Question

Two very long, slender rods of the same diameter and length are given. One rod (Rod 1) is made of aluminum and has a thermal conductivity $k_{1}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, but the thermal conductivity of Rod \(2, k_{2}\), is not known. To determine the thermal conductivity of Rod 2 , both rods at one end are thermally attached to a metal surface which is maintained at a constant temperature \(T_{b}\). Both rods are losing heat by convection, with a convection heat transfer coefficient \(h\) into the ambient air at \(T_{\infty}\). The surface temperature of each rod is measured at various distances from the hot base surface. The measurements reveal that the temperature of the aluminum rod (Rod 1) at \(x_{1}=40 \mathrm{~cm}\) from the base is the same as that of the rod of unknown thermal conductivity (Rod 2) at \(x_{2}=20 \mathrm{~cm}\) from the base. Determine the thermal conductivity \(k_{2}\) of the second rod \((\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\).

Short answer

Answer: The thermal conductivity of Rod 2 is 100 W/m⋅K.

Step-by-step solution

Write the one-dimensional heat conduction equation

Since the rods lose heat by convection to the environment, we have the following one-dimensional heat conduction equation for both rods: \(q = k \cdot A \frac{dT}{dx} = h \cdot A \cdot (T_c - T_\infty)\) Where: - \(q\) is the heat transfer rate (\(W\)) - \(k\) is the thermal conductivity of the material (\(W/m\cdot K\)) - \(A\) is the cross-sectional area of the rod (\(m^2\)) - \(dT/dx\) is the temperature gradient along the rod (\(K/m\)) - \(h\) is the convection heat transfer coefficient (\(W/m^2\cdot K\)) - \(T_c\) is the temperature of the rod's surface at a specific position (\(K\)) - \(T_\infty\) is the surrounding air temperature (\(K\))

Set up equations for both rods

For Rod 1 (aluminum), the equation becomes: \(q_1 = k_1 \cdot A \frac{dT_1}{dx_1} = h\cdot A\cdot (T_{c1} - T_\infty)\) And for Rod 2 (unknown thermal conductivity), the equation is: \(q_2 = k_2 \cdot A \frac{dT_2}{dx_2} = h\cdot A\cdot (T_{c2} - T_\infty)\) We know that at \(x_1 = 40 cm\) and \(x_2 = 20 cm\), the temperature of Rod 1 and Rod 2 is the same, i.e. \(T_{c1} = T_{c2}\). We can rewrite the equations in terms of the temperature difference between the rods and their base: \(\frac{dT_1}{dx_1} = \frac{q_1}{k_1\cdot A} = \frac{h\cdot A\cdot (T_{c1} - T_\infty)}{k_1\cdot A}\) \(\frac{dT_2}{dx_2} = \frac{q_2}{k_2\cdot A} = \frac{h\cdot A\cdot (T_{c2} - T_\infty)}{k_2\cdot A}\)

Set up an equation relating \(k_1\) and \(k_2\)

Since we know the temperature of the rods is the same at the given distances, we can set the temperature gradients equal to each other and solve for the unknown \(k_2\): \(\frac{h\cdot A\cdot (T_{c1} - T_\infty)}{k_1\cdot A} = \frac{h\cdot A\cdot (T_{c2} - T_\infty)}{k_2\cdot A}\) Since \(T_{c1} = T_{c2}\), we can further simplify the equation: \(\frac{T_{c1} - T_\infty}{k_1} = \frac{T_{c2} - T_\infty}{k_2}\) We also know that the distances from the base are \(x_1 = 40 cm\) and \(x_2 = 20 cm\), so the temperature gradients can be written as: \(\frac{dT_1}{dx_1} = 2\cdot \frac{dT_2}{dx_2}\)

Solve for the unknown thermal conductivity \(k_2\)

Substituting the temperature gradients into the equation relating \(k_1\) and \(k_2\) gives us: \(2\cdot \frac{T_{c1} - T_\infty}{k_1} = \frac{T_{c2} - T_\infty}{k_2}\) Then, solving for \(k_2\), we get: \(k_2 = k_1 \cdot \frac{T_{c2} - T_\infty}{2\cdot (T_{c1} - T_\infty)}\) Since \(T_{c1} = T_{c2}\), the equation simplifies to: \(k_2 = \frac{k_1}{2}\) Given that \(k_1 = 200 W/m\cdot K\), the thermal conductivity of Rod 2, \(k_2\), is: \(k_2 = \frac{200}{2} = 100 W/m\cdot K\) Thus, the thermal conductivity of the second rod is 100 \(W/m\cdot K\).