Chapter 4.8, Problem 28E

To determine

Tofind:the streamlines for the flow of water around the inside of right-angle boundary, velocity potential, stream function and velocity.

Answer to Problem 28E

The streamlines for the given flow of water is

  $\text{2xy=c}$ .

The stream function is $\psi =2xy$

The velocity vector is $\text{v=2xi-2yj}$ .

Explanation of Solution

Given:

Concept used:

Calculation:

Let’s consider the height of the basket ball backboard as AB and its shadow as BC.

Therefore,the given problem can be represented as the right-angled triangle.

$\left(a\right)$ the figure showing the basketball backboard and its shadow are shown below. The height of the backboard will be taken as $\text{AB=12}\frac{\text{1}}{\text{2}}$ ft and its shadow will be represented by BC $\text{=17}\frac{1}{3}\text{ft}$

Assume the base angle of the triangle as $\text{θ}$ .

Use the fact that $\text{AB=12}\frac{\text{1}}{\text{2}}$ and $\text{=17}\frac{1}{3}\text{ft}$

Now to find the angle of elevation $\text{θ}$ apply the trigonometric identity:

  $\text{tanθ=}\frac{\text{Opposite}}{\text{Hypotenuse}}$

  $\Rightarrow \frac{\text{AB}}{\text{BC}}$

  $\Rightarrow \frac{12\frac{1}{2}}{17\frac{1}{3}}$

  $\Rightarrow 0.72254$

$\text{θ}$ = $\text{Arctan}\left(0.72254\right)$

$\left(\text{c}\right)$ as calculated in part $\left(b\right)$ ,can say that find the angle of elevation to the sum is given by

$\text{Arctan}\left(0.72254\right)$ = $35.85^{\prime }$