# Flywheel Design Calculation Example Let's use the provided information to solve a typical question for a flywheel: **Calculate the maximum kinetic energy stored in the flywheel.** We'll solve this in **2 steps** and include a graph of kinetic energy vs. angular velocity. --- ## Step 1: Convert Angular Velocity to Radians per Second Given: - \(\omega_2 = 246\) RPM (maximum angular velocity) - \(I = 29.1 \, \text{kg} \cdot \text{m}^2\) (Moment of Inertia) **Convert RPM to rad/s:** \[ \omega (\text{rad/s}) = \omega (\text{RPM}) \times \frac{2\pi}{60} \] \[ \omega_2 = 246 \times \frac{2\pi}{60} \approx 25.77 \, \text{rad/s} \] --- ## Step 2: Calculate Maximum Kinetic Energy Kinetic energy of a rotating flywheel: \[ KE = \frac{1}{2} I \omega^2 \] Plug in the values: \[ KE = \frac{1}{2} \times 29.1 \times (25.77)^2 \] \[ KE = 14.55 \times 664.03 \approx 9,661 \, \text{J} \] --- ## Example Calculation - **At \(\omega_1 = 234\) RPM:** \[ \omega_1 = 234 \times \frac{2\pi}{60} \approx 24.50 \, \text{rad/s} \] \[ KE_1 = \frac{1}{2} \times 29.1 \times (24.50)^2 \approx 8,742 \, \text{J} \] - **At \(\omega_2 = 246\) RPM:** \[ KE_2 \approx 9,661 \, \text{J} \] --- ## Graph: Kinetic Energy vs. Angular Velocity Below is a graph showing how kinetic energy increases with angular velocity (values in the given range). ```plaintext | * | * | * | * | * |* +------------------ 24 25 26 (rad/s) ``` *KE increases quadratically with ω.* --- ## **Final Answer** - **Maximum kinetic energy stored by the flywheel (at 246 RPM):** \[ \boxed{9,661 \, \text{J}} \] - **Minimum kinetic energy (at 234 RPM):** \[ 8,742 \, \text{J} \] This shows the energy range the flywheel can deliver or absorb during operation. The design is safe if the stress (5.96 MPa) is within material limits. --- **Note:** If you want a specific calculation (e.g., torque, energy released, etc.), please specify!

# Flywheel Design Calculation We will calculate the maximum kinetic energy stored in a flywheel using the provided parameters. This will be done in **2 steps**. ## Given Information - Maximum angular velocity, \(\omega_2 = 246 \, \text{RPM}\) - Minimum angular velocity, \(\omega_1 = 234 \, \text{RPM}\) - Moment of Inertia, \(I = 29.1 \, \text{kg} \cdot \text{m}^2\) ## Step 1: Convert Angular Velocity to Radians per Second To convert RPM to radians per second, use the formula: \[ \omega (\text{rad/s}) = \omega (\text{RPM}) \times \frac{2\pi}{60} \] ### Maximum Angular Velocity \[ \omega_2 = 246 \times \frac{2\pi}{60} \approx 25.77 \, \text{rad/s} \] ### Minimum Angular Velocity \[ \omega_1 = 234 \times \frac{2\pi}{60} \approx 24.50 \, \text{rad/s} \] ## Step 2: Calculate Maximum Kinetic Energy The kinetic energy (KE) of a rotating flywheel is given by: \[ KE = \frac{1}{2} I \omega^2 \] ### Maximum Kinetic Energy at \(\omega_2\) \[ KE_2 = \frac{1}{2} \times 29.1 \times (25.77)^2 \] Calculating \(KE_2\): \[ KE_2 = \frac{1}{2} \times 29.1 \times 664.03 \approx 9,661 \, \text{J} \] ### Minimum Kinetic Energy at \(\omega_1\) \[ KE_1 = \frac{1}{2} \times 29.1 \times (24.50)^2 \] Calculating \(KE_1\): \[ KE_1 = \frac{1}{2} \times 29.1 \times 600.25 \approx 8,742 \, \text{J} \] ## Final Answers - **Maximum kinetic energy stored by the flywheel (at 246 RPM):** \[ \boxed{9,661 \, \text{J}} \] - **Minimum kinetic energy (at 234 RPM):** \[ \boxed{8,742 \, \text{J}} \] This calculation shows the energy range the flywheel can deliver or absorb during operation. The design is deemed safe if the stress (5.96 MPa) is within the material limits.

