Fundamentals of Fluid Mechanics Questions and Answers  
Question by AME536A Student
Dr. Parent, for A5Q4, can we assume that the problem is 2D and that the height of the snow plow is negligible?
No, absolutely not. This is a 3D problem, and it's the three-dimensionality of the problem that makes this particularly hard to solve. If you assume 2D, you'll be missing the point of this question. Why would you want to assume the height of the plow is negligible?
Question by AME536A Student
Dr. Parent, I am having trouble understanding how to find an expression for the velocity at 2 for A5Q4. I understand that the velocities are not simply going to be related by a combination of sine and cosine terms, and although I think I can visualize the velocity in 3D, I am wondering whether there is another way to look at the problem. I have been spending a few days on this part of the problem, and I would greatly appreciate some input on that. Thank you so much!
Split the problem into simpler 2D parts. For instance, in the top view, the angle indicated shows that $u_2=-\cos(30^\circ)\cdot (q_2)_{xz}$. But here $(q_2)_{xz}=u_2^2+w_2^2$. Similarly, in the front view.. Then link the $(q_2)_{xz}$ and $(q_2)_{xy}$ together somehow.
Question by AME536A Student
Following to the thread above, if I solve the 2 equations I get dividing the problem as discussed, I still have v and w in my final equations: $$(q_2)^2_{xz}=\frac{w_2^2}{1+cos(30)}$$ $$(q_2)^2_{xy}=\frac{v_2^2}{1-cos(150)}$$ and if I make $$q_2^2 = (q_2)^2_{xy} + (q_2)^2_{xz}$$ I still have v and w in the expression. Those terms do not seem to cancel out if I do: $$q_2^2 = u_2^2 + v_2^2 + w_2^2$$ and set the two equations equal to each other. Am I missing something? I know that $$q_2^2 = q_1^2$$ but that does not seem to help at this stage.
No, $q_2^2$ is not equal to $(q_2)^2_{xy} + (q_2)^2_{xz}$.
Question by AME536A Student
Is there a way on the question answer portion of your website to display more than 12 Q/As at a time?
The number of Q/As per page is fixed to 12 only because more would lead to a too long rendering time (when processing the math in ).
Question by AME536A Student
During class, you mentioned that you needed to make updates to the table. Has that been done already?
Not yet. Thanks for the reminder, will do this ASAP.
Question by AME536A Student
For Q4A6, I understand that I cannot assume a steady-state condition, but I find myself stuck with this expression: $$\rho \frac{\partial H}{\partial t} - \frac{dP}{dt} + \rho v_{y}\frac{\partial H}{\partial y} + \rho v_{x}\frac{\partial H}{\partial x} = 0 $$ Could you provide me with some feedback as to what I should think about next?
If you must assume steady state to prove that $H$ is conserved on a streamline then do so.
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