Fundamentals of Fluid Mechanics Questions and Answers
 Question by AME536A Student My bad, I forgot to mention that I was referring to part b, or the proof for the particle path. I was already able to complete part A.
 10.10.21
For part (b), ask yourself this question: how would the PDE need to look like for $H$ to be conserved on a particule path? Then, set this new PDE to the energy conservation equation and equate terms and see where this leads you.
 Question by AME536A Student Is it a fair assumption to neglect the speed of the fluid flowing out of the tank for A6Q4? (We are given the volumetric flow rate, but it is not specified whether the liquid is water or something else, so the mass flow rate cannot be obtained from the volumetric flow rate, given the absence of a specified density.)
 10.11.21
I guess you mean A6Q3. If you can't determine velocity because not enough information is given, you'll need to assume that the kinetic energy is small. Just make sure you state your assumptions correctly.
 Question by AME536A Student For A6Q3 part b, after I expand the differentials and regroup the common terms, I end up with this equation: $$dm_{a}(e_{a}-h_{R}) + dm_{w}(e_{w}+h_{w}) + m_ade_a + m_wde_w = 0$$ This would imply that there are 4 integrals to be solved, and that they will all contain terms that cannot be integrated in the respective integrals (e.g. $e_a$ will be present in the integrals of $dm_a$. $dm_w$ and $de_w$). I think I am a bit confused if I am on the right track as that seems mathematically incoherent. Would you be able to provide any feedback? Thank you
 10.12.21
Hint. For a liquid, $e=h=cT$ with $c$ a constant. For a liquid, the temperature can not change in time unless there is heat transfer to the liquid because the liquid is incompressible. Thus, in this case because there is no heat transfer, the temperature of the liquid will not vary.
 Question by AME536A Student I see. That would get rid of the $dcT$ terms, but I still have an equation that includes multiple terms in the wrong integrals. If I substitute h=e=cT, I get: $$dm_a(e_a-h_R)+dm_w(2cT) + m_a de_a + m_wd(cT) = 0$$ I understant that the dcT term will go to zero as $T_1 = T_2$, but if I now multiply everything by $$\frac{1}{(e_a-h_R)(2cT)(m_a)(m_w)}$$, I get unwanted terms like $e_a$ or $cT$ in different integrals, and that does not allow me to evaluate all of them as they will act as constants in all the integrals that are in terms of other variables. For example in this $$\int_{m_{a1}}^{m_{a2}} \frac{dm_a}{(2cT)(m_a)(m_w)}$$ $m_w$ would be treated as constant and it would never cancel out with other terms in different integrals. Am I missing something else?
Some terms looks off. After simplifications are done, the equation to integrate is not so different as the one I integrated in class today.
 Question by AME536A Student The only extra term I have in that equation that differs from the part a) of the problem is $dm_w(2cT)$, but we do see a difference in the mass of the water portion of the control volume, so I think that that term would not need to be neglected or canceled.
There's a mistake somewhere because this should cancel out with another term.
 Question by AME536A Student Dr. Parent, for problem 3, assignment 6, should the $e_wdm_w$ term cancel out with the $h_wdm_w$ term? I am trying to get rid of the $dm_w$ terms, but due to the water flowing in the direction of the normal vector to the surface S, I have a $(e_w + h_w)dm_w$ term in my equation.
Not quite but.. it's something along those lines.
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