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Two aligned parallel rectangles with dimensions \(6 \mathrm{~m} \times\) \(8 \mathrm{~m}\) are spaced apart by a distance of \(2 \mathrm{~m}\). If the two parallel rectangles are experiencing radiation heat transfer as black surfaces, determine the percentage of change in radiation heat transfer rate when the rectangles are moved \(8 \mathrm{~m}\) apart.

Short Answer

Expert verified
Answer: The percentage change in radiation heat transfer rate when the distance between the rectangles is increased from 2m to 8m is 33.33%.

Step by step solution

01

Calculate the total surface area of both rectangles

To calculate the total surface area of both rectangles, we simply multiply the dimensions of the rectangles: \(A_1 = 6 \cdot 8 = 48 \text{~m}^2\) Since there are two rectangles, the combined surface area is: \(A_{1\&2} = 2 \cdot A_1 = 2 \cdot 48 = 96 \text{~m}^2\)
02

Determine the shape factor between the two rectangles

The shape factor is a dimensionless number that represents the proportion of radiant energy leaving one surface that arrives at the second surface. In this case, the shape factor can be determined through the following formula: \(F_{1 \rightarrow 2} = 1 - \frac{1}{1 + \frac{A_{1\&2}}{A_1} \frac{d}{L_{1 \times 2}}}\) where \(d\) is the distance between rectangles, and \(L_{1 \times 2}\) represents the length of the emitting surface onto the receiving surface.
03

Calculate the radiation heat transfer rate in both cases

The radiation heat transfer rate between two black surfaces can be defined as follows: \(Q = c \cdot A_1 \cdot F_{1 \rightarrow 2} \cdot \Delta Temp\) In this exercise, the temperature difference and the constant \(c\) remain the same, so the only thing affected by the change in distance between rectangles is the shape factor. Thus, we need to calculate the shape factor for both cases. a) When the rectangles are 2m apart: \(d_1 = 2m\) We have: \(F_{1 \rightarrow 2}(2m) = 1 - \frac{1}{1 + \frac{96}{48} \frac{2}{8}} = 1 - \frac{1}{2} = 0.5\) b) When the rectangles are 8m apart: \(d_2 = 8m\) We have: \(F_{1 \rightarrow 2}(8m) = 1 - \frac{1}{1 + \frac{96}{48} \frac{8}{8}} = 1 - \frac{1}{1 + 2} = \frac{2}{3}\) Now, knowing the shape factors for both cases, we can calculate the radiation heat transfer rates. \(Q_1 = c \cdot A_1 \cdot F_{1 \rightarrow 2}(2m) = k \cdot 48 \cdot 0.5\) \(Q_2 = c \cdot A_1 \cdot F_{1 \rightarrow 2}(8m) = k \cdot 48 \cdot \frac{2}{3}\)
04

Determine the percentage change in radiation heat transfer rate

Finally, we can find the percentage change between the two radiation heat transfer rates: \(Percentage\,change = \frac{Q_2 - Q_1}{Q_1} \cdot 100\) \(Percentage\,change = \frac{k \cdot 48 \cdot \frac{2}{3} - k \cdot 48 \cdot 0.5}{k \cdot 48 \cdot 0.5} \cdot 100\) \(Percentage\,change = \frac{\frac{2}{3} - 0.5}{0.5} \cdot 100\) The percentage change in radiation heat transfer rate when the rectangles are moved to 8m apart is: \(Percentage\,change = \frac{1}{3} \cdot 100 = 33.33\%\) Thus, the radiation heat transfer rate increases by 33.33% when the distance between the rectangles is increased from 2m to 8m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Shape Factor
The concept of shape factor, also known as view factor or configuration factor, is crucial in understanding radiation heat transfer. It refers to the fraction of radiation leaving a surface that directly strikes another surface. The shape factor depends solely on the geometry and relative positioning of the two surfaces involved.

