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Chapter 3: Problem 139

What is a conduction shape factor? How is it related to the thermal resistance?

Short Answer

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Question: Explain the relationship between conduction shape factor and thermal resistance. Answer: The conduction shape factor and thermal resistance are related through an equation derived from Fourier's law of heat conduction. The thermal resistance (R) is inversely proportional to the product of the conduction shape factor (S) and the material's thermal conductivity (k). As the conduction shape factor increases, the thermal resistance decreases, making the material conduct heat more easily. Conversely, as the conduction shape factor decreases, the thermal resistance increases, making the material a more effective heat insulator.

Step by step solution

01

Definition of Conduction Shape Factor

In heat transfer, the conduction shape factor (S) is a dimensionless parameter that represents the effect of the geometry of a conducting body on the rate of heat conduction. It depends on the material's shape and is used to simplify complex heat conduction problems. It is especially useful when dealing with non-uniform heat flow through conducting materials.
02

Definition of Thermal Resistance

Thermal resistance (R) is a property of a material that quantifies its resistance to heat flow. It is usually denoted by the letter 'R' and is calculated using the formula: R = \frac{螖T}{Q} Where 螖T is the temperature difference between the two sides of the material and Q is the rate of heat flow through the material (measured in watts).
03

Functional Relationship between Conduction Shape Factor and Thermal Resistance

The conduction shape factor and thermal resistance are related through the following equation, which is derived using Fourier's law of heat conduction: Q = S * k * 螖T Where: - Q is the rate of heat flow through the object (measured in watts) - S is the conduction shape factor (dimensionless) - k is the thermal conductivity of the material (measured in watts per meter-kelvin, W/m.K) - 螖T is the temperature difference between the two sides of the object (measured in kelvin, K) This equation shows that the rate of heat transfer through an object is directly proportional to the conduction shape factor and the thermal conductivity of the material. The conduction shape factor accounts for the effects of the object's geometry on heat transfer, while the thermal conductivity represents the material's intrinsic ability to conduct heat. To find the thermal resistance (R), we can rearrange the equation as follows: R = \frac{螖T}{Q} = \frac{1}{S * k} This equation shows that the thermal resistance is inversely proportional to the product of the conduction shape factor and the material's thermal conductivity. Thus, as the conduction shape factor increases, the thermal resistance decreases, and the material will conduct heat more easily. Conversely, as the conduction shape factor decreases, the thermal resistance increases, and the material becomes a more effective heat insulator.

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Key Concepts

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

Thermal Resistance
Thermal resistance is like a hurdle for heat trying to pass through a material. It describes how well a material resists the flow of heat, much like electrical resistance describes how well a material resists the flow of electricity. Imagine wrapping a cozy blanket around yourself; the blanket's thermal resistance slows down the loss of body heat to the cooler air around you.
The formula for thermal resistance is given by:
  • \( R = \frac{\Delta T}{Q} \)
Here, \( \Delta T \) is the temperature difference across the material, and \( Q \) is the rate of heat flow leaving or entering the material. High thermal resistance means the material won't let heat pass through easily. Conversely, low thermal resistance means heat can pass through quickly. This concept is crucial in designing systems where you want to control heat flow, like in building insulation or electronic gadget cooling systems.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or substance to another. It always flows from a region of higher temperature to a region of lower temperature, trying to reach thermal equilibrium, meaning the temperatures become equal.
There are three main modes of heat transfer:
  • Conduction: This occurs when heat is transferred through direct contact. For example, when a metal spoon gets hot after being left in a pot of boiling water.
  • Convection: This is the heat transfer due to fluid motion, such as the rising of warm air and the sinking of cool air in the atmosphere.
  • Radiation: This method does not require a medium, such as when the sun's heat warms your face even through the vacuum of space.
Each mode has its applications and significance, but in the context of conduction and thermal resistance, understanding heat transfer through materials is essential for everything from heat shielding materials to everyday cooking.
Fourier's Law
Fourier's Law is like the guiding rule for heat conduction. It provides a simple formula to calculate the rate at which heat moves through a material. This principle is foundational in understanding how the conduction shape factor and thermal resistance work together.
The law states:
  • \( Q = -k \cdot A \cdot \frac{dT}{dx} \)
Where:
  • \( Q \) is the heat transfer per unit time (in watts).
  • \( k \) is the thermal conductivity of the material.
  • \( A \) is the cross-sectional area through which heat flows.
  • \( \frac{dT}{dx} \) is the temperature gradient across the material.
Fourier's Law helps in calculating the heat transfer by conduction, especially when the temperature difference, material conductivity, and cross-sectional dimensions are known. This is the basis for understanding how changes in shape or material affect heat transfer rate, as seen with the conduction shape factor.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. Think of it as the opposite of thermal resistance. A higher thermal conductivity means a material is better at allowing heat to pass through, whereas lower thermal conductivity means it's not as good.
A few key points about thermal conductivity include:
  • Materials like metals, such as copper and aluminum, have high thermal conductivities. That's why they feel cold to the touch 鈥� they quickly draw heat away from your hand.
  • Materials like wood or rubber have low thermal conductivities, making them good insulators.
  • Thermal conductivity is denoted by the symbol \( k \) and is measured in watts per meter-kelvin (W/m.K).
Understanding thermal conductivity is crucial in engineering and design for applications like designing heat sink materials for electronics or insulating buildings to maintain temperature efficiently. High thermal conductivity materials are used where heat dissipation is important, while low-conductivity materials are used where insulation is key.

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Most popular questions from this chapter

Consider two walls, \(A\) and \(B\), with the same surface areas and the same temperature drops across their thicknesses. The ratio of thermal conductivities is \(k_{A} / k_{B}=4\) and the ratio of the wall thicknesses is \(L_{A} / L_{B}=2\). The ratio of heat transfer rates through the walls \(\dot{Q}_{A} / \dot{Q}_{B}\) is (a) \(0.5\) (b) 1 (c) \(2 \quad(d) 4\) (e) 8

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

Superheated steam at an average temperature \(200^{\circ} \mathrm{C}\) is transported through a steel pipe \(\left(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{o}=8.0 \mathrm{~cm}\right.\), \(D_{i}=6.0 \mathrm{~cm}\), and \(L=20.0 \mathrm{~m}\) ). The pipe is insulated with a 4-cm thick layer of gypsum plaster \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulated pipe is placed horizontally inside a warehouse where the average air temperature is \(10^{\circ} \mathrm{C}\). The steam and the air heat transfer coefficients are estimated to be 800 and \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) the daily rate of heat transfer from the superheated steam, and \((b)\) the temperature on the outside surface of the gypsum plaster insulation.

Consider the conditions of Example \(3-14\) in the text except that the ambient air is at a temperature of \(30^{\circ} \mathrm{C}\). A person with skin/fat layer thickness of \(0.003 \mathrm{~m}\) is doing vigorous exercise which raises the metabolic heat generation rate from 700 to \(7000 \mathrm{~W} / \mathrm{m}^{3}\) over a period of time. Calculate the perspiration rate required in lit/s so as to maintain the skin temperature at \(34^{\circ} \mathrm{C}\). Use the perspiration properties to be the same as that of liquid water at the average surface skin temperature of \(35.5^{\circ} \mathrm{C}\).

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a \(15-\mathrm{cm}-\) thick wall with a thermal conductivity of \(k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)
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