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FromTheMoon [43]
3 years ago
5

An open 1- -m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the

center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed? No water is spilled from the tank.

Engineering
2 answers:
aalyn [17]3 years ago
8 0

Answer:

The answer to the question is

At an angular velocity of 7.4106 s⁻¹ the bottom of the tank of water will be first  exposed with no water spilling from the tank

Explanation:

The rotational motion of the fluid in the tank is

P(r, z) = \rho(\frac{r^{2} \omega^{2}  }{2}-gz)+c

where r = radius,

g = Acceleration due to gravity

P = Pressure

ρ = Fluid density

c = constant

We have the boundary condition giving by

at r = 0,  z = z₀ and P = P₀

Therefore as we move from  the center outwards towards the wall of the cylinder we have

P(r, z) = P₀ + \rho(\frac{r^{2} \omega^{2}  }{2}-g(z-z_{0} ))

When P(r, z) = P₀ we have  \rho(\frac{r^{2} \omega^{2}  }{2}-g(z-z_{0} )) = 0 from which

\frac{r^{2} \omega^{2}  }{2} = g(z-z_{0} ) and

z-z_{0} = \frac{r^{2} \omega^{2}  }{g2} which is the equation for paraboloid revolution

The fluid height at the wall of the cylinder where r = R is then

z_{R}  = z_{0} + \frac{R^{2} \omega^{2}  }{g2}

For a paraboloid circumscribed in a circle, its volume = half the volume of the cylinder thus

\frac{1}{2}(z_{R}-z_{0})R^{2} \pi= (h-z_{0})R^{2}.\pi from which, we have by combining with the previous equation

z_{0} = h-\frac{R^{2} \omega^{2}  }{g2} therefore z_{R} = h+\frac{R^{2} \omega^{2}  }{g2}

At the point the fluid spills from the cylinder we have

Critical condition, \omega =\omega_{l} , z_{R} = H this results in

\omega_{l} = \frac{2}{R}\sqrt{g(H-h)} the condition when the bottom of the tank will be exposed is given by h = H/2 hence

\omega_{l} = \frac{2}{R}\sqrt{g(H-\frac{H}{2} )} =\omega_{l} = \frac{2}{R}\sqrt{(\frac{gH}{2} )}

Solving we have for R = 0.5 m, H = 0.7 m and g = 9.81 m/s²

\omega_{l} = 7.4106 s⁻¹

Ira Lisetskai [31]3 years ago
7 0

Answer:

Explanation:

To get the angular velocity with which the bottom of the tank first be exposed - The careful and detailed step by step analysis is as shown in the attached file

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BabaBlast [244]

Answer:

a) since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

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now since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b)

we know that the velocity of water at the surface contact is zero

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so from the equation

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