OBD II Data for HVB

[larryh]

I tried the same experiment again.  This time I increased the time resolution of the data, i.e. logged data faster to get more accurate results.  I also removed the power being consumed by accessories.

 

Acceleration   Energy (kWh)

slow             0.134

moderate         0.134

fast             0.137

 

I get similar results except for moderate acceleration--something was wrong the first time I did it.  There is very little difference between the measurements.  I'm not sure how to control all the variables to get more accurate results.  But I suspect fast acceleration is wasteful as indicated by the data and the analysis in my previous post. 

Edited July 12, 2014 at 12:59 PM by larryh

[larryh]

The following is a plot of the voltage variation between the HVB cells for my 60 mile commute this morning.  The plot on the left shows the variation vs. SOC of the HVB.  The right plot shows the variation vs. power output of the HVB, where negative power results from regen.   I'm not sure how correct this is, but I suspect that large voltage variation is bad and is indicative of stress on the battery.  You want all the cells to be balanced with less than 0.01 volt variation.  According to the graphs below, large voltage variation occurs when the SOC is low and during high power output of the HVB.  I'm not sure how long it takes the cells to recover and become rebalanced after the balance is perturbed.  The worst thing you can do for the balance is to run the HVB at a low SOC.  It increases significantly below 17% SOC.  That is about 50% charge for the hybrid battery icon. Perhaps we should stay in EV later mode and maintain the charge level above 20% until the end of the trip.  At the end, then let it enter hybrid mode and use up the remaining charge of the battery.  Try to stay out of hybrid mode as long as possible.  This may help preserve HVB life.

 

For further discussion of this and analysis of the HVB, see the following thread:

 

"http://fordcmaxenergiforum.com/topic/2794-video-explaining-what-happens-to-mpge-and-soc-charging-while-powered-on/?p=21685"

Edited July 14, 2014 at 12:42 AM by larryh

[larryh]

The following is a plot of SOC vs. Cell Voltage for the HVB made from the data collected during my commute on Friday.  The HVB has 84 cells in series.  In order to make this plot, I had to assume the HVB had an internal resistance of 0.11 Ohms.  If I just plotted the data and assumed it had no internal resistance, the dots would be scattered all over the chart.  Ideally, the voltage measurement should be made with no load on the HVB.  But that is not practical when driving--I'm not about to pull over to the side of the road every mile and turn off the car so I can make a measurement. 

 

The lowest voltage observed of any cell during the trip was 3.38 V.  When there was no load on the HVB, the lowest cell voltage observed was 3.44 Volts.  The maximum cell voltage at that time was 3.48 V. 

 

 

 

Edited July 14, 2014 at 01:09 AM by larryh

[larryh]

On my commute home yesterday, I allowed the car to operate normally in Hybrid Mode, turning the ICE on and off, to observe the HVB cell voltage variation when the SOC of the HVB is low.    The HVB cell variation generally remained less than 10 mV.  It is only at the end that it started rising close to 20 mV when I tried to prevent the ICE from coming on and use up the remaining charge in the HVB, forcing the HVB to provide slightly more power to propel the car.  So low SOC, i.e. Hybrid Mode, does not seem to cause any undo stress on the battery if you allow the car turn on and off the ICE as it is programmed to do.  The problem seems to occur when you try to override its programming and prevent the ICE from coming on, discharging the battery for a long period of time without allowing the ICE to share the load or allowing the HVB to recover by running the ICE to recharge the battery and rebalance the cells.  The HVB seems to be in a more weakened state, it cannot recover as easily from the stress of providing high power output.  That's probably one of the reasons why the power output of the HVB is limited to 2 bars in the Empower screen while in hybrid mode. 

 

Hybrid mode begins when the SOC falls below 21.5% at about time 4:38.  The purple line shows the ICE power output.  The ICE is on when the power is non-zero.  The green line shows the power output of the HVB.  Negative power means regen or the ICE is running the motor/generator to generate electricity.   Most of the time the HVB was assisting the ICE in propelling the car.  However, at times, the ICE was generating electricity for the HVB. 

