Optimization of Flywheel based energy recovery system( - PowerPoint Presentation

Slide1

Optimization of Flywheel based energy recovery system(Flybrid)

By Naga Abhishek

Bollapragada

Vishal

Chandrasekhar

Vincent

Victor

Sanjai

Mohammad

HejaziSlide2

Flybrid system What is it??

Flybrid KERS

is one such system where instead of a battery, a mechanical flywheel is used. The system aims at storing the energy lost due to braking and use it to aid in accelerating the vehicle later. This is done by a flywheel which is a metal/composite disk rotating at very high speed and store rotational energy

Flybrid Technology Systems uses Flywheel based Kinetic Energy Recovery System (KERS). In conventional KERS system battery and other associated systems are used, but this uses a new low weight flywheel system which has delivers greater power and fuel efficiency than electronic systems. Slide3

Motivation for the project

Stringent pollution policies

This

technology might be the answer to stringent pollution control norms of the future.Fuel saving urgeThe flywheel hybrid system for road cars is expected to offer 20% CO2 fuel savings under New European Driving Cycle test conditions, and up to 30% in real-world conditions.

Reliable and efficient replacement for BatteryDesigned to last 250,000 km,

this technology

makes it possible to store more energy during short braking periods dramatically increasing system effectiveness. The systems are also very efficient with up to 70% of braking energy being returned to the wheels to drive the vehicle back up to speed.

Flybrid KERS technology is around 1/3 the cost of an equivalent power electric hybrid system

.Slide4

Overall objective and the subsystem Objective: Maximize the kinetic energy stored in flywheel.

The

overall objective is achieved by fragmenting it into

subsystems which have their own objective and constraints Maximizing Kinetic energy of flywheel by improving inertiaMinimizing Stator-Rotor losses in the flybrid

Minimizing coupling losses in the flybrid Minimizing mechanical losses in flybridSlide5

Structural subsystem

Overview on the subsystem

The objective of this subsystem is to maximize

the Kinetic energy storage in the Flywheel

by

maximizing the Moment of inertia of the

flywheel.

The moment of inertia of the flywheel is the sum of the moment of inertia of the rim and the moment of inertia of the web.

The flywheel is a taken as a uniform thickness rotating disk

The

stress at a point in the disk is

in three

stress states: the radial stress , tangential stress , and axial stress . Because the surface of the disk is a free surface in the z direction, =0.

 Slide6

Design Variables

: Inner radius of the flywheel

: Outer radius of the flywheel

r :

radius

of the shaft

: Web thickness

H : Thickness of the

flywheel

Design Parameters

r

is

based on the joined rotor

shaft

h depends

on the stress

and stiffness of the materials is selected within the maximum outer radiusThe energy density and specific energy equations are used to generate a plot to determine the radius ratio.All the geometric parameters obtained must be simultaneously analyzed and optimizedThe maximum stress is always at the inner radius of the flywheel rotor. The maximum outer diameter decreases as inner diameter increases.

 Slide7

Objective Function

Polar Moment of inertia

=

+

=

(

+

)

+

(

+

)

=

(

ρπ

(

- ) + ρπ (

-

))/2

- Moment of inertia of the flywheel - Moment of inertia of the web - Moment of inertia of the rimKinetic Energy stored in the flywheel= -]

 Slide8

Constraints

Stress Constraint based on Tresca Failure Criterion

–

=

=

ρ

(

+

) <

[

]

Volume Constraint

Linear Geometric Constraints

=

constant

<

0

0.25H

-

-r -0.33H +  Slide9

Optimization Results

Optimization was done using

fmincon

toolbox

AlgorithmNo. of iterations

(m)

(m)

R(m)

(m)

H(m)

Moment

of Inertia,

(kg.)SQP110.086

0.176

0.02

0.044

0.133

0.546

Active Set80.0860.1760.020.0440.1330.546Interior Point230.0860.1760.020.0440.1330.546Algorithm

No. of

iterations

R(m)H(m)SQP110.0860.1760.020.0440.1330.546Active Set80.0860.1760.020.0440.1330.546Interior Point230.0860.1760.020.0440.1330.546Slide10

Vibrational Subsystem

Objective : Reducing energy loss due to vibration in coupling

Energy of the CVT at (t = 0)

