Fundamentals of Fluid Mechanics Questions and Answers  
Question by AME536A Student
For HMWk 11. Can we say that the virtual mass will only depend on the linear displacement of the rotating cylinder ($\hat{i}$) because the virtual mass generated by the rotation of the cylinder is very small compared to the one generated by the linear displacement?
12.09.20
If there is a virtual mass caused by the rotation of cylinder, then this has to be included in the calculation whether small or large.
Question by AME536A Student
Just want to clarify. In class, we derived virtual mass because of the acceleration of the cylinder, which accelerates the fluid around it. Hence, the virtual mass concept needs to be taken into consideration only if the cylinder is accelerating? However, if the cylinder wouldn't be accelerating ($\frac{dU}{dt} = 0$), then the virtual mass concept won't be applied?
Yes the virtual mass concept only applies when there is an acceleration of the cylinder.
Question by AME536A Student
For Homework 11 Question 1 Part c: I am having trouble getting the correct answer. I am using the equation I derived in part b that solves for the virtual mass and plugging that into the following equation:
$F_{ext}=(m_{cyl}+m_{virtual})a_{cyl}$
Is this the wrong approach?
This is not in vector form. Acceleration is a vector and you should use the same force vector you found in part (a).
Question by AME536A Student
How do we use the same force vector from part A if there is an $\hat{i}$ and $\hat{j}$ component in part A, and the force in part c is explicitly said to be in the $\hat{i}$ direction only?
I rephrased the question, have a look at it again.
Question by AME536A Student
Hello Dr. Parent, I am looking for your guidance on Homework #11. I am having problems with properly identifying the hydraulic mass vs. the actual mass in Part (b), such that I am struggling to find the correct answer in Part (c). I was able to derive an equation for the hydraulic force acting on the cylinder in Part (a) as follows: $$ \overrightarrow{\rm F_H}=-(2{\pi}R^2L\rho\frac{{\partial}U_\infty}{{\partial}t})\overrightarrow{\rm i}+({\rho}L{\Gamma}U_\infty)\overrightarrow{\rm j} $$ With the setup of the cylinder relative to the freestream as shown in the below diagram:
PXL_20201210_063233869.jpg
With the assumption that my $\overrightarrow{\rm F_{H}}$ expression is correct, since virtual mass only occurs as an effect of acceleration in a flow, I am thinking that we only need to focus on the $x$ term of the hydraulic force: $$ \overrightarrow{\rm F_{Hx}}=-(2{\pi}R^2L\rho\frac{{\partial}U_\infty}{{\partial}t})\overrightarrow{\rm i} $$ given that it contains the acceleration term: $$ \frac{{\partial}U_\infty}{{\partial}t} $$ If so, the resulting total effective mass would be: $$ m_{Eff}=2{\pi}R^2L\rho $$ where: $$ m_{Eff}=m_{Hydraulic}+m_{Actual} $$ In class, the example we went through of a non-rotating accelerating cylinder gave us an identical expression for total effective mass of: $$ m_{Eff}=2{\pi}R^2L\rho $$ In that in-class example, we were able to simply assert that the hydraulic mass and actual mass were equivalently 1/2 of the effective mass, such that: $$ m_{Hydraulic}=m_{Actual}=\frac{1}{2}m_{eff}={\pi}R^2L\rho $$ However, in class, we (implicitly) assumed that the density of the fluid and of the cylinder were equivalent, such that no distinction had to be made between the two densities in the effective mass equation: $$ m_{Eff}=2{\pi}R^2L\rho $$ However, in our case when we get to Part (c), the fluid and the cylinder no longer have the same density, such that our clean 1/2 break of the effective mass no longer seemingly applies, as the definition of the proper density value becomes ill-defined. Thus, this discrepancy implies that more generally, the effective mass is composed of the cylinder volume times the cylinder density to give the actual mass, and the cylinder volume times the fluid density to give the hydraulic mass, with this coming out to simply 2 times the actual mass when the two densities are equivalent. In our homework problem's case, we know the actual mass of our steel cylinder is confidently defined as: $$ m_{Actual}={\pi}R^2L\rho_{Steel} $$ or the volume of the cylinder multiplied by the density of the cylinder. Does it then follow that the hydraulic mass for our cylinder is defined as the volume of the fluid displaced by the cylinder (i.e. the cylinder volume) multiplied by the density of the fluid? If this is true, then would the hydraulic mass in our case simply be: $$ m_{Hydraulic}={\pi}R^2L\rho_{Water} $$ such that our effective mass we use in Part (c) is: $$ m_{Eff}={\pi}R^2L\rho_{Water}+{\pi}R^2L\rho_{Steel} $$ and then our expression for $\overrightarrow{\rm F_{H}}$ that we apply in Part (c) becomes: $$ \overrightarrow{\rm F_H}=-(({\pi}R^2L\rho_{Water}+{\pi}R^2L\rho_{Steel})\frac{{\partial}U_\infty}{{\partial}t})\overrightarrow{\rm i}+({\rho_{Water}}L{\Gamma}U_\infty)\overrightarrow{\rm j} $$ Is this the correct logic or am I missing something? If I am missing something, could you please provide feedback as to what step and what assumptions I am making a mistake on? Any and all feedback/guidance would be greatly appreciated.
12.10.20
Yes this seems correct. However, there is a mistake in the answer given for part (c), and I didn't have time to correct it. If you don't get the right answer, it doesn't mean your logic is wrong. I'll update the answer when I get some free time on Friday or over the weekend.
Question by AME536A Student
Dr. Parent, Thank you for the quick feedback...that is a relief to here I am on the right track and I will proceed accordingly!
OK.
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