Chapter 4, Problem 112P

To determine

The flow velocity along the floor and location of the maximum speed in the flow field.

Answer to Problem 112P

The flow speed along the floor is $\frac{-\dot{V}x}{\pi L\left(x^2+b^2\right)}$.

The location of the maximum velocity is $x=+b$ and $x=-b$.

Explanation of Solution

Given information:

The velocity component along the x direction is $\frac{-\dot{V}x}{\pi L}\frac{x^2+y^2+b^2}{x^4+2x^2{y}^2+2x^2{b}^2+y^4-2y^2{b}^2+b^4}$ and the velocity component along the y direction is $\frac{-\dot{V}y}{\pi L}\frac{x^2+y^2-b^2}{x^4+2x^2{y}^2+2x^2{b}^2+y^4-2y^2{b}^2+b^4}$.

Write the expression for the velocity component along x direction.

  $u=\frac{-\dot{V}x}{\pi L}\frac{x^2+y^2+b^2}{x^4+2x^2{y}^2+2x^2{b}^2+y^4-2y^2{b}^2+b^4}$   ...... (I)

Here, the distance of the attachment above the floor is $b$, the volume flow rate along the x direction is ${\dot{V}}_x$, constants are $a$, and $c$, variable in the x direction is $x$ and the variable in the y direction is $y$.

Write the expression for the velocity component along y direction.

  $v=\frac{-\dot{V}y}{\pi L}\frac{x^2+y^2-b^2}{x^4+2x^2{y}^2+2x^2{b}^2+y^4-2y^2{b}^2+b^4}$   ...... (II)

The flow is assumed to be steady and incompressible.

Write the expression for the maximum speed along x direction.

  $\frac{\partial u}{\partial x}=0$  ...... (III)

Calculation:

Substitute $0$ for $y$ in Equation (I).

  $\begin{array}{c}u=\frac{-\dot{V}x}{\pi L}\frac{x^2+{\left(0\right)}^2+b^2}{x^4+2x^2{\left(0\right)}^2+2x^2{b}^2+{\left(0\right)}^4-2{\left(0\right)}^2{b}^2+b^4}\\ =\frac{-\dot{V}x}{\pi L}\frac{x^2+b^2}{x^4+2x^2{b}^2+b^4}\\ =\frac{-\dot{V}x}{\pi L}\frac{x^2+b^2}{{\left(x^2+b^2\right)}^2}\\ =\frac{-\dot{V}x}{\pi L\left(x^2+b^2\right)}\end{array}$

Therefore, the flow speed along the floor is $\frac{-\dot{V}x}{\pi L\left(x^2+b^2\right)}$.

Substitute $0$ for $x$ in Equation (II).

  $\begin{array}{c}v=\frac{-\dot{V}\left(0\right)}{\pi L}\frac{x^2+y^2-b^2}{{\left(0\right)}^4+2{\left(0\right)}^2{y}^2+2{\left(0\right)}^2{b}^2+y^4-2{\left(0\right)}^2{b}^2+b^4}\\ =0\end{array}$

Substitute $\frac{-\dot{V}x}{\pi L\left(x^2+b^2\right)}$ for $u$ in Equation (III).

  $\begin{array}{l}\frac{\partial }{\partial x}\left(\frac{-\dot{V}x}{\pi L\left(x^2+b^2\right)}\right)=0\\ \frac{-\dot{V}}{\pi L}\left(\frac{1}{x^2+b^2}+x\frac{-2x}{{\left(x^2+b^2\right)}^2}\right)=0\\ \frac{-\dot{V}}{\pi L}\left(\frac{2x^2}{{\left(x^2+b^2\right)}^2}-\frac{1}{x^2+b^2}\right)=0\\ \frac{2x^2}{{\left(x^2+b^2\right)}^2}-\frac{1}{x^2+b^2}=0\end{array}$

  $\begin{array}{l}\frac{2x^2}{{\left(x^2+b^2\right)}^2}-\frac{1}{x^2+b^2}=0\\ \frac{1}{x^2+b^2}\left(\frac{2x^2}{x^2+b^2}-1\right)=0\\ \frac{2x^2}{x^2+b^2}-1=0\\ \frac{2x^2}{x^2+b^2}=1\end{array}$

  $\begin{array}{l}2x^2=x^2+b^2\\ x^2=b^2\\ x=+b\\ x=-b\end{array}$

Substitute $0$ for $x$ and $0$ for $y$ in Equation (I).

  $\begin{array}{c}u=\frac{-\dot{V}\left(0\right)}{\pi L}\frac{{\left(0\right)}^2+{\left(0\right)}^2+b^2}{{\left(0\right)}^4+2{\left(0\right)}^2{\left(0\right)}^2+2{\left(0\right)}^2{b}^2+{\left(0\right)}^4-2{\left(0\right)}^2{b}^2+b^4}\\ =0\end{array}$

Substitute $0$ for $x$ and $0$ for $y$ in Equation (II).

  $\begin{array}{c}v=\frac{-\dot{V}\left(0\right)}{\pi L}\frac{{\left(0\right)}^2+{\left(0\right)}^2+b^2}{{\left(0\right)}^4+2{\left(0\right)}^2{\left(0\right)}^2+2{\left(0\right)}^2{b}^2+{\left(0\right)}^4-2{\left(0\right)}^2{b}^2+b^4}\\ =0\end{array}$

At the origin, the velocity components are zero thus, the vacuum cleaner is not good at the origin.

Conclusion:

The flow speed along the floor is $\frac{-\dot{V}x}{\pi L\left(x^2+b^2\right)}$.

The location of the maximum velocity is $x=+b$ and $x=-b$.