91Ó°ÊÓ

The total heat transfer involves both convection and radiation. The heat transfer due to convection can be calculated using the formula \(q_{conv} = hA(T_{surf} - T_{air})\), where \(q_{conv}\) is the convective heat transfer, h is the heat transfer coefficient, A is the surface area of the plate glass, \(T_{surf}\) is the surface temperature of the glass and \(T_{air}\) is the air temperature. For radiation, we can use the formula \(q_{rad} = \epsilon \sigma A (T_{surf}^4 - T_{surroundings}^4)\), where \(q_{rad}\) is the radiative heat transfer, \(\epsilon\) is the surface emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\)), and \(T_{surroundings}\) is the temperature of the surroundings (which is equal to \(T_{air}\)). The total heat transfer will be the sum of these two, i.e., \(q_{total} = q_{conv} + q_{rad}\).

Fourier's law relates the heat transfer in a solid to the temperature gradient: \(q_{total} = -kA\frac{dT}{dx}\), where k is the thermal conductivity of the glass, and \(\frac{dT}{dx}\) is the temperature gradient. Rearranging for the temperature gradient, we have \(\frac{dT}{dx} = -\frac{q_{total}}{kA}\).

We have to ensure the temperature gradient does not exceed the given limit of \(15^{\circ} \mathrm{C} / \mathrm{mm}\). Set the maximum temperature gradient equal to the limit, and solve for the corresponding air temperature: \(\frac{dT}{dx} = 15^{\circ} \mathrm{C} / \mathrm{mm} \rightarrow T_{air} = T_{surf} - \frac{k_{glass}}{hA}\frac{15^{\circ} \mathrm{C} / \mathrm{mm}}{q_{total}}\)

Substitute the given values into the equation derived in Step 3 and solve for the air temperature: \(T_{air} = 600^{\circ} \mathrm{C} - \frac{1.4 \mathrm{W/m\cdot K}}{5 \mathrm{W/m^2\cdot K}}\frac{15^{\circ} \mathrm{C} / \mathrm{mm}}{q_{total}}\). Since the glass thickness was not given, we assume A = 1 m², that simplifies the expression to \(T_{air} = 600^{\circ} \mathrm{C} - \frac{1.4}{5} \frac{15}{q_{total}}\). Now, the minimum allowed \(T_{air}\) is when \(q_{total}\) is maximum, and since both \(q_{conv}\) and \(q_{rad}\) are positive, then the maximum heat transfer that occurs will be when there are no radiation heat transfer, i.e., \(q_{total} = hA(T_{surf} - T_{air})\). Thus, the expression becomes \(T_{air} = 600^{\circ} \mathrm{C} - \frac{1.4}{5} \frac{15}{h(600^{\circ} \mathrm{C} - T_{air})}\). Solve for \(T_{air}\), you'll get \(T_{air} \approx 579^{\circ} \mathrm{C}\). Therefore, the lowest air temperature that can initially be used for cooling without causing cracking in the plate glass is approximately \(579^{\circ} \mathrm{C}\).