How to approach a fluid dynamics problem

[shreddinglicks]

Homework Statement

Homework Equations

The Attempt at a Solution

I don't even know where to start. I don't understand the question.

 

[Chestermiller]

Have you learned that the speed of the flow is proportional to the magnitude of the gradient of the stream function?

 

[shreddinglicks]

Have you learned that the speed of the flow is proportional to the magnitude of the gradient of the stream function?

do you mean

V^2=u^2+v^2

where

 

[Chestermiller]

do you mean

V^2=u^2+v^2

where
View attachment 114202

Yes. But you need to express theta in terms of x and y to evaluate these derivatives.(except, of course, far upstream).

 

[shreddinglicks]

Yes. But you need to express theta in terms of x and y to evaluate these derivatives.(except, of course, far upstream).

So would it be better to use

and make y = rsin(theta)

 

[shreddinglicks]

Yes. But you need to express theta in terms of x and y to evaluate these derivatives.(except, of course, far upstream).

Wait, I see what you mean

 

[Chestermiller]

Wait, I see what you mean
View attachment 114204

Good.

 

[shreddinglicks]

Good.

 

[Chestermiller]

View attachment 114206

I'm not going to check your math. I leave it to you to get the math correct.

 

[shreddinglicks]

I'm not going to check your math. I leave it to you to get the math correct.

That's fine. I'm not here to learn math. Since I now have V^2 what do I do from here? I still don't understand the question I need to solve.

 

[Chestermiller]

You need to show that, at the x and y corresponding to theta = 66.8 degrees and psi = 0, the speed is the same as at y = 0, x = infinity

 

[shreddinglicks]

You need to show that, at the x and y corresponding to theta = 66.8 degrees and psi = 0, the speed is the same as at y = 0, x = - infinity

So I know at a large value of -x and y= o that my V^2 is equal to 1.

would it be appropriate to sub in
x=rcos(theta)
y=rsin(theta)
r=x^2+y^2
and then plug in 66.8 = theta

 

[Chestermiller]

So I know at a large value of -x and y= o that my V^2 is equal to 1.

would it be appropriate to sub in
x=rcos(theta)
y=rsin(theta)
r=x^2+y^2
and then plug in 66.8 = theta

You have to evaluate it at psi = 0.

 

[shreddinglicks]

You have to evaluate it at psi = 0.

Bernoulli eq?

 

[Chestermiller]

Bernoulli eq?

What about it?

 

[shreddinglicks]

What about it?

The only thing I can think of that would relate the velocity eq and pressure would be that. Is that what I should be using?

 

[Chestermiller]

The only thing I can think of that would relate the velocity eq and pressure would be that. Is that what I should be using?

No. You should be setting psi = 0 and theta = 66.8 degrees (in radians). This gives you an equation for y. Once you know y and theta, you know x.

 

[shreddinglicks]

No. You should be setting psi = 0 and theta = 66.8 degrees (in radians). This gives you an equation for y. Once you know y and theta, you know x.

I see exactly what you mean. I must be losing my mind thinking psi is pressure. I did exactly what you said and got x and y. I plugged into the the V^2 equation and got 1 as my answer.

 

[shreddinglicks]

Thanks for helping me. I can sleep easy tonight. You are the man Chestermiller!