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Calculation of pressure drag with pressure coefficient distribution

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Old   February 5, 2018, 17:26
Default Calculation of pressure drag with pressure coefficient distribution
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Xilef Smith
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Dear,

I am trying to calculate the pressure drag with an obtained pressure coefficient distribution according to: Cd_p(i) = Cp(i) * dy(i)

I have a geometry described by x and y coordinates, and dy is the difference in y direction between each point.
I therefore first calculate Cp*dy and calculate the area under the curve, which should be equal to the integral of Cp*dy. (Attached is the graph of the curve)


However, when I increase the number of points describing my geometry, automatically dy will be smaller as the distance between two points becomes smaller. Therefore, the whole Cp*dy curve shifts downwards leaving me with a smaller value of integral, hence smaller value of Cd which should not be the case.

I know I'm going wrong somewhere but can't seem to figure out where. Any help or advise is appreciated! Cheers!
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Last edited by Springroove; February 6, 2018 at 05:43.
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Old   February 6, 2018, 00:35
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Uwe Pilz
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I don't understand fully hat you try to calculate. But you have to use the same area (or line, in 2D) for calculation of the pressure. For a flow past a body you may use the whole surface. It doesn't count then how small the elements are.
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Old   February 6, 2018, 05:42
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Xilef Smith
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Thank you for your answer. I am trying to calculate the drag coefficient which is would be Cp*dy. But my area under the curve always changes with different dy, because I have Cp*dy on my y axis.

If I understand correctly, the drag coefficient is the integral of Cp*dy from the front to the back right
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Old   February 6, 2018, 06:25
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Quote:
Originally Posted by Springroove View Post
Thank you for your answer. I am trying to calculate the drag coefficient which is would be Cp*dy. But my area under the curve always changes with different dy, because I have Cp*dy on my y axis.

If I understand correctly, the drag coefficient is the integral of Cp*dy from the front to the back right
The numerical integral has smaller terms, but more of them.
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