Question 10.3: Using Karman’s relation l = X [du/dy / d²u/dy²], show that t...
Using Karman’s relation l=\chi\left|\frac{ d \bar{u} / d y}{ d ^{2} \bar{u} / d y^{2}}\right|, show that the universal velocity distribution in a fully developed channel flow (Fig. 10.11) is given by
\frac{U_{\max }-\bar{u}}{u_{\tau}}=-\frac{1}{\chi}\left[\ln \left(1-\sqrt{\frac{y}{h}}\right)+\sqrt{\frac{y}{h}}\right]where, 2h is the height of the channel, y is the distance measured from the centre line of the channel and χ is an empirical constant. The pressure gradient in flow direction is – (dp/dx).
From Reynolds equation, we get
\frac{\partial\left(\tau_{t}\right)}{\partial y}=\frac{\partial \bar{p}}{\partial x}
or \tau_{t}=\left(\frac{\partial \bar{p}}{\partial x}\right) y+C_{1}
At y = 0, \tau_{t}=0, \text { that makes } C_{1} = 0
Thus, \tau_{t}=\left(\frac{\partial \bar{p}}{\partial x}\right) y
and \tau_{w}=\left(\frac{\partial \bar{p}}{\partial x}\right) h
we get \frac{\tau_{t}}{\tau_{w}}=\frac{y}{h}
From Karman’s relation, we can write
\tau_{t}=\frac{\rho \chi^{2}(\partial \bar{u} / \partial y)^{4}}{\left(\partial^{2} \bar{u} / \partial y^{2}\right)^{2}}
Then \tau_{w}=\frac{h \rho \chi^{2}(\partial \bar{u} / \partial y)^{4}}{y\left(\partial^{2} \bar{u} / \partial y^{2}\right)^{2}}
and \tau_{w}=\left(\frac{\partial \bar{p}}{\partial x}\right) h
we get \frac{\tau_{t}}{\tau_{w}}=\frac{y}{h}
From Karman’s relation, we can write
\tau_{t}=\frac{\rho \chi^{2}(\partial \bar{u} / \partial y)^{4}}{\left(\partial^{2} \bar{u} / \partial y^{2}\right)^{2}}
Then \tau_{w}=\frac{h \rho \chi^{2}(\partial \bar{u} / \partial y)^{4}}{y\left(\partial^{2} \bar{u} / \partial y^{2}\right)^{2}}
u_{\tau}^{2}=\frac{h \chi^{2}(\partial \bar{u} / \partial y)^{4}}{y\left(\partial^{2} \bar{u} / \partial y^{2}\right)^{2}}
Thus, \frac{\partial^{2} \bar{u}}{\partial y^{2}}=\pm \frac{\chi}{u_{\tau}} \sqrt{\frac{h}{y}}\left(\left|\frac{\partial \bar{u}}{\partial y}\right|\right)^{2}
Substituting for m=\frac{\partial \bar{u}}{\partial y} and integrating,
-\frac{1}{m}=\pm 2 \frac{\chi}{u_{\tau}} \sqrt{h y}+C_{2}
\text { at } y=h, m \rightarrow \infty \text { and } C_{2}=\pm 2 \frac{\chi}{u_{\tau}} h
Now, we know that \frac{\partial \bar{u}}{\partial y} \leq 0 \text { for } y>0 and we write
d \bar{u}=-\frac{u_{\tau}}{2 \chi h} \int \frac{ d y}{1-\sqrt{y / h}}
Substituting for \xi=1-\sqrt{\frac{y}{h}} and integrating,
\bar{u}=\frac{u_{\tau}}{\chi}[\ln \xi-\xi]+C_{3}
or \bar{u}=\frac{u_{\tau}}{\chi}\left[\ln \left(1-\sqrt{\frac{y}{h}}\right)-\left(1-\sqrt{\frac{y}{h}}\right)\right]+C_{3}
\text { at } y=0, \bar{u}=U_{\max }
C_{3}=U_{\max }+\frac{u_{\tau}}{\chi}
Finally, \frac{U_{\max }-\bar{u}}{u_{\tau}}=\frac{1}{\chi}\left[\ln \left(1-\sqrt{\frac{y}{h}}\right)-\sqrt{\frac{y}{h}}\right]