Fundamentals of Fluid Mechanics Questions and Answers  
Question by AME536A Student
From today's class, you showed that you can have circulation in an irrotational flow. Hence,$\nabla \times \vec{V}$ is not a condition to have circulation. However, can you say, that every rotational flow has circulation? Or in other words, could you have a rotational flow that doesn't have circulation?
As I mentioned at the end of the class we'll cover the physical significance of circulation and critical circulation next week.
Question by AME536A Student
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Would it be possible to get some feedback please? I don't know what your x refers too.
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Feedback is limited if you don't follow the rules. No two equal signs on the same line. Follow the instructions written at the top of the assignment.
Question by AME536A Student
Prob 3b) Can you elaborate now? thanks! Recall, for an irrotational and incompressible flow: $$ \nabla \phi = \vec{V}$$ $$ \nabla^2 \phi = 0 $$ \indent Recall continuity equation $$ \dot{\rho} + \nabla\cdot(\rho\vec{V}) = 0 $$ From continuity and assuming a uniform density: $$ \nabla\cdot\vec{V} = 0 $$ Hence, the fluid is incompressible $$ \therefore \frac{\partial(\nabla\phi)}{t} + \frac{1}{2}\nabla(\vec{V}\cdot\vec{V}) + \frac{\nabla P}{\rho} = 0 $$ Factorizing: $$ \nabla ( \frac{\partial\phi}{\partial t} + \frac{1}{2}\vec{V}\cdot\vec{V} + \frac{P}{\rho} ) = 0 $$ Hence, this will only be true if the terms are uniform across the flow at one instant
Resubmit your assignment following the instructions and I will give you more feedback.
Question by AME536A Student
For Assignment 9 Question 1 B, after integrating, I arrived at $F=\frac{1}{3}\rho U^2R$
While troubleshooting the answer, I noticed that I had not integrated for the surface integral, only the arc length that was resultant of the radius. Is the dimension for the length of the circle into the page 8? If not, what is it?
I don't understand. What do you mean the “length of the circle into the page 8”?
Question by AME536A Student
Continuation from the last question;
We know:
Given by the problem we can make the following conversion:
$dS=Rd\theta dL$
where L is the length of the shape normal to the surface of the page. The problem does not give the dimension of L, so can we therefore assume it to be 8? If not, what is it?
Why should this be 8? This doesn't make sense to me. 8 what?
Question by AME536A Student
The problem does not give dimensions for the radius so let's just say 8[units]. I think it should be 8 because the answer does not have units of length in it so therefore this length must be incorporated in the integral. As the problem currently stands, this length is not defined so we cannot integrate over the entire surface.
This is a mess and makes no sense. The radius is denoted by $R$ in the question and can not be substituted arbitrarily by 8. You have to keep it as $R$.
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