Question

For a flywheel, if $\omega_{Max}$ is the maximum speed and $\omega_{Min}$ is the minimum speed, then the coefficient of fluctuation of speed will be:

• A. $\frac{(\omega_{Max} - \omega_{Min})}{(\omega_{Max} + \omega_{Min})}$
• B. $\frac{(\omega_{Max} + \omega_{Min})}{(\omega_{Max} - \omega_{Min})}$
• C. $\frac{2(\omega_{Max} - \omega_{Min})}{(\omega_{Max} + \omega_{Min})}$
• D. $\frac{2(\omega_{Max} + \omega_{Min})}{(\omega_{Max} - \omega_{Min})}$

Hint:

Remember the basic concept: Fluctuation divided by Mean. Since the mean speed has a "2" in its denominator ($\frac{a+b}{2}$), it moves to the numerator when you divide, resulting in the factor of 2 at the top.

Answer 1

The Correct Option is C

The coefficient of fluctuation of speed ($C_s$) is a dimensionless parameter used in the design of flywheels to quantify the variation in speed during a machine's cycle. 1. Definition and Mathematical Formula: The coefficient of fluctuation of speed is defined as the ratio of the maximum fluctuation of speed to the mean speed of the cycle.

• Maximum Fluctuation of Speed: This is the difference between the maximum and minimum speeds: $(\omega_{Max} - \omega_{Min})$.

• Mean Speed ($\omega_{mean}$): This is the average of the maximum and minimum speeds: $\omega_{mean} = \frac{\omega_{Max} + \omega_{Min}}{2}$.

2. Deriving the Coefficient ($C_s$): Substituting these into the ratio: $$C_s = \frac{\omega_{Max} - \omega_{Min}}{\omega_{mean}}$$ $$C_s = \frac{\omega_{Max} - \omega_{Min}}{\frac{\omega_{Max} + \omega_{Min}}{2}}$$ $$C_s = \frac{2(\omega_{Max} - \omega_{Min})}{\omega_{Max} + \omega_{Min}}$$

3. Engineering Significance: A smaller value of $C_s$ indicates a more stable machine with less speed variation. For instance, in internal combustion engines, the flywheel must limit speed fluctuations to ensure smooth power delivery and prevent structural vibrations.