Russael Posted April 16, 2014 at 02:52 PM Report Share Posted April 16, 2014 at 02:52 PM In Hybrid Drive, do you mean the ICE just floats the battery and propels the car by itself? If so, I wonder what the speed break even point is in efficiency for having the powertrain start flipping back and forth between an EV mode and charging the battery vs just having the ICE propel the car... Quote Link to comment Share on other sites More sharing options...
larryh Posted April 16, 2014 at 07:09 PM Author Report Share Posted April 16, 2014 at 07:09 PM (edited) I am referring to the power flow display on MFT accessed by selecting the EV Info button and then the power tab. The Status indicates either “Hybrid Drive” or “Charging HV Battery” when the ICE is on while driving. When it indicates “Hybrid Drive”, then either the generator or the motor is generating electricity while the other is consuming it. Usually, the motor is consuming combined power from the ICE and generator to generate electricity, while the generator is consuming electricity to assist the ICE in providing power to the wheels and the motor (acting as a generator). In this configuration, the motor may generate more electricity than the generator consumes. The surplus electric power is used to charge the HVB. Alternatively, the generator may use more electricity than the motor generates. The electricity power deficit is supplied by the HVB. Usually the surplus or deficit in electricity is only a few kW. I have not observed both the generator and motor acting together, along with the ICE, to provide power to the wheels. I have only observed the generator and motor to act together to generate electricity during regenerative braking. When the car decides to aggressively charge the HVB, the status changes to “Charging HV Battery”. Now both the motor and the generator are consuming power from the ICE to charge the HVB. They may produce 10 kW or more of electric power. Determining whether less gas will be consumed by constantly running the ICE in “Hybrid Drive”, or by repeatedly running the ICE in “Charging HV Battery” mode and then turning off the ICE while allowing EV operation to consume the energy generated while in “Charging HV Battery” mode is a very complex task. Hopefully, the car has been programmed to do the right thing. Edited April 16, 2014 at 07:12 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 18, 2014 at 07:52 PM Author Report Share Posted April 18, 2014 at 07:52 PM (edited) I updated my post showing the efficiency of the motor/generator with better data at: http://www.fordfusionenergiforum.com/topic/1880-obd-ii-data-for-ice/?p=13034 During normal operation in EV mode, the generator is generating electricity and the motor is consuming it. The amount of electricity consumed by the generator is a function of generator rpms. The faster the generator spins, the more torque it consumes to generate electricity up to 6000 rpm and then it levels off. From the Motor/Generator map, you can see that efficiency increases significantly with increasing torque. I assume the engineers are trying to increase the torque on the electric motor to force it to operate more efficiently. I would have expected them to both work together to power the car rather than opposing each other. During regenerative braking, they both work together to generate electricity. Edited April 18, 2014 at 07:55 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 18, 2014 at 08:42 PM Author Report Share Posted April 18, 2014 at 08:42 PM (edited) The following plot is a Motor/Generator Map that shows the efficiency of the motor/generator for various RPMs and Torque during regenerative braking. Generally, the efficiency is from 85% to 98%, i.e. if 1 kW of mechanical power consumed by the motor/generator, the motor/generator will produce between 0.85 and 0.98 kW of electrical power. This map is a rough approximation of the true map. I am unable to get the measurements that are required for an accurate map and have to make several assumptions/approximations. The efficiency increases with increasing Torque. The blue markers show actual operating points of the motor/generator during my trips. The power output required for the upper right portion of the map is beyond the capabilities of the motor/generator. The maximum electrical power that can be applied to the HVB through regenerative braking is 35 kW. If you brake harder than that, the friction brakes absorb the additional power. The curved lines in the map show the mechanical power consumed by the motor/generator. The lines range from 5 kW to 40 kW from left to right. A motor speed of 8400 rpm corresponds to about 60 mph. Note I can't isolate the generator from the motor, they both work together to stop the vehicle. The motor and generator work together to consume mechanical power and produce electrical power. The electric motor does most of the work. The optimal strategy to maximize the energy stored in the HVB during regenerative braking is to apply the brakes so that the generator/motor consume as much power (up to 35 kW) for as long as possible. But that would require very precise timing. You would be applying gentle braking at the beginning of the stop and increasing the amount of braking until the car comes to a stop. If you don't get the timing right, you won't stop where you planned, i.e. running into the car in front of you. When you get 100% braking score, it means that the friction brakes were not used. It doesn't mean that you maximized the amount of energy that could have been stored in the HVB. The curved line of markers clumped near the bottom is regen resulting when you remove your foot from the accelerator. The efficiency of this regen is about 75%. The markers above this line result from regen when pressing the brake pedal. Edited April 18, 2014 at 08:50 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
Hybridbear Posted April 19, 2014 at 09:03 PM Report Share Posted April 19, 2014 at 09:03 PM So basically your chart shows that taking your foot off the gas & coasting is an inefficient use of the kinetic energy of the vehicle because less of it is stored in the HVB. A better braking strategy is to wait until the last possible moment and then brake near the 35 kW limit for regen for a short period. I guess this really shows how coasting in Neutral helps improve MPGs by removing that inefficient regen from coasting. Quote Link to comment Share on other sites More sharing options...
