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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online