/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 One-dimensional flow in a horizo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One-dimensional flow in a horizontal conduit is illustrated in Figure \(6.6,\) where the velocity profile is symmetric about the centerline of the conduit and varies with time. The differential equation describing the velocity profile, \(u(r, t),\) is given by $$\rho \frac{\partial u}{\partial t}=\mu\left(\frac{1}{r} \frac{\partial u}{\partial r}+\frac{\partial^{2} u}{\partial r^{2}}\right)$$ where \(\rho\) is the density of the fluid, \(t\) is time, \(\mu\) is the dynamic viscosity of the fluid, and \(r\) is the radial distance from the conduit centerline. Consider the case where \(R\) is the radius of the conduit and the velocity fluctuates such that the maximum velocity is \(U\) and the temporal frequency of the oscillation is \(\omega\). Using \(R, \omega,\) and \(U\) as reference variables, express the governing equation in normalized form.

Short Answer

Expert verified
The normalized equation is \( \frac{\partial u^*}{\partial t^*} = \frac{1}{\alpha^2} \left( \frac{1}{r^*} \frac{\partial u^*}{\partial r^*} + \frac{\partial^2 u^*}{\partial r^{*2}} \right) \).

Step by step solution

01

Define Dimensionless Variables

To normalize the equation, we introduce dimensionless variables based on our reference variables. Let \( r^* = \frac{r}{R} \), \( u^* = \frac{u}{U} \), and \( t^* = \omega t \). These substitutions will transform the problem to a dimensionless form, which can simplify analysis and understanding.
02

Substitute Dimensionless Variables into Equation

Substitute the dimensionless variables into the differential equation. This means replacing \( r \) with \( r^* R \), \( u \) with \( u^* U \), and \( t \) with \( t^* / \omega \). This gives us:\[\rho \frac{U}{\omega} \frac{\partial u^*}{\partial t^*} = \mu \left( \frac{1}{r^* R} \frac{\partial (u^* U)}{\partial (r^* R)} + \frac{\partial^2 (u^* U)}{\partial (r^* R)^2} \right)\]
03

Simplify Partial Derivatives

Compute the derivatives for the dimensionless and scaled terms. These transformations give:- \( \frac{\partial u}{\partial t} = U \omega \frac{\partial u^*}{\partial t^*} \)- \( \frac{\partial u}{\partial r} = \frac{U}{R} \frac{\partial u^*}{\partial r^*} \)- \( \frac{\partial^2 u}{\partial r^2} = \frac{U}{R^2} \frac{\partial^2 u^*}{\partial r^{*2}} \)
04

Insert Simplified Derivatives into Equation

Substituting these derivatives back into the main equation gives:\[\rho U \omega \frac{\partial u^*}{\partial t^*} = \mu \left( \frac{1}{r^* R} \cdot \frac{U}{R} \frac{\partial u^*}{\partial r^*} + \frac{U}{R^2} \frac{\partial^2 u^*}{\partial r^{*2}} \right)\]
05

Non-dimensionalize

Divide through the entire equation by \( \rho U \omega \), and simplify each component using the dimensionless variables:\[\frac{\partial u^*}{\partial t^*} = \frac{\mu}{\rho U \omega} \left( \frac{1}{r^*} \frac{\partial u^*}{\partial r^*} + \frac{1}{R} \frac{\partial^2 u^*}{\partial r^{*2}} \right)\]Introduce the Womersley number, \( \alpha^2 = \frac{\rho \omega R^2}{\mu} \).
06

Final Normalized Equation

The normalized governing equation becomes:\[\frac{\partial u^*}{\partial t^*} = \frac{1}{\alpha^2} \left( \frac{1}{r^*} \frac{\partial u^*}{\partial r^*} + \frac{\partial^2 u^*}{\partial r^{*2}} \right)\]This equation is now expressed in normalized form using the dimensionless Womersley number, \( \alpha \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

