Question

A 90 -cm-diameter flat black disk is placed in the center of the top surface of a \(1-m \times 1-m \times 1-m\) black box. The view factor from the entire interior surface of the box to the interior surface of the disk is (a) \(0.07\) (b) \(0.13\) (c) \(0.26\) (d) \(0.32\) (e) \(0.50\)

Short answer

Based on the calculations of the view factor for each choice, the highest value of the view factor is achieved in option (e) with a value of 0.50. Therefore, the correct answer is (e) 0.50.

Step-by-step solution

Find the surface area of the disk

The diameter of the disk is 90 cm, so the radius is 45 cm or 0.45 m. The surface area of the disk can be calculated using the formula for the area of a circle: $$A_{disk} = \pi (radius)^2 = \pi (0.45)^2 = 0.6362 m^2$$

Find the surface area of the interior surface of the black box

The black box has dimensions \(1 \, m \times 1 \, m \times 1 \, m\). As it is a cube, each face has an area of \(1 m^2\), and since it has six faces, the total surface area is: $$A_{box} = 6(1) = 6 \, m^2$$ Now we can use the view factor formula and given choices to find the correct view factor.

Check each choice

We will check each choice one by one using the view factor formula: (a) \(F_{1 \to 2} = 0.07\) $$A_1 F_{1 \to 2} = A_{disk} = 0.6362$$ $$A_2 = \frac{A_1 F_{1 \to 2}}{F_{1 \to 2}} = \frac{0.6362}{0.07} = 9.0886$$ If (a) is correct, the surface area of the box should be equal to \(9.0886 \, m^2\). However, this is not true as the surface area of the box is \(6 \, m^2\). So (a) is incorrect. (b) \(F_{1 \to 2} = 0.13\) $$A_2 = \frac{A_1 F_{1 \to 2}}{F_{1 \to 2}} = \frac{0.6362}{0.13} = 4.8938$$ This is close to the surface area of the black box but it is not equal so then this is also incorrect. (c) \(F_{1 \to 2} = 0.26\) $$A_2 = \frac{A_1 F_{1 \to 2}}{F_{1 \to 2}} = \frac{0.6362}{0.26} = 2.4469$$ This option is also not equal to the surface area of the black box, so it is incorrect. (d) \(F_{1 \to 2} = 0.32\) $$A_2 = \frac{A_1 F_{1 \to 2}}{F_{1 \to 2}} = \frac{0.6362}{0.32} = 1.9881$$ This choice is also not equal to the surface area of the black box, so it is incorrect. (e) \(F_{1 \to 2} = 0.50\) $$A_2 = \frac{A_1 F_{1 \to 2}}{F_{1 \to 2}} = \frac{0.6362}{0.50} = 1.2724$$ This choice is also not equal to the surface area of the black box, so it is incorrect. However, we made a mistake! We should have used the formula: $$F_{1 \to 2} = \frac{A_1 F_{1 \to 2}}{A_2}$$

Correctly check each choice

When checking each choice, we should directly compute the \(A_1 F_{1 \to 2}\). Here, \(A_1 = A_{disk} = 0.6362\). Let's redo the calculation: (a) \(F_{1 \to 2} = 0.07\) $$A_1 F_{1 \to 2} = A_{disk} \times 0.07 = 0.6362 \times 0.07 = 0.0445$$ (b) \(F_{1 \to 2} = 0.13\) $$A_1 F_{1 \to 2} = A_{disk} \times 0.13 = 0.6362 \times 0.13 = 0.0827$$ (c) \(F_{1 \to 2} = 0.26\) $$A_1 F_{1 \to 2} = A_{disk} \times 0.26 = 0.6362 \times 0.26 = 0.1654$$ (d) \(F_{1 \to 2} = 0.32\) $$A_1 F_{1 \to 2} = A_{disk} \times 0.32 = 0.6362 \times 0.32 = 0.2036$$ (e) \(F_{1 \to 2} = 0.50\) $$A_1 F_{1 \to 2} = A_{disk} \times 0.50 = 0.6362 \times 0.50 = 0.3181$$ Now, we have the \(A_1F_{1 \to 2}\) values in each case. As we can see, our previously computed values were not relevant to the view factor. Now, we need to find the choice where the ratio between \(A_1F_{1 \to 2}\) and \(A_2\) is maximum, i.e., the highest value of \(F_{1 \to 2}\) is achieved. In this case, the highest value is for option (e) \(F_{1 \to 2} = 0.50\). So, the correct answer is (e) 0.50.