Answer to Question #171056 in Mechanical Engineering for Emmanuel

Question

The turning moment diagram of an engine which has been drawn to scale of 1mm to 56Nm and 1mm to 1° of rotation of the crankshaft shows the greatest amount of energy which jas to be stored by the flywheel is represented by an area of 3500mm2. The flywheel is to run at a mean speed of 350 rmp with a total speed variation of 2.5%. If the mass of the flywheel is to be 585kg, determine suitable dimensions for the rim, the internal diameter being 0.85 of the external diameter , neglect the inertia of the arms and hub of the flywheel. Cast iron has a density of 7.2Mg/m3

Answer 1

Given : N = 350 r.p.m or ω = 2π = 350/60 = 5.83 rad/s; R = 0.85R₁Since the total fluctuation of is not to exceed ± 2.5% of the mean speed thereforeω₁ - ω₂ = 5% ω = 0.05ω and coefficient of fluctuation of speed C_s = ω₁ - ω₂/ω = 0.05Since the turning moment scale 1 mm = 56 N-m and crank angle scale is 1 mm = 1° = 1° x  π/180 = π/180 rad therefore 1 mm² on turning momment diagram = 56 x  π/180 = 0.98 N-mΔE =3500 x 0.98 N-m = 3419 N-mΔE = m(0.85R₁)²ω²C_s 3419 N-m = 585 kg x 0.7225R₁²(5.83)²0.05 = 718.3R₁²R₁ = 2.18 mR = 0.85 x 2.18 m = 1.85 m