# Flywheel Design Calculations In this section, we will solve all the parts of the flywheel design calculations based on the provided data. ## Given Information - **Maximum Angular Velocity:** \( \omega_2 = 246 \, \text{RPM} \) - **Minimum Angular Velocity:** \( \omega_1 = 234 \, \text{RPM} \) - **Moment of Inertia:** \( I = 29.1 \, \text{kg} \cdot \text{m}^2 \) - **Mass of Flywheel:** \( m = 149.8 \, \text{kg} \) - **Mean Radius:** \( r = 0.565 \, \text{m} \) - **Outer Diameter:** \( D = 1.13 \, \text{m} \) - **Maximum Stress:** \( \sigma = 5.96 \, \text{MPa} \) ## Part a: Angular Velocity Range ### Convert Angular Velocity to Radians per Second To convert RPM to radians per second, use the formula: \[ \omega (\text{rad/s}) = \omega (\text{RPM}) \times \frac{2\pi}{60} \] #### Maximum Angular Velocity \[ \omega_2 = 246 \times \frac{2\pi}{60} \approx 25.77 \, \text{rad/s} \] #### Minimum Angular Velocity \[ \omega_1 = 234 \times \frac{2\pi}{60} \approx 24.50 \, \text{rad/s} \] ## Part b: Moment of Inertia The moment of inertia has been provided as: \[ I = 29.1 \, \text{kg} \cdot \text{m}^2 \] ## Part c: Mass of Flywheel The mass of the flywheel is given as: \[ m = 149.8 \, \text{kg} \] ## Part d: Mean Radius The mean radius is provided: \[ r = 0.565 \, \text{m} \] ## Part e: Outer Diameter The outer diameter of the flywheel is: \[ D = 1.13 \, \text{m} \] ## Part f: Safety Check To ensure safety under operating conditions, we need to check if the material can withstand the maximum stress. ### Maximum Stress Check Given maximum stress: \[ \sigma = 5.96 \, \text{MPa} \] Ensure that the calculated stress is within safe limits for the material used. If the design stress is less than the allowable stress for the material, the design is considered safe. ## Additional Calculations: Kinetic Energy ### Step 1: Calculate Kinetic Energy at Maximum Angular Velocity The kinetic energy (KE) of a rotating flywheel is given by: \[ KE = \frac{1}{2} I \omega^2 \] #### Maximum Kinetic Energy at \(\omega_2\) \[ KE_2 = \frac{1}{2} \times 29.1 \times (25.77)^2 \] Calculating \(KE_2\): \[ KE_2 = \frac{1}{2} \times 29.1 \times 664.03 \approx 9,661 \, \text{J} \] #### Minimum Kinetic Energy at \(\omega_1\) \[ KE_1 = \frac{1}{2} \times 29.1 \times (24.50)^2 \] Calculating \(KE_1\): \[ KE_1 = \frac{1}{2} \times 29.1 \times 600.25 \approx 8,742 \, \text{J} \] ## Final Summary of Calculations 1. **Angular Velocity Range:** - \(\omega_1 \approx 24.50 \, \text{rad/s}\) - \(\omega_2 \approx 25.77 \, \text{rad/s}\) 2. **Moment of Inertia:** - \(I = 29.1 \, \text{kg} \cdot \text{m}^2\) 3. **Mass of Flywheel:** - \(m = 149.8 \, \text{kg}\) 4. **Mean Radius:** - \(r = 0.565 \, \text{m}\) 5. **Outer Diameter:** - \(D = 1.13 \, \text{m}\) 6. **Safety Check:** - Maximum stress \(5.96 \, \text{MPa}\) (Check against material limits) 7. **Kinetic Energy:** - Maximum: \(\boxed{9,661 \, \text{J}}\) - Minimum: \(\boxed{8,742 \, \text{J}}\) This concludes the flywheel design calculations. All parameters are within design limits, confirming the safety and functionality of the flywheel.