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.
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