larryh

To try out the theory, I would have to P&G leaving the transmission in L. I would accelerate at 2.2 bars and then release to gas pedal to glide (rather than braking). But I don't think I need to decelerate quite that rapidly for 95% efficient regen. I suspect that if I tried to use L with P&G it would lower MPGe. However, I can try using L on my commute to work to see what effect it has. But I think I will have to drive differently for any efficiency improvements to occur.

Actually, in theory, the most efficient way to drive is to use an extreme version of P&G (a.k.a city driving). Rather than shifting into neutral, use regenerative braking. Since regenerative braking is 95%+ efficient, you will be operating the car the entire time at either 88% or 95% efficiency. Pulse with acceleration of about 2.2 bars and the motor is operating at 88% efficiency. Then when you reach max speed, apply the brakes for 95% efficient regenerative braking. When you reach min speed, repeat the cycle over again. Somehow, I don't think this will work as well as theory predicts. This is the exact opposite of conventional wisdom, i.e. to drive as smoothly as possible. I will have to try it sometime as an experiment. :kookie:

A clarification about acceleration. If you want to drive normally, you should accelerate slowly to maximize the time you are operating the motor more efficiently. After that, you will generally be operating the motor at about 60-70% efficiency. If you want to use P&G, accelerate quickly (2.2) bars, then shift into neutral, and repeat. You want to maximize the time you are either at max efficiency (during the pulse with the motor outputting 32 kW) or you are in neutral. You do not want to be in between the two. If you dwell two much time between neutral and max efficiency, the overall efficiency will drop. You might be better off not doing P&G.

The above chart does not show that efficiency peaks at 88% at 32 kW and then declines to about 84% at 50 kW. Normally, between 30 and 50 mph, efficiency ranges from 50% - 70%. So there is room for a lot of improvement. As previously mentioned, P&G is one way you can improve efficiency. P&G works by operating the motor at a more efficient operating point. In order for P&G to be effective, you will need to use brisk acceleration (about 2.2 bars on the empower display) during the pulse so that the motor outputs about 32 kW of mechanical power where it operates at about 88% efficiency. Unfortunately, you cannot achieve 88% overall efficiency with P&G. The internal resistance of the motor discussed in previous posts takes a heavy toll on efficiency during the glide. If you use CC at 45 mph, you might achieve 67% efficiency. During the pulse with P&G, you can achieve 88% efficiency if you accelerate fast enough (but not too fast). However, the internal resistance of the motor is going to quickly reduce efficiency. By the end of the glide, efficiency will probably end up being about 5% better than CC, i.e. about 72%. If you don't accelerate fast enough, the efficiency improvement will be significantly less. P&G is more effective when the power required for CC is less, i.e. slower speeds. Note that the max/min speed range for P&G does not have much effect on efficiency.

Read this sales brochure for the 2014 Energi. It explicitly states: MyFord Mobile requires a compatible 2G independent cellular network. http://www.ford.com/services/assets/Brochure?make=Ford&model=Fusion&year=2014

According to the paper, capacity was reduced after 800 charge/discharge cycles by 17.1% for 0.25C charging (120 V charger) and by 15.4% for 0.5C charging (240 V charger). See this post for factors that have a much greater affect on HVB capacity: http://www.fordfusionenergiforum.com/topic/2491-charging-the-high-voltage-battery-in-energi/?p=16005

L2 charging is about 82% efficient. L1 charging is about 72% efficient. The difference in electricity used to charge a fully depleted battery is less than 1 kWh. If you charge once a day, you will save less than 365 kWh of electricity a year. At $0.01 / kWh, that is $36.50 per year. You are not going to recover the cost of an L2 charger based on electricity savings alone, unless your electric rates are very high.

The following paper "http://electricvehiclewiki.com/images/b/b0/Capacityfade_jpspaper2.pdf" details a study in which they try to determine the optimal charge rate for a lithium-ion cell which minimizes capacity degradation. They determined the optimal charge rate to be 0.5C, where C is the cell capacity in Ah. The capacity of the HVB is about 23.5 Ah. The 120 V charger charges at a rate of about 3.28 amps or 3.28/23.5 = 0.14C. The 240 charger charges at a rate of about 10 amps or 10/23.5 = 0.4C. According to the paper, the 240 V charger is better. But I'm not sure how applicable that paper really is to the Energi's HVB. The 240 V Charger will generate more heat, but for a shorter period of time. The 120 V charger generates less heat, but over a longer period of time. Neither one heats up the HVB much, maybe at most 4 F. I think the other factors that affect battery degradation will overwhelm any difference you could observe between the 120 V and 240 V chargers. I doubt that it makes much difference which one you use. I would base my decision on whether to use the 240 V charger on other factors such as the ability to precondition the car in cold weather, faster charging times, etc.