larryh Posted April 19, 2014 at 09:45 PM Author Report Share Posted April 19, 2014 at 09:45 PM (edited) So basically your chart shows that taking your foot off the gas & coasting is an inefficient use of the kinetic energy of the vehicle because less of it is stored in the HVB. A better braking strategy is to wait until the last possible moment and then brake near the 35 kW limit for regen for a short period. I guess this really shows how coasting in Neutral helps improve MPGs by removing that inefficient regen from coasting.That is correct. You want to apply maximum braking up to the 35 kW regen limit to capture as much of the lost kinetic energy as possible. But that requires very precise timing and judgement. I would need a lot of practice to get that right. You would have to shift to neutral to coast and then start applying the brakes at the proper distance from a red stop light with the right pressure. If you manage to stay near the 35 kW regen limit, you will have captured at least 90% of the energy available to the motor/generator. At 55 mph, when I place the car in neutral, the motor and generator consume 2.85 kW of mechanical power and generate about 0.2 kW of electricity which is not distinguisable from noise. But that is much less than the 10 kW of mechanical power consumed by the motor and generator to produce about 7.6 kW of electrical power when coasting in drive. I placed the car in neutral this afternoon to determine how much power to the wheels is required to maintain 55 mph. When I shifted to neutral, deceleration was -0.2064 m/s. Assuming the car and contents weigh about 1850 kg, that amounts to about 9.16 kW of power to overcome regen, rolling resistance, and aerodynamic drag. Regen consumed 2.85 kW, so that means the power required to overcome rolling resistance, aerodynamic drag, etc., is about 6.32 kW. The regen/friction from the car's drivetrain is a significant fraction of the total power loss, about 30%. Rolling resistance, aerodynamic drag, etc. make up the remaining 70%. I looked at the power output of the motor/generator to maintain 55 mph. It was about 8.0 kW. This implies that about 80% of the power from the motor/generator makes it to the wheels. Based on this new information, I updated the following post: http://www.fordfusionenergiforum.com/topic/1880-obd-ii-data-for-ice/?p=13038 Whatever the car is reporting for power at the wheels can't be correct. It can't require 14.3 kW to overcome aerodynamic drag and rolling resistance. It should be around 9.2 kW. Edited April 21, 2014 at 11:23 AM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 20, 2014 at 12:17 AM Author Report Share Posted April 20, 2014 at 12:17 AM (edited) When in Low at 55 mph after lifting my foot from the accelerator, I observed 29 kW of regen. The regen limit is 35 kW. So Low does not provide the maximum possible deceleration via regen. You can still apply the brakes and keep the regen power within the 35 kW limit. Regen efficiency was about 94%. Note a better strategy to maximize regen might be to shift into Low early before the stop light and gently apply the brakes. After the speed drops significantly, shift back to drive, coast, and apply the brakes normally as required. Edited April 20, 2014 at 03:35 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 20, 2014 at 11:30 AM Author Report Share Posted April 20, 2014 at 11:30 AM (edited) The following chart shows the regen power applied to the HVB I observed in Low for various speeds: mph regen (kW)55 -28.8226 50 -28.5724 45 -27.4783 40 -26.1635 35 -24.5969 30 -22.1688 25 -19.9991 20 -16.4015 15 -13.097 10 -10.6333 5 -2.58271 0 0 Edited April 20, 2014 at 11:32 AM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 21, 2014 at 11:26 AM Author Report Share Posted April 21, 2014 at 11:26 AM Apparently, when the car is in neutral you do not get any regen. If you apply the brakes when in neutral, the friction brakes are used to stop the car and none of the kinetic energy is stored in the HVB. Quote Link to comment Share on other sites More sharing options...
