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Consider a surface of area \(A\) at which the convection and radiation heat transfer coefficients are \(h_{\text {conv }}\) and \(h_{\mathrm{rad}}\), respectively. Explain how you would determine \((a)\) the single equivalent heat transfer coefficient, and \((b)\) the equivalent thermal resistance. Assume the medium and the surrounding surfaces are at the same temperature.

Short Answer

Expert verified
Question: Calculate the single equivalent heat transfer coefficient and the equivalent thermal resistance for a surface with a convection heat transfer coefficient of \(h_{\text {conv }} = 10 \, W/m^2K\) and a radiation heat transfer coefficient of \(h_{\mathrm{rad}} = 5 \, W/m^2K\). The surface area is \(A = 2 \, m^2\). Answer: The single equivalent heat transfer coefficient, \(h\), is the sum of the convection and radiation heat transfer coefficients, so \(h = h_{\text {conv }} + h_{\mathrm{rad}} = 10 + 5 = 15 \, W/m^2K\). To find the equivalent thermal resistance, \(R_{\text{eq}}\), use the formula \(R_{\text{eq}} = \frac{1}{h \cdot A} = \frac{1}{15 \cdot 2} = 0.0333 \, K/W\). Therefore, the single equivalent heat transfer coefficient is \(15 \, W/m^2K\) and the equivalent thermal resistance is \(0.0333 \, K/W\).

Step by step solution

01

Understand the heat transfer coefficients

Heat transfer coefficients represent the efficiency of heat transfer in convection and radiation. In this case, \(h_{\text {conv }}\) represents the convection heat transfer coefficient, and \(h_{\mathrm{rad}}\) represents the radiation heat transfer coefficient.
02

Calculate the combined heat transfer coefficient

The combined heat transfer coefficient, \(h\), is obtained by adding the individual heat transfer coefficients. The equation for the combined heat transfer coefficient is: \[h = h_{\text {conv }} + h_{\mathrm{rad}}\] Substitute the given values of \(h_{\text {conv }}\) and \(h_{\mathrm{rad}}\) in the equation to get the equivalent heat transfer coefficient, \(h\).
03

Calculate the equivalent thermal resistance

The thermal resistance (\(R_{\text{eq}}\)) is defined as the ratio of the temperature difference between the surface and the environment to the heat transferred through the surface. The equation for the thermal resistance is: \[R_{\text{eq}} = \frac{1}{h \cdot A} \] Substitute the calculated value of \(h\) from Step 2 and given area \(A\) in the equation to obtain the equivalent thermal resistance, \(R_{\text{eq}}\).

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer involves the movement of heat due to the flow of fluid, which could be liquid or gas, over a surface. This is a common mechanism observed in everyday activities, like boiling water or feeling the breeze on your skin. Convection occurs in two forms:
  • Natural Convection: This happens when fluid movement is caused by buoyancy forces that occur due to temperature differences in the fluid, without any external influence like fans or pumps. For example, the natural circulation of air in a room with a heater.
  • Forced Convection: This is achieved when an external force, such as a fan or a pump, moves the fluid over the surface to enhance heat transfer. An example would be the cooling system in a car engine.

In convection, the heat transfer coefficient, denoted as \(h_{\text{conv}}\), quantifies the rate of heat exchange per unit area and temperature difference between the surface and the fluid. A higher coefficient means more efficient heat transfer.
To optimize convection heat transfer, it is crucial to understand the properties of the fluid, as well as the geometry and orientation of the surface involved.
Radiation Heat Transfer
Radiation heat transfer occurs through electromagnetic waves, primarily in the infrared spectrum. This method does not require a medium, meaning it can occur in a vacuum. It is the way the Sun heats the Earth through space. All objects emit radiation energy, and the rate is influenced by their temperature and surface characteristics.
The efficiency of radiation heat transfer is represented by the radiation heat transfer coefficient, \(h_{\mathrm{rad}}\), which depends on the emissivity of the surface, its surface area, and the temperatures of the surface and surroundings.
  • Emissivity: This is a measure of a material's ability to emit thermal radiation compared to a perfect black body. A surface with high emissivity is more efficient in radiating heat.
  • Surface Area and Temperature: Larger areas and higher temperatures generally increase the rate of radiation heat transfer.
Understanding these properties helps in managing heat transfer in systems where radiation plays a significant role, such as in satellite thermal controls or in architectural designs promoting energy efficiency.
Thermal Resistance
Thermal resistance is a measure of a material's ability to resist the flow of heat. It is an essential concept in engineering, especially in thermal management of mechanical and electrical systems. The thermal resistance \(R_{\text{eq}}\) is similar to electrical resistance but deals with heat instead of electricity.
The thermal resistance for a heat transfer process is calculated using the formula:\[ R_{\text{eq}} = \frac{1}{h \cdot A} \]where \(h\) is the combined heat transfer coefficient (sum of convection and radiation coefficients) and \(A\) is the surface area. A smaller \(R_{\text{eq}}\) signifies a system that readily allows the flow of thermal energy.
  • Importance in Design: In designing systems like heat exchangers or insulating materials, minimizing thermal resistance helps to ensure efficient heat transfer.
  • Balance in Systems: Proper thermal management involves balancing thermal resistance to maintain optimal temperatures, avoiding overheating or excessive cooling.
By carefully calculating and managing thermal resistance, engineers can design systems that maintain energy efficiency while meeting functional requirements.

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Most popular questions from this chapter

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

A transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of \(70^{\circ} \mathrm{C}\) is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of \(1 \mathrm{~mm}\) and thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the solid plate is \(13 \mathrm{~mm}\) thick and has a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Inside the heated chamber, the convection heat transfer coefficient is \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the bottom surface of the solid plate is maintained at \(52^{\circ} \mathrm{C}\), determine the temperature inside the heated chamber and the surface temperature of the transparent film. Assume thermal contact resistance is negligible.

Why are the convection and the radiation resistances at a surface in parallel instead of being in series?

The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Consider two surfaces pressed against each other. Now the air at the interface is evacuated. Will the thermal contact resistance at the interface increase or decrease as a result?

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