Apportioning and Mitigation of losses in a Flywheel ... - IEEE Xplore

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Mumbai, India [email protected]. Abstract— A typical Flywheel Energy Storage (FES) system consists of a flywheel, an electrical machine and bidirectional.
Apportioning and Mitigation of losses in a Flywheel Energy Storage system S.R.Gurumurthy1 Archana Sharma3 Satyabrata Sarkar4 Homi Bhabha National Institute, Mumbai, India [email protected] Abstract— A typical Flywheel Energy Storage (FES) system consists of a flywheel, an electrical machine and bidirectional converter/controller. Between the flywheel (which stores the energy) and the load (which consumes the energy) there are different systems like, electrical machine and bi-directional power converter. A portion of extracted energy from the flywheel is dissipated as loss in these devices before it is delivered to the load. These losses can be categorized as mechanical losses (drag, Bearing friction), electrical losses (hysterisis, eddy current, copper) and power converter losses (switching, conduction). Magnitude of these losses depends on the operating conditions like, motor speed, dc bus voltage, switching frequency, load current etc. It is necessary to find out the sources of these losses and quantify them, which will help us to adopt suitable loss mitigation techniques. One such system was built, tested; various sources of losses are found out and quantified. Also the tests were conducted to find the performance parameters such as backup time, boost converter gain at various loads, energy efficiency etc and the results are presented.

Vivek Agarwal2 Indian Institute of Technology Mumbai, India [email protected] flywheel when dc power is available and back to dc bus when it fails.

Keywords— Flywheel Energy Storage, Energy efficiency and Power Conversion. TABLE I.

NOTATIONS USED IN THIS DOCUMENT

--------------------------------------------------------------------------Pmech Generated mechanical power (Watts) PLoad Electrical power available (Watts) PLoss Power which is lost (Watts) ω Angular velocity (Rad/s) Vdc DC voltage (V) Idc DC current (A) Tbu Power back up time (Seconds) J Moment of Inertia (Kgm2) N Flywheel Speed(RPM) Vg Generator voltage (V) D Duty cycle of switching pulses RS Source Resistance RL Load Resistance --------------------------------------------------------------------------I.

ENERGY CONVERSION PROCESS IN A FLYWHEEL ENERGY STORAGE (FES) SYSTEM

A typical FES system consists of a flywheel, an electrical machine and bidirectional converter/controller as shown in Fig.1. This system is connected in parallel to the dc source, which is feeding a dc load. This system stores the energy in the

Figure 1. Block diagram of typical FES system. A. Energy conversion process Please refer to the Fig.1. The flywheel is coupled to the rotor of an electrical machine. This machine is used to accelerate the flywheel in charging mode to store energy in the flywheel [1]. The same machine discharges the flywheel in deceleration mode to feed power back to the dc bus. A bidirectional converter along with a controller will act as high frequency drive for motoring action and voltage-regulating system for generating action of the above mentioned electrical machine [2]. B. Sources of losses The devices like, flywheel, electrical machine, bidirectional power converter and their sources of losses [3] are as shown in the Fig.2. It is possible to find the sources of losses, quantify them by computation and their relative contributions to the total loss. By knowing the sources and their contribution to total losses, one can adopt different techniques to reduce the losses.

II.

EXPERIMENTAL SETUP

The complete system has been built using a Permanent Magnet (PM) Brushless DC (BLDC) machine, IGBT inverter and a mechanical flywheel. Block diagram of test setup is as shown in Fig.1. Motorola make DSP chip DSP56F805 is used as the controller. Motor current, rotor position and dc bus voltage are sensed using Hall effect sensors and fed back to the DSP. These signals are used by DSP for the generation of control pulses for IGBT and voltage regulation. Specifications of the system built were as given below: • Input voltage : 440volts DC • Output power : 5.0 kW • Rated speed : 15000 RPM • Machine Type : 3Φ, 4 Pole, BLDC • Weight of the flywheel : 42 kg • Moment of Inertia (J) : 0.681 Kg-m2 • Backup time : 60 Sec • Controller : Motorola DSP56F805 • Power Switch (IGBT) : SKM100GB123D • Drive Switching Freq : 16 kHz A. Test/Experiments conducted Machine coupled to the flywheel is made to accelerate to the rated speed using input dc power. After reaching full speed, if the input power supply to the motor is switched off, the flywheel will continue to run due to its inertia driving the generator till all stored energy is consumed. If an electrical load is connected across the generator it draws current and absorbs the energy stored in the flywheel [4]. This makes the flywheel to decelerate. If electrical load is not connected, the stored mechanical energy in the flywheel is dissipated as losses in mechanical components, electrical machine and the power converter. Total power input to the system is during the acceleration/running is computed using the following relationship: Power input = Vdc* Idc Mechanical losses are assumed be negligible in the case when the flywheel is not coupled to the rotor (of motor) as compared to the case when flywheel is coupled to the rotor. This assumption is made because, the mass of the rotor (8.8kg) is 1/5th as compared to mass of the flywheel (45.3kg)

