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Chapter 6: Problem 47

Consider the flow of a liquid of viscosity \(\mu\) and density \(\rho\) down an inclined plate making an angle \(\theta\) with the horizontal. The film thickness is \(t\) and is constant. The fluid velocity parallel to the plate is given by $$V_{x}=\frac{\rho t^{2} g \cos \theta}{2 \mu}\left[1-\left(\frac{y}{t}\right)^{2}\right]$$ where \(y\) is the coordinate normal to the plate. Calculate \(\Phi\) and \(\Psi\) for this flow and show that neither satisfies Laplace's equation. Why not?

Short Answer

Expert verified
\(\Phi = \frac{\rho t^{2} g \cos\theta}{2 \mu}\left[y-\frac{y^{3}}{3t^{2}}\right]\) and \(\Psi = \frac{\rho t^{2} g \cos \theta}{2 \mu} x \left[1-\left(\frac{y}{t}\right)^{2}\right]\). Neither \(\Phi\) nor \(\Psi\) satisfies the Laplace's equation because the given flow velocity is not a potential flow.

Step by step solution

01

Understand the given

The velocity component \(V_{x}\) for the fluid flow is given. The parameters \(g\) (acceleration due to gravity), \(\theta\) (angle of inclination), \(t\) (film thickness), \(\mu\) (viscosity), \(\rho\) (density) are all defined.
02

Derive the stream function (\(\Phi\))

The stream function \(\Phi\) is defined as that function which gives us the velocity components when differentiated partially w.r.t to the spatial coordinates. Thus, for a 2D incompressible flow, we use the following relations: \(V_{x} = \partial\Phi / \partial y\) and \(V_{y} = - \partial\Phi / \partial x\). Since the fluid is flowing parallel to the plate, there will be no movement in the y-direction. Thus the y-component of velocity \(V_{y}\) will be zero. Integrating \(V_{x}\) w.r.t \(y\) gives us the stream function \[\Phi =\int V_{x} dy = \frac{\rho t^{2} g \cos\theta}{2 \mu}\left[y-\frac{y^{3}}{3t^{2}}\right]\].
03

Derive the velocity potential (\(\Psi\))

The velocity potential \(\Psi\) is that function which when differentiated w.r.t to space coordinates provides the velocity components. The relations are \(V_{x} = \partial\Psi / \partial x\) and \(V_{y} = \partial\Psi / \partial y\) . Taking \(V_{y} = 0\) (since there's no movement in y-direction), integrating \(V_{x}\) w.r.t \(x\) (keeping \(y\) constant) will yield \(\Psi\). But since \(V_{x}\) does not depend on \(x\), we get \(\Psi = V_{x} \cdot x\), or \[\Psi = \frac{\rho t^{2} g \cos \theta}{2 \mu} x \left[1-\left(\frac{y}{t}\right)^{2}\right]\].
04

Check if \(\Phi\) and \(\Psi\) satisfy Laplace's equation

Laplace's equation states that the sum of the second order partial derivatives of a function w.r.t. the spacial coordinates should be equal to zero. If we calculate this for both \(\Phi\) and \(\Psi\), we will see that neither of them satisfies Laplace's equation. This is due to the nature of flow, the thickness of the film and the inclination of the plane.

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Most popular questions from this chapter

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