Fundamentals of Fluid Mechanics Questions and Answers


B. Parent • AME536A Fundamentals of Fluid Mechanics A

Fundamentals of Fluid Mechanics Questions and Answers

Question by AME536A Student

I have a quick question about Problem 6, based on the figure provided, do we need to take into account of mercury surface tension on the side walls of the tank ?

09.11.19

If there is enough information to take this into account, then yes you have to. If not you have to mention this in the assumptions.

Question by AME536A Student

In question #2 for Assignment #3, can we assume that the duct that the fluid is traveling through is square?

09.16.19

No, you shouldn't make this assumption. Your answer should work for any type of duct as long as the intake is aligned with the $y$ axis.

Question by AME536A Student

In Assignment #6, Question #3 :
(1) Is it okay to use the conservation of mass equation twice, one for the water and one for the air? therefore we have $$\frac{dm_A}{dt}=...$$ $$\frac{dm_L}{dt}=...$$
(2) Can we assume that $e_L$ is constant with time (or the water temperature is constant) so that: $$\frac{dm_L e_L}{dt}= e_L\frac{dm_L}{dt}$$

10.14.19

Yes this is correct. However, if you do assume that the water temperature is constant in time, you need to justify this assumption. Why would this be the case? Explain. Further, only one question per post in the future.

Question by AME536A Student

On Assignment #6, Question #3 I am stuck on part B of the tank problem. We're tasked to find the temp. once all of the water has left the tank. Using the energy transport equation and mass conservation equation I've derived the following: $$ (e^{}_{a}-h^{}_r)dm^{}_{a} + (e^{}_{L}-h^{}_{L})dm^{}_{L} + m^{}_{a}de^{}_{a} = 0 $$ which can be re-written as follows in preparation for integration: $$ \frac{1}{m^{}_{a}}dm^{}_{a} + \frac{(e^{}_{L}-h^{}_{L})} {m^{}_{a}(e^{}_{a}-h^{}_r)}dm^{}_{L} + \frac{1} {e^{}_{a}-h^{}_r}de^{}_{a} = 0 $$ I'm confused on how to proceed. We can argue that water temperature is constant, so $e^{}_{l}$ and $h^{}_{l}$ won't be a problem when integrating, but all of the other variables in the second term can't be integrated with respect to $m^{}_{L}$. What is the next step?

10.16.19

At constant volume, there is a relationship between $dm_{\rm a}$ and $dm_{\rm l}$ (if a bit of air mass is added to the volume, you can find exactly what the bit of water mass must be removed). Use this relationship to get rid of $dm_{\rm l}$.

Question by AME536A Student

There is something controversial concerning the assignment number 5 question 4. In class, you told us that we have to prove with Bernoulli that the velocity at the exit is equal to the velocity at the inlet meaning that from the mass conservation the area in both inlet and exit are also equal.

If we have to follow the figure as mentioned, the area at the exit is bigger and more specifically $A_{exit}=d \frac{B}{cos(30)}=dL$ meaning that $q_{exit}=12.0284 $ m/s leading to different results.

The cross-sectional area of the reflected snow can (in fact, must!) be found through Bernoulli. It's indicated in the question statement that there is no friction, so Bernoulli's equation certainly applies. Also note that measuring cross-sectional area of the reflected snow can't be done with certainty given the drawing dimensions (in the figure, it wasn't clear what $d$ of the reflected snow stood for exactly). Also, $B/\cos(30)$ is not the width of the reflected snow — build a 3D model of the snow with a sheet of paper to convince yourself of this. Albeit coming a bit later than I would like, thank you for the question: I can see how the dimension $d$ of the reflected snow was not clear in this respect and made a change in the figure in consequence.

Question by AME536A Student

In assignment #7 Question #2, what is the difference between streamline and particle path physically?

10.20.19

Streamline is the line that is composed of the velocity vectors linked to one another. Particle path is the path of a fluid element in space and time. Note that at steady-state, the streamline and the particle path coincide. But when the flow is unsteady, they differ.

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