V DC. 3-phase inverter/ rectifier 2. Battery. Energy storage. ED2 Traction

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1 3. Series Hybrid lectric Drive Train Figure 3.1 shows configuration of a simple series hybrid electric drive train. Although details are significantly more complicated, Chevrolet Volt is an example of a plug-in hybrid electric vehicle (PHV) based on the series hybrid architecture. v F v Fuel IC n 1 T 1 lectric motor/ generator 1 3-phase inverter/ rectifier 1 D1 Battery charging (alternator) IC starting V DC Battery nergy storage 3-phase inverter/ rectifier D Traction lectric motor/ generator Regenerative braking n T Transmission Wheels (radius r v ) n v T v Figure 3.1. Series hybrid electric drive train architecture. In the Chevy Volt example, the IC is a 6 kw (83 hp), 1.4 L engine, D1 is a 55 kw electric drive, D is a 111 kw (149 hp) electric drive, and the tery has a total of 16 kwh capacity out of which 65% (1.4 kwh) is usable. After the tery is fully charged from the electric grid, Volt operates as an electric vehicle with an all-electric range of 4-8 km (5-5 miles). After that, the IC is engaged to maintain the tery state of charge and Volt operates as a hybrid electric vehicle. In the series architecture, all system components are connected in series: IC, electric drive 1 (D1), tery, electric drive (D), and transmission to wheels. Main functions of the system components are as follows: The vehicle is propelled by the electric drive (D) consisting of the 3-phase inverter/rectifier and the electric motor/generator. In the traction mode, at the input of the inverter, power comes from the tery and/or the IC via generator 1 and rectifier 1. From the DC tery voltage V DC, inverter generates variablefrequency 3-phase voltages and currents for motor. Motor shaft turns at n rpm and produces traction torque T. A single-gear transmission turns the wheels at n v rpm with torque T v. The resulting traction force T v Fv (3.1) rv propels the vehicle at speed v, where r v is the wheel radius. The traction power propelling the vehicle forward is P v = F v v = T v n v (/6). The mechanical power at the output of generator is P m P v Tn / 6 (3.) where t is the transmission efficiency. lectrical power supplied by the tery and/or rectifier 1 is P P m i m t (3.3) where i is the efficiency of inverter, and m is the efficiency of motor. CU Boulder, DM Fall 1 1

2 When negative traction power is required, i.e. when the vehicle is decelerating or descending, the power flows in the opposite direction: from the wheels back to the tery. This mode of operation is called regenerative braking. In the hybrid electric mode, generator 1 takes mechanical power P ice = T 1 n 1 (/6) and via rectifier 1 produces electrical power P 1 P1 g1 r1 P, (3.4) where g1 and r1 are the efficiencies of generator 1 and rectifier 1, respectively. lectrical power P 1 charges the tery and/or supplies a part of or the entire traction power P. lectric drive 1 (D1) serves two main functions: controls the tery state of charge (SOC), and sets the operating point of IC. Note the IC operating point can be set independent of the traction power requirement. Hence, it is possible to improve efficiency by setting the IC operating point close to its maximum efficiency point. D1 also serves as an IC starter. During IC start-up, power flows in the opposite direction, from tery via inverter 1 and motor 1 to the IC crankshaft. A system controller in the series hybrid electric drive train has two main objectives: 1. Control traction and braking based on the driver command. This is accomplished by electronically controlling inverter to deliver the requested torque T at the output of motor. Upon braking command, D performs regenerative braking as long as the requested braking torque and power are within the capabilities of electric drive, and as long as the tery state of charge (SOC) is below an upper limit (SOC) max. Any excess braking is performed by conventional mechanical (hydraulic) brakes.. Control tery state of charge so that it remains between target limits ice ( SOC) SOC ( SOC. (3.5) min ) max The limits (SOC) max and (SOC) max are system design parameters, which are decided based on tery size, cost, and tery life trade-offs. This will be discussed in more detail in the section on teries. The tery SOC control is accomplished via D1 and IC. One simple SOC control strategy is to turn IC on and operate it at the maximum efficiency point whenever SOC drops to (SOC) min, and then turn IC off whenever SOC reaches (SOC) max. 3.1 Sizing of series hybrid electric drive components This section discusses basic ideas behind sizing components in the series hybrid electric drive, i.e. selecting the power ratings for IC, D1 and D, as well as the power rating and the energy storage capacity for the tery. To illustrate the ideas, a numerical example is considered in this section based on xample 1.1. Vehicle parameters are as follows: Vehicle mass M v = 15 kg CU Boulder, DM Fall 1

