Fundamentals of Fluid Mechanics B Questions and Answers
Question by AME536B Student
In class, you said that you would see us on Thursday - just a friendly reminder that this Thursday is a reading day.
 02.23.21
Yes, I noticed that and posted an announcement on the D2L. The next lecture in on Tuesday.
 Question by AME536B Student Dr.Parent, I did not realize that there was a deadline on the resubmission of the homework. I had a question about Assignment 2 - Question 1-b. I am still unclear as to how to proceed to prove that $S_{xy}$ becomes 0 for pure rotation. Would it be sufficient to say that $-\beta = \alpha$. Therefore, $$S_{xy} = \lim_{\Delta t\to 0} \frac{1}{(2 \Delta t)} \Bigg\{ \frac{y_{A^\prime}-y_{O^\prime}}{\xi} + \frac{x_{B^\prime}-x_{O^\prime}}{\eta} \Bigg\}$$ can be written as, $$S_{xy} = \lim_{\Delta t\to 0} \frac{1}{(2 \Delta t)}\Bigg\{ \frac {OA \sin(\alpha)}{\xi} + \frac {OB \sin(\beta)}{\eta} \Bigg \}$$ Since $OA = \xi$ and $OB = \eta$, the equation becomes, $$S_{xy} = \lim_{\Delta t\to 0} \frac{1}{(2 \Delta t)}\{ \sin(\alpha) + \sin(\beta) \}$$ Since, $$-\beta = \alpha \\ \sin(\alpha) = sin(-\beta) \\ \sin(-\beta) = -sin(\beta)$$ We can show that $S_{xy} = 0$ for pure rotation? Or would it be right to take another approach showing, like we did in class that, $$S_{xy} = \frac{1}{2} \Bigg\{ \frac {\partial u}{\partial y} + \frac{\partial v}{\partial x} \Bigg\}$$ and saying that $\beta = -\beta$ by convention since the angle is taken counterclockwise. So we can prove that $-\beta = \frac {\partial u}{\partial y}$ and $\alpha = \frac {\partial v}{\partial x}$. Now for pure rotation, $\beta$ will be positive and $\alpha = \beta$. Hence we can show that $S_{xy} = 0$. Thank you.
 02.26.21
You're on the right track, the first explanation you offer is sound but is missing a schematic to make it clear why $\beta$ should be equal to $-\alpha$ for pure rotation. On the other hand, your second explanation is confusing. For instance, you mention in a sentence that $\beta=-\beta$. This is an impossibility. Another part that is not right in your explanation is when you mention that for pure rotation $\beta$ will be positive. This is incorrect. For pure rotation $\beta$ can be either positive or negative. What matters is not the sign of $\beta$ but how $\beta$ varies with respect to $\alpha$.
 Question by AME536A Student Dr. Parent. When will the midterm exam be performed?
 03.03.21
I'd like to schedule it after 5 assignments are submitted (there are 10 assignments scheduled in this course). Let's discuss this tomorrow.
 Question by AME536B Student Dr. Parent- Is there any way you could make the assignments due at 11:59 pm on the dates they are due rather than right before class?
 03.08.21
No, because you could have a quiz related to the assignment on the due date. Thus, the assignment must be due at the latest at the start of the lecture.
 Question by AME536A Student Dr.Parent. I submitted again the revised HW3 into HW3 session in D2L. Could you please review it?
 03.09.21
Yes, you don't need to notify me about the resubmissions because I get a notice on the D2L automatically. I'll review your resubmission soon.
 Question by AME536B Student Dr. Parent, in Assignment 3, Question 4, we arrive at the equation from the $\theta - MOM$ equation as, $$\frac{1}{r} \frac{\partial }{\partial r} \Big( r\frac{\partial V_\theta}{\partial r} \Big) = \frac{V_\theta}{r^2}$$ I am not sure what a convincing explanation is for getting rid of the $\frac{V_\theta}{r^2}$ term. I assume it has something perhaps to do with the fact that $` r \textrm'$ will always lie in the annulus between $R_1$ and $R_2$, hence it will always be greater than $R_1$. While on the LHS the denominator $\partial r$ will be $R_2 - R_1$ which was given as $0.25 R_1$. From this can we say that LHS $>>$ RHS? Thank you.
You don't need to get rid of the term on the RHS. It's possible to solve the problem by rewriting the term differently and combining it with the other and then integrating. However, you can also get rid of this term if you wish in the way you mention. But in that case, you'll find that the LHS is not much greater than the RHS but about 10 times greater. Show how much greater it is first, and then provide an estimate for the error present in your final answer.
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