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What is the relation \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\) known as? Under what conditions is it valid? What is the practical importance of it? \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) and what are the names of the variables in it? Under what conditions is it valid? What is the importance of it in engineering?

Short Answer

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#tag_title# Short Answer The Chilton-Colburn analogy, expressed as \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\), is a useful relation for estimating heat, mass, and momentum transfer in fluid systems, provided that the flow is turbulent, steady-state, fully developed, and the fluid properties are constant. The term \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) is known as the Chilton-Colburn mass transfer Stanton number, and it provides an alternative method for estimating mass transfer coefficients under similar conditions. This analogy and the Stanton number are important in engineering for efficiently designing heat exchangers, cooling systems, mass transfer processes, and other equipment involved in heat and mass transfer.

Step by step solution

01

Relation \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\)

The relation \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\) is known as the Chilton-Colburn analogy or Chilton-Colburn j-factor analogy. This relation is useful in determining heat, mass, and momentum transfer in fluid systems.
02

Conditions for the Chilton-Colburn Analogy

The Chilton-Colburn analogy is valid under the following conditions: 1. The fluid flow is turbulent and steady-state. 2. The flow is fully developed in a pipe or over a flat plate. 3. The fluid properties, such as viscosity, density, thermal conductivity, and diffusivity, are constant.
03

Practical Importance of the Chilton-Colburn Analogy

The Chilton-Colburn analogy is important in engineering applications because it provides a useful tool for estimating heat and mass transfer coefficients in fluid systems. This helps engineers to efficiently design heat exchangers, cooling systems, and mass transfer processes.
04

Definition of \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\)

The term \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) is known as the Chilton-Colburn mass transfer Stanton number (denoted as \(\mathrm{St}_{\text {mass }}\)). It includes the following variables: 1. \(\mathrm{Sc}\) : Schmidt number, the ratio of kinematic viscosity to mass diffusivity. 2. \(\mathrm{St}_{\text {mass }}\) : Mass transfer Stanton number, the ratio of mass transfer coefficient to average fluid velocity.
05

Conditions for the Validity of \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\)

The term \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) is valid under certain conditions: 1. The fluid flow is turbulent. 2. The fluid properties, such as viscosity, density, thermal conductivity, and diffusivity, are constant. 3. The flow is fully developed in a pipe or over a flat plate.
06

Importance of \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) in Engineering

The Chilton-Colburn mass transfer Stanton number is important in engineering because it provides an alternative method for estimating mass transfer coefficients in fluid systems. It can be used to design mass transfer equipment, such as absorbers, evaporators, and distillation columns, by analyzing fluid flow and mass transfer behavior. This helps engineers optimize equipment efficiency and performance by accurately predicting mass transfer rates.

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Key Concepts

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

Heat Transfer
Understanding heat transfer is essential for a variety of engineering applications. Heat transfer is the movement of thermal energy from one object or medium to another. It occurs through three basic modes: conduction, convection, and radiation. In engineering, being able to predict and control heat transfer can be crucial for the design of systems such as heat exchangers, radiators, and HVAC systems.

When it comes to convection, which is the transfer of heat through fluid motion, the Chilton-Colburn analogy plays a key role. This analogy helps us estimate the convective heat transfer coefficient by relating it to the fluid’s momentum transfer. The convective heat transfer coefficient is necessary for calculating the rate of heat transfer in a system, which is vital for ensuring the appropriate functioning and safety of thermal devices.
Mass Transfer
Mass transfer involves the movement of particles from one location to another, often driven by a concentration gradient. This process is central to numerous engineering applications like separation processes, chemical reactors, and membrane technology. Analogous to how heat transfer is quantified, mass transfer also relies on coefficients that assess how effectively a species is transported in a fluid system.

The Chilton-Colburn analogy is useful for estimating these mass transfer coefficients, especially when dealing with turbulent flow conditions. Understanding and using the mass transfer Stanton number, coupled with the Schmidt number, allows engineers to predict mass transfer rates accurately. Such predictions are essential for the design and optimization of equipment like distillation towers, scrubbers, and bioreactors, where precise control of mass transfer is critical.
Fluid Mechanics
Fluid mechanics is the study of fluids in motion and at rest. It is a cornerstone of many engineering fields, playing a crucial role in understanding the behavior of liquids and gases in various applications. Core concepts include flow dynamics, pressure, viscosity, and turbulence -- all crucial for describing how fluids interact with their environments.

The Chilton-Colburn analogy stems from this domain as it pertains to turbulent flow, a common occurrence in engineering systems. It draws parallels between fluid momentum and heat and mass transfer phenomena. Grasping these relationships is vital for engineers when designing and analyzing systems like pipelines, pumps, and turbines, where fluid behavior directly impacts performance and efficiency.
Engineering Applications
Engineering applications of the Chilton-Colburn analogy are far-reaching. This principle aids in the design and analysis of thermal and mass transfer equipment, offering a simplified approach to complex calculations. It streamlines processes in industries such as chemical manufacturing, environmental engineering, and energy systems.

For example, the analogy is particularly useful for heat exchanger design, allowing engineers to estimate transfer coefficients under turbulent flow conditions, which otherwise would require extensive experimentation or complex numerical methods. Additionally, it supports the design of mass transfer equipment by providing a way to correlate mass transfer rates to easily measurable fluid properties. This analogy is a powerful tool in the hands of engineers, enabling them to innovate and improve system designs across a multitude of platforms.

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Most popular questions from this chapter

Write down the relations for steady one-dimensional heat conduction and mass diffusion through a plane wall, and identify the quantities in the two equations that correspond to each other.

A sphere of ice, \(5 \mathrm{~cm}\) in diameter, is exposed to \(50 \mathrm{~km} / \mathrm{h}\) wind with 10 percent relative humidity. Both the ice sphere and air are at \(-1^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\). Predict the rate of evaporation of the ice in \(\mathrm{g} / \mathrm{h}\) by use of the following correlation for single spheres: Sh \(=\left[4.0+1.21(\mathrm{ReSc})^{2 / 3}\right]^{0.5}\). Data at \(-1^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}: D_{\text {air- } \mathrm{H}, \mathrm{O}}=2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}^{3}\), kinematic viscosity (air) \(=1.32 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), vapor pressure \(\left(\mathrm{H}_{2} \mathrm{O}\right)=\) \(0.56 \mathrm{kPa}\) and density (ice) \(=915 \mathrm{~kg} / \mathrm{m}^{3}\).

A thick wall made of natural rubber is exposed to pure oxygen gas on one side of its surface. Both the wall and oxygen gas are isothermal at \(25^{\circ} \mathrm{C}\), and the oxygen concentration at the wall surface is constant. Determine the time required for the oxygen concentration at \(x=5 \mathrm{~mm}\) to reach \(5 \%\) of its concentration at the wall surface.

Define the penetration depth for mass transfer, and explain how it can be determined at a specified time when the diffusion coefficient is known.

You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of \(0.1\)-mm-thick soft rubber and has a diameter of \(15 \mathrm{~cm}\) when inflated. The pressure and temperature inside the balloon are initially \(110 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The permeability of rubber to helium, oxygen, and nitrogen at \(25^{\circ} \mathrm{C}\) are \(9.4 \times 10^{-13}, 7.05 \times 10^{-13}\), and \(2.6 \times 10^{-13} \mathrm{kmol} / \mathrm{m} \cdot \mathrm{s} \cdot\) bar, respectively. Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first \(5 \mathrm{~h}\) assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\).

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