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Show that the effective viscosity \(\mu_{\text {eff }}\) for the channel flow of a power-law fluid is given by $$ \mu_{\mathrm{eff}} \equiv \frac{\tau}{d u / d y}=\left(\frac{p_{1}-p_{0}}{L}\right) \frac{h^{2}}{4(n+2) \bar{u}}\left(\frac{2 y}{h}\right)^{1-n} $$ or $$ \frac{\mu_{\text {eff }}}{\mu_{\text {eff, wall }}}=\left(\frac{2 y}{h}\right)^{1-n} $$ where \(\mu_{\text {eff,wall }}\) is the value of \(\mu_{\text {eff }}\) at \(y=\pm h / 2\). Plot \(\mu_{\text {eff }} / \mu_{\text {eff, wall as a function of }} y / h\) for \(n=1,3\), and 5 .

Short Answer

Expert verified
Effective viscosity ratio is \(\left(\frac{2y}{h}\right)^{1-n}\). Plot varies with \(n\).

Step by step solution

01

Understand the Problem

We have a power-law fluid flowing in a channel. Our task is to derive expressions for the effective viscosity, \(\mu_{\text{eff}}\), and its ratio to the wall value, \(\mu_{\text{eff, wall}}\). We also need to plot the ratio as a function of the normalized position \(y/h\) for various values of the power-law index \(n\).
02

Define Key Variables and Equations

For a power-law fluid, the shear stress \(\tau\) and shear rate \(\frac{du}{dy}\) relationship is given by \(\tau = K \left(\frac{du}{dy}\right)^n\), where \(K\) is the consistency index, and \(n\) is the flow behavior index. We will use this to explore how \(\mu_{\text{eff}} = \frac{\tau}{du/dy}\) varies across the channel.
03

Derive Expression for Effective Viscosity \(\mu_{\text{eff}}\)

From the power-law relation, we express \(du/dy\) in terms of \(\tau\): \[ du/dy = \left(\frac{\tau}{K}\right)^{1/n} \]. Replace \(\tau\) with the pressure-driven flow expression in a channel: \(\frac{p_{1} - p_{0}}{L} \frac{y}{h}\). Substituting these into the definition of \(\mu_{\text{eff}}\) gives us the expression to solve for \(\mu_{\text{eff}}\).
04

Calculate \(\mu_{\text{eff}}/\mu_{\text{eff, wall}}\)

Using the derived expression for \(\mu_{\text{eff}}\), find the value of \(\mu_{\text{eff, wall}}\) by setting \(y = \frac{h}{2}\) (at the wall). The ratio can be found by dividing the general expression of \(\mu_{\text{eff}}\) by its wall value.
05

Simplify and Verify Expression

Verify that substituting \(y = \pm h/2\) into \(\mu_{\text{eff}}\) simplifies to the wall value, confirming that \(\frac{\mu_{\text{eff}}}{\mu_{\text{eff, wall}}} = \left(\frac{2y}{h}\right)^{1-n}\)\ holds true.
06

Plot \(\mu_{\text{eff}}/\mu_{\text{eff, wall}}\) vs \(y/h\)

For \(n=1, 3,\) and 5, use software or graphing tools to plot \(\frac{\mu_{\text{eff}}}{\mu_{\text{eff, wall}}}\) as a function of \(y/h\) ranging from -1 to 1. Analyze the behavior for different \(n\) values, noting the power-law effect on fluid behavior in a channel.

