91Ó°ÊÓ

Convection heat transfer coefficients for a hot horizontal surface facing upward may be determined by a gage whose specific features depend on whether the temperature of the surroundings is known. For configuration A, a copper disk, which is electrically heated from below, is encased in an insulating material such that all of the heat is transferred by convection and radiation from the top surface. If the surface emissivity and the temperatures of the air and surroundings are known, the convection coefficient may be determined from measurement of the electrical power and the surface temperature of the disk. Configuration B is used in situations for which the temperature of the surroundings is not known. A thin, insulating strip separates semicircular disks with independent electrical heaters and different emissivities. If the emissivities and temperature of the air are known, the convection coefficient may be determined from measurement of the electrical power supplied to each of the disks in order to maintain them at a common temperature. (a) In an application of configuration A to a disk of diameter \(D=160 \mathrm{~mm}\) and emissivity \(\varepsilon=0.8\), values of \(P_{\text {elec }}=10.8 \mathrm{~W}\) and \(T=67^{\circ} \mathrm{C}\) are measured for \(T_{\infty}=T_{\text {sur }}=27^{\circ} \mathrm{C}\). What is the corresponding value of the average convection coefficient? How does it compare with predictions based on a standard correlation? (b) Now consider an application of configuration \(B\) for which \(T_{\infty}=17^{\circ} \mathrm{C}\) and \(T_{\text {sur }}\) is unknown. With \(D=160 \mathrm{~mm}, \varepsilon_{1}=0.8\), and \(\varepsilon_{2}=0.1\), values of \(P_{\text {elect, } 1}=9.70 \mathrm{~W}\) and \(P_{\text {elec, } 2}=5.67 \mathrm{~W}\) are measured when \(T_{1}=T_{2}=77^{\circ} \mathrm{C}\). Determine the corresponding values of the convection coefficient and the temperature of the surroundings. How does the convection coefficient compare with predictions by an appropriate correlation?

Step by step solution

01

Find Heat Transfer Rate from Surface by Radiation

First, we need to find the heat transfer from the surface by radiation. We can do this by using the Stefan-Boltzmann Law: \[Q_{rad} = \varepsilon \sigma A_s (T^4 - T_{\infty}^4)\] where \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (5.67 x 10^{-8} W/m^2K^4), \(A_s\) is the surface area, and \(T\) and \(T_{\infty}\) are the temperatures in Kelvin. For our given data, \(D = 160\) mm, which means the surface area (\(A_s\)) of the disc is: \[A_s = \pi\frac{(D/2)^2}{1000^2}\] \[A_s = \pi\frac{(80/1000)^2}{1} = 0.0201\text{ m}^2\] Convert temperatures to Kelvin: \(T = 67^{\circ}\text{C} + 273.15 = 340.15\text{K}\) and \(T_{\infty} = 27^{\circ}\text{C} + 273.15 = 300.15\text{K}\) Now, calculate \(Q_{rad}\): \[Q_{rad} = 0.8 \times 5.67 \times 10^{-8} \times 0.0201 \times (340.15^4 - 300.15^4) = 4.06\mathrm{~W}\]

02

Find Heat Transfer Rate from Surface by Convection

Next, we can determine the heat transfer rate from the surface by convection: \[Q_{conv} = P_{elec} - Q_{rad}\] where \(P_{elec} = 10.8\) W is the measured electrical power. \[Q_{conv} = 10.8 - 4.06 = 6.74\mathrm{~W}\]

03

Determine the average convection coefficient

Now, we can calculate the average convection coefficient (\(h_{avg}\)) using the formula: \[Q_{conv} = h_{avg} A_s (T - T_{\infty})\] Rearrange to solve for \(h_{avg}\): \[h_{avg} = \frac{Q_{conv}}{ A_s (T - T_{\infty})}\] \[h_{avg} = \frac{6.74}{0.0201 \times (340.15 - 300.15)}\] \[h_{avg} = 8.84\mathrm{~W/m^2K}\] We have determined the average convection coefficient for Configuration A as 8.84 W/m²K. #Part B: Convection Coefficient and Temperature of Surroundings for Configuration B#

04

Solve for the convection coefficient

First, let's find the convection coefficient (\(h\)). According to the data we have, \(P_{elec, 1} + P_{elec, 2} = 9.7\mathrm{~W} + 5.67\mathrm{~W} = 15.37\mathrm{~W}\) should equal the heat transfer rates from both disks by convection and radiation: \[Q_{conv,1} + Q_{conv,2} + Q_{rad,1} + Q_{rad,2} = 15.37\mathrm{~W}\] We know that for both disks, the convection heat transfer can be expressed as follows (since they share the same \(h\) and \(\Delta T = T - T_{\infty}\)): \[Q_{conv,i} = h A_{s,i} (T - T_{\infty})\] \[Q_{rad,i} = \varepsilon_i \sigma A_{s,i} (T^4 - T_{\text {sur}}^4)\] Using the given data, we can find expressions for \(Q_{conv,1}\), \(Q_{conv,2}\), \(Q_{rad,1}\), and \(Q_{rad,2}\). By combining these expressions and rearranging terms, we can find the value of \(h\). This calculation is complex but eventually results in: \[h = \frac{P_{elec,1}- P_{elec,2} (\frac{\varepsilon_2}{\varepsilon_1})}{A_s \frac{T - T_{\infty}} {(\frac{\varepsilon_2}{\varepsilon_1} - 1)}}\] From the previous part of the exercise, we know that \(A_s = 0.0201\text{ m}^2\). Plug in the values given for this problem: \[h = \frac{9.7 - 5.67 (\frac{0.1}{0.8})}{0.0201 \frac{77 - 17}{(0.125 - 1)}} = \frac{9.7- 0.709}{0.0201 \times 60} = 7.63\mathrm{~W/m^2K}\] So, the convection coefficient for Configuration B is 7.63 W/m²K.

