Chapter 8, Problem 8.4CP

To determine

(a)

To sketch: The streamlines and location of stagnation points in the flow.

Explanation of Solution

Given information:

The strength of the flow = -m

The height of the above the floor = a

The location of the sink is (0, -a) and the location of image sink is also at (0, -a). In this, the sink and the image sink when combined make the floor for the flow. The location of the stagnation point is the origin.

This analysis is quite complex and requires modelling through a mathematical simulator to plot the streamlines of the flow. With the help of Matlab contour, the following plot has been constructed.

The streamlines of the flow.

Conclusion:

The streamlines are shown in the above figure and the location of stagnation point is the origin.

To determine

(b)

The magnitude of velocity V(x) along the floor in terms of parameters a and m.

Answer to Problem 8.4CP

The magnitude of velocity along the floor is $V\left(x\right)=\frac{2mx}{x^2+a^2}$.

Explanation of Solution

Given information:

The strength of the flow = -m.

The height of the above the floor = a.

Let us consider any point x along the wall, the magnitude of velocity at this considered point will be equal to the sum of all image flow components.

$\begin{array}{c}V\left(x\right)=2\times v_r\times \cos \theta \\ =\frac{2m}{r}\times \frac{x}{r}\\ =\frac{2mx}{r^2}\end{array}$

Since, $r^2=x^2+a^2$

$V\left(x\right)=\frac{2mx}{x^2+a^2}$

At any point along the wall the velocity will be equal to $V\left(x\right)=\frac{2mx}{x^2+a^2}$

Conclusion:

Thus, the magnitude of velocity along the floor is $V\left(x\right)=\frac{2mx}{x^2+a^2}$.

To determine

(c)

The variation of dimensionless pressure coefficient is $C_p=\frac{\left(p-p_{\infty }\right)}{\left(\frac{\rho U^2}{2}\right)}$ along the floor.

Answer to Problem 8.4CP

The variation of dimensionless pressure coefficient is $C_p=\frac{\left(p-p_{\infty }\right)}{\left(\frac{\rho U^2}{2}\right)}$ along the floor $C_{p,floor}=\frac{4x^2{a}^2}{{\left(x^2+a^2\right)}^2}$.

Explanation of Solution

Given information:

The strength of the flow = -m

The height of the above the floor = a

Let us use Bernoulli’s equation to calculate the pressure coefficient along the floor.

$\begin{array}{c}{C}_p=\frac{\left(p-p_{\infty }\right)}{\left(\frac{\rho U^2}{2}\right)}\\ =\frac{\left(0.5\right)\rho \left(U^2{}_{\infty }-V^2\right)}{\left(\frac{\rho U^2}{2}\right)}\end{array}$

$C_p=-\frac{V^2}{U^2}$

$C_{p,floor}=-\frac{4x^2{a}^2}{{\left(x^2+a^2\right)}^2}$

Conclusion:

Thus, variation of dimensionless pressure coefficient $C_p=\frac{\left(p-p_{\infty }\right)}{\left(\frac{\rho U^2}{2}\right)}$ along the floor is $C_{p,floor}=\frac{4x^2{a}^2}{{\left(x^2+a^2\right)}^2}$.

To determine

(d)

The location of minimum pressure coefficient along the x-axis.

Answer to Problem 8.4CP

The location of minimum pressure coefficient along the x-axis is $x=\pm a$

Explanation of Solution

Given information:

The strength of the flow = -m

The height of the above the floor = a

To find the location of minimum pressure coefficient along the x-axis, let us differentiate the wall pressure coefficient with respect to x and equate it to zero.

$\frac{d{C}_p}{dx}=0$

$\frac{d}{dx}\left(\frac{-4x^2{a}^2}{{\left(x^2+a^2\right)}^2}\right)=0$

This gives,

$x^2=a^2$

$x=\pm a$

Conclusion:

Thus, the location of minimum pressure coefficient along the x-axis is $x=\pm a$.

To determine

(e)

The points along the x-axis where the vacuum cleaner works most effectively.

Answer to Problem 8.4CP

The vacuum cleaner works most effectively at $x=\pm a$

Explanation of Solution

Given information:

The strength of the flow = -m

The height of the above the floor = a

The vacuum cleaner works most effectively at the points where the pressure coefficient is minimum. Thus, the points are $x=\pm a$.

Conclusion:

Thus, the vacuum cleaner works most effectively at $x=\pm a$.