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
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
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 \ ohmIf 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.