Respuesta :
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.
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,
