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Both Fourier's law of heat conduction and Fick's law of mass diffusion can be expressed as \(\dot{Q}=-k A(d T / d x)\). What do the quantities \(\dot{Q}, k, A\), and \(T\) represent in \((a)\) heat conduction and \((b)\) mass diffusion?

Short Answer

Expert verified
Answer: In the context of Fourier's law of heat conduction, \(\dot{Q}\) represents the heat flow rate, \(k\) represents the thermal conductivity, \(A\) represents the cross-sectional area, \(T\) represents the temperature, and \(\frac{dT}{dx}\) represents the temperature gradient. In the context of Fick's law of mass diffusion, \(\dot{Q}\) represents the mass flow rate, \(k\) represents the mass diffusion coefficient, \(A\) represents the cross-sectional area, \(T\) represents the concentration, and \(\frac{dT}{dx}\) represents the concentration gradient.

Step by step solution

01

(a) Fourier's Law of heat conduction

Fourier's Law of heat conduction is given by: \(\dot{Q} = -kA \frac{dT}{dx}\) Here, \(\dot{Q}\) represents the heat flow rate, which is the amount of heat transferred through the material per unit time (in watts). \(k\) represents the thermal conductivity of the material (in watts per meter-kelvin, W/(m·K)). It denotes the ability of the material to conduct heat. \(A\) represents the cross-sectional area through which heat is being transferred (in square meters). \(T\) is the temperature of the material at a given position x (in kelvin). \(\frac{dT}{dx}\) represents the temperature gradient in the material (in temperature units per distance unit, like K/m). It demonstrates how the temperature changes with respect to distance.
02

(b) Fick's Law of mass diffusion

Fick's Law of mass diffusion can also be represented by the same equation: \(\dot{Q} = -k A\frac{dT}{dx}\) In this context, \(\dot{Q}\) represents the mass flow rate, which is the amount of mass transferred through the material per unit time (in kg/s). The variable \(k\) represents the mass diffusion coefficient of the substance (in square meters per second, m²/s) that determines the ability of the substance to diffuse through another material. \(A\) represents the cross-sectional area through which mass is being transferred (in square meters). For Fick's law of mass diffusion, \(T\) represents the concentration of the substance being diffused at a given position x (in kg/m³ or other concentration units). \(\frac{dT}{dx}\), in this case, represents the concentration gradient in the substance (in concentration units per distance unit, like kg/(m⁴)). It demonstrates how the concentration changes with respect to distance.

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Key Concepts

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

Fourier's Law
Fourier's Law offers a foundational understanding of how heat moves through materials. It's described by the equation:\[ \dot{Q} = -kA \frac{dT}{dx} \]Here,
  • \(\dot{Q}\) is the heat flow rate, indicating how much heat passes through a material over time. It tells us how efficiently heat energy is transferred, much like how water flows through a pipe.
  • \(k\), the thermal conductivity, is critical. It measures how well a material can conduct heat. Higher \(k\) means better heat conduction, similar to how metals transmit heat faster than wood.
  • \(A\) is the cross-sectional area. Imagine it as the "window" through which heat moves. Larger windows let more heat through.
  • Lastly, \(\frac{dT}{dx}\) shows how temperature changes over distance within the material. It’s crucial for knowing where and how quickly heat spreads.
These components together form a picture of heat transfer that helps engineers design everything from cozy homes to powerful engines.
Fick's Law
Fick’s Law provides insight into how substances spread out over time due to concentration differences. It's expressed quite similarly to Fourier's Law:\[ \dot{Q} = -k A \frac{dT}{dx} \]In Fick's Law,
  • \(\dot{Q}\) represents the mass flow rate, showing how quickly mass moves from one place to another.
  • Instead of thermal conductivity, we use \(k\) for the mass diffusion coefficient. It measures how easily one substance spreads through another.
  • \(A\), the area, helps understand how much space the substance has to move through.
  • Lastly, \(T\) is about concentration rather than temperature. We are interested in the concentration gradient \(\frac{dT}{dx}\), revealing how concentration changes with distance.
Fick's Law aids in processes like pollutant dispersion in air and nutrient absorption in biology, impacting various scientific and industrial fields.
Thermal Conductivity
Thermal conductivity, denoted as \(k\), is a critical factor in the study of heat transfer. It tells us how easily heat can travel through a material. Imagine a metal rod heated at one end: if it has high thermal conductivity, the other end heats up quickly.

Understanding thermal conductivity involves:
  • Knowing that materials with high \(k\) are great conductors. Metals usually top the list, making them great for cooking pots and radiators.
  • Recognizing that low \(k\) materials, like wool or fiberglass, are better insulators, keeping heat in or out depending on the use.
Factors affecting \(k\) include the material structure, temperature, and even moisture content. Engineers manipulate these properties to design better insulation systems or improve energy efficiency in buildings.
Mass Diffusion Coefficient
The mass diffusion coefficient, often represented as \(k\) in the context of Fick's Law, measures how quickly a substance spreads within another. It plays a crucial role in many natural and industrial processes.

Key points about the mass diffusion coefficient include:
  • It varies depending on the medium and the substances involved. For example, gases diffuse faster in air than in liquids due to fewer collisions in gas.
  • Temperature, pressure, and medium properties strongly influence its value. Higher temperatures often increase diffusion rates as molecules move faster.
Understanding this coefficient helps in areas like designing chemical reactors, where uniform mixing of reactants is crucial, or in environmental science for predicting pollutant spread.

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Most popular questions from this chapter

When handling corrosive and toxic substances, chemical resistant gloves should be worn. When selecting gloves to handle a substance, the suitability of the gloves should be considered. Depending on the material of the gloves, they could be easily permeable by some substances. An employee is handling tetrachloroethylene solution for a metal-cleaning process. Dermal exposure to tetrachloroethylene can cause skin irritation, and long-term exposure to it can have adverse neurological effects on humans. As a protective measure, the employee wears rubber-blend gloves while handling the tetrachloroethylene solution. The average thickness of the gloves is \(0.67 \mathrm{~mm}\), and the mass diffusivity of tetrachloroethylene in the gloves is \(3 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\). Estimate how long can the employee's hand be in contact with the tetrachloroethylene solution before the concentration of the solution at the inner glove surface reaches \(1 \%\) of the concentration at the outer surface. Is this type of glove suitable for handling tetrachloroethylene solution?

The solubility of hydrogen gas in steel in terms of its mass fraction is given as \(w_{\mathrm{H}_{2}}=2.09 \times 10^{-4} \exp (-3950 / T) P_{\mathrm{H}_{2}}^{0.5}\) where \(P_{\mathrm{H}_{2}}\) is the partial pressure of hydrogen in bars and \(T\) is the temperature in \(\mathrm{K}\). If natural gas is transported in a 1-cm-thick, 3-m-internal-diameter steel pipe at \(500 \mathrm{kPa}\) pressure and the mole fraction of hydrogen in the natural gas is 8 percent, determine the highest rate of hydrogen loss through a 100 -m-long section of the pipe at steady conditions at a temperature of \(293 \mathrm{~K}\) if the pipe is exposed to air. Take the diffusivity of hydrogen in steel to be \(2.9 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\).

A tank with a 2-cm-thick shell contains hydrogen gas at the atmospheric conditions of \(25^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\). The charging valve of the tank has an internal diameter of \(3 \mathrm{~cm}\) and extends \(8 \mathrm{~cm}\) above the tank. If the lid of the tank is left open so that hydrogen and air can undergo equimolar counterdiffusion through the 10 -cm- long passageway, determine the mass flow rate of hydrogen lost to the atmosphere through the valve at the initial stages of the process.

Under what conditions will the normalized velocity, thermal, and concentration boundary layers coincide during flow over a flat plate?

Consider a carbonated drink in a bottle at \(37^{\circ} \mathrm{C}\) and \(130 \mathrm{kPa}\). Assuming the gas space above the liquid consists of a saturated mixture of \(\mathrm{CO}_{2}\) and water vapor and treating the drink as water, determine \((a)\) the mole fraction of the water vapor in the \(\mathrm{CO}_{2}\) gas and \((b)\) the mass of dissolved \(\mathrm{CO}_{2}\) in a 200-ml drink.

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