Question

The turning moment diagram for a multicylinder engine has been drawn to a scale of 1 mm =

4500 N-m vertically and 1 mm = 2.4° horizontally. The intercepted areas between output torque curve

and mean resistance line taken in order from one end are 342, 23, 245, 303, 115, 232, 227, 164 mm2,

when the engine is running at 150 r.p.m. If the mass of the flywheel is 1000 kg and the total fluctuation

of speed does not exceed 3% of the mean speed, find the minimum value of the radius of gyration.​

Answer 1

Answer:

Given : N = 600 r.p.m or ω = 2 π = x 600/60 = 62.84 rad/s; R = 0.5 m Since the total fluctuation of is not to exceed ± 1.5% of the mean speed therefore ω1 - ω2 = 3% ω = 0.03ω and coefficient of fluctuation of speed, Cs = ω1 - ω2/ω = 0.03 The turning diagram is shown Since the turning moment scale 1 mm = 600 N-m and crank angle scale is 1 mm = 3° = 3° x π/180 = π/60rad,therefore 1 mm2 on turning momment diagram = 600 x π/60 = 31.42 N-m Let the total energy at A = E, We known that maximum fluctuation of energy, ΔE = maximum energy - minimum energy = (E + 52) - (E - 120) = 172 = 172 x 1.42 = 5404 N-m Let m = Mass of the flywheel in kg. We know that maximum fluctuation of energy(ΔE) Read more on Sarthaks.com - https://www.sarthaks.com/491135/the-turning-moment-diagram-for-multicylinder-engine-has-been-drawn-scale-600-vertically