Question

A Single cylinder double acting steam engine develops 150 kw at a mean speed of 80 rpm. The co-efficient of fluctuation of energy is 0.1 and fluctuation of speed is 2% of the mean speed. If the diameter of the flywheel rim is 2 meters and the hub and spokes provided 5% of the rotational inertia of the flywheel. Find the mass and cross sectional area of the flywheel rim. Assume density of flywheel as 7200 kg/m3.

Answer 1

Question #81297, Engineering / Mechanical Engineering

A Single cylinder double acting steam engine develops $150\,\mathrm{kw}$ at a mean speed of 80 rpm. The co-efficient of fluctuation of energy is 0.1 and fluctuation of speed is $2\%$ of the mean speed. If the diameter of the flywheel rim is 2 meters and the hub and spokes provided $5\%$ of the rotational inertia of the flywheel. Find the mass and cross sectional area of the flywheel rim. Assume density of flywheel as $7200\,\mathrm{kg/m^3}$.

Solution

$$P = 150\,\mathrm{kW} = 150000\,\mathrm{W};\ N = 80\,\mathrm{rpm}\,\mathrm{or}\,\,\omega = \frac{2\pi 80}{60} = 8.4\,\frac{\mathrm{rad}}{\mathrm{s}}$$ $$C_E = 0.1;\ D = 2\,\mathrm{m}\,\mathrm{or}\,\,\mathrm{R} = 1\,\mathrm{m};\ \rho = 7200\,\frac{\mathrm{kg}}{\mathrm{m^3}}$$

Total fluctuation of speed:

$$\omega_1 - \omega_2 = 4\%\omega = 0.04\omega$$

Coefficient of fluctuation of speed:

$$C_s = \frac{\omega_1 - \omega_2}{\omega} = 0.04$$

1. The mass of the flywheel rim

Work done per cycle:

$$W = \frac{P60}{N} = 150000\,\frac{60}{80} = 112500\,\mathrm{Nm}$$

Maximum fluctuation of energy:

$$\Delta E = W C_E = (112500)(0.1) = 11250\,\mathrm{Nm}.$$ $$\Delta E = I\omega^2 C_E = I(8.4)^2(0.04) = 2.8224I$$ $$I = \frac{11250}{2.8224} = 3986\,\mathrm{kg/m^2}.$$

Since the hub and spokes provided $5\%$ of the rotational inertia of the flywheel, mass moment inertia of the flywheel rim is

$$I_{\mathrm{rim}} = 0.95I = 0.95(3986) = 3787\,\mathrm{kg/m^2}.$$ $$m = \frac{I_{\mathrm{rim}}}{k^2} = \frac{3787}{1^2} = 3787\,\mathrm{kg}$$

2. Cross sectional area of the flywheel rim

$$m = \rho V = 2\pi RA\rho$$ $$3787 = 2\pi(1)(7200)A = 45245A.$$ $$A = \frac{3787}{45245} = 0.084\,\mathrm{m^2}.$$