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Chapter 6: Problem 78

Two horizontal, infinite, parallel plates are spaced a distance \(b\) apart. A viscous liquid is contained between the plates. The bottom plate is fixed, and the upper plate moves parallel to the bottom plate with a velocity \(U\). Because of the no-slip boundary condition (see Video \(V 6.12\) ), the liquid motion is caused by the liquid being dragged along by the moving boundary. There is no pressure gradient in the direction of flow. Note that this is a so-called simple Couette flow discussed in Section 6.9 .2 (a) Start with the Navier-Stokes equations and determine the velocity distribution between the plates. (b) Determine an expression for the flowrate passing between the plates (for a unit width). Express your answer in terms of \(b\) and \(U\).

Short Answer

Expert verified
The velocity distribution of the fluid between the plates is given by \(u(y) = \frac{Uy}{b}\), and the flow rate per unit width is \(\frac{U}{2}\).

Step by step solution

01

Derive the velocity distribution using Navier-Stokes equations

We can simplify the Navier-Stokes equation since the flow is steady, there is no pressure gradient (\(\frac{{dp}}{{dx}} = 0\)), and there is no motion perpendicular to the plates (\(v = w = 0\)). We get: \(\mu \frac{{d^2u}}{{dy^2}} = 0\), where \(u\) is the velocity along the plates and \(y\) is the perpendicular distance from the bottom plate. Solving this gives us the velocity profile: \(u(y) = C_1y + C_2\). The boundary conditions are \(u(0) = 0\) and \(u(b) = U\). Substituting these conditions, we find \(C_2 = 0\) and \(C_1 = \frac{U}{b}\), thus \(u(y) = \frac{Uy}{b}\).
02

Calculate the flow rate

The flow rate \(Q\) is obtained by integrating the velocity distribution over the cross-sectional area. Since the problem specifies unit width, the area just becomes the gap distance \(b\). Hence, \(Q = \int_0^b u(y) dy = \int_0^b \frac{Uy}{b} dy\). Solving this integral, we find \(Q = \frac{U}{2}\), the flowrate per unit width.

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Most popular questions from this chapter

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