Question

During forward stroke of the piston of the double acting steam engine, the turning moment has the maximum value of 2000 N-m when the crank makes an angle of 80° with the inner dead centre. During the backward stroke, the maximum turning moment is 1500 N-m when the crank makes an angle of 80° with the outer dead centre. The turning moment diagram for the engine may be assumed for simplicity to be represented by two triangles.

If the crank makes 100 r.p.m. and the radius of gyration of the flywheel is 1.75 m, find the coefficient of fluctuation of energy and the mass of the flywheel to keep the speed within ±0.75% of the mean speed. Also determine the crank angle at which the speed has its minimum and maximum values.

Answer 1

Given : 

N = 100 r.p.m. or ω = 2π x 100/60 = 10.47rad / s;k = 1.75m

Since the fluctuation of speed is ± 0.75% of mean speed, therefore total fluctuation of speed,

ω₁ - ω₂ = 1.5% ω

and coefficient of fluctuation of speed,

C_S = \(\frac{ω_1-ω_2}{ω}\) = 1.5% = 0.015

Coefficient of fluctuation of energy

The turning moment diagram for the engine during forward and backward strokes. The point O represents the inner dead centre (I.D.C.) and point G represents the outer dead centre (O.D.C). We know that maximum turning moment when crank makes an angle of 80°(or 80 x π/180 = 4π / 9 rad) with I.D.C.,

∴ AB = 2000 N-m

and maximum turning moment when crank makes an angle of 80° with outer dead centre (O.D.C.) or 180° + 80° = 260° = 260 x π/180 = 13π/9rad with I.D.C.,

LM = 1500 N-m

Let T_mean ​ = EB = QM = Mean resisting torque.

We know that work done per cycle

= Area of triangle OAG + Area of triangle GLS

= 1/2 x OG x AB + 1/2 x GS x LM

= 1/2 x π x 2000 + 1/2 x π x 1500 = 1750π N-m ---(1)

We also know that work done per cycle

= T_mean x 2π N-m ----(ii)

From equations (i) and (ii),

T_mean = 1750π/2π = 875N-m

From similar triangles ACD and AOG,

CD/AE = OG/AB

or CD = OG/AB x AE = OG/AB (AB-EB) = π/2000 (2000-875) = 1.764 rad

∴ Maximum fluctuation of energy,

ΔE = Area of triangle ACD = 1/2​ x CD x AE

= 1/2 x CD(AB - EB) = 1/2 x 1.764(2000 - 875) = 992 N-m

We know that coefficient of fluctuation of energy,

C_E​ = Max. fluctuation of energy/ Work done per cycle = 992/1750π = 0.18 or 18%

Mass of the flywheel

Let  m = Mass of the flywheel.

We know that maximum fluctuation of energy (ΔE),

992 = m . k² . ω² . C_s . = m x (1.75)² x (10.47)² x 0.015 = 5.03 m

∴  m = 992 / 5.03 = 197.2 kg

Crank angles for the minimum and maximum speeds

We know that the speed of the flywheel is minimum at point C and maximum at point D

Let θ_C​ and θ_D​ = Crank angles from I.D.C., for the minimum and maximum speeds.

From similar triangles ACE and AOB,

CE/OB = AE/AB

Or CE = AE/AB x OB = AB-EB/AB x OB = 2000 - 875/2000 x 4π/9 = π/4 rad

∴ θ_C = 4π/9 - π/4 = 7π/36 rad = 7π/36 x 180/π = 35°

Again from similar triangles AED and ABG,

ED/BG = AE/AB

or ED = AE/AB x BG = AB-EB/AB (OG-OB)

= 2000 - 875 / 2000 (π -4π/9) = 2.8π/9 rad

∴ θ_D = 4π/9 + 2.8π/9 = 6.8π/9 rad = 6.8π/9 x 180/π = 136°