In an experiment to measure convection heat transfer coefficients, a very thin metal foil of very low emissivity (e.g., highly polished copper) is attached on the surface of a slab of material with very low thermal conductivity. The other surface of the metal foil is exposed to convection heat transfer by flowing fluid over the foil surface. This setup diminishes heat conduction through the slab and radiation on the metal foil surface, while heat convection plays the prominent role. The slab on which the metal foil is attached to has a thickness of \(25 \mathrm{~mm}\) and a thermal conductivity of \(0.023 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). In a condition where the surrounding room temperature is \(20^{\circ} \mathrm{C}\), the metal foil is heated electrically with a uniform heat flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\). If the bottom surface of the slab is \(20^{\circ} \mathrm{C}\) and the metal foil has an emissivity of \(0.02\), determine \((a)\) the convection heat transfer coefficient if air is flowing over the metal foil and the surface temperature of the foil is \(150^{\circ} \mathrm{C}\); and \((b)\) the convection heat transfer coefficient if water is flowing over the metal foil and the surface temperature of the foil is \(30^{\circ} \mathrm{C}\).

Step by step solution

01

Understand Provided Information

We have the following information: - Metal foil surface temperature: \(T_1 = 150^{\circ} \mathrm{C}\) for air and \(T_2 = 30^{\circ} \mathrm{C}\) for water. - Surrounding room temperature: \(T_\infty = 20^{\circ}\mathrm{C}\). - Heat flux: \(q'' = 5000 \mathrm{~W} / \mathrm{m}^{2}\). - Emissivity of metal foil surface: \(\epsilon = 0.02\).

02

Calculate Heat Radiation

Since the foil is heated electrically, the total heat flux should be equal to the sum of heat radiation and heat convection. We first calculate the heat radiation (\(q_{rad}\)) for both air and water scenarios using the Stefan-Boltzmann law: \(q_{rad1} = \epsilon \sigma(T_1^4-T_\infty^4)\) for air, \(q_{rad2} = \epsilon \sigma(T_2^4-T_\infty^4)\) for water. We have: - \(\epsilon = 0.02\) - \(\sigma = 5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}^4\) - \(T_1 = 423 \mathrm{K}\) - \(T_2 = 303 \mathrm{K}\) - \(T_\infty = 293 \mathrm{K}\) Now, substituting the values: \(q_{rad1} = 0.02 \times 5.67 \times 10^{-8} (423^4 - 293^4) \approx 1.91 \mathrm{~W}/\mathrm{m}^2\) \(q_{rad2} = 0.02 \times 5.67 \times 10^{-8} (303^4 - 293^4) \approx 0.58 \mathrm{~W}/\mathrm{m}^2\)

03

Calculate Heat Conduction

Now, we will calculate the heat convection for both scenarios. Since we have neglected heat conduction, we can write: \(q_{conv1} = q'' - q_{rad1} = 5000 - 1.91 \approx 4998.09 \mathrm{~W} / \mathrm{m}^2\) \(q_{conv2} = q'' - q_{rad2} = 5000 - 0.58 \approx 4999.42 \mathrm{~W} / \mathrm{m}^2\)

04

Calculate Convection Heat Transfer Coefficients

Using Newton's law of cooling, we can find the convection heat transfer coefficient \((h)\). We have: \(h_{1} = \frac{q_{conv1}}{A (T_1 - T_\infty)}\) \(h_{2} = \frac{q_{conv2}}{A (T_2 - T_\infty)}\) For both equations, we can cancel out A, as we are looking for the coefficient per unit area. Then, we have: \(h_{1} = \frac{4998.09}{(423 - 293)} \approx 38.45 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) \(h_{2} = \frac{4999.42}{(303 - 293)} \approx 499.94 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\)

05

Present the Results

The convection heat transfer coefficient \((h)\) is: - For air flow over the metal foil: \(h_{1} = 38.45 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\). - For water flow over the metal foil: \(h_{2} = 499.94 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\).

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Key Concepts

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

Heat Radiation

Heat radiation, also known as thermal radiation, is a process by which energy is emitted by a heated surface in the form of electromagnetic waves. In our daily lives, we experience this phenomenon when we feel the warmth of the sun or a fireplace without being in direct contact with the source.

In the context of the given exercise, heat radiation is a form of heat loss from the metal foil. The highly polished metal has an emissivity value, a measure of how effectively a surface emits heat radiation. Since the emissivity value of the metal foil is low (\( \text{0.02} \)), it means the foil is not a good radiator and most of the heat transfer occurs through convection - the principle focus of the experiment.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in thermodynamics that describes the power radiated from a black body in terms of its temperature. Specifically, it states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature.

Mathematically, it's presented as:
\( q_{rad} = \text{\textepsilon} \text{\textsigma} T^4 \) where \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \text{W/m}^2\text{K}^4 \) ), and \( \text{\textepsilon} \) is the emissivity of the material.

While the foil in the experiment is not a perfect black body, the law still applies to estimate the heat radiation by using the given emissivity, reflecting that only a portion of the heat is radiated away.

Newton's Law of Cooling

Newton's law of cooling is a model that describes the rate of heat loss from a body as proportional to the difference in temperature between the body and its surroundings. It can be succinctly stated as:
\( q_{conv} = hA(T_{surface} - T_{ambient}) \)
where:

• \( q_{conv} \) is the convective heat transfer
• \( h \) is the convection heat transfer coefficient
• \( A \) is the area through which heat is being transferred
• \( T_{surface} \) is the temperature of the body's surface
• \( T_{ambient} \) is the temperature of the ambient environment

In the exercise, this law helps us to deduce the convection heat transfer coefficient by rearranging the formula to solve for \( h \) using the known values of the temperatures and the convective heat transfer calculated.

Thermal Conductivity

Thermal conductivity is a property of a material that indicates its ability to conduct heat. It is defined as the amount of heat that passes in unit time through a plate of particular area and thickness when its opposite faces differ in temperature by one Kelvin.

For instance, metals typically have high thermal conductivity, thus they are good heat conductors. Conversely, materials with low thermal conductivity, such as the slab in the experiment, act as insulation, preventing heat flow through them.

This concept explains why a slab with very low thermal conductivity is used in the experiment – to ensure the predominant heat transfer mode measured is convection across the metal foil surface, not conduction through the slab.