A composite slab consists of two plates of thicknesses $$L1$$ and $$L2$$ and thermal conductivities $$K1$$ and $$K2$$. The $$cross-sectional$$ areas are equal. The equivalent thermal conductivity is:

Correct option is 1K1K2(L1+L2)K2L1+K1L2

Questions

A composite slab consists of two plates of thicknesses L₁ and L₂ and thermal conductivities K₁ and K₂. The cross-sectional areas are equal. The equivalent thermal conductivity is:

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K1K2(L1+L2)K2L1+K1L2

K2L1+K1L2K1K2(L1+L2)

L1L2(K1+K2)(K2L1+K1L2)

K2L1+K1L2L1L2(K1+K2)

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detailed solution

Correct option is A

In series connectionKeq=∑li∑liKi=L1+L2L1K1+L2K2=K1K2(L1+L2)K2L1+K1L2

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A composite slab consists of two plates of thicknesses L1 and L2 and thermal conductivities K1 and K2. The cross-sectional areas are equal. The equivalent thermal conductivity is:

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Ready to Test Your Skills?

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A composite slab consists of two plates of thicknesses L₁ and L₂ and thermal conductivities K₁ and K₂. The cross-sectional areas are equal. The equivalent thermal conductivity is: