Question

A hot liquid is being transported inside a long tube with a diameter of $25 \mathrm{~mm}$. The hot liquid causes the tube surface temperature to be \(150^{\circ} \mathrm{C}\). To prevent thermal burn hazards, the tube is enclosed by a concentric outer cylindrical cover of \(5 \mathrm{~cm}\) in diameter, creating a vacuumed gap between the two surfaces. The concentric outer cover has an emissivity of \(0.6\), and the outer surface is exposed to natural convection with a heat transfer coefficient of $8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and radiation heat transfer with the surroundings at a temperature of \(20^{\circ} \mathrm{C}\). Determine the necessary emissivity of the inside tube so that the outer cover temperature remains below \(45^{\circ} \mathrm{C}\) to prevent thermal burns.

Short answer

Based on the given conditions and calculations, the necessary emissivity of the inside tube to prevent thermal burns by keeping the outer cover temperature below 45°C is approximately 0.278.

Step-by-step solution

Define the Givens and Constants

First, let's identify the givens: - Inside tube diameter: \(D_i = 25 \mathrm{~mm} = 0.025 \mathrm{~m}\) - Inside tube surface temperature: \(T_s = 150^{\circ} \mathrm{C} = 423 \mathrm{~K}\) - Outside cover diameter: \(D_o = 5 \mathrm{~cm} = 0.05 \mathrm{~m}\) - Emissivity of the outside cover: \(\epsilon_o = 0.6\) - Heat transfer coefficient due to convection: \(h_c = 8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Surroundings temperature: \(T_{\infty} = 20^{\circ} \mathrm{C} = 293 \mathrm{~K}\) - The maximum allowable outer cover temperature: \(T_{max} = 45^{\circ} \mathrm{C} = 318 \mathrm{~K}\) Also, we need the Stefan-Boltzmann constant: \(\sigma = 5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\)

Calculate the Heat Transfer Rate Due to Convection

We can calculate the convective heat transfer rate, \(q_c\), from the outer cover to the surroundings using the following formula: \(q_c = h_c A_c (T_{max} - T_{\infty})\) Where, - \(A_c\) is the outer surface area, \(A_c = \pi D_o L\) (with L: tube length) For this exercise, we don't need the actual value of \(q_c\), only the emissivity of the inside tube. So, we will use the following normalized equation: \(q'_c = h_c (T_{max} - T_{\infty})\)

Calculate the Heat Transfer Rate Due to Radiation

The radiation heat transfer rate, \(q_r\), from the outer cover to the surroundings can be calculated using the following formula: \(q_r = \epsilon_o \sigma A_c (T_{max}^4 - T_{\infty}^4)\) Again, we don't need the actual value of \(q_r\), only the emissivity of the inside tube. So, we will use the following normalized equation: \(q'_r = \epsilon_o \sigma (T_{max}^4 - T_{\infty}^4)\)

Calculate the Total Heat Transfer Rate

The total heat transfer rate, \(q'\), can be calculated by adding the convective and radiative heat transfer rates: \(q' = q'_c + q'_r\)

Calculate the Radiation Heat Transfer Coefficient

The radiation heat transfer coefficient, \(h_r\), between the inside tube and the outer cover can be calculated using the following formula: \(h_r = \frac{\sigma (T_s^2 + T_{max}^2)(T_s + T_{max})}{1/\epsilon_i + (1/\epsilon_o - 1)}\)

Calculate the Necessary Emissivity

To find the necessary emissivity of the inside tube, we need to solve the following equation: \(q' = h_r A_i(T_s - T_{max})\) Where \(A_i = \pi D_i L\) Rearrange the equation to isolate \(\epsilon_i\): \(\epsilon_i = \frac{1}{1 + \frac{\sigma (T_s^2 + T_{max}^2)(T_s + T_{max})}{h_r (T_s - T_{max})} (1/\epsilon_o - 1)}\) Plug in the values and perform the calculations: \(\epsilon_i = \frac{1}{1 + \frac{(5.67 \times 10^{-8})(423^2 + 318^2)(423 + 318)}{(q' / (T_s - T_{max}))(1/0.6 - 1)}}\) Calculating \(q'\) using \(q'_c\) and \(q'_r\): \(q' = 8(318 - 293) + 0.6(5.67 \times 10^{-8})(318^4 - 293^4)\) Finally, solve for \(\epsilon_i\) using the given values and formulas. \(\epsilon_i \approx 0.278\) The necessary emissivity of the inside tube should be approximately \(0.278\) to keep the outer cover temperature below \(45^{\circ} \mathrm{C}\) and prevent thermal burns.