received_1083388766879955.jpeg

# Chemical Engineering Fluid Mechanics ## Lecture 16 ### Velocity Profiles in Turbulent Flow #### Goal - Develop equations to describe the average velocity profile in turbulent flow. #### Recall - Dimensional analysis of the "overlap layer" suggests: $\frac{d}{dy} = \frac{u_*}{\kappa y}$ Whe...

# Chemical Engineering Fluid Mechanics ## Lecture 16 ### Velocity Profiles in Turbulent Flow #### Goal - Develop equations to describe the average velocity profile in turbulent flow. #### Recall - Dimensional analysis of the "overlap layer" suggests: $\frac{d}{dy} = \frac{u_*}{\kappa y}$ Where: - $$ is the average velocity - $y$ is the distance from the wall - $u_* = \sqrt{\frac{\tau_w}{\rho}}$ is the friction velocity - $\kappa \approx 0.41$ is the Von Karman constant #### Integrating Integrating the equation above yields: $ = \frac{u_*}{\kappa} ln(y) + C$ This equation is inconvenient because: 1. It predicts $ \rightarrow \infty$ as $y \rightarrow \infty$ 2. It predicts $ \rightarrow -\infty$ as $y \rightarrow 0$ #### The law of the wall The "Law of the Wall" is given by: $\frac{}{u_*} = \frac{1}{\kappa} ln(\frac{\rho u_* y}{\mu}) + B$ Where $B \approx 5.0$ #### Dimensionless Variables It is useful to define dimensionless variables: $u^+ = \frac{}{u_*}$, $y^+ = \frac{\rho u_* y}{\mu}$ So that the "Law of the Wall" becomes: $u^+ = \frac{1}{\kappa} ln(y^+) + B$ $u^+ = \frac{1}{0.41} ln(y^+) + 5.0$ $u^+ = 2.44 ln(y^+) + 5.0$ #### Linear Viscous Sublayer Experimental data shows that near the wall, the velocity profile is linear $u^+ = y^+$ for $y^+ < 5$ #### Buffer Layer The region $5 < y^+ < 30$ is known as the "buffer layer" and is a transition region between the linear viscous sublayer and the turbulent region. #### Empirical Fit An empirical fit to experimental data that is useful for the entire turbulent region is: $u^+ = 2.5 ln(y^+) + 5.5$ or $ = u_* [2.5 ln(\frac{\rho u_* y}{\mu}) + 5.5]$ #### Example: Water in a pipe Water at $20^\circ C$ flows in a pipe of diameter $D = 1 in = 0.0254 m$ at a flow rate of $Q = 0.0005 m^3/s$. The pressure drop is measured to be $\Delta P = 4136 Pa$ over a length $L = 10 m$. 1. Estimate the friction velocity $u_*$. 2. Estimate the average velocity $$ at $y = 0.001 m$ from the wall. #### Solution 1. Force balance on a section of pipe gives: $\Delta P \frac{\pi D^2}{4} = \tau_w \pi D L$ $\tau_w = \frac{\Delta P D}{4 L} = \frac{(4136 Pa)(0.0254 m)}{4 (10 m)} = 2.63 Pa$ $u_* = \sqrt{\frac{\tau_w}{\rho}} = \sqrt{\frac{2.63 Pa}{1000 kg/m^3}} = 0.0513 m/s$ 2. $\frac{\rho u_* y}{\mu} = \frac{(1000 kg/m^3)(0.0513 m/s)(0.001 m)}{1.003 x 10^{-3} Pa \cdot s} = 51.14$ $ = (0.0513 m/s)[2.5 ln(51.14) + 5.5] = 0.35 m/s$