2) Performance analysis tests: In these tests the input power supply is switched off and the flywheel is allowed to decelerate. The time duration for which the dc bus voltage is maintained constant at 400V is recorded at various loads. Following tests were conducted to find various performance parameters like backup time, boost converter gain, maximum energy extracted etc. a) Power backup test (To find overall performance) b) Power /Energy balance test c) Boost converter performance test. The losses in the system are measured, tabulated and plotted as a function of speed using data taken from above mentioned tests as shown in the Fig.3. Speed Vs Losses 10000 8000

Losses in Watts

Figure.2. Block diagram representing various losses in a FES System

and surface area of rotor is only 1/8th (0.036m2) as compared to the surface area of flywheel (0.2883m2). Hence, when the flywheel is not coupled to the rotor shaft, the loss due to the bearing friction as well as drag can be neglected and entire measured loss is due to the iron loss of the electrical machine used. Following are the tests conducted; 1) Loss Analysis Tests (To find out the source of losses): In order to evaluate the mechanical losses in the machine the classical retardation test is done. The machine along the flywheel is made to run up to a speed of 15000 RPM. Power is cut off and machine is allowed to decelerate. The speed at regular intervals of time are recorded and plotted. The tests are conducted with the following conditions. a) Retardation test without flywheel b) Retardation test with flywheel, without load c) Retardation test with flywheel, with load

With Flywheel (Total Loss)

6000

Plot 2

4000

Without Flywheel (Electrical Loss)

2000 Plot 1

0 0

5000

10000

15000

20000

Speed in RPM

Figure.3. Losses with and without flywheel as a function of speed.

IV.

APPORTIONING OF VARIOUS LOSSES AND THEIR RELATIVE CONTRIBUTION TO THE TOTAL LOSS

B. Relationship between the flywheel speed and various losses The relation between various losses as a function of speed is found out using curve-fitting method. If PLoss is the total loss in the system and "N" is the operating speed in RPM of the machine, then the equation thus obtained are given as below:

A. Various Losses in generation mode of operation: The individual losses are computed at various speeds using (1) and (2). These values have been tabulated in TABLE II and plotted as a function of rotor speed as shown in Fig.4 & Fig .5

1) Without flywheel (Plot-1 of Fig.3): PLoss = 0.00000862464N 2 + 0.0388244N (1) (Eddy current loss) (Hysterisis loss) 2) With flywheel (Plot-2 of Fig.3): 0.00004N x + 0.0498244N (2) PLoss = (Drag loss) (Bearing friction loss) (“x” varies from 1.91 to 1.985 as “N” varies from 0 to 15000 RPM). It is found that the right hand side of the equation contains two terms, one is proportional to the speed and the other is proportional to the square of speed. This is in the expected lines. From the theory, it is known that the bearing friction loss in the flywheel and hysterisis loss in machine are proportional to speed and Drag loss in the flywheel and eddy current loss in the machine are proportional to square of the speed. In case of drag loss, it may be noticed that the power of speed is not constant “2”, but it varies from 1.9 to 1.985 as the speed varies from 0 to 15000 RPM.

TABLE II.

Speed in kRPM

Total input power (Watts)

VARIOUS LOSSES IN THE SYSTEM Electrical machine Loss (Watts)

Mechanical Loss (Watts)

B/F

Drag

Hyste resis

Eddy Copper current

Power converter Loss (Watts) Switch -ing

Conduc -tion

1.3

185

15

24.3

51

12

0.3

71.6

11.2

4.0

578

44

219

156

113

1.0

23.9

20

8.3

1923

91

978

321

479

3.4

11.9

37

12.1

4723

133

3005

470

1028

11.2

8.4

66

15.0

8670

165

6216

582

1575

24.9

7.0

99

Losses in the Electro-mechanical System 7000.0 6000.0 5000.0

A. Testing without flywheel: As explained in Para II A, when the flywheel is not coupled to the rotor shaft, the loss due to the bearing friction as well as the drag can be neglected. Therefore, the measured loss during this test can be entirely due to the iron loss of the machine. B. Testing with flywheel: As explained in Para II A, when flywheel is coupled to the rotor shaft, loss computed is the sum total of all the losses in the machine. They are the losses due to bearing friction, drag, hysteresis and eddy current. Subtracting the square component of loss obtained from test “without flywheel” from square component of loss obtained from the test “with flywheel” gives the drag loss in the system. Similarly, subtracting the linear component of loss of test “without flywheel" from the linear component of loss obtained from the test “with flywheel” gives the bearing friction loss of the machine.

