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Chapter 6: Problem 69

The velocity of a fluid particle moving along a horizontal streamline that coincides with the \(x\) axis in a plane, two-dimensional. incompressible flow field was experimentally found to be described by the equation \(u=x^{2}\). Along this streamline determine an expression for (a) the rate of change of the \(v\) component of velocity with respect to \(y,(\mathrm{b})\) the acceleration of the particle. and (c) the pressure gradient in the \(x\) direction. The fluid is Newtonian.

Short Answer

Expert verified
The rate of change of \(v\) with respect to \(y\) is 0, the acceleration of the particle is 0, and there is no pressure gradient in the \(x\) direction.

Step by step solution

01

Determine the \(v\) Component

The velocity field of the flow is two dimensional, so \(u\) and \(v\) are the only contributors to the velocity vector \(\vec{V}\). As the flow is incompressible and there is no motion in the \(y\) direction, we can assume that the fluid velocity in the \(y\) direction or 'v' can be zero.
02

Find the Rate of Change of \(v\) with Respect to \(y\)

The rate of change of \(v\) with respect to \(y\) is simply the derivative of \(v\) with respect to \(y\). As \(v = 0\), the derivative of \(v\) with respect to \(y\) is also 0.
03

Determine the Acceleration of the Particle

The acceleration of the particle is the time derivative of the velocity. Since the velocity \(u\) does not depend on time, the acceleration \(a\) is zero.
04

Determine the Pressure Gradient in the \(x\) Direction

For an incompressible Newtonian fluid in steady flow, the equation of motion in the \(x\) direction can be expressed as: \[0 = -\frac{dp}{dx} + \mu \nabla^2 u\] where \(p\) is the pressure and \(\mu\) is the dynamic viscosity of the fluid. Given that \(u = x^2\) and \(\nabla^2 u = 0\) (since \(u\) depends only on \(x\)), it follows that \(-dp/dx = 0\). Hence, there is no pressure gradient in the \(x\) direction.

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Most popular questions from this chapter

For each of the following stream functions, with units of \(\mathrm{m}^{2} / \mathrm{s},\) determine the magnitude and the angle the velocity vector makes with the \(x\) axis at \(x=1 \mathrm{m}, y=2 \mathrm{m} .\) Locate any stagnation points in the flow field. (a) \(\psi=x y\) (b) \(\psi=-2 x^{2}+y\)

Air at \(25^{\circ} \mathrm{C}\) flows normal to the axis of an infinitely long cylinder of 1.0 -m radius. The approach velocity is \(130 \mathrm{km} / \mathrm{hr}\). Find the rotational velocity of the cylinder so that a single stagnation point occurs on the lower-most part of the cylinder. Assume potential flow.

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The two-dimensional velocity field for an incompressible Newtonian fluid is described by the relationship $$\mathbf{V}=\left(12 x y^{2}-6 x^{3}\right) \hat{\mathbf{i}}+\left(18 x^{2} y-4 y^{3} \hat{\mathbf{j}}\right.$$ where the velocity has units of \(\mathrm{m} / \mathrm{s}\) when \(x\) and \(y\) are in meters. Determine the stresses \(\sigma_{x x}, \sigma_{y y},\) and \(\tau_{x y}\) at the point \(x=0.5 \mathrm{m}\) \(y=1.0 \mathrm{m}\) if pressure at this point is \(6 \mathrm{kPa}\) and the fluid is glycerin at \(20^{\circ} \mathrm{C}\). Shew these stresses on a sketch.

(a) Show that for Poiseuille flow in a tube of radius \(R\) the magnitude of the wall shearing stress, \(\tau_{r}\), can be obtained from the relationship $$\left|\left(\tau_{r_{z}}\right)_{\mathrm{will}}\right|=\frac{4 \mu Q}{\pi R^{3}}$$ for a Newtonian fluid of viscosity \(\mu .\) The volume rate of flow is \(Q\) (b) Determine the magnitude of the wall shearing stress for a fluid having a viscosity of \(0.004 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) flowing with an average velocity of \(130 \mathrm{mm} / \mathrm{s}\) in a \(2-\mathrm{mm}\) -diameter tube.
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