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alexandr402 [8]
3 years ago
3

A 1.2-m-diameter, 3-m-high sealed vertical cylinder is completely filled with gasoline whose density is 740 kg/m3 . The tank is

now rotated about its vertical axis at a rate of 70 rpm. Determine (a) the difference between the pressures at the centers of the bottom and top surfaces and (b) the difference between the pressures at the center and the edge of the bottom surface.
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer:

a) The difference between the pressures at the centers of the bottom and top surfaces is 21756 N/m²

b) The difference between the pressures at the center and the edge of the bottom surface is 7156.69 N/m²

Explanation:

Given:

D = diameter of the cylinder = 1.2 m

z₂ - z₁ = height of the cylinder = 3 m

ρ = density of gasoline = 740 kg/m³

N = rotational speed = 70 rpm

The angular velocity is equal:

w=\frac{2\pi N}{60} =\frac{2\pi 70}{60} =7.33rad/s

a) The difference pressure between two points is:

P_{2} -P_{1} =\frac{\rho w^{2} }{2} (r_{2}^{2}  -r_{1}^{2}  )-\rho g(z_{2} -z_{1} )

r₂ - r₁ = 0

P_{2} -P_{1} =0-\rho g(z_{2} -z_{1} )=-740*9.8*3=-21756N/m^{2} =21756N/m^{2}

b) The difference pressure is:

P_{2} -P_{1} =\frac{\rho w^{2}(R_{2}^{2}-0)-0   }{2} =\frac{740*7.33^{2}*0.6^{2}  }{2} =7156.69N/m^{2}

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