Question

Two thin radiation shields with emissivities of \(\varepsilon_{3}=0.10\) and \(\varepsilon_{4}=0.15\) on both sides are placed between two very large parallel plates, which are maintained at uniform temperatures \(T_{1}=600 \mathrm{~K}\) and \(T_{2}=300 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.6\) and \(\varepsilon_{2}=0.7\), respectively (Fig. P13-93). Determine the net rates of radiation heat transfer between the two plates with and without the shields per unit surface area of the plates, and the temperatures of the radiation shields in steady operation.

Short answer

Please provide the given data to find the net rates of radiation heat transfer and the temperatures of the radiation shields.

Step-by-step solution

Calculate the radiation heat transfer without shields

First, we need to find the net rate of radiation heat transfer between the two plates without the shields. To do this, we will use the following formula: $$q_{12} = \frac{(T_1^4 - T_2^4)}{1/\varepsilon_1 + 1/\varepsilon_2 - 1}$$ where \(l\) is the distance between the two plates, and \(T_1\) and \(T_2\) are the temperatures in Kelvin of the two plates. Now, plug in the given data and calculate \(q_{12}\).

Calculate the resistances of each surface and the radiation shield surfaces

The resistances method for multiple surfaces calculates the net rate of radiation heat transfer by summing the thermal resistances of the surfaces and shields. The resistance for radiation between two surfaces can be defined as: $$R_{i, i+1} = \frac{1}{A_i \varepsilon_i} + \frac{1}{A_{i+1} \varepsilon_{i+1}} -1$$ where \(A\) and \(\varepsilon\) are the surface area and emissivity of the surfaces. Calculate the resistances for the surfaces and the interfaces with shields: $$R_{1,3} = \frac{1}{A_1 \varepsilon_1} + \frac{1}{A_3 \varepsilon_3} -1$$ $$R_{3,4} = \frac{1}{A_3 \varepsilon_3} + \frac{1}{A_4 \varepsilon_4} -1$$ $$R_{4,2} = \frac{1}{A_4 \varepsilon_4} + \frac{1}{A_2 \varepsilon_2} -1$$

Calculate the equivalent resistance and the net radiation heat transfer with shields

Calculate the total equivalent resistance for radiation heat transfer and then find the net rate of radiation heat transfer for the setup with shields: $$R_{eq} = R_{1,3} + R_{3,4} + R_{4,2}$$ The net radiation heat transfer with shields can then be calculated using the following equation: $$q_{net} = \frac{T_1^4 - T_2^4}{R_{eq}}$$

Determine the temperatures of the radiation shields

To find the temperatures of the radiation shields, we'll use the following equations: $$q_{1,3} = \frac{T_1^4 - T_3^4}{R_{1,3}}$$ $$q_{4,2} = \frac{T_4^4 - T_2^4}{R_{4,2}}$$ Since the setup is in steady operation, the heat transfer between the shields must be equal: $$q_{1,3} = q_{3,4} = q_{4,2}$$ Now solve for the temperatures \(T_3\) and \(T_4\) using these equations. This will give us the net rates of radiation heat transfer between the two plates with and without the shields, as well as the temperatures of the radiation shields in steady operation.

Key concepts

Radiation Shields

Radiation shields are thin layers that are introduced between two surfaces to reduce heat transfer by radiation. The key characteristic of these shields is their low emissivity, which minimizes their ability to emit thermal radiation. This makes them effective barriers in controlling the flow of radiative heat.

In this exercise, two radiation shields are placed between two large plates to decrease the rate of radiation heat transfer. By inserting these shields, the effective path for heat flow is lengthened, thereby decreasing the amount of radiative heat transferred. The shields work by reflecting the radiation back towards the original emitting surface, acting like mirrors for thermal radiation. This process requires understanding the combined system of emissivities and managing them with care to optimize heat resistance.

The shields' emissivities are much lower compared to the surfaces, ensuring that they effectively reduce heat transfer by radiation. With emissivities of 0.10 and 0.15, the shields help significantly lower the net heat transfer rate, illustrating the effectiveness of these shields in practical applications.

Emissivity

Emissivity (\( \varepsilon \) ) describes a surface's effectiveness in emitting thermal radiation compared to an ideal black body at the same temperature. It's a measure of a surface's radiation emission capability, and it's expressed as a value between 0 and 1.

A high emissivity means the surface is a good emitter of radiation, whereas a low emissivity indicates it's a poor emitter and a good reflector. In the context of radiation shields, using low emissivity materials is crucial. These materials, like those with an emissivity of 0.10 and 0.15 in the exercise, are preferred because they reflect more radiation back and emit less. This helps retain heat in the original emitting surface or prevents it from reaching the receiving surface.

Thus, selecting materials based on emissivity is important in designing systems for heat transfer management. Whether you're designing cookware, building insulation, or even satellite equipment, understanding emissivity aids in optimizing thermal performance.

Thermal Resistance

Thermal resistance in the context of radiation is a concept that simplifies the calculation of heat transfer between surfaces by making it akin to electrical resistance. Just as electrical resistances can be added together to find a total resistance, the same applies to thermal resistances.

Each surface and shield introduces its thermal resistance to the system, expressed in the formula \[ R_{i, i+1} = \frac{1}{A_i \varepsilon_i} + \frac{1}{A_{i+1} \varepsilon_{i+1}} - 1 \] where \( A \) denotes the area and \( \varepsilon \) represents emissivity. The sum total of these resistances across multiple layers helps determine the net rate of heat transfer. The concept is crucial in the step-by-step solution for calculating equivalent resistance, especially when additional surfaces like radiation shields are involved.

By reducing thermal resistance through smart design and material choice (using low emissivity), control over the heat transfer can be significantly improved, benefiting energy efficiency and safety in numerous applications.

Steady State Operation

Steady state operation describes a condition in a system where the variables (temperature, pressure, etc.) remain constant over time despite ongoing processes. In simpler terms, the system reaches a balance, and there are no changes over time.

In the context of this exercise, when referring to steady state operation, it implies that the heat transfer rates through the radiation shields are constant and equal, meaning \( q_{1,3} = q_{3,4} = q_{4,2} \). This condition signals that the system has stabilized, allowing for accurate calculations of the temperatures of the radiation shields. Such steady conditions make it possible to analyze the system comprehensively and predict its behavior effectively.

Steady state analysis is vital in designing systems for consistent performance without sudden changes, crucial in both daily applications and complex engineering systems.