An alternative to the ICE is a fuel-cell power source, typically operating with hydrogen as the fuel. A key advantage is that emissions of carbon are eliminated as well as criteria pollutants. There are a few important distinctions to be made for fuel-cell hybrids. First, the fuel cell, like the battery, generates DC electrical power; therefore, the fuel-cell hybrid is an all-electrical vehicle. The architecture is invariably a series hybrid as shown in Figure 12.16, where all of the traction power passes through the electrical motor. Electrical power from the fuel cell can either be used to drive the electrical motor or to charge the battery. As with any series configuration, the electric motor must be sized large enough to deliver the maximum power to the wheels.
Figure 12.16 Typical architecture for a fuel-cell hybrid system.
The efficiency characteristics of a fuel-cell system are markedly different from those of an ICE. Recall from Chapters 9 and 10 that the efficiency of a fuel cell is roughly proportional to its operating potential, with higher efficiencies at high voltages and low levels of power. This trait is easy to comprehend in terms of cell polarizations. The current increases with increasing power, resulting in increases in the ohmic, kinetic, and mass-transfer polarizations. Each of these polarizations represents a loss in efficiency loss; consequently, the efficiency of the fuel cell itself increases as the power decreases.
The efficiency of a fuel-cell system, however, does not continue to increase indefinitely with decreasing power. Instead, as shown in Figure 12.17, there is a maximum in the system efficiency. We can understand this maximum by remembering that these data represent system efficiency given by
(12.7)
Some ancillary power is always needed to operate the fuel-cell system, irrespective of power level. For instance, a blower may be needed to provide air to the fuel cell. This ancillary power, as well as any electrical losses, must be subtracted from the gross electrical output of the fuel-cell stack (IV) to get the net power. The losses attributed to these ancillary devices dominate at low power and the efficiency of the fuel-cell system goes through a maximum.
Figure 12.17 Fuel-cell system efficiency (solid line) as a function of fuel-cell power. The bars represent the frequency of time spent at each power level for an example driving schedule.
Notably, this maximum in efficiency occurs at low power levels for a properly designed fuel-cell system. Compare this behavior with the engine map shown in Figure 12.11, which shows decreased efficiency at low power. Also shown in Figure 12.17 is a Pareto chart for the required traction power corresponding to a typical driving schedule. For the vast majority of driving, the vehicle is operating at part power; furthermore, the most frequent power level is well below half of the maximum power. This outcome is characteristic of nearly all driving schedules. Thus, the fuel-cell hybrid is particularly well suited for personal vehicles, which have a lot of dynamic load changes, but rarely require peak power.
A common way of operating the fuel-cell hybrid is shown in Figure 12.18, which shows the power profile for the vehicle and the fuel-cell system. The output of the fuel cell is maintained nearly constant, and the battery provides the small peaks in demand as well as being available for regenerative braking. Both the fuel cell and the battery can be used fully to meet large power demands. In contrast to the ICE, the FC has high efficiency at part power.
ILLUSTRATION 12.5
Compare the power requirements for of the engine, fuel cell, and battery for three systems: (i) ICE only, (ii) ICE battery hybrid, and (iii) fuel-cell battery hybrid. The hybrids are charge-sustaining. The average power to complete the driving schedule is 10 kW, and a maximum power of 90 kW is required. For the hybrids, size the ICE and the fuel cell to achieve maximum efficiency while supplying the average 10 kW. Use data in Figure 12.11 for the ICE, and the plot below for the fuel cell.
In systems where the ICE is the only source of power, the engine must be sized to meet the maximum power required. Energy storage is not an issue since it comes in the form of a hydrocarbon fuel with very high energy density. In this case, the engine must provide 90 kW, which corresponds roughly to the maximum value in Figure 12.11. Its maximum efficiency is near 35 kW, and at 5 kW, specific fuel consumption (300 g·Wh−1) compared to 225 g·Wh−1 at 35 kW, corresponding to a significantly lower efficiency.
