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vivado [14]
3 years ago
12

A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =

0 at the surface (y = 0 ) to the freestream velocity, U, at the boundary-layer edge (y = δ). The equation for the profile is u = Uy/δ, where δ = cx1/2 and c is a constant. (a) What is the simplest expression for the y component of velocity (use x, y, u)? (b) Evaluate the maximum value of the ratio v/U, at a location where x = 0.5 m and δ = 6.0 mm.
Physics
1 answer:
slega [8]3 years ago
7 0

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:  

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:  

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):  

v=∫δv/δy(dy)=U*y/4cx^1/2

 =y/x*(U*y/4cx^1/2)

 =u*y/4x

which is exactly what we needed to demonstrate.  

Also, using u = U*y/δ in the last equation we can obtain:  

v/U=u*y/4*U*x

     =y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

                =δ/4x

                =2.5 * 10^-3

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shusha [124]

Answer:

- 3.72 Ns.

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Explanation:

mass of racket, M = 1000 g = 1 kg

mass of ball, m = 60 g = 0.06 kg

initial velocity of racket, U = 11 m/s

initial velocity of ball, u = 18 m/s

final velocity of ball, v = - 44 m/s

Let the final velocity of the racket is V.

(a) Momentum is defined as the product of mass and velocity of the ball.

initial momentum of the ball = m x u = 0.06 x 18 = 1.08 Ns

Final momentum of the ball = m x v = 0.06 x (- 44) = - 2.64 Ns

Change in momentum of the ball = final momentum - initial momentum

                                                        = - 2.64 - 1.08 = - 3.72 Ns

Thus, the change in momentum of the ball is - 3.72 Ns.

(b) By use of conservation of momentum

initial momentum of racket and ball = final momentum of racket and ball

1 x 11 + 0.06 x 18 =  1 x V - 0.06 x 44

12.08 = V - 2.64

V = 9.44 m/s

Thus, the final velocity of the racket afetr the impact is 9.44 m/s .

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Answer:

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Explanation:

Generally the thermal efficiency is mathematically represented as

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