Question

The turning moment curve for an engine is represented by the equation, T = (20 000 + 9500 sin 2 θ – 5700 cos 2θ) N-m, where θ is the angle moved by the crank from inner dead centre. If the resisting torque is constant, 

find: 

1.  Power developed by the engine ; 

2.  Moment of inertia of flywheel in kg-m², if the total fluctuation of speed is not exceed 1% of mean speed which is 180 r.p.m; and 

3.  Angular acceleration of the flywheel when the crank has turned through 45° from inner dead centre. 

Answer 1

Given : T = (20 000 + 9500 sin 2θ) - 5700 cos 2θ) N-m; N = 180 r.p.m. or ω = 2π x 180/60 = 18.85 rad/s

Since the total fluctuation of speed (ω₁ - ω₂) is 1 of mean speed (ω) , therefore coefficient of fluctuation of speed,

C_s = ω₁ - ω_2/ω = 1 % = 0.01

1. Power developed by the engine

We know that work done per revolution

and mean resisting torque torque of the engine,

T_mean = Wok done per revolution/2π = 40 000/2π = 20000 N-m

 We known that power developed by the engine

T_mean = ω = 20 000 x 18.85 = 377 000 W = 377kW

2. Moment of inertia of the flywheel

Let I = Moment of inertia of the flywheel in kg- m².

The turning moment diagram for one stroke (i.e half revolution of the crankshaft) is shown in fig. Since at points B and D, the torque exerted on the crankshaft is equal to the mean resisting torque on the the flywheel, therefore,

T= T_mean

Maximum fluctuation of energy,

we know that maximum fluctuation of energy (Δ, E)'

11078 = I.ω².CS = I x (18.85)2 x 0.01 = 3.35I

I = 11078/3.55 = 3121 kg-m²

3. Angular acceleration of the flywheel

Let α = Angular acceleration of the flywheel, and

θ = Angle turned by the crank from inner dead centre = 45° ...(Given)

The Angular acceleration in the flywheel is produced by the excess torque over the mean torque.

We known that excess torque at any instant,

We also know that excess torque

= I.α = 3121 x α

From equations (i) and (ii),

α = 95000/3121 = 3.044 rad/s²