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 rate of heat transfer between the plates per unit surface area for each case is: (a) Filled with atmospheric air: 364 W/m² (b) Evacuated: 578 W/m² (c) Filled with fiberglass insulation: 280 W/m² (d) Filled with superinsulation: 1.05 W/m²

Step-by-step solution

Identify the thermal conductivity of air

The thermal conductivity of atmospheric air at room temperature is approximately \(k_{air} = 0.026 \mathrm{~W/m \cdot K}\).

Calculate the heat transfer by conduction

Using the equation for heat transfer by conduction, \(q = \frac{k_{air}A(T_1 - T_2)}{L}\), where \(L = 0.02\mathrm{~m}\) (converted from 2 cm), \(T_1=290\mathrm{~K}\), and \(T_2=150\mathrm{~K}\), determine the heat transfer rate per unit surface area: \(q_{air} = \frac{0.026 (290-150)}{0.02} = \frac{0.026 \cdot 140}{0.02} = \mathrm{364\,W/m^2}\) ##Case (b): Evacuated##

Calculate the heat transfer by radiation

As the gap is evacuated, there is no conduction, and we must consider only radiation for heat transfer. Using the equation \(q = \sigma \varepsilon A (T_1^4 - T_2^4)\), where \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2 \cdot K^4}\) (Stefan-Boltzmann constant) and \(\varepsilon = 1\), determine the heat transfer rate per unit surface area: \(q_{rad} = \sigma (T_1^4 - T_2^4) = 5.67 \times 10^{-8}(290^4 - 150^4) = \mathrm{578\,W/m^2}\) ##Case (c): Filled with fiberglass insulation##

Identify the thermal conductivity of fiberglass insulation

The thermal conductivity of fiberglass insulation is approximately \(k_{fib} = 0.04 \mathrm{~W/m \cdot K}\).

Calculate the heat transfer by conduction

Using the equation for heat transfer by conduction, \(q = \frac{k_{fib}A(T_1 - T_2)}{L}\), determine the heat transfer rate per unit surface area: \(q_{fib} = \frac{0.04 (290-150)}{0.02} = \frac{0.04 \cdot 140}{0.02} = \mathrm{280\,W/m^2}\) ##Case (d): Filled with superinsulation##

Identify the thermal conductivity of superinsulation

The thermal conductivity of superinsulation is given as \(k_{sup} = 0.00015 \mathrm{~W/m \cdot K}\).

Calculate the heat transfer by conduction

Using the equation for heat transfer by conduction, \(q = \frac{k_{sup}A(T_1 - T_2)}{L}\), determine the heat transfer rate per unit surface area: \(q_{sup} = \frac{0.00015 (290-150)}{0.02} = \frac{0.00015 \cdot 140}{0.02} = \mathrm{1.05\,W/m^2}\) The rate of heat transfer between the plates per unit surface area for each case is: (a) Filled with atmospheric air: \(\mathrm{364\,W/m^2}\) (b) Evacuated: \(\mathrm{578\,W/m^2}\) (c) Filled with fiberglass insulation: \(\mathrm{280\,W/m^2}\) (d) Filled with superinsulation: \(\mathrm{1.05\,W/m^2}\)

Key concepts

Thermal Conductivity

Thermal conductivity is a fundamental property of materials that measures their ability to conduct heat. It's often denoted by the symbol k and is defined as the amount of heat that passes in a unit of time through a unit of surface area with a unit of temperature gradient. Essentially, it tells us how well a material can transfer heat by conduction.

For instance, in the given problem, the thermal conductivity of air helps us determine the heat transfer rate through the air-filled space between the plates. Materials with higher thermal conductivity, like most metals, are better at transferring heat, whereas materials with lower conductivity, such as fiberglass insulation or superinsulation, are used to restrict heat flow.

Stefan-Boltzmann Constant

The Stefan-Boltzmann constant, represented by σ, is imperative in the world of thermodynamics as it factors into the Stefan-Boltzmann Law, which is used to calculate the heat radiated from a black body in unit time per unit surface area. It's equal to approximately 5.67 × 10⁻⁸ W/m²·K⁴.

When the plates are in a vacuum in the problem, radiation is the sole method of heat transfer. Here, the constant enables the determination of the rate of heat transfer by radiation between the plates based on their temperatures to the fourth power, understanding that hotter objects emit more thermal radiation than cooler ones.

Heat Transfer by Conduction

Heat transfer by conduction occurs when heat moves through a solid material or between objects in direct contact. The rate at which heat is transferred by conduction is governed by the Fourier's Law, represented as q = (kAΔT)/L, where q is the heat transfer per unit time, k is the material's thermal conductivity, A is the cross-sectional area through which heat is flowing, ΔT is the temperature difference between the two sides, and L is the thickness of the material.

For each of the scenarios with different fillings between the plates, we can calculate the conductive heat transfer rate by applying this equation with the material's specific thermal conductivity.

Heat Transfer by Radiation

Heat transfer by radiation is the process by which heat is emitted as electromagnetic waves and does not require a medium for transmission. This type of heat transfer is significant, especially in a vacuum, where conductive and convective heat transfer cannot occur. The radiative heat transfer rate between two bodies can be determined using the equation q = σϵA(T_1^4 - T_2^4), where ϵ is the emissivity of the materials, A is the area, and T_1 and T_2 are the absolute temperatures of the bodies. The exercise illustrates such a scenario in case (b), where the vacuum between the plates allows only radiative heat transfer.