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

Design a heating element for heating a metal specimen in a hot stage microscope.

Specimen size      25 mm diameter 15 mm thick

Maximum temperature      1500°C

Atmosphere      Air or Helium/Argon

Power estimate for hot chamber      800 W

Step-by-Step
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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.

 Metal/Alloy Density g/c$m^3$ Melting Point °C Max. Useful Temp. °C Resistivity $\times 10^6$ ohm.cm (Temp. coef. $\alpha \times 10^{-3}$ ) Forms Refractory stability °C Surface load W/c$m^2$ T>1200° Atmosphere Pure Platinum 21.45 1768 1400 (1550 short time) 10.65 (3.9) Wire, Mesh, Foil, Strip $Al_2O_3−1800 \\ MgO_2−2400 \\ (Oxidizing) \\ ZrO_2−2300 \\ ThO_2−2600$ 3-5 Oxidizing only, Air. Avoid contact with Si$O_2$ (>900°C) reducing atm. Platinum 10% Rh. 20 1825 1550, 1600 18.5 Wire do do do Platinum 40% Rh. 16.6 1900 1750 16.8 Wire do do do

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