For the ICE/battery hybrid, the engine can be reduced in size to provide the average power at maximum efficiency. To do this, the engine is scaled from the one shown in Figure 12.11 so that the maximum efficiency is at 10 kW. Assuming linear scaling, the maximum power of this ICE is
which is not sufficient to meet the maximum power requirement. Since, the combined power of the battery and engine must still be 90 kW,
A fuel-cell system is designed to achieve a maximum in system efficiency at roughly 10 kW as shown in Figure 12.17. The net power of the same system is shown below. The maximum power of this fuel-cell system is about 43 kW. Thus, the required battery power is
While greatly simplified, this example illustrates trade-offs between the different types of vehicles.
Figure 12.18 Typical method of operating a fuel-cell hybrid. Fuel-cell power is constant.
Closure
In this chapter, we considered hybrid vehicles, with particular emphasis on vehicles that use batteries as a rechargeable energy storage system. Efficiency gains made possible through the use of hybrids were discussed for several different hybrid strategies and architectures. The connection between the driving schedule, vehicle characteristics, and power requirements was also examined in the context of a driving schedule and a vehicle model. Hybrid vehicles that utilize combustion engines offer significant efficiency advantages as a result of energy recovery through regenerative braking and increased engine efficiency through decoupling of the engine speed from that of the vehicle. The use of electrochemical capacitors and fuel cells in hybrid vehicles was also examined. Finally, the power and energy requirements of the energy storage system were determined for different levels of hybridization. Because of the increased efficiency of hybrid vehicles, they will likely play an important role in future transportation.
Note
- Sometimes the start-stop hybrid is referred to as a micro-hybrid. The micro-hybrid is differentiated by allowing recovering of energy from braking, where a start-stop hybrid does not include energy recovery.
Further Reading
Ehsani, M., Gao, Y., and Emadi, A. (2009) Modern Electric, Hybrid Electric, and Fuel Cell Vehicles, CRC Press, Boca Raton.
Husain, I. (2011) Electric and Hybrid Vehicles: Design Fundamentals, CRC Press, Boca Raton.
Scrosati, B. Garche, J. and Tillmet, W., eds., (2015) Advances in Battery Technologies for Electric Vehicles, Woodhead Publishing.
Problems
12.1. What are some of the reasons that hybrid and all-electric vehicles don’t rely solely on regenerative braking?
12.2. Using the same vehicle information as found in Illustration 12.1, but rather than constant power, assume that the rate of deceleration is constant. Find the braking power as a function of time. If the maximum charge rate of the battery is 5 C, and the motor/generator is limited to 50 kW, what fraction of the kinetic energy can be recovered?
12.3. A small SUV has a mass of 1750 kg and a battery with a 2 kWh capacity. If the energy stored in the battery is converted with 100% efficiency to potential energy, determine the change in elevation associated with the 2 kWh of energy.
12.4. In Section 12.4, a rule of thumb was provided: 1 kWh propels a vehicle for about 6 km. Using data for aerodynamic drag, rolling resistance, and accessory power, calculate the vehicle speed where the vehicle uses exactly 1 kWh of energy to travel 6 km. What other factors will affect this value?
12.5. Use the rule-of-thumb of 6 km·kWh−1, configure a battery for an all-electric vehicle with a range of 150 km. The design voltage of the battery is 300 V, which is made from 2S-4P modules. Each of the cells has a nominal voltage of 3.8 V and a cutoff potential of 3.2 V. How many modules are required? What is the required capacity of each cell? If the peak power is 75 kW, what is the maximum resistance of each cell in mΩ?
12.6. A battery for a power-assist HEV is required to deliver 25 kW and 300 Wh. The characteristics of the cell are a capacity of 12.2 A·h·m−2, a resistance of 2 mΩ·m2, and an open-circuit potential given by U = 3.8-B (1-SOC). Determine the separator area required by the battery if B = 0.2 V and the allowable change in SOC is 0.3. What if the permitted change in SOC is only 0.2? Explain the results.