4000.0 3000.0 2000.0

Eddy

1000.0

Hysterisis B/F

0.0 0

2000

4000

6000

8000

10000

12000

14000

16000

Speed in RPM

Figure.4. Electro-mechanical losses in the system as a function of speed. Losses in the Power Converter System 80.0 70.0

Losses in Watt

DATA ANALYSIS

Losses in Watt

Drag

III.

Switching loss @ 16KHz

60.0 50.0 40.0

Conduction loss

30.0 20.0

Copper loss

10.0 0.0 0

2000

4000

6000

8000

10000

12000

14000

16000

Speed in RPM

Figure.5. Power converter losses in the system as a function of speed.

Generator Induced/Output voltage v/s Time

B. Relative contributions of various losses to the total loss.

TABLE III.

VARIOUS LOSSES AND THEIR CONTRIBUTIONS TO THE TOTAL LOSS

Speed in RPM 2670 5400 10530 15000

Type of Losses in Percentage (%) Drag

B/F

Eddy Current

28 43 60 72

9 7 3 2

15 23 23 18

Hyste -resis

Switch -ing

Condu -ction

Copper

32 23 12 7

11 2.0 0.28 0.10

3 2 1.5 1

0.20 0.17 0.20 0.30

From the experiments and studies, which are reflected in the TABLE III, we can make observation as follows: • Bearing friction and hysteresis losses increase linearly with the rotor speed. Drag and eddy current losses increase in square law with rotor speed. • Loss due to drag and eddy current dominates at higher rotor speeds. • The loss can reduce to the large extent if the drag and eddy current losses are reduced at higher speeds. This is due to the fact that the drag and eddy current losses dominate at higher rotor speeds. • Most dominant loss in the system at higher speed is drag followed by eddy current loss. • Switching losses are high at low speed due to high input current.

Output Voltages

Gen / Output Voltage in Volt.

Various losses and their contributions to the total loss at different speeds are given in the table TABLE.III.

500 400

RL= 51.2 Ω

RL=39.4 Ω

RL=60 Ω

300 200

Induced Voltages @ 39.4 Ω

100

@60 Ω

0 0

20

40

60

@51.2Ω

POWER BACKUP TEST

This test was conducted to find out the following parameters. a) Time duration for which the system supplies the power to the load (in the absence of input dc power) at desired constant voltage. b) Minimum speed (or induced voltage) up to which the system maintains the output dc bus voltage. c) To verify energy balance and power balance equations. Input supply is switched off and the flywheel is allowed to decelerate. The time up to which the dc bus voltage is maintained at 400V is recorded. The test is conducted with various load currents, at constant dc bus voltage 400V. Fig.5 shows the variation of input and output voltages with time. It may be noted that, even though the generator voltage reduces with time (as speed is reduced), the dc bus voltage is maintained constant at 400V for a duration depending up on the load. It is also observed that the minimum speed (induced voltage) up to which the system maintains the output dc voltage is a function of load resistance.

100

Time in Seconds

Figure5. Output voltage at various load as a function of time VI.

POWER /ENERGY BALANCE EQUATIONS

When input power is not available, flywheel continues to run and driving the generator. Power available at load will be equal to power generated minus the losses in the system. The losses are taking place in flywheel, machine and power converter. A. Power balance equations in generation mode Mechanical power generated, Power drawn by load and the power lost in the conversion process are computed as given below:

Pmech = J ∗ ω ∗

dω dt

(3)

P Load = V dc ∗ I dc V.

80

-100

(4)

P Loss = 0.00004 N + 0.0498244 N x

(5)

B. Power Balance Test A resistive load of 39.2 Ohms was connected across the dc bus. Input power was switched off and the flywheel was allowed to decelerate. A 4.0kW of power was drawn continuously from the dc bus. Total loss in the system, power delivered to the load are measured at various speeds every five minutes interval as the flywheel decelerated. Power generated from the flywheel is also computed from the retardation test data for this period. All these values are tabulated and plotted as given in the Fig.6.

VII. BOOST CONVERTER PERFORMANCE TEST

Power balance graphs

Topology of the boost converter used is as shown in Fig.1. The top speed of the flywheel, load voltage, output power and back up time (with load resistance as the parameter) are noted down. The readings taken from this test are tabulated which is shown in the TABLE V.

7000 6000 Generated power

Power in Watts

5000 4000

Load power

3000

TABLE V.

2000 1000

Loss power

0 0

2000

4000

6000

8000

10000

12000

Speed in RPM

LOAD TESTING WITH BOOST CONVERTER AT 5.0KW

Minimum Output Load Backup Set Output Top speed Max input Resistance power in time in voltage in of flywheel Voltage voltage gain in Ohms watts seconds volts in RPM in volts 130

1230

35

185

400

11100

11.43

Figure 6. Power at various stages as a function of speed

82

2254

45

130

430

11370

9.56

C. Energy balance Equation Mechanical energy extracted from the flywheel (E1), averaged generated energy (E2), energy lost in the various power processing components (E3), energy delivered to the load (E4) and energy efficiency (η) during the energy extraction process are computed as follows

40

4775

80

70

437

11250

5.46

27

5925

82

25

403

7800

4.91

22

11170

110

10

8670

4.52

(

1 ∗ J ∗ ω 2max − ω 2min 2 dω ∗ T bu E2 = J ∗ω ∗ dt

E1 =

(6)

)

(7)

E 3 = P Loss ∗ T bu

(8)

E4 =V

(9)

η =

dc

∗ I dc ∗ T bu

E4 E1

TABLE IV. Sr.No

It is observed that the voltage gain of the boost converter has reduced as the load power is increased. This results in the increase of the minimum speed (induced voltage) up to which the boost converter can maintain the output dc voltage. This is expected lines as the voltage gain is related to ratio of source resistance to load resistance by following relationship [5] Vdc 1 1 = × Vg (1 − D) 1+ Rs / RL (1 − D)2

MEASURED/COMPUTED ENERGY AT VARIOUS STAGES Sources of Energy

Energy in Joules

1

Extracted Mechanical Energy (By computation)

327882

2

Generated Mechanical Energy (By computation)

336227

3

Energy Lost by in various levels (By Measurement)

82086

4

Electrical Energy available at load (By measurement)

246000

5

Energy efficiency

Drag loss can be reduced by providing the vacuum enclosure [6] for the rotating parts of the system. Both eddy current and hysteresis loss can be reduced by the using a two pole machine (with this the supply frequency becomes half) as well as by using low specific loss core material [7]. Bearing friction loss can be reduced by using an active magnetic bearings or ceramic/hybrid bearings [8]. With magnetic bearings, the system will become complicated and very expensive. Using a two pole machine will be a cheaper option compared to costly low specific loss core material. Usage of two pole machine will reduce the flux dependent loss as well as armature current dependent eddy current loss. As per the available literature[9], keeping the rotating parts in a enclosure with vacuum of 0.1 mb, the loss due to drag will get reduce to the extent of 75 percent (compared to atmosphere). Therefore, implementing the system with vacuum enclosure and two pole machine will reduce the overall loss to a large extent. Various options available for loss reduction are given in the TABLE VI.

75 %

There is a slight difference in values between total energy generated and sum of energy lost and the energy utilized by the load. This might be due to measurement errors and the assumptions made.

(11)

VIII. LOSS REDUCTION TECHNIQUES

(10)

D. Energy balance test. Measurement and computation of energy at various subsystems and levels are carried out. They are shown in the TABLE IV given below. The measured value and the computed values are in expected lines.

498

TABLE VI. Drag Loss

LOSS REDUCTION TECHNIQUES

Eddy current Bearing Loss friction Loss Two pole Magnetic machine bearing Vacuum Low loss core Ceramic / enclosure material like NiHybrid Fe Bearing

Hysteresis Switching Conduction Loss Loss loss Two pole ZVT machine Low VCEsat Switches Low loss ZCT machine

IX.

CONCLUSIONS

From the experiments and studies, the conclusions are made as follows: • Low cost and easy solutions for improving the efficiency of the system are vacuum enclosure for rotating parts and usage of two-pole machines. With these techniques the efficiency can be increased to a value as high as 80 percent. • Boost converter gain is an inverse function of load resistance, at higher loads (low RL) the energy harvested will reduce. This is due to the fact that, lowest speed up to which one can harvest energy will go up. • Efficiency of boost converter will suffer due to the higher input current while maximizing the harvested energy. This can be improved by ZVT/ZCT technique. • Maximum energy harvested depends on maximum running speed of flywheel and the maximum voltage gain of boost converter (i.e. lowest speed up to which the boost converter will be able to maintain output voltage constant). Therefore gain of the boost converter plays an important role in maximizing the harvested energy. • Power backup time is a function of top speed and the connected load to satisfy energy balance and power balance equations. ACKNOWLEDGEMENT The first author would like to thank Mr. T K Bera, Dr. Kallol Roy and Mr. H A Balasubramanya for their encouragement and support. Special thanks to Mr A Nandakumar for his motivation and guidance during conceptual design and fabrication of the system. We thank S Sathesh kumar for his ever enthusiastic approach and continuous help in fabrication, testing of hardware and documentation of the work. The author sincerely acknowledges Ramya P for her suggestions in conceptualizing the ideas and fruitful discussions. REFERENCES [1]

[2]

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[5]

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