HEV Optimization. Ganesh Balasubramanian Grad. Berrin Daran Grad. Sambasivan Subramanian Grad. Cetin Yilmaz Grad.

Transcription

1

2 HEV Optimization By Ganesh Balasubramanian Grad. Berrin Daran Grad. Sambasivan Subramanian Grad. Cetin Yilmaz Grad. ME Winter 2001 Final Report Abstract The design project is the optimization study of a hybrid electric vehicle system. The first part of this study is the component-based optimization approach, which focuses on each subsystem individually to obtain its optimum specifications. The second part is the system level optimization, where the optimal control strategy is determined besides the design variables. The HEVs are composed of subsystems that have quite different characteristics. There are only a few linking variables that are considered in system integration. So the optimization approach in each subsystem need not be similar. Therefore, various optimization analysis techniques are utilized in this study. Moreover, this system involves time dependence, which complicates the analysis. Therefore, time dependence is handled besides the design optimization. It is anticipated that at moderate power requirements, the electric motor will be the main propulsion system. On the other hand, at higher power requirements, the engine becomes dominant. The power rating allocation between the electric motor and the engine, and the optimum operating points for the engine, the motor and the continuously variable transmission are determined through a system level optimization study. Besides, optimum planetary transmission system ratio, and the optimum capacity for the battery are obtained.

3 Table of Contents Table of Contents 2 Introduction 4 Nomenclature Internal Combustion Engine 5 Transmission 7 Electric Motor 9 Battery 10 System 11 Internal Combustion Engine Problem Statement 13 Mathematical Model 16 Model Analysis 27 Numerical Results 29 Discussion of Results 39 Transmission System Introduction 40 Mathematical Model 44 Results 55 Continuous Variable Transmission 56 Electric Motor Problem Statement 61 Mathematical Model 63 Model Analysis 67 Numerical Results 72 Battery Introduction 74 Mathematical Model 75 Model Analysis 84 2

4 Numerical Results 86 System Integration 88 All-at-Once Approach Summary 91 System Decomposition 95 Optimization of The Master Problem 97 Numerical Results 99 Parametric Study 105 Discussion of Results and Conclusions 108 Future Work 109 Acknowledgements 109 References 110 Appendix 112 3

5 Introduction Problem Statement A hybrid electric vehicle (HEV) combines an electric drive train with an auxiliary power unit that is an internal combustion engine in this study. HEVs combine the advantages of their combined power plants while reducing their disadvantages. HEVs are seen as the solution for reducing the pollution emissions and improving fuel economy. Optimizing the engine operation for reduced fuel consumption and emissions and recovering kinetic energy by regenerative braking, HEVs can perform better than a conventional vehicle. In order to get the minimum fuel consumption in a HEV, the optimum component sizes have to be determined and an appropriate control strategy has to be implemented. In this study the HEV system with subsystems such as the spark ignition engine, planetary transmission system, electric motor and battery are going to be optimized for minimum fuel consumption and maximum energy stored in the battery. The vehicle is of the parallel type. Parallel hybrids have mechanical connections to the wheels from both the electric motor and the engine allowing the vehicle to accelerate faster than a series HEV. In a parallel HEV, the electric motor assists the engine during start-up and acceleration. In this study the first issue is determining the optimum component specifications. First of all, the drive cycle determines the component sizes. The design optimization and control strategy are uncoupled for the engine, motor and battery whereas they are coupled for the planetary transmission system (PTS). Therefore, the former components are optimized based on their design variables and the PTS is optimized based on both its design variables and control variables. When the subsystem optimization procedure is completed, the optimal control strategy will be selected to minimize the fuel economy while maximizing the energy stored in the batteries. To do this, the drive cycle will be discretized; hence time dependence will be removed from the optimization study. Thus the velocity and acceleration of the vehicle will be constant 4

6 for each discretized interval. Since the engine and the motor will be used simultaneously, there will be a compromise between the operating points for each of them, which is also dependent on the characteristics of the battery and planetary transmission system. Moreover, CVT ratio will be selected to obtain the optimum operating point. Therefore, the optimum operating conditions will be calculated for each discretized interval. In the literature, when HEVs are optimized, more emphasis is given on the control strategy. The subsystems are often not optimized or optimized via a simulation program. But in this study, balanced attention will be paid on both component optimization and control strategy. Comprehensive mathematical models are derived for each subsystem; hence rigorous optimization analysis can be performed. The engine and the motor are connected by a planetary transmission system (PTS). The linking variables between them are the engine s speed and torque, the motor s speed and torque. The PTS is connected to the wheels by a continuously variable transmission system (CVT). The variables linking these components are PTS s output speed and torque, which are the inputs to the CVT; and the CVT s output speed and torque, which are inputs to the wheels. There is a controller between the battery and the motor for adjusting voltage and current drawn by the motor. The linking variables between the motor and the controller are the battery current and the battery voltage whereas the linking variables between the controller and the electric motor are the input voltage and current to the motor. When the HEV is in drive mode, the battery supplies the necessary energy to the motor. While regenerative braking, the motor will act like a generator charging battery. As a summary, the subsystems are the internal combustion engine (by Berrin Daran), the planetary transmission system (by Ganesh Balasubramanian), the electric motor (by Sambasivan Subramanian) and the battery (by Cetin Yilmaz). 5

7 Nomenclature Internal Combustion Engine Brake Specific Fuel Consumption Fuel Conversion Efficiency Engine Mechanical Efficiency Lower Heating Value of Fuel Compression Ratio Equivalence Ratio Surface to Volume Ratio Engine Speed Engine Speed Engine Bore Engine Stroke Friction Mean Effective Pressure Indicated Mean Effective Pressure Volumetric Efficiency Density of Air Air-Fuel Ratio Base Volumetric Efficiency Mach Index Port Discharge Coefficient Pumping Mean Effective Pressure Specific Constants Ratio of Air Corrected Indicated Mean Effective Pressure Number of Intake Valves per Cylinder Inlet Valve diameter Atmospheric Pressure Inlet Manifold Pressure Bsfc [g/kw.h] h f h m Q LHV [MJ/kg] C r f S v [1/mm] N [rpm] N [rad/s] b [mm] s [mm] fmep [kpa] imep [kpa] h v r [kg/m3] A/F h vb Z n C s pmep [kpa] g imep c [psi] G d i [mm] Pa [psia] p 1 [psia] 6

8 Inlet Manifold Vacuum Mean Piston Speed Mechanical Friction Mean Effective Pressure Number of Rings per Cylinder Bearings Friction Coefficient Required Engine Power Displacement Volume Number of Cylinders Exhaust Manifold Pressure Exhaust Valve Diameter Maximum Engine Torque Maximum Engine Power Engine Speed at Maximum Power Volumetric Efficiency at Max Power Fuel Conversion Efficiency at Maximum Power p i [psia] S p [m/min] tmep [kpa] R N k P e [kw] V d [liters] n c P e [psia] d e [mm] T max,e [N.m] P max,e [kw] N Pmax [rpm] n v, Pmax n f, Pmax 7

9 Nomenclature Transmission Torque in the Engine T N m Torque in the electric motor T em N m Torque in arm T a N m (Torque in the arm is the output torque for the transmission and Input torque to CVT) Torque in the Sun T s N m Torque in the Ring T r N m Torque in the Planet T p N m Speed in arm N a rpm Speed in the Engine N rpm Speed in the electric motor N em rpm Speed in the Sun N s rpm Speed in the Ring N r rpm Speed in the Planet N p rpm Transmission ratio R - (Radius of Ring gear/ Radius of sun gear) Maximum Transmission ratio Rmax - Minimum Transmission ratio Rmin - Efficiency of CVT Eff - 8

10 : Nomenclature Electric Motor Symbol Name Units P Power Requirement W T Torque N-m ω Speed of Armature Shaft Rad / s V Voltage V I Current A R Resistance Ω DI/DT Constant - L Inductance Henry η Efficiency - ρ Resistivity Ω-m L A Length of Armature Slot M A Area of Armature Slot M 2 N No: of Armature Turns - P o No: of Poles - Φ Flux Wb C Rigidity Modulus M 4 ϑ Permissible Twist Rad J Mass Moment of Inertia Kg-m 2 Y Length of Armature M D Diameter of Armature M R A Resistance of Armature at 25 o C Ω θ A Allowable Temperature Rise o C 9

11 Nomenclature Battery Mt M b P h V oco Total mass of the vehicle (kg) Battery mass (kg) Power necessity of HEV (W) Open circuit voltage of one battery module (V) V oco Average open circuit voltage of one battery module (V) V oc SOC SOC i IR do IR co C IR d IR c υ I M v C d Open circuit battery voltage (V) State of Charge Initial State of Charge Internal resistance for reference battery capacity in discharging mode (mω) Internal resistance for reference battery capacity in charging mode (mω) Battery capacity (Ah) Internal resistance in the discharging mode (mω) Internal resistance in the charging mode (mω) Velocity of the HEV (m/s) Current (A) Vehicle mass without the battery (kg) Coefficient of drag of the vehicle A Frontal area of the vehicle (m 2 ) µ Coefficient of rolling resistance µ tr Transmission efficiency µ mot Electric motor efficiency µ cvt Continuously variable transmission efficiency µ cont Controller efficiency θ M bo C o N o Road inclination Mass of the reference battery (kg) Reference battery capacity (Ah) Number of Battery modules 10

12 ρ Density of air (kg/m 3 ) g Acceleration due to gravity (m/s 2 ) Nomenclature System Variables: P h = Power necessity of HEV (W) T w = Torque on Wheels (Nm) ω w = Speed of rotation of wheels (rad/sec) T cvt = Torque output from the CVT (Nm) ω cvt = Speed output from the CVT(rad/sec) R cvt = Transmission ratio at the CVT Eff cvt = Efficiency of the CVT Τ tr =Torque output from the planetary transmission (Nm) ω tr = Speed output from the planetary transmission (rad/sec) R tr = Transmission ratio at the planetary transmission T ic = Torque demand from the IC engine (Nm) ω ic = Speed of the IC engine (rad/sec) P maxic = Maximum power rating of the IC engine (W) m fuel = Mass rate of consumption of fuel (g/sec) T em = Torque demand from the Electric motor (Nm) ω em = Speed of the electric motor (rad /sec) P maxem = Maximum power rating of the electric motor (W) Eff em = Efficiency of the electric motor P bat = Power output of the battery (W) C = Battery capacity (Ah) M b = Battery mass (Kg) IR d0 = Internal resistance for reference battery capacity in discharging mode (Ω) IR c0 = Internal resistance for reference battery capacity in charging mode (Ω) 11

13 IR d = Internal resistance for battery in discharging mode (Ω) IR c = Internal resistance for battery in charging mode (Ω) V oc = Open circuit battery voltage (V) V oco = Open circuit voltage of one battery module (V) SOC = State of Charge I = Current (A) Parameters: υ = Velocity of the HEV (m/s) a = acceleration of the HEV (m/s 2 ) M v = Vehicle mass without the battery (kg) θ = Road inclination (rad) C d = Coefficient of drag of the vehicle A = Frontal area of the vehicle (m 2 ) r tire = Tire radius(m) R r = Radius of the ring gear (m) P tot = Total power of propeller systems (W) Eff cont = Efficiency of controller M bo = Mass of the reference battery (kg) N 0 = Number of battery modules C 0 = Reference battery capacity (Ah) V oco = Average open circuit voltage of one battery module (V) T = Discrete Time Interval (s) SOC i =Initial state of charge Constants: ρ = Density of air (kg/m 3 ) g = Acceleration due to gravity (m/s 2 ) Q= Lower Heating Value of Gasoline (MJ/kg) 12

14 Internal Combustion Engine Problem Statement The internal combustion engine is one of the components in the hybrid electric vehicle. The engine as a subsystem is going to be optimized to get minimum fuel consumption while providing the maximum required power. The engine that is used in this study is a spark ignition engine, which utilizes gasoline as the fuel. Performance of the engine is determined when it is operating at full load conditions, which is at wide-open throttle conditions. Engine fuel consumption however, is connected with part load conditions. As the load increases at a constant engine speed, the fuel consumption decreases. However, it is complex to model an engine with equations by taking into account all of these effects, especially when the size of the engine is undetermined. Since the size of the engine is unknown and going to be determined as a result of this subcomponent optimization study, some assumptions are made. In this part of the study, the engine is assumed to work at wide-open throttle conditions. The engine that has the minimum fuel consumption at WOT conditions is assumed to have the minimum fuel consumption at part load conditions as well. Once the size of the engine is determined, the fuel map is going to be obtained and for the rest of the study the part load conditions will be taken into account. The engine is going to be operated at the optimum operation point for the required power because the operating conditions of the engine determine the performance, and fuel efficiency. The required maximum engine power, which depends on the acceleration resistance, aerodynamic drag, rolling resistance and gradient resistance, is determined from the cycle. The power that the engine has to supply is affected by the individual efficiency factors within the drive train. For the subcomponent design problem the maximum power that the engine has to supply is taken as 50 kw at 5500 rpm. Since the electric motor is also going to be used as a source of power the actual power needed from the engine can decrease depending on the control strategy. 13

