Question

An engine flywheel has a mass of 6.5 tones and the radius of gyration is 2m. If the maximum and minimum speed are 120 r.p.m. and 118 r.p.m. respectively, find maximum fluctuation energy .
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Answer 1

Answer:

Given : Mass = 6.5

tonnes = 6500 kg

Radius of gyration = k = 1.8 m

Fluctuation energy = ∆E

= 56 KN-m = 56\times10^{3}N-m56×103N−m

N = 120 rpm

let,

 N_1 and N_2N1 and N2 = Maximum and minimum speeds

.

Since,

 \triangle\:E=\dfrac{\pi^{2}}{900}\times\:mk^{2}N(N_1-N_2)△E=900π2×mk2N(N1−N2)

\therefore\:56\times10^{3}=\dfrac{\pi^{2}}{900}\times6500\times(1.8)^{2}\times120\times(N_1-N_2)∴56×103=900π2×6500×(1.8)2×120×(N1−N2)

= 27715 (N_1-N_2)(N1−N2)

= (N_1-N_2)= \dfrac{56\times10 ^{3}}{27715}(N1−N2)=2771556×103

= (N_1-N_2)(N1−N2) = 2 rpm ...... (i)

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

\implies⟹ 120=\dfrac{(N_1+N_2)}{2}120=2(N1+N2)

or, (N_1-N_2)(N1−N2) = 240 rpm ...... (ii)

from equation (i) and (ii) we get

N_1N1 = 121 rpm

N_2N2 = 119 rpm