3 Front area A v =.16 m Aerodynamic drag coefficient C d =.6 Tire rolling resistance f rr =.1. Wheel radius r v =.3 m The performance objectives are: Acceleration time t a = 11 s from to v f = 1 km/h (6 mph). Maximum speed v max = 16 km/h (1 mph). Maximum continuous cruising speed v cmax = 13 km/h (8 mph). Gradeability: 5% at 1 km/h (6 mph) lectric motors have a maximum speed n = 5 rpm, and a maximum to base speed ratio x = n max /n b = 5. Transmission efficiency is t = 9%. fficiencies of electric motor/generators and inverter/rectifiers are assumed to be 95% each. Sizing of D components In the series hybrid of Fig. 1, the entire traction power must be provided by D. Therefore, motor/generator must be sized according to the maximum traction power requirement, which follows from the acceleration performance specification. In the considered numerical example, assuming single-gear transmission efficiency of t =.9, the required power rating for the motor/generator has been found in xample 1.1 to equal 71.5 kw. The US6 driving cycle test resulted in the maximum power requirement of 71.5/.9 = 79.4 kw. We select P 8 kw (17 hp). (3.6) mg The maximum speed of the motor/generator (n max = 5 rpm) should match the maximum vehicle speed (v max = 16 km/h = 44.4 m/s). Since n vmax = (3/)(v max /r v ) = 1413 rpm, the required gear ratio is 5/1413 = The motor/generator base speed is n b = n max /x = 1 rpm. The maximum torque T max that the motor/generator should produce for < n < 1 rpm is Pmg T max (3 / ) 764 Nm. (3.7) n Inverter/rectifier should be rated at P mg / g, which assuming 95% efficient motor/generator, gives the required inverter/rectifier power rating, b P (8 kw)/ kw. (3.8) ir Sizing of IC The internal combustion engine does not need to supply the maximum traction power. Instead, IC should be sized so that the vehicle can meet the maximum (continuous) CU Boulder, DM Fall 1 3

4 cruising speed and the gradeability performance requirements. Using (1.), the maximum cruising speed v cmax of 8 mph requires traction power of P v =.7 kw. The gradeability (5% at 6 mph) requires P v = 3. kw. Taking the larger of the two, and taking into account the assumed 95% efficiency for the series components (generator 1, rectifier 1, inverter, motor, transmission), the required IC power rating is: 4 P (3. kw)/(.9.95 ) 41kW (55 hp). (3.9) IC Note that the required IC power rating is significantly lower than the maximum required traction power. The Chevy Volt example (see Fig. 3.1) also illustrates this point. The engine downsizing is one of the HV advantages. It should also be noted that, even though each component in series is assumed to have relatively high efficiency, the cumulative effect of losses in the series HV drive results in relatively significant increase in the required IC power rating (from about 3 kw to about 4 kw in this example). This is considered one of disadvantages of the series HV drive train. Alternative parallel or parallel/series HV drive configurations will be discussed in a later section. Sizing of D1 D1 components are sized based on the IC power rating, P mg1 = mg1 P IC P IC, P ir1 = mg1 ir1 P IC P IC. HV Battery sizing In a hybrid electric vehicle (HV), tery ratings include a power rating P, i.e. the ability of the tery to supply power P while keeping the output voltage V DC above a minimum threshold, and an energy capacity rating based on the ability to supply (or absorb) energy while its state of charge SOC remains within the limits (SOC) min and (SOC) max. Charging or discharging of the tery depends on the driving cycle and the system control strategy. A simple example is considered here assuming the US6 driving cycle and the same test vehicle as before. The tery power P b equals the difference between the power P 1 supplied by IC via D1, and the traction power P delivered to the wheels via D and transmission, P b P 1 P. (3.1) Suppose that the system controller adjusts IC to provide the average required power, i.e. suppose that P 1 = (P ) avg. With the driving cycle starting at t =, and ending at t = t trip, the total energy absorbed by the tery is then b ( t trip ) t trip t trip P P d P P d 1 avg, (3.11) neglecting electric drive losses. Under the assumptions stated above, the net change in energy stored over the entire driving cycle equals zero. This means that the tery state of charge SOC(t trip ) equals the state of charge SOC() at the beginning of the trip. During the driving cycle, however, SOC changes in time as CU Boulder, DM Fall 1 4

