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January 3, 2012, 06:11 |
BL seperation of flow over cylinder
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#1 |
New Member
Vishnu Prasad
Join Date: Sep 2010
Location: Goa
Posts: 19
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Can someone give the the solution/direct me to a solution ( I need a karman-pohlhausen solution, not a cfd solution) of point of seperation in a laminar steady flow over a cylinder?
I tried solving as given BLT by Schlichting (Chp.12), but I'm getting point of seperation as 90 degree, when its told that it should be 76. Can someone give me a the correct solution?? ( steady flow assumption, third order approximation of velocity, U(s)=2Uinf sin(),=r ) PS: If possible, when you are refering papers, limit them to ASME/Sciencedirect. My institute doesnt subscribe to many journals. |
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January 4, 2012, 13:39 |
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#2 |
Senior Member
duri
Join Date: May 2010
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Laminar flow over cylinder starts to separate at 90deg. Due to upstream propagation of pressure disturbance separation point moves upstream to some extant. Boundary layer equations are parabolic in nature, upstream propagation is not possible with boundary layer equations.
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January 4, 2012, 13:43 |
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#3 |
New Member
Vishnu Prasad
Join Date: Sep 2010
Location: Goa
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err...thats not correct. Theoretical answers are varying for some reason from 104 to 76 :|. I found this paper [ classic ] on cylinders http://naca.central.cranfield.ac.uk/...report-527.pdf that gives 104 degree seperation point but in another published about 2 years later on by NACA its about 80, and in BLT text, i found it as 76 degree. Can someone verify why the results are varying?? and what is the experimental value??
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January 4, 2012, 14:46 |
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#4 |
Senior Member
duri
Join Date: May 2010
Posts: 245
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Don't get confused by experimental results. Paper you given is over ellipse and not on cylinder. Try to understand the mathematical properties of BL equation. In favorable pressure gradient, boundary layer equation won't give zero shear stress.
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January 4, 2012, 14:48 |
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#5 |
New Member
Vishnu Prasad
Join Date: Sep 2010
Location: Goa
Posts: 19
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thank you for your help. If you take the limiting case for eccentricity = 1, what you get is the cylinder solution.
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January 4, 2012, 16:11 |
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#6 |
Senior Member
Join Date: Jul 2009
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For a circular cylinder, laminar flow will separate at roughly 80 deg. The laminar boundary equation solutions using a Pohlhausen type approximation and an inviscid pressure distribution give results around 108 deg. The reason for the discrepancy is that the pressure distribution is far from the inviscid profile past the separation point, and laminar flow is very susceptible to small adverse pressure gradients. Under ideal condtions, the flow may not separate until 108 deg, but in practice it will separate well ahead of 90 deg.
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