Two cylindrical rods are identical, except that one has a thermal conductivity k

Question

Two cylindrical rods are identical, except that one has a thermal conductivity k1 and the other has a thermal conductivity k2. As the drawing shows, they are placed between two walls that are maintained at different temperatures TW (warmer) and TC (cooler). When the rods are arranged as in part a of the drawing, a total heat Q' flows from the warmer to the cooler wall, but when the rods are arranged as in part b, the total heat flow is Q. Assuming that the conductivity k2 is twice as great as k1 and that heat flows only along the lengths of the rods, determine the ratio Q'/Q. Express your answer as a number with no units.

Explanation / Answer

since a single slab of material ?Q / ?t = - ? A ?T / ?x theresince e, since part (a) ?Q1 / ?t1 = - ?1 A ( Tc – Tw ) / L ==> ?Q1 = - ?1 A ?t1 ( Tc – Tw ) / L ?Q2 / ?t2 = - ?2 A ( Tc – Tw ) / L ==> ?Q2 = - ?2 A ?t2 ( Tc – Tw ) / L Q ' = ?Q1 + ?Q2 = - [ ?1 A ?t1 ( Tc – Tw ) / L ] – [ ?2 A ?t2 ( Tc – Tw ) / L ] ……… = - [ ?1 ?t1 + ?2 ?t2 ] A ( Tc – Tw ) / L since part (b) … ?Q1 / ?t1 = - ?1 A ( Tm – Tw ) / L … Tm = temperature at the junction of the two rods … ?Q2 / ?t2 = - ?2 A ( Tc – Tm ) / L … … assuming steady heat flow … ?Q1 / ?t1 = ?Q2 / ?t2 … thus that … … - ?1 A ( Tm – Tw ) / L = - ?2 A ( Tc – Tm ) / L … … ?1 ( Tm – Tw ) = ?2 ( Tc – Tm ) ==> ?1 Tm – ?1Tw = ?2 Tc – ?2 Tm … … [ ?1 + ?2 ] Tm = ?1Tw + ?2 Tc ==> Tm = [ ?1Tw + ?2 Tc ] / [ ?1 + ?2 ] … … Tm – Tw = { [ ?1Tw + ?2 Tc ] / [ ?1 + ?2 ] } – Tw = ?2 [ Tc – Tw ] / [ ?1 + ?2 ] … ?Q1 / ?t1 = - ?1 A ( Tm – Tw ) / L = - ?1 A { ?2 [ Tc – Tw ] / [ ?1 + ?2 ] }

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