5 b ( t) 1 SOC( t) SOC() SOC() P avg P d. (3.1) where is the tery rated energy storage capacity. From (3.1), the maximum change in SOC over the driving cycle is t b max b min SOCmax. (3.13) Given a tery specification SOC max, (3.13) can be used to find the required energy storage capacity, b max b min. (3.14) SOC Fig. 3. shows the waveforms obtained for the test vehicle in the US6 driving cycle: the vehicle speed v, the required vehicle traction power P v, and the change in energy b (t), neglecting electric drive losses. In this case, bmax bmin is found to be about.5 kwh. Assuming SOC max = 3% is allowed by the tery system design to ensure sufficiently long tery life, (3.14) gives the required tery capacity rating (in kwh), max b max b min.5 kwh 1.7 kwh. (3.15) SOC.3 max It is interesting to note that this simplified example gives a tery rating requirement close to the actual tery energy storage capacity in Toyota Prius. This tery capacity (3.15) is worth only 1.7 kwh/(1.7 kwh/kg) = 134 g of gasoline. The fact that the tery only supplies the difference between the IC power and the traction power results in very modest tery size requirements. Importantly, HVs can achieve sizable efficiency and therefore fuel economy improvements with relatively small teries. It is instructive at this point to briefly examine the tery capacity requirements in a pure tery electric vehicle (V), or plug-in HV (PHV). Battery sizing in V or PHV vehicle In a pure electric vehicle, IC is not available, and the tery must provide the total traction energy. A block diagram of a pure tery electric vehicle (V) is the same as the series HV in Fig. 3.1, except that IC and D1 are not present, and a way of charging the tery from the electric grid would be provided. Considering the same test vehicle in the US6 driving cycle, we find that the total traction energy required over the entire cycle is trip = 1. kwh. The trip distance is l trip = 8 miles, which means that trip /l trip =.15 kwh/mile of traction energy is required. Assuming that the entire traction energy is supplied by the tery, assuming the same SOC max specification (3%), and neglecting losses in the electric drive train, the vehicle with the tery capacity found in (3.13) would have an all-electric driving range l range of only CU Boulder, DM Fall 1 5

6 SOCmax.5 l range miles 3.3 miles (3.16) ( / l ).15 trip trip before the tery reaches the minimum SOC. From (3.16) it is easy to see that the range of a pure electric vehicle could be improved by deeper discharge cycles, i.e. by allowing a larger SOC max (which would however result in reduced tery life), or by increasing the tery capacity. Very similar considerations hold for plug-in electric vehicles (PHV) that are expected to operate as pure electric vehicles over a limited range of X miles (PHVX), and then have an extended driving range as needed operating as HV, with the tery recharged from an IC. In the example above, a PHV, i.e. a plug-in hybrid with miles of all-electric range would require about /3.3 = 6 times larger tery storage compared to the HV. The Chevy Volt, which is considered a PHV4, i.e. a plug-in hybrid electric vehicle with about 4 miles of electric range, has a 16 kwh tery with about 1 kwh of usable energy storage (SOC max = 65%), which means that it is expected to spend about.5 kwh per mile. CU Boulder, DM Fall 1 6

7 1 8 v [mph] time [s] 1 5 P v [kw] time [s].6.4 [kwh] time [s] Figure 3. Vehicle speed v, traction power P v, and the required change in stored energy b in the series HV example neglecting electrical drive losses and assuming that the IC supplies the average traction power at all times. CU Boulder, DM Fall 1 7