13. Design of Flywheel free study note

  • Reduces fluctuation in speed
  • Stores excess energy
  • Supplies energy during peak load
  • Ensures smooth operation
Design of flywheel

Mathematically,

CE=ฮ”EEmeanC_E = \frac{\Delta E}{E_{mean}}

where:

Cs=Nmaxโˆ’NminNmeanC_s = \frac{N_{max} – N_{min}}{N_{mean}}

  • Indicates permissible speed variation

E=12Iฯ‰2E = \frac{1}{2} I \omega^2

Where:

  • II = mass moment of inertia
  • ฯ‰\omega = angular velocity

Energy Fluctuation Relation

ฮ”E=Iฯ‰2Cs\Delta E = I \omega^2 C_s

9. Design of Flywheel

(a) Determination of Mass Moment of Inertia

I=ฮ”Eฯ‰2CsI = \frac{\Delta E}{\omega^2 C_s}

(b) Rim Design

  • Most energy stored in rim
  • Assume rim carries ~90% of total energy

(c) Stresses in Flywheel Rim

Centrifugal Stress

ฯƒc=ฯv2\sigma_c = \rho v^2Where:

  • ฯ\rho = density
  • vv = peripheral velocity

Bending Stress

  • Due to restraint of arms

(d) Design of Arms

  • Subjected to bending
  • Usually elliptical cross-section

(e) Design of Shaft and Hub

  • Shaft designed for torque
  • Hub fitted to shaft with key

10. Materials for Flywheel

  • Cast Iron (most common)
  • Steel (high-speed applications)

Properties required:

  • High strength
  • Good fatigue resistance
  • Good castability

11. Advantages of Flywheel

  • Smoothens speed fluctuations
  • Improves efficiency
  • Reduces vibration

12. Limitations

  • Cannot control cyclic speed variations completely
  • Adds weight and space
  • Initial cost

13. Flywheel vs Governor

14. Numerical Examples

Problem: A flywheel stores 5000 J of kinetic energy at its mean speed. During one cycle, the maximum fluctuation of energy is 250 J. Find the coefficient of fluctuation of energy.

Given:

  • Energy stored, E=5000 JE = 5000\ \text{J}E=5000ย J
  • Maximum fluctuation of energy, ฮ”E=250 J\Delta E = 250\ \text{J}ฮ”E=250ย J

Solution:CE=ฮ”EEC_E=\frac{\Delta E}{E}=2505000=\frac{250}{5000}CE=0.05\boxed{C_E=0.05}โ€‹

Answer: The coefficient of fluctuation of energy is 0.05, or 5%.


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