Fundamentals of Fluid Mechanics Questions and Answers

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.

Question by AME536A Student I have a question on homework #10, question #1. From the problem statement, we know that the flow around the circular part of the body follows potential flow theory, meaning we can find velocity components using potential theory definitions. I'm confused on how to include the blunt edge into a pressure coefficient drawing. Using the bernoulli equation, we can derive Cp as the following: $$ C^{}_{p} = 1 - \frac{u^{2}_{}}{u^{2}_{\infty}} = \frac{P-P^{}_{\infty}}{\frac{1}{2}\rho u^{2}_{\infty}} $$ If we can find the flow velocity, we should be able to find the Cp. The flow at the top and bottom of the circular body should be purely horizontal (and from potential flow theory we find the velocity), but I'm not sure what the velocity does after those points. We can't assume that the top and bottom of the circular body with a blunt edge are stagnation points, correct? I am stuck on how to figure out what happens to the flow velocity in the blunt region of the body ( $\theta = \frac{3\pi}{2}$ to $\frac{\pi}{2}$).

11.24.19

I guess that “blunt edge” here means the right side of the half cylinder. No, the top and the bottom of the half cylinder are not stagnation points. Hint: for the air to flow horizontally at the top and bottom edges, and to continue in that direction, the net force along $y$ on the flow at those locations must be zero. Think about what this means with respect to the pressure on the right side of the object.

Question by AME536A Student We are having trouble on Homework #10, Question #2 part B. We originally found the max velocity to be at the top of the bump with $r = 2R$ and $\theta = \frac{\pi}{2}$. We are stuck on how to find the max velocity, if it's not at this point. We are trying to use the streamline function to find velocity components, but can't find them because we don't know the radius on the bump at any point other than the top. We can't assume that the bump is a cylinder and so there isn't any information on how to find the radius at at given theta on the bump. What is the best way to look at this problem?

11.25.19

You can find $q$ as a function of $\theta$ on the bump by setting $\psi$ to a constant which is determined from $r=\infty,y=h$.

Question by AME536A Student Hi Professor, would it be possible to get an extension on homework 10?

Hm, OK. It will be due Tuesday after the Thanksgiving. Note that you will have two homeworks due on that week.

Question by AME536A Student In Assignment #10, problem #3, part (b): I am struggling to prove that $\int_S \rho \vec{v} (\vec{v}\cdot\vec{n}) dS = \frac{3}{2}\rho Q U \vec{i}$. I tried to find the integral and use the Momentum equation without any result. Is there any hint to proceed in this problem ?	12.03.19

But, can you prove that the force has no component in $\vec{j}$? This is easier to do and will guide you towards the right solution to find the component along $\vec{i}$.

Question by AME536A Student On assignment #12 on question #1, I think there is a mistake on the equation for the supersonic potential flow.	12.08.19

That's right. It has been corrected.

Question by AME536A Student For homework #12 question #2, are we able to solve for lift and drag on the different sections of the air foil and sum them together? In addition, should the question include the value for L?

Question #2 is clear, just answer the question.

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