Chapter 4.8, Problem 32E

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 32E

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:

The sine of the angle is given by the ratio of length of the perpendicular to the hypotenuse, therefore in right angled triangle $\Delta \text{ACB}$$\text{sinA=}\frac{\text{BC}}{\text{AB}}$

Put the values

  ${\text{sin60}}^0\text{=}\frac{h}{30}$

  ${\text{h=30×sin60}}^{\text{0}}$

  $\text{h=25}\text{.98}$

Thus, the height of h of the side is 25.98 feet.

Since the tangent of the angle is given by the ratio of the length of the perpendicular to the base, therefore in right angled triangle $\Delta \text{DEF}$

  $\text{tanD=}\frac{\text{ED}}{\text{DE}}$

Put the values

  $\text{tanD=}\frac{\text{d}}{\text{h}}$

  $\Rightarrow \frac{\text{d}}{25.98}$

Since ${\text{D=90}}^{\text{0}}\text{-θ}$ , therefore

  $\text{tan}\left({\text{90}}^{\text{0}}\text{-θ}\right)=\frac{d}{25.98}$

$\text{cotθ}=\frac{d}{25.98}$ .... (i) since $\text{tan}\left({\text{90}}^{\text{0}}\text{-θ}\right)=\text{cotθ}$

  ${\text{θ=cot}}^{-1}\left(\frac{d}{25.98}\right)$

Hence the angle of depression from the top of the side to the end of the side at the ground is

  ${\text{θ=cot}}^{-1}\left(\frac{d}{25.98}\right)$

Assume that the angle o f depression of the ride is bounded by safety restrictions to be less than ${25}^0$ and not more than ${30}^0$

When the minimum angle of depression is ${25}^0$

  $\text{cotθ}=\frac{d}{25.98}$

  $d=25.98\cot {25}^0$

  $d=25.98\times 2.1445$

  $\Rightarrow 55.71$

When the maximum angle of depression is ${30}^0$ then from (i)

  $\text{cotθ}=\frac{d}{25.98}$

  $d=25.98\cot {30}^0$

  $d=44.99$

  $d=45$

Hence, the rider travels horizontally between $\text{45feet to 55}\text{.71feet}$ .