murphy Posted April 21, 2014 at 12:38 PM Report Share Posted April 21, 2014 at 12:38 PM By definition Neutral means there is no connection between the power source (ICE/electric motor) and the wheels. The wheels can't turn the generator if there is no connection. ;) Quote Link to comment Share on other sites More sharing options...
larryh Posted April 21, 2014 at 01:02 PM Author Report Share Posted April 21, 2014 at 01:02 PM (edited) I'm not sure if neutral actually terminates the connection to the wheels. The car reports the generator and motor rpms proportional to the wheel rpms regardless of the gear selected, even neutral. The relationship is Generator/motor rpms = 139.65 x vehicle speed The sign for generator rpms is negative and for motor rpms it is positive. In addition, the car reports a small amount of torque at both the generator and motor in neutral. In EV mode, the generator torque is always a function of generator rpms, regardless of gear. In neutral only, motor torque is a function of motor rpms. I'm not quite sure how to interpret the data. Edited April 21, 2014 at 03:49 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 21, 2014 at 09:22 PM Author Report Share Posted April 21, 2014 at 09:22 PM (edited) According to the following article, http://wikicars.org/en/Hybrid_Synergy_Drive, when in neutral, the electrical connection to the both the generator and motor are broken so that they may now spin freely. So I am not sure why I am observing torque at both the generator and the motor. They are not generating any electrical power. Both the generator and the motor are consuming up to about 1.3 kW of mechanical power and not doing anything with it. I'm not sure if the car is simply measuring energy losses due to friction, transmission fluid viscosity, and other sources. I have read that a CVT transmission is less efficient than standard transmissions. A standard transmission is about 80% efficient. Also, according to the article, the generator rotates freely when in EV mode. It doesn't consume or produce any electricity. I observe an identical power loss at the generator as I observe when the car is in neutral. In all the charts that I have presented, I have included this loss in the calculations. If I did not include it, then the regen efficiency wouldn't make any sense. It would appear that the car is generator more electrical power than it is consuming from mechanical power for an efficiency exceeding 100%. Edited April 21, 2014 at 09:24 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
viajero Posted April 21, 2014 at 11:22 PM Report Share Posted April 21, 2014 at 11:22 PM By definition Neutral means there is no connection between the power source (ICE/electric motor) and the wheels. Not with the power split device. It's not really a transmission, continuous or otherwise. No gear ratios ever change. As larryh has observed, whether in D, N, or even R the mechanical connections are the same. All the "shift lever" does is tell the computer what the driver wants to do. The computer then decides how much electrical load or power to apply to the motor/generators to make that happen. The MG2 can spin backwards to make the car go backwards, but the connection between the motor and the wheels is always the same. This has been linked a few times before, but a good animation to play around with and see how it works. http://eahart.com/prius/psd/ Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 22, 2014 at 09:24 PM Author Report Share Posted April 22, 2014 at 09:24 PM (edited) I observed the following power measurements this morning for my commute on the Freeway going about 65 mph in EV Later mode with the power flow screen showing Hybrid Drive: Mechanical Power ConsumedAverage power applied to wheels: 19.1 kW.Average power applied to motor: 7.4 kWTotal: 26.5 kW Mechanical Power GeneratedAverage power output from ICE: 21.0 kWAverage power output from generator: 5.4 kWTotal: 26.4 kW Electrical Power ConsumedAverage power consumed by accessories: 0.5 kWAverage power applied to HVB: 0.6 kWTotal: 1.1 kW An detailed explanation of how the FFE's powertrain works can be found here: "http://prius.ecrostech.com/original/PriusFrames.htm". A concise summary follows: In order for the ICE to transfer power to the wheels, the generator must resist the force transferred to it through the planetary gear system by the ICE, otherwise the power from the ICE would simply all go to the generator and not to the wheels. Without any resistance from the