Question

Consider two rectangular surfaces perpendicular to each other with a common edge which is \(1.6 \mathrm{~m}\) long. The horizontal surface is \(0.8 \mathrm{~m}\) wide and the vertical surface is \(1.2 \mathrm{~m}\) high. The horizontal surface has an emissivity of \(0.75\) and is maintained at \(400 \mathrm{~K}\). The vertical surface is black and is maintained at \(550 \mathrm{~K}\). The back sides of the surfaces are insulated. The surrounding surfaces are at \(290 \mathrm{~K}\), and can be considered to have an emissivity of \(0.85\). Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings.

Short answer

Answer: The net rate of radiation heat transfer between the two surfaces is -2167.12 W (heat transfer from the vertical surface to the horizontal surface). The net radiation heat transfer between the horizontal surface and the surroundings is -3637.71 W (heat transfer from the horizontal surface to the surroundings).

Step-by-step solution

Calculate the areas of the surfaces

First, we need to find the areas of the surfaces. For the horizontal surface, its length is 1.6 m and its width is 0.8 m. For the vertical surface, its height is 1.2 m and its length is 1.6 m. Calculate the areas as follows: \(A_\text{horizontal} = 1.6 \times 0.8 = 1.28 \mathrm{~m^2}\) \(A_\text{vertical} = 1.6 \times 1.2 = 1.92 \mathrm{~m^2}\)

Calculate view factors

Next, we need to calculate the view factor between the two surfaces. Since they are perpendicular to each other with a common edge, the view factor for radiation heat transfer can be calculated using the following formula: \(F_{1 \to 2} = F_{2 \to 1} = \frac{1}{2}\) This view factor accounts for the fact that these surfaces are perpendicular and have a common edge.

Calculate heat transfers for each surface due to their own emissivity

Using the given temperatures and emissivities, we can calculate the radiation heat transfer from each surface due to their own emissivity. We'll use the Stefan-Boltzmann law: \(q_i = \epsilon_i \sigma A_i T_i^4\) For the horizontal surface (i=1): \(q_1 = 0.75 \cdot (5.67\times10^{-8}) \cdot 1.28 \cdot 400^4 = 263.66\mathrm{~W}\) For the vertical surface (i=2): \(q_2 = 1.00 \cdot (5.67\times10^{-8}) \cdot 1.92 \cdot 550^4 = 4597.90\mathrm{~W}\)

Calculate net radiation heat transfers between the two surfaces

Now, using the view factors and the individual heat transfers calculated, we can find the net heat transfers between the two surfaces: \(q_\text{net}^{1 \to 2} = (q_1 - q_2) \cdot F_{1 \to 2}\) \(q_\text{net}^{1 \to 2} = (263.66 - 4597.90) \cdot \frac{1}{2} = -2167.12\mathrm{~W}\) So the net radiation heat transfer between the two surfaces is -2167.12 W (heat transfer from the vertical surface to the horizontal surface).

Calculate heat transfer between horizontal surface and surroundings

We need to find the net heat transfer between the horizontal surface and the surroundings, using the surrounding temperature (290 K) and emissivity (0.85) as follows: \(q_\text{net}^{1 \to s} = \epsilon_\text{horizontal} \sigma A_\text{horizontal}(T_1^4 - T_\text{s}^4)\) \(q_\text{net}^{1 \to s} = 0.75 \cdot (5.67\times10^{-8}) \cdot 1.28 \cdot (400^4 - 290^4) = -3637.71\mathrm{~W}\) So the net radiation heat transfer between the horizontal surface and the surroundings is -3637.71 W (heat transfer from the horizontal surface to the surroundings).

Key concepts

Understanding the Stefan-Boltzmann Law

When studying radiation heat transfer, one of the most crucial concepts to grasp is the Stefan-Boltzmann Law. This law provides the foundation for understanding how much thermal radiation is emitted by a body. According to this law, the energy radiated per unit surface area of a black body is proportional to the fourth power of the body's absolute temperature. Mathematically, it is expressed as:

\[ E = \big( \epsilon \cdot \sigma \big) A T^4 \]
where \(E\) is the radiant energy, \(\epsilon\) is the emissivity of the surface, \(\sigma\) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \mathrm{\ W \ m^{-2} \ K^{-4}}) \), \(A\) is the area of the emitting surface, and \(T\) is the absolute temperature in Kelvin.

This formula indicates that even small changes in temperature can result in significant changes in thermal radiation power due to the fourth power dependence. In the exercise, these principles are applied to calculate the radiation heat transfer for both the vertical and horizontal surfaces. The larger exponent for the vertical surface's temperature in the Stefan-Boltzmann equation reflects why the vertical surface, which is at a higher temperature, emits substantially more radiation compared to the horizontal surface.

The Role of Emissivity in Radiation Heat Transfer

Emissivity is a measure of a material's ability to emit thermal radiation compared to a perfect black body, which has an emissivity of 1 (or 100%). Materials with high emissivity are efficient at emitting radiation, while those with low emissivity reflect much of the incident radiation. Importantly, emissivity affects the rate at which an object can cool down or heat up through thermal radiation. The concept of emissivity is embedded in the Stefan-Boltzmann law's equation mentioned above, and it's important to note that the emissivity value must be between 0 and 1.

For instance, in the given exercise, the horizontal surface has an emissivity of 0.75, meaning it does not emit as much energy as a black body would at the same temperature. This is factored into the heat transfer calculations to find the accurate net radiation heat transfer from the surface. Understanding emissivity is critical when evaluating real-world problems because most materials are not perfect black bodies, which is clearly illustrated in the exercise's solution.

View Factor and Its Implications on Heat Transfer

Another concept that is essential in radiation heat transfer is the view factor, also known as the configuration factor or shape factor. This dimensionless quantity represents the proportion of radiation that leaves one surface and directly impinges upon another. It is dependent on the geometry and relative orientation of the surfaces involved. A view factor of 1 means all radiation from the first surface hits the second, while a view factor of 0 means no radiation from the first surface reaches the second.

In the exercise, we are told that two surfaces are perpendicular with a common edge, which simplifies their view factor to \(\frac{1}{2}\), because the geometry results in half of the radiation from each surface reaching the other. This calculation is significant when determining the net radiation exchange between the two surfaces, as it helps to accurately quantify the amount of energy being transferred. Consequently, understanding view factors is crucial for calculating and optimizing thermal systems in real-world applications such as building design or electrical component placement where heat emission can affect performance and lifespan.