For the following velocity profiles, determine whether the flow has separated or on the verge of separation or will be attached with the surface.

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written 3.0 years ago by

RakeshBhuse • 3.2k

• modified 3.0 years ago

Solution :

Given :

$1^{st}$ Velocity Profile

$\begin{aligned} \frac{u}{U} &=\frac{3}{2}\left(\frac{y}{\delta}\right)-\frac{1}{2}\left(\frac{y}{\delta}\right)^{3} \\ \quad u &=\frac{3 U}{2}\left(\frac{y}{\delta}\right)-\frac{U}{2}\left(\frac{y}{\delta }\right)^{3} \end{aligned}$

Differentiating w.r.t. $y$ ,

$\frac{\partial u}{\partial y}=\frac{3 U}{2} \times \frac{1}{\delta }-\frac{U}{2} \times 3\left(\frac{y}{\delta }\right)^{2} \times \frac{1}{\delta }$

At $y=0$ ,

$\begin{aligned}\left(\frac{\partial u}{\partial y}\right)_{y=0} &=\frac{3 U}{2\delta }-\frac{3 U}{2}\left(\frac{0}{\delta }\right)^{2} \times \frac{1}{\delta }\\ &=\frac{3 U}{2\delta }\end{aligned}$

As $\left(\frac{\partial u}{\partial y}\right)_{y=0}$ is positive.

Hence flow will not separate or flow will remain attached with the surface.

$2^{nd}$ Velocity Profile

$\begin{aligned} \frac{u}{U} &=2\left(\frac{y}{\delta}\right)^{2}-\left(\frac{y}{\delta}\right)^{3} \\ u &=2 U\left(\frac{y}{\delta}\right)^{2}-U\left(\frac{y}{\delta}\right)^{3} \end{aligned}$

Differentiating w.r.t. $y$ ,

$\frac{\partial u}{\partial y}=4 U \left(\frac{y}{\delta}\right) \times \frac{1}{\delta}-3U \left(\frac{y}{\delta}\right)^{2} \times \frac{1}{\delta}$

At $y=0,$

$\begin{aligned}\left(\frac{\partial u}{\partial y}\right)_{y=0} &=4 U \times \left(\frac{0}{\delta}\right) \frac{1}{\delta}-3U \times \left(\frac{0}{\delta}\right)^{2} \frac{1}{\delta}\\ &=0 \end{aligned}$

As $\left(\frac{\partial u}{\partial y}\right)_{N=0}=0$ , the flow is on the verge of separation.

$3^{rd}$ Velocity Proflle

$\begin{aligned} \frac{u}{U} &=-2\left(\frac{y}{\delta}\right)+\left(\frac{y}{\delta}\right)^{2} \\ u &=-2 U\left(\frac{y}{\delta}\right)+U\left(\frac{y}{\delta}\right)^{2}\end{aligned}$

Differentiating w.r.t. $y$ ,

$\frac{\partial u}{\partial y} =-2 U\left(\frac{1}{\delta}\right)+2 U\left(\frac{y}{\delta}\right) \times \frac{1}{\delta}$

at $y=0,$

$\begin{aligned}\left(\frac{\partial u}{\partial y}\right)_{y=0} &=-\frac{2 U}{\delta}+2 U\left(\frac{\partial}{\delta}\right) \times \frac{1}{\delta} \\ &=-\frac{2 U}{\delta}\end{aligned}$

As $\left(\frac{\partial u}{\partial y}\right)_{y=0}$ is negative the flow has separated.

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