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Question 4.8: A pulse with an energy 10 J is incident on a thin steel plat......

A pulse with an energy 10 J is incident on a thin steel plate. The focal spot is small and can be treated as a point.

Density of steel ρ = 7.86 gm/cm³

Specific heat c = 0.5 J/gm°C

Thermal diffusivity α = 0.15 cm²/sec

If the plate is to be joined to a small spring at about 1000°C, determine the time within which the process has to be carried out to obtain a suitable temperature at a depth 0.1 mm.

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For an instantaneous point source the expression for temperature is

t_{z, \tau}=\frac{Q}{8 \rho c(\pi \alpha \tau)^{3 / 2}} \exp -\frac{z^2}{4 \alpha \tau} \\ Given \ Q=10 \mathrm{~J}, \quad \tau=0.05 \mathrm{~s}, \quad z=0.01 \ \mathrm{~cm} \\ \begin{aligned}t_{(0.01,0.05)}= & \frac{10}{8 \times 7.86 \times 0.5} \times \frac{1}{(\pi \times 0.15 \times 0.05)^{3 / 2}} \\& \times \exp -\frac{0.01^2}{4 \times 0.15 \times 0.05} \\= & \frac{10}{31.4} \times \frac{1}{3.62 \times 10^{-3}} \times 0.997 \\ & =87^{\circ} \mathrm{C}\end{aligned}

For the same spot at τ = 0.025

t=\frac{10}{31.4} \times \frac{1}{1.28 \times 10^{-3}} \exp -\frac{0.01^2}{4 \times 0.15 \times 0.05} \\ =250 °C

Similarly

For \tau = 0.012 sec  t = 740°C

and for \tau = 0.005 sec

\begin{aligned}t & =\frac{10}{31.4} \times \frac{1}{(\pi \times 0.15 \times 0.005)^{3 / 2}} \exp \frac{0.01^2}{4 \times 0.15 \times 0.005} \\& =\frac{10}{31.4} \times \frac{1}{1.14 \times 10^{-4}} \times 0.967 \\& =2700^{\circ} \mathrm{C}\end{aligned}

Thus the joining has to be carried out within about 10 m/sec after the pulse.

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