Flywheel Design

Design Brief

An air compressor vital to operations is experiencing severe vibrations, causing equipment malfunction and disrupting workflow. This critical situation necessitates immediate action to design a new drive system capable of handling the tremors and ensuring smooth operation. Engineering expertise is urgently needed to develop a solution and restore optimal functionality before further operational issues and potential financial losses occur.

• The compressor must run at an average speed of 180 rpm.

• A flywheel needs to be attached to a suitable location.

• The compressor will be running for 10 hours a day, 6 days a week, and 51 weeks a year.

• The drive system should be designed to be as compact as possible.

1. State the function of the flywheel in the system.

2. Determine the required moment of inertia for the flywheel.

3. Select a suitable off-the-shelf flywheel or design a customised flywheel.

Solution

Function

A flywheel is a heavy disc used to dampen fluctuations in power within a system. Flywheels store kinetic energy generated by a motor, acting as a mechanical battery. Newton’s first law of motion, an object in motion will stay in motion, is demonstrated through the flywheel where the flywheel is able to keep rotating after the motor has stopped. This can be used to provide energy to a system even when power production to the system has ceased or at a higher energy than the energy source.

The use of the flywheel can be better understood through looking at the fluctuations in torque throughout the compressor cycle, this is shown in Figure 1. The flywheel is used to smoothen the differentiation in torque throughout the cycle. The cycle can be described as the air is pulled, suction, as the torque is low, when the air is being pressurised, compression, the torque increases dramatically until it reaches a peak and the air is delivered, power, and as the remaining air is released, exhaust, the torque returns to zero. The flywheel function in this system is to both store energy and to deliver energy at a higher rate than the source.

Design

Energy in Output Shaft

The first consideration in the design of the flywheel is calculating the energy that can be delivered during one cycle. The cycle that is used is shown in Figure 8 representative of the torque on the output shaft attached to the compressor. The fluctuations in torque are caused from the vibrations from the compressor. The total energy that can be delivered is the integral of the graph, or the area underneath the curve. As shown in Appendix 1:

Mean Torque in Output Shaft

The total energy used in the system can be divided by the angular displacement during the cycle to find the mean torque in Nm. This is the torque the system will aim to provide at a constant rate, with the flywheel being able to compensate for fluctuations. As shown in Appendix 2:

Location

The location of the flywheel can either be on the shaft with the largest angular displacement, or the shaft closest to the vibrations. For this case of a transmission system for an air compressor, a step-down system is used. This is where the input shaft has high speed, low torque and the output shaft has low speed, high torque. The vibrations in the system are a result of the air compressor at the output shaft, the shaft closest to the vibrations. The shaft with the largest angular displacement is the input shaft, so the flywheel was selected to be attached at this point. Because it is spinning as fast as possible in this position, this enables the flywheel to retain its maximum possible energy for the given power delivery system.

Mean Torque in Input Shaft

The mean torque in the input shaft directly corresponds to the torque outputted from the motor. The operating speed for the motor is 1.96 Amps, with the corresponding torque value at this location being 3.069 Nm (Appendix 2: Graph 1). This is representative of the mean torque outputted by the motor.

Comparing the mean torque at the output shaft, 28.33 Nm, to the input shaft, 3.069 Nm, a relationship can be established where the torque in the input shaft is 1/9.23 of the torque in the output shaft.

Change in System Energy

The largest fluctuation in energy that the flywheel must absorb occurs at 210 degrees, as shown on Figure 8. To find the change in relative energy, the mean is subtracted from the torque curve. The torque curves remain constant throughout the whole system, meaning the values for input and output shaft are linked by the relationship established in Mean Torque in Input Shaft.

Appendix 3 contains the relevant calculations for the system energy changes, in summary:

Energy in Flywheel

The energy in the flywheel is the product of speed fluctuation, the flywheels moment of inertia and the square of its angular velocity. This can also be multiplied by the efficiency of the flywheel.

As the operating speed of the motor is set to 3524 rpm, a low amount of speed fluctuation is going to be present. As such, a value of 0.1 or 10% is assumed. This means ±10% variation in speed is expected at the input shaft.

The flywheel is set to being 85% efficient.

Using Equation 1 the energy were solved simultaneously for the moment of inertia, resulting in 𝐼 = 0.001225 kgm² (Appendix 4)

Flywheel Sizing

Using the moment of inertia required, arbitrary geometric values were used and iterated upon to find a functional and cost-effective design. The final flywheel has a radius of 50 mm and a thickness of 20 mm. The flywheel is shown below.

Justification and Explanation

Mean Torque in Input Shaft and Change in System Energy

The torque curve that is given for the compressor and the output shaft can also be used to represent the torque being transmitted to the input shaft. The torque variation in the system is constant. As the mean torque at the output was calculated to be 28.33 Nm by finding the matching torque at the input, a relationship can be established between the step-up in torque. The system is a step-down system where the rpm is decreased from input to output and torque increases from input to output. From the motor specifications, the torque at the operating current of 1.96 Amps is 3.069 Nm.

