Show that the Poiseuille field of flow is rotational

[Ben2]

Homework Statement

From Halliday & Resnick, "Physics for Students of Science and Engineering", Problem 18.21: "The so-called Poiseuille field of flow is shown in Fig. 18-20. The spacing of the streamlines indicates that although the motion is rectilinear, there is a velocity gradient in the transverse direction. Show that such a flow is rotational."

Relevant Equations

A_1*v_1 = A_2*v_2 (Equation of continuity)
p_1 + (1/2)\rho*v_1^2 + \rho*g*y_1 = p_2 + (1/2)\rho*v_2^2 + \rho*g*y_2

I tried using these equations, but it's not clear if we should hold y_1 = y_2. A transverse velocity vector would produce a flow at some angle to the horizontal, but How do they known there's such a vector?

 

[TSny]

I tried using these equations, but ....

Is there a relevant equation from the textbook related to "rotational flow"?

A transverse velocity vector would produce a flow at some angle to the horizontal, but How do they known there's such a vector?

Note that the homework statement says that there is a velocity gradient in the transverse direction. This does not imply that the velocity itself has a transverse component. You can have a velocity gradient in the transverse direction even though the velocity at every point is horizontal.

Can you describe Fig 18-20?

 

[pasmith]

Set $\mathbf{u}(x,y,z) = u(y)\mathbf{e}_x$ and compute the curl. Does it vanish identically?

 

[TSny]

I had a look at an old copy of Halliday's text. See if problem 18.20 contains information about how to check that a velocity field is irrotational or rotational by considering the line integral of the velocity field around closed paths: $\oint \vec v \cdot \vec {ds}$. Then apply that to problem 18.21.

 

[Ben2]

Is there a relevant equation from the textbook related to "rotational flow"?
Note that the homework statement says that there is a velocity gradient in the transverse direction. This does not imply that the velocity itself has a transverse component. You can have a velocity gradient in the transverse direction even though the velocity at every point is horizontal.

Can you describe Fig 18-20?

Thanks for your timely response! I've not previously heard of a velocity gradient. Figure 18-20 features ten horizontal streamlines, where the spacing narrows from top and bottom to the middle three. Theorem 10, Chapter 13 of Stewart's "Calculus" gives the curvature k(t) = |r'(t) x r"(t)|/|r'(t)|^3. But if there's no vector in the transverse direction, I don't see how to prove the flow is rotational.

 

[TSny]

Thanks for your timely response! I've not previously heard of a velocity gradient. Figure 18-20 features ten horizontal streamlines, where the spacing narrows from top and bottom to the middle three. Theorem 10, Chapter 13 of Stewart's "Calculus" gives the curvature k(t) = |r'(t) x r"(t)|/|r'(t)|^3. But if there's no vector in the transverse direction, I don't see how to prove the flow is rotational.

I’m not sure what edition of the textbook you have. I found a very early edition. In section 18-1 it gives a qualitative description of rotational and irrotational flow. Then, in problem 18.20, a mathematical test is described for rotational flow. It says,

“A flow is a potential flow (hence irrotational) if $\oint \vec V \cdot \vec{ds} =0$ for every closed path in the field.”

In problem 18.21, you want to show that the flow shown in Fig. 18-20 is rotational. So, you need to show that there exists a closed path for which $\oint \vec V \cdot \vec{ds} \neq 0$.

In the figure, the flow is horizontal everywhere. From the spacing of the lines of flow, you can see that the speed changes as you move transversely to the direction of flow. This is the "velocity gradient in the transverse direction" mentioned in the problem statement. Can you visualize a closed path for which $\oint \vec V \cdot \vec{ds} \neq 0$?

 

[Ben2]

Thanks to Tsny and pasmith for help with this! Will do Problem 18.20 as suggested.
Ben2