 

Edited July 14, 2014 at 10:22 AM by larryh

[rspray]

Hi Larry,

 

Just curious what PID codes you are using for the cell voltages.  Are these posted anywhere?  I found the standard ones:

http://fordcmaxhybridforum.com/topic/2757-obd2-%E2%80%93-elm-327-%E2%80%93-torque/?do=findComment&comment=33374

 

EDIT: NM, just found all the info on Scangauge, so I'll bet you're using that. Curious if I can find PID codes for the cell voltages so I can get the data from ELM627 Wifi...

Edited July 15, 2014 at 08:12 PM by rspray

[larryh]

The X-Gauge Codes are listed here:

 

"http://fordfusionhybridforum.com/topic/6503-scangauge-ii-x-gauge-codes/"

 

You can translate them using this document:

 

"http://www.scangauge.com/wp-content/uploads/XGaugeCoding.pdf"

[snedecr]

I tried the same experiment again.  This time I increased the time resolution of the data, i.e. logged data faster to get more accurate results.  I also removed the power being consumed by accessories.

 

Acceleration   Energy (kWh)

slow             0.134

moderate         0.134

fast             0.137

 

I get similar results except for moderate acceleration--something was wrong the first time I did it.  There is very little difference between the measurements.  I'm not sure how to control all the variables to get more accurate results.  But I suspect fast acceleration is wasteful as indicated by the data and the analysis in my previous post. 

 

 

Thanks for trying it!  It pretty much validated your initial assumptions.  I did not think there would be much wind drag at 40 mph, and indeed there isn't.

 

I want to give you some extended thanks for your posts in this thread!  It makes my head hurt sometimes from trying to digest all the data, but it's helped me to learn a lot about how the car works, and about what the engineers were thinking when they programmed it  the way that they did.  It has some glitches (i.e. the way you have to fake out the climate system on engaging the A/C compressor), but all-in-all, it's a well-thought out system.  Coming from a Toyota Camry Hybrid (the company who first widely implemented the PSD) the improvements in both hardware and software are staggering!

 

Thanks again!

[larryh]

The following is a plot I made showing the energy consumed from the HVB (kWh) vs. distance for three different accelerations:  about 1.5 bars (slow), 2 bars (medium), and 3 bars (fast).  The total distance traveled is about 0.25 miles and the final speed is 40 mph.  The x-axis shows GPS longitude traveling straight East.  The y-axis shows the amount of energy consumed from the HVB by the motor.  (I have subtracted out the power consumed by accessories.)  The curves are not smooth due to errors in GPS longitude measurement. 

 

With 1.5 bar acceleration, I reach 40 mph at the end of 0.25 miles.  With the faster accelerations, I reach 40 mph sooner and then used cruise control to maintain 40 mph. 

 

The faster the acceleration, the more energy that is consumed.  However, the differences are small.  Fast acceleration consumes about 0.012 kWh more energy than slow acceleration.  With fast acceleration, you initially use a lot of energy to get up to speed and then the rate of energy usage drops to what is required to maintain constant speed (the purple curve).  With slow acceleration, you will have a constant rate of energy usage until reaching the final speed (the blue curve).  The faster the acceleration, the more the curve shifts to the left and up.  It looks like the optimal acceleration is to apply constant power until reaching the final speed.  The resulting curve will be a straight line from the start to the terminal point.  Any faster acceleration will result in a curve bowed outward shifting up and to the left. 

 

The observed efficiency of the motor for slow acceleration was 79.6%, moderate acceleration was 80.7% and 79.2%, and fast acceleration was 78.6%--there was no significant difference. 

 

 

Edited July 18, 2014 at 10:21 PM by larryh

[meyersnole]

Thanks for checking that larryh, I would not have expected those results... especially since quicker acceleration really dings your driving score... and shows up negatively on the driving coach screen.