(Constant term)

Energy stored in flywheel

at desired time

Energy stored in CVT

at desired time

( Not important for us)

Usual time duration that CVT and flywheel are engaged = 6.67 sec

Find optimal values for spring constant(K) and damping coefficient (C) of the couplingSlide11

To find the optimal values for K and C, following coupled system of 2

nd

order ODEs should be solved

Where

is the moment of inertia of the CVT and is already defined

To solve this system, we will find the optimal dimensions of the flywheel

 Slide12

Variables

:

C = Equivalent damping coefficient of the coupling

K = Equivalent spring coefficient of the coupling

= Thickness of the web

H =

Length of the flywheel

= Inner radius of the flywheel

= Outer radius of the flywheel

r = Radius of the shaft

Constraints

:

:

C>0

:

K>0

Initial Guesses:

C

= 0.45 (Kg.

)/sec

K =

200(

N.m

)/rad

=

0.04 m

H =

0.04 m

=

0.08 m

=

0.1 m

r =

0.03 m

Optimal results:

C

= 0.8345 (Kg.

)/sec

K = 170.002 (

N.m

)/rad

= 0.0459 m

H = 0.1438 m

= 0.0976 m

= 0.1991 m

r = 0.0300 m

 Slide13

Stator-Rotor losses

Details of the subsystem

Requirement of a flywheel is to store as much energy as possible.

This requires high operational speeds.

Results in friction in the bearings.

These frictional losses can be higher for such high speeds.

So magnetic bearings are used.

Stator - rotor subsystem is associated with the magnetic bearings.

Objective

To minimize the losses in the stator and the rotor, while maintaining the optimal design of the system.

If we consider only the stator – rotor system, the minimization of losses will lead us to the corresponding dimensions of the stator and the rotor.

But when this is to be integrated into the flywheel system, additional geometric constraints need to be satisfied.Slide14

Objective

Function

Two objective functions.

Multi

objective problem, so we consider the first one as the objective and use the second one as the constraint.

Split Ratio x(1)

Aspect Ratio x(2)

Rotor Length x(3)

Shaft diameter x(4)

Stator outer diameter x(5)

Flywheel inner radius x(6)

Web Thickness

x(7)

Flywheel

outer radius

x(8)

Length

of the flywheel x(9)

Design Variables

Split Ratio:

Aspect Ratio:

 Slide15

Nonlinear Constraints

Volume Constraint

:

Geometric Constraints:

 Slide16

Linear Constraints and Lower and Upper bounds on variables

A = [0,0,0,0.5,-1,0,0,0,0; b = [0; -0.0025; -0.00025]

0,0,0,0,0.5,-1,0,0,0;

0,0,1,0,0,0,0,0,-0.5];

Aeq

= [0,0,0,0.5,0,-1,1,0,0];

beq

= 0

Lower bounds: [0,1,0,0.05,0,0,0,0,0];

Upper bounds: [1,inf,inf,inf,inf,inf,inf,0.25,inf];Slide17

Optimization Results

MATLAB optimization toolbox is used.

Optimization is carried out using the

fmincon and SQP algorithms

.Minimum value of the function is 753.64W

Optimum

Values are given below:

X(1)

0.355

X(2)

2.222

X(3)

0.116X(4)0.050X(5)0.154X(6)0.080X(7)0.055X(8)0.166X(9)0.232Slide18

Mechanical

losses in flybrid

Mechanical losses accounts to the highest loss in flywheel system, which primarily focuses on the drag resistance of rotating flywheel in the container.

Here, both drag force and frictional losses on the surface of the web have been dealt with.Slide19

Objective function

Power loss due to drag force

P

drag =

Where

Power loss due to air friction on the web

P

air =

Where

and

Total

Power loss =P

drag

+ P

air

 Slide20

Optimization parameters

Variables

= Inner radius of the flywheel

= Outer radius of the flywheel

H = Length of the flywheel

r = Radius of the

shaft

d= Distance from web to the top surface

Constraints

Lower and Upper bounds on variables

Order: [

,

,

, r, d]

Lower bounds

: [

0,0,0,0.1,0];

Upper bounds:

[inf,0.3,inf,inf,inf];

Why Lower and Upper bounds on r, H variables

For H

, if

upper bound is not given it makes the H value too large and

very small in order to satisfy my volume constraints, but in real

time this is not

feasible

For r, if lower bound is not given it makes Ri very small (or zero) in order to make the cylinder attached to shaft so that there is no air friction loss.

