Question

Two square plates, with the sides \(a\) and \(b\) (and \(b>a\) ), are coaxial and parallel to each other, as shown in Fig. P13-142, and they are separated by a center-to-center distance of \(L\). The radiation view factor from the smaller to the larger plate, \(F_{a b+}\) is given by $$ F_{a b}=\frac{1}{2 A}\left\\{\left[(B+A)^{2}+4\right]^{0.5}-\left[(B-A)^{2}+4\right]^{0.5}\right\\} $$ where, \(A=a / L\) and \(B=b / L\). (a) Calculate the view factors \(F_{a b}\) and \(F_{b a}\) for $a=20 \mathrm{~cm}\(, \)b=60 \mathrm{~cm}\(, and \)L=40 \mathrm{~cm}$. (b) Calculate the net rate of radiation heat exchange between the two plates described above if \(T_{a}=800^{\circ} \mathrm{C}\), $T_{b}=200^{\circ} \mathrm{C}, \varepsilon_{a}=0.8\(, and \)\varepsilon_{b}=0.4$. (c) A large, square plate (with the side $c=2.0 \mathrm{~m}, \varepsilon_{c}=0.1$, and negligible thickness) is inserted symmetrically between the two plates such that it is parallel to and equidistant from them. For the data given above, calculate the temperature of this third plate when steady operating conditions are established.

Short answer

Question: Calculate the view factors F_ab and F_ba, the net rate of radiation heat exchange between the two plates, and the temperature of the third plate when steady operating conditions are established. Answer: The view factors F_ab and F_ba are: - \(F_{ab} = 2(2\sqrt{2}-\sqrt{5})\) - \(F_{ba} = \frac{2(2\sqrt{2}-\sqrt{5})\times(20)^2}{(60)^2}\) The net rate of radiation heat exchange between the two plates is approximately 11371.35 W. The temperature of the third plate when steady operating conditions are established is approximately 743.08 K.

Step-by-step solution

Calculate A and B

First, we will calculate values for A and B using the given formulas before finding view factors. Recall that \(A = \frac{a}{L}\) and \(B = \frac{b}{L}\), where \(a = 20 cm\), \(b = 60 cm\), and \(L = 40 cm\). \(A = \frac{20}{40} = 0.5\) \(B = \frac{60}{40} = 1.5\)

Calculate the radiation view factor F_ab

Now, we can use the provided formula for the radiation view factor \(F_{ab}\) by substituting the values of A and B obtained above. \(F_{ab} = \frac{1}{2A} \left\\{ \left[(B+A)^{2}+4\right]^{0.5}-\left[(B-A)^{2}+4\right]^{0.5} \right\\} = \frac{1}{2(0.5)}\left\\{\left[(1.5+0.5)^{2}+4\right]^{0.5}-\left[(1.5-0.5)^{2}+4\right]^{0.5}\right\\} = 2\left( \sqrt{4+4}-\sqrt{1+4}\right) = 2(2\sqrt{2}-\sqrt{5})\)

Calculate the other view factor F_ba using energy balance

As per energy balance, the following equation holds true: \(F_{ba} A_2 = F_{ab} A_1\) \(F_{ba} = \frac{F_{ab} A_1}{A_2} = \frac{F_{ab} a^2}{b^2} = \frac{2(2\sqrt{2}-\sqrt{5})\times(20)^2}{(60)^2}\) For part (a), the view factors \(F_{ab}\) and \(F_{ba}\) are: \(F_{ab} = 2(2\sqrt{2}-\sqrt{5})\) \(F_{ba} = \frac{2(2\sqrt{2}-\sqrt{5})\times(20)^2}{(60)^2}\)

Calculate net rate of radiation heat exchange between the two plates

In order to compute the net rate of radiation heat exchange (Q) between the two plates, we first need to find the radiosity factor. We use the formula: \(Q = A_1 \sigma F_{ab} (T_a^4 - T_b^4)\) We are given \(\sigma = 5.67 \times 10^{-8} W/(m^2 K^4)\), \(T_a=800^{\circ}C\), \(T_b=200^{\circ}C\). We convert \(T_a\) and \(T_b\) to Kelvin: \(T_a = 800 + 273.15 = 1073.15K\) \(T_b = 200 + 273.15 = 473.15K\) Then, we calculate the net rate of radiation heat exchange: \(Q = (0.2^2) \times (5.67 \times 10^{-8}) \times 2(2\sqrt{2}-\sqrt{5}) (1073.15^4 - 473.15^4) \approx 11371.35 W\) For part (b), the net rate of radiation heat exchange between the two plates is approx \(11371.35 W\).

Calculate the temperature of the third plate when steady operating conditions are established

This is represented by the energy balance equation: \(A_1 \sigma F_{ac} (T_a^4 - T_c^4) = A_1 \sigma F_{cb} (T_c^4 - T_b^4)\) In this case, the view factors F_ac and F_cb can be calculated as \(F_{ac} = F_{ab}\frac{A_1}{A_1 + A_3}\), and \(F_{cb} = F_{ab}\frac{A_2}{A_2 + A_3}\), where A_3 is the area of the third plate. We are given the third plate's side (\(c=2.0m\)). Therefore, \(A_3 = 2^2 = 4m^2\). Therefore, the view factors can be calculated as follows: \(F_{ac} = 2(2\sqrt{2}-\sqrt{5})\frac{0.2^2}{0.2^2+4}\) \(F_{cb} = 2(2\sqrt{2}-\sqrt{5})\frac{0.6^2}{0.6^2+4}\) Finally, the energy balance equation becomes: \((0.2^2) \times (5.67 \times 10^{-8}) \times F_{ac} (1073.15^4 - T_c^4) = (0.2^2) \times (5.67 \times 10^{-8}) \times F_{cb} (T_c^4 - 473.15^4)\) By solving for \(T_c\), we get: \(T_c \approx 743.08 K\) For part (c), the temperature of the third plate when steady operating conditions are established is approx \(743.08 K\).