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Chapter 6: Problem 22

A liquid layer separates two plane surfaces as shown. The lower surface is stationary; the upper surface moves downward at constant speed \(V\). The moving surface has width \(w,\) perpendicular to the plane of the diagram, and \(w \gg L .\) The incompressible liquid layer, of density \(\rho,\) is squeezed from between the surfaces. Assume the flow is uniform at any cross section and neglect viscosity as a first approximation. Use a suitably chosen control volume to show that \(u=V x / b\) within the gap, where \(b=b_{0}-V t .\) Obtain an algebraic expression for the acceleration of a fluid particle located at \(x .\) Determine the pressure gradient, \(\partial p / \partial x,\) in the liquid layer. Find the pressure distribution, \(p(x) .\) Obtain an expression for the net pressure force that acts on the upper (moving) flat surface.

Short Answer

Expert verified
The solution involves deriving mathematical expressions for fluid velocity, particle acceleration, pressure gradient, pressure distribution across the liquid layer, and the net pressure force acting on the upper moving surface using concepts from fluid dynamics and calculus. The specific expressions will depend on the exact values given for the fluid properties, speed of the upper surface, width and initial spacing of the plane surfaces, and initial fluid volume.

Step by step solution

01

Velocity Distribution

Using the conservation of mass, a suitable control volume can be defined within the gap. For the liquid exiting the control volume on the left side, the velocity is zero as the lower surface is stationary. For the liquid in the control volume, the velocity \(u\) is often taken as a proportion of \(V\), the speed of the upper surface, and the proportionality considers \(x\), the distance from the stationary bottom, and \(b\), the spacing between the two planes which is given as \(b = b_0 - Vt\). Thus, \(u = Vx / b\).
02

Fluid Acceleration

Acceleration of the fluid particle is determined as the rate of change of velocity with respect to time. Differential calculus can be used here to find the rate of change of the velocity expression obtained from step 1 with respect to time. Since \(\(b = b_0 - Vt\)\), its derivative with time gives \(db/dt = -V\). Upon substituting in the differential of \(u = Vx / b\), we get a mathematical expression for particle acceleration.
03

Pressure Gradient

Pressure gradient, denoted as \(\( \partial p / \partial x\)\), can be obtained by applying Newton's second law in the horizontal direction. This gives the sum of the horizontal forces equals density times acceleration. The horizontal forces are due to pressure and, hence, the pressure gradient can be expressed in terms of density and acceleration.
04

Pressure Distribution

Integrating the expression of the pressure gradient will give the pressure distribution. The constant of integration can be evaluated depending on the boundary conditions mentioned.
05

Pressure Force on Moving Surface

The pressure force on the upper (moving) surface can be achieved by integrating the distribution of pressure found in Step 4 over the area of the upper surface.

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