Question

The turning moment diagram for a multicylinder engine has been drawn to a scale 1 mm = 600 N-m vertically and 1 mm = 3° horizontally. The intercepted areas between the output torque curve and the mean resistance line, taken in order from one end, are as follows : + 52, – 124, + 92, – 140, + 85, – 72 and + 107 mm² , when the engine is running at a speed of 600 r.p.m. If the total fluctuation of speed is not to exceed ± 1.5% of the mean, find the necessary mass of the flywheel of radius 0.5 m.

Answer 1

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,

C_s = ω₁ - ω_2/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)