15 When the previous studies on engine modeling for optimization are examined, one realizes that most of the studies use look up tables and maps for the engine since the engine size in those studies are already known. [Georgiopoulos et al. 1999] Wu (1998) models the engine fuel consumption but again a specific engine is used, therefore the mathematical model is formulated in such a way that the specific engine data is used in the equations. In Wagner s (1993) and Anastassopoulos s (1986) study the engine is optimized for maximum brake power. Since the engine has the maximum brake power at full load conditions, the design problem is formulated for WOT conditions. Hubbard (1996) models the drivetrain of a hybrid electric vehicle from a system level approach. The dynamic models are implemented into a simulation and a control algorithm is further developed. Rizzoni et al. (1999) uses a simulation program to model conventional, electric and hybrid drivetrains. Boyd et al. (1998) applies convex optimization to hybrid vehicle optimization. The problem of optimizing fuel efficiency is formulated as a nonlinear convex optimization problem. The problem is then approximated as a large linear program and as a result globally minimum fuel consumption over the given drive cycle is determined. Assanis, Papalambros et al. (1999) uses a simulation program for the optimization of a hybrid electric vehicle. The simulation program ADVISOR (Cuddy and Wipke, 1997; Senger, 1997) is used in that study as the hybrid electric system simulation model. For the engine model, a turbocharged diesel engine simulation code (TDES) that is developed by Assanis and Heywood (1986) is used and integrated with the ADVISOR simulation model. The coupled ADVISOR+TDES simulation is then integrated with the optimization algorithms for the optimization approach to hybrid electric propulsion systems. Some of the other optimization studies also use simulation programs for modeling. It is really complex to model an engine with equations, especially when the engine size is undetermined. One has to do lots of assumptions and the effects of some of the phenomena have to be neglected. Therefore; the results obtained from this kind of a study should not be expected to give the most accurate results. Simulations are developed in years and their ability to predict the real engine process is proven by comparing the simulation results with the actual engine data obtained in the laboratories through experiments. Most of the equations used in this study are empirical and there are lots of assumptions, ie. the effects of spark timing, valve overlap and timing, etc. on performance and fuel consumption are not considered. As a result, simulations 14

16 give better results for engine modeling. In this study the engine is simplified and only the events that are thought to have the major effects on the performance and economy are considered. One has to evaluate the results obtained by taking this fact into consideration. The model basically takes into account the heat losses, time losses, the engine friction, the volumetric efficiency and the thermal efficiency by the use of empirical equations. The objective of the study is to get the minimum fuel consumption; therefore both the mechanical efficiency and the fuel conversion efficiency have to be increased. As the engine speed increases, the friction mean effective pressure increases and as a result the mechanical efficiency decreases. However; at the same time the thermal efficiency increases as the engine speed is increased; therefore there is a trade-off between these two quantities. Compression ratio is one of the other variables in this model. It has a similar effect on the efficiencies like speed. As the compression ratio is increased the fuel conversion efficiency increases as well. However, again the mechanical efficiency decreases. There is again a trade-off. Decreasing the bore decreases both the thermal efficiency and the mechanical efficiency; therefore it is expected to get the maximum bore length for the engine, while satisfying all the constraints. The more important ratio is the b/s (bore to stroke) ratio, because increasing this ratio increases the heat losses and decreases the engine friction. One of the other design variables is the inlet valve diameter that decreases the mechanical efficiency when increased. Briefly, the optimum values for the bore, stroke, compression ratio, speed, the inlet and exhaust valve diameters are going to be determined in this part of the study. Based on these results the engine size is going to be determined and the corresponding fuel map will be obtained through a simulation program. This map is going to be used for determining the engine s optimum operation point for the required power. 15

17 Internal Combustion Engine Mathematical Model Objective The objective for the engine design is to get minimum fuel consumption for an engine that will give the maximum required power as one of the constraints. The brake specific fuel consumption of an engine can be determined as follows: bsfc = η 3600 η f Q LHV m The objective function of this part of the study is Minimize = 3600 η f Q η LHV m Where η f is the fuel conversion efficiency, η m is the mechanical efficiency and Q LHV is the lower heating value of the fuel used which is gasoline in this study. When the heat losses and time losses taken into account the thermal efficiency of the engine can be expressed as [ref.5]; η f = 0.9(1 C (1 γ ) r )( ϕ ) ( S v 1500 )( N ) 0.5 for φ<= 1.05 [1] η f = 0.9(1 C (1 γ ) r )( ϕ ) ( S v 1500 )( ) N 0.5 for φ> 1.05 [2] In this optimal design study the engine is assumed to operate at stoichiometric conditions, therefore the equivalence ratio (φ) is 1. The chamber geometry studied in this design is a flat head design. The surface to volume ratio can be expressed as [ref.4]; 16

18 0.83(12s + ( Cr 1)(6b + 4s)) S v = [3] ( bs(3 + C 1)) r The mechanical efficiency can be determined as follows; fmep η m = 1 [4] imep where fmep is the friction mean effective pressure and imep is the indicated mean effective pressure. Indicated mean effective pressure can be expressed as, η 10 3 f ηvρqlhv imep = [5] ( A / F ) where η v is the volumetric efficiency, ρ is the density of air and (A/F) is the air-fuel ratio. It is assumed that the density of air is constant and since the engine is run at stoichiometric conditions the air-fuel ratio is set as The volumetric efficiency is a function of gas velocity, inlet pressure, temperature, density in the intake manifold, sonic velocity, inlet viscosity, exhaust pressure, coolant temperature, fuel-air ratio, and engine design ratios such as the compression ratio and all the other ratios that are required to describe the whole gas-flow through the engine. For simplicity in this study the volumetric efficiency is expressed as a function of engine speed, bore, stroke, and intake valve diameter. Based on the previous studies of Taylor, one can find curves for the volumetric efficiency as a function of Mach Index. However these curves are drawn for a fixed compression ratio, therefore rather than interpolating Taylor s curve for the volumetric efficiency, an empirical equation obtained by Wagner (1993) is used. 5 2 ηvb (1 + (7.72(10 ) N ) ) η v = [6] 2 (1 + Z ) n Z (9.428(10 ) Ns( b / d ) ) 8 2 i n = [7] Cs 17

19 where η vb is the base volumetric efficiency. In this empirical formula, port and chamber design and the effect of engine speed are taken into consideration. Wagner (1993) using a curve fitting technique formulated the base volumetric efficiency of a best-in class engine empirically as follows; 3 ( N (10 ) 5.25) η vb = e for N 5250 [8] η vb = N ( N10 ) ( N10 ) for N < 5250 Friction mean effective pressure can be formulated using Bishop s model. Engine friction is composed of pumping friction, accessory friction and rubbing friction. Pumping friction mean effective pressure (pmep) can be formulated as follows; 2 imepc N 1.7 ( b / 25.4) ( s / 25.4) 1.28 pmep = ( (1.3( ) ( ( ) )))6.895 [9] G( d / 25.4) i imep = 12.8( p p c a ) [10] p1 = p a p i [11] where p i is the inlet port pressure. The inlet port pressure is determined by curve fitting (Taylor, 1960) and the empirical equation is as follows; p i 8 2 = p (1 1.12(10 )( S / ) ) [12] a p where S p is the mean piston speed and can be expressed as ; 3 S p = 2Ns10 [13] 18

20 Mechanical friction mean effective pressure can be described as follows; s( R tmep = (( 2 b N ) Pe bnk + ( + P1 ) s pa p s + ( (0.088C b r C ( Sp) r )) + + pa p1 (1.72C r ( C r N )( ) N ) ( ) N G( di + (30 )( 1000 / 25.4) sb ))6.895 [14] where R N is the number of rings per cylinder, k is the bearings friction coefficient, G is the number of intake valves per cylinder. These parameters are set as 3,0.85 and 2 for this engine design. The expression for the total friction mean effective pressure is then; fmep = tmep + pmep [15] Constraints The optimum engine should be able to provide the required maximum power at WOT conditions. The required power is determined by the aerodynamic drag, rolling resistance, gradient resistance and the acceleration resistance, which are functions of vehicle, mass, velocity, acceleration, and rolling friction. The power that the engine has to supply is affected by the individual efficiency factors within the drive train such as transmission efficiency, differential efficiency, and wheel bearing efficiency. The required maximum engine power can be described as; G1: C1 = P max, e η = f, P max η v, Pmax N Pmax V d 120( A/ F ) Q LHV ρ10 3 where V d is the total displacement volume, N Pmax is the engine speed at maximum power and n f,pmax and n v,pmax are the fuel conversion efficiencies at maximum power and volumetric efficiency at maximum power respectively. 19

21 V d πb = sn c 4 [16] where n c is the number of cylinders that is 3 for this design. Since this design is for a spark ignition engine, there should be an upper limit on the compression ratio in order to prevent knocking. Knock limited compression ratio can be determined from Taylor s graphs. The knock limited compression ratio is determined in terms of bore length by curve fitting to data in Taylor s (1960) study. (Internal Combustion Engine Appendix A-1): The following equation (equation 17) is obtained; however Taylor s graph is for 82-octane fuel. Today, octane ratings for available fuels are higher; therefore it is more appropriate to use Heywood s (1980) curves that are given for 98-octane fuel. The equations for knock limited compression ratio are as follows; Taylor s equation (Appendix A-1): b = C r [17] Heywood s equation for knock-limited compression ratio: G b = C r There are flow losses related to the compressibility of the fuel-air charge in the induction process. In order to keep these losses below a certain point the Mach Index of the port and chamber design has to be less than a certain limit. This limit is given by Taylor (1960) as 0.6. The constraint can be stated as follows [ref.4]: 8 2 (9.428(10 ) Ns( b/ di ) ) G3: Zn = 0.6 = C2 C s Some of the equations used in the modeling are valid only for specific ranges. For model validity, the bore to stroke ratio should be limited as follows [ref.4]; 20

22 b G4: C 3 = = C4 s The current design practice for spark ignition engines constraints the displacement per cylinder as follows; 2 3 πb s 3 G5: C 5 = 400(10 ) 600(10 ) = C6 4 The upper limit for the engine speed should be; G6: N 8000 = C7 There are also packaging constraint related to the geometry of the combustion chamber. There is an engine height limit of 200 mm. This constraint can be stated as follows: G9: 2s 200 = C10 Engine block has to have a maximum length of 400 mm. There must be a distance separating the cylinders as well and this distance should be 20% greater than the bore dimension. The constraint is ; G10: 1.2n c b 400 = C11 For the design of the flat cylinder head combustion chamber due to the geometric constraints the intake and exhaust valve diameters has to meet the following constraints [ref.4]; G11: d i + de 0. 82b d e G13: C12 = = C13 di The model is different from the previous study of Wagner. First of all, the objective function is totally different, and this leads to a different approach in the model development. 21

23 Mechanical efficiency is introduced to the system and related to that more detailed friction model is used that takes into account the accessory friction, the mechanical friction and the pumping friction. Since the objective function is to minimize fuel consumption, one of the constraints is for the required maximum power output. The engine should maximize fuel economy while maintaining the required maximum power constraint. For the constraint associated with the knock phenomenon, Taylor s data is also investigated, however it is determined that Heywood s data is more up-to-data and valid; therefore it is used. The physical constraints are the same with Wagner s study. 22

24 Design Variables and Parameters Parameters Parameter Value Parameter Value Parameter Value Q LHV 44 MJ/kg p a 14.2 psi G 2 g 1.33 R N 3 A/F 14.6 f 1 k 0.85 r(air) kg/m 3 C s 0.44 p e 15 psia C5 100 *10 3 mm 3 nc 3 C3 0.7 C6 600 *10 3 mm3 C1 60 kw C4 1.3 C rpm C2 0.6 C C C mm C N Pmax 5500 rpm P max,e 50 kw Table 1: List of Parameters for the Internal Combustion Engine 23

25 Design Variables The design variables for the engine are the compression ratio, bore, stroke, inlet valve diameter, exhaust valve diameter, and the engine speed at which minimum fuel consumption is obtained. The number of cylinders and the number of intake valves could also be set as variables, however in that case the model would have discrete variables. For simplicity, these variables are set as parameters. The degrees of freedom for this design is 5, since there is an equality constraint for the maximum power requirement. Set of values satisfying the constraints stated above are determined using Excel, with a trial and error method. These values show that there is at least one set of variables satisfying this design. Variable Name Determined Value Bore - b (mm) 88 Stroke - s (mm) 68 Compression Ratio - Cr 4 Engine Speed N (rpm) 4000 Intake Valve Diameter d i (mm) 31.5 Exhaust Valve Diameter d e (mm) 27 Brake Specific Fuel Consumption [g/kwh] Table 2 : Set of values satisfying the engine model 24

26 Model Summary Minimize f = 3600 η f 44η m [in g/kw h ] Where η S f v = 0.9(1 C 1 γ r )( ) ( S 0.83(12s + ( Cr 1)(6b + 4s)) = ( bs(3 + C 1)) r v 1500 )( N ) 0.5 η m = 1 fmep imep imep η v η = η = vb f η v (44) (1 + (7.72(10 (1 + Z 2 n 5 ) ) N) 2 ) Z n 8 (9.428(10 ) Ns( b / di ) = ) ( N* ) ( 3 η = e ) for N 5250 vb η vb = N ( N10 ) ( N10 ) for N < 5250 fmep = tmep + pmep pmep = ( imep 163 c N (1.3( ) ( b / 25.4) ( s / 25.4) ( ( )) 2 G( d / 25.4) i 1.28 ))6.895 imep c = 12.8( p p ) a p p 1 i = = p a p i p (1 1.12(10 a 8 )( S p / ) 2 ) S p = 2Ns s( R tmep = (( 2 b N ) Pe bnk + ( + P1 ) s pa p1 + ( s (0.088C 2 b r C ( Sp) r )) + 25

27 + pa p1 (1.72C r ( C r N )( ) N ) ( ) N G( di + (30 )( 1000 / 25.4) sb ))6.895 subject to 3 η f, P maxηv, Pmax NP maxvdqlhv ρ10 e1= C1 = Pmax, e = 0 120( A / F ) g2= b C r 8 2 (9.428(10 ) Ns( b/ d i ) ) g3= C g4= b 1.3s 0 g5= 0.7s b g6= π b s 2400(10 ) g7= 1600(10 ) π b s 0 g8= N g9= 2s g10= 1.2 b n c g11= ( d + d ) 0.82b 0 i g12= d 0.87d 0 e g13= 0.83d d 0 i e and X={all variables positive} e s i There are 12 inequality constraints, 6 variables and 1 equality constraint. The equalities given other than the P max,e are for the intermediate variables and just used to define the intermediate variables. 26

28 Model Analysis Minimize f = The objective function for this part of the study is; 3600 η f 44η m and by inspection ; f = f n f, n ) ( m If the intermediate variables are plugged into the objective function, the objective function becomes too complicated and it is impossible to perform a monotonicity analysis; therefore the preferred procedure in this study is to determine the monotonicities of each variable, for every intermediate function. The fuel conversion efficiency for an equivalence ratio of 1, in the objective function can be described as in equations 1 and γ 0.5 η f = 0.9(1 Cr )( ) ( Sv )( ) [1] N 0.83(12s + ( Cr 1)(6b + 4s)) S v = [3] ( bs(3 + C 1)) r To describe the monotonicity of S v with respect to variables b, s and C r, equation 3 is rearranged as follows; S 0.83(12) 0.83( Cr 1) 6 4 = + ( ) [18] b( C + 2) ( C + 2) s b v + r r Cr 1 d ( ) Cr = dc ( Cr + 2) r 2 0 and since all the variables are positive so can conclude that Sv = Sv( Cr, b, s ) [19] 27

29 ds dc v r 0.83(6b + 4s) 0.83(12s + ( Cr 1)(6b + 4s)) = Rearranging the terms the derivative can be 2 bs(2 + C ) bs(2 + Cr) r written as: [20] ds dc v r b = ( ) Since all variables are positive this derivative is always positive. 2 b(2 + Cr) s Therefore the final form is: S v + r = S ( C, b, s ) [21] v Then the fuel conversion efficiency can be stated as follows: γ 0.5 η f = 0.9(1 Cr )( ) ( Sv )( ) [22] N Since γ is 1.33, the power of C r is always negative, and the monotonicity statement is as follows; + + n f = n f ( Cr, Sv, N ) [23] The mechanical efficiency term in the objective function can be described as in equation 4. fmep + η m = 1 ; by inspection nm = nm ( fmep, imep ) [24] imep The indicated mean effective pressure is described as in equation 5. 3 η fηv1.225(44) imep = ; by inspection imep = imep( n f, nv ) [25] 14.6 The volumetric efficiency can be described as in equations 6,7 and 8. 28

30 5 2 ηvb (1 + (7.72(10 ) N ) ) η v = Since n 2 vb (baseline volumetric efficiency) is positive, by (1 + Z ) n + v = v vb n + inspection n n ( n, N, Z ) [26] 3 ( N* ) η vb = e for N 5250 ; by inspection for this case nvb = nvb ( N ) for N 5250 [27] η vb = N ( N10 ) ( N10 ) for N < 5250 To determine the trend of N, the equation for N < 5250 case is plotted. It is observed from the plot that n vb is increasing with respect to N for this case. n vb N Figure 1 : Base Volumetric Efficiency versus Engine Speed Therefore; + nvb = nvb ( N ) for N < 5250 [28] equation 7. The last variable in the volumetric efficiency equation is Z n that is described as in Z n 8 2 (9.428(10 ) Ns( b / di ) ) + + = ; by inspection Zn = Zn ( N, s, b, di ) [29]

31 To determine the monotonicity of Z n with respect to N and d i ; d ( Z d( d n i ) ) b = 3 d i 2 Ns Since all the variables are positive. [30] d ( Z n ) d ( N ) b = 2 d i 2 s Since all the variables are positive, therefore it is concluded that : n = n i Z Z ( N, s, b, d ) [31] The monotonicity analysis of the volumetric efficiency can be summarized as follows: + v = v vb n + n n ( n, N, Z ) [32] + nvb = nvb ( N ) for N < 5250 [33] nvb = nvb ( N ) for N 5250 [34] n = n i Z Z ( N, s, b, d ) [35] It is observed that for both of the cases the trend of N is undetermined, one has to determine which N term dominates over the other in order to determine the final monotonicity of the volumetric efficiency n v with respect to N. The equations are examined; however the dominant terms could not be determined. The final forms can be stated as follows: + v = v i n n ( N, b, s, d ) [36] The friction mean effective pressure can be described as in equations 9, 15 and fmep = tmep + pmep ; by inspection fmep = fmep( tmep, pmep ) [37] Substituting equations 10,11,12 into equation 9, pmep can be expressed as: pmep b s = N 10 ( )( ( N 10 s )) [38] 2 di For the design space the term in the brackets is always positive, therefore by inspection; = i + pmep pmep( N, b, s, d ) [39] 30

32 Substituting equations 11,12 and 13 into equation 14, gives the tmep expression as follows: tmep 1.75 N di (30 ) bn s 6.895( N b s s b = ( s C ( C ) N )( N 10 ) 0.4 r r Ns (0.088Cr Cr 14.7( N s 10 ) + 2 b ) s( N 2 s ) ) + ( r tmep = tmep N, s, b, d i, C ) for N < 5250 [40] Even tough the derivatives of tmep with respect to b, s, C r, and N are each examined separately, the monotonicity of tmep with respect to these variables couldn t be determined. The friction mean effective pressure can be described as in equation 9. As explained before; + + fmep = tmep + pmep ; by inspection fmep = fmep( tmep, pmep ) [41] Tmep is increasing with respect to d i, however pmep is decreasing with respect to d i, therefore again the term that dominates should be determined. However, dominant terms could not be determined. The second approach to determine the monotocity of fmep with respect to the variables is to directly substitute tmep and pmep equations, and to state fmep in terms of N, b, s, C r, d i. The final form of the fmep equation is not presented here since this approach also did not provide any results. The equation became even more complex, and the monotonicity of fmep with respect to the variables are still undetermined. If the constraints are examined: The maximum power requirement constraint, when the values for n vmax power, n fmaxpower, N max power are substituted into constraint e1; 31

33 (12s + ( Cr 1)(6b + 4s)) (733.23b s(0.8595(1 Cr ) )) b(2 + Cr ) s P max, e = 0 [42] b 10 s 1+ 4 d i The maximum power constraint is not substituted into the objective function in order to eliminate one of the variables since the equations are going to become more complicated. The monotonicities for the inequality constraints are: + + g2= C r b g 2 = g 2( Cr, b ) [43] 8 2 (9.428(10 ) Ns( b/ d i ) ) g3= C s = i g 3 g3( N, s, b, d ) [44] + g4= b 1.3s 0 g 4 = g4( b, s ) [45] + g5= 0.7s b 0 g 5 = g5( s, b ) [46] 2 g6= 2400( π b s ) 0 g 6 = g6( b, s ) [47] 3 2 g7= 1600(10 ) π b s 0 g 7 = g7( b, s ) [48] + g8= N g 8 = g8( N ) [49] + g9= 2s g 9 = g9( s ) [50] + g10= 1.2n c b g 10 = g10( b ) [51] + + i e g11 = g11( d, de, b ) g11= ( d + d ) 0.82b 0 i [52] g12= d 0.87d 0 g 12 g12( d i, d ) [53] e i + = e g13= 0.83d d 0 g 13 g13( d i, d ) [54] i e + = e 32

34 Monotonicity Table N b s C r d i d e F U U U U U g2 + + g g4 + - g5 - + g6 + + g7 - - g8 + g9 + g10 + g g g e1 U U U U Table 3 : Monotonicity Table for the Internal Combustion Engine Subsytem Since the monotonicity of the objective function with respect to variables N, C r, b, s, and d i cannot be determined; the active constraints cannot be determined. However, if these variables are monotonic, then one can say which of the constraints will be active. follows; For example, if the objective function is monotonic with respect to each variable such as + = r i + + f f ( N, C, b, s, d ) [55] 33

35 The equality constraint should be directed to be able to perform a monotonicity analysis, however the monotonicity of the variables in the objective function cannot be determined, therefore it is not possible to direct the equality constraint. However, if the equality constraint could be directed then one could at least determine which of the constraints would be active if the variables in the objective function are monotonic. If we now put aside the equality constraint, which in fact cannot be done, we can at least state which of the constraints can be active alone or together. One should remember that if the equality constraint could be directed, it should also be stated as one of the active constraints in the following analysis. From the first monotonicity principal with respect to N, at least one of the constraints g3 or g8 is active. With respect to C r, g2 is active. With respect to b, at least one of g5, g7 or g11 are active. With respect to s, at least one of g4 or g7 is active. With respect to d i, at least one of g3 or g12 is active and with respect to d e, g13 is active. There is one more point that we can emphasize at this point. From the second monotonicity principle, d e variable that does not occur in the objective function is either irrelevant or relevant and bounded by 2 active constraints with opposite signs. For example, if g11 is active with respect to d i, then g11 and g13 are active with respect to d e. If g2 is active with respect to d i, then g12 and g13 are active with respect to d e. As a result, the activities cannot be determined at this point; however the activities of the constraints will be examined once more after the numerical results are obtained. In fact, some of the variables may not even be monotonic. Also as explained above the equality constraint could not be either eliminated or directed. The discussion presented here is true for a variable if that variable is monotonic. 34

36 Numerical Results The model presented in the mathematical model section is used with the constraints given. The model has 1 equality constraint and 12 inequality constraints. There are 6 variables that are C r, b, s, N, d i and d e. Since there is only one equality constraint, the degrees of freedom of this system is 5. The mathematical model is coded and numerical results are obtained using Sequential Quadratic Programming algorithm (SQP) in Matlab. The model is also solved by Generalized Reduced Gradient (GRG) algorithm and the same results are obtained. The outputs are presented in the Internal Combustion Engine Appendix A-2, and it is summarized below for the 6 design variables: Variable Optimum Value Units b* 71.4 mm s* 100 mm N* 2868 rpm C r * 10 d i * 24.4 mm d e * 21.2 mm bsfc * [g/kwh] Table 4 : Numerical Results 35

37 Lagrange Multipliers g Active g3 0 g4 0 g5 0 g6 0 g Active g8 0 g Active g10 0 g11 0 g12 0 g13 0 e Active Table 5 : Table of Lagrange Multipliers The numerical results show that constraintsg2, that is the constraint on compression ratio, g9 that is the constraint on stroke, and g7 that is the constraint on bore and stroke are active. The equality constraint is satisfied. For the validity of the model constraints g4 and g5 should not be active, and in fact it is determined that these constraints are not active. It is observed that the knock phenomenon constraints the compression ratio and provides and upper bound for the C r variable. When the constraint associated with the compression ratio is removed, the optimum changes; therefore one can conclude that this constraint is active. The numerical results also show that the constraint g12 associated with the inlet and exhaust valve diameters; is satisfied strictly, but in fact it is not active. When this constraint is removed from the system the optimum does not change. Therefore, the variable d e is in fact irrelevant, and can be removed from the system with all of its constraints, since it does not also appear in the objective function. When this is done and the problem is solved once more numerically, the optimum does not change. 36

38 Therefore the constraints g11, g12 and g13 can be removed and the only bound on d i comes from the constraint g3. On the other hand, when the constraint associated with the stroke is removed, the optimum of the system changes; therefore this constraint is active. When the constraint associated with the compression ratio is moved, the optimum changes and at the new optimum the new constraint related to the compression ratio is satisfied strictly, and it is active. Therefore, one can conclude that the objective function decreases monotonically with the compression ratio. Another important aspect is that the design is bounded by physical constraints that become active at the minimum such as the stroke constraint. The reason for termination in the algorithm is convergence. The optimality conditions are satisfied with a constraint tolerance of epsilon. The objective function value converges and there is no improvement in its value. The algorithm terminates at iteration where there is no improvement in the value of the objective function from the previous iteration. We can state this termination criterion as follows: f k f k+1 ε [56] where ε is a very small number determined by the user. While this criterion is satisfied for termination, the maximum constraint violations are also less than the user-defined amount. Therefore both the function value converged and the maximum constraint violation was less than ε. The SQP algorithm is run several times to determine the effect of different starting points on the optimum. In the following table, the different starting points and the corresponding optimum values obtained are presented. One can observe that for different starting points the algorithm gives the same results or the optimum, so there is a robust solution. Starting Points (b,s,n,c r,di) Optimum Values (b,s,n,c r,di,bsfc) 40,40,1000,4, ,100,2868,10,24.4, ,1,1,1,1 71.4,100,2868,10,24.4, ,150,5000,15,30, ,100,2868,10,24.4, ,200,500,1,40, ,100,2868,10,24.4,230.2 Table 6 : Optimum Values with Different Starting Points 37

39 In the second part of the study, based on the engine size results corresponding fuel map is obtained from ADVISOR (Cuddy and Wipke, 1997; Senger, 1997). Using the data from ADVISOR; the engine fuel map is scaled for the engine size.the engine size is a variable in the system integration and determined by the system optimization study. The subsystem study performed in this section is just used to get an initial value of the engine size for the required power output, so that an engine map closer to this subsystem optimization results is used. Scaling of the maps can be performed unless the scaling factor causes more than 50 % change in the engine size. Therefore the initial subsystem optimization study is required and the scaling factor is set as a variable in the system optimization study with a constraint of maximum 50% change on the initial size of the engine. The map is used to determine engine s optimum operating point for the required power of a given interval of a discretized cycle. Since the size of the engine is determined assuming WOT conditions, part load conditions have not been taken into account. However; the operating conditions of the engine determine the fuel efficiency. Both the engine speed and torque variations, therefore load variations are taken into account by using the maps, to determine the best operating point. The engine speed, the torque and therefore the power are the linking variables in the system. According to the cycle power demand and individual efficiencies of the drivetrain components, and to the control strategy, the size and the maximum power output of the engine may be altered. The ADVISOR data used in the system integration is presented in Appendix A-3. The data for 41 kw 1liter engine is scaled for the system integration. The maximum torque values are also scaled and the following equation is obtained by curve fitting. The graph of maximum torque versus engine speed is presented in Appendix A-4. 2 T = N N [57] max + where N is in rad/s and T max is in N.m. The T max equation obtained here is for a 41kW-1 liter engine and when the scaling factor is also included the maximum torque constraint that is going to be utilized in the system optimization study becomes: 2 T ( e N + N ) * s 0 [58] max, where s is the scaling variable that is 38

40 P max, e s = [59] 41 where 41 is the power of the engine that is used from ADVISOR. The more detailed explanations of the constraints used will be given in the system integration section of the report. The parametric study is not performed in this section because it is more convenient to do it after the system integration is done, since the subsystem model is not used completely in the system integration. Discussion of Results The results obtained for the subsystem optimization study indicate that the minimum fuel consumption is obtained at 2868 rpm when the engine operates at wide-open throttle conditions. The engine provides 50 kw maximum power with a 1.2-liter displacement. The compression ratio is determined to be 10, and this ratio seems to be reasonable for a small displacement engine to provide the required maximum power. The minimum brake specific fuel consumption value determined is small compared to the brake specific fuel consumption of an engine with the same power output and displacement volume. Since the mathematical model is created with lots of assumptions and simplifications as described at the beginning, this result is anticipated. The physical constraints, the packaging constraints are active and this implies that if the packaging constraints are not an issue the optimum values change, for example if the engine height does not have a limit the stroke can be increased more. However, this means that the engine compartment dimensions and the combustion chamber dimensions have to be altered. Also the compression ratio constraint limits the design, that constraint is active. If the compression ratio can be increased without knock, then the optimum values determined here change, with an increased compression ratio more power can be obtained for a smaller displacement. Increasing the compression ratio without knock decreases the fuel consumption. This can be available with producing fuels with a higher octane rating. To improve the answers determined here as suggested at the beginning of the report, an engine simulation can be integrated with an optimizer, thus eliminating the use of lots of assumptions in the mathematical model development. 39

41 Transmission System Introduction: In the vehicle model that is being proposed we have 3 components in the transmission system. 1. The power output from the engine and the electric motor needs to be combined and a common output needs to be taken. This is achieved with the help of planetary gear train system. In this system the output from the electric motor is given as an input to the ring gear and the output from the IC engine is given as an input to the sun. The common out put is taken from the arm. 2. Another planetary transmission system could be utilized in order to recover the idling losses of the IC engine. It would be optimized for only running at a constant speed (idling speed). But this additional system will not be modeled and used for this study. 3. The third component in the transmission system is a CVT (continuous variable transmission). The CVT receives the combined output from IC engine and the electric motor through the planetary gear train system and the output is given to the wheels. The CVT offer high degree of flexibility by offering infinite transmission ratios between the input and the output. Each component is described in detail in the following pages. 40

42 Analysis of Each Subsystem in the Transmission System and Problem Statement 1. Planetary gear train system combining the output from the electric motor and the IC engine: A schematic diagram of the planetary gear train system is shown below. The components in a planetary gear train system are a sun gear (where the input from the electric motor is given), ring gear (where the input from the IC engine is given), arm (from which the output is taken) and set of planet gears, which revolves about the axis of the sun and a rotate about its own axis. This system is a 1 degree of freedom system with respect to torque, meaning that once the torque on any one of the components is set we can find the torques on the other components by means of some basic set of equations. But this system is a 2-degree of freedom system with respect to speed meaning that to determine the speed of all the components we need at least 2 speeds to be known. Now according to the driving cycle along which vehicle will be 41

43 driven we have set of torque and speed requirement placed at the output. That is there a requirement on the torque and speed is placed at the arm. Since the system is a one-degree of freedom system with respect to the torque, the torque requirement on the sun and the ring of the planetary gear train system will be set. From the torque efficiency graph of an IC engine for torque load placed in the ring gear there will a particular speed at which the IC would have to operate for it to have a maximum efficiency. If the engine operates at this speed then the speed at which the motor has to rotate become set. But this need not be the most efficient case for the motor. Because from the torque efficiency diagram of a motor for the torque load placed at the sun gear there will be a particular speed at which the electric motor is very efficient. This speed need not match up with speed requirement placed at the sun gear end. So this will give raise to a situation in which the engine will be operating an efficient point for the load requirement but the motor will have very poor efficiency. Similarly we can argue and say that if we allow the motor to operate at an efficient point the IC engine will be operating at an inefficient point. This is further explained with help the following graphs. The following graphs show an approximate torque efficiency relation of a motor and an engine for a particular speed. So if we were to vary speed we will get a three dimensional plot between speed, torque and efficiency. For a given speed the engine will most efficient if it were to operate at point A. Similarly the electric motor will be most efficient if it were to work at point B. For a particular torque requirement at the output the torque at the IC engine and the electric motor is set. For the particular torque requirement at the IC engine we have to find the speed at which 42

44 the engine operates at the most efficient condition. This will set up the speed at which the electric motor should operate and this speed need not correspond to the maximum efficiency at which the motor should operate. So we have to trade off between the efficiency at which the engine should operate and the efficiency at which the motor should operate. Our optimization goal is to find out a transmission ratio in the planetary gear train system at which the overall efficiency at the system is maximum. An algorithm has been developed to solve this. 43

45 Mathematical model development Algorithm: From the drive cycle we have the requirement of torque and speed placed at the arm of the planetary gear train system. Let, T represents the torque requirement N represents the speed requirement Ta = [T1, T2, T3,....Tn] Na = [N1, N2, N3,.Nn] The above data is got from the driving cycle. We have to determine the transmission ratio (R tr ) in the planetary gear train at which this system will operate at maximum efficiency. So we start our iteration by changing the value of R min to R max Where R is the transmission ratio. It is the ratio of the radius of the ring gear to the radius of the sun gear R = R r / R s R s is the radius of the sun gear R r is the radius of the ring gear R min is the minimum transmission ratio R max is the maximum transmission ratio For a particular R, for each torque requirement placed in the arm (T a ) the torque required from the IC engine (T) and the torque required from the electric motor (Tem) is set. This is because the system has one degree of freedom with respect to the torque. But the given a speed of the 44

46 arm we cant determine the speed of the sun and the ring. So we start iterating on the speed of the Electric motor from 0 to N em max the corresponding speed of the IC engine (N) can be calculated. Now at present an approximate relation between the torque, speed and efficiency has been assumed and a model has been developed for this. From this model, for a given speed and torque on the IC engine the efficiency of the engine (which has been evaluated in terms of fuel consumption) is calculated. Corresponding to the engine speed the speed on the motor is set and from the torque speed, efficiency model developed for an electric motor we have efficiency of the electric motor. Then the weighted efficiency of the system is developed using the following relation. Efficiency of the System = (1/Total Power)(Power from IC engine * Efficiency of IC engine + Power of Electric Motor * Efficiency of Motor) The maximum efficiency for each Torque and speed combination is calculated and is summed over the entire driving cycle. This is got for a particular transmission ratio. The transmission ratio at which the sum total of the efficiencies is maximum is taken as the optimum ratio. 45

47 Objective Function Definition: To find the optimal transmission ratio in a planetary gear train system having an objective function dependent on a matrix of power. T 1 N 1 T 1 N 2 T 1 N 3 P = T 2 N 1 T 2 N 2 T 2 N 3 T n N m Maximize F = (Eff (T 1 N 1 ) + Eff (T 1 N 2 ) + Eff (T 1 N 3 ) +Eff (T n N m )) Or Minimize F = - (Eff (T 1 N 1 ) + Eff (T 1 N 2 ) + Eff (T 1 N 3 ) +Eff (T n N m )) Subject to the following Constraints R lies between a minimum value and a maximum value R min R tr R max In the negative null form g1 : R min R tr 0 g2 : R tr R max 0 where R tr is the transmission ratio. R min is the minimum transmission ratio R max is the maximum transmission ratio 46

48 The Torque T should lie between a maximum and a minimum value T min T tr T max In the negative null form g3: T min T tr 0 g4: T tr T max 0 T tr is the torque transmitted. T min is the minimum torque transmitted. T max is the maximum torque transmitted. The Speed N should lie between a maximum and a minimum value N max N N max g5: N min N 0 g6: N N max 0 N is the speed N min is the minimum speed N max is the maximum speed 47

49 Speed Analysis in an Epicyclic gear train: As stated in the previous section, an epicyclic gear train system has 2 degree of freedom with respect to speed meaning that we need speed of 2 components to determine the speed of all the other components. The speed analysis is done using the tabular method of speed determination. We shall go ahead building the table for a simple epicyclic gear train. Sun Planet Arm Ring Motion with x x x x arm Motion N s x -(N s x) R s /R p 0 -(N s x) R s /R r relative to arm Total motion N s -(N s x) R s /R p + x x -(N s x) R s /R r + x Step 1. Motion with Arm: In this case the Sun, Planet and the Ring are fixed and arm is rotated but x rotations. The all the components rotate by the same x in the same direction. Hence the first row has a x under each column. Where, x is the speed of the arm Step 2: Motion relative to the Arm: In this case the motion is considered relative to the arm. Or in other words the arm is fixed and the sun is rotated by a fixed number of rotations (N s x) rotations. When the sun rotates by (N s x) rotations the planet rotates according to the gear ratio between the sun the planet. 48

50 Where, x is the speed of rotation of arm N s is the speed of the sun gear N p = - (N s x) R s / R p Where, N p is the speed of the planet gear R s is the radius of the sun gear R p is the radius of the planet gear The negative sign indicates that the direction of rotation of the planet is opposite to the direction of rotation of the sun. Similarly the speed of rotation of the ring is computed using the following relation N r = - (N s x) R s / R r Where, N p is the speed of the planet gear R s is the radius of the sun gear R r is the radius of the ring gear The negative sign indicates that the sun gear and the ring gear rotate in the opposite directions. Step 3 Combined motion: The combined motion is given adding the results from the step 1 and 2 So the equations of the speeds of various components are N s =Speed of sun 49

51 N p = Speed of planet = - (N s x) R s / R p + x N r = Speed of ring = - (N s x) R s / R r + x The transmission ratio has been defined as R tr = transmission ratio = R r / R s = Radius of Ring gear / Radius of sun gear The transmission ration between the sun gear and the planet gear is R s / R p = Radius of the sun gear / Radius of the planet gear Radius of ring gear = Radius of sun gear + 2 * Radius of planet gear R r / R s = (R s + 2 R p ) / R s = (1 + 2 R p /R s ) R = (1 + 2 R p /R s ) R p / R s = (R 1) / 2 (or) R s / R p = (R 1) / 2 The above equation gives the relation between the transmission ratio between the sun and planet and the transmission ratio between sun and the ring 50

52 Torque analysis in an Epicyclic gear train: The epicyclic gear train has one degree of freedom with respect to the torques. Meaning that once the torque on any one of the components is set the toque on all the other components gets set. According to the driving cycle requirements the speed and torque requirements at the output (arm) are set. Now for the given torque requirement at the arm the torque on the ring and the sun gets set. Lets now calculate, for a given torque on the arm what will be the torque on the sun and the ring gear. Let, T a = Torque requirement in the arm Torque in arm = Force in arm * Radius of arm where, F a = Force in arm = F a R a R a = Radius of arm = R s + R p Where R s = Radius of sun gear R p = Radius of planet gear There fore Force on arm = T a / R a 51

53 This force in the arm gets equally divided in to the ring and sun as F a / 2 There fore F s = Force in sun = F a / 2 F r = Force in ring = F a / 2 Calculation of torque in the ring gear: The torque on the ring gear is equal to the force on the ring gear multiplied by the radius of the ring gear. T r = Torque on the ring gear = Force on ring gear * Radius of the ring gear = F r R r = (F a / 2) R r = (T a / 2 R a ) R r = (T a / 2) ( R r / (R s + R p ) ) = (T a / 2) (2 R r / (R r + R s )) = T a ((R r + R s ) / R r ) -1 = Ta ( 1 + R -1 tr ) -1 = Ta R tr / (R tr + 1) Since the ring is connected to the IC engine the torque in the IC engine is must be equal to the Torque in the ring T r = T = Ta R tr / (R tr + 1) Where, 52

54 T = Torque on the IC engine T r = Torque in the ring Calculation of torque in the sun gear: The torque on the sun gear is equal to the force on the sun gear multiplied by the radius of the sun gear. T s = Torque on the sun gear = Force on sun gear * Radius of the sun gear = F s R s = (F a / 2) R s = (T a / 2 R a ) R s = (T a / 2) ( R s / (R s + R p ) ) = T a R s / (R r + R s ) = T a ((R r + R s ) / R s ) -1 = T a ( R tr + 1) -1 = T a / ( R tr + 1) Since the sun is connected to the electric motor the torque in the motor is should be the torque in the sun. T s = T em = T a /(R tr + 1) Where, T em = Torque on the electric motor T s = Torque in the sun 53

55 Torque Efficiency Speed Model for an IC engine: An approximate model at present has been developed to catch the trend for understanding the variation of the speed, torque and efficiency for an IC engine. The following relation connects the fuel consumed per KWH, Torque and speed. f = (7/50) (T 50) 2 + (1.25/100000) (N 3000) Where, f = Fuel consumption per KWH T = Torque from the IC engine N = Speed of the IC engine (Efficiency) IC engine α 1 / f Torque Efficiency Speed Model for an Electric Motor: Again another model torque speed efficiency model has been assumed for the electric motor. The efficiency has been assumed to be of the following form. (Efficiency) electric motor = (-1/9) 10-5 (N em ) (T em 100) Where, N em = Speed of the electric motor T em = Torque of the electric motor 54

56 Optimization Program and Results Based on the algorithm developed for solving the optimization problem, a Matlab code was developed using the torque and speed relations developed in the previous section. The Matlab code has been attached at the end of the report. The program progressively iterates to find the optimal R. Since we don t know the final driving cycle on which we will be working on, some values of speed and torque has been selected in order to cover the entire range of operation of the vehicle. Their values are as follows: T a = [10, 50, 100, 150, 200] N a = [2500,3000,3500,4000,4500] The objective function is calculated based on every combination in the speed and torque so that the optimum solution to these values can be assumed to be an indication as a global solution. Based on the output of the program it is observed the system is most efficient when the transmission ratio is minimum. The code actually iterates for various values of R and combinations of torque and speed and finds the transmission ratio R at which the efficiency of the transmission is maximum. 55

57 2. Continuous Variable Transmission (CVT) There will be a CVT between the transmission and the wheels. The reason for the necessity of a CVT is to have degrees of freedom in adjusting the torque and the speed that are input to the transmission. The degrees of freedom associated with the CVT is beneficial in finding a better optimum in the sense that there will be more degrees of freedom in selecting the individual operating speeds for the engine and the motor. The three variables associated with a CVT, which are considered in this model, are torque, speed and the transmission ratio. From the work done by William we have the following threedimensional plot relating the speed, torque and efficiency at a particular transmission ratio. The above plot has been done for a transmission ratio of 0.6 to 1. 56

58 So for sets of different transmission ratio s we have a set of different plots. An approximate method has been devised to get the plot over a range of transmission ratios. Since it is a three-dimensional graph we have a set of 2 two-dimensional graphs placed on top each other. First is a torque efficiency curve and the second is a speed efficiency curve. By carefully studying the above graph the slope in the torque efficiency curve has the following equation. Where Ta is the torque. Slope = 6 T tr Further investigating we get a relation between the efficiency drop over a given range of speed and torque to be Efficiency drop = 6 N tr T tr 5000 Where N tr = Input speed to the CVT T tr = Input torque to the CVT This efficiency drop is measures with respect to the efficiency measured at 0 rpm as the input speed. Thus the model of relating the overall efficiency to all the parameters namely speed, torque, transmission ratio has been developed to be Overall Efficiency = (Efficiency of Transmission at 0 rpm Efficiency Drop) * Transmission Ratio multiplier 57

59 Determination of the Transmission Ratio Multiplier: The transmission efficiency at 0 rpm is calculated from the graph below from the work done by Georg Sauer. The equation of the above curve was determined using the method of least squares. Based on the output from the model the equation has been determined to be Eff = R cvt R 2 cvt Where Eff represents the Efficiency R cvt represents the transmission ratio The maximum efficiency (0.92) occurs at a transmission ratio of 1. There fore the transmission ratio multiplier is set to 1 at this point and others ratio multipliers are calculated with respect to this point. 58

60 Transmission ratio multiplier = Eff / 0.92 = ( R cvt R cvt 2 ) / 0.92 Efficiency of Transmission at 0 rpm: The efficiency of transmission at 0 rpm is got from a relation developed from the following curve. The curve has been got from the work on Continuous Variable Transmission for Tractor Drive Line by Georg Sauer. The equation governing the above has been approximately derived as - K (Ta + 11) Eff cvt = 0.95 e Where, Eff cvt = Efficiency of the CVT 59

61 T tr = Input Torque to the CVT K = a constant which has been determined to be equal to There fore the efficiency equation is given by Eff cvt = 0.95 e 0.095(T+11) Summary An optimization problem of finding the speed and torque map between the IC engine and electric motor for a given torque and speed has been defined in the planetary gear train system used in the transmission system. The objective function is as follows Maximize F = (Eff (T 1 N 1 ) + Eff (T 1 N 2 ) + Eff (T 1 N 3 ) +Eff (T n N m )) Or Minimize F = - (Eff (T 1 N 1 ) + Eff (T 1 N 2 ) + Eff (T 1 N 3 ) +Eff (T n N m )) Where T and N are the torque from the driving cycle. An approximate model for the efficiency, which depends on torque and speed, has been developed for the IC engine and the electric motor. Then, the problem has been solved using the optimization algorithm developed. A complete model of the CVT has been developed. 60

62 DC Machine - Optimization Problem Statement: There is always an inherent power requirement from the DC Machine depending on the Driving Cycle. This Power requirement can be expressed as various combinations of Torque and Speed and the combinations of which would have different efficiencies. These combinations of power and torque are not necessarily the same as those required by the wheel drive. The point to be noted here is that since the DC Machine is coupled to the wheel drive through a transmission system, that combination of torque and speed which give the DC Machine its highest efficiency can be translated to the one required by the wheel drive through this transmission system without any compromise anywhere (other than transmission losses). Thus the problem now deals with identifying the proper torque and speed of the DC Machine for the given power requirement which would give the highest efficiency. Mathematical Model: The mathematical model is re-formulated thus: From the Driving Cycle, there is a Power (P) requirement from the DC Machine. But this power can be expressed as various combinations of Torque (T) and Speed (ω). Thus, P = T em ω em The Power to run the DC Machine comes from the Battery via the controller unit. Assuming the parameters of the Power Input to the DC Machine as Voltage (V) and Current (I), we can now highlight the losses that will be considered. Again some of the losses noted in the previous report have been discarded because they were redundant as they were always expressed as a percentage of the rating. 61

63 There are two losses that were salient and thus considered The copper losses and the Inductance Losses in the armature. This is also because of the configuration of the application selected i.e. DC Machine with Permanent Magnets. The copper losses are given off as heat energy and can be expressed as follows: I 2 cont R Where R is the Resistance of the copper armature windings The Inductance Losses need to be considered from the proper perspective. One can argue that there are no inductance losses in DC Circuits but when we have DC Voltage requirement changing with time (like during acceleration and deceleration), we have a varying quantity viz. DI/DT Now, if we consider the Inductance of the copper winding to be L, then we have inductance loss occurring whenever the acceleration is required by the DC Machine and also at all times during re-generative braking and which is L (DI/DT) I cont Thus the summation of the losses would be I cont 2 R + L (DI/DT) I cont The efficiency of the DC Machine can now be summarized as η = (V cont I cont ) (I 2 cont R + L (DI/DT) I cont ) / (V cont I cont ) = 1 - (I 2 cont R + L (DI/DT) I cont ) / (V cont I cont ) The following quantities affect the above equation: Diameter of the Armature Wire 62

64 Number of turns on the armature The Resistance of the Armature Wire The behavior of these can be inferred from the following diagrams: Fig 1 In Figure 1, it can be seen that since the diameter of the armature is small, more turns can be accommodated on the Armature Shaft when compared to that in Figure 2, where the larger diameter accommodates lesser number of turns. Fig 2 63

65 Thus increase in area of armature decreases the number of turns. But this decrease in the number of armature turns causes torque to decrease, which is not favorable. At the same time decrease in armature area causes resistance offered by the armature wire to passage of electricity to increase. Thus a proper tradeoff is required between the area of the armature coil and the torque generated. Thus optimization is implemented whose objective function would be to increase efficiency of the motor as stated in the Problem Statement above. The equations relating Area (A) and Resistance (R) of Armature coil, Number of Armature Turns and the Torque are as follows: R = ρl A /A Where ρ = Resistivity of Copper L A = Length of Armature Slot T = NP O ΦI / 2Π A (Adapted from derivation in previous report) Where N Number of Turns P o Number of Poles Φ - Flux per Pole There is a constraint on the Maximum Resistance that the copper wires can posses. This is based on the fact that there is a maximum allowable temperature for the motor to operate. Thus, R R A (θ A ) / Where R A Resistance at ambient temperature 64

66 θ A Allowable temperature rise There is also a constraint on the maximum torque that can be generated which is limited by the failure point of the Armature shaft due to excessive twist. Thus, T CϑJ/Y Where C Modulus of material of shaft ϑ - Permissible twist angle J Mass Moment of Inertia Y - Length of shaft Further, Mass Moment of Inertia of the Shaft is given by: J = ΠD 4 /32 Where D Diameter of Shaft Also, Q = M S T Where Q = Heat Loss M = Mass of the armature coil S = Specific heat of the armature material T = Rise in temperature Here the T represents the maximum rise in temperature that is allowable for the material. The equation is divided by time to get the power equation. And, I cont 2 R + (Other Losses) Q max Where Q max is the maximum power loss 65

67 Other losses include the minor losses other than the copper loss and equals 0.5% of Motor Rating Other constraints that are also included are: Voltage supplied is never greater than the Battery Capacity (G (3)); the Inductance of the Copper Wires is always greater than 0.025e-3 (G (4)); the Resistance of the Copper Wires is always greater than 0.1 Ω (G (5)). There the objective function is to maximize efficiency or minimize losses. Thus the objective function is Maximize η = (V cont I cont ) (I 2 cont R + L (DI/DT) I cont ) / (V cont I cont ) = 1 - (I 2 cont R + L (DI/DT) I cont ) / (V cont I cont ) (Or) Minimize (I 2 cont R + L (DI/DT) I cont ) Subject to The Equality Constraints H (1): P em - Tω em = 0 H (2): R - ρl A /A = 0 H (3): T em NP O ΦI cont / 2Π A = 0 H (4): J - ΠD 4 /32 = 0 The In-Equality Constraints G (1): T em - CϑJ/Y 0 66

68 G (2): R - R A (θ A ) / G (3): V 300 G (4): 0.025e-3 - L 0 G (5): R 0 G (6): I 2 cont R + (Other Losses) - Q max 0 Model Analysis: I cont R V cont L P T em ω em A N F G (1) + G (2) + G (3) + G (4) - G (5) - G (6) + + H (1) H (2) + - H (3) H (4) Here H (1), H (2), H (3) and H (4) is assumed 0 67

69 During the solving process the known values that are substituted are as follows: Name Value Inductance Resistivity 0.025e-3 Henry 1.56e-8 Ω-m No: of Poles 4 Flux 1 Wb Modulus of Rigidity 80e9 N/m 2 Permissible Slip 0.7 Length of motor Shaft Radius of Armature Coil 0.2 m 0.1 m By MP1 we see that wrt L, G (4) is active, therefore substituting L = 0.025e-3 Henry we get the objective function as: F = I cont R e-3 (DI/DT) / (V cont ) Again, by MP1 we see that wrt I, H (3) is active, from H3 we have T em NP O ΦI cont / 2Π A = 0 Model Calculation to demonstrate robustness: For a Torque out put of 50 Nm We have T em = = (NP O Φ/2Π) (I cont N / A) Now, we have A = Π r 2 Where, r = radius of the armature coil 68

70 N = number of armature coils = Π R / (2 r) (from diagram above) R = Radius of the shaft Substituting numerical values and above expressions, we get I cont / r 3 = Now, From H (2) we have R - ρl A /A = 0 69

71 Where L A = length of the armature coil (from diagram) = (2(r + R) + Ls) Π R / r A = Π r 2 Therefore R = ρ ((2(r + R) + Ls) Π R / r)/ Π r 2 Simplifying R = ρ (2(R + r) + Ls) Rs / r 3 Therefore the Objective function becomes F = I 2 cont R + I cont L DI/DT 2 = I cont ρ (2(R + r) + Ls) Rs / r 3 + I cont 0.025e-3 DI/DT Substituting the constants assumed in the objective function we have Minimize F = r 3 ( r) 70

72 So we have to find the minimum radius of the armature shaft for minimizing the objective function. To get the lower bound on r we use the heat equation as the radius of the armature ( r ) and the resistance are inversely proportional. I 2 cont R + (Other Losses) = Q max Q max = M S T M = mass of the armature shaft = Π r 2 ((2(r + R) + Ls) Π R / r ) (Density of Copper) Density of copper = 8800 kg / m 3 Substituting and simplifying M = r ( r +.2 ) S = specific heat of copper = 385 J /kg K T/dt = 1.5 e -3 Therefore Q = ( r (r + 2)) (385) (1.5 e 3 ) Q = I 2 cont R + minor losses = I 2 cont R + 20 Therefore I 2 cont R + 20 = ( r (r + 2)) (385) (1.5 e 3 ) Putting I cont / r 3 = We have 71

73 ( ) 2 r 6 (1.7 e -8 ) (2(.1 + r) +.2) (1/r 3 ) = r ( r + 0.2) (8800) (385) (1.5e -3 ) A matlab code was written to solve the above equation and the value of the armature coil radius was calculated as follows Torque in Nm Radius in m e e e-0031 We thus see that the solutions for the armature radius are similar and thus it is robust. Numerical Results: When the Power requirement is known and so also the input Voltage and Current, the above model is optimized and appropriate values of Torque and Speed are calculated for maximum efficiency. Also, the value of DI/DT is obtained from the Driving Cycle and expressed as a constant throughout the problem. The Software used to solve is: Microsoft Excel. The following sample values were set: P em = W V cont = 100V I cont = 352 A DI/DT = The solution converged for the above values and the feasible solution found for maximum efficiency were: 72

74 T em = 50 Nm ω em = 86.3 rad/sec Area of cross section of Armature was = 9.59 mm 2 Number of turns = 10 73

75 Battery Battery is the electrical energy storage element in the Hybrid Electrical Vehicle. Although there are various types of batteries, lead acid (PbA) type batteries will be investigated in this modeling and optimization study. An optimum battery has high energy capacity, high power capacity, low mass, low energy loss during charging and discharging. Also it should have long cycle life, low power leakage. Some of the properties cited above cannot be changed during the operation of a battery. On the other hand, a battery can be operated optimally by minimizing the energy losses during charging and discharging. Moreover, the battery can be optimized to store maximum amount of energy via determining the optimum capacity, which is dependent on several variables such as internal resistance, battery mass, etc. In some battery models, some linear assumptions are made in order to reduce the complexity of an actual battery (Kellermeyer, 1998). With these assumptions, the model becomes poor in representing an actual battery. Moreover, it is very hard to model a battery based upon only physical laws because of the charge and mass transfer, which are dependent on the State of Charge of the battery as well as the current drawn or delivered to it. So a model, which is both based on experimental measurements and physical laws can more appropriately represent an actual battery (Johnson and Keyser, 1999). 74

76 Mathematical Model Objective of this optimization study will be to maximize the energy stored in the battery. In order to quantify the energy transfer in charging and discharging, some physical laws and experimental data will be used. The first two variables to be introduced are the Open Circuit Voltage (V oc ) and the State of Charge (SOC). Open circuit voltage is the potential difference between the terminals of the battery when there is no current flowing through the terminals. State of Charge is the percentage of energy in the battery to the maximum amount of energy that can be stored in the battery. The relation between V oc and SOC can be determined experimentally. PbA batteries have different internal resistances when charging and discharging. Furthermore, they both are functions of SOC. The equations related to internal resistance and SOC can be determined from experimental data. If V denotes voltage, I denotes current and R denotes resistance; V = I R, from Ohm s Law (1) Moreover, if P denotes the electrical power, then; P = V I = I 2 R (2) Since the battery will be used to supply the necessary power when driving with the electric motor or will be charged while regenerative braking, the power related to change in kinetic energy of the HEV, P h, should be derived. P h = [(M t υ dυ/dt) + (C d A ρ υ 3 / 2) + (µ M t g υ Cosθ) + (M t g υ Sinθ)] / [µ tr µ mot µ cvt µ cont ] (3) where M t = Total mass of the HEV υ = Velocity of the HEV 75

77 C d = Coefficient of drag of the vehicle A = Frontal area of the vehicle ρ?= Density of air µ = Coefficient of rolling resistance g = Acceleration due to gravity θ = Road inclination µ tr = Transmission efficiency µ mot = Electric motor efficiency µ cvt = Continuously variable transmission efficiency µ cont = Controller efficiency M t = M v + M b (4) where M v = Vehicle mass without the battery M b = Battery mass When P h is positive, the battery is in discharging mode. The corresponding power balance equation is as follows: P h = V oc I I 2 IR d (5) where IR d = Internal resistance in the discharging mode When P h is negative, the battery is in the charging mode. The corresponding power balance equation is as follows: P h = V oc I I 2 IR c (6) where IR c = Internal resistance in the charging mode It can be solved from equations (5,6), with roots that has the correct magnitude and sign: If P h >0 I = [V oc (V oc 2 4 IR d P h ) 1/2 ] / [2 IR d ] (7) If P h <0 I = [-Voc + (Voc 2 4 IR c P h ) 1/2 ] / [2 IR c ] (8) 76

78 Internal resistance is inversely proportional to the capacity of the battery. Although batteries are manufactured in discrete capacity ratings, either by using the same cell size and using them in parallel configuration or manufacturing the cells in prescribed sizes, they can be manufactured in any capacity by changing the cell size. Moreover, by using multiple modules in series increases the internal resistance. IR d = N o IR do Co / C (9) IR c = N o IR co Co / C (10) where C o = Reference battery capacity C = Battery capacity IR do = Internal resistance for reference battery capacity in discharging mode IR co = Internal resistance for reference battery capacity in charging mode N o = Number of battery modules The relationship between the mass of the battery and its capacity is assumed to be linear. M b = N o M bo C / Co (11) where M bo = Mass of the reference battery When the battery is being discharged the energy that is delivered is as follows: oc disch arg ing ( V I) dt (12) When the battery is being charged the energy that is being stored is as follows: ( V I) dt (13) oc ch arg ing State of Charge, SOC, can be calculated by integrating the energy charged and discharged then dividing the quantity by the capacity C in appropriate terms and finally adding the result to the initial state of charge. This can be formulated as follows: SOC = SOC i + ( V charg ing oc I) dt discharging C 3600(sec/ h) N ( V o oc V I) oco dt (14) 77

79 In the above formula capacity is converted from Ah units to Joules in order to be consistent with the numerator of the rational expression. Here V oco is the average open circuit voltage of one battery module, which is a parameter. It can also be stated with the following equation. 1 Voco = VocodSOC (15) 0 Finally the objective function can be stated as finding the optimum capacity value that will yield in maximizing the energy stored in the battery. The objective function can be formulated as: Min C ch arg ( V + oc I) dt ( Voc I) dt (16) ing disch arg ing Discretization of the problem If the problem were to be solved without discretizing, the SOC function would be in the form: t SOC( t) = SOC i + f ( SOC ( t), Ph ( t)) dt (17) 0 By taking time derivatives of both sides the following form can be obtained: dsoc ( t) = f ( SOC ( t), Ph ( t)) (18) dt If f ( SOC ( t), Ph ( t)) were separable in the sense that, f ( SOC ( t), Ph ( t)) = g( SOC ( t)) Ph ( t) (19) Then the differential equation would take the form: dsoc( t) = Ph ( t) dt (20) g( SOC ( t)) Thus, SOC(t) could be obtained by integrating both sides, SOC ( t ) dsoc ( t) = g( SOC ( t)) SOC t i 0 P ( t) dt h (21) But in this problem f ( SOC ( t), Ph ( t)) is not separable so SOC(t) can not be obtained from the differential equation explicitly. 78

80 To handle this problem and remove the dependency of time, the cycle will be discretized. The objective function will be maximized for the discretized cycle. Hence the velocity and acceleration will be constant for each discretized interval. The acceleration will be approximated as follows: dυ/dt [υ(i) - υ(i-1) ] / T (22) where T is the discretized time interval. Then, the energy stored or delivered will be approximated as: ch arging ( Voc I) dt ( Voc I ) dt ([V oc (i) I(i)]T) ([Voc (j) I(j)]T) (23) discharging ch arging disch arging Hence SOC can be approximated as: charging ([V oc (i) I(i)] T) SOC SOC i + (24) C 3600(sec/ h) N V o ([V oc discharg ing Finally, the objective function can be approximated by: Min C ch arg ing oco (j) I(j)] T) ([V oc (i) I(i)]T) + ([Voc (j) I(j)] T) (25) disch arg ing As a result, time dependency is being removed from the problem. 79

81 Constraints Practical Constraints The battery life is highly reduced if SOC becomes higher than 100% and lower than 20%. g1 = SOC 1 0 (26) g2 = 0.2 SOC 0 (27) If the current on the battery is very high the heat generation due to high current will permanently affect the battery materials, which should be avoided by current limitation. g3 = I / C 10 0 (28) Mass of the battery should not exceed 300kg. g4 = M b (29) Physical Constraints The following equality constraints are determined from equations (3, 4, 11). They will be used to eliminate some of the variables. h1 = M b (N o M bo C / C o ) = 0 (30) h2 = M t (M v + M b ) = 0 (31) h3 = P h - [(M t υ dυ/dt) + (C d A ρ υ 3 / 2) + ( µ M t g υ Cosθ) + (M t g υ Sinθ)] / [µ tr µ mot µ cvt µ cont ] = 0 (32) V oco, IR do and IR co assumed to be only a function of SOC. To determine their functions relating to SOC, curve fitting is done to experimental data gathered from (Johnson and Keyser, 1999). Least squares method is utilized in order to get explicit solutions that can be easily differentiated. V oco = SOC 3.50 SOC SOC 3 (33) where V oco = Open circuit voltage of one battery module 80

82 IR do = SOC SOC SOC 3 (34) IR co = SOC SOC 2 (35) Hence the following equality constraints are formed: h4 = V oco ( SOC 3.50 SOC SOC 3 ) = 0 (36) h5 = IR do ( SOC SOC SOC 3 ) = 0 (37) h6 = IR co ( SOC SOC 2 ) = 0 (38) The open circuit voltage of the whole battery is found by multiplying V oco by number of modules, since all the modules will be in series. V oc = N o V oco (39) So the equality constraint is as follows: h7 = V oc N o V oco = 0 (40) The following equality constraints are determined from equations (7, 8, 9, 10). They will also be used to eliminate some of the variables. h8 = IR d (N o IR do C o / C) = 0 (41) h9 = IR c (N o IR co C o / C) = 0 (42) h10 = I - [V oc (V 2 oc 4 IR d P h ) 1/2 ] / [2 IR d ] = 0 for P h > 0 (43) h11 = I - [-V oc + (V 2 oc 4 IR c P h ) 1/2 ] / [2 IR c ] = 0 for P h < 0 (44) The last equality constraint is obtained from equation (18) h12 = SOC SOC i + charging ([V oc (i) C I(i)] T) 3600(sec/ h) N o ([V oc discharging V oco (j) I(j)] T) = 0 (45) The HEV will be either in charging or discharging mode depending on each discretized intervals of the drive cycle. The 12 equality constraints are not utilized at the same time. More precisely, h1, h2, h3, h4, h6, h7, h9, h11, h12 will be utilized in charging mode and h1, h2, h3, h4, h5, h7, h8, h10, h12 will be utilized in discharging mode. 81

83 Design Variables and Parameters Variables M t = Total mass of the vehicle M b = Battery mass P h = Power necessity of HEV V oco = Open circuit voltage of one battery module V oc = Open circuit battery voltage SOC = State of Charge IR do = Internal resistance for reference battery capacity in discharging mode IR co = Internal resistance for reference battery capacity in charging mode C = Battery capacity IR d = Internal resistance in the discharging mode IR c = Internal resistance in the charging mode I = Current Parameters The speed and acceleration of the vehicle will be a parameter in each discretized interval of the drive cycle. υ = Velocity of the HEV dυ/dt = Acceleration of the HEV The following values are for a typical family automobile M v = Vehicle mass without the battery = 1400kg C d = Coefficient of drag of the vehicle = 0.32 A = Frontal area of the vehicle = 2 m 2 µ = Coefficient of rolling resistance = The efficiencies of the components of the HEV are set as parameters. µ tr = Transmission efficiency = 0.95 µ mot = Electric motor efficiency =

84 µ cvt = Continuously variable transmission efficiency = 0.90 µ cont = Controller efficiency = 0.90 Road inclination can be set as a parameter, such as equal to zero. θ = Road inclination = 0 The reference battery mass and capacity are set as parameters, their values are adapted from (Johnson and Keyser, 1999). M bo = Mass of the reference battery = 5kg C o = Reference battery capacity = 13Ah V oco = Average open circuit voltage of one battery module = 12V Also the number of battery modules, N o, is a parameter determined by the motor s maximum voltage rating. Typically a HEV DC motor will operate at 300V, and since typical number of modules is 25. V oco is 12V, the N o = Number of Battery modules = 25 Constants ρ?= Density of air = kg/m 3 g = Acceleration due to gravity = 9.8m/s 2 There are 10 variables for charging mode: M b, M t, P h, V oco, IR co, V oc, IR c, I, SOC, C. Similarly there are 10 variables for discharging mode: M b, M t, P h, V oco, IR do, V oc, IR d, I, SOC, C. Moreover there are 9 equality constraints for each mode. So there is one degree of freedom for each mode. 83

85 Selection of the Drive Cycle A drive cycle is selected for the HEV that will only run on the electric motor, which is powered by the battery. The time interval, T, is selected to be 1 sec and the corresponding velocity values are as follows: υ 0 = 0, υ T = 2, υ 2T = 6, υ 3T = 10, υ 4T = 10, υ 5T = 8, υ 6T = 4, υ 7T = 2, υ 8T = 0 Then υ(i) is formed by: υ(i) = (υ it + υ (i-1)t ) / 2 (45) The acceleration is defined by: dυ (i) = (υ it - υ (i-1)t ) / T (46) dt Model Analysis After discretizing the objective function, all the variables in both charging and discharging mode except C are removed through direct elimination using the equality constraints. Hence the problem is reduced to a one variable problem with only inequality constraints. min F(C) g1 = SOC(C) g2 = 20 SOC(C) 0 g3 = I(C) / C 10 0 g4 = M b (C) The form of the objective function depends on the number of discretized intervals. Hence the second derivative of the objective function with respect to the capacity also depends on the number of intervals. For a single interval the second derivative of the objective function is of the form: 84

86 C C 3 3 k 3 n m C k 3 n m + C for discharging mode for charging mode where k, m, n are positive real numbers that are used for substitutes for the battery variables in order to make simplification and C > 0 n It can be seen that if C >, then the second derivative is real and also positive for the m discharging mode, provided that C is finite. Moreover, the second derivative is always positive for the charging mode again for finite C. Thus, the objective function is convex for these single intervals. Although the form of the objective function and its second derivatives change for different number of intervals, the forms that are presented above are repeated. They are either multiplied with positive real numbers or divided or multiplied by powers of C. Thus, the objective function is convex for any number of intervals. Since the function is convex, there will be a unique minimum for the objective function for the bounded problem. The inequality constraint g4 will be considered separately from g1, g2, g3. The latter inequalities will be utilized in the system integration. The reason is that, if one of them becomes active, they will limit the power transfer hence the cycle cannot be tracked. But for the system case, if one of them becomes active, then the cycle can be tracked by adjusting the power on the engine. Then, the further reduced optimization problem becomes: min F(C) g4 = M b (C)

87 Numerical Results For a single interval, which can be selected as the first interval of the cycle, the objective function can be plotted. F(C) C For this case the 5optimum point 10 can be 15determined 20by setting the first derivative of the objective function to zero. If the 5 th interval of the cycle is selected, then the objective function is as follows: F(C) g4(c) For this case g4 will be active and optimum will be found at the boundary. These results can be physical interpreted as follows: C If the vehicle is accelerating hence the battery is discharging, the mass of the battery will be a burden on the increasing the speed. On the other hand, capacity is directly proportional to mass and inverse proportional to internal resistance. So, for high capacity, the electrical energy losses 86

88 will be low. So, electrical energy and kinetic energy losses are competing quantities, which cause the non-monotonic behavior of the objective function for this case. For the regenerative braking case, the increased mass of the vehicle is an advantage since the kinetic energy is going to be converted into electrical energy. Moreover, for higher mass the electrical energy losses will be low. Thus, the problem is monotonic in the sense that the objective function decreases with increasing capacity. The monotonicity of the objective function cannot be known without knowing the driving cycle. If more charging will occur than discharging, then the objective function will be monotonic. For more intervals, the computation time required for plotting the objective function increases drastically. Moreover, any calculation that is related to taking first or second derivatives of the function takes long time. So the optimum can be determined by approximating the objective function by a quadratic function, then finding the optimum for the approximation, and then this optimum is assumed to be the approximate optimum for the objective function. After that, the objective function is approximated by a new quadratic function that uses the previous approximate optimum. This procedure is continued until the optimum value does not change more than a specified termination value. Here is again the first interval of the cycle, but now the optimum solution is given in numerical values by the quadratic approximation approach. C* = F* = Moreover, if the inequality constraint were to be violated through the iteration procedure, then it would become active. So the solution would be on the active constraint. This would be the case for the charging case i.e., 5 th interval of the cycle. 87

89 System Integration The most challenging part of this study is the system integration. It requires a physically reasonable objective function, moreover, well-developed and integrable mathematical models. Thus, the subsystems were modeled accordingly at the initial subsystem optimization stage. Finally, the system objective is formed as minimization of fuel consumption while maximizing the energy stored in the battery. This objective has an effect on every component of the HEV. To quantify the objective function and to run an optimization study, the vehicle is assumed to go through a driving cycle, which determines the torque and power requirements on the components. Every component of the vehicle is modeled. Starting from the wheels including the rest of the drivetrain, that contains the differential, and the axles, up to the two energy sources, namely the fuel tank and the battery, the vehicle is modeled. Hence, the effect of the cycle data, which is the speed and the acceleration of the vehicle, can mathematically be reflected to every subsystem. There are various assumptions made in this whole integration. First of all, the drive cycle, which is normally a continuous function of time, is discretized and the vehicle is assumed to cruise at an average speed and acceleration at every discretized interval. Furthermore, inertia effects on the components are neglected. In other words, a quasi-static approach is considered. There can always be additions to the current system in order to increase the complexity but the aim of this study is to pose a reasonable problem and solve it by applying both rigorous analytical methods and by making use of high computing power. Although powerful computers can handle quite large scale and complex problems, they often fail if the mathematical models are not well formulated or some of the constraints are highly non-linear. Thus, the complexity of the mathematical models are chosen to be sufficient enough to capture the general trends and simple enough to be handled analytically or numerically. 88

90 The sketch of the system is presented in order to give a clear explanation about how the components are linked in the system. υ, a Wheels T w, ω w Differential T cvt, ω cvt ω tr = ω cvt * R cvt CVT Τ tr = Τ cvt / (R cvt * Eff cvt ) T tr, ω tr Transmission T ic, ω cvt T em, ω cvt IC Engine Electric Motor m fuel V cont, I cont Fuel Controller V bat, I bat Battery 89

91 The driving cycle determines the total power rating of the HEV that is the sum of engine s and motor s power ratings. Power allocation between the engine and the motor is a part of the optimization routine. Since the cycle is discretized, the speed and the torque on the wheels of the vehicle, and hence the power demand for each discretized interval are known. The CVT varies the individual speed and torque values while maintaining the power demand. The CVT s speed and torque output are the inputs to the PTS. The PTS s ratio is a design variable. Since the PTS input torque is determined, the individual torques on the engine and motor are only functions of PTS s ratio. PTS is a two-degree of freedom system with respect to speed, therefore the individual speeds of the engine and the motor cannot be determined by knowing the input speed to the PTS. The individual motor and engine speeds are control variables. The term control variables are used for variables related to operation of the vehicle. There is an optimization routine to determine the optimum engine and motor speeds for optimizing the objective. Both the engine and the motor cannot operate at their optimum operating points because of the cycle requirements. Therefore, there is a trade off in choosing their operating points. The capacity of the battery, which is a design variable, is proportional to its mass. High mass is a burden in the acceleration of the vehicle. On the other hand, high capacity battery yields in less electric energy losses. Therefore, there is a trade-off in selecting the optimum capacity. There are various equality constraints that define the operation of components. Some of them are linear and can be solved explicitly; some of them involve square roots of some variables and need to be analyzed very carefully in order to avoid complex numbers. First of all, the all-at-once approach is tried in order to solve the optimization problem. Therefore, the equality constraints that are relevant to the system integration are formed. Along with these constraints the number of variables are determined. In this approach (all-at-once) there are 27 variables and 22 equality constraints. There are also 22 inequality constraints most of which are due to physical limitations in the components. Here is the mathematical summary of the all-at-once approach: 90

92 All-at-Once Approach Summary Objective Function: Minimize E g + E b where E g and E b are defined as follows, both in Joules; Eg = m fuel TQ Eb = Σ Voc(i) I(i) T - Σ Voc(j) I(j) T Equality Constraints: h1: P h { [(M v + M b ) υ a] + [C d A ρ υ 3 / 2] + [µ (M v + M b ) g υ Cosθ] + [(M v + M b ) g υ Sinθ]} = 0 h2: ω w - υ / r tire = 0 h3: T w - P h / ω w = 0 h4: T w - R r T cvt = 0 h5: ω w - ω cvt / R r = 0 h6: ω tr (ω cvt R cvt )= 0 91

93 h7: Τ tr (Τ cvt / (R cvt Eff cvt ))= 0 h8: Eff cvt {[0.95 e (Ttr + 11) (6 ω tr / T tr 5000)] * ( R cvt R cvt 2 ) / 0.92} = 0 h9: T ic - (R tr / (R tr + 1)) T tr = 0 h10: T em - (1 / (R tr + 1) ) T tr = 0 h11: ω tr [ω ic (R tr / (R tr + 1)) + ω em / (R tr + 1)] = 0 h12: m fuel - [(7 / 50) (T ic / P maxic 50) 2 + ( ) (ω ic 300) ] * (T ic ω ic ) / ( ) = 0 h13: Eff em [(-1/9) 10-5 (ω em -300) (T em?45000 / P maxem 100) ] = 0 h14: P maxic + P maxem - P tot = 0 h15: P bat Eff cont Eff em - T em ω em = 0 h16: M b (N o M bo C / C 0 ) = 0 h17: V oco [ SOC 3.5 SOC SOC 3 ] = 0 h18: IR d0 [ SOC SOC SOC 3 ] / 1000= 0 IR C0 [ SOC SOC 2 ] / 1000s = 0 h19: V oc - N o V oco = 0 h20: IR d - N 0 IR d0 C 0 / C = 0 92

94 IR c - N 0 IR c0 C 0 / C = 0 h21: I - [V oc [V oc 2 4 IR d (P h / Eff cont Eff em Eff cvt )] 1/2 ] / 2 IR d = 0 I - [V oc [V oc 2 4 IR c (P h / Eff cont Eff em Eff cvt )] 1/2 ] / 2 IR c = 0 h22: SOC SOC i + charging ([V oc (i) C I(i)] T) 3600(sec/ h) N o ([V oc discharging V oco (j) I(j)] T) = 0 Inequality Constraints: g1: 0.5 R cvt = 0 g2: R cvt 2 = 0 g3: 0.2 R tr = 0 g4: R tr 5 = 0 g5: 10 - T tr = 0 g6: P maxic P tot = 0 g7: P maxic = 0 g8: P maxem = 0 93

95 g9: P h P tot = 0 g10: ω ic 800 = 0 g11: - ω ic = 80 g12: ω em 800 = 0 g13: -ω em = 0 g14: [Τ em / P maxem ] 200 = 0 g15: Τ ic [( ω ic ω ic ) (P maxic / 41000)] = 0 g16: Τ ic ω ic - P maxic = 0 g17: Τ em ω em - P maxem = 0 g18: M b 500 = 0 g19: [4 IR d (P h / Eff cont Eff em Eff cvt )] - V oc 2 = 0 g20: I 10 C = 0 g21: SOC 1 = 0 g22: 0.2 SOC = 0 In the all-at-once approach, 18 of the equality constraints can be solved explicitly. The remaining equality constraints are h7, h8, h15 and h21. There are 27 variables in the system. By solving 18 of the equality constraints, the number of variables is reduced to 9. The remaining 94

96 variables are C, R cvt,r tr,p maxic,ω ic, Τ tr, ω em, I, P bat. Although the elimination said to be done explicitly, the remaining form of the equality constraints is very complex. That s why they are not written here. In fact, the elimination is done in an algorithm. Nevertheless, with the highly complex forms of the equality constraints, the SQP algorithm in MATLAB could not yield in real number results. System Decomposition The system is decomposed into two subsystems in order to eliminate some of the nonlinear equality constraints. Since the equality constraints were written in the order of the power demand from the wheels to the energy sources, it was not hard to make the decision of decomposition. The equality constraints up to h15 are grouped in one subsystem and the remaining constraints are grouped in another subsystem. Physically, the battery is one subsystem and the rest of the vehicle is the other one. This decomposition eliminated two more equality constraints, namely h15 and h21. So the variables P bat and I are solved in terms of the other 7 variables. Then, two objective functions are determined for the two subsystems. The objective for the battery is to maximize the electrical energy stored and the objective for the rest of the HEV is to maximize fuel efficiency and minimize the power demand from the battery. Rest of the HEV s objective function is very similar to the original whole systems objective function. Therefore, it has an appropriate physical meaning. But the battery s objective function only involves its own variables and does not contain a variable that will cause a trade off in the system. So the equality constraint h21 is modified slightly. Previously it was: h21: I - [V oc [V oc 2 4 IR d P bat ] 1/2 ] / 2 IR d = 0 I - [V oc [V oc 2 4 IR c P bat ] 1/2 ] / 2 IR c = 0 95

97 Then it is modified to: h21: I - [V oc [V oc 2 4 IR d (P h / Eff cont Eff em Eff cvt )] 1/2 ] / 2 IR d = 0 I - [V oc [V oc 2 4 IR c (P h / Eff cont Eff em Eff cvt )] 1/2 ] / 2 IR c = 0 This modification made an indirect linking with the battery energy to fuel consumption. The reason is as follows: P h depends on M b and M b depends on C. Moreover, m fuel depends on P h. So when the battery capacity changes, the fuel consumption changes. With this modification a well-posed objective function is obtained. However, the SQP algorithm in MATLAB still did not avoid complex numbers in solving the equality constraints h7 and h8. As a result, one more modification is made. h7 and h8 both depend on T tr and R cvt. Since the relation of these variables in these two equations is non-linear, the algorithm has the chance to fail by diverging to complex numbers while searching the optimum. Although the domain of these variables constrain them to stay real, since SQP does not stay feasible during iterations, the algorithm finds complex numbers. Thus, h8 is modified to have a simpler relation. It was previously of the form: h8: Eff cvt {[0.95 e (Ttr + 11) (6 ω tr / T tr 5000)] * ( R cvt R 2 cvt ) / 0.92} = 0 Then it is modified to: h8: Eff cvt [0.94 (6 ω tr / T tr 5000)] = 0 It seems like the equation is simplified too much. But physically, what is being done is that, first the dependence of efficiency to R cvt is removed. Actually, within the limits of R cvt, its effect on the efficiency is at most 4%. So this modification does not have much effect. Then the exponential dependence of T tr is removed. This modification has large effect if T tr takes small values at the optimum, i.e first 20% of the torque range. 96

98 This modification enabled to solve for T tr and ω em. So, the problem is reduced to a 5 variable problem with only inequality constraints. The variable C belongs to the battery and the other 4 of the variables namely, R cvt,r tr,p maxic,ω ic belong to the rest of the HEV. This simplified problem has no chance of giving complex results. As anticipated, the SQP algorithm was able yield in feasible solutions for different driving cycle intervals and different starting points. After the solutions are obtained, the value of T tr is checked. It has been seen that it is high enough to make the exponential term virtually zero if the term appeared in the equation. But if the cycle power requirements were too low to make T tr a small value, then the validity of the modified problem will be less. To sum up, this model for HEV is valid for moderate to high power ratings. Also, solving the battery optimization subproblem was completed successfully. Optimization of the Master Problem The decomposed subsystems have parallel objective functions. The rest of the HEV s objective function can be taken as the master objective function with a slight modification. HEV Maximize fuel economy and energy stored in the battery Battery Maximize energy stored in the battery Rest of the HEV Maximize fuel economy and minimize power demand from battery 97

99 If battery and the rest of the HEV are optimized separately, they will contribute positively to the system optimization. Therefore, a sequential approach is being tried to solve the master problem. The scheme of the approach is as follows: Battery M b Eff cvt Eff em Rest of the HEV The decomposition of the system yielded in 3 linking variables, namely, Eff cvt, Eff em, M b. In the sequential optimization approach the starting component is not important. If battery is selected to be optimized first, then, Eff cvt and Eff em are taken as parameters. The optimization routine results in a value of M b. Then, this value is used in the rest of the HEV as a parameter for the optimization of the rest of the HEV. Since both of the subsystems objective functions are not competing, convergence of the solution is expected in this sequential approach. In other words, both objective functions have same kind of trade-off. Therefore, if a solution exists, this approach will yield in finding that solution. This approach is tried to solve the master problem numerically. Convergence is obtained when a feasible solution existed. In the next section, these numerical results are investigated in detail. 98

100 Numerical Results The sequential approach is utilized for various cycle data. As long as all the cycle intervals remain feasible for the problem, this method will converge to the discrete intervals optima. In other words, the individual cycle intervals are independent and if all of them yield in feasible solutions, then a general solution exists for the overall cycle. On the other hand, if any one of the intervals yields in infeasible solution, then, no solution exists. In the numerical analysis, only discharging intervals are investigated in detail. Because, in the charging mode, as proved in battery section, the mass constraint of the battery (h16) will be active by MP1. So C is determined without any computation. Moreover, W ic, P maxic and R tr will be bounded by their lower limits again by MP1. A physical proof for the last statement can be stated as follows: Kinetic energy of the vehicle is converted into electrical energy and the objective function is dependent on both electrical energy and fuel energy. Since the fuel energy cannot be converted into kinetic energy then back to electrical energy with more than 100% efficiency, the objective function value will increase for any increase in usage of the engine. Then the objective function will said to be monotonic with respect to W ic, P maxic and R tr. Therefore, by MP1, they will be bounded by their lower limits. Therefore, there is also no computation necessary to determine these variables. The only variable that is to be found is R cvt. So this simple case of optimization is not investigated in detail in this numerical study. Here are some example cycle intervals for discharging: 99

101 Interval Properties V=20 m/s a= 1 m/s 2 Variables Subsystem 1 (Engine & Motor & Transmission) CVT Ratio-R cvt Transmission Ratio-R tr Power Rating of Engine-P maxic (W) Engine Operating Speed-W ic (rad/s) Subsystem 2 (Battery) Battery Capacity-C Linking Variables Subsystem 1 (Engine & Motor & Transmission) Motor Efficiency- Eff em CVT Efficiency- Eff cvt Subsystem 2 (Battery) Battery Mass-M b For this interval, the initial input to the first subsystem is the battery mass of 500 kg as a parameter. The starting points and the initial input are summarized below. Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 500 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff em Results for Eff cvt These results for the electric motor efficiency and the CVT efficiency, from the first subsystem are then given as inputs to the second subsystem. The results obtained from the second subsystem can be summarized as follows: Subsystem 2 (Battery) Initial Inputs (Eff em, Eff cvt ) , Starting Points ( C ) 150 Results for M b 345 kg Now, the new value for the battery mass is used in the subsystem 1 to obtain the new values of the linking variables; 100

102 Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 345 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff em Results for Eff cvt The new values for the efficiencies are again input to the second subsystem to obtain the new value of the battery mass. Subsystem 2 (Battery) Initial Inputs (Eff em, Eff cvt ) , Starting Points ( C ) 150 Results for M b 318 kg The procedure is followed and the following iteration steps are observed: Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 318 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff em Results for Eff cvt Subsystem 2 (Battery) Initial Inputs (Eff em, Eff cvt ) , Starting Points ( C ) 150 Results for M b 320 kg Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 320 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff em Results for Eff cvt

103 Subsystem 2 (Battery) Initial Inputs (Eff em, Eff cvt ) 0.910, Starting Points ( C ) 150 Results for M b 320 kg So the sequential algorithm converged, and this point the values obtained for the variables are summarized below: Variables Optimum Value for the Units given interval R cvt 1.2 R tr 0.2 P max,ic 19 kw w ic 80 rad/sec C 30.7 Ah Local Optimum Value for the Units Variables given interval M b 320 kg m fuel 302 g/kwh Constraints Lagrange Multipliers g3 7.78(10 4 ) g11 44 Rest of the ineq. 0 Constraints Active constraints in this interval are the lower bound for the engine speed and the transmission ratio. There are 2 active constraints whose Lagrange multiplier values are presented above. 102

104 Interval Properties V=10 m/s a= 4 m/s 2 Variables Subsystem 1 (Engine & Motor & Transmission) CVT Ratio-R cvt Transmission Ratio-R tr Power Rating of Engine-P max,ic (W) Engine Operating Speed-W ic (rad/s) Subsystem 2 (Battery) Battery Capacity-C Linking Variables Subsystem 1 (Engine & Motor & Transmission) Motor Efficiency- Eff mot CVT Efficiency- Eff cvt Subsystem 2 (Battery) Battery Mass-M b For this interval, the initial input to the first subsystem is the battery mass of 500 kg again as a parameter. The starting points and the initial input is summarized below. Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 500 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff mot Results for Eff cvt These results for the electric motor efficiency and the CVT efficiency, from the first subsystem are then given as inputs to the second subsystem, as before. The results obtained from the second subsystem can be summarized as follows: Subsystem 2 (Battery) Initial Inputs (Eff mot, Eff cvt ) , Starting Points ( C ) 150 Results for M b 498 kg Now, the new value for the battery mass is used in the subsystem 1 to obtain the new values of the linking variables; 103

105 Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 498 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff mot Results for Eff cvt The new values for the efficiencies are again input to the second subsystem to obtain the new value of the battery mass. Subsystem 2 (Battery) Initial Inputs (Eff em, Eff cvt ) , Starting Points ( C ) 150 Results for M b 498 kg Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 320 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff mot Results for Eff cvt So the sequential algorithm converged more quickly this time, and at this point the values obtained for the design variables are summarized below: Variables Optimum Value for the Units given interval R cvt 2 R tr 0.2 P max,ic 36 kw w ic 80 rad/sec C 47.8 Ah Local Optimum Value for the Units Variables given interval M b 498 kg M fuel 322 g/kwh 104

106 Constraints Lagrange Multipliers g2 2.6(10 4 ) g11 71 g Rest of the inequality Constraints Active constraints in this interval are the lower bound for the engine speed and the upper bound for the CVT ratio. The maximum torque constraint for the internal combustion engine is also active. There are 3 active constraints whose Lagrange multiplier values are presented above. Parametric Study In this part of the study, only the interval with a velocity of 20 m/s and an acceleration of 4 m/s 2 is used. The upper limit for the battery mass limit is changed from 500 kg to 300 kg. The battery inequality constraint: g18: M b 500 = 0 is changed to g18: M b 300 = 0 to see the effect of battery mass, on the optimum values of the system. Subsystem 2 (Battery) Initial Inputs (Eff mot, Eff cvt ) 0.9,0.9 Starting Points ( C ) 150 Results for M b 300 kg Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 300 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff mot Results for Eff cvt

107 Subsystem 2 (Battery) Initial Inputs (Eff mot, Eff cvt ) , Starting Points ( C ) 150 Results for M b 300 kg The convergence is observed and the results are: Variables Optimum Value for the Units given interval R cvt 1.3 R tr 0.21 P max,ic 15 kw w ic 80 rad/sec C 28.8 Ah Constraints Lagrange Multipliers g11 40 g Rest of the inequality Constraints Two of the inequality constraints become active in this case. The second parameter value changed is the lower limit of the transmission ratio. This parameter is chosen for investigation due to its high Lagrange multiplier. The inequality constraint; g3: 0.2 R tr = 0 is changed to g3: 0.18 R tr = 0. Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 320 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff mot Results for Eff cvt

108 Subsystem 2 (Battery) Initial Inputs (Eff mot, Eff cvt ) , Starting Points ( C ) 150 Results for M b 335 kg Subsystem 1 (Engine & Motor & Transmission) Initial Input (M b ) 335 kg Starting Points 1,1,10000,150 (R cvt,r tr,p maxic,w ic ) Results for Eff mot Results for Eff cvt Subsystem 2 (Battery) Initial Inputs (Eff mot, Eff cvt ) , Starting Points ( C ) 150 Results for M b 320 kg The program cycles between two near mass values. The subsystem 2 returns battery mass of 335 kg however the subsystem 1 using this input returns linking variable values that causes the second subsystem to give 320 kg as the battery mass output which is the initial battery mass input used in the first subsystem. But the objective function value does not change very much at the two cyclic points. Quantitatively, F = J for M b = 335kg and F = J for M b = 335kg. The objective function value was for R tr = 0.2. These two values are quite similar to the value at R tr = 0.2 and it is hard to make judgement about improvement of the optimum although it is expected to improve when the active constraint is relaxed. On the other hand, if the lower bound R tr = 0.1 is used for the parametric study, the sequential approach converges successfully. And as expected, the optimum is improved. Numerically, F became Variables Optimum Value for the Units given interval R cvt 1.2 R tr 0.18 P max,ic 16 kw w ic 80 rad/sec C Ah 107

109 Constraints Lagrange Multipliers g3 7.77(10 4 ) g Rest of the inequality Constraints The active constraints are the lower bound on the engine speed and the lower bound on the transmission ratio. Discussion of Results and Conclusion One can conclude that if there is no limitation on the electric motor, the vehicle optimally operates on it. Hence for this optimization study in which only a few discrete intervals of the driving cycle are tried, the inequality constraints on the SOC didn t become active, i.e., g21 and g22. As a result, for a given total power rating of the vehicle, i.e, 100 kw, the optimum size of the electric motor is quite large compared to the engine. The ratio of the power ratings is very dependent on the interval selected, however the ratio of maximum electric motor power to the maximum engine power varies from 1/3 to 1/6. In this study, every discrete interval of the cycle is studied independently. However; if a complete cycle is considered, there should be a way of combining all the independent solutions. This can be devised as follows: If every interval has a feasible solution, then the overall cycle solution will also be feasible. To compare the individual solutions, weighting functions that are proportional to the discrete interval s power requirement can be utilized. F overall = Σ c P hi F i where P hi is the i th interval s power requirement, F i is the i th interval s objective function value, and c is a global constant for normalization. F overall is the overall cycle s normalized objective function value. There are some equations that have some scaling factors to enable them to be used for different ranges than they are defined for such as the engine s fuel and the motor s efficiency 108

110 maps. These scaling operations do not violate the general trends of the equations, however their reliability becomes less at high scaling. Future Work In this study, the reduced system with 4 non-linear equality constraints, 22 inequality constraints with 5 degrees of freedom is investigated only numerically due to its complex form. However; monotonicity analysis can be further exploited to find the active constraints analytically. Furthermore, instead of using the approximations for the engine and motor maps, the Bezier surface (Bezier Surface is defined in the System Appendix) defined by the original data points can be utilized to obtain more accurate results. Acknowledgements The author would like to thank John Whitehead for his help and support in the internal combustion engine scaling 109

111 References Internal Combustion Engine 1.J.B. Heywood, Internal Combustion Engine Fundamentals, Mc-Graw Hill Inc, New York, C.F. Taylor, The Internal Combustion Engine In Theory And Practice, John Wiley and Sons, New York, S.Anastassopoulos, Engine Design For Maximum Brake Horsepower, P.Y. Papalambros, D. J. Wilde, Principles of Optimal Design, Cambridge University Press, New York, W. Wu, M. Ross, Spark-Ignition Engine Fuel Consumption Modeling, SAE Paper , D.Assanis, G.Delagrammatikas, R.Fellini, Z.Fellini, Z.Filipi, J.Liedtke, N.Michelena, P.Papalambros, D.Reyes, D.Rosenbaum, A.Sales, M.Sasena, An Optimization Approach to Hybrid Electric Propulsion System Design, Automotive Research Center University of Michigan, G. Abramovich, P. Georgiopoulos, Optimization Study of Hybrid Electric Vehicle- A Dynamic Approach, Department of Mechanical Engineering and Applied Mechanics- University of Michigan, E.Tate, S. Boyd, Finding Ultimate Limits of Performance For Hybrid Electric Vehicles, SAE Paper 00FTT-50, G. Hubbard, Modeling and Control of A Hybrid Electric Vehicle Drivetrain, Department of Mechanical Engineering-MIT, 1996 Transmission 1. Georg Sauer, Continuously Variable Transmissions for Tractor Drive Line Institute of Agricultural Machinery, Technical University of Munich, Germany 2. William F. Kellermeyer III, Development and Validation of a Modular Hybrid Electric Vehicle Simulation Model, Master of Science Thesis, West Virginia University, May

112 Electric Motor [1] Electric Machinery Fundamentals, Stephen J Chapman, McGraw Hill, 1999 [2] Handbook of Electric Motors, Engelmann & Middendorf, Dekker, 1995 [3] Electric Machines and Power Systems, Vincent Del Toro, Prentice Hall, 1985 Battery Johnson, Valerie, and Keyser, Matthew, Testing, Analysis, and Validation of a Hawker Genesis Lead Acid Battery Model in ADVISOR, Center for Transportation Technologies and Systems National Renewable Energy Laboratory, March 1999 Kellermeyer, William F., Development and Validation of a Modular Hybrid Electric Vehicle Simulation Model, MSME Thesis, West Virginia University, May 1998 System R.L. Hodkinson, The Hybrid Electric Solution, Electro Technology, May

113 Internal Combustion Engine Appendix A-1 Compression Ratio versus Cylinder Bore Graph (Taylor 1960) The following equation is obtained by curve fitting: C r = b