Off-Peak Flywheel Energy Storage
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m3 , in the shape of a uniform disk with a thickness of 13.9 cm .
Using off-peak hours to generate and store energy in large flywheels can indeed be a viable method for managing peak loads in power plants. Flywheels are mechanical devices that store energy in the form of rotational motion and can release it when needed. Below, is the calculation of the parameters and properties of the flywheel described above.
Given data:
- Density of iron (ρ) = 7800 kg/m³
- Shape: Uniform disk
- Thickness (t) = 13.9 cm = 0.139 m
- First, we need to find the volume (V) of the flywheel:
Volume (V) = Area of the disk (A) × Thickness (t)
For a uniform disk, the area is given by:
Area (A) = π × (Radius)^2
- We also need to find the mass (m) of the flywheel:
Mass (m) = Density (ρ) × Volume (V)
- Next, we can calculate the moment of inertia (I) of the flywheel:
Moment of Inertia (I) = (1/2) × Mass (m) × (Radius)^2
- Finally, let’s consider the energy (E) stored in the flywheel:
Energy (E) = (1/2) × Moment of Inertia (I) × (Angular velocity)^2
Note: The energy stored in the flywheel is proportional to the square of its angular velocity.
Proceeding with the calculations:
- Finding the volume (V): The radius (R) of the disk is not provided, so we can’t calculate the exact volume. Let’s consider a hypothetical value for the radius (e.g., R = 1 meter) to demonstrate the calculations.
Assuming R = 1 meter, the area (A) of the disk is:
A = π × (1 meter)^2 ≈ 3.14159 m²
Now, calculating the volume (V):
V = A × t V = 3.14159 m² × 0.139 m ≈ 0.437 m³
- Finding the mass (m): m = ρ × V m = 7800 kg/m³ × 0.437 m³ ≈ 3408.6 kg
- Finding the moment of inertia (I): For a uniform disk, the moment of inertia is given by:
I = (1/2) × m × R^2
Using our hypothetical value R = 1 meter:
I = (1/2) × 3408.6 kg × (1 meter)^2 ≈ 1704.3 kg·m²
- Finding the energy (E): Assuming the angular velocity (ω) is in radians per second, the energy stored in the flywheel can be calculated using:
E = (1/2) × I × ω^2
Notably, the above calculations are based on the assumption of a specific radius (R = 1 meter). Additionally, the practical implementation of such flywheel-based energy storage systems involves various engineering challenges, but the concept of using off-peak hours to store energy for peak demand times is valid and utilized in various energy storage solutions.