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Key Concepts

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

Effective Viscosity
In fluid mechanics, viscosity describes how thick or sticky a fluid is, which affects its flow. For power-law fluids, which are non-Newtonian, the viscosity is not constant. Instead, it changes depending on the shear rate, the rate at which neighboring layers of fluid move relative to each other. This is where the concept of effective viscosity, \( \mu_{\text{eff}} \), comes into play. The effective viscosity is essentially a running measure of how the fluid's resistance to flow varies under different conditions. In mathematical terms, \( \mu_{\text{eff}} = \frac{\tau}{dou/dy} \), where \( \tau \) is the shear stress, and \( du/dy \) represents the shear rate. Unlike Newtonian fluids, where viscosity is constant, the effective viscosity of power-law fluids adapts to changes in shear rate. This has significant implications in industrial processes where uniform and predictable flow is necessary. Understanding effective viscosity helps engineers and scientists tailor the flow properties of fluids to meet specific requirements. It ensures processes involving non-Newtonian fluids are efficient and effective.
Power-Law Fluid Dynamics
Power-law fluid dynamics govern the behavior of certain non-Newtonian fluids. These fluids do not have a constant relationship between shear stress and shear rate. Instead, they follow a power-law model defined by the equation \( \tau = K (du/dy)^n \), where \( K \) is the consistency index, and \( n \) is the flow behavior index.
  • If \( n = 1 \, \) the fluid behaves like a Newtonian fluid (constant viscosity).
  • If \( n < 1 \, \) the fluid is pseudoplastic (shear-thinning).
  • If \( n > 1 \, \) the fluid is dilatant (shear-thickening).
When dealing with power-law fluids, understanding \( n \) is crucial. It indicates how the fluid's viscosity responds to changes in shear rate. This dynamic nature requires more complex considerations in calculations and predictions. Power-law fluids offer practical examples such as paints, food products, and even some biological fluids like blood, that demand a tailored approach in controlling viscosity for desired applications.
Shear Stress
Shear stress is a pivotal concept in fluid dynamics, measuring the force per unit area acting parallel to the surface within a fluid. For power-law fluids, shear stress provides insight into how these materials might respond under various conditions. The formula for shear stress, \( \tau = K (du/dy)^n \, \) is central to understanding how power-law fluids function. It implies that the strain rate (or shear rate) influences the amount of shear stress experienced by the fluid. In practical terms, shear stress helps to predict and optimize the behavior of fluids under different mechanical influences. For engineers, this means they can design systems that accommodate the varying responses of power-law fluids, whether in a pipeline or an industrial process.
Channel Flow Analysis
Channel flow analysis involves understanding fluid behavior as it moves through a confined space, like a pipe or channel. For power-law fluids, channel flow offers specific challenges due to their non-uniform viscosity. The effective viscosity formula derived for channel flows helps determine how the fluid will behave from the center to the edges of the channel. It uses parameters like the pressure difference across the channel \( (\Delta p = p_1 - p_0) \), the channel height \( h \), and the average velocity \( \bar{u} \). This complex, yet insightful expression reveals how the flow profile changes, allowing engineers to predict and manage these variations. Channel flow analysis is used extensively in areas such as chemical engineering, hydraulics, and environmental assessments, where knowing the precise behavior of differing fluids can optimize the performance and safety of fluid transport systems.

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Most popular questions from this chapter

Consider a long circular cylinder of elastic-perfectly plastic material that is subjected to a torque \(T\) at its outer surface \(r=a\). The state of stress in the cylinder can be characterized by an azimuthal shear stress \(\tau\). Determine the torque for which an elastic core of radius \(c\) remains. Assume that the yield stress in shear is \(\sigma_{0}\). In the elastic region the shear stress is proportional to the distance from the axis of the cylinder \(r\). What is the torque for the onset of plastic yielding? What is the maximum torque that can be sustained by the cylinder?

According to the law of Dulong and Petit the specific heats of solids should differ only because of differences in \(M_{a}\). Calculate \(M_{a}\) and \(c\) for \(\mathrm{MgSiO}_{3}\) and \(\mathrm{MgO} .\) The measured values of \(c\) at standard conditions of temperature and pressure are 815 \(\mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1}\) for \(\mathrm{MgSiO}_{3}\) and \(924 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\) for \(\mathrm{MgO} .\) A MATLAB solution to this problem is provided in Appendix \(D\).

Consider the state of stress \(\sigma_{x x}=\sigma_{y y}=\sigma_{z z}=\sigma\) and \(\sigma_{x y}=\sigma_{y x}=\tau, \sigma_{x z}=\sigma_{z x}=\sigma_{y z}=\sigma_{z y}=0\) Determine the yield conditions on the basis of the Tresca and von Mises criteria. How does hydrostatic loading affect plastic yielding?

Obtain an order of magnitude estimate for the spring constant \(\bar{k}\) associated with the interatomic forces in a silicate crystal such as forsterite by assuming \(\bar{k} \sim E b\), where \(E\) is Young's modulus and \(b\) is the average interatomic spacing. Young's modulus for forsterite is \(1.5 \times 10^{11}\) Pa. Obtain a value for \(b\) by assuming \(b^{3}\) is the mean atomic volume. The density of forsterite is \(3200 \mathrm{~kg} \mathrm{~m}^{-3}\). Estimate the maximum amplitude of vibration of an atom in a forsterite crystal at a temperature of \(300 \mathrm{~K}\). How does it compare with the mean interatomic spacing? What is the Einstein frequency at this temperature? The spring constant may also be estimated from the compressibility of forsterite using \(\bar{k} \sim 3 b / \beta,\) where the factor of 3 arises from the relation between fractional volume changes and fractional changes in length. How does this estimate of \(\bar{k}\) compare with the previous one? The compressibility of forsterite is \(0.8 \times 10^{-11} \mathrm{~Pa}^{-1}\).

Compute the binding energy of \(\mathrm{CsCl}\). Use \(\beta_{0}=\) \(5.95 \times 10^{-11} \mathrm{~Pa}^{-1}, \rho_{0}=3988 \mathrm{~kg} \mathrm{~m}^{-3},\) and \(A=\) 1.7627. The molecular weight of \(\mathrm{CsCl}\) is 168.36 , and thermodynamic data give \(-U_{0}=660 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

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