Answer to Question #170748 in Mechanical Engineering for Katinghan David

Question

A double acting steam engine runs at 210 rpm. A curve of turning moment plotted on the crank angle base showed the following areas alternatively above and below the mean turning moment line: 880,500,620,720,360,560,440,520mm²

The scales used where 1mm= 520Nm and 1mm= 1^o crank angle. If the total fluctuation in speed is limited to 1.85% of the ,mean speed, determine the mass of flywheel necessary if the radius gyration is 1.4m

Answer 1

Maximum energy =$880+E$, where E is the minimum energy

$\triangle E=E_{max}-E_{min}=E +880-E=880 mm^2$

$\triangle E=880 \times520 \times \frac{\pi}{180}=7986.62666 Nm$

We know that $\triangle E =mR^2w^2C_s$

$w_1=210 \times1.85 \times \frac{2 \pi}{60}=40.68$

$w_2=210 \times \frac{2 \pi}{60}=10.995$

$w=\frac{w_1+w_2}{2}=\frac{40.68+10.995}{2}=25.8375$

$C_s=\frac{w_1-w_2}{w}=\frac{40.68-10.995}{25.8375}=1.1489$

$\triangle E =mR^2w^2(\frac{w_1-w_2}{w})=mR^2w(w_1-w_2)$

$m=\frac{\triangle E}{R^2w(w_1-w_2)}$

$m=\frac{7986.62666}{1.4^2 \times 25.8375(40.68-10.995)}=5.313 kg$