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Chapter 6: Problem 125

Consider the flow field formed by combining a uniform flow in the positive \(x\) direction with a sink located at the origin. Let \(U=50 \mathrm{m} / \mathrm{s}\) and \(q=90 \mathrm{m}^{2} / \mathrm{s} .\) Use a suitably chosen control volume to evaluate the net force per unit depth needed to hold in place (in standard air) the surface shape formed by the stagnation streamline.

Short Answer

Expert verified
To get the net force per unit depth needed to maintain the surface shape in place, perform the steps as outlined above. Calculate the velocity at each point of the flow field, apply Bernoulli's equation to get the pressure, calculate the differential force on an element of the droplet surface. Integrate this quantity over all the points in the surface to yield the net force.

Step by step solution

01

Calculate Velocity

Calculate the velocity of the flow at an arbitrary point (r, theta) by adding the velocities due to the sink and the uniform flow. \[V = \sqrt{(U\cos(\theta) - \frac{q}{2\pi r})^{2} + (U\sin(\theta))^{2}} \] This equation represents a field in polar coordinates where r is the distance between the point and the origin, theta is the angle in radians between the positive x-axis and the line joining the point and the origin.
02

Apply Bernoulli's equation

Use Bernoulli's equation to find the variation of pressure on the droplet. Assuming the system to be steady and inviscid, Bernoulli's equation becomes a constant along a streamline. The equation is \[p + \frac{1}{2} \rho V^{2} + \rho g z = Constant\] As the streamline under consideration (the stagnation streamline) is symmetric about the x-axis, every dA area element has a counterpart on the opposite side, leading to the vertical pressure force components cancelling out. Thus, we don't consider the 'rho g z' part of the equation. Substitute V from the step 1 into this equation, which gives us the pressure across the fluid.
03

Find the force components

The differential force on an element of the droplet surface is \[dF = -p dA n\] Where n is the unit vector normal to its differential area dA, -p dA is the differential force due to pressure. Writing the differential force in components, and observing the symmetry of the problem, we deduce that the vertical components of the pressure force on each side of the x-axis cancel out. So, we need to calculate just the horizontal component of the force: \[dF_{x} = -p dx\] The negative sign indicates that the net force is in the negative x direction (i.e., opposite to the flow).
04

Integrate to find the net force

By integrating the force over all the points in the surface, we find the net force per unit depth: \[F_{x} = -\int_{-R}^{R} p dx = -\int_{-R}^{R} (P_{\infty} - \frac{1}{2} \rho U^{2} + \frac{\rho q^{2}}{8 \pi^{2} x^{2}}) dx\] where R is the an arbitrary radius, P_infinity is the pressure when x tends to infinity. After completing this integration, the result is the net horizontal force per unit depth required to hold the shape in place.

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