12.7. Describe the main differences between series- and parallel-hybrid drive trains. Typically, series-drive trains have larger batteries and smaller engines compared to parallel architectures, why? Which would be preferred in city stop-and-go driving? Highway driving?
12.8. Design an EDLC system that would be appropriate to absorb the energy required to stop a 1325 kg vehicle moving at 50 km·h−1? Assume that multiple cells are place in series to achieve a maximum voltage of 450 V. The capacitance per unit superficial area is 800 F·m−2, and the voltage per cell is 3.0 V.
12.9. Using a single EDLC module described in Illustration 12.3, estimate the standby time permissible if there is a self-discharge current of 3 mA. At the beginning of the standby period, the potential was 13 V and the starting power required is 5 kW.
12.10. A key difference between a start–stop hybrid and a microhybrid is that energy is recovered. How much kinetic energy does a vehicle with a mass of 1520 kg traveling at 45 km·h−1 have? Express your answer in Wh. The RESS is a number of EDLC modules of 230 F arranged in parallel, with a maximum voltage of 15 V. The braking event takes 10 seconds and at the start the potential of the EDLC is 8.8 V. If the regenerated power is constant, what is the maximum current? How many modules are needed to recover this energy?
12.11. In Illustration 12.3, the lead–acid battery is replaced with a lithium-ion battery that has a voltage of 42 V, a turnover capacity of 400, and a specific energy of 130 W·h·kg−1. What is the size of the battery required in kg?
12.12. Identify three advantages and three disadvantages of full-hybrid vehicles. How would improvements in energy storage technology mitigate these disadvantages?
12.14 Cells have an average potential of 3.8 V, a loading of 31 A·h·m−2, and a resistance of 40 mΩ·m2. The maximum battery voltage is 300 V, and the cutoff potential for the cells is 3.1 V. Finally, the cell size is limited to 30 A·h. Configure a battery for a vehicle with an all-electric range of 100 km. What is the maximum power of the battery? Would there be an advantage to increasing the maximum battery voltage?
12.14. A fuel-cell hybrid vehicle is required to complete a specified driving schedule. A simplified representation of the power needed for traction and parasitic power is shown in the table. Assume that the system is operating in charge-sustaining mode, and that the cycle is repeated many times. The battery is limited to being discharged at 5 C, and the rate of charging is limited to 2 C. Suggest how to size the battery and fuel cell to accomplish this profile. Comment on the effectiveness of the control strategy outlined in Figure 12.18. There is not a “right” answer, but you need to justify your choices.
Time [s] Power [kW]
30 3
10 40
58 25
14 −10
34 8
16 60
40 20
18 −5
38 15
14 30
26 10
12.15. Using data from the engine map (Figure 12.11), sketch the efficiency of the engine as a function of power. Use a heating value of the fuel of 44 MJ·kg−1 Compare this efficiency versus power with that for a fuel-cell system (Figure 12.17). How might hybrid designs and operation differ?
12.16. Calculate the additional power needed to sustain 90 km·h−1 up a 6% grade compared to level ground. Neglect the change in rolling resistance. The vehicle mass is 1700 kg.
12.17. A vehicle requires 12 kW of average power, and 70 kW maximum power to complete a typical driving schedule. If 22 kW additional power is required to sustain 90 km·h−1 up a 6% grade, what is the maximum degree of hybridization?
12.18. Using the start–stop information from Illustration 12.3, for two EDLC modules how many attempts to start the vehicle can be made after a period of stopping for 45 seconds?
12.19. Using data from the table, determine power to keep vehicle moving at different speeds. Express these answers in terms of km·kWh−1 of battery power assuming 80% efficiency. The frontal area is 0.6 m2, Cd = 0.50, and fr = 0.016. Assume parasitic power is 1 kW.