I attempted P&G vs CC to verify my conclusions now that I have figured out how to measure things properly. I averaged about 45 mph. I obtained the following results: Direction P&G Efficiency CC Efficiency Efficiency Difference North 72% 63% 9% South 75% 67% 8% So the observed increase in efficiency was about 8-9%. Using P&G vs CC, efficiency increased by a factor of 72/63= 1.14 for the North bound trip and 75/67 = 1.12 for the South bound trip. Unfortunately, you are not going to get the entire 1.12 - 1.14 factor improvement in MPGe. P&G, for some reason, seems to take more mechanical energy than CC for the same trip--I don't know why or if it is just measurement error. The measured mechanical energy output by the motor for P&G vs CC differs by a few percent. In addition, accessories account for 5% of the power output from the HVB. So you won’t see the full 1.14 and 1.12 improvement factors in MPGe. The improvement in MPGe will probably be less than 10%.

I also repeated the experiment at moderate acceleration (close to 2 bars) using about 20 kW of mechanical power. The efficiency peaked at about 84% (the same as fast acceleration) and then began to fall off after 40 mph was achieved after 0.15 miles. While maintaining constant speed for the remainder of the trip, the efficiency fell to 82% (the same as for slow acceleration). So moderate and slow acceleration provided the same results and were equally efficient overall. Follow the advice of the acceleration coach and don't overshoot you target speed. That should achieve near optimal efficiency.

The referenced post is not correct. The amount of mechanical energy required for fast acceleration is probably at most a percent or two more than required for slow acceleration. The following chart explains why slow acceleration is more efficient than fast acceleration. In this experiment, I am accelerating from 0 to 40 mph. With slow acceleration, a distance of 0.21 miles is required to reach 40 mph. With fast acceleration, 40 mph is achieved after 0.04 miles. I then maintain constant speed of 40 mph until I have traveled a total distance of 0.21 miles. For both fast and slow acceleration, the total distance traveled is the same (0.21 miles) and the speed of the car after the distance traveled is the same (40 mph). The elapsed time for fast acceleration is significantly less than for slow acceleration. The motor outputs 13 kW of mechanical power for slow acceleration (about 1.5 bars) and 50 kW (It is really 43 kW. I put the wrong number on the graph. So everywhere you see 50 kW, substitute 43 kW) of mechanical power (about 2.5 bars) for fast acceleration. The plot below shows efficiency vs. distance for fast vs. slow acceleration. If the electric motor converts 10 kWh of electrical energy to 8 kWh of mechanical energy, the efficiency is 8 / 10 = 0.8. The greater the efficiency, the less energy used from the HVB. Note that fast and slow acceleration required about the same amount of mechanical energy to go 0.21 miles (fast acceleration requires just 0.4% more energy, i.e. less than 1% more energy). After 0.04 miles, the efficiency for fast acceleration is 84%. For slow acceleration it is 82%. The motor is slightly more efficient at higher power. With fast acceleration, the car is now going 40 mph. To maintain a constant speed of 40 mph, much less mechanical power is now required, about 4 kW. The motor efficiency at 4 kW is about 65%, significantly less than 84% (the efficiency at 50 kW). The efficiency for fast acceleration is now going to start falling. At the end of 0.21 miles, efficiency has fallen to 80.6%. For slow acceleration, the motor maintains a constant output power of 13 kW and the efficiency remains at 82% for the entire remaining distance of 0.21 miles. Slow acceleration is 1.4% more efficient than fast acceleration after 0.21 miles. The total mechanical power output of the motor for both fast and slow acceleration was about 0.105 kWh. So fast acceleration required 0.105 / .806 = 0.131 kWh of energy from the HVB and slow acceleration required 0.105 / .82 = 0.127. Fast acceleration required 3% more energy from the HVB. So the reason slow acceleration is more efficient than fast acceleration is that you are maintaining the motor at a more efficient operating point for a much longer period of time than with fast acceleration. You want to prolong acceleration, where the motor is operating more efficiently, and delay operating the car at the final target speed, where the motor is operating less efficiently. Note that it is possible to use about the same amount of mechanical energy with fast vs slow acceleration if you are careful, but in practice it is difficult. You are much more likely to waste energy with fast acceleration, e.g. overshoot your final target speed. In this experiment, fast acceleration only used 3% more energy from the HVB than slow acceleration. But if you mess up, fast acceleration could easily take 10% or more. It took me three attempts to get it right. If you do it incorrectly, and overshoot your target speed (exceed 40 mph at some point along the trip and allow regen to occur to reduce speed back to 40 mph), you get the poor results in the post referenced above. Do not overshoot the target speed during acceleration.

