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

Yakamoz · June 16, 2012, 14:22

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, 14:35

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

June 16, 2012, 14:22	how to calculate the viscous dissipation rate for a turbulent pipe flow	#1
Yakamoz New Member Haoran Yu Join Date: Mar 2012 Location: The Netherlands Posts: 2 Rep Power: 0	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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June 16, 2012, 14:35		#2
Yakamoz New Member Haoran Yu Join Date: Mar 2012 Location: The Netherlands Posts: 2 Rep Power: 0	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