Fundamentals of Fluid Mechanics Questions and Answers

Question by AME536A Student Following the above question, if we assume constant volume, we get: $$\frac{dV_a}{dt} + \frac{dV_w}{dt} = 0$$ with $$V = \frac{m}{\rho}$$ If we cancel out the dt, we get: $$d(\frac{m_a}{\rho_a}) + d(\frac{m_w}{\rho_w}) =0$$ thus, $$d(m_w) = -\rho_w d(\frac{m_a}{\rho_a})$$ but this seems to complicate things more as now we still have the $-\rho_wd(\frac{m_a}{\rho_a})(e_w + h_w)$ term with the added densities for both fluids. Am I missing something?

10.12.21

I think I gave more than enough hints for this question. You should be able to figure the rest out on your own.

Question by AME536A Student Is the integral form of the energy equation written correctly in the tables?

Yes, I see no problem.

Question by AME536A Student For A4Q4, would the velocity of the control volume be equal to zero in the cart reference frame? Previously, when I solved the problem I set $$v_{cv} = U(t)$$ and I’m not sure if that implies that I am solving the problem in the cart’s reference frame or the ground’s.	10.20.21

When using the accelerating control volume form of the momentum equation, it's simpler to set the reference frame on the ground as we did for the rocket problem. Then $v_{\rm cv}$ and $\vec{v}$ are measured with respect to the ground.

Question by AME536A Student For assignment 8, how will submission work, given that the 11th is an observed holiday by the university?	11.06.21

Correct. I'll change the submission date to the following Tuesday.

Question by AME536A Student For A8Q1 a, when determining the stagnation points setting $v_r$ and $v_{\theta}$ equal to zero, I find a location for the stagnation point set by $$x=\frac{-Q}{2\pi U}$$ and $$y=0,\pi$$ I think the expression for x makes very little sense as the units would be meters squared, if I assume Q is in terms of volumetric flow rate. In the example problem you set Q = -Uh, so I was wondering if that can be the case for this problem too, or not.

Yes, the units for $Q$ will be the same either in the example problem in class or in this problem.

Question by AME536A Student Follow-up to the question above: I know the units of Q are the same but what I was confused about was whether Q= -Uh in this problem too.	11.07.21

Is it mentioned in the problem statement that $Q=-Uh$? If not, then it is most probably not equal to this.

Question by AME536A Student Dr. Parent, will you still have additional office hours on Mondays from 5-6 pm?	11.17.21

Yes, from now the office hours are every Monday from 1-2 pm and from 5-6 pm.

Question by AME536A Student Hi Dr. Parent. I'm working on the first problem for assignment 9 and am having a hard time getting started. My initial approach has been as follows: Find $\Psi$ as the sum of $$(\Psi) _{uniform} + (\Psi)_{Doublet}$$ along the streamline where $$r = R$$ and then using this equation to find $v_r$ and $v_{\theta}$. I am doing all of this so I can find an expression for P in terms of $\theta$ using Bernoulli's equation. This answer, however, yields me a Pressure term in terms of r only as the final answer for this becomes: $$q^2 = U_{\infty}(1-{R^2}/{r^2})^2(cos^2({\theta})+sin^2({\theta}))$$ which simplifies to $$q^2 = U_{\infty}(1-{R^2}/{r^2})^2$$ I'm assuming there is a flaw in my initial logic but I'm not sure where it is. Is there any way you can nudge me in the right direction? Thanks!

11.21.21

The pressure will change at $r=R$ because $v_{\theta}$ changes. First find $v_{\theta}$ at $r=R$.

Question by AME536A Student For assignment 9 Question 1, I find that the drag is equal to $$\vec{F} = \frac{4}{3}\rho U_{\infty}R$$ I am off by a factor of 2, and I was wondering whether that might be because I used the wrong streamfunction solution. I used the cylinder fundamental solution to find the velocities, and plug them into Bernoulli's equation, but I am not sure if I should have used the uniform fundamental solution too, in addition to the cylinder one. Any feedback is appreciated!

Using the cylinder fundamental solution is ok to find the pressure on some parts of the body, but not on all.

Question by AME536A Student I am working on question 2 part a and am unable to get the correct answer. So far, I have $\Psi_1$ and have set it equal to my streamline function, but when I try to solve this equation for r I end up with a quadratic function that cannot be simplified any further. If I try to solve for y I end up with a function in terms of r and everything cancels out when I try to assess my point of interest. is there another option I am missing?

11.22.21

Try solving it in cylindrical coordinates instead of cartesian. You should be able to find what $r$ is.

Question by AME536A Student Dr. Parent, We would like to ask that problem 4 from assignment 1 be excluded from being in the pool of questions drawn for the final. Our justification for this is that the problem more of an surface integration question and not a fluids question. Thank you for your consideration	12.02.21

The difficulty of this problem is not so much doing the integration but setting up the integral properly. And this becomes a fluid problem.. You need good understanding of fluid physics to know how to set up such an integral properly. So no, we should keep this problem for the final exam.

Question by AME536A Student Dr. Parent, We would like to ask that problem 4 from assignment 2 be excluded from being in the pool of questions drawn for the final. Our justification for this is that this problem is a more of a "know how to integrate over a half a cylinder kind of problem". Problems 3 (water plate) and 6 (mercury pool) do a better job at testing the fluids portion in our opinion. Thank you for your consideration

No, this problem is kept and is important. You should know how to integrate pressure forces over a portion of a cylinder.

Question by AME536A Student Dr. Parent, We would like to ask that problem 3 from assignment 3 be excluded from being in the pool of questions drawn for the final. Our justification for this is is a really long problem. We are of the opinion that the same content can be assessed with other problems more expediently. Thank you for your consideration

This is not a proper reason to exclude a problem and I don't think this is a very lengthy problem. You should be able to easily answer this within 30-35 minutes because you should have worked it out already.

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