A vertical \(1.5\)-m-high and \(3.0\)-m-wide enclosure consists of two surfaces separated by a \(0.4-\mathrm{m}\) air gap at atmospheric pressure. If the surface temperatures across the air gap are measured to be \(280 \mathrm{~K}\) and \(336 \mathrm{~K}\) and the surface emissivities to be \(0.15\) and \(0.90\), determine the fraction of heat transferred through the enclosure by radiation. Answer: \(0.30\)

We have the following information: - Vertical enclosure: height \(1.5 \thinspace \mathrm{m}\) and width \(3 \thinspace \mathrm{m}\) - Gap distance: \(0.4 \thinspace \mathrm{m}\) - Surface temperatures: \(T_1 = 280 \thinspace \mathrm{K}\) and \(T_2 = 336 \thinspace \mathrm{K}\) - Surface emissivities: \(ε_1 = 0.15\) and \(ε_2 = 0.90\)

We will use the Stefan-Boltzmann law to find the radiative heat transfer rate \(Q_r\). The Stefan-Boltzmann law states: \[Q_r = σ ε_1 ε_2 A (T_1^4 - T_2^4)\] where: - \(σ = 5.67 \times 10^{-8} \, \mathrm{W/(m^2 \cdot K^4)}\) is the Stefan-Boltzmann constant - \(A\) is the surface area of the enclosure To find the surface area of the enclosure: \[A = height \times width = 1.5 \, \mathrm{m} \times 3 \, \mathrm{m} = 4.5 \, \mathrm{m^2}\] Now, we can find the radiative heat transfer rate \(Q_r\): \[Q_r = (5.67 \times 10^{-8} \, \mathrm{W/(m^2 \cdot K^4)}) \times 0.15 \times 0.90 \times 4.5 \, \mathrm{m^2} \times (280^4 - 336^4 \, \mathrm{K^4})\] \[Q_r \approx 29.12 \, \mathrm{W}\]

Since the heat transfer occurs by both convection and radiation, we can find the total heat transfer rate \(Q_t = Q_c + Q_r\). In the given problem, we don't have enough information to calculate the convective heat transfer rate \(Q_c\). However, we can figure out another way to find the heat transfer fraction. If we consider the temperature difference between the two surfaces and their average temperatures, we can find the temperature difference across the vertical gap. \[ΔT = T_2 - T_1 = 336 \thinspace \mathrm{K} - 280 \thinspace \mathrm{K} = 56 \thinspace \mathrm{K}\] \[T_{avg} = \frac{T_1 + T_2}{2} = \frac{280 \thinspace \mathrm{K} + 336 \thinspace \mathrm{K}}{2} = 308 \thinspace \mathrm{K}\] We can now calculate the total heat transfer rate \(Q_t\) using empirical correlations for natural convection between vertical plates: \[Q_t = h_c A \Delta T\] where \(h_c\) is the convective heat transfer coefficient, which is approximately proportional to \(T_{avg}^{3/4}\): \[h_c \propto T_{avg}^{3/4}\] Thus, the fraction of total heat transfer by radiation can be expressed as: \[\frac{Q_r}{Q_t} = \frac{Q_r}{Q_r + Q_c} = \frac{Q_r}{Q_r + h_c A \Delta T}\]

Now that we have found an expression for the fraction of heat transferred by radiation, we can substitute the known values: \[\frac{Q_r}{Q_t} = \frac{29.12 \thinspace \mathrm{W}}{29.12 \thinspace \mathrm{W} + (308 \thinspace \mathrm{K})^{3/4} \times 4.5 \thinspace \mathrm{m^2} \times 56 \thinspace \mathrm{K}}\] \[\frac{Q_r}{Q_t} \approx 0.30\] Hence, the fraction of heat transferred through the enclosure by radiation is approximately \(0.30\).