Question

The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along the y axis. What is the angle between the velocity vector and the x axis at points 15, 52, and 10, 52

Answer 1

Answer:

Using the formula

V =20y/(x^2+y^2)^1/2 - 20x/(x^2+y^2)^1/2

Hence fluid speed at x axis =20x/(x^2+y^2)^1/2

While the fluid speed at y axis =20y/(x^2+y^2)^1/2

Now the angle at 1, 5

We substitute into the formula above

V= 20×5/(1+25)^1/2= 19.61

For x we have

V = 20× 1/(1+25)^1/2= 3.92

Angle = 19.61/3.92= 5.0degrees

Angel at 5, and 2

We substitute still

V = 20×5/(2+25)^1/2=19.24

At 2 we get

V= 20×2/(2+25)^1/2=7.69

Dividing we get 19.24/7.69=2.5degrees

At 1 and 0

V = 20/(1)^1/2=20

At 0, v =0

Angel at 2 and 0 = 20degrees

At 5 and 2

V = 100/(25+ 4)^1/2=18.56

At x = 2

40/(√29)=7.43

Angle =18.56/7.43 = 2.49degrees.

Answer 2

There is a part of the question missing and it says;

The velocity field of a flow is given by V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s, where x and y are in feet.

Answer:

A) At (1,5),angle is -11.31°

B) At (5,2),angle is -68.2°

C) At (1,0), angle is -90°

Explanation:

From the question ;

V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s

Let us assume that u and v are the flow velocities in x and y directions respectively.

Thus we have the expression;

u = [20y/(x² + y²)^(1/2)]

and v = - [20x/(x² + y²)^(1/2)]

Thus, V = √(u² + v²)

V = √[20y/(x² + y²)^(1/2)]² + [-20x/(x² + y²)^(1/2)]²

V = √[400y²/(x²+y²)] +[400x²/(x²+y²)

V = √(400x² + 400y²)/(x²+y²)

Now for the angle;

tan θ = opposite/adjacent

And thus, in this question ;

tan θ = v/u

tan θ = [-20x/(x² + y²)^(1/2)]/ [20y/(x² + y²)^(1/2)]

Simplifying this, we have;

tan θ = - 20x/20y = - x/y

so the angle is ;

θ = tan^(-1)(-x/y)

So let's now find the angle at the various coordinates.

At, 1,5

θ = tan^(-1)(-1/5) = tan^(-1)(-0.2)

θ = -11.31°

At, 5,2;

θ = tan^(-1)(-5/2) = tan^(-1)(-2.5)

θ = -68.2°

At, 1,0;

θ = tan^(-1)(-1/0) = tan^(-1)(-∞)

θ = -90°

At,