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how to calculate the viscous dissipation rate for a turbulent pipe flow

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Old   June 16, 2012, 13:22
Question how to calculate the viscous dissipation rate for a turbulent pipe flow
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Haoran Yu
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The transport equation for the mean kinetic energy reads as follows:

0 = -\frac{\bar{u_x}}{\rho} \frac{\partial \bar{p}}{\partial x} - \frac{1}{r} \frac{\partial}{\partial r}(r \bar{u_x}\bar{u_x^\prime u_r^\prime }) + \bar{u_x^\prime u_r^\prime} \frac{\partial \bar{u_x}}{\partial r}

if we apply Boussinesq hypothesis, the equation can be rewritten as:

0 = -\frac{\bar{u_x}}{\rho} \frac{\partial \bar{p}}{\partial x} + \frac{\mu_t}{r}\frac{\partial}{\partial r}(r \frac{\partial (\frac{\bar{u_x}^2}{2})}{\partial r}) - \mu_t (\frac{\partial \bar{u_x}}{\partial r})^2

and for a turbulent pipe flow, the dissipation rate can be defined as:

\epsilon_b = \int^{D/2}_{0}2 \pi r \epsilon(r) dr /(\pi D^2/4)

In which D is the diameter of the pipe, \rho is the density of water, these parameter are known. My question is \epsilon(r) equals to which term in the equation, is it the last term?

and how can we compute it ?

thank you in advance,

Haoran
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Old   June 16, 2012, 13:35
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Haoran Yu
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Now i get

\epsilon_b = - \frac{8 \mu_t}{D^2} \int^{D/2}_{0} r (\frac{\partial \bar{u_x}}{\partial r})^2 dr

and does anyone know how to compute this integration?

Haoran
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