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Chapter 6: Problem 29

Air flow over a stationary circular cylinder of radius \(a\) is modeled as a steady, frictionless, and incompressible flow from right to left, given by the velocity field \\[ \vec{V}=U\left[\left(\frac{a}{r}\right)^{2}-1\right] \cos \theta \hat{e}_{r}+U\left[\left(\frac{a}{r}\right)^{2}+1\right] \sin \theta \hat{e}_{\theta} \\] Consider flow along the streamline forming the cylinder surface, \(r=a .\) Express the components of the pressure gradient in terms of angle \(\theta .\) Obtain an expression for the variation of pressure (gage) on the surface of the cylinder. For \(U=75 \mathrm{m} / \mathrm{s}\) and \(a=150 \mathrm{mm},\) plot the pressure distribution (gage) and explain, and find the minimum pressure. Plot the speed \(V\) as a function of \(r\) along the radial line \(\theta=\pi / 2\) for \(r>a\) (that is, directly above the cylinder), and explain.

Short Answer

Expert verified
The pressure is minimum at the top and bottom of the cylinder directly facing and opposing to the flow. The speed decreases as we move radially outward from the surface of the cylinder.

Step by step solution

01

Calculate the velocity when r=a

We are tasked with deriving the flow's expression along the streamline that forms the surface of the cylinder, at r=a. We use the given velocity function to substitute r=a into expression and it simplifies to \(\vec{V}_{a}=0 \cdot \hat{e}_{r}+2 U \sin \theta \hat{e}_{\theta}\).
02

Derive expression for the pressure gradient components

Next, we use Euler鈥檚 equation, which relates the pressure gradient to the velocity field. This yields two components of the pressure gradient: \(\frac{\partial P}{\partial r}=-\rho U^{2} \frac{2 a^{2} \sin ^{2} \theta}{r^{3}}\) and \(\frac{1}{r} \frac{\partial P}{\partial \theta}=-\rho U^{2} \frac{2 a^{2} \sin \theta \cos \theta}{r^{3}}.\) But along the surface of the cylinder r = a, so these simplify to: \(\frac{\partial P}{\partial r}=-\rho U^{2} \frac{2 a \sin ^{2} \theta}{a^{3}}\) and \(\frac{1}{a} \frac{\partial P}{\partial \theta}=-\rho U^{2} \frac{2 a \sin \theta \cos \theta}{a^{3}}.\)
03

Derive the pressure distribution expression

The variation of pressure P can be obtained by integrating the above gradient expressions. P would be: \(P=P_{0}+\frac{\rho U^{2}}{2}\left[\left(\frac{a^{2}}{r^{2}}-1\right)^{2}-2 \cos (2 \theta)\right]\) Assuming atmospheric pressure at far field, \(P_{0}=P_{\infty}=0,\) we get the pressure difference between the cylinder surface and the far field: \(P-P_{\infty}=\frac{\rho U^{2}}{2}\left[\left(\frac{a^{2}}{r^{2}}-1\right)^{2}-2 \cos (2 \theta)\right]\)
04

Plotting the pressure distribution and finding minimum pressure

By substituting the given values (a=150 mm and U=75 m/s) into the equation obtained in Step 3, we can plot the pressure variation on the cylinder's surface as a function of \(\theta.\) By observing the graph, it is clear that the pressure is lowest at the top (胃=蟺/2) and bottom (胃=3蟺/2) of the cylinder, corresponding to the points directly facing the oncoming and opposite to the flow.
05

Plot the flow speed in radial direction above the cylinder

The speed over a point located above the cylinder (胃=蟺/2) in radial direction (r>a) is defined as: \(V_{r}=U\left[\left(\frac{a^{2}}{r^{2}}-1\right)\right]\) This shows that as we move away from the surface of the cylinder, the respective speeds decrease.

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Most popular questions from this chapter

Water flows steadily up the vertical 1-in.-diameter pipe and out the nozzle, which is 0.5 in. in diameter, discharging to atmospheric pressure. The stream velocity at the nozzle exit must be \(30 \mathrm{ft} / \mathrm{s}\). Calculate the minimum gage pressure required at section (1). If the device were inverted, what would be the required minimum pressure at section (1) to maintain the nozzle exit velocity at \(30 \mathrm{ft} / \mathrm{s} ?\)

Consider the flow represented by the stream function \(\psi=A x^{2} y,\) where \(A\) is a dimensional constant equal to 2.5 \(\mathrm{m}^{-1} \cdot \mathrm{s}^{-1}\). The density is \(1200 \mathrm{kg} / \mathrm{m}^{3}\). Is the flow rotational? Can the pressure difference between points \((x, y)=(1,4)\) and (2,1) be evaluated? If so, calculate it, and if not, explain why.

A liquid layer separates two plane surfaces as shown. The lower surface is stationary; the upper surface moves downward at constant speed \(V\). The moving surface has width \(w,\) perpendicular to the plane of the diagram, and \(w \gg L .\) The incompressible liquid layer, of density \(\rho,\) is squeezed from between the surfaces. Assume the flow is uniform at any cross section and neglect viscosity as a first approximation. Use a suitably chosen control volume to show that \(u=V x / b\) within the gap, where \(b=b_{0}-V t .\) Obtain an algebraic expression for the acceleration of a fluid particle located at \(x .\) Determine the pressure gradient, \(\partial p / \partial x,\) in the liquid layer. Find the pressure distribution, \(p(x) .\) Obtain an expression for the net pressure force that acts on the upper (moving) flat surface.

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.)

An incompressible liquid with negligible viscosity and density \(\rho=1.75\) slug/ft \(^{3}\) flows steadily through a horizontal pipe. The pipe cross- section area linearly varies from 15 in \(^{2}\) to 2.5 in \(^{2}\) over a length of 10 feet. Develop an expression for and plot the pressure gradient and pressure versus position along the pipe, if the inlet centerline velocity is \(5 \mathrm{ft} / \mathrm{s}\) and inlet pressure is 35 psi. What is the exit pressure? Hint: Use relation \\[ u \frac{\partial u}{\partial x}=\frac{1}{2} \frac{\partial}{\partial x}\left(u^{2}\right) \\]
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