Question 10.2: The well known scientist Theodore von Karman suggested the m...
The well known scientist Theodore von Karman suggested the mixing length to be l=\chi\left|\frac{ d \bar{u} / d y}{ d ^{2} \bar{u} / d y^{2}}\right|. Using this relation drive the velocity profile near the wall of a flat-plate boundary layer flow.
We know
\tau_{t}=\mu_{t} \frac{\partial \bar{u}}{\partial y}, \text { where } \mu_{t}=\rho l^{2}\left|\frac{\partial \bar{u}}{\partial y}\right|
So, \tau_{t}=\rho l^{2}\left|\frac{\partial \bar{u}}{\partial y}\right|^{2}
Substituting von Karman’s suggestion, we get
\tau_{t}=\frac{\rho \chi^{2}( d \bar{u} / d y)^{2}( d \bar{u} / d y)^{2}}{\left( d ^{2} \bar{u} / d y^{2}\right)^{2}}=\frac{\rho \chi^{2}( d \bar{u} / d y)^{4}}{\left( d ^{2} \bar{u} / d y^{2}\right)^{2}}
or \left(\frac{ d \bar{u}}{ d y}\right)^{4}=\frac{\tau_{t}}{\rho} \frac{1}{\chi^{2}}\left(\frac{ d ^{2} \bar{u}}{ d y^{2}}\right)^{2}
Assuming \tau_{t}=\tau_{w} \text { and considering } u_{\tau}=\sqrt{\tau_{w} / \rho}
\left(\frac{ d \bar{u}}{ d y}\right)^{4}=\frac{u_{\tau}^{2}}{\chi^{2}}\left(\frac{ d ^{2} \bar{u}}{ d y^{2}}\right)^{2}
Taking the square root, and applying physical argument that (d \bar{u} / d y) cannot be imaginary, we obtain
\left(\frac{ d \bar{u}}{ d y}\right)^{2}=\pm \frac{u_{\tau}}{\chi}\left(\frac{ d ^{2} \bar{u}}{ d ^{2}}\right)
Let m= d \bar{u} / d y
then d m / d y=\pm \frac{\chi}{u_{\tau}} m^{2}
Integration yields,
-\frac{1}{m}=\pm \frac{\chi}{u_{\tau}} y+C_{1}
\text{Using}m\Rightarrow\infty\text{as}y\Rightarrow0\text{,weget}C_{1}=0
Then, \frac{ d \bar{u}}{ d y}=\frac{u_{\tau}}{\chi y}, \quad \text { since } \frac{ d \bar{u}}{ d y} \geq 0
Integrating, we obtain,
\bar{u}=\frac{u_{\tau}}{\chi} \ln y+C_{2}
At some value y=y_{0}, \bar{u}=0
Invoking this, C_{2}=-\frac{u_{\tau}}{\chi} \ln y_{0}
Thus, \frac{\bar{u}}{u_{\tau}}=\frac{1}{\chi} \cdot \ln \left(y-y_{0}\right)
Let us substitute y_{0}=\beta \frac{v}{u_{t}} order of which is same as viscous sublayer and β is an arbitrary constant.
We shall get, thus,
\frac{\bar{u}}{u_{\tau}}=\frac{1}{\chi}\left(\ln \frac{u_{\tau} y}{v}-\ln \beta\right)
or \frac{\bar{u}}{u_{\tau}}=A_{1} \ln \eta+D_{1}
This is the universal velocity profile.