Question

Consider steady heat transfer between two large parallel plates at constant temperatures of \(T_{1}=290 \mathrm{~K}\) and \(T_{2}=150 \mathrm{~K}\) that are \(L=2 \mathrm{~cm}\) apart. Assuming the surfaces to be black (emissivity \(\varepsilon=1\) ), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is \((a)\) filled with atmospheric air, \((b)\) evacuated, \((c)\) filled with fiberglass insulation, and \((d)\) filled with superinsulation having an apparent thermal conductivity of \(0.00015 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short answer

Answer: The rates of heat transfer per unit surface area for each case are: (a) Filled with atmospheric air: 181.95 W/m², (b) Evacuated: 231417.46 W/m², (c) Filled with fiberglass insulation: 280 W/m², and (d) Filled with superinsulation: 1.05 W/m².

Step-by-step solution

Find the temperature difference

First, we need to determine the temperature difference between the two plates, which will be given by: ΔT = T1 - T2 = 290K - 150K = 140K

Calculate thermal resistance of air (a)

The thermal resistance of atmospheric air can be calculated using the formula: R_a = L / k_a Where L is the distance between the plates (0.02m) and k_a is the thermal conductivity of the air, which is approximately 0.026 W/(m·K). Therefore, R_a = 0.02 / 0.026 = 0.7692 W/m.K

Calculate the thermal resistance when evacuated (b)

When the gap between the plates is evacuated, we don't have conduction or convection of heat. Therefore, we will only have radiation. The thermal resistance for radiative heat transfer is given by: R_b = 1 / (A * σ * (T1 + T2) * (T1^2 + T2^2)) Where A is the surface area, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/(m^2·K^4)). Since the heat transfer rate is per unit surface area, we can ignore A in the formula: R_b = 1 / (σ * (290 + 150) * (290^2 + 150^2)) = 1 / 1654.14

Calculate the thermal resistance of fiberglass insulation (c)

The thermal resistance of fiberglass insulation can be calculated similarly to that of air: R_c = L / k_c Where k_c is the thermal conductivity of fiberglass insulation, which is approximately 0.04 W/(m·K). Therefore, R_c = 0.02 / 0.04 = 0.5 W/m.K

Calculate the thermal resistance of superinsulation (d)

For superinsulation, which has an apparent thermal conductivity of 0.00015 W/m.K, the thermal resistance is: R_d = L / k_d = 0.02 / 0.00015 = 133.33 W/m.K

Calculate the rate of heat transfer per unit surface area for each case

Using the formula Q = ΔT / R, we can find the rate of heat transfer per unit surface area for each case: Q_a = ΔT / R_a = 140 / 0.7692 = 181.95 W/m^2 Q_b = ΔT / R_b = 140 / (1 / 1654.14) = 231417.46 W/m^2 Q_c = ΔT / R_c = 140 / 0.5 = 280 W/m^2 Q_d = ΔT / R_d = 140 / 133.33 = 1.05 W/m^2 Therefore, the rate of heat transfer per unit surface area for each case is: (a) Filled with atmospheric air: 181.95 W/m² (b) Evacuated: 231417.46 W/m² (c) Filled with fiberglass insulation: 280 W/m² (d) Filled with superinsulation: 1.05 W/m²