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Chapter 2: Problem 31

A flow is described by velocity field \(\vec{V}=a y^{2} \hat{i}+b j\) where \(a=1 \mathrm{m}^{-1} \mathrm{s}^{-1}\) and \(b=2 \mathrm{m} / \mathrm{s}\). Coordinates are measured in meters. Obtain the equation for the streamline passing through point \((6,6) .\) At \(t=1 \mathrm{s}\), what are the coordinates of the particle that passed through point (1,4) at \(t=0 ?\) At \(t=3 \mathrm{s},\) what are the coordinates of the particle that passed through point (-3,0) 2 s earlier? Show that pathlines, streamlines, and streaklines for this flow coincide.

Short Answer

Expert verified
The equation for the streamline through point (6,6) is \(x - (1/3) y^3 = -72\). At \(t=1 \mathrm{s}\), the coordinates of the particle that passed through point (1,4) at \(t=0 \) are approximately (19.33, 6). At \(t=3 \mathrm{s}\), the coordinates of the particle that passed through point (-3,0) 2s earlier are (-3,6). The pathlines, streamlines, and streaklines do coincide as the flow is steady.

Step by step solution

01

Equation for the Streamline

From the given velocity field, we get two equations \(dx/dt = ay^2\) and \(dy/dt = b\). We can solve these by separating variables and integrating. Thus, from \(dx/dt = ay^2\), we get \(dx = ay^2 dt\). Now, we need to integrate this equation. The integral of \(dx\) is \(x\) and for \(ay^2 dt\), it's \(a/3 y^3\). Thus, with the constant integration \(c\) (which will vanish in the next steps), we get \(x - a/3 y^3 = c\). Knowing we pass through (6,6), we can solve for c. When y=6, x=6. Thus, \(6 - (1/3) * (6)^3 = c\) which gives \(c = -72\). Thus, the equation for the streamline passing through (6,6) is \(x - (1/3) y^3 = -72\).
02

Coordinates of the Particle passing through (1,4) at \(t =1 \)s

We have the initial condition of \(x(0) = 1\) and \(y(0) = 4\). Since \(dy/dt = b\), We can integrate it to find \(y(t) = bt + y(0)\). Thus, \(y(1) = 2*1 + 4 = 6\). Similarly, for \(x(t)\) we know that \(x(t) = \int_{0}^{t} a*(y(s))^2 ds + x(0)\). Now this integral becomes \(x(1) = \int_{0}^{1} (4s^2 + 8s + 16) ds + 1\) which results in \(x(1) = 4/3 + 8/2 + 16 + 1 = 19.33\). Thus, at t = 1s, the coordinates of the particle that passed through (1,4) will be about (19.33, 6).
03

Coordinates of the Particle passing through (-3,0) 2s earlier

We know that the particle passed through (-3,0) at \(t = 1\)s. Following the same procedure from previous step, we can find \(y(3) = 2*3 + 0 = 6\) and \(x(t) = \int_{1}^{3} (-6s^2) ds - 3\). Solving the integral we get \(x(3) = -18/3 + 18 - 3 = -3\). So, the coordinates of the particle that passed through (-3,0) 2s earlier at \(t = 3\)s are (-3,6).
04

Verifying Pathlines, Streamlines, Streaklines Coincidence

Pathlines, streamlines, and streaklines coincide if the flow is steady, that is, the fluid properties at a point in the system do not change over time. For this to be true, \(dx/dt\) and \(dy/dt\) should be independent of \(t\). Looking at our velocity field, it is clear that \(dx/dt = a*y^2\) is dependent on \(y\) but not on \(t\), likewise \(dy/dt = b\) is a constant. Hence, it can be concluded that it is a steady flow, and therefore, pathlines, streamlines, and streaklines coincide.

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