Comparing the mean torque at the output shaft, 28.33 Nm, to the input shaft, 3.069 Nm, a relationship can be established where the torque in the input shaft is 1/9.23 of the torque in the output shaft. This means that for all torque values at the output shaft have a corresponding input torque 1/9.23 of the value.

When calculating the change in system energy, the mean torque was subtracted from the largest differential in torque. Using the torque angular displacement curve in Figure 8 and multiplying the maximum displacement (the trapezium on the right-hand side) by 1/9.23 and subtracting the mean torque on the input shaft, the maximum energy change in the system was able to be calculated.

Location

The flywheel had two possible attachment points of: the point with the largest angular velocity, input shaft, or the point with the largest vibration, output shaft. The design decision was made to place the flywheel on the input shaft, as the larger angular velocity results in a smaller flywheel being needed. The smaller required flywheel results in an easier to manufacture product as less material is needed and as less material is needed the cost for the smaller flywheel is comparatively less.

At the output shaft a moment of inertia of 4.33 kgm² (Appendix 5) is present as the energy within the system is at a much higher 130.91 J compared to at the input shaft where the energy is 14.18J and the moment of inertia of the flywheel is only required to be 0.001225 kgm². The larger the moment of inertia, the larger the required flywheel must be, as the force of inertia is more easily stored in the fast-moving shaft compared to the slow moving one. The effect of this when applied to Equation 2 results in a vastly different energy within the flywheel and moments of inertia, as shown above.

As the system is a step-down system, the energy taken and supplied by the flywheel is still passed on to the compressor regardless of its location.

Energy in Flywheel

A coefficient of speed fluctuation of 0.1 or 10% was assumed. As the flywheel is attached to the input shaft and thus the motor, only a small fluctuation in angular velocity needs to be accounted for. Assuming a value of 10% ensures that all rotational velocity ±10% of the operating velocity are considered.

The efficiency of the flywheel was assumed to be 85%. A study by J.K Kaldellis and a second study by Arash Moradzadeh found that the efficiency of the flywheel as an energy storage system resulted in efficiencies between 80% and 90% (Kaldellis, 2010) (Moradzadeh, 2021). Taking the mean of these values allows for an accurate assumption of the efficiency of the flywheel in the system.

Flywheel Sizing

The minimum flywheel sizing is based off the minimum required moment of 0.001225 kgm². A radius of 50 mm and a thickness of 20 mm were chosen. This results in a moment of inertia of 0.001532.

Appendix 6 considers this design for both minimum radius and thickness for each of these values. The minimum required thickness for a flywheel of radius 50 mm is 15.997 mm, and the minimum required radius for a flywheel of thickness 20 mm is 47.285 mm. This means both values allow each other to be feasible.

The flywheel is assumed to be a solid disc as it is connected to the shaft. If the flywheel was considered as a ring where the hole for the shaft is considered as the inner radius (4.3725 mm). The resultant required thickness is 15.998 mm. This is marginally larger than the required size for the solid disk, and thus is negligible. Appendix 8 contains the relevant calculations.

Stress Considerations

Stress within the flywheel needs to be considered as if the radial or tangential stress exceeds the yield strength of steel the flywheel will tear itself apart. Appendix 7 contain these calculations. The results in Table 2 show that the stresses within the flywheel do not exceed the yield strength of steel (MatWeb, 2023).

Manufacturing and Cost

The flywheel is to be a custom-made part, it is to be made from steel and machined through laser cutting. The maximum thickness of steel that can be laser cut is 20-25 mm (KOMASPEC, 2021). This means the disc can be manufactured from a single piece of sheet metal. The flywheel also contains a keyway to mount to the shaft, the use of a laser cutter allows this to be machined easily without the need for additional parts to be made to aid in the manufacturing process and for the accuracy to be high as the laser cutter can operate at an accuracy within 0.1 mm.

The cost of 20 mm thick sheet metal is $770 per ton (TJ Emerson Steel, 2023). Laser cutting steel costs $3-$13 per minute (McHone Industries, 2019). A ton of steel has a volume of 1.133 m3 and the flywheel has a volume (including the hole and keyway) of 0.000157 m³, resulting in an equivalent cost of $0.11. A high-powered steel laser cutter cuts at a speed of 0.6-0.9 mm/min for 20 mm thick steel (Artizono, 2023). The total distance the laser cutter must travel 380 mm, therefore the time to cut (assuming a speed of 0.9 mm/min) is 422 minutes and thus a cost ($3/min) of $1266 to manufacture as a one-off part. Depending on the budget of the project, can this solution be deemed cost-effective. If the flywheels were to be mass manufactured, the cost would be greatly reduced and could be cost-effective.

Other options would be to find a gear or pulley of similar size and density and utilise this as a flywheel, this will yield the same results at a much lower cost due to mass manufacturing of the part and its availability for off the shelf purchase.