Is variable ‘d’ important what is its influence

There are innumerable reasons why distance d is important, though the loss due to air friction is negligible compared to drag loss.

The main reason is construction manner, symmetrySlide21

Optimization results

Matlab

optimization toolbox was used as main optimization toolbox

Different algorithms like Active set, SQP were used and the global minimum was found as below:Optimal results:

=

0.138 m

H

=

0.3

m

=

0.068

md= 0.112 mr = 0.058 m Slide22

Overall optimization results

For the complete optimization we took, the maximizing kinetic energy as overall objective and power losses due to other subsystem as constraint

The variables which are not common between different subsystems were filled in with the optimized value from subsystem like ‘d’ and rotor stator length.

The overall objective now becomesMaximizing overall kinetic Energy stored in the flywheel=

-

]

 Slide23

Objective function for integrated system:

Constraints in

integrated system

:

Stator rotor loss

-((

33330*((2*x(3)+0.0074)/(2*x(1)-0.006))^4)-(3333*((2*x(3)+0.0074)/(2*x(1)-0.006))^3)+(916.7*((2*x(3)+0.0074)/(2*x(1)-0.006))^2)-(5342*((2*x(3)+0.0074)/(2*x(1)-0.006)))+2150)+750;

mechanical loss: (((

4*10^11)/real((log(100.80*10^7*(x(2)^2)))^2.58))*((0.4*(x(2)^5))+(x(4)*(x(2)^4)))+((62.69*10^6)*(x(1)^5-x(3)^5))/(((0.112/x(1))^(1/6))*((x(1)^(1/2)))))+

35718.9

Coupling loss:

((((

3+mu)/4)*(rho*((omega)^2))*((x(2)^2+((1-mu)/(3+mu))*x(3)^2)))-455000000); -((1/2)*((1/2)*rho*pi)*((0.0459*(x(1)^4 - x(3)^4)) + (x(4)*(x(2)^2-x(1)^2))) * theta1dot^2)]Slide24

Comparison with actual model

Speed

Energy

results

Dimensional results

Obtain

ed results

Actual results

Obtained

Results

Actual Results

20000

I=0.4593Energy=0.161 KJRo=0.1379 mRi=0.0676 mr= 0.0561H= 0.3012Tw= 0.045 mRo= 0.14 mRi= 0.07 mr= 0.06 mH= 0.30 m30000I=0.4593Energy=0.816 MJI=0.45Energy=0.8 MJ Percentage of increase in energy is 2% might not look significant but the increase of energy is 16 KJ which is appreciable. Slide25

Conclusion and outcome

This project has thrown us some insight on influence on variation of important parameters of flywheel

This project gave us knowledge on how to break a complex system in subparts and recombine them to optimize with a particular objective.

Being a very new technology it has very limited recourses about the details, if more constraints are added a better optimization can be done Slide26

Reference

Jerome

Tzeng

, Ryan Emerson, Paul Moy “Composite flywheels for energy storage” Composites Science and Technology 66 (2006) 2520–2527.“An Assessment of Flywheel High Power Energy Storage Technology for Hybrid Vehicles “ Oak Ridge National Laboratory, Department of Energy Johan Abrahamsson, Janaína

Gonçalves de Oliveira, Juan de Santiago, Johan Lundin and Hans Bernhoff “On the Efficiency of a Two-Power-Level Flywheel-Based

All-Electric Driveline”

Energies 2012, 5,

2794-2817.

Baoquan

Kou,

Haichuan

Cao, Da Zhang, Weili Li and

Xiaochen Zhang, “Structural Optimization of High Speed Permanent Magnet Synchronous Motor for Flywheel Energy Storage”, Electromagnetic Launch Technology (EML), 2012. E. Maleki Pour, S. Golabi “Design of Continuously Variable Transmission (CVT) with Metal Pushing Belt and Variable Pulleys” International Journal of Automotive Engineering Vol. 4, Number 2, June 2014