Question

an engine flywheel has a mass of 1000 tonne and the radius of gyration is 5m. The maximum and minimum speed are 220 r.p.m and 120 respectively. Find the coefficient of fluctuation of speed. a)0.5 b)0.58 c)0.75 d)0.68​

Answer 1

Given : Mass = 6.5 tonnes = 6500 kg

Radius of gyration = k = 1.8 m

Fluctuation energy = ∆E

= 56 KN-m =

N = 120 rpm

= Maximum and minimum speeds.

N_1and N_2

let,

\triangle\:E=\dfrac{\pi^{2}}{900}\times\:mk^{2}N(N_1-N_2)

Since,

\therefore\:56\times10^{3}=\dfrac{\pi^{2}}{900}\times6500\times(1.8)^{2}\times120\times(N_1-N_2)

(N_1-N_2)

(N_1-N_2)= \dfrac{56\times10 ^{3}}{27715}

= 2 rpm ...... (i)

(N_1-N_2)

we also know that mean speed (N) is given as,

120=\dfrac{(N_1+N_2)}{2}

(N_1-N_2)

or,

= 240 rpm ...... (ii)

from equation (i) and (ii) we get

= 121 rpm

N_1