One-dimensional flow
In fluid dynamics, one-dimensional flow refers to a simplified flow model where the velocity field is considered to vary with one spatial dimension only, while essentially ignoring variations in other dimensions. This model is extremely useful when analyzing flows within geometries like pipes or conduits. Here, the velocity is only a function of the radial position.
This allows us to make the analysis less complex and concentrate on the core aspects of flow behavior, especially when movement varies with time or along the centerline of the conduit.
In our exercise scenario, one-dimensional flow is used to evaluate the velocity profile within a horizontal conduit, assuming it varies in time but is symmetric around the centerline.
Velocity profile
A velocity profile in fluid dynamics describes how velocity varies across a section of a flow channel, in our case, across the radius of the conduit. It gives insight into how fluids move at different positions within the conduit,
  • The velocity is typically maximum at the centerline and decreases toward the edges due to no-slip boundary conditions.
  • This pattern ensures a smooth, symmetrical velocity distribution which is crucial for predicting how fluids behave under different conditions.
In time-dependent analyses like the one in the exercise, understanding the velocity profile helps us foresee how drastic or subtle changes in velocity can impact the system. Accurately computing it requires considering factors like fluid viscosity and inertial forces.
Womersley number
The Womersley number is a dimensionless parameter, vital in oscillatory flow analysis. It provides a measure of the relative importance of viscous effects compared to inertial effects in pulsatile flows such as blood through arteries.
The Womersley number is calculated as \[\alpha = \left( \frac{\rho \omega R^2}{\mu} \right)^{0.5}\]where:
  • \(\rho\) is the fluid density,
  • \(\omega\) is the angular frequency of pulsation,
  • \(R\) is the radius of the conduit,
  • \(\mu\) is the dynamic viscosity of the fluid.
In the context of this exercise, the Womersley number assists in scaling the equations, allowing simplification of complex oscillatory flow phenomena and making the equation dimensionless.
Dimensionless analysis
Dimensionless analysis involves the process of converting physical equations into a form that retains the valid relationships and balances among the variables, without units. This conversion is achieved using dimensionless numbers and variables.
In our exercise, using dimensionless variables
  • \(r^* = \frac{r}{R},
  • u^* = \frac{u}{U},
  • t^* = \omega t\).
helps simplify and generalize the governing differential equations.
The main advantage here is that dimensionless analysis reduces the problem’s complexity and often highlights the fundamental physics by focusing on the ratios of key forces or effects in a system. It provides deeper understanding and broader applicability without being limited to specific units or conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A pump impeller of diameter \(D\) contains a mass \(m\) of fluid and rotates at an angular velocity of \(\omega\). Use dimensional analysis to obtain a functional expression for the centrifugal force \(F\) on the fluid in terms of \(D, m,\) and \(\omega\).

A 1: 60 scale model of a ship is used in a water tank to simulate a ship speed of 10 \(\mathrm{m} / \mathrm{s}\). What should be the model speed? If a towing force of \(10 \mathrm{~N}\) is measured in the model, what force is expected on the prototype? Neglect viscous effects.

Consider the case where a particle of diameter \(d\) and density \(\rho_{\mathrm{s}}\) rests on the bottom of an open channel where a liquid with density \(\rho\) and viscosity \(\mu\) flows over the particle. The particle begins to move when the fluid velocity is \(V_{\mathrm{c}},\) and in this analysis, gravity, \(g,\) is a relevant variable. (a) Determine a dimensionless group of the given variables that measures the ratio of the fluid force tending to move the particle to the weight of the particle. Take into account that the particle does not weigh the same when submerged in a liquid as it does when not submerged. (b) Using this result, perform a dimensional analysis on the relevant variables to determine a functional relationship between dimensionless groups that can be used to analyze measurements in a scale model to measure the critical fluid velocity, \(V_{\mathrm{c}},\) to move a particle of a given size, \(d\). In your dimensional analysis, incorporate the fact that \(\rho_{\mathrm{s}}\) and \(g\) can be combined into the single variable \(\left(\rho_{\mathrm{s}}-\rho\right) g .(\mathrm{c})\) If a scale model is to be built that is twice the size of the actual system, what dimensionless group(s) would you make the same in the model and the prototype? If model measurements using a 3-mm-diameter particle show a critical velocity of \(25 \mathrm{~cm} / \mathrm{s}\), what is the corresponding critical velocity and particle size in the prototype?

When an object of a given shape falls very slowly in a fluid, the steady-state velocity of the object is observed to depend only on the size of the object and the density and viscosity of the fluid. Use dimensional analysis to determine the functional relationship between the steady-state velocity and the influencing variables.

An orifice in the side of an open storage tank discharges the stored liquid at an average velocity of \(V\) when the depth of liquid above the orifice is \(h\). The liquid has a density and viscosity of \(\rho\) and \(\mu\), respectively, and the acceleration due to gravity is \(g\). (a) Show by dimensional reasoning that if viscous effects are negligible, the density of the fluid does not influence the average discharge velocity. (b) If viscous effects are negligible, use dimensional analysis to determine the functional relationship between \(V\) and the influencing variables.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.