Question

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

Short answer

#Question#: Given the following information: (a) Two square parallel plates with sides a and b, separated by distance L. a = 20 cm, b = 60 cm, L = 40 cm. (b) Equation for the view factor F_ab. (c) Temperatures and emissivities of the plates: T_a = 800 °C, T_b = 200 °C, ε_a = 0.8, ε_b = 0.4. Calculate the following: i) The view factors F_ab and F_ba. ii) The net rate of radiation heat exchange between the given plates. iii) The temperature of a third plate inserted between the given plates for steady operation. #Answer#: i) The view factors are: F_ab = 3 - √5 F_ba = (1/3)(3 - √5) ii) The net rate of radiation heat exchange between the two plates is approximately 1064.52 W. iii) The temperature of the third plate is approximately 773K.

Step-by-step solution

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: $$ F_{ab} = 3 - \sqrt{5}\\ F_{ba} = \frac{1}{3}(3 - \sqrt{5}) $$

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 \(1064.52 W\).

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 \(773K\).

Key concepts

View Factor Calculation

View factors play a crucial role in radiation heat transfer. They altogether determine how much energy emitted from one surface reaches another. For instance, between two parallel plates, view factors are used to calculate the fraction of radiation leaving one surface and arriving at the other. To find the view factor from the smaller to the larger plate (\( F_{ab} \)), we use the given formula:

• \( A = \frac{a}{L} \)
• \( B = \frac{b}{L} \)
• Substitute these into the view factor equation.

Substituting the values for A and B, you will find \( F_{ab} = 3 - \sqrt{5} \). Understanding the conservation of energy helps us find the reverse view factor (\( F_{ba} \)), using the relationship:

• \( A F_{ab} = B F_{ba} \)
• Solve for \( F_{ba} \), which gives \( \frac{1}{3}(3 - \sqrt{5}) \).

This symmetry ensures energy balance in radiation transfer between the two plates.

Radiative Heat Exchange

Radiative heat exchange is essential to understand how energy moves between surfaces due to temperature differences. It includes factors like emissivity and temperature. The net rate of radiation heat exchange between two plates can be expressed with a specific formula:

• Calculate the temperatures in Kelvin.
• Substitute the temperatures and emissivities into the heat exchange formula.

By substituting \( T_{a} = 1073K \) and \( T_{b} = 473K \), along with their corresponding emissivities (\( \varepsilon_a = 0.8 \), \( \varepsilon_b = 0.4 \)), we solve for \( Q_{ab} \):

• The answer is approximately \( 1064.52 W \).

This power denotes the energy radiatively exchanged between the plates, emphasizing the significance of thermal radiation in heat transfer calculations.

Finite Surfaces Energy Balance

The concept of energy balance for finite surfaces ensures that energy entering a system equals the energy leaving it. In our exercise, the third plate's temperature is determined using this balance. The third plate acts as an intermediate surface between the two plates:

• With low emissivity (\( \varepsilon_c = 0.1 \)), it reflects most of the energy.
• Its temperature stabilizes around the mean temperature of the other plates.

For a quick estimation in scenarios with similar conditions:

• \( T_c \approx \frac{T_{a} + T_{b}}{2} \)
• Substitute known temperatures to find \( T_c \approx 773K \).

This approximation derives from a reflective surface being less effective at emitting radiation itself, maintaining a balance based on surrounding temperatures.