Question

the mass of flywheel of an engine in 6.5 tonnes and the radius of gyration is 1.8 it is found from the turning moment diagram that the fluctuation of energy is 56 KN-m and if the mean speed of the engine is 120 RPM find the maximum and minimum speed ​

Answer 1

Solution :

Given : Mass = 6.5 tonnes = 6500 kg

Radius of gyration = k = 1.8 m

Fluctuation energy = ∆E

= 56 KN-m = $56\times10^{3}N-m$

N = 120 rpm

let, $N_1and N_2$ = Maximum and minimum speeds.

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

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

= 27715 $(N_1-N_2)$

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

= $(N_1-N_2)$ = 2 rpm ...... (i)

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

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

or, $(N_1-N_2)$ = 240 rpm ...... (ii)

from equation (i) and (ii) we get

$N_1$ = 121 rpm

$N_</strong><strong>2</strong><strong>$ = 119 rpm