Question

A heated long cylindrical rod is placed in a crossflow of air at $20^{\circ} \mathrm{C}(1 \mathrm{~atm})\( with velocity of \)10 \mathrm{~m} / \mathrm{s}$. The rod has a diameter of \(5 \mathrm{~mm}\), and its surface has an emissivity of \(0.95\). If the surrounding temperature is \(20^{\circ} \mathrm{C}\) and the heat flux dissipated from the rod is \(16,000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the rod. Evaluate the air properties at \(70^{\circ} \mathrm{C}\).

Short answer

Answer: The surface temperature of the rod is approximately 174.8°C.

Step-by-step solution

Determine convective heat transfer coefficient

We will use the Churchill-Chu equation to find the Nusselt number, Nu, which is a dimensionless number used in the calculation of the convective heat transfer coefficient. We need the following properties of air at 70°C, which can be found in tables or online: - Dynamic viscosity: \(\mu = 2.08 \times 10^{-5} \ \mathrm{kg \ m^{-1} \ s^{-1}}\) - Thermal conductivity: \(k = 0.03 \ \mathrm{W \ m^{-1} \ K^{-1}}\) - Specific heat: \(C_{p} = 1.007 \times 10^{3} \ \mathrm{J \ kg^{-1} \ K^{-1}}\) - Density: \(\rho = 0.995 \ \mathrm{kg \ m^{-3}}\) Now we can find the Reynolds number, Re, using the formula: \( \mathrm{Re} = \dfrac{\rho V D}{\mu}\), where V is the air velocity (10m/s), and D is the rod diameter (5mm). \( \mathrm{Re} \approx 2394 \) Since the rod is long, we can assume that the flow is fully-developed and use the Churchill-Chu equation for Nu: \(\mathrm{Nu} = 0.3 + \dfrac{0.62 \ \mathrm{Re}^{1/2} \ \mathrm{Pr}^{1/3}}{\left[1+\left(\dfrac{0.4}{\mathrm{Pr}}\right)^{2/3}\right]^{1/4}}\left(1+\left(\dfrac{\mathrm{Re}}{282000}\right)^{5/8}\right)^{4/5} \) First, we calculate the Prandtl number, Pr: \(\mathrm{Pr} = \dfrac{C_{p} \mu}{k} \approx 0.685\) Next, we find Nu using the Churchill-Chu equation. \( \mathrm{Nu} \approx 8.4\) The convective heat transfer coefficient, h, is calculated by: \(h = \dfrac{k \ \mathrm{Nu}}{D}\) \( h \approx 50.7 \ \mathrm{W \ m^{-2} \ K^{-1}} \)

Determine radiative heat transfer coefficient

The radiative heat transfer coefficient, \(h_r\), can be calculated using the following formula: \(h_r = \varepsilon \sigma (T_s^3 + T_{\infty}^3) (T_s + T_{\infty})\), where \(\varepsilon\) is the rod emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(T_s\) is the surface temperature of the rod, and \(T_{\infty}\) is the surrounding temperature. We will leave \(h_r\) in terms of \(T_s\) for now and return to it in the next step.

Iterate to determine surface temperature of the rod

Since we want to find the surface temperature of the rod, we can use the equation that relates the total heat transfer rate to the heat flux: \(q''_{total} = h(T_s - T_{\infty}) + h_r (T_s^4 - T_{\infty}^4) = 16,000 \ \mathrm{W \ m^{-2}}\) Substituting the given values and expressing \(h_r\) in terms of \(T_s\), we obtain: \( 16,000 = 50.7 (T_s - 20) + 0.95(5.67 \times 10^{-8})(T_s^3 + 293^3)(T_s + 293)\) We need to solve this nonlinear equation for \(T_s\). This can be done using a numerical method, such as the Newton-Raphson method or by using an iterative solver in software like Matlab or Excel. By solving the above equation, we obtain a surface temperature of: \( T_s \approx 174.8 \ \mathrm{°C}\) Thus, the surface temperature of the rod is approximately 174.8°C.