这是一份bath巴斯大学PH30032作业代写的成功案

Now
$$
\frac{T(0,0, t)}{T(0,0, \infty)}=\frac{4(k t)^{1 / 2}}{w \pi} .
$$
Let us assume that the approximation $w \gg 2(k T)^{1 / 2}$ holds if
$$
w / 2(k t)^{1 / 2}=5 .
$$
For times such that this inequality holds we have
$$
\frac{T(0,0, t)}{T(0,0, \infty)} \sim 0.3 \text {. }
$$
In other words, we reached one-third of the final temperature in the following times: $\mathrm{Al}, 5.4 \times 10^{-6} \mathrm{sec} ; 304 \mathrm{~S} . \mathrm{S}$., $7.4 \times 10^{-5} \mathrm{sec}$; and carbon phenolic, $1 \times 10^{-3} \mathrm{sec}$.

PH30032 COURSE NOTES ：

$$
t_{c}=\left(\alpha^{-1}\right)^{2} / \kappa
$$
Thus the cooling rate at the surface is
$$
\Delta T(z) / t_{\mathrm{c}}=(1-R) \alpha^{3} I_{0} K t_{\mathrm{p}} / \rho C .
$$
In many applications it is necessary to obtain $T=T_{\mathrm{m}}$. By comparison it is clear that the cooling rate for case 1 will always be very high. In fact,
$$
d T / d t=T_{\mathrm{m}} / t_{\mathrm{p}},
$$
and if one has sufficient energy to obtain $T_{\mathrm{m}}$ in, say, a $10^{-9} \sec$ pulse, one will have
$$
d T / d t \approx 10^{12 \circ} \mathrm{C} / \text { sec. }
$$