Question

8. For a semi-infinite solid buried in soil (a=0.15 x 106 m²/s and k=0.4 W/m." minimum depth of burial to cause a temperature difference gradient, 1-8 (x,1), 0.4 @ h = o) for two months is a. 1.1 m b. 2.2 m c. 0.8 m d. 30 cm e. All of the above f. None of the above

Answer 1

$\textbf{Given data}\rightarrow$

hvaT h - = 00

$\theta \left ( x,\tau \right )=\frac{T\left ( x,\tau \right )-T_{\infty }}{T_{i}-T_{\infty }}$

1-0 (x, T) = 0.4

0.4 = 0.6 0 (x, T) = 1 -

$\textup{Thermal diffusivity}\left ( \alpha \right )=0.15*10^{-6}\textup{ m}^{2}\textup{/s}$

$\textup{Time}\left ( \tau \right )=60*24*3600=5184000\textup{ sec}$

$\textbf{Solution}\rightarrow$

$\textup{Calculate z corresponding to erf(0.6)}$

z	erf(z)
0.55	0.5633
0.6	0.6039

$\textup{By interpolation}$

$\frac{0.6039-0.6}{0.6-z}=\frac{0.6039-0.5633}{0.6-0.55}$

$z=0.5952$

$\textup{For semi-infinite solid body}$

$z=\frac{x}{2\sqrt{\alpha \tau }}$

$x=2*z*\sqrt{\alpha \tau }$

$x=2*0.5952*\sqrt{0.15*10^{-6}*5184000}$

$x=1.05\textup{ m}$--------------------- Ans

$\textup{Take nearest value as x=1.1 m}$

$\textup{Correct choice is A}$