Calculating Flywheel Weight for a Steam Engine

This problem requires us to determine the weight (or mass in this context) of a flywheel needed for a horizontal cross-compound steam engine based on its power output, speed, coefficient of energy fluctuation, desired speed fluctuation, and radius of gyration. The flywheel helps in smoothing out the variations in torque and speed during the engine cycle.

Understanding Key Parameters and Concepts

• Power (P): The rate at which work is done by the engine. Given as 300 kW.
• Speed (N): The rotational speed of the crankshaft in revolutions per minute (rpm). Given as 90 rpm.
• Angular Speed (ω): The speed in radians per second. Related to N by $\omega = \frac{2\pi N}{60}$.
• Coefficient of Fluctuation of Energy ($C_E$): The ratio of the maximum fluctuation of energy ($\Delta E$) to the work done per cycle. Given as 0.1.
• Fluctuation of Speed: The variation in speed above and below the mean speed. Given as ± 0.5% of the mean speed.
• Coefficient of Fluctuation of Speed ($C_s$): The ratio of the maximum fluctuation of speed ($\Delta \omega$) to the mean speed ($\omega_{mean}$). Total fluctuation is $2 \times 0.5\% = 1\%$. So $C_s = 0.01$.
• Radius of Gyration (k): A measure of how the mass of the flywheel is distributed relative to its axis of rotation. Given as 2 metres.
• Moment of Inertia (I): A measure of an object's resistance to changes in its rotational motion. For a flywheel with mass m and radius of gyration k, $I = m k^2$.
• Maximum Fluctuation of Energy ($\Delta E$): The difference between the maximum and minimum energy stored in the flywheel during a cycle. This is related to the coefficient of energy fluctuation and the work done per cycle.

Step-by-Step Calculation of Flywheel Weight

First, let's calculate the mean angular speed (ω):

Next, we determine the work done per cycle. For a steam engine, one cycle typically corresponds to one revolution of the crankshaft in the context of the turning moment diagram used for fluctuation analysis. The work done per cycle is related to power and speed:

Given P = 300 kW = $300 \times 10^3$ W and N = 90 rpm:

The maximum fluctuation of energy ($\Delta E$) is related to the coefficient of fluctuation of energy ($C_E$) and the work done per cycle:

Given $C_E = 0.1$ and Work done per cycle = $200 \times 10^3$ J:

The fluctuation of speed is given as ± 0.5% of the mean speed. This means the total fluctuation from minimum to maximum speed is $2 \times 0.5\% = 1\%$ of the mean speed. The coefficient of fluctuation of speed ($C_s$) is this total fluctuation expressed as a fraction:

The maximum fluctuation of energy ($\Delta E$) is also related to the moment of inertia (I) of the flywheel, the mean angular speed ($\omega$), and the coefficient of fluctuation of speed ($C_s$) by the formula:

The moment of inertia I is related to the mass (m) and radius of gyration (k) by $I = m k^2$. Substituting this into the energy fluctuation equation:

We need to find the mass (m) of the flywheel. Rearranging the formula to solve for m:

Now, we substitute the calculated values:

• $\Delta E = 20,000$ J
• $k = 2$ m
• $\omega = 3\pi$ rad/s
• $C_s = 0.01$

The calculated mass of the flywheel is approximately 5628.6 kg. Comparing this value to the given options, 5630 kg is the closest value.

The required weight (mass) of the flywheel is approximately 5628.6 kg, which rounds to 5630 kg based on the provided options.