AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The conduction of heat through series combinations of rods is shown below. The expression of the rate of heat flow for series conduction is, $\frac{d Q_1}{dt} = \frac{d Q_2}{dt} = \frac{d Q_3}{dt} = \frac{T_i - T_f}{k_{eq}}$ $\frac{k_1 A_1 (T_i - T_1)}{l_1} = \frac{k_2 A_2 (T_1 - T_2)}{l_2}$ $= \frac{k_3 A_3 (T_2 - T_f)}{l_3}$ Substitute $A_1 = A_2 = A_3$ and $l_1 = l_2 = l_3$ in above expression, $2k(100 - T_1) = k(T_1 - T_2)$ $= 0.5(T_2 - 0)$ ...... (I) The equivalent coefficient of thermal conductivity, $\frac{1}{k_{eq}} = \frac{1}{2k} + \frac{1}{k} + \frac{2}{k}$ $k_{eq} = \frac{2k}{7}$ The heat current equation is, $\frac{dQ}{dt} = \frac{dQ_1}{dt}$ $\frac{k_{eq} A (T_i - T_f)}{l} = \frac{k_1 A_1 (T_i - T_1)}{l_1}$ ...... (II) The equivalent length is, $l = l_1 + l_2 + l_3 = 3l_1$ The equivalent area is, $A = A_1 + A_2 + A_3 = 3A_1$ Substitute values in equation (II) $\frac{2k}{7} \frac{3 A_1 (100-0)}{3 l_1} = \frac{2k A_1 (100 - T_1)}{l_1}$ $\frac{1}{7} \times 100 = 100 - T_1$ $T_1 = \frac{600^{\circ}\text{C}}{7}$ Substitute values in equation (I), $2k \left(100 - \frac{600}{7}\right) = 0.5 k T_2$ $T_2 = \frac{400^{\circ}\text{C}}{7}$