Flywheels Introduction

The major source of heat within a subway system is the vehicle. The bulk of this heat is generated when the train is braking to a stop. During a typical station-to-station run, the total amount of heat generated by the train is roughly equivalent to the maximum level of kinetic energy achieved by the train, since to bring the train to a halt, all that energy must be dissipated as heat. Methods of storing this dissipated energy, for later use, were being explored. One of the more promising methods being tested during revenue service (see Ref. 5) consists of using flywheel units on board each car of a train. The flywheel receives energy from the train during braking, and returns energy to the train propulsion system when the train is motoring. A typical flywheel unit consists of a disk mounted on a shaft, a speed

reduction unit, and an electric motor/generator and the necessary controls.

Basically, energy is stored in the flywheel by spinning the disk faster and faster. In other words, the translational kinetic energy of the train is converted to the rotational kinetic energy of the flywheel.

The amount of energy stored in the flywheel assembly at any instant is related to the rotational speed of the disc by the following equation.

K.E. = 1/2 (J ω²) (8.10.1)

where

K.E. = Kinetic Energy stored in flywheel

J = Polar moment of inertia of the flywheel assembly ω = Rotational speed of the disk

The polar moment of inertia of the flywheel assembly is a measure of the flywheel's energy storage capacity. For a disk, it is proportional to the size and weight of the flywheel. A typical flywheel unit would be designed to accept and store an amount of energy equivalent to the kinetic energy of a car at top speed and under “crush” passenger load (assuming one flywheel per car). The flywheel design must also satisfy system weight and space requirements.

Flywheels are normally designed to operate within a prescribed rotational speed range. Rather than allowing the flywheel to be “drained” of all its energy by bringing it to a stop, a minimum energy level is set by selection of a minimum rotational speed. For example, the prototype described in Reference 5 operates between 9,800 and 14,000 RPM. Taking 14,000 RPM to be the maximum rotational speed, this represents an operating range from 70% to 100% of maximum rotational speed. In terms of energy

SES Flywheel Model

The model developed for use with the SES assumes the flywheel operates within the bounds of a maximum and minimum rotational speed limit. During a braking cycle, the flywheel stores energy by spinning faster and faster until the maximum rotational speed is reached. The remaining train energy (if any) is dissipated by conventional means; i.e., through resistor grids. Similarly, during acceleration, the train draws propulsion energy from the flywheel until the flywheel slows down to its minimum rotational speed. When this point is reached, the train switches over and derives its propulsion energy from the third rail. The flywheel motor will also draw sufficient power from the third rail to overcome any internal losses (due to windage, bearings, etc.) in order to maintain its minimum rotational speed.

The model assumes that the flywheel has been properly matched with the traction motors and has been properly sized for the anticipated braking rates. These two assumptions imply that the flywheel can accept and return energy at the same rates at which the train generates and draws energy.

Flywheel Input Data

An entry of 2.0 for the Onboard Flywheel Simulation Option on Form 9H instructs the program that flywheels are to be simulated for this train. The description of the flywheel unit is done on Forms 9K and 9L.

Input on Form 9K includes the polar moment of inertia (defined as J in equation (8.10.1) above) and number of flywheels per car. The polar moment of inertia is defined differently for various geometric shapes, for a solid disk revolving around a central axis:

J = 1/2 Mr² (8.10.2)

where

M = Mass of Disk in lb r = Radius of Disk in ft

The corresponding units of J are lb-ft². For example, for a flywheel with a 20-inch diameter and 8- inch thickness constructed of steel:

M = π(20/12x2)² x (8/12) x 487 lb/ft³

= 708.3 lb

J = 1/2 (708.3) x (10/12)² = 246 lb-ft²

assuming a density for steel of 487 lb/ft³.

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The value of J input in the program should be the sum of the J of the flywheel disk and equivalent J's of the motor alternator and other components of the unit. The other entries on Form 9K are minimum allowable, maximum allowable, and initial rotational speed. These should be entered in units of

revolutions per minute (RPM). The initial rotational speed of the flywheel must lie between the limits established by the minimum and maximum allowable rotational speeds.

Form 9L is used to input efficiency of power transfer from train to flywheel and from flywheel to train. Both these values are products of the efficiencies of the individual steps of power transfer. For example, suppose the efficiency of energy transfer from the train wheel to the train traction motor is 90%, and the efficiency from the traction motor to the alternator is 85%, and from the alternator to the flywheel it is 75%. Then the value that should be input for Efficiency of Power Transfer from Train to Flywheel should be: (0.90 x 0.85 x 0.75) x 100 = 57.4%. The efficiency of any particular step may be different in one direction than in the other.

The last five items of input on Form 9L describe the equation of internal losses of the flywheel.

This value varies with flywheel rotational speed. The form of the equation is:

F(ω) = Aω^d+ Bω^e + C (8.10.3)

The variables A, B, and C must be entered in scientific notation (in Fortran Programming this is known as E-Format). That is they should be entered in two parts, the first being the mantissa which can be any number greater than zero, and the second being the power of 10 to which this number should be raised. For example, if A = 0.00006718 it should be entered as 6.7180E-05 which is equivalent to 6.718 x 10^-5. As can be seen from Input Form 9L, the letter E and the signed power of 10 to which the number is to be raised occupies the last 4 spaces of the entry. If these spaces are left blank, the program will multiply the mantissa by 1. The power of 10 to which the values of A, B, and C are raised should not be confused with the exponents d and e. The exponents d and e indicate how many times ω is to be multiplied by itself before being multiplied by A or B, respectively. If losses are directly proportional to rotational speed, the value of the exponent would be 1. If the losses are proportional to the square or square root of rotational speed, the exponent would be 2 or 0.5, respectively. Exponents may vary from 0.

to 3.

Note: The internal time unit of the program is seconds. Therefore the time units of variables A and B must be in revolutions per second (RPS), and since the units of F(ω) must be in KW, the specific units for A, B and C are:

For the case of the flywheel discussed in Reference 5, it was determined that at maximum rotational speed (14,000 RPM = 233.3 RPS) internal losses were equal to 2.94 KW. From Reference 6, it was determined that general internal losses were proportional to the 2.5 power of the speed. Solving for A gave:

F (ω) = Aω^d + Bω^e + C (8.10.3)

2.94 KW = A(233.3 RPS)^2.5

A = 2.94 KW/233.3 RPS^2.5 = 3.536 x 10^-6 KW/RPS^2.5

As can be seen, these are the units specified above for A, and this would be entered as 3.5360E-06, as explained above.

It should also be noted that only one variable and one exponent were necessary to describe the internal losses of this flywheel. Depending on the flywheel being simulated and the arrangement of its components, two or all three variables may be necessary.

While the input to the model is in units associated with the flywheels, it is possible to simulate other energy storage devices by appropriate manipulation of the input. This could be done by calculating the ranges of energy storage capacity and internal losses of the particular device and relating them to fictitious values of ω by means of formula (8.10.1) and (8.10.3) above.