In the context of two parallel rectangles, the shape factor is determined by their size and the distance between them. For instance, when the rectangles are closer, more radiation from one rectangle reaches the other, resulting in a higher shape factor. Conversely, if they are further apart, less radiation reaches, leading to a decrease in the shape factor. The mathematical representation for this is:

\[F_{1 \rightarrow 2} = 1 - \frac{1}{1 + \frac{A_{1\&2}}{A_1} \frac{d}{L_{1 \times 2}}}\]

Here, \(d\) is the distance between the rectangles, \(A_{1\&2}\) is the total area, and \(L_{1 \times 2}\) is the linear dimension between them. Understanding and calculating the shape factor is key to predicting how efficiently surfaces transfer heat to each other through radiation.
Blackbody Radiation
Blackbody radiation is a fundamental concept in radiative heat transfer. A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. At a given temperature, it emits radiation at the maximum possible rate per unit area.

This concept is crucial for our exercise since the rectangles are assumed to be black surfaces. This means they are perfect emitters and absorbers of radiation. Therefore, they follow Planck’s law of blackbody radiation, which defines the spectral density of the radiation emitted. The intensity and distribution of this radiation depend solely on the temperature of the body, making blackbody models straightforward for calculations where surface emissivity doesn’t vary.

Using blackbody principles helps simplify the problem as there is no need to consider variations in emissive power. Instead, the focus shifts to geometric factors, like shape factor, and how they affect the heat exchange between the surfaces.
Percentage Change in Heat Transfer Rate
The percentage change in heat transfer rate quantifies how the efficiency of radiation exchange between two surfaces changes when conditions alter. In this problem, this involves altering the distance between the rectangles.

When calculating the heat transfer rate for the initial and altered positions, the essential factor to consider is the shape factor, as it dynamically influences the radiation exchange. For example, when the rectangles are closer, they have a lower shape factor, resulting in a different rate of heat transfer compared to when they are further apart. The exercise shows how:
  • At 2m apart, the shape factor is 0.5.
  • At 8m apart, the shape factor changes to \(\frac{2}{3}\).
To determine how much the rate changes, we calculate the rates \(Q_1\) and \(Q_2\) and find their percentage difference using the formula:

\[Percentage\,change = \frac{Q_2 - Q_1}{Q_1} \cdot 100\]

In this exercise, as the rectangles move from 2m to 8m apart, the heat transfer rate increases by approximately 33.33%. This increase reflects how geometrical arrangements in radiative setups can significantly alter the thermal behavior of the system.

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Most popular questions from this chapter

Give examples of radiation effects that affect human comfort. 13-84 A thin aluminum sheet with an emissivity of \(0.15\) on both sides is placed between two very large parallel plates, which are maintained at uniform temperatures \(T_{1}=900 \mathrm{~K}\) and \(T_{2}=650 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.8\), respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result with that without the shield.Give examples of radiation effects that affect human comfort.

Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}\), \(600 \mathrm{~K}\), and \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer between the base and the side surfaces is (a) \(22.5 \mathrm{~kW}\) (b) \(38.6 \mathrm{~kW}\) (c) \(60.7 \mathrm{~kW}\) (d) \(89.8 \mathrm{~kW}\) (e) \(151 \mathrm{~kW}\)

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.

Consider two black coaxial parallel circular disks of equal diameter \(D\) that are spaced apart by a distance \(L\). The top and bottom disks have uniform temperatures of \(500^{\circ} \mathrm{C}\) and \(520^{\circ} \mathrm{C}\), respectively. Determine the radiation heat transfer coefficient \(h_{\text {rad }}\) between the disks if they are spaced apart by \(L=D\).

Two parallel disks of diameter \(D=0.6 \mathrm{~m}\) separated by \(L=0.4 \mathrm{~m}\) are located directly on top of each other. Both disks are black and are maintained at a temperature of \(450 \mathrm{~K}\). The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at \(300 \mathrm{~K}\). Determine the net rate of radiation heat transfer from the disks to the environment.

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