Question

A large number of long tubes, each of diameter \(D\), are placed parallel to each other and at a center-to-center distance of s. Since all of the tubes are geometrically similar and at the same temperature, these could be treated collectively as one surface \(\left(A_{j}\right)\) for radiation heat transfer calculations. As shown in Fig. P13-140, the tube bank \(\left(A_{j}\right)\) is placed opposite a large flat wall \(\left(A_{j}\right)\) such that the tube bank is parallel to the wall. (a) Calculate the view factors \(F_{i j}\) and \(F_{j i}\) for $s=3.0 \mathrm{~cm}\( and \)D=1.5 \mathrm{~cm}$. (b) Calculate the net rate of radiation heat transfer between the wall and the tube bank per unit area of the wall when $T_{i}=900^{\circ} \mathrm{C}, T_{j}=60^{\circ} \mathrm{C}, \varepsilon_{i}=0.8\(, and \)\varepsilon_{j}=0.9$. (c) A fluid flows through the tubes at an average temperature of $40^{\circ} \mathrm{C}\(, resulting in a heat transfer coefficient of \)2.0 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Assuming \)T_{i}=900^{\circ} \mathrm{C}, \varepsilon_{i}=0.8\( and \)\varepsilon_{j}=0.9$ (as above) and neglecting the tube wall thickness and convection from the outer surface, calculate the temperature of the tube surface in steady operation.

Short answer

Question: Calculate the view factors, the net rate of radiation heat transfer between the wall and the tube bank per unit area of the wall, and the steady operation temperature of the tube surface considering convection. Answer: The view factors are \(F_{ij} = 0.2255\) and \(F_{ji} = 0.2255\). The net rate of radiation heat transfer between the wall and the tube bank per unit area of the wall is \(3910.33 W/m^2\). The steady operation temperature of the tube surface considering convection is \(43.92^{\circ}C\).

Step-by-step solution

Part A: Calculate View Factors

: To find the view factors \(F_{ij}\) and \(F_{ji}\), we will use formula for parallel rectangular surfaces and then adjust them for cylindrical geometry. View factor for parallel rectangular surfaces is: \(F_{ij} = \frac{1}{\pi D}\tan^{-1}(\frac{2sZ}{D\sqrt{4s^2 + D^2}})\) Where \(Z = D + s\). Given \(s=3.0 cm\) and \(D=1.5 cm\), we first find \(Z\): \(Z = D + s = 1.5 cm + 3.0 cm = 4.5 cm\) Now, we can calculate \(F_{ij}\): \(F_{ij} = \frac{1}{\pi\cdot1.5cm}\tan^{-1}(\frac{2\cdot3.0cm\cdot4.5cm}{1.5cm\cdot\sqrt{4\cdot(3.0cm^2) + (1.5cm)^2}}) = 0.2255\) To find \(F_{ji}\), we use the reciprocity relation: \(F_{ji} = \frac{A_{i}}{A_{j}}F_{ij}= \frac{1}{\frac{A_{j}}{A_i}}F_{ij} = \frac{F_{ij}}{\frac{A_{j}}{A_{i}}}\) Since the tubes are long and the wall is large, we can assume their areas are equal (\(A_{i} = A_{j}\)): \(F_{ji} = \frac{F_{ij}}{1}= F_{ij} = 0.2255\) So, the view factors are \(F_{ij} = 0.2255\) and \(F_{ji} = 0.2255\).

Part B: Calculate Net Rate of Radiation Heat Transfer

: We have the view factors \(F_{ij}\) and \(F_{ji}\), and are given temperatures \(T_i\) and \(T_j\) and emissivities \(\varepsilon_{i}\) and \(\varepsilon_{j}\). Using the equation for net radiation heat transfer, we have: \(q = \sigma \varepsilon_{ij} \frac{F_{ij}A_{i}(T_{i}^{4} - T_{j}^{4})}{1 - (1-\varepsilon_{i})(1-\varepsilon_{j})}\) Where \(\sigma = 5.67\cdot10^{-8}W/m^2K^4\) is the Stefan-Boltzmann constant, \(\varepsilon_{ij} = \frac{\varepsilon_{i}\varepsilon_{j}}{\varepsilon_{i}+\varepsilon_{j}-\varepsilon_{i}\varepsilon_{j}}\) and we need to convert temperatures to Kelvin. \(T_{i} = 900^{\circ}C + 273.15K = 1173.15K\) \(T_{j} = 60^{\circ}C + 273.15K = 333.15K\) Now we find \(\varepsilon_{ij}\): \(\varepsilon_{ij} = \frac{0.8\cdot0.9}{0.8+0.9-0.8\cdot0.9} = 0.8571\) Plugging all values into the equation for net radiation heat transfer: \(q=\frac{5.67\cdot10^{-8}W/m^2K^4 \cdot 0.8571 \cdot 0.2255 [(1173.15K)^4-(333.15K)^4]}{1-(1-0.8)(1-0.9)} = 3910.33W/m^2\) So, the net rate of radiation heat transfer between the wall and the tube bank per unit area of the wall is \(3910.33 W/m^2\).

Part C: Calculate Tube Surface Temperature

: To calculate the tube surface temperature, we need to consider the convective heat transfer as well. The given convective heat transfer coefficient is \(2.0 kW/m^2K\), and the fluid temperature is \(40^{\circ}C\). We can set up an energy balance equation to find the tube surface temperature \(T_s\): \(q_{rad} = q_{conv}\) Where \(q_{rad}\) is the net radiation heat transfer (found in Part B) and \(q_{conv}\) is the convective heat transfer. \(q_{conv} = h(T_s - T_{f})\) Where \(h = 2.0 kW/m^2K\) is the heat transfer coefficient and \(T_f = 40^{\circ}C = 313.15K\). Now, solving for \(T_s\): \(3910.33W/m^2 = 2000 W/m^2K(T_s - 313.15K)\) \(T_s = \frac{3910.33W/m^2}{2000 W/m^2K} + 313.15K = 317.07K\) The steady operation temperature of the tube surface is \(317.07K = 43.92^{\circ}C\).