91Ó°ÊÓ

In order to calculate the inside air temperature, we need to determine the overall heat transfer coefficient (U) for the walls of the igloo. We can find it using the formula: \( \frac{1}{U_{wall}} = \frac{1}{h_{in}} + \frac{t}{k} + \frac{1}{h_{out}}\) Where \(h_{in}\) is the inside surface heat transfer coefficient, \(h_{out}\) is the outside surface heat transfer coefficient, \(t\) is the wall thickness, and \(k\) is the thermal conductivity. Plugging in the given values, we get: \( \frac{1}{U_{wall}} = \frac{1}{6} + \frac{0.5}{0.15} + \frac{1}{15}\) Calculating \(U_{wall}\), we have: \(U_{wall} = 0.2443 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

We can use a similar formula to calculate the overall heat transfer coefficient for the floor, considering the ice cap as a semi-infinite medium: \( \frac{1}{U_{floor}} = \frac{1}{h_{in}} +\sqrt{ \frac{k}{2 \alpha \tau}}\) Where \(\tau\) represents an appropriate reference time (say 1 day or 86400 seconds) and \(\alpha\) is the thermal diffusivity. Since we are given that the thermal conductivity for the ice cap is the same as that of the compacted snow, we can use the relation: \(\alpha = \frac{k}{\rho c_p}\) For ice, the density (\(\rho\)) can be taken as 920 kg/m³, and specific heat (\(c_p\)) can be taken as 2090 J/(kg·K). Now we can find \(\alpha\) and subsequently calculate \(U_{floor}\): \(\alpha = \frac{0.15}{920 \times 2090} = 7.41\times10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) \( \frac{1}{U_{floor}} = \frac{1}{6} +\sqrt{ \frac{0.15}{2 \times 7.41\times10^{-7}\times 86400}}\) Calculating \(U_{floor}\), we have: \(U_{floor} = 1.4145 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Now we can determine the heat transfers through walls and floor by considering the temperature difference between inside and outside air: \(Q_{wall} = A_{wall} U_{wall}(T_{in} - T_{\infty})\) \(Q_{floor} = A_{floor} U_{floor}(T_{in} - T_{icecap})\) Since the igloo is a hemisphere, the surface area of the walls, \(A_{wall}\) is equal to 2 times the surface area of the hemisphere base, and the surface area of the floor, \(A_{floor}\) is given by the surface area of the hemisphere base: \(A_{wall} = 2 \pi r_{int}^2 = 2 \pi (1.8)^2 \mathrm{~m}^{2}\) \(A_{floor} = \pi r_{int}^2 = \pi (1.8)^2 \mathrm{~m}^{2}\)

Given that the occupants' body heat provides a continuous source of 320 W, we can use energy balance: \(Q_{body} = Q_{wall} + Q_{floor}\) Substituting the expressions for heat flow, we have: \(320 = A_{wall} U_{wall}(T_{in} - T_{\infty}) + A_{floor} U_{floor}(T_{in} - T_{icecap})\) Plugging in the values, we can solve for the inside air temperature: \(320 = 20.371 \times 0.2443(T_{in} - (-40^{\circ} \mathrm{C})) + 10.1785 \times 1.4145(T_{in} - (-20^{\circ} \mathrm{C}))\) Solving for \(T_{in}\), we have: \(T_{in} = 17.25^{\circ} \mathrm{C}\) Thus, the inside air temperature when the outside air temperature is -40°C is approximately 17.25°C. (b) Perform a parameter sensitivity analysis To perform a parameter sensitivity analysis, we can study the response of the inside air temperature to changes in values of parameters like outside convection coefficient and wall thickness, keeping other variables constant. This can be done using numerical methods or by running a series of calculations with perturbed values of the parameters and observing the changes in the response variable (inside air temperature). If we double or triple the outside convection coefficient, we will observe that the overall heat transfer coefficient through the walls will increase. This, in turn, will cause a decrease in the inside air temperature. We can also check how much variation in the inside air temperature we would observe if the walls were half or twice as thick. This way, we can conclude if it makes sense to construct the igloo with walls of different thicknesses.