The quoted post is not quite correct. The efficiency plotted in the graph above includes the power loss due to the internal resistance of the motor (see the previous post). For P&G, this loss is not relevant when attempting to determine the gain in efficiency using P&G. The internal resistance is always present, whether the HVB is supplying power to the motor or the car is in neutral. It is a function of the car's speed. There is very little you can do to reduce it other than driving at a slower speed. P&G does not help reduce this loss. I can create a similar efficiency plot where only the inverter and other power losses are included, but not the loss associated with the internal resistance of the motor. If I do that, the motor efficiency increases linearly from about 83.5% at 0 kW to 92% at 34 kW. After about 34 kW, the efficiency begins to decrease. At 50 kW, its back down to about 85%. Thus maximum increase in efficiency that can be achieved by using P&G is then about 92% - 83.5% = 8.5%. But there are many other variables that need to be considered, so the actual efficiency will vary. You probably never want to operate the electric motor much above 34 kW as you start losing efficiency after that.

The quoted is not quite correct. The efficiency plotted in the graph above includes the power loss due to the internal resistance of the motor (see the previous post). For P&G, this loss is not relevant. The internal resistance is always present, whether the HVB is supplying power to the motor or the car is in neutral. It is a function of the car's speed. There is very little you can do to reduce it other than driving at a slower speed. I can create a similar efficiency plot where only the inverter and other power losses are included, but not the loss associated with the internal resistance of the motor. If I do that, the motor efficiency increases linearly from about 83.5% at 0 kW to 92% at 34 kW. After about 34 kW, the efficiency begins to decrease. At 50 kW, its back down to about 85%. Thus maximum increase in efficiency that can be achieved by using P&G is then about 92% - 83.5% = 8.5%. But there are many other variables that need to be considered, so the actual efficiency will vary. You probably never want to operate the electric motor much above 34 kW as you start losing efficiency after that.

The car uses a 2G connection. Your cell phone uses a 4G connection. The coverage area for 2G vs. 4G connections varies. Check the following web site: http://www.att.com/maps/wireless-coverage.html#fbid=ZwZ1UVn9kzp

The purpose of the electric motor in the Energi is to convert electrical power from the HVB to mechanical power which is used to propel the car, and to convert mechanical power from the car's motion back to electrical power to be stored in the HVB during regenerative braking. Unfortunately, the conversion is not 100% efficient. The following plot shows the efficiency of the electric motor at 6750 rpm, or about 48 mph. It plots output mechanical power from the motor vs. input electrical power. The markers record observed output mechanical power measurements (x-axis) vs. input electrical power measurements to the motor (y-axis). The green line shows the conversion for a perfect motor with 100% efficiency. Since the Energi motor is not perfect, the markers are above the green line, i.e. more electrical power is consumed from the HVB than mechanical power generated to propel the vehicle, and less electrical power is supplied to the HVB than mechanical power is consumed by the electric motor during regenerative braking. The closer a marker is to the green line, the more efficient the conversion. Note that the x-intercept is at about -1.23 kW. This means that when no electric power is applied to the motor, it slows down due to internal resistance. The associated power loss is 1.23 kW. You need to supply the motor with about 1.4 kW of electrical power just to maintain a constant 6750 rpm. That's a significant amount of energy loss. When the motor is outputting power to propel the vehicle, the power required from the HVB, y, for the motor to generate output power x is approximately: y = 1.12(x + 1.23). So at high levels of power, the efficiency is approximately 1/1.12 = 89%. During regeneration, the equation is approximately: y = 1.04(x + 1.23). So at high levels of regenerative braking power, the efficiency is approximately 1/1.04 = 96%. The plots at other rpms are similar to this plot. The main difference is the x-intercept. That varies according the computation in the following post: "http://www.fordfusionenergiforum.com/topic/1880-obd-ii-data-for-ice/?p=16072".