[larryh]

The driving coach is properly coaching you to accelerate moderately.  If you limit acceleration to two bars, you will get a good acceleration score.  Limiting acceleration to 2 bars results in near minimal energy consumption from the HVB.  Fast acceleration wastes energy. 

Edited July 19, 2014 at 12:09 AM by larryh

[larryh]

I have had my car for over a year now and have been wondering how to determine how much capacity has been lost from the HVB during the first year of ownership, i.e. the state of health (SOH) of the HVB.  In the C-Max Energi forum, a member from Texas has been experiencing decreased range from their HVB.  The HVB battery charges normally, i.e. takes two hours using a 240 V charger, and the display shows 100% SOC after it is fully charged.  However, when doing a comparison with another fully charged Energi, the plug-in energy used reported by the car's display was 4.4 kWh after entering hybrid mode, whereas the other Energi, after several more miles, reported 5.5 kWh before entering hybrid mode.  The dealer measured the SOC and voltage of the HVB after it was fully charged and depleted.  The values were normal.

 

There is a PID named Energy To Empty (ETE) that measures the current energy stored in the HVB.  When the bad HVB was fully charged to 100% SOC, ETE for was only 6.39 kWh.  My car reports ETE is 7.14 kWh.  If 7.14 kWh is the capacity for a HVB, that is SOH is 100%, then the SOH of the bad HVB was only 6.39/7.14 = 89%.

 

So it appears you can track SOH of the HVB from ETE and SOC.  Compute ETE/SOC, this is the capacity of the HVB.  Then divide the capacity by 7.14 (assuming 7.14 is the capacity of a HVB that has 100% SOH), you can determine the SOH.  For example, this morning, ETE was 7.05 and SOC was 98.70%.  So the capacity of the HVB was 7.05/.9870 = 7.14 kWh.  The SOH is then 7.14/7.14 = 100%.  I don't know the actual capacity of a HVB with 100% SOH, so I'm not sure what the true SOH of my battery is. 

Edited July 24, 2014 at 08:58 PM by larryh

[larryh]

The following are some results that I obtained trying to use Pulse & Glide to improve MPGe during EV operation.

 

I made measurements along the same stretch of highway in both directions using both pulse and glide and cruise control.  The measurements for pulse and glide began at the start of the 50 mph pulse to 60 mph and ended at the end of the glide from 60 mph to 50 mph in neutral.  The average speed for all the measurements in the table below was 53.6 mph.   The motor output is mechanical power output by the motor.  Motor efficiency measures how efficiently the motor coverts electrical energy to mechanical energy. 

 

                           Motor       HVB          Motor
                           Output      Output       Efficiency
Method  Direction   MPGe   (kWh/mile)  (kWh/mile)   (%)
CC      North       145.3  0.173       0.232        74%
P&G     North       156.0  0.180       0.216        84%
CC      South       120.6  0.218       0.279        79%
P&G     South       122.2  0.225       0.276        83%

 

I repeated the experiment, this time at 45 mph rather than 55 mph.  For P&G, I pulsed to 50 mph and then let the speed drop to 40 mph.  I tried two variations of P&G.  In P&G1, I shifted to neutral for the glide.  In P&G2, I left the shift lever in drive.  However, I applied slight pressure to the accelerator so the power being supplied to/drawn from the HVB was about 0 kW. 

 

                           Motor       HVB         Motor
                           Output      Output      Efficiency
Method  Direction   MPGe   (kWh/mile)  (kWh/mile)  (%)
CC      North       164.7  0.130       0.205       64%
P&G1    North       183.8  0.131       0.183       72%

P&G2    North       182.9  0.159       0.190       69%

 

CC      South       138.3  0.165       0.244       68%
P&G1    South       147.9  0.190       0.228       83%

P&G2    South       146.0  0.199       0.230       72%

 

P&G works better at 45 mph than 55 mph.  MPGe improved by 7% to 12% using P&G.  At 55 mph, the improvement was between 2% and 8%.  There are two ways to do P&G.  During the glide, you can either shift into neutral.  Alternatively, you can leave it in drive, and apply slight pressure to the accelerator to prevent regen.  If you do it right, the current or power into/out of the HVB should be around 0 amps/kW.  A scanner is useful to monitor the power/current into/out of the HVB.  There was a small amount of regen taking place so the MPGe could have been slightly higher--the same as for P&G1. 

