Velocity field calcs

Homework Help
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a velocity field is given by u=3y^2, v=2x, w=0. Is the flow steady or unsteady. is it 2D or 3D. At \vec{P}=2\hat{x}+\hat{y}, calculate the local velocity, acceleration, and convective acceleration.

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This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:
The velocity field is given by

u=3y^2,\quad v=2x,\quad w=0\,.

  1. Steadiness and Dimensionality
    • There is no explicit time dependence in the velocity components, so the flow is steady (i.e. \partial/\partial t=0).
    • The velocity has no z component and there is no dependence on z, so the flow is two-dimensional.

  2. Local Velocity at \vec{P}=2\hat{x}+\hat{y}
    At x=2 and y=1:

u(2,1)=3(1)^2=3,\quad v(2,1)=2(2)=4,\quad w(2,1)=0.

Thus, the velocity vector is

\vec{u}(2,1)=3\hat{x}+4\hat{y}+0\hat{z}\,.
  1. Local (Unsteady) Acceleration
    The local acceleration is given by the partial time derivative of \vec{u}:
\vec{a}_{\text{local}}=\frac{\partial \vec{u}}{\partial t}\,.

Because the flow is steady,

\frac{\partial \vec{u}}{\partial t}=0\,,

so

\vec{a}_{\text{local}}=\vec{0}\,.
  1. Convective Acceleration
    The convective acceleration is
\vec{a}_{\text{conv}}=\vec{u}\cdot\nabla\vec{u}\,,

and its components can be calculated as follows.

For the x-component:

\begin{aligned} \frac{\partial u}{\partial x} &= 0 \quad\text{(since $u=3y^2$ depends only on $y$)}\,,\\[1mm] \frac{\partial u}{\partial y} &= 6y\,. \end{aligned}

Thus,

a_{\text{conv},x} = u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} = (3y^2)(0) + (2x)(6y)=12xy\,.

At \vec{P} (x=2, y=1):

a_{\text{conv},x} = 12(2)(1)=24\,.

For the y-component:

\begin{aligned} \frac{\partial v}{\partial x} &= 2\,,\\[1mm] \frac{\partial v}{\partial y} &= 0 \quad\text{(since $v=2x$ depends only on $x$)}\,. \end{aligned}

Thus,

a_{\text{conv},y} = u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y} = (3y^2)(2) + (2x)(0)=6y^2\,.

At \vec{P}:

a_{\text{conv},y} = 6(1)^2=6\,.

For the z-component: Since w=0 everywhere and there is no z dependence,

a_{\text{conv},z}=0\,.

Thus, the convective acceleration is

\vec{a}_{\text{conv}}=24\hat{x}+6\hat{y}\,.
  1. Total Acceleration
    The total acceleration is the sum of local and convective accelerations:
\vec{a}=\vec{a}_{\text{local}}+\vec{a}_{\text{conv}}=24\hat{x}+6\hat{y}\,.

To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked their relevance below. If they are marked as relevant, I encourage you to take a look at them:

Textbook sections:

  1. Chapter 9.3: Why We Use Irrotational Flow Analysis Part 5 (HIGHLY RELEVANT)
  2. Appendix F.3: Volumetric Viscous Force Part 1 (HIGHLY RELEVANT)
  3. Chapter 5 Engineering Models: Quasi-steady flow (HIGHLY RELEVANT)
  4. Chapter 11.2: Couette Flow Part 2 (HIGHLY RELEVANT)
  5. Chapter 10.3: Boundary Layers: Overview Part 1 (HIGHLY RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.

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I kind of hate these problems that are in other professor’s textbooks, because they lose the connection to anything real. Real flowfields have units, so writing out a flowfield in terms of x and y is dimensionally consistent only if we assume there are hidden unit corrections all over the place. To me it is a calculus exercise that has had the engineering and physics stripped away. Anyway how do I insert a grumpy emoji… wait here we go :confused: