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