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sweet-ann [11.9K]

2 years ago

4

In fully-developed laminar pipe flow, consider the rate of work done on an annulus of thickness dr (hint: consider, for each fac

e of the annulus, rate of work = power = force × velocity).
(i) Find an expression for the power (per unit volume) dissipated by the flow in the fluid annulus, and show that it is equal to µ(du/dr)

(ii) By using u(r) from 3(ii) above, and integrating this expression, show that the power dissipated across a length of pipe is Q∆P

1 answer:

Naddik [55]2 years ago

6 0

Answer:

We write the share stress relationship for the pipe

t = u du/dy = u du/dr

Now calculating the force, we have

F = tA = udu/dr( A)

Cal. The power

∆P= Fdu

=udu/dr (Adu)

Now we cal. The power per unit volume

∆P= u(du/dr)^2=Adr

Now power per unit length will be

Fdu =udu/dr (Adu)

=udu/dr (Adu)

=Q∆P

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A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6

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1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

5 0

2 years ago