Question

3. A single cylinder double acting steam engine develops 150kW at a mean
speed of 80 rp.m. the coefficient of fluctuation of energy is 0.1 and the
fluctuation of speed is +3% of the mean speed. If the mean diameter of
the flywheel rim is 2 meters and the hub and spokes provide 5% of the
rotational inertia of the wheel, find the mass of the flywheel and cross-
sectional area of the rim Assume density of the material as 7500kg/m²
20
OTO​

Answer 1

Answer:

The answer will be  3787kg and 0.0080 m^2

Step-by-step explanation:

According to the problem  it is given that,

Power , p = 150 KW = 150000W, N = 80 rpm

therefore the angular velocity, ω = 80 x 2π/60 = 8.4 rad/s

The coefficient of energy fluctuation C = 0.1  diameter d = 2 m or r = 1 m

ρ = 7500kg /m^2

Now the total fluctuation of speed is given as 3%

therefore,

ω1-ω2 = 3% = 0.03ω

Therefore , the coefficient fluctuation of speed, C(speed) = ω1-ω2/ω  = 0.03

Now the mass will be:

Work done = p x 60 /N = 150000 x 60 /80 = 112500 Nm

Now the maximum energy fluctuation = delta E = W x C = 11250Nm

again, delta E = I ω^2C = 2.8224I

Therefore I = 3986 kg m^2

As the hub and spoke provide 5 % inertia,

I = 0.95I = 0.95 x 3986 = 3787kg m^2

now mass ,m = I/k^2 = 3787/1^2 = 3787kg

Now the cross sectional area:

m = ρv = 2πrAρ[ A is area]

=> 3787 = 2π x 1  x 7500 A

=> A = 0.0080 m^2