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January 30, 2007, 22:11
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turbulent kinetic energy in 2D k-e model
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Guest
Posts: n/a
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Dear All,
I have a problem about the turbulent kinetic energy in 2D k-e model. I would like to ask when considering 2D cases, the turbulent kinetic energy is ((u')^2+(v')^2)/2 or ((u')^2+(v')^2+(w')^2)/2 ? Thanks, pH |
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January 31, 2007, 08:55
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Re: turbulent kinetic energy in 2D k-e model
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#2 |
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Guest
Posts: n/a
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turbulent kinetic energy is just the trace of the reynolds stress tensor so sould be, ((u')^2+(v')^2)/2
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