Chapter 4.8, Problem 34E

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 34E

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:

Consider the following figure which describe the path of the airplane traveling at the speed of 275 feet per second and having an angle of ascent as ${\text{18}}^0$ .

Since the distance is given by the product of the speed and time, therefore the distance travelled by the airplane in 1 minute is

  $\text{s=v×t}$

  $\Rightarrow \text{275}\times \text{60}$

  $s=16500$

Note that the above triangle where hypotenuse represents the distance x since sine of the angle is given by the ratio of the length of the perpendicular of the hypotenuse therefore;

  $\sin \left({18}^0\right)=\frac{h}{x}$

  $h=x\sin \left({18}^0\right)$

  $h=16500\times \sin \left({18}^0\right)$

  $h=5098.78$

Therefore, the planes altitude after 1 minute is $5098.78\text{ feets}$ .

Since sine of the4 angles is given by the ratio of the length of the perpendicular to the hypotenuse, therefore when $h=10000\text{ ft}$

  $\sin \left({18}^0\right)=\frac{h}{s}$

  $s==\frac{h}{\sin \left({18}^0\right)}$

  $s=32360.68$

Since the distance given by the product of the speed and time, therefore the time taken for the plane to climb to an altitude of 10,000 feet is

  $t=\frac{32360.68}{\sin \left({18}^0\right)}$

  $t=117.68$ this, the time taken to climb to altitude of b=10000 ft is 117.68 seconds that is 1.96minute.