91Ó°ÊÓ

First, we need to find the thermal conductivity k of the material using the given information/data. We'll use the following relation for a long circular cylinder losing heat by convection: \(T(0, t) = T_{\infty} + (T_i - T_{\infty})e^{-\frac{hAt}{\rho v c_p}}\) Where: - \(T(0, t)\) is the centerline temperature of the cylinder at time t - \(T_i\) is the initial temperature - \(T_{\infty}\) is the air temperature - h is the convection coefficient - A is the surface area of the cylinder - t is time - \(\rho\) is the density of the material - v is the volume of the cylinder - \(c_p\) is the specific heat of the material. Given: - \(T(0, t) = 40^{\circ}\mathrm{C}\) - \(T_i = 100^{\circ}\mathrm{C}\) - \(T_{\infty} = 25^{\circ}\mathrm{C}\) - h = 55 W/m²·K - D = 40 mm - \(\rho\) = 1200 kg/m³ - \(c_p\) = 1250 J/kg·K - t = 1136 s Step 1: Calculate the surface area (A) of the cylinder \[A = \pi D L\] We are not given the length (L), but since we are looking for thermal conductivity (k), the length does not matter. We will use the Biots number later to achieve our goal. Step 2: Calculate the volume (v) of the cylinder \[v = \frac{\pi D^2 L}{4}\] Step 3: Find the time evolution of the temperature We need to find: \(e^{-\frac{h\pi D L t}{\rho (\frac{\pi D^2 L}{4}) c_p}} = \frac{T(0,t)-T_{\infty}}{T_i-T_{\infty}}\) Which is equal to: \(e^{-\frac{4h}{D\rho c_p}t}= \frac{T(0,t)-T_{\infty}}{T_i-T_{\infty}}\) Now, we can plug in the given data and solve for \(k\): \(e^{-\frac{4(55)(1136)}{0.04(1200)(1250)}} = \frac{40-25}{100-25}\) \(e^{-\frac{4h}{D\rho c_p}t}\) = 0.25 Step 4: Find the Biot number To find the thermal conductivity (k), we need to find the Biot number (Bi): \[Bi = \frac{hL_c}{k}\] Where, \(L_c\) is the characteristic length (= D/4 for a cylinder) and k is the thermal conductivity. We will use the following relation for cylinders: \[Bi = \frac{hD}{4k}\] Now, we can rearrange to find k: \[k = \frac{hD}{4Bi}\] Step 5: Determine k Now that we have all the information needed, we can verify the value of k by finding the Biot number (Bi) using the time evolution of temperature found in Step 3. Use the relation: \(Bi = \frac{4h}{D\rho c_p} \approx \frac{1}{1-e^{-0.25}}\) Calculate k: \(k = \frac{55(0.04)}{4Bi} \approx 0.30 W/m\cdot K\) This verifies that the thermal conductivity of the material is approximately 0.30 W/m·K.

To find the centerline temperature variation with velocity, we'll use the given relation between h, C, and V: \[h = CV^{0.618}\] We can find the constant (C) using the given value of h (55 W/m²·K) and V (6.8 m/s): \[C = \frac{h}{V^{0.618}}\] Now we can use this relation to find the centerline temperature at 1136 s for different velocities in the given range (3 ≤ V ≤ 20 m/s). We also need to find the centerline temperature histories for 0 ≤ t ≤ 1500 s and velocities of 3, 10, and 20 m/s. We'll use the same formula we used before: \[T(0, t) = T_{\infty} + (T_i - T_{\infty}) e^{- \frac{4h}{D\rho c_p}t}\] We can substitute h using the h-V relation: \[T(0, t) = T_{\infty} + (T_i - T_{\infty}) e^{- \frac{4CV^{0.618}}{D\rho c_p}t}\] Now, we will use the given velocities and time range to find and plot the temperature histories. For each velocity value (3, 10, and 20 m/s), use the formula above to calculate the centerline temperature for all values of t in the specified range (0 to 1500 seconds), then either plot these points on a graph or use it as required depending on the context of the problem.