05

Determine the temperature of surroundings

Now, let's find the temperature of the surroundings (\(T_{\text {sur}}\)). Re-examine the previous equation: \[Q_{conv,i} = h A_{s,i} (T - T_{\infty})\] \[Q_{rad,i} = \varepsilon_i \sigma A_{s,i} (T^4 - T_{\text {sur}}^4)\] To find \(T_{\text {sur}}\), we will need to solve for \(Q_{rad, 1}\) using: \[P_{elec, 1} = Q_{conv, 1} + Q_{rad, 1}\] \[Q_{rad, 1} = P_{elec, 1} - Q_{conv, 1}\] Plug in known values along with our calculated \(h\) and surface area: \[Q_{rad, 1} = 9.7 - (7.63 \times 0.0201 (77 - 17)) = 1.53\mathrm{~W}\] Now, we can rearrange the expression for \(Q_{rad, 1}\) to find \(T_{\text {sur}}\): \[T_{\text {sur}} = \sqrt[4]{\frac{Q_{rad,1}}{\varepsilon_1 \sigma A_s} + T^4}\] Plug in known values and the calculated \(Q_{rad, 1}\): \[T_{\text {sur}} = \sqrt[4]{\frac{1.53}{0.8 \times 5.67 \times 10^{-8} \times 0.0201} + 77^4} = 285.67\text{K}\] Convert the temperature of surroundings back to Celsius: \[T_{\text {sur}}^{\circ}C = 285.67\text{K} - 273.15 = 12.52^{\circ}\text{C}\] So, the temperature of the surroundings for Configuration B is 12.52°C.

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Key Concepts

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

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a principle in physics that describes how a body emits thermal radiation. It states that the power radiated per unit area of a surface is proportional to the fourth power of the temperature of the body. Mathematically, this can be expressed as \[Q_{rad} = \varepsilon \sigma A_s (T^4 - T_{\infty}^4)\]where:

• \(\varepsilon\) is the emissivity of the surface, representing how well it emits thermal radiation compared to a perfect black body.
• \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \ \frac{\text{W}}{\text{m}^2 \text{K}^4}\).
• \(A_s\) is the surface area of the body through which radiation occurs.
• \(T\) is the absolute temperature of the body in Kelvin.
• \(T_{\infty}\) is the absolute temperature of the surroundings in Kelvin.

In the context of the exercise, the law helps determine the portion of heat transfer that occurs through radiation, which is critical for calculating the remaining heat transferred via convection.

emissivity

Emissivity (\(\varepsilon\)) is a material property that quantifies a surface's efficiency in emitting thermal radiation. It is dimensionless and ranges from 0 to 1, where 1 indicates a perfect black body that emits the maximum possible radiation and 0 means the body does not emit radiation at all. Most real-world materials have emissivity values between these extremes. Factors affecting emissivity include the material's surface roughness, coating, and temperature. For example, in the exercise, two different materials are used in Configuration B at different emissivities to accommodate unknown temperature conditions. The higher the emissivity, the greater the heat lost as radiation. Knowing the emissivity is essential when analyzing heat transfer problems as it influences the amount of energy that is radiated compared to what is absorbed. Understanding emissivity is necessary to solve problems involving combined modes of heat transfer, like radiation and convection.

convection coefficient

The convection coefficient (\(h\)) is a parameter that characterizes the heat transfer between a solid surface and a fluid (such as air or water) moving over the surface. It is usually expressed in \(\text{W/m}^2\text{K}\) and depends on several factors such as the fluid's properties, the flow regime, surface roughness, and temperature difference between the surface and the fluid. In the given exercise, the convection coefficient is crucial to calculate the convective heat transfer part. When direct calculations are made using configurations A and B, this coefficient provides insights into how efficiently heat is being removed from the disk's surface by convection compared to other modes of heat transfer. This is determined using the relationship:\[Q_{\text{conv}} = h A_s (T - T_{\infty})\] Meaning that the higher the convection coefficient, the more efficient the convective heat transfer process is. Comprehension of this concept helps in designing systems where temperature control is vital.

heat transfer rate

The heat transfer rate is a measure of the amount of heat energy transferred per unit time. In thermal systems, it indicates how effectively heat moves from one body or medium to another. It can occur via conduction, convection, or radiation.For convection, the heat transfer rate is typically calculated with:\[Q_{\text{conv}} = P_{\text{elec}} - Q_{\text{rad}}\] This formula illustrates that any residual heat, after radiation is accounted for, is assumed to be transferred through convection. Identifying and calculating heat transfer rates allows engineers and scientists to optimize thermal systems for energy efficiency and control measures. In practical applications like those found in the exercise, understanding how the heat is distributed can lead to better system designs and improve temperature control methodologies.