Results

Individual mark for flywheel component: 10/10

Overall group mark: 74.5/100

Personal reflection after results

This assignment could be improved through the addition of FEA. It would allow the stress considerations to be further verified.

Appendix

1 Total Energy in the Output Shaft

The total energy in the output shaft system during operation can be expressed as: 

Where T is torque in Nm, θ is angular displacement in rad.

Figure 21 displays the torque in relation to the crank angle within the system.

Finding the total energy in the shaft system: 

2 Mean Torque in the Output Shaft

The mean torque can be found by dividing the total energy in the shaft system by the angular displacement of the cycle. For this cycle the angular displacement in 2π. 

Applying this to Figure 21:

The flywheel is located on the other end of the transmission system on the input shaft, whereas these values correspond to the output shaft.

The mean torque value was sourced from the motor specifications. Where the motor will be operating at 1.96 Amps, operating current, a resultant torque will be produced of 0.313 kgfm. This is equivalent to 3.069Nm.

Comparing the motor output torque, 3.069 Nm, to the shafts mean torque, 28.33 Nm, a relationship can be established.

Therefore, a factor of 9.23 should be applied to the torque curve.

3 Change in Energy

Subtracting the mean torque from the torque curve and integrating again to find the change in energy relative to the mean:

As the flywheel is located on the input shaft and not the output shafts, the torque must be adjusted: 

4 Energy in Flywheel

The velocity of the input shaft is: 3524 rpm, this is equivalent to an angular velocity of 369 rad/s.

Thus, 𝜔 = 369 rad/s

The coefficient of speed fluctuation is assumed as: 𝐶ₛ = 0.1

And the efficiency of the system is 85%.

5 Moment of Inertia

For the input shaft: 

14.18 = 85% × 0.1 × 𝐼 ×(369)²

Solving for I: 

𝐼 = 0.001225 kgm²

For the output shaft: 

𝜔 = 180 rpm =18.85 rad/s

130.91 = 85% × 0.1 × 𝐼 × (18.85)²

Solving for I: 

𝐼 = 4.335 kgm²

6 Required Flywheel Size

For a solid disk:

Taking a thickness of 20mm and a radius of 50mm made of solid steel with density 7800 kg/m³: 

𝐼 = 0.5 × 7800 × 𝜋 × (0.02) × (0.05)⁴ = 0.001532 kgm²

As the moment of inertia is lower than the Moment of Inertia of the system this a valid design.

The minimum thickness for a radius of 50mm: 

The minimum radius for a thickness of 20mm: 

Both values within the ranges

7 Stress Considerations

Where ν = Poisson’s Ratio = 0.25 (MatWeb, 2023) and 𝜎 is the maximum radial stress.

Where ν = Poisson’s Ratio = 0.25 (MatWeb, 2023) and 𝜎 is the maximum tangential stress.

The yield strength of steel is 350 MPa (MatWeb, 2023). The flywheel will not yield at operating speeds.

8 Flywheel as a Ring

If the flywheel was considered as a hole, the moment of inertia is:

Where R2 is the outer radius and R1 is the inner radius.

Rearranging for thickness: 

Solving with the outer radius of 50mm and the inner radius as the radius of the shaft as 4.3275mm.

References

Artizono. (2023). Retrieved from Laser Cutting Thickness & Speed Chart: https://artizono.com/fiber-laser-cutting-thickness-speed-chart/

KOMASPEC. (2021, 8 16). An Engineer's Guide to Laser Cutting. Retrieved from KOMASPEC: https://www.komaspec.com/about-us/blog/an-engineer-s-guide-to-laser-cutting/#:~:text=Generally%2C%20when%20laser%20cutting%20metals,cut%20thicker%20plates%20than%20this.

MatWeb. (2023). Steel, General Properties. Retrieved 2023, from MatWeb: https://www.matweb.com/search/datasheet.aspx?bassnum=MS0001&ckck=1

McHone Industries. (2019, 11 7). How Much Does Laser Cutting Steel Cost. Retrieved from https://www.mchoneind.com/how-much-does-laser-cutting-steel-cost/#:~:text=Total%20Costs,%2413%2D%2420%20per%20hour.

Moradzadeh, A. (2021). Chapter 2 - Energy storage fundamentals and components. In Energy Storage in Energy Markets (pp. 23-39).

TJ Emerson Steel. (2023). 20mm Thick Steel Plate Price Per Ton Mild Steel Sheet Coils Hot Rolled Steel Plate. Retrieved from Made-in-China: https://tjemersonsteel.en.made-in-china.com/product/RKoxUgXvHbpY/China-20mm-Thick-Steel-Plate-Price-Per-Ton-Mild-Steel-Sheet-Coils-Hot-Rolled-Steel-Plate.html

USA Roller Chain & Sprockets. (n.d.). 40BS19 Sprocket. Retrieved from USA Roller Chain: https://www.usarollerchain.com/product-p/40bs19-1-sprocket.htm

Woodford, C. (2023, 4 31). Flywheels. Retrieved from Explain That Stuff: https://www.explainthatstuff.com/flywheels.html