# Question 9.6: Brass slugs are to be heated to 600°C in about 1.0 sec by ca......

Brass slugs are to be heated to 600°C in about 1.0 sec by capacitive discharge.

Slug dimensions      10 mm diameter, 15.0 long

Sp. heat of brass (C)      0.385 J/gm K

Density (r)      8.72 g/cc

Ambient temperature      30°C

Assume a charging time      5 sec

Now determine:

1. Heat required

2. Condenser required

3. Voltage required

4. Resistance required in charging and discharging circuits

Step-by-Step
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Slug size 10 mm diameter, 15 mm long
Material-brass

Volume = $\frac{\pi \times 3.5^2 \times 50}{4}$ = 1.18 cc

Density = 8.72 g/cc
Weight = 8.72 × 1.18
= 18.3 gm

Specific heat C = 0.385 J/g °C
Initial temperature = 30°C
Final temperature = 600°C
Heat required = 18.3 × 0.385(600 − 30)
= 4.015 kJ

Thermal losses and contact losses assumed = 25%, i.e.,
0.25 × 4.015 = 1.0 kJ
Total heat required
= 4.015 + 1.00
$\simeq 5 \ \mathrm{~kJ}$

The capacitor bank has a 3000 μF capacity. If the charging voltage is V the energy stored is

= $\frac{1}{2}$ CV² J

This should be equal to the energy required for heating

\begin{aligned} & =\frac{1}{2} \times 3000 \times 10^{-6} V^2=5 \times 10^3 \\ V & =\sqrt{\frac{2 \times 5 \times 10^3}{3000 \times 10^{-6}}} \\ & =1800 \ \mathrm{~V} \end{aligned}

The slug is to be heated in about 1.0 sec
Let the time constant λ = C $R_D$ = 0.1 sec

$3000 \times 10^{-6} \times R_D=0.1 \\ R_D=\frac{0.1}{3000 \times 10^{-6}} \\ = 30 \ ohm$

If the charging is to be done in 5.0 sec, then assume a time constant λ = C $R_C$ = 0.2 sec

$3000 \times 10^{-6} \times R_D = 0.1 \\ R_D=\frac{0.2}{3000 \times 10^{-6}} \\ = 66 \ ohm$

If we use a step-down transformer with a few turns (1–3) on secondary, the heating current will be increased and voltage will be decreased.

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