Question 9.5: Steel billets of 35 mm diameter and 500 mm length are to be ......

Volume of billet

= =\frac{\pi \times 3.5^2 \times 50}{4}=480 \ \mathrm{~cm}^3
Surface area (curved surface)
= π × 3.5 × 50
= 550 cm²
= 0.055 m²
Weight
= 480 × 7.6
= 3648 gm
= 3.648 kg
Resistance

\begin{aligned} \sigma \frac{\ell}{A} & =\frac{61 \times 10^{-6} \times 50}{38.48} \\ & =80 \times 10^{-6} \ \Omega \end{aligned}
Heat required to raise the temperature from 30 to 1200°C = mass × sp. heat × temp. difference
= 3.648 × 10³ × 0.576 × (1200 − 30)
= 2.458 × 106 J
If heating is done in 30 sec

Power = \frac{2.458 \times 10^6}{30}

= 85 kW

Heat radiated
Assume a mean surface temperature 800°C and an emissivity 0.8

Total power
= Heating load + radiated loss
= 85 + 12
= 97 kW~100 kW

Note : There may be additional losses in grips.

If a 100 V DC power supply is available

Current required = \frac{100 \times 10^3}{1000}

= 1000 A

The radiation loss is very high. It can be reduced by keeping the billet in an enclosure or heating for a shorter duration.
Note that the calculated radiation loss is at 1200°C surface temperature which is attained at the end of the heating time.
If the power available is AC at 50 Hz, the effective resistance will change from R_{DC} to R_{AC}.
R_{DC} = 80 × 10^{-6} Ω (as calculated earlier)
\rho ⋅ (30–760°C) = 57× 10^{-6} ohm.cm
\rho ⋅ (760–1200°C) = 120 ×10^{-6} ohm.cm
\mu_o=4 \pi \times 10^{-7} \quad \mu_r=100
We can get R_{AC} by using the formula
R_{AC} = \rho \frac{\ell}{A}

where a is the area of annulus through which most of the current flows. This will be circular ring of outer diameter 3.5 cm and inner diameter = 3.5 − 2δ, where δ is the depth of penetration.
For the temperature range 30–760°C

= 0.53 cm or 5.3 mm

Inner diameter = 3.5-2(2 \times \delta)

= 3.5 – 2 (2× 0.53)

= 1.38 cm

Area = \frac{\pi\left(3.5^2-1.38^2\right)}{4}

= 8.125 cm²

or    = 35 × 10^{-5} ohm (30-760°C).