generator, increasing ICE RPMs would simply cause the generator to rotate faster in the positive direction, the same direction as the ICE. Power transfer takes the path of least resistance. It takes a lot of power to maintain RPMs at the wheels due to rolling resistance, aerodynamic drag, etc. If the generator did not offer any resistance, it would get all the RPMs and power the ICE has to offer instead. If the generator were rotating at -5000 rpm before the ICE turned on, it would now rotate at maybe -1000 rpm instead with the ICE running if it offered no resistance. We need to force the generator to rotate faster in the negative direction, and work against the ICE, to transfer the power from the ICE to the wheels. The only way to do that is to have the generator act like a motor and push harder in the negative direction, i.e. make RPMs more negative. The generator will need a constant supply of electricity to do that. We can't rely on the HVB to provide that power since it would eventually run out of energy. Instead, the motor acts as a generator to provide that electricity. In the data from my commute, you can see the generator is acting as a motor to generate 5.4 kW of mechanical power consuming about 6 kW of electrical energy to increase generator RPMs in the negative direction The motor consumed 7.4 kW of mechanical power to generate 6.7 kW of electricity (and reducing RPMs at the wheels). This is the typical mode of hybrid operation on the highway. If the ICE rotates fast enough and you don't go too fast, eventually the ICE will force the generator RPMs to become positive. Now in order to resist the ICE and rotate faster in the negative direction, we can simply let the generator act like a generator, reducing the RPMs of the generator towards zero, and generate electricity. This normally only happens when going up a hill, passing, or the car decides it needs to charge the HVB. You will hear the engine surge as the ICE increases RPMs. Edited April 22, 2014 at 09:58 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 24, 2014 at 11:25 PM Author Report Share Posted April 24, 2014 at 11:25 PM (edited) The following is a summary of my observations of the Fusion Energi’s powertrain based on the OBD II data that I have observed. RPM Relationships I am able to measure the rotation speed of the motor, the generator, the ICE, and the wheels. Define these quantities as follows: M = rotation speed of motor (rpm)G = rotation speed of generator (rpm)E = rotation speed of ICE (rpm)W = rotation speed of wheels (rpm) The relationships that I observe between these quantities are as follows: M + G = 3.55*EM = 10.394*W The sum of the rotational speeds of the motor and generator is 3.55 times the rotation speed of the ICE. The rotation speed of the motor is 10.394 times the rotation speed of the wheels (actually, the more precise equation is 0.9948*M + G = 3.55*E). For a picture of powertrain components for a Prius, see http://prius.ecrostech.com/original/PriusFrames.htm. The Fusion Energi should be similar. However, it does not have a chain drive like the Prius. From these equations, we can compute the final drive ratio by assuming the generator speed is zero, i.e. G = 0. We then get: 3.55*E = M + 0 = M = 10.394*W, or E = (10.394/3.55)*W = 2.93*W, which is close the actual final drive ratio of 2.91. The rpm of the motor is related to the vehicle's speed by motor rpm = 139.65 * vehicle mph. Torque Relationships I am able to measure the torque supplied by the generator, motor, ICE, and the torque at the wheels. From these, I can indirectly compute the torque at the ring gear of the “Power Split Device” (as Toyota calls it) which attaches to the motor (see http://prius.ecrostech.com/original/PriusFrames.htm). Define these quantities as follows: Tm = torque of motor (Nm)Tg = torque of generator (Nm)Te = torque of ICE (Nm)Tr = torque at ring gear (Nm)Tw = torque at wheels (Nm) Te = -3.53*TgTe = 3.56*Tr10.394*(Tr + Tm) = Tw Ideally, I would have expected Te = -3.55*Tg and Te = 3.55*Tr. They are close. For the Prius, the relationship between the RPMs is as follows: G + 2.6*M = 3.6*E,M = 3.905*W The Fusion’s motor rotates about 2.66 times faster than the Prius’ motor. Note that when the ICE is off, i.e. E = 0, the Fusion’s generator and motor RPMs are equal, but opposite in sign. For the Prius, the generator rotates 2.6 times faster than the motor (the signs are also opposite). For the Prius, that is because the sun gear, which is directly attached to the generator, is