A cylinder 6 ft in diameter and 9 ft high is completely filled with glycerin, sp gr 1.60, under a pressure of 35.2 psi at the top. The steel plates that form the cylinder are 1/2 in thick and can withstand an allowable unit stress of 12,000 psi. What maximum speed in rpm can be imposed on the cylin

Question

A cylinder 6 ft in diameter and 9 ft high is completely filled with glycerin, sp gr 1.60, under a pressure of 35.2 psi at the top. The steel plates that form the cylinder are [latex]\frac{1}{2} [/latex] in thick and can withstand an allowable unit stress of 12,000 psi. What maximum speed in rpm can be imposed on the cylinder?

Accepted Answer

From the specifications for the tank and the hoop tension formula [latex]\sigma =p^{′}r/t [/latex] ,

[latex]p^{′}_{A}=\sigma t/r =(12,000)\left(\frac{1}{2} ight)/36=167 [/latex] psi

Also , p′_A = ∑ pressures (35.2 imposed + due to 9 ft glycerin + due to rotation)

or [latex]167=35.2+\frac{1.60 imes 62.4 imes 9}{144}+\left(\frac{\omega ^2}{2 imes 32.2} ight) (3^2)\left(\frac{1.60 imes 62.4}{144} ight) [/latex] psi

Solving, ω = 36.0 rad/sec or 344 rpm.
The pressure conditions are shown graphically in Fig. 5-9, not to scale. Line RST indicates a pressure head of 50.8 ft of glycerin above the top of the tank before rotation. The parabolic pressure curve with vertex at S is caused by the constant angular velocity of 36.0 rad/sec. If the vessel were full but not under pressure, vertex S would coincide with the inside top of the vessel.

Question 5.12: A cylinder 6 ft in diameter and 9 ft high is completely fill...

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