Qstnid -Iv-22-a Composite Slab Is Prepared By Pasting Two Plates Of Thickness L<sub>1</sub> And L<sub>2</sub> And Thermal Conductivites K<sub>1</sub> And K<sub>2</sub>. The Slabs Have Equal Cross-sectional Area. Find The Equivalent Conductivity Of The Composite Slab. - 28: Heat Transfer - Neet-xii-physics

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NEET-XII-Physics

28: Heat Transfer

with Solutions - page 4

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#22
A composite slab is prepared by pasting two plates of thickness L₁ and L₂ and thermal conductivites K₁ and K₂. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.
Ans :
It is equivalent to the series combination of 2 resistors.
∴ R_S = R₁ +R₂
Resistance of a conducting slab, `` R=\frac{L}{KA}``
R_S = R₁ + R₂
`` \frac{{L}_{1}+{L}_{2}}{{K}_{\,\mathrm{\,s\,}}.A}=\frac{{L}_{1}}{{K}_{1}A}+\frac{{L}_{2}}{{K}_{2}A}``
`` \frac{{L}_{1}+{L}_{2}}{{K}_{\,\mathrm{\,S\,}}}=\frac{{L}_{1}}{{K}_{1}}+\frac{{L}_{2}}{{K}_{2}}``
`` \frac{{L}_{1}+{L}_{2}}{{K}_{\,\mathrm{\,S\,}}}=\frac{{L}_{1}{K}_{2}+{L}_{2}{K}_{1}}{{K}_{1}\times {K}_{2}}``
`` {K}_{\,\mathrm{\,S\,}}=\frac{\left({L}_{1}+{L}_{2}\right)\left({K}_{1}{K}_{2}\right)}{{L}_{1}{K}_{2}+{L}_{2}{K}_{1}}``
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