smaller than the ring gear, which attaches to the motor. For this configuration to work in the Fusion, the sun gear and the ring gear would have to be the same size for the RPMs to be equal. However, that is impossible. The sun gear must be smaller than the ring gear for it to fit inside the ring gear. The Power Split Device for the Fusion must be more complex than for the Prius. There must be additional gears involved. Edited April 25, 2014 at 06:18 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 25, 2014 at 12:54 AM Author Report Share Posted April 25, 2014 at 12:54 AM (edited) Hybrid Drive Using the equations in the previous post, we can determine how the car operates at 65 mph on the freeway in hybrid drive. In order to go 65 mph, the wheels will need to rotate at 870 rpms given the car’s tire size. In addition, to maintain that speed, the car will need to apply 192 Nm of torque at the wheels. That’s a total power output of (2*pi*(870/60)*192)/1000 = 17.5 kW to the wheels. The motor will need to rotate at 10.394*870 = 9043 rpm. The car computes the optimal RPMs and torque to run the ICE. For this particular drive, it has chosen 2000 rpm at 98 Nm of torque, for a power output of 20.5 kW. The generator speed is 3.55*2000 - 9043 = -1943 rpm. The generator torque is -98/3.53 = -27.8 Nm. The generator must push the against the ICE’s positive rotational force to maintain its -1943 rpm with a force of 27.8 Nm, acting as a motor by consuming electricity to output 5.6 kW of mechanical power. Both torque and rpms are negative. That means the generator needs to supply mechanical power, attempting to increase the negative rpms. The torque at the ring gear is 98/3.56 = 27.5 Nm. The motor must supply the following amount of torque: 192/10.394 – 27.5 = -9.0 Nm. The motor must resist the positive rotational force from the ring gear to maintain speed, acting as a generator by consuming 8.5 kW of mechanical power to generate electricity. In this instance, the rpms are positive and the torque is negative. That means the motor is consuming mechanical power, attempting to slow down the positive rpms. So we have the following: Source RPMs Mechanical PowerMotor 9043 -8.5 kWGenerator -1943 5.6 kWICE 2000 20.5 kW The total power supplied by the sources is 17.6 kW, which matches the power required at the wheels of 17.5, taking into account roundoff errors. -8.5 + 5.6 = -2.9 kW of mechanical energy from the motor and generator is being used to generate electricity. In this instance, about 2 kW of electric power are generated and efficiency is about 70%. Edited April 25, 2014 at 06:21 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 25, 2014 at 01:16 AM Author Report Share Posted April 25, 2014 at 01:16 AM (edited) EV Drive When in EV mode, the ICE is fixed and does not rotate. That implies that the generator and motor rotate at the same RPMs, but in opposite directions. Ideally, the torque on the ICE, generator, and ring gear are zero. The generator should be rotating freely and not offering any resistance. It is neither consuming nor generating electricity. All the power from the motor should be going power to the wheels. In actuality, the following happens at 55 mph: W = 744 (Wheel rpm)M = 7732 (Motor rpm)G = -7700 (Generator rpm)E = 0 (ICE rpm) Tm = 21.8 Nm (Torque at motor)Tg = 2.2 Nm (Torque at generator)Te = 0 Nm (Torque at ICE) There is non-zero torque at the generator, i.e. the generator is not spinning freely. Since the generator RPMs are negative and the torque is positive, the generator is resisting rotational speed, trying to reduce RPMs and consuming 1.8 kW of mechanical power, probably wasting it in the form of heat. The motor will have to generate that much more power to overcome the power drain of the generator. Edited April 25, 2014 at 08:48 AM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 25, 2014 at 08:27 AM Author Report Share Posted April 25, 2014 at 08:27 AM (edited) Regenerative Braking During regenerative braking, the ICE is usually off. So again, the generator and motor rotate at the same RPM, but in opposite directions. Only the motor is used to generate electricity. Ideally, the generator is just spinning freely. But in actuality, as in EV mode, the generator is resisting rotational speed due to friction and other losses, resulting in less power to the motor for regen. If the generator were to generate electricity during regenerative braking, since the RPMs are negative, the torque would have to be positive, slowing down the