 12 W/cm²
Volume of wire = area × length
= 7.8 × 10^{−3} × 210 = 1.64 cm³
Weight of wire at 20.0 gm/cm³
= 20 × 1.64 = 32 gm
We can repeat the calculations for different power ratings and element weight reduction. It is important to design the hot chamber properly in order to reduce heat loss. The power supply terminals (usually copper rods) will have to be water-cooled.