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Standard VII

Physics

Conduction Law: Differential Form

A cylindrical...

Question

A cylindrical steel rod, $0.01 m$ in diameter and $0.2 m$ in length is first heated to $750^{\circ} C$ and then immersed in water bath at $100^{\circ} C .$ The heat transfer coefficient is $250 W / m^{2} K .$ The density, specific heat and thermal conductivity of steel are $ρ = 7801 k g / m^{3}, c = 473 J / k g K,$ and $k = 43 W / m K,$ respectively. The time required for the rod to reach $300^{\circ} C$ is seconds

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Solution

Given data:

$D = 0.01 m; L = 0.2 m; T_{1} = 750^{\circ} C$
$T_{\infty} = 100^{\circ} C;$ $h = 250 W / m^{2} K; ρ = 7801 k g / m^{3}$
$c = 473 J / k g K$

$\frac{T - T_{\infty}}{T_{i} - T_{\infty}} = e^{- \frac{h A t}{ρ V c}}$

where $\frac{V}{A} = \frac{π D^{2} L}{4 \times π D L} = \frac{D}{4} = \frac{0.01}{4}$

$= 2.5 \times 10^{- 3} m$
$∴ \frac{300 - 100}{750 - 100} = e^{\frac{- 250 \times t}{7801 \times 2.5 \times 10^{- 3} \times 473}}$
$\frac{200}{650} = e^{- 0.0271 t}$
$0.30769 = e^{- 0.0271 t}$

Taking $l o g_{e}$ both sides, we get
$l o g_{e} 0.30769 = - 0.0271 t l o g_{e} e$
$(∵ l o g_{e} e = 1)$
$- 1.1786 = - 0.027 t$
or $t = \frac{1.1786}{0.027} = 43.5 s$

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