negative rotational speed. But if the generator torque were positive, then ICE torque would have to be negative. The ICE can only rotate forwards and provide positive torque. The only other alternative is to lock the ICE and planet carrier driving the ICE in place so that they don’t rotate and the generator could function as a generator. But that doesn’t appear to be what the car is doing. When the car is in neutral, the electrical connection to the motor and generator is broken. Neither can consume nor generate electrical power. When the ICE is off, I observe the torque at the generator to be solely a function of generator rpms and resisting the rotation, which is most likely due to the friction and other losses. I observe the same torque at the generator as a function of generator rpms no matter what gear the car is in or whether I am applying the brakes or not. If the car were using the generator during regenerative braking or while in Drive or Low, then the torque on the generator should be different from what it is in neutral, but this is not the case. Edited April 25, 2014 at 06:29 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 25, 2014 at 10:59 AM Author Report Share Posted April 25, 2014 at 10:59 AM (edited) Generator I’m not sure why the generator is called a generator. Its main purpose seems to be starting the ICE and controlling torque delivered by the ICE. It is not really used as a generator—the motor does the regenerative braking. In addition, the generator has no useful function when the ICE is off. During hybrid drive, the generator is usually acting as a motor to control torque from the ICE. It will, however, act as a generator if the ICE rpm is great enough or the car is moving slowly enough. But it is only doing so because that is what is required to control the torque from the ICE. In the previous example, the ICE was rotating at 2000 rpm, the motor at 9043 rpm, and the generator at -1943 rpm. If we increase the rpm of the ICE to 9043/3.55 = 2547 rpm or more, then the generator rpm would now be positive. For example, if the ICE rpm were 2600, then the generator rpm would be +187 rpm. The generator torque must still remain negative. However, now the torque is opposing the direction of rotation. The generator now acts like a generator consuming mechanical power to convert that power to electrical power. High ICE rpm usually only occurs when the car needs to charge the HVB, during acceleration, or when climbing hills. The same effect can be achieved by driving at slower speeds. Instead of driving 65 mph, if I drive at 35 mph, I observed the following: Source RPMs Mechanical PowerMotor 5443 -10.5 kWGenerator 366 - 1.2 kWICE 1636 19.8 kW Now that the car is going slow enough that the generator rpm is positive (without excessive ICE rpm). The generator is generating electricity. Note, that the motor is generating the majority of the electricity in this instance. In the case of the ICE idling at 1500 rpm when the car is not moving, all the power from the ICE will be directed to the generator to generate electricity. The generator rpm will be 3.55*1500 = 5325 rpm and the motor rpm will be 0. The motor receives no power and neither generates nor consumes electrical power. Edited April 26, 2014 at 11:48 AM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 25, 2014 at 02:10 PM Author Report Share Posted April 25, 2014 at 02:10 PM (edited) The following table shows for various vehicle speeds the minimum ICE rpm that will force the generator to act as generator and generate electricity. If the ICE rpm is below this threshold, then the generator will instead be acting as a motor and consume electricity. motor ICEmph rpm rpm10 1397 39320 2793 78730 4190 118040 5586 157450 6983 196760 8379 236070 9776 275480 11172 3147 If you travel at or below 40 mph, the ICE rpm will likely be above the threshold. If you go faster than 60 mph, the ICE does not often reach the required threshold. Edited April 25, 2014 at 02:15 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 25, 2014 at 08:33 PM Author Report Share Posted April 25, 2014 at 08:33 PM (edited) The Fusion’s motor rotates about 2.66 times faster than the Prius’ motor. Note that when the ICE is off, i.e. E = 0, the Fusion’s generator and motor RPMs are equal, but opposite in sign. For the Prius, the generator rotates 2.6 times faster than the motor (the signs are also opposite). For the Prius, that is because the sun gear, which is directly attached to the generator, is smaller