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Chapter 6: Problem 17

For a certain two-dimensional flow field $$\begin{array}{l}u=0 \\\v=V\end{array}$$ (a) What are the corresponding radial and tangential velocity components? (b) Determine the corresponding stream function expressed in Cartesian coordinates and in cylindrical polar coordinates.

Short Answer

Expert verified
The radial and tangential velocity components are \(V\) and \(0\) respectively. The stream function is \(0\) in Cartesian coordinates and \(Vr\) in cylindrical polar coordinates.

Step by step solution

01

Calculate Radial and Tangential Velocity Components

For the given flow field with \(u=0\) and \(v=V\), the radial and tangential velocity components in polar coordinates can be obtained using the transformation:Radial Component \[ v_r= u cos(\theta) + v sin(\theta) \]andTangential Component \[ v_ \theta= v cos(\theta) - u sin(\theta) \]Since for the given situation, \(u=0\) and \(v=V\), \(\theta=90\) degrees. If we substitute these values in the above two equations, the radial and tangential velocity components will be:\[v_r=V \]\[v_ \theta = 0\]
02

Derive Stream Function in Cartesian Coordinates

For Cartesian coordinates, the stream function \(\psi\) is defined by:\[u= \frac{\partial \psi}{\partial y} \]\[v= -\frac{\partial \psi}{\partial x} \]Integrating the second equation w.r.t x gives:\[\frac{\partial \psi}{\partial y}= constant\]Choosing the constant as zero (since it doesn’t affect the flow field), we obtain:\[\psi = 0\]
03

Fnd Stream Function in Polar Coordinates

For cylindrical (2D) polar coordinates, the stream function \(\psi\) is defined by:\[ u = \frac{1}{r} \frac { \partial \psi } { \partial \theta } \]\[ v = \frac { \partial \psi } { \partial r } \] Integrating the second equation w.r.t r gives:\[\frac { \partial \psi } { \partial r } = V\]Thus, giving the stream function in polar coordinates as:\[\psi = V*r + constant\]The constant is chosen as zero and thus we can write:\[\psi = V*r]\]
04

Summarizing the Results

The radial velocity is \(V\) and tangential velocity is 0. The stream function in Cartesian coordinates is 0 and in cylindrical polar coordinates is \(V*r\)

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Most popular questions from this chapter

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 \(U\).

A liquid of constant density \(\rho\) and constant viscosity \(\mu\) flows down a wide, long inclined flat plate. The plate makes an angle \(\theta\) with the horizontal. The velocity components do not change in the direction of the plate, and the fluid depth, \(h,\) normal to the plate is constant. There is negligible shear stress by the air on the fluid. Find the velocity profile \(u(y),\) where \(u\) is the velocity parallel to the plate and \(y\) is measured perpendicular to the plate. Write an expression for the volume flow rate per unit width of the plate.

By considering the rotational equilibrium of a fluid mass element, show that \(\tau_{x y}=\tau_{y x}\).

The stream function for an incompressible. two-dimensional flow field is $$\psi r=3 x^{2} y+y$$ For this flow field, plot several streamlines.

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.
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