Question 10.4: A length of 8 cm at one end of a 0.3% carbon steel rod is to......
A length of 8 cm at one end of a 0.3% carbon steel rod is to be heated for upsetting. The bar diameter is 1.0 cm and total length 25.0 cm. The temperature desired is 1000°C. Considering the demand, it is decided that induction heating will be used for the job.
Now determine:
1. The thermal power required
2. Suitable frequency
3. Approximate heating time
4. Coil turns
5. Coil current
6. Pipe size for coil
7. Copper loss
8. Total power required and the recommended generator
Data
1. Specific heat of steel (c) = 0.56 kJ/kg°C
2. Density of steel = 7.85 g/cm³
3. Resistivity of steel (average) = 66.4 × 10^{-6} ohm.cm
4. Emissivity of steel = 0.8
5. Relative permeability of steel (30 − 760°C) = 15
6. Resistivity of copper = 2.2 × 10^{-6} ohm.cm
7. Ambient temperature = 27°C
Volume of steel to be heated
= \frac{\pi \times 1^2}{4} \times 6.3 \ cm^3
Weight of heated portion
= 6.3 × 7.85 = 50 gm (0.05 kg)
Area of the radiating surface
= Area of the radiating surface + Area of one end
= \pi \times 1 \times 8 \times \frac{\pi \times 1^2}{4} \\ = 26 \ cm^2
Heat required for heating from 27 to 1000°C
= weight × sp . heat × temp difference
= 50 × 0.56(1000 – 27)
= 27.2 kJ
Heat radiated at 1000°C
\begin{aligned}& =5.67 \times 10^{-8} \times 26 \times 10^{-4} \times 0.8\left(1273^4-300^4\right) \\& \simeq 300 \ \mathrm{w}\end{aligned}For quick heating we consider a power input
P = 4 × 300 = 1200 w
The work is to be heated in a magnetic condition up to 760°C and in a nonmagnetic condition from 760–1000°C. Relative permeability μr = 0.15. The work diameter is small and frequency 10 kHz is suitable for heating in the whole temperature range.
Average depth of penetration up to 760°C
Depth of penetration from 760–1000°C
\begin{aligned}\delta_{1000} & =5030 \sqrt{\frac{66.4 \times 10^{-6}}{1 \times 10^4}} \\& =0.41\ \mathrm{~cm} \ (4.1\ \mathrm{~mm}) \\& =\frac{\gamma}{\delta_{760}}=4.5 \mathrm{~F}=1.0 \\& =\frac{\gamma}{\delta_{1000}}=1.22 \mathrm{~F}=0.3\end{aligned}For heating up to 760°C
\begin{aligned}T u_{760} & =\sqrt[4]{\frac{P}{K \epsilon A}+T_1} \\& =\sqrt[4]{\frac{1.2 \times 10^3}{5.67 \times 10^{-8} \times 0.8 \times 26 \times 10^4}+300^4} \\& =1860 \mathrm{~K}\left(1587^{\circ} \mathrm{C}\right) \\\frac{c M}{K \epsilon A T u^3} & =\frac{0.56 \times 10^3 \times 0.05}{5.67 \times 10^{-8} \times 0.8 \times 26 \times 10^4 \times 6.43 \times 10^9}=37 \end{aligned} \\ \tau_{760}=37\left[\begin{array}{l}\tan ^{-1} \frac{1033}{1860}-\tan ^{-1} \frac{300}{1860}+\frac{2.3}{2} \\\times \log \frac{(1033+1860)(1860-300)}{(1860-1033)(1860+300)} \end{array}\right] \\ = 37(0.507 – 0.16 + 0.630) = 34 \ secFor heating from 760 to 1000°C. It is assumed that power input is kept constant at 1.2 kW
\begin{aligned}\tau_{1000} & =37\left[\begin{array}{l}\tan ^{-1} \frac{1273}{1860}-\tan ^{-1} \frac{1033}{1860}+1.5 \\\times \log \frac{(1273+1860)(1860-1033)}{(1860-1273)(1860+1033)}\end{array}\right] \\& =37(0.600-0.506+0.192)=11 \ \mathrm{sec}\end{aligned}Some time will be taken by the magnetic transformation at 760°C. As these calculations are beyond the scope, the time is arbitrarily estimated at 5 sec.
Thus, the total heating time at 1.2 kW is
The length of the heated part is 0.08 m.
Power per meter of work
= \frac{1200}{0.08} = 1.5 \times 10^4 \ w/m
Up to 760°C \gamma / \delta=4.5 and F = 1.0
Power per meter
\begin{aligned}P & =8 \pi H_o^2 \rho \times \frac{\gamma}{\delta} \times F \\1.5 \times 10^4 & =8 \pi H_o^2 66.4 \times 10^{-8} \times 4.5 \times 1 \\H_o & =\sqrt{\frac{1.5 \times 10^4}{8 \pi \times 66.4 \times 10^{-8} \times 4.5}} \\& =14.14 \times 10^3 \ \mathrm{AT} / \mathrm{m}\end{aligned}From 760–1000°C \gamma / \delta=1.22 and F = 0.3
The power input is the same
\begin{aligned}H_o & =\sqrt{\frac{1.5 \times 10^4}{8 \pi \times 66.4 \times 10^{-8} \times 1.22 \times 0.3}} \\& =49 \times 10^3 \ \mathrm{AT} / \mathrm{m}\end{aligned}These AT are much more than those required to be heated up to 760°C, hence, power will have to be reduced to about 300 w to maintain the same AT
Consider H_o = 20 \times 10^3 \ AT/m throughout
Assume that we use a 6 mm internal diameter copper pipe for the coil. The wall thickness should be about the depth of penetration for copper (δ_c). The resistivity of copper is 2.2 × 10^{-6} ohm.cm
\delta c=5030 \sqrt{\frac{2.2 \times 10^{-6}}{1 \times 10^4}}=7444 \times 10^{-5}=0.07 \ \mathrm{~cm}(0.7 \ \mathrm{~mm})Outer diameter of the copper pipe (d_p)
= 6.0 + 2 × 0.07 \simeq 7.6 mm
About 8–9 turns can be accommodated in a length 8–8.5 cms
Coil current =\frac{A T}{m} \times \ell_c / N=\frac{20 \times 10^3 \times 0.08}{8} \\ I_c = 200 \ Amp
Coil diameter
\begin{aligned}d_c & =d_w+\text { Air gap } \\& =10+50=60 \ \mathrm{~mm}(6.0 \ \mathrm{~cm})\end{aligned}Pipe length = Length of one turn × Number of turns
= \pi \times 6 \times 9 = 170 \ cm
Cross-sectional area of pipe
= \pi / 4 (0.76^2 – 0.6^2) = 0.171 \ cm^2Resistance of coil at \rho = 2.2 \times 10^{-6} \ ohm.cm \\ R_c=\frac{2.2 \times 10^{-6} \times 170}{0.171}=2.19 \times 10^{-3} \ \mathrm{ohm}
Copper loss in coil
=I_c^2 R_c=200^2 \times 2.19 \times 10^{-3}=87 \ \mathrm{~W}Generator power
= Work power + Copper loss = 1200 + 87 = 1.3 kW
To this we will have to add transmission loss, etc. To heat faster, the work power may be increased to 1500–2000 W.
A step-down transformer and a tank circuit will be required. For 10 kHz a solid state generator is recommended.