SolSec 5.7

# SolSec 5.7 - 5 69 Problems and Solution Section 5.7(5.74...

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5- 69 Problems and Solution Section 5.7 (5.74 through 5.80) 5.74 A 100-kg compressor rotor has a shaft stiffness of 1.4 × 10 7 N/m. The compressor is designed to operate at a speed of 6000 rpm. The internal damping of the rotor shaft system is measured to be ζ = 0.01. (a) If the rotor has an eccentric radius of 1 cm, what is the rotor system's critical speed? (b) Calculate the whirl amplitude at critical speed. Compare your results to those of Ex. 5.7.1. Solution: (a) The critical speed is the rotor's natural frequency, so ω c k m == × 14 10 100 374 2 7 . . rad/s 3573 rpm (b) At critical speed, r = 1, so from Eq. (5.81), () m 5 . 0 01 . 0 2 01 . 0 2 = = = ζ α X So a system with higher eccentricity and lower damping has a greater whirl amplitude (see Ex. 5.7.1). 5.75 Redesign the rotor system of Problem 5.74 such that the whirl amplitude at critical speed is less than 1 cm by changing the mass of the rotor. Solution: From Problem 5.74, k = 1.4 × 10 7 N/m, m = 100 kg, ζ = 0.01, and α = 0.01m. Since the whirl amplitude at critical speed must be less than 0.01 m, the value of ζ that would satisfy this is, from Eq. (5.81), 5 . 0

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## This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 5.7 - 5 69 Problems and Solution Section 5.7(5.74...

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