Simple Differential Equation

  • Thread starter Menisto
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  • #1
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Homework Statement



Suppose a fluid is flowing down a pipe that has a circular cross-section of radius a. Assuming that the velocity V of the fluids depends only on its distance from the centre of the pipe, the equation satisfied by V is

(1/r)(d/dr)(r dV /dr) = -P where P is a positive constant

Find the expression for velocity given that:

1. The velocity should be finite at all point in the pipe/
2. Fluid "sticks" to boundaries (V(a) = 0 )

Show that:

V(r) = P/4 (a^2 - r^2)

The Attempt at a Solution



I integrate the first time to get:

dV/dr = -Pr/2 + c/r

and integrate again to get (FTC)

V(r) - V(a) = (-Pr^2 / 4) + C ln r [with lower bound a and upper r]

To show what I needed, it seems I only needed to get rid of that second term, but I'm unsure what assumption can justify that and where it applies.
 

Answers and Replies

  • #2
you have to apply the boundary condition at r = 0 and r = a.....
1. The velocity should be finite at all point in the pipe/
2. Fluid "sticks" to boundaries (V(a) = 0 )
Since log r blow up at r = 0, it cannot exist, otherwise it violate the first condition...
Apply the second condition yourself in order to get the constant term correct.

Good luck
 
  • #3
18
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Thank you, that works out well.
 

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