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shreddinglicks
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Homework Statement
Homework Equations
The Attempt at a Solution
I don't even know where to start. I don't understand the question.
do you meanChestermiller said:Have you learned that the speed of the flow is proportional to the magnitude of the gradient of the stream function?
Yes. But you need to express theta in terms of x and y to evaluate these derivatives.(except, of course, far upstream).shreddinglicks said:
So would it be better to useChestermiller said: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 meanChestermiller said:Yes. But you need to express theta in terms of x and y to evaluate these derivatives.(except, of course, far upstream).
Good.shreddinglicks said:Wait, I see what you mean
View attachment 114204
Chestermiller said:Good.
I'm not going to check your math. I leave it to you to get the math correct.shreddinglicks said:
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 said:I'm not going to check your math. I leave it to you to get the math correct.
So I know at a large value of -x and y= o that my V^2 is equal to 1.Chestermiller said: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
You have to evaluate it at psi = 0.shreddinglicks said: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
Bernoulli eq?Chestermiller said:You have to evaluate it at psi = 0.
What about it?shreddinglicks said:Bernoulli eq?
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 said:What about it?
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 said: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?
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.Chestermiller said: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.
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