As per the given figure, two plates A and B of thermal conductivity K and 2K are joined together of form a compound plate. The thickness of plates are 4.0cm and 2.5cm respectively and the area the area of cross-section is 120 cm² for each plate. The equivalent thermal is 120 cm² for each plate. The equivalent thermal conductivity of the compound plat is $(1 + \frac{5}{α}) K,$ then the value of a will be________.

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

k_{eq=
$\frac{L_{1} + L_{2}}{\frac{L_{1}}{k_{1}} + \frac{L_{2}}{k_{2}}} = \frac{4 + 2.5}{\frac{4}{k} + \frac{2.5}{2 k}}$}

_{keq=
$\frac{6.5}{\frac{8 + 2.5}{2 k}} = \frac{6.5}{10.5} \times 2 k = \frac{13 \times 2}{21} k$}

_{=
$(1 + \frac{5}{51}) k$}

_{=
$(1 + \frac{5}{α})$
=21}

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