Answer to Question #180225 in Mechanical Engineering for nnaemeka

Question

A vertical diesel engine running at 420 rpm develops 650kw and has 4 impulses per revolution. If the fluctuation of energy is 28% of the work done during each impulse, estimate the cross sectional area of the rim

Answer 1

w = (2$\pi$*420)/60 = 14$\pi$ rad/s

Work done per cycle = (650 000 x 60)/180 = 216 667 J

e = 0.28 x 216 667 J = 60 667 J

60 667 = 1/2 * I * (w₁² - w₂²) = I * w² * ((w1 - w₂)/w) = I * (12$\pi$)² ∗ 0.01

I = 4273 kg m²

Centrifugal stress, σ = ρ * V² = ρ * ω² * R²

where R is the mean rim radius, 

5.5* 10⁶ = 7200 * 144π² * R²

R = 0.732 m

I = mk² = ρ * A * 2πR * R²

4273 = 7200 * A * 2$\pi$ *(0.732 m)²

A = 0.176 m²