/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A flow field is given by \(\vec{... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 15

A flow field is given by \(\vec{V}=A x i+2 A y\), where \(A=2 s^{-1}\) Verify that the parametric equations for particle motion are given by \(x_{p}=c_{1} e^{A t}\) and \(y_{p}=c_{2} e^{2 u_{l}} .\) Obtain the equation for the pathline of the particle located at the point \((x, y)=(2,2)\) at the instant \(t=0 .\) Compare this pathline with the streamline through the same point.

Short Answer

Expert verified
Also, the parametric equations given were verified. The equation obtained for the pathline is \(x_p = 2 e^{2t}\) and \(y_p = 2 e^{4t}\) and for the streamline is \(x y^2 = 8\), which is different than the one for pathline, indicating different directions for the path of the particle and the flow at that point.

Step by step solution

01

Verify the Parametric Equations

The given velocity vector is \(\vec{V}=2x\hat{i}+4y\hat{j}\). Starting with \(x\), the motion equation is \(\frac{dx}{dt} = 2x\). Solving this differential equation, we'll find \(x(t)=c_1 e^{2t}\), which matches with \(x_p\) given. Now proceeding with \(y\), the motion equation is \(\frac{dy}{dt} = 4y\). Solving this, we'll find \(y(t) = c_2 e^{4t}\), which matches \(y_p\). This verifies the given parametric equations.
02

Obtain Equation for Pathline

To find the equations of the pathline, we substitute \(c_1\) and \(c_2\) using the initial conditions at \(t=0\), where \((x, y) = (2,2)\). Hence, \(c_1 = x(0) = 2\) and \(c_2 = y(0) = 2\). Then we can obtain the equations for the pathline as \(x_p = 2 e^{2t}\) and \(y_p = 2 e^{4t}\).
03

Obtain Equation for Streamline

The streamline equation can be obtained by integrating the formula \(dy/dx = V_y/V_x = 4y/2x = 2y/x\). Simplifying gives us the streamline equation \(x y^2 = k\), where \(k\) is an arbitrary constant. Applying the initial condition \((x, y) = (2,2)\) gives \(k = 8\). So the equation for the streamline is \(x y^2 = 8\).

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Most popular questions from this chapter

The velocity for a steady, incompressible flow in the \(x y\) plane is given by \(\vec{V}=i A / x+i A y / x^{2},\) where \(A=2 \mathrm{m}^{2} / \mathrm{s},\) and the coordinates are measured in meters. Obtain an equation for the streamline that passes through the point \((x, y)=\) (1,3). Calculate the time required for a fluid particle to move from \(x=1 \mathrm{m}\) to \(x=2 \mathrm{m}\) in this flow field.

The velocity distribution for laminar flow between parallel plates is given by \\[\frac{u}{u_{\max }}=1-\left(\frac{2 y}{h}\right)^{2}\\] where \(h\) is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at \(15^{\circ} \mathrm{C},\) with \(u_{\max }=0.10 \mathrm{m} / \mathrm{s}\) and \(h=0.1 \mathrm{mm} .\) Calculate the shear stress on the upper plate and give its direction. Sketch the variation of shear stress across the channel.

A concentric cylinder viscometer may be formed by rotating the inner member of a pair of closely fitting cylinders. For small clearances, a linear velocity profile may be assumed in the liquid filling the annular clearance gap. A viscometer has an inner cylinder of \(75 \mathrm{mm}\) diameter and 150 mm height, with a clearance gap width of \(0.02 \mathrm{mm}\) A torque of \(0.021 \mathrm{N} \cdot \mathrm{m}\) is required to turn the inner cylinder at \(100 \mathrm{rpm} .\) Determine the viscosity of the liquid in the clearance gap of the viscometer.

A 73 -mm-diameter aluminum \((\mathrm{SG}=2.64)\) piston of 100 \(\mathrm{mm}\) length resides in a stationary 75 -mm-inner-diameter steel tube lined with SAE \(10 \mathrm{W}-30\) oil at \(25^{\circ} \mathrm{C}\). A mass \(m=2 \mathrm{kg}\) is suspended from the free end of the piston. The piston is set into motion by cutting a support cord. What is the terminal velocity of mass \(m ?\) Assume a linear velocity profile within the oil.

Crude oil, with specific gravity \(\mathrm{SG}=0.85\) and viscosity \(\mu=2.15 \times 10^{-3}\) lbf \(\cdot \mathrm{s} / \mathrm{ft}^{2},\) flows steadily down a surface inclined \(\theta=45\) degrees below the horizontal in a film of thickness \(h=0.1\) in. The velocity profile is given by \\[u=\frac{\rho g}{\mu}\left(h y-\frac{y^{2}}{2}\right) \sin \theta\\] (Coordinate \(x\) is along the surface and \(y\) is normal to the surface.) Plot the velocity profile. Determine the magnitude and direction of the shear stress that acts on the surface.
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