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Old   November 15, 2005, 09:30
Default A simple question
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We all know that in k-epsilon model of turbulence the turbulent viscosity is calculated using the follwing expression given by launder and jones


can anyone of you guys please tell me how did he come to this expression for calculating the turbulent viscosity
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Old   November 15, 2005, 23:45
Default Re: A simple question
Harry Fulmer
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Somewhere along the line it was realized that the turbulent viscosity had to be based on a length scale multiplied by a velocity scale. k is a velocity scale, k/e is a length scale. As to why these properties were chosen... I've no idea but would like to find out as well
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Old   November 16, 2005, 02:57
Default Re: A simple question
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The dimensions of turbulent viscosity is same as that of kinematic viscosity(m2/s). Since for an eddy we have a velocity and length scale available. We get correct dimensions for turbulent viscosity if the length and velocity scales are multiplied.

turbulent viscosity proportional to U * l

This is reasonable, because it is these scales which are responsible for most of the transport by turbulent diffusion.

The turbulent length scale is taken as the square root of turbulent kinetic energy k(m2/s2) , i.e

turbulent viscosity proportional to sqrt(k) * l

for finding k a transport equation will be used . We could also use a transport equation for the turbulent length scale l, but it has been found that it is better to solve an equation for the dissipation epsilon (epsilon= u*u*u/l) . Thus we want to express turbulent viscosity in k and epsilon, i.e

turbulent viscosity = cmu*kaεb ( a and b are the respective powers)

a and b are determined by dimensional analysis.

turbulent viscosity= m2/s k= m2/s2 ε= m2/s3.

thus the values of a=2 and b= -1

so, turbulent viscosity = cmu*k2/ε .


Pavitran D
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Old   November 16, 2005, 04:37
Default Re: A simple question
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Dear Pavitran but do you really think this is based upon dimensional analysis(did you read it somewhere) Can you give me the refernces. Q
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