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Chapter 6: Problem 97

A viscous fluid is contained between two infinitely long, vertical, concentric cylinders. The outer cylinder has a radius \(r_{o}\) and rotates with an angular velocity \(\omega\). The inner cylinder is fixed and has a radius \(r_{i} .\) Make use of the Navier-Stokes equations to obtain an exact solution for the velocity distribution in the gap. Assume that the flow in the gap is axisymmetric (neither velocity nor pressure are functions of angular position \(\theta\) within the gap) and that there are no velocity components other than the tangential component. The only body force is the weight.

Short Answer

Expert verified
The velocity distribution in the gap between the two cylinders is given by \( u_{θ}(r) = ωr_{i}r_{o} \frac{ln(\frac{r_{o}}{r})}{r_{o}-r_{i}}\)

Step by step solution

01

Start with governing equations

Since we have a rotational system, the most convenient coordinate system to use is cylindrical coordinates. We can write the Navier-Stokes equations in cylindrical coordinates as: \[ \rho \left( \frac{\partial u_{r}}{\partial t} + u_{r} \frac{\partial u_{r}}{\partial r} +\frac{u_{θ}}{r} \frac{\partial u_{r}}{\partial θ} - \frac{u_{θ}^{2}}{r} \right) = -\frac{\partial p}{\partial r} + \mu \frac{1}{r^{2}} \frac{\partial }{\partial r} \left(r^{2} \frac{\partial u_{r}}{\partial r} \right) - \frac{\mu u_{r}}{r^{2}}\] Here \(\mu\) is the dynamic viscosity, \(u_{r}\) is the radial velocity, and \(u_{θ}\) is the tangential velocity.
02

Apply Boundary Conditions

The problem states that the velocity has only tangential component, which means \(u_{r} = 0\) . The boundary conditions at \(r = r_{i}\) and \(r = r_{o}\) are \(u_{θ}(r_{i}) = 0\) and \(u_{θ}(r_{o}) = ωr_{o}\) respectively.
03

Simplify The Equation

Considering the problem context and imposed assumptions, the derivative terms involving \(u_{r}\) can be removed, and the Navier-Stokes equation for our case simplifies to \( \mu \frac{1}{r} \frac{d}{dr}\left(r \frac{du_{θ}}{dr}\right) = 0\) .
04

Solve The Equation

Upon integrating this simplified equation once with respect to \(r\) , we get \( \mu r \frac{du_{θ}}{dr} = A\) , where \(A\) is the constant of integration. This gives \( \frac{du_{θ}}{dr} = \frac{A}{r \mu} \) . Upon integrating this equation again with respect to \(r\) , we get \(u_{θ} = -\frac{A}{\mu}ln(r) + B \) Here \(B\) is the constant of integration.
05

Evaluate The Constants

Using the boundary condition \(u_{θ}(r_{o}) = ωr_{o}\) gives \( B = ωr_{o} +\frac{A}{\mu}ln(r_{o}) \) . Using the boundary condition \(u_{θ}(r_{i}) = 0\) gives \( 0 = -\frac{A}{\mu}ln(r_{i}) + ωr_{o} +\frac{A}{\mu}ln(r_{o}) \) . Solving this for \(A\) , the result is \( A = \muωr_{o}(ln(r_{i}) - ln(r_{o})) \) . Substituting this value into equation for \(B\) , we find \( B = ωr_{i}r_{o} \frac{ln(\frac{r_{o}}{r_{i}})}{r_{o}-r_{i}}\) .
06

Write The Final Solution

Substituting these values of \(A\) and \(B\) back into the equation for \(u_{θ}\) , we get the velocity distribution: \( u_{θ}(r) = ωr_{i}r_{o} \frac{ln(\frac{r_{o}}{r})}{r_{o}-r_{i}}\)

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