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wio
 one year ago
Best ResponseYou've already chosen the best response.1Umm, irrotational? What do you mean?

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0Ya....is that all I have to do?

wio
 one year ago
Best ResponseYou've already chosen the best response.1Hold on, I'm not sure what irrotational is. Can you give me a definition?

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0Well my textbook it as flows for which no particle rotation occurs

wio
 one year ago
Best ResponseYou've already chosen the best response.1Okay so curl is 0... meaning we need to find if it has a potential function.

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0This is impossible in reality because all fluids have viscosity, but flows can be assumed irrotational in certain cases

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta) \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta) \]This let's us solve for \(g(\theta)\)

wio
 one year ago
Best ResponseYou've already chosen the best response.1there might be another method, I'm not completely sure right now.

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0Okay. Well, I think it has to do with the gradient (del) cross velocity function (V)

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0The only thing I am confused about is whether there is a different between the irrotational flow and the incompressible flow. It seems like if I find that there is an incompressible flow, it's automatically going to give me an irrotational flow

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0My textbook defines the curl as (1/2)(del) X V.....but in rectangular coordinates it's simply defined as (1/2) (dv/dx  du/dx)

wio
 one year ago
Best ResponseYou've already chosen the best response.1What's confusing me is if we take into consideration that it is in polar coordinates, and just treat it like Cartesian coordinates.

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0Ya. I mean every example I see in my book seems to look different ha....for example, for Vr = 0 and Vtheta = f(r) (1/r)* d/dr(r*Vtheta) = 0 for irrotational flow

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0I feel like I kind of understand what is going on, but just not quite able to piece it together.

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta)\\ g'(\theta) =(5r^2/2+r) (\sin\theta)\implies g = (5r^2/2+r) (\cos\theta)+C \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{\partial f}{\partial r} =5r\cos(\theta) \implies f = \frac{5}{2}r^2\cos(\theta) +g(\theta)\\ f = 5r^2\cos(\theta)+r\cos(\theta) \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1It has a potential function, that means it's curl is 0, it's conservative, irrotational, etc.

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0Another example my prof. gave in class was this: Given: velocity field V= \[\frac{ q }{2 \Pi r } e _{r} + \frac{ K }{ 2 \Pi r } e _{}\] is it irrotational

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0and what he did was gradient x V

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0and got 0.....so I guess that is the same thing as you are saying essentially, right?

wio
 one year ago
Best ResponseYou've already chosen the best response.1Whoops, did a bit of algebra wrong there.

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{\partial f}{\partial \theta} = \frac{5}{2}r^2\sin(\theta) +g'(\theta)=5r\sin(\theta)\\ g'(\theta) =(5r^2/25r) (\sin\theta)\implies g(\theta) = (5r^2/2+r) (\cos\theta)+C \]

wio
 one year ago
Best ResponseYou've already chosen the best response.1I keep messing up trying to find the damn potential function. Maybe it doesn't have one.

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0I think it is \[del = e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} \]

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0and then that cross the the velocity vector of vr and vtheta

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0\[e _{_{r}} \frac{\partial}{\partial r} + e _{_{\theta}} \frac{ 1 }{ r } \frac{\partial}{\partial \theta} X [ e _{_{r}} 5\cos(\theta) + e _{_{\theta}} 5\sin(\theta)]\]

wio
 one year ago
Best ResponseYou've already chosen the best response.1yeah, I don't know those formula unfortunately. The fact you bring them up makes me think we're dealing with polar coords though.

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0We are definitely dealing with polar coordinates, but I think it's still the curl

wio
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{dx}{dt} = \frac{dx}{dr}\frac{dr}{dt} \]

flumech2
 one year ago
Best ResponseYou've already chosen the best response.0Okay, thanks for your help
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