Fundamentals of Fluid Mechanics Questions and Answers  
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 ?
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?
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.
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?
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 $\frac{\pi}{2}$).
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.
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