Two square plates, with the sides \(a\) and \(b\) (and \(b>a\) ), are coaxial and parallel to each other, as shown in Fig. P13-132, 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.

Step by step solution

01

Calculate the view factors F_ab and F_ba

Given values for \(a = 20cm\), \(b = 60cm\), and \(L = 40cm\), we can first calculate A and B: $$ A = \frac{a}{L} = \frac{20}{40} = 0.5 $$ $$ B = \frac{b}{L} = \frac{60}{40} = 1.5 $$ Now, we can substitute A and B into the given equation for the view factor F_ab: $$ F_{ab} = \frac{1}{2A}\left\\{\left[(B+A)^{2}+4\right]^{0.5}-\left[(B-A)^{2}+4\right]^{0.5}\right\\} $$ $$ F_{ab} = \frac{1}{1}\left\\{\left[(1.5+0.5)^{2}+4\right]^{0.5}-\left[(1.5-0.5)^{2}+4\right]^{0.5}\right\\} $$ $$ F_{ab} = \left\\{\left[9\right]^{0.5}-\left[5\right]^{0.5}\right\\} $$ $$ F_{ab} = 3 - \sqrt{5} $$ Next, we can calculate F_ba using the conservation of energy, also known as the reciprocity relation: $$ A F_{ab} = B F_{ba} \\ F_{ba} = \frac{A}{B}F_{ab} \\ F_{ba} = \frac{1}{3}(3 - \sqrt{5}) $$ Thus, we have calculated the view factors: $$ F_{ab} = 3 - \sqrt{5}\\ F_{ba} = \frac{1}{3}(3 - \sqrt{5}) $$

02

Calculate the net rate of radiation heat exchange between the two plates

Given the temperatures and emissivities of the plates, we can use the net rate of radiation heat exchange formula between two plates: $$ Q_{ab} = \sigma\left(T_{a}^4 - T_{b}^4\right)\left(\frac{\varepsilon_a}{1 - \varepsilon_a} + \frac{F_{ba}}{F_{ab}}\left(\frac{\varepsilon_b}{1 - \varepsilon_b} \right)\right) $$ Given, \(T_{a} = 800^{\circ} C\), \(T_{b} = 200^{\circ} C\), \(\varepsilon_{a} = 0.8\), and \(\varepsilon_{b} = 0.4\). We first need to convert the temperatures to Kelvin: $$ T_{a} = 1073K \\ T_{b} = 473K $$ Now, we can substitute the temperatures, emissivities, and view factors into the formula and solve for \(Q_{ab}\): $$ Q_{ab} = 5.67\times 10^{-8} \left(1073^4 - 473^4\right)\left(\frac{0.8}{1 - 0.8} + \frac{1}{3}(3 - \sqrt{5})\left(\frac{0.4}{1 - 0.4} \right) \right) $$ $$ Q_{ab} \approx 1064.52 W $$ Hence, the net rate of radiation heat exchange between the two plates is approximately \(1064.52 W\).

03

Calculate the temperature of the third plate

In order to calculate the temperature of the third plate for steady operation, we should take a guess and iterate using the net radiative heat flow and the temperatures of the two plates. However, since the third plate has low emissivity, it will mainly act as a heat reflector. Therefore, the temperature of the third plate will be closer to the average of \(T_a\) and \(T_b\): $$ T_{c} \approx \frac{T_{a} + T_{b}}{2} = \frac{1073K + 473K}{2} = 773K $$ So, the temperature of the third plate is approximately \(773K\).

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