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-   -   how to calculate the viscous dissipation rate for a turbulent pipe flow (http://www.cfd-online.com/Forums/main/103300-how-calculate-viscous-dissipation-rate-turbulent-pipe-flow.html)

Yakamoz June 16, 2012 13:22

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

Yakamoz June 16, 2012 13:35

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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