Question

A flywheel whose moment of inertia is 0.62 kg m² rotates at a speed 2500 r.p.m. in a large heat insulated system, the temperature of which is 20°C. 

(i) If the K.E. of the flywheel is dissipated as frictional heat at the shaft bearings which have a water equivalent of 1.9 kg, find the rise in the temperature of the bearings when the flywheel has come to rest. 

(ii) Calculate the greatest possible amount of the above heat which may be returned to the flywheel as high-grade energy, showing how much of the original K.E. is now unavailable. What would be the final r.p.m. of the flywheel, if it is set in motion with this available energy ?

Answer 1

Moment of inertia of the flywheel, I = 0.62 kg m²

Initial angular velocity of the flywheel, ω₁ = \(\cfrac{2\pi N_1}{60}\) 

= \(\cfrac{2\pi \times2500}{60}\) = 261.8 rad/s.

Temperature of insulated system, T₀ = 20 + 273 = 293 K 

Water equivalent of shaft bearings = 1.9 kg 

(i) Initial available energy of the flywheel,

(K.E.)_initial = \(\cfrac12\) Iω₁²

\(\cfrac12\) × 0.62 × (261.8)² = 2.12 × 10⁴ N.m = 21.2 kJ

When this K.E. is dissipated as frictional heat, if ∆t is the temperature rise of the bearings, we have 

Water equivalent of bearings × rise in temperature = 21.2 

i.e., (1.9 × 4.18) ∆t = 21.2

∆t = \(\cfrac{21.2}{1.9\times4.18}\) = 2.67°C

Hence, rise in temperature of bearings = 2.67°C.  

∴ Final temperature of the bearings = 20 + 2.67 = 22.67°C. 

(ii) The maximum amount of energy which may be returned to the flywheel as high-grade energy is,

The amount of energy rendered unavailable is 

U.E. = (A.E.)_initial – (A.E.)_{returnable as high grade energy} 

= 21.2 – 0.096 = 21.1 kJ. 

Since the amount of energy returnable to the flywheel is 0.096 kJ, if ω₂ is the final angular velocity, and the flywheel is set in motion with this energy, then

0.096 × 10³ = \(\cfrac12\) x 0.62 × ω₂²

ω₂² = \(\cfrac{0.096\times10^3\times2}{0.62}\) = 309.67 or ω2 = 17.59 rad/s.

If N₂ is the final r.p.m. of the flywheel, then

ω₂ = \(\cfrac{2\pi N_1}{60}\) or N₂ = \(\cfrac{17.59\times60}{2\pi}\) = 168 r.p.m.

Final r.p.m. of the flywheel = 168 r.p.m.