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.

Question by AME536A Student For problem #1 on homework #7, we are asked to solve for the pressure inside a satellite as a function of time. You gave us the speed and mass flux in terms of the molecules, so we can solve with the energy equations. We are a bit confused on the starting point. Do we start from the isentropic relation equation and solve in conjunction with the momentum equation (like we found compressible stagnation pressure), $$ \frac{P}{\rho^{\gamma}_{}} = constant $$ or do we start the solution from the energy transport equation? $$ \frac{d}{dt}\int_{}^{}\rho EdV + \int_{}^{} \left(\rho H \overrightarrow{\rm v} \right) \cdot \overrightarrow{\rm n} ds = \int_{}^{} \dot{Q} dV $$

10.22.19

Start from the total energy transport equation. Find the answer without assuming an isentropic path. If you solve the problem assuming an isentropic path, would you get the same answer? Why or why not?

Question by AME536A Student For problem #1 on Homework #7 I am facing a problem in finding the exact expression for the pressure. Starting from the energy transport equation $\frac{d}{dt}\int_{}^{}\rho EdV + \int_{}^{} \left(\rho H \overrightarrow{\rm v} \right) \cdot \overrightarrow{\rm n} ds = 0 $ with $H=h+0.5u_{exit}^2$ where $u_{exit}=0.25\sqrt{\frac{8RT}{\pi}}$ and $E=e$ because the kinetic energy is considerable only near the hole I got $T(t)=T_i \left(\frac{\rho(t)}{\rho_i}\right)^{\frac{4\pi+1}{10\pi}}$. By substituting to the mass conservation equation I got an expression without $\gamma$. Is my process correct or I am missing something?

10.23.19

Yes you are missing a term! Don't make an assumption and remove terms unless you have no other option. Keep your expression as complete as possible.

Question by AME536A Student For Assignment 7, #1: We have gotten to this point, and despite our best efforts have not been able to get to the solution provided at the end of the assignment. Would it be possible to go over this question in class and/or receive an extension (since many of us have had at least one midterm this week)? $$ [1 - \frac{1}{20} \frac{A}{V} \sqrt{\frac{8RT}{\pi}}(\frac{8}{32\pi}+1)]^{\frac{2\gamma - 2}{\gamma}}_{} = \frac{P}{P^{}_{i}} $$

I'll give you an extension until Tuesday 11:00. The quiz will take place on Tuesday, not tomorrow.

Question by AME536A Student For Assignment 7, #1: I got the following as a solution: $$\frac{P}{P_i}=\left( 1-\frac{bA}{8V} t \sqrt{\frac{8RT_i}{\pi}} \right)^{\frac{2-2b}{b}}$$ However, the solution is : $$\frac{P}{P_i}=\left( 1-\frac{bA}{8V} t \sqrt{3RT_i} \right)^{\frac{2-2b}{b}}$$ Is there any hint for what I miss in solving this question ?	10.26.19

You have the right answer: there was a typo and it is now fixed.

Question by AME536A Student While studying for the midterm I face a problem concerning the integral form of the energy equation in a moving control volume. I derived that equation and it's the following: ​ $\frac{d}{dt} \int_{V}^{} \rho E dV +\int_{S}^{} \rho H (\vec{v}-\vec{v_{cv}}) \cdot \vec{n} dS +\int_{S}^{} P (\vec{v_{cv}} \cdot \vec{n}) dS=\dot Q$ ​ I think that the pressure term should vanish but I can't understand why. Is it possible to explain that?

10.30.19

I am not sure why you think the pressure term should vanish. You should explain your reasoning.

Question by AME536A Student I solved Question #3 Assignment #6 by using the above equation without the pressure term and I got the correct result. If I understand well that term is the work per time done by pressure, but we have already taken into account this.

The work due to pressure in the ground frame scales with the product between pressure and velocity of surface where pressure is applied (i.e. velocity of the surface with respect to the ground). If you expand $H$ in terms of $P$ in your equation and rewrite, you'll find that this is exactly the case. I see nothing wrong with the equation you posted.	10.31.19

Question by AME536A Student I have a question for Assignment #7 Question #2 Part b. After manipulation of the energy equation I arrive at $\frac{DH}{Dt}=\frac{1}{\rho} \frac{\partial P}{\partial t}$. After applying the chain rule $\frac{\partial P}{\partial t}=\frac{\partial P}{\partial s}\frac{\partial s}{\partial t}=q\frac{\partial P}{\partial s}$ where q is the velocity magnitude and s the stream line coordinates the equation becomes $\frac{DH}{Dt}=q\frac{1}{\rho} \frac{\partial P}{\partial s}$. If we want the total enthalpy to be conserved, we should have zero pressure gradient along the streampath. Is this correct?

11.03.19

You can't apply the chain rule here. You can only apply the chain rule when converting a total derivative, not a partial derivative.

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