than the ring gear, which attaches to the motor. For this configuration to work in the Fusion, the sun gear and the ring gear would have to be the same size for the RPMs to be equal. However, that is impossible. The sun gear must be smaller than the ring gear for it to fit inside the ring gear. The Power Split Device for the Fusion must be more complex than for the Prius. There must be additional gears involved. If you look at page 8 of the document at this link https://www.motorcraftservice.com/vdirs/diagnostics/pdf/OBDSM1303_HEV.pdf, you can see how the generator, ICE and (traction) motor are connected to the planetary gear set. The generator is directly connected to the sun gear (in blue). The ICE is directly connected to the planet carrier (connecting the red gears). The motor is indirectly connected to the ring gear (the large green gear) via intermediate gears (also shown in green). This is the same as the Prius, except that the motor in the Prius is directly connected to the ring gear. It appears for the FFE, the intermediate gears connecting the ring gear and the motor reduce the gear ratio for the motor (relative to the ICE) to be the same as the gear ratio for the generator. The ring gear is much larger than the sun gear. Probably the sun gear and gear attached directly to the motor have the same radius. That page also describes the Positive and Negative Split Modes of operation. Positive Split Mode is when the generator rpm is positive and the generator is consuming some of the mechanical power output from the ICE to generate electricity. So the ICE is producing more power than it needs to power the wheels. Negative Split Mode is when the generator rpm is negative and the generator consumes electrical energy to generate mechanical power. The planetary gear set combines the power from the generator and the ICE and feeds it to the motor. Since the generator is producing some of the power, the ICE does not have generate all of it and can run at reduced power. Well that is not quite true, since power from the planetary gear set must provide sufficient power to run the motor as a generator to generate enough electricity to run the generator and to power the wheels. Ultimately, this configuration will result in the ICE producing more power than is required to directly power the wheels. But hopefully, it is operating in a more efficient operating region and using less fuel than would be required for a direct connection to the wheels.. According to the document referenced above, this allows the ICE to run at reduced speed (and consequently higher torque). I suspect the power flow screen indicates the mode of operation. When it says "Charging HV Battery", it is operating in the Positive Split mode and the generator rpm is positive. When it says "Hybrid Drive", it is operating in Negative Split mode and the generator rpm is negative. Edited April 25, 2014 at 10:55 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 26, 2014 at 11:48 AM Author Report Share Posted April 26, 2014 at 11:48 AM (edited) Perhaps a better description for the role of the generator is that it used to “shift gears on the transmission” (see the following article http://prius.ecroste...s/Overdrive.htm). Recall, that the ICE rpm is determined as follows: E = (M + G)/3.55, where E = rpm of ICE, M = rpm of motor, and G is rpm of generator. If you want a lower gear with the ICE delivering power with higher rpm and lower torque, increase the rpm of the generator in the positive direction. If you increase the generator rpm in the positive direction by 35.5 rpm, the ICE rpm will increase by 35.5/3.55 = 10 rpm. The lowest gears are achieved by making the generator rpm positive (called Positive Split mode of operation). In this mode, the generator runs like a generator, resisting the forward torque applied to it from the ICE, converting mechanical energy to electrical energy, and allowing the ICE to provide power to the wheels. If you want a higher gear with the ICE delivering power with lower rpm and higher torque, increase the rpm of the generator in the negative direction. If you increase the generator rpm in the negative direction by 35.5 rpm, the ICE rpm will decrease by 35.5/3.55 = 10 rpm. The highest gears, including overdrive, are achieved by making the generator rpm negative (called Negative Split mode of operation). In this mode, the generator runs like a motor, providing power to the planetary gear set, converting electrical energy to mechanical energy, and allowing the ICE