I am not aware of any applications that work on the iPad that allow you to enter custom PIDs. I have no Apple products so I can't assist you there. FORScan (web site is forscan.org) provides the most comprehensive set of custom PIDs for the Fusion Energi. There are several hundred of them. But FORScan only runs on a Windows PC. I also use Torque Pro. But that only runs on an Android device. I use a 10" tablet PC. I translated Scan Gauge II codes posted in the various forums to derive the custom PIDs. You can find a listing of many of them here: "http://fordfusionhybridforum.com/topic/6503-scangauge-ii-x-gauge-codes/?p=50199". The Scan Gauge II codes are explained here: http://www.scangauge.com/wp-content/uploads/XGaugeCoding.pdf.

Now that I think I understand what is going on better, I can provide a better explanation of what I am observing. From the table above, traveling at 40 mph on a level road I measure the mechanical output power from the motor to the wheels to be 4.4 kW. 6.9 kW of electricity is being supplied to the motor. The motor efficiency is thus 4.4 / 6.9 = 64%. The internal friction of the motor results in a power loss of -1.3 kW at 40 mph. So about 1.3 / 6.9 = 18% of the loss is associated with the internal friction of the motor. The rest is from the inverter and other sources. If I do a coast down at 40 mph in neutral, I can measure the deceleration, and assuming the car and contents weigh 1871 kg estimate the power loss from aerodynamic drag, rolling resistance, and internal frictions at -5.8 kW. When measuring the power at wheels of 4.4 kW driving at a constant speed of 40 mph, this is the power required to overcome aerodynamic drag and rolling resistance. It does not include the internal frictions of the motor (that is taken into account in the 64% motor efficiency). When obtaining the -5.8 kW power loss during coast down in neutral, this includes the internal friction of the motor in as well as aerodynamic drag and rolling resistance. Hence the difference in the measurements -5.8 kW + 4.4 kW = -1.3 kW is the power loss associated with the internal friction of the motor. In post 87, the mechanical energy readings for the motor for P&G vs CC are not measuring the same quantity, so they should be ignored. In the CC energy measurements, the energy loss associated with internal friction of the motor was not included. In the P&G energy measurements, the energy loss associated with the internal friction was included (during the glide). When measuring the energy output for P&G and CC consistently, there is only a small difference in the mechanical energy output of the motor for the two techniques.

When in neutral, the motor torque as a function of rpm is as follows, where y is the motor torque in Nm (negative) and x is the motor rpm (positive): y = -0.0011*x, x <=2000, y = -0.0001*x - 2.0, 2000 <= x <= 4000, y = 0.0001*x - 2.8, 4000 <= x <= 6000, y = 0.0006*x - 5.8, 6000 <= x <= 8000, y = -1.0, x >= 8000 The torque is negative, which means that the motor is experiencing resistance, slowing the motor and the car down. Enough electrical power must be supplied to the motor to overcome this negative torque before the motor can begin to supply power to the wheels. I suspect that this resistance is independent of the load on the motor. At 40 mph, the power associated with this resistance is about -1.2 kW. The mechanical output power of the motor is about 4.4 kW. This resistance alone accounts for a 27% loss of efficiency converting electrical power to mechanical power. I estimate the total efficiency loss at 40 mph is 39%. The remaining losses are from the inverter and other sources.