 

Finally, I repeated the experiment at 35 mph.  For P&G, I pulsed to 40 mph and then let the speed drop to 30 mph.  I obtained  similar results 

 

                           Motor       HVB         Motor
                           Output      Output      Efficiency
Method  Direction   MPGe   (kWh/mile)  (kWh/mile)  (%)
CC      North       202.9  0.104       0.166       62%
P&G     North       223.0  0.119       0.151       79%

 

There was a 10% improvement in MPGe.  Note that motor efficiency increased by a factor of 79%/62% = 1.27, however, the motor has to output more energy to propel the vehicle using P&G by a factor of 0.119/0.104 =1.15.  Thus the overall improvement factor was only 1.27/1.15 = 1.099, i.e. 10% improvement.

Edited August 16, 2014 at 06:45 PM by larryh

[larryh]

P&G works by operating the motor at a more efficient operating point than cruise control.  The motor produces power more efficiently at higher power output, during the pulse (and, of course, consumes no power during the glide).  Although P&G requires the motor to produce slightly more mechanical energy to propel the car than cruise control, the increase in motor efficiency out weighs the increased mechanical energy required to power the wheels.   

 

The plot below shows the efficiency of the electric motor vs. mechanical power output by the motor.   At about 30 kW, the motor is converting approximately 85% of the power received from the HVB to mechanical power.  Not all of this mechanical power makes it to the wheels.  There are additional energy losses through the transmission.  I did not subtract the approximately 300 watts of power used by accessories when making this chart.  That should have minimal impact.

 

The red line is an approximation for the motor efficiency curve

 

e = 1/(1.12 + 2.09/p),

 

where e is the efficiency as a fraction, and p is the power output of the motor in kW. 

 

The following table shows the approximate power and efficiency required by CC at various speeds.  The actual power and efficiency will vary greatly depending on grade of the road, wind, and other factors.

 

mph Power (kw) Efficiency

10     0.97     0.30
20     2.45     0.51
30     4.34     0.63
40     6.53     0.70
50     8.89     0.74
60    11.34     0.77
70    13.74     0.79
80    15.99     0.80
 

The best you can hope to achieve with P&G is to capture the difference in efficiency between max efficiency of 85% and the efficiency of the motor at the selected speed using CC.  For example, going 40 mph, the best you can achieve is capturing the difference in motor efficiency of 85% - 70% = 15%.  I'm not sure that you can get it all, but that all depends on your technique.  If you could capture it all, you would increase MPGe by a factor of 85%/70% = 1.21.  P&G is going to be more effective at slower speeds, when cruise control is most inefficient. 

 

 

Edited August 16, 2014 at 06:59 PM by larryh

[larryh]

The following plot shows where the motor mechanical energy losses are occurring for P&G relative to CC.  The losses are occurring between the motor and the wheels during the pulse portion of P&G.  Only about 85% of the power from the motor is making it to the wheels during the pulse (this is not quite correct--see post 95).  With CC, it looks like most of the power from the motor is making it to the wheels.

 

The purple line shows the mechanical power output from the motor using P&G.  During the glide, the power, as reported by the car, is negative.  I'm not exactly sure what this represents.  If this is supposed to be an internal drag due to friction and other motor phenomena, I don't know.  But if it is, it doesn't seem to match what I would expect.  The red line shows mechanical power output from the motor using CC.  I am going straight North, so latitude along the x-axis corresponds to distance. 

 

The cyan line is the power at the wheels for P&G.  For the pulse, the power is positive, i.e. power is applied to the wheels.  For the glide, the power is negative due to aerodynamic drag and rolling resistance.  However, I plotted power as positive during the glide so that it could more easily be correlated with the power from the motor used by CC.  The power at the wheels is computed based upon changes in vehicle speed, assuming that the car and contents weigh 1871 kg, and is, consequently, a rather noisy measurement since speed is not updated frequently enough. 

 

During the pulse, you can see the cyan line is about 85% the height of the purple line, meaning only 85% of the power from the motor is making it to the wheels.  I can't be sure that 85% is accurate since I had to make some assumptions when computing power at wheels.  I am not accounting for all the power required at the wheels.  It is actually much better--probably in the upper 90s.  During the glide, the cyan and red lines match rather closely, meaning that most of the power from the motor is making it to the wheels for CC.  I notice that there seems to be slightly more power consumed by aerodynamic drag, rolling resistance, and internal frictions during the first half of the glide than is being output by the motor, i.e. the cyan line is slightly higher than the red line for the first half of the glide.  Since the car is going faster during the first half of the glide using P&G vs. CC, that is to be expected.  Similarly, at the end of the glide, the cyan line tends to fall below the red line--the car is now traveling slower using P&G vs CC.  It takes more power to maintain constant speed at higher speeds. 

 

Edited September 1, 2014 at 03:14 PM by larryh

[larryh]

I recently charged the HVB at a public charging station.  I notice the power supplied to the car's internal charger is different than at home.

 

Charging   Input    Input      Input

State      Voltage  Current    Power

Home       235 V     14 A      3285 W

Public     205 V     15 A      3080 W

 

So at the public charging station, it is going to take slightly longer to charge the car.  At home the power from the wall is about 3435 W.   For the public charging station, I estimate it is about 3230 W. 

 

I wonder why there is a difference.  Could it be because the public charging station uses commercial three phase power rather than residential two phase power?

Edited August 22, 2014 at 05:19 PM by larryh

[murphy]

Very probable, 3 phase power is 208 volts.

[larryh]

The following is a plot I made showing the energy consumed from the HVB (kWh) vs. distance for three different accelerations:  about 1.5 bars (slow), 2 bars (medium), and 3 bars (fast).  The total distance traveled is about 0.25 miles and the final speed is 40 mph.  The x-axis shows GPS longitude traveling straight East.  The y-axis shows the amount of energy consumed from the HVB by the motor.  (I have subtracted out the power consumed by accessories.)  The curves are not smooth due to errors in GPS longitude measurement. 

 

With 1.5 bar acceleration, I reach 40 mph at the end of 0.25 miles.  With the faster accelerations, I reach 40 mph sooner and then used cruise control to maintain 40 mph. 

 

The faster the acceleration, the more energy that is consumed.  However, the differences are small.  Fast acceleration consumes about 0.012 kWh more energy than slow acceleration.  With fast acceleration, you initially use a lot of energy to get up to speed and then the rate of energy usage drops to what is required to maintain constant speed (the purple curve).  With slow acceleration, you will have a constant rate of energy usage until reaching the final speed (the blue curve).  The faster the acceleration, the more the curve shifts to the left and up.  It looks like the optimal acceleration is to apply constant power until reaching the final speed.  The resulting curve will be a straight line from the start to the terminal point.  Any faster acceleration will result in a curve bowed outward shifting up and to the left. 

 

The observed efficiency of the motor for slow acceleration was 79.6%, moderate acceleration was 80.7% and 79.2%, and fast acceleration was 78.6%--there was no significant difference. 

 

 

 

After further analysis, I don't think the information posted below is correct.  When done correctly, the mechanical energy loss is probably only a percent or two with faster acceleration--not 8% or 20%.  But it is hard to do correctly, so you may end up with poor results as in this post. 

 

I performed some additional experiments today.  It seems the car does not transmit power from the motor to the wheels to propel the car as efficiently with fast acceleration vs. slow acceleration.  Maximum acceleration requires about 20% more mechanical energy from the motor than 1.5 bar acceleration to provide the increase in kinetic energy necessary to accelerate the car to 40 mph.  Acceleration at 2 bars requires about 8% more mechanical energy than 1.5 bar acceleration.  Fast acceleration is inefficient.  I'm not quite sure why this is true.  I wouldn't have expected much difference in efficiency when it comes to transmitting power from the motor to the wheels to propel the car.  The coaching assistant is correctly advising you to accelerate slowly. 

 

Note the previous post, quoted above, plotted HVB energy required to accelerate at different speeds.  The main reason that more HVB energy is required for faster acceleration, is because more mechanical power is wasted with faster acceleration. 

 

Effectively, maximum acceleration has the same impact on efficiency as getting an 82% braking score (I don't know the exact number).  Similarly, accelerating at 2 bars has the same impact as a 93% braking score.   In the case of acceleration, you are taking energy from the HVB to increase the kinetic energy of the car.  For braking, you are storing the kinetic energy lost by the car in the HVB.  I may be advisable to focus as much attention on efficienct acceleration as on efficient braking.  The should provide an acceleration coach similar to the braking coach. 

Edited September 6, 2014 at 10:19 AM by larryh

[Hybridbear]

Fascinating observations! Thanks for all your time spent on this!

Edited September 1, 2014 at 02:50 PM by Hybridbear

[larryh]

I did a series of experiments to determine if the power loss from the motor reported by the car was real when in neutral while performing P&G (see the purple line in post 89 which shows negative power when the car is in neutral).  I first drove the car at constant speed and measured the mechanical power output of the motor.  Next, I placed the car in neutral and let it coast down from a few mph above the constant speed to a few mph below the constant speed.  Assuming the car and contents weigh 1871 kg, I can estimate the power loss from aerodynamic drag, rolling resistance, and internal frictions for a given speed.   This should be the power required by the motor to maintain constant speed (assuming 100% of the power from the motor is transmitted to the wheels--in reality, probably somewhere around 98% of the power makes it to the wheels).  The following are my results:

            

Speed (mph)  Motor Output (kW)  Power Loss (kW)  Difference (kW)  Reported Loss (kW)

50           8.1                -8.9             -0.9             -1.1

40           4.4                -5.8             -1.3             -1.3

30           2.6                -3.6             -1.0             -1.0

 

So at 30 mph, the motor must output 2.6 kW of mechanical power to maintain that speed.  When in neutral, the power drain at 30 mph from aerodynamic drag, rolling resistance, internal frictions, and a loss from the motor when in neutral is -3.6 kW.  That implies that when the car is in neutral, the motor is a drag on the momentum of the car, draining away about 2.6 - 3.6 = -1.0 kW of power in addition to the aerodynamic drag, rolling resistance, and internal frictions.  This matches the mechanical power from the motor that is reported by the car at 30 mph. 

 

So when performing P&G, there appears to be some headwinds to overcome when trying to operate the motor at a more efficient operating point.  The motor provides an additional drag on the momentum of the car when in neutral that is not present when in drive (I don't know understand the reason why this appears to be true).  In addition, less power from the motor is transmitted to the wheels during acceleration than at constant speed.  For these two reasons, you cannot obtain the full increase in efficiency with P&G that would be expected by operating the motor at a more efficient operating point.  The loss during the glide from the motor when in neutral is far larger than the loss during the pulse. 

Edited September 1, 2014 at 03:11 PM by larryh

[larryh]

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). 

Edited September 1, 2014 at 03:14 PM by larryh

[larryh]

I did a series of experiments to determine if the power loss from the motor reported by the car was real when in neutral while performing P&G (see the purple line in post 89 which shows negative power when the car is in neutral).  I first drove the car at constant speed and measured the mechanical power output of the motor.  Next, I placed the car in neutral and let it coast down from a few mph above the constant speed to a few mph below the constant speed.  Assuming the car and contents weigh 1871 kg, I can estimate the power loss from aerodynamic drag, rolling resistance, and internal frictions for a given speed.   This should be the power required by the motor to maintain constant speed (assuming 100% of the power from the motor is transmitted to the wheels--in reality, probably somewhere around 98% of the power makes it to the wheels).  The following are my results:

            

Speed (mph)  Motor Output (kW)  Power Loss (kW)  Difference (kW)  Reported Loss (kW)

50           8.1                -8.9             -0.9             -1.1

40           4.4                -5.8             -1.3             -1.3

30           2.6                -3.6             -1.0             -1.0

 

So at 30 mph, the motor must output 2.6 kW of mechanical power to maintain that speed.  When in neutral, the power drain at 30 mph from aerodynamic drag, rolling resistance, internal frictions, and a loss from the motor when in neutral is -3.6 kW.  That implies that when the car is in neutral, the motor is a drag on the momentum of the car, draining away about 2.6 - 3.6 = -1.0 kW of power in addition to the aerodynamic drag, rolling resistance, and internal frictions.  This matches the mechanical power from the motor that is reported by the car at 30 mph. 

 

So when performing P&G, there appears to be some headwinds to overcome when trying to operate the motor at a more efficient operating point.  The motor provides an additional drag on the momentum of the car when in neutral that is not present when in drive (I don't know understand the reason why this appears to be true).  In addition, less power from the motor is transmitted to the wheels during acceleration than at constant speed.  For these two reasons, you cannot obtain the full increase in efficiency with P&G that would be expected by operating the motor at a more efficient operating point.  The loss during the glide from the motor when in neutral is far larger than the loss during the pulse. 

 

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. 

Edited September 3, 2014 at 08:37 PM by larryh

[larryh]

I did a series of experiments to determine if the power loss from the motor reported by the car was real when in neutral while performing P&G (see the purple line in post 89 which shows negative power when the car is in neutral).  I first drove the car at constant speed and measured the mechanical power output of the motor.  Next, I placed the car in neutral and let it coast down from a few mph above the constant speed to a few mph below the constant speed.  Assuming the car and contents weigh 1871 kg, I can estimate the power loss from aerodynamic drag, rolling resistance, and internal frictions for a given speed.   This should be the power required by the motor to maintain constant speed (assuming 100% of the power from the motor is transmitted to the wheels--in reality, probably somewhere around 98% of the power makes it to the wheels).  The following are my results:

            

Speed (mph)  Motor Output (kW)  Power Loss (kW)  Difference (kW)  Reported Loss (kW)

50           8.1                -8.9             -0.9             -1.1

40           4.4                -5.8             -1.3             -1.3

30           2.6                -3.6             -1.0             -1.0

 

So at 30 mph, the motor must output 2.6 kW of mechanical power to maintain that speed.  When in neutral, the power drain at 30 mph from aerodynamic drag, rolling resistance, internal frictions, and a loss from the motor when in neutral is -3.6 kW.  That implies that when the car is in neutral, the motor is a drag on the momentum of the car, draining away about 2.6 - 3.6 = -1.0 kW of power in addition to the aerodynamic drag, rolling resistance, and internal frictions.  This matches the mechanical power from the motor that is reported by the car at 30 mph. 

 

 

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. 

[CJr]

I have a wifi OBDII port interface and use my iPad to monitor it using custom PIDs in the OBD fusion software. Normal PIDs work but not the custom Energi PIDs IHave found. Does anyone have a listing of a set of custom Energi PIDs that work on their software that I can port to OBD Fusion. I believe I can load Torque PIDs. Thanks for any pointers...

[larryh]

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.

 

Edited September 4, 2014 at 08:13 AM by larryh

[larryh]

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".

 

 

Edited September 5, 2014 at 04:54 PM by larryh