to provide power to the wheels. The motor is run as a generator to power the generator. Edited April 26, 2014 at 12:02 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 26, 2014 at 06:40 PM Author Report Share Posted April 26, 2014 at 06:40 PM (edited) I found the following in a textbook describing planetary gears. It verifies the model I proposed earlier, derived empirically from recorded data, is correct. Here is the derivation for those who are interested in all the technical details: Sun Gear. Ring Gear Planet Carrier. Planet Gear. The generator is directly attached to the sun gear. The ICE is directly attached to the planet carrier, which connects the planet gears. The motor is connected to the ring gear via a series of intermediate gears. Definez1: Number of teeth on sun gear.z2: Number of teeth on ring gear.k = z2/z1: Base gear ratiom: Gear ratio of ring gear to gear attached to electric motorG: Rotation speed of generator (rpm)E: Rotation speed of ICE (rpm)R: Rotation speed of ring gear (rpm)M: Rotation speed of motor (rpm)Tg = torque of generator acting on sun gear (Nm)Te = torque of ICE acting on planet carrier (Nm)Tr = torque acting on ring gear from motor (Nm)Tm = torque of motor acting on gear connected to planetary gear set (Nm) Then, from the textbook, the following equations describe the operation of the planetary gears: G + kR – (1+k)E = 0,Tg = Tr/k,Te = -((k+1)/k)Tr In addition, we have: M = mR We can combine the first and last equations into: G + (k/m)M = (1+k)E From the empirical relationship derived from recorded data, G + M = 3.55E, we deduce: k = m = 2.55 The base gear ratio is 2.55. I don’t know the actual number of teeth on any of the gears, but it is the base gear ratio that determines how the planetary gear system operates. For the Prius, the base gear ratio is 2.6. From the second two equations, we can derive: Te = -(k+1)Tg = -3.55Tg Which is consistent with the equation derived empirically (Te = -3.53*Tg). Finally, Te = -((k+1)/k)Tr = -((k+1)/k)mTm = -(k+1)Tm = -3.55Tm, which is also consistent with the equation derived empirically (Te = 3.56*Tr). Well not quite. I didn’t know that there were a series of gears between the ring gear and the motor when I wrote the original post. And Tr, as originally defined, was the torque acting on the motor rather than torque from the motor acting on ring. Edited April 26, 2014 at 08:35 PM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 26, 2014 at 11:09 PM Author Report Share Posted April 26, 2014 at 11:09 PM (edited) The following plot shows a clear relationship between generator speed (rpm) vs. the car's speed (mph). When the car travels at increasingly slower speeds, the generator speed increases in the positive direction. High positive rotation of the generator corresponds to low gear ratios for the ICE to power the wheels. When the car is traveling at increasingly higher speeds, the generator speed increases in the negative direction. High negative rotation of the generator corresponds to high gear ratios for the ICE to power the wheels. So we see a continuous shift in the gear ratio from low ratios to a high ratios with increasing speed. Every once in a while at the higher speeds, the car decides to charge the HVB and shifts to a lower gear ratio as evidenced by the cluster of markers corresponding to generator rpm near or above 0 around 65 mph. Otherwise, at 65 mph the car runs in "overdrive" with the generator rpm around -1500. Edited April 26, 2014 at 11:11 PM by larryh Hybridbear and TX NRG 2 Quote Link to comment Share on other sites More sharing options...
larryh Posted April 27, 2014 at 11:00 AM Author Report Share Posted April 27, 2014 at 11:00 AM (edited) Definez1: Number of teeth on sun gear.z2: Number of teeth on ring gear.k = z2/z1: Base gear ratiom: Gear ratio of ring gear to gear attached to electric motorG: Rotation speed of generator (rpm)E: Rotation speed of ICE (rpm)R: Rotation speed of ring gear (rpm)M: Rotation speed of motor (rpm)Tg = torque of generator acting on sun gear (Nm)Te = torque of ICE acting on planet carrier (Nm)Tr = torque acting on ring gear from motor (Nm)Tm = torque of motor acting on gear connected to planetary gear set (Nm) The most accurate data that I have indicates that m = 2.5432 and k = 2.53. So the equation involving RPMs is: 0.9948*M + G = 3.53*E Edited April 27, 2014 at 11:04 AM by larryh Hybridbear 1 Quote Link to comment Share on other sites More sharing options...
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