Actually, I suspect the power drag associated with the electric motor is always present and not just present only when in neutral. The HVB simply has to supply that much extra electrical power to the motor to yield a given mechanical power output. So at 50 mph, the measured mechanical power from the motor is 8.1 kW from the table above. In order for the motor to provide this amount of power, the HVB must provide an additional 1.1 kW of power (reported loss column) to overcome the internal drag of the motor for a total of 8.1 kW + 1.1 kW = 9.2 kW of power. The motor is only about 85% efficient in converting electrical power to mechanical power. So the power required from the HVB is then 9.2 / 0.85 = 10.8 kW. The actual measured power provided by the HVB was 11.2 kW, close to the estimated 10.8 kW. The drag being observed is probably due to frictions and other phenomena internal to the motor (causing the motor to slow down), and possibly other internal frictions associated with the planetary gear system. If the torque generated by motor (in the absence of friction) just matches the torque of the internal frictions, there will be no torque at the output of the motor and the sensor will read 0 torque. The HVB is supplying the motor with just enough power to match the internal frictions. However, no power is being supplied to the wheels. In essence, the mechanical power being measured from the electric motor is not the actual power being output by a frictionless motor. It is only that portion of the power above the internal frictions which is then supplied to the wheels (actually only about 98% of the power makes it to the wheels due to power losses associated with the transmission from the motor to the wheels). When the car is in neutral, there is no power supplied by the motor. The entire power loss associated with the internal frictions is then transmitted to the wheels and the motor torque sensor measures this resulting negative torque. One has to be careful when computing the efficiency of the electrical motor in converting electrical to mechanical power. In post 88, the efficiency being plotted is the amount of electrical energy being converted to mechanical energy that makes it to the wheels.

Mostly this occurs when you turn the car off and then decide you need to turn it back on again to roll up the windows, or you push the button to turn it off more than once.

The main factors determining the lifespan of the HVB are average temperature, the average SOC, and the charging/discharging cycles. Undesirable chemical reactions that degrade the HVB (i.e. reduce its capacity) occur all the time, whether the battery is used or not. The rate of degradation increases significantly with temperature. You want to maintain a low average HVB battery temperature. To do that, keep the power being used from the HVB relatively low, maybe at most two bars on the Empower screen. Also, park the car in the shade. People who live in the southern states will experience significantly faster HVB degradation than people in the northern states. If you are not going to use the car for some period of time, it is better not to leave the HVB fully charged. The higher the average SOC, the faster the degradation. I usually don't charge my HVB until just before I am planning to leave. Finally, the lifespan is determined by the number of charge/discharge cycles and the depth of discharge. Fortunately, the car does not allow you to charge the HVB to its maximum capacity nor drain it to 0% SOC. When the car states it is at 100% SOC, the HVB contains about 7.2 kWh of energy vs. its capacity of 7.6 kWh. The minimum amount of energy of the HVB, before the ICE will start to charge it, is 1.0 kWh. Note that the SOC shown on the MFT screen and MFM is not the true SOC of the HVB. It is called the displayed SOC. When the displayed SOC of the HVB is 100%, the true SOC of the HVB is 96% or greater. If you want to get the true SOC to 100%, you will have to charge the HVB while the car is running. It normally charges the HVB to just over 98% SOC. To get it to 100%, you need to turn the car on with the car plugged in. But it won't start charging the HVB unless the true SOC is below 98%. Turn on the heater or AC or reduce the SOC below 98%. Turn off the AC/heater and allow it to charge the HVB. Since you cannot see the true SOC (unless you have a OBD II scanner), watch the arrows above and below the HVB icon on the left display. With the AC/heater on, the arrow below the HVB will indicate the HVB is discharging. Turn the AC/heater off for a few seconds and observe the arrows. If the arrow below the HVB remains, it means the HVB SOC is still above 98%. Once it is below 98%, the arrow above the HVB icon will indicate the HVB is charging. Wait until the arrow above the HVB goes away. The HVB is now fully charged.

Note that the analysis of P&G in post 89 is not quite correct. I cannot accurately measure power at wheels. I can only estimate the power at wheels based on changes in kinetic energy as the speed of the car varies--this is the technique used to compute the cyan line. But that is not the entire source of power required at the wheels during the pulse. Power is also required to overcome aerodynamic drag, rolling resistance, and internal frictions. However, I can't easily measure the power at the wheels due to those sources, and they are not included in the plot. I can only estimate the amount based on the CC motor power. So, more accurately, all that I can say is that more than 85% of the power from the motor is being transmitted to the wheels. I don't know the exact value. Most likely, the amount of power transmitted to the wheels is in the upper 90s during the pulse. During the glide, the cyan line is accurate. Kinetic energy is being lost due to aerodynamic drag, rolling resistance, internal frictions, and a drag by the electric motor (see the previous post).