OBD II Data for HVB

[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. 

 

 

 

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. 

[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. 

 

 

 

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. 

Edited September 5, 2014 at 07:27 PM by larryh

[larryh]

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. 

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. 

 

Edited September 6, 2014 at 11:58 AM by larryh

[larryh]

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. 

Edited September 6, 2014 at 12:11 PM by larryh

[larryh]

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%. 

Edited September 7, 2014 at 02:32 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 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. 

 

 

 

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. 

Edited September 7, 2014 at 01:45 PM by larryh

[larryh]

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. 

Edited September 7, 2014 at 04:22 PM by larryh

[larryh]

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:

Edited September 7, 2014 at 05:14 PM by larryh

[murphy]

Try driving in L.  It's not really low like in a standard car, it maximizes regeneration without having to hit the sweet spot on the brake pedal.  The only downside is the brake lights don't come on while it is reducing your speed.

[larryh]

Try driving in L.  It's not really low like in a standard car, it maximizes regeneration without having to hit the sweet spot on the brake pedal.  The only downside is the brake lights don't come on while it is reducing your speed.

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. 

Edited September 7, 2014 at 06:13 PM by larryh

[larryh]

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:

 

Actually, for my city commute to work, I average about 33 mph.  The efficiency of the motor at 33 mph is around 60%.  I measure the efficiency of the motor for my commute to work at around 65%.  So city driving does actually make the conversion from electrical to mechanical energy more efficient.  Unfortunately, it don't see a comparable improvement in MPGe.  For constant 33 mph, mileage is more than 220 MPGe.  For my commute to work, mileage is at most 165 MPGe. 

 

For a perfect car, all the energy from the motor should be transferred to the wheels and road to propel the car.  For real cars, that does not happen.  It appears the transfer of mechanical energy to the road to propel the car is far less efficient for city driving than cruising at constant speed.  The power transfer at constant speed is nearly 100%.  For city driving, this efficiency drops to around 75%.  Apparently, there are significant losses associated with changes in the car's speed.  Not all the energy associated with the change makes it from/to the motor. 

[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. 

 

The following is an explanation of why driving in L is generally inefficient.  Assume regen is 100% