Vehicle Propulsion Systems Electric & Hybrid Electric Propulsion Systems Part III 1
Planning of Lectures and Exercises: Week Lecture, Friday, 8:15-10:00, ML F34 Book chp. 38, 21.09.2018 Introduction, goals, overview propulsion systems and options Exercise, Friday, 12:00-13:30, CHN E46 1 Introduction 39, 28.09.2018 Fuel consumption prediction I 2 Exercise I, Milestone 1 40, 05.10.2018 Fuel consumption prediction II 2 Exercise I, Presentation 41, 12.10.2018 IC engine propulsion systems I 3 Exercise II, Milestone 1 42, 19.10.2018 IC engine propulsion systems II 3 Exercise II, Milestone 2 43, 26.10.2018 ICE III & Case Study Hybrid Pneumatic Engine 4 Exercise II, Presentation 44, 02.11.2018 Hybrid electric propulsion systems I 4 Exercise III, Milestone 1 45, 09.11.2018 Hybrid electric propulsion systems II 4 Exercise III, Milestone 2 46, 16.11.2018 Hybrid electric propulsion systems III 5 Exercise III, Presentation 47, 23.11.2018 Supervisory Control Algorithms I 7 Exercise IV, Milestone 1 48, 30.11.2018 Supervisory Control Algorithms II 7 Exercise IV, Milestone 2 49, 07.12.2018 Supervisory Control Algorithms III 7 Exercise IV, Milestone 3 50, 14.12.2018 Non-electric hybrid propulsion systems Exercise IV, Presentation 51, 21.12.2018 Tutorial Lecture, Q & A 2
Today 1. Supercaps Overview Equivalent Circuit Model Comparison with Batteries 2. Electric Power Links 3. Torque Couplers 4. Power Split Devices 5. Continuously Variable Transmissions 3
Capacitor 4
Charge separation leads to induced voltage U = Q C Capacitor Energy stored in electric field E = 1 2 CU2 5
Capacitor Capacity is a function of area, separation width and the dielectric C = ε 0ε r A d Large capacity needs large area, small separation width ε 0 = 8.85 10 12 F m powerful dielectric ε r 6
Supercaps = electric double layer capacitor (EDLC) 7
Supercaps Large area Long collector foil Porous carbon electrode Very thin separator Charge separation on molecular basis in electrolyte (water + ions) Electrolyte doubles as a dielectric Water: ε r 80 -> very high capacitance! 8
Supercaps 9
Supercaps Energy stored in electric field, by charge separation. No chemical reaction involved. Very high power density Robust against aging and high current Rather low energy density Electrolyte sensitive to temperature > 60 C 10
Today 1. Supercaps Overview Equivalent Circuit Model Comparison with Batteries 2. Electric Power Links 3. Torque Couplers 4. Power Split Devices 5. Continuously Variable Transmissions 11
Supercaps Relation between charge and energy: Charge: Q b Q Energy: E b = b 0 U oc Q dq = 1 CU 2 oc 2 12
Supercaps Choices as a state variable: Charge: Q b, dq b dt Energy: E b = 1 CU 2 oc 2 de b dt duoc Voltage: U oc dt = I b = I bu oc = I b C 13
ሶ Normalization People like to think in normalized values State of charge: x SOC = Q CU max = U oc U max State of energy: x SOE = U 2 oc U2 max xሶ SOC = I b CU max x SOE = I bu oc 1 2 CU 2 max 14
Supercaps Usually the voltage of a supercap is limited to a certain range: U oc [ 1 2 U max, U max ] This results in limitations for charge, energy, state of charge and state of energy. 15
Quick Check Calculate amount of energy that can be stored in a supercap between U b,max and 1 2 U b,max. Compare with total energy capacity. E eff = E b@ub,max E b@ub,min = 1 2 C U 2 b,max 1 2 U b,max = 3 4 E b,max 2 16
Voltage limitations Discharging the supercap to voltage levels below 1 2 U b,max leads to Higher current for the same power Higher ohmic losses Lower efficiency Only 1 4 more energy capacity Usually the voltage range is chosen U b [ 1 2 U b,max, U b,max ] 17
Quick Check Calculate the current necessary to deliver 200kW, when the supercap (12.5F, 0.1Ohm) is fully charged (620V) compared to when it is only half charged. I b = 1 2R U oc U 2 oc 4P b R b I b 200 kw@620 V 340 A I b 200 kw@310 V 915 A P loss 200 kw@620 V 11.5 kw P loss 200 kw@310 V 83.7 kw 18
Quick Check Find the current necessary to achieve maximum power output of a supercap 2 P b = I b U oc RI b dp b = U di b oc 2RI b = 0 I b,max = U oc 2R = U oc RI b,max = U oc U b,@ Ib,max P b,max = I b,max U b,@ Ib,max 2 2 = U oc 4R 19
Maximum Current Maximum power of a supercap P b,max = U oc 4R Usually, this number is very high and will lead to overheating of cell, especially if stacked with other cells. Therefore manufacturers specify maximum current much lower I b,max U oc 2R 2 20
Supercaps How to control a supercap? DC/DC-converter Usually DC-Link has a fixed voltage Converter needs be able to handle both throttle and boost mode 21
Supercaps Kirchhoff Law U b = U oc RI b Power P b = I b U oc RI b 2 Inverse model for current I b = 1 2R U oc U 2 oc 4P b R b 22
Today 1. Supercaps Overview Equivalent Circuit Model Comparison with Batteries 2. Electric Power Links 3. Torque Couplers 4. Power Split Devices 5. Continuously Variable Transmissions 23
Equivalent Circuit Battery: Open circuit voltage can be approximated by affine function of SOC U oc t Q b t ሚC + U 0 Internal resistance is approximately constant 24
Charge/Discharge Cycle 25
Equivalent Circuit Supercap: Open circuit voltage is a linear function of SOC U oc t Q b t C Internal resistance is approximately constant Voltage U oc = Q C Charge 26
Comparison of Discharge Modes 27
Supercap vs Battery Energy stored in battery Q b,maxuoc E = න Q dq Q b,min 28
Quick Check Calculate the energy capacity and the internal resistance of this 10 Ah battery cell. κ 1 = 0.5V/10Ah U cell 80%, 4A = 2.16V κ 0 = 2 V U cell 80%, 100A = 2V 30
Quick Check From graph: U oc Q = κ 0 + κ 1 Q, with κ 0 = 2 V, and κ 1 = 0.5V/10Ah = 0.05 V/Ah Lower estimate 10 Ah * 2 V = 20 Wh Upper estimate 10 Ah * 2.5 V = 25 Wh Q Exact integration: E b = max 0 U(Q)dQ = κ 0 Q max + 1 κ 2 2 1Q max = 2 V * 10 Ah + 0.5*0.05V/Ah*(10Ah)^2 = 20+2.5 Wh = 22.5 Wh Internal resistance: U cell 80%, 4A = 2.16V, U cell 80%, 100A = 2V R i = U = 0.16V = 1.6mΩ I 96A 31
Today 1. Supercaps 2. Electric Power Links 3. Torque Couplers 4. Power Split Devices 5. Continuously Variable Transmissions 32
Example: DC-Link of AHEAD Vehicle 33
Example: Multi-Layer Power Link 34
Power Balance of a DC-Link P loss P 1 P 2 P m P m+1 «load» QSS: Energy conservation: m P m+1 t = P j t P loss t j 35
Dynamics of a DC-Link U DC I loss Q DC I 1 I 2 I m I m+1 «load» Dynamic: Charge conservation: m d dt Q DC = C d dt U DC = j I j t I loss t I m+1 t 36
Example U DC P loss Q DC P aux P gen P bat P mot QSS: Dynamic: Auxiliaries Gen-Set Battery Traction Motor P mot t + P aux t + P loss t = P bat t + P gen t C d dt U DC = I bat t + I gen t I mot t I aux t I loss t 1 DOF 1 DOF 37
Example U DC P loss C P aux P gen P bat P mot Auxiliaries Gen-Set Battery Traction Motor High-level control: Decide on how to distribute power via reference values for active components (dc-converter / inverter) Low-level control: Introduce deviations from reference values such that DC-link voltage is stabilized 38
Example Control Law ΔV 1 P nom ΔV 0 V nom ΔP max V DC 39
Example V DC Priority 0 Priority 1 Priority 2 P i P nom ΔP max P generator P battery ΔP min 40
Quick Check DC-Link with a capacitance of C = 1 10 3 F, calculate the time for the voltage to increase by 10 V when σ i I i = 1A. d dt U DC = σ i I i C = 1C/s 1 10 3 C/V = 1000 V s t 40V = 10V 1000V/s = 0.01s 41
Today 1. Supercaps 2. Electric Power Links 3. Torque Couplers 4. Power Split Devices 5. Continuously Variable Transmissions 42
Torque Couplers Example: Motor connected to crankshaft via a single gear with gear ratio γ 43
Torque Couplers In mechanical parallel hybrids Usually two or more (N) devices connected to the final crank shaft via a torque coupler Output speed and torque (final crank shaft) ω f Gear ratio for i-th connected component γ i Resulting speed ω i = γ i ω f Final torque T f = σ 1 N γ i T i T loss Degrees of freedom: N 1 44
Quick Check Estimate the distance driven per engine cycle in the shortest and longest gear of a passenger car. α eng = Nπ α wh = Nπ/γ tot d veh = Nπr wh /γ tot With N = 4, r wh = 31.7cm, γ tot,1 13, γ tot,6 2.6 d veh,1 = 0.3m d veh,2 = 1.5m 45
Today 1. Supercaps 2. Electric Power Links 3. Torque Couplers 4. Power Split Devices 5. Continuously Variable Transmissions 46
Power Split Devices use a planetary gear set allow for a continuously variable transmission ratio introduce an additional degree of freedom (speed) can be used to optimize the operating point of an engine 47
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Idle / Standstill Engine turning carrier Generator = Zero Torque Ring «blocked» by heavy vehicle 50
Pure Electric / Engine off Engine off, planets blocked Generator Zero Torque Ring speed = Vehicle speed 51
Pure Engine Mode Planets speed = Engine speed Sun gear blocked Ring speed = Vehicle speed 52
CVT Mode: ω engine ω vehicle = 1 Planets speed = Engine speed Generator speed = Engine Speed Ring speed = Vehicle speed = Engine speed 53
CVT Mode: ω engine ω vehicle < 1 Planets speed = Engine speed Generator speed < Engine Speed Ring speed = Vehicle speed = 54
CVT Mode: ω engine ω vehicle > 1 Planets speed = Engine speed Generator speed > Engine Speed Ring speed = Vehicle speed = 55
Lever Analogy a. Pure electric ω e = 0 b. Engine on while ω f = 0 c. ω g = 0 while driving d. High ω e for acceleration 56
Planetary Gear Set Willis formula ω r ω c ω s ω c = z, where z = n s n r 57
Quick Check Assuming ω e = ω c, ω g = ω s, ω f = ω m = ω r express the engine speed as a function of the final drive speed (determined by driver) and generator speed (control input). ω f ω e ω g ω e = z, where z = n s n r ω e = ω f + zω g 1 + z 58
Power Split Device The transmission ratio from engine to the final drive shaft can be controlled by the generator speed. What does it cost us? ω e = ω f + zω g 1 + z 59
Power Split Device Applying torque and power balances yields the following: To maintain its speed, the generator needs to deliver a torque proportional to T e T g = z 1 + z T e The torque at the final crankshaft is T f = T m + 1 1 + z T e 60
Power Split Device By controlling generator speed, we can control the transmission ratio between final drive and engine However, when T g = 0, then T f = T m! 61
Quick Check Derive the expressions for generator and engine torque. P f = P m + P e P g T f ω f = T m ω m + T e ω e T g ω g ω f + zω g T f ω f = T m ω f + T e T 1 + z g ω g ω f T f T m T e 1 + z + T g T ez 1 + z ω g = 0 ω g, ω f = 0 = 0 T e = 1 + z z T g T m = T f 1 z T g 62
Quick Check With your Prius you drive 50km/h and you want your engine to run at the fuel optimal operating point (i.e. 2300rpm, 100Nm). Calculate the torque and speed of both electric machines. (z=0.385 and final drive ratio = 4.6, r = 0.3). Evaluate the power balance to check for correctness. F t = 1 2 ρc da f v 2 + mgc r = 1 2 1.2 0.7 13.82 + 1500 10 0.013 = 80 + 190 = 270N, T f = F tr γ = 270 0.3 4.6 = 17.61Nm ω f = vγ = 50 4.6 rad = 213.0 r 3.6 0.3 s 1 ω g = w e 1 + z w f 2130rpm z 0.385 T g = T e = 100 = 27.80Nm 1+z 1.385 z = 2300 2π/60 1.385 213.0 1 0.385 T m = T f T g 27.80 = 17.61 = 54.60Nm z 0.385 T m ω m = 54.60 213.0 = 11.63kW (charging the battery) T f ω f = 17.61 213.0 = 3.75kW T e ω e = 100 2300 2π/60 = 24.09kW T g ω g = 27.80 313.2 = 8.48kW (charging the battery) Sum = 24.09 3.75 8.48 11.63 = 0.23 kw (ca zero!) = 313.2 rad s 63
Today 1. Supercaps 2. Electric Power Links 3. Torque Couplers 4. Power Split Devices 5. Continuously Variable Transmissions 64
Continuously Variable Transmission 65
Continuously Variable Transmission 66
Example: Variomatik 67
Variomatik 68
Variator w eng F c = kω eng F c F s = cδx 69
Variomatik ω eng Weights fully outside ω eng v veh = γ min r wh Stabilization Speed = f(weight) Weights moving towards outside ω eng v veh = γ max r wh Weights fully retracted v veh 70
Maximum Power vs Speed http://scooterxxl.de/rollertuning/variomatik-abstimmen.php 71
Test with 3 different Vario-Weights http://scooterxxl.de/rollertuning/variomatik-abstimmen.php 72
CVT Transmission ratio γ = ω 1 ω 2 Transmitted torque T 1 = 1 γ T 2 73
ሶ More detailed model More detailed model including losses, inertia, actuator dynamics T 1 = 1 γ T 2 + T loss + Θ 1γ 2 + Θ 2 γ γ = u CVT ωሶ 2 + γω ሶ 2 Often, losses or static efficiency is tabulated as a function of torque and speed 74
Engine and Clutch In a graph in the vehicle speed vs traction force plane, explain how the fastest possible acceleration to top-speed can be achieved. F t T e,max γ 1 r wh T e,max γ 2 r wh T e,max γ 3 r wh F t v = P e,max 4 5 Vehicle Engine 1. Standstill 2. Speed up engine 3. Apply clutch 4. Synchronize 5. Shift 6. 3 T e,max γ 4 r wh 1 2 T e,max γ 5 r wh v = 0 ω idle r wh γ 1 ω max r wh γ 1 v 75