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Chapter 6: Problem 37

The \(x\) component of velocity in a two-dimensional, incompressible flow field is given by \(u=A x y ;\) the coordinates are measured in meters and \(A=2 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1} .\) There is no velocity component or variation in the \(z\) direction. Calculate the acceleration of a fluid particle at point \((x, y)=(2,1) .\) Estimate the radius of curvature of the streamline passing through this point. Plot the streamline and show both the velocity vector and the acceleration vector on the plot. (Assume the simplest form of the \(y\) component of velocity.)

Short Answer

Expert verified
The acceleration of a fluid particle at point P (2,1) is \(8 \, \mathrm{m/s^2}\). The radius of curvature of the streamline passing through this point is 4 meters. The velocity vector has a magnitude of \(4 \, \mathrm{m/s}\) and the acceleration vector has a magnitude of \(8 \, \mathrm{m/s^2}\). Both vectors point in the positive x direction. The sketch will illustrate that the streamline, velocity vector and acceleration vector are located at the point P(2,1)

Step by step solution

01

Calculate the Velocity of the Fluid at the Point P (2,1)

Firstly, we need to calculate the velocity of the fluid at the point \((2,1)\). This could be done using the equation for the x-component of the velocity given: \(u = Axy\). The constant A is given as 2 \(\mathrm{m^{-1} \cdot s^{-1}}\) and at point P, x itself equals 2m and y equals 1m. So the x-component of velocity, \(u\), can be calculated by substituting these values into the given equation: \(u = 2(2)(1) = 4 \, \mathrm{m/s}\).
02

Calculate Acceleration of the Fluid Particle

In this step, we apply the formula for fluid particle acceleration in a steady flow, which is: \(a = u \, (du/dx)\); \(du/dx\) is the derivative of \(u\) with respect to \(x\) and can be calculated based on the formula for \(u\). Differentiating the expression \(u = 2xy\), treating y as a constant: \(du/dx = 2y = 2(1) = 2 \, \mathrm{m^{-1}}\), hence, \(a = u \, (du/dx) = 4 \times 2= 8\, \mathrm{m/s^2}\).
03

Estimation of the Radius of Curvature of the Streamline

The formula for the radius of curvature, \(R\), of the streamline at point \((x, y)\) in a two-dimensional flow is given by: \(R = u^2/(du/dy)\). To find \(du/dy\), we differentiate the equation \(u = 2xy\) with respect to \(y\) (this time treating \(x\) as a constant): \(du/dy = 2x = 4 \, \mathrm{m^{-1}}\). Thus the radius of curvature at point \((x, y)\) is: \(R = u^2/(du/dy) = (4)^2/4 = 4 \, \mathrm{m}\).
04

Plotting the Streamline

Plotting the streamline, velocity vector and acceleration vector requires graphing knowledge and tools. The streamline passes through the point 2,1 with slope 1/2 (calculated from \(du/dy\)). The velocity vector being tangent to the streamline at point P, going in the positive x direction with magnitude 4 m/s. The acceleration vector points in the same direction with magnitude 8 m/s^2.

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