InterviewSolution
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InterviewSolution
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Two plates eachb of area A, thickness `L_1 and L_2` thermal conductivities `K_1 and K_2` respectively are joined to form a single plate of thickness `(L_1+L_2)`. If the temperatures of the free surfaces are `T_1 and T_2`. Calculate. (a) rate of flow of heat (b) temperature of interface and (c) equivalent thermal conductivity. |
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Answer» Correct Answer - A::B (a) If the thermal resistance of the two plates are `R_1 and R_2` respectively. Plates are in series. `R_s = R_1 + R_2 = (L_1/(AK_1)) + (L_2/(AK_2))` as `R = L/(KA)` and so `H = (dQ)/(dt) = (DeltaT)/R` `= ((T_1-T_2)/(R_1+R_2)) = (A(T_1 -T_2))/[(L_1/K_1) + (L_2/K_2)]` (b) If T is the common temperature of interface, then as in series rate of flow of heat remains same, i.e. `H = H_1 (=H_2)` `((T_1 - T_2)/(R_1 + R_2)) = (T_1 - T)/R_1` i.e. `T = ((T_1R_2 + T_2R_2)/(R_1 + R_2))` or ` T = ( [T_1 L_2/K_2] + T_2[L_1/K_1])/([L_1/K_1 + L_2/K_2] ) (as R = L/(KA))` (c) If K is the equivalent conductivity of composite slab, i.e. slab of thickness `L_1 + L_2` and cross - sectional area A, then as in series. `R_S = R_1 + R_2 or (L_1+L_2)/(AK_(eq)) = R_1 + R_2` i.e. `K_(eq) = (L_1 + L_2)/(A(R_1 + R_2)) = (L_1 + L_2)/([L_1/K_1 + L_2/K_2])` (as `R = L/(KA)` ). |
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