Speed, km·h−1 60 60 90
Mass, kg 1300 1400 1400
12.20. How much power is needed to accelerate a 1500 kg vehicle from 0 to 90 km·h−1 in 10 seconds? What about power to stop the vehicle in 4 seconds.
12.21. A battery for a hybrid electric vehicle has 80 cells connected in series. Treat the battery as ohmically limited with U = 3.8 V, and Vmax = 4.2 V. What is the maximum value of resistance that would still allow recovery of 80 kW during braking? What is the corresponding value of joule heating during this braking?
12.22. A large hybrid-electric transit bus is being designed. It is desired to recover the kinetic energy during stopping. The fully loaded mass is 20,000 kg, and the bus is estimated to make 200 stops per day from 50 km·h−1, each in 10 seconds. Using information provided below, first size a battery and then an EDLC for 1 year of operation, and complete the table below. The maximum allowable voltage is 600 V.
For the battery, the . For the EDLC, assume that the size must be increased by 20% to meet life requirement.
Appendix: Primer on Vehicle Dynamics
The fundamental equation is Newton’s second law of motion:
(A.1)
F is the net force acting on the vehicle of mass Mv. The forces acting on the vehicle include the tractive force, road resistance (including the effect of the grade), and aerodynamic drag. The tractive force Ft is related to the torque produced by the internal combustion engine or the electric motor:
(A.2)
Tp is the torque, ig is the gear ratio, ηt is the mechanical efficiency of the drivetrain, and rd is the effective radius of the wheels. At steady speed, this tractive force is balanced against the frictional force associated with rolling the wheels, gravity, and aerodynamic drag. Both the forces of rolling resistance, Ff, and gravity, Fg, depend on the weight of the vehicle and are conveniently combined. For a vehicle of mass, Mv, climbing a grade with an angle α, the road resistance is
(A.3)
fr is the coefficient of rolling resistance and g is the acceleration of gravity. The aerodynamic drag is
(A.4)
ρa is the density of air, CD is the drag coefficient, and Af is the projected area of the vehicle.
The sum of these forces are used with Equation A.1:
(A.5)
Equation A.5 forms the basis of a dynamic model that can be used to simulate a driving schedule. When the velocity is constant, the net force is zero, and the required torque can be calculated:
(A.6)
From Equation A.5 we can calculate the time, energy, and instantaneous power for a prescribed vehicle velocity versus time. Below we consider a highly simplified case. The time required to accelerate uniformly from speed 1 to 2. Since the force is constant, Equation A.5 is integrated.
(A.7)
The distance travel during this period of constant acceleration is
(A.8)
Thus, if the distance S, the two speeds (v1 and v2), and the grade (α) are specified, the required torque of the ICE or electric motor can be determined. Since the torque is a function of engine or motor speed, which is related to vehicle speed and gear ratio, the integration of Equation A.8 must be done numerically.
ILLUSTRATION A.1
A 1350 kg passenger vehicle is moving at 90 km·h−1 on a level grade.
Determine the power needed to overcome aerodynamic drag and rolling resistance. The following data are provided: Af = 0.58 m2, ρa = 1.188 kg·m−3, CD = 0.25, and fr = 0.015. The speed of 90 km·h−1 is 25 m s−1.
Using Equation A.4,
The power to overcome aerodynamic drag is Fwv = 1.35 kW. The force due the road is given by Equation A.3.
The associated power is Frdv = 4.96 kW.
How much less power is needed to maintain this speed going down a grade of 6%?
Again, using Equation A.3
The difference is
In this case, the force from the acceleration of gravity is greater than the forces of rolling resistance and aerodynamic drag. You would need to brake (and recover energy with a hybrid) to maintain a speed of 90 km·h−1. Multiplying the right side of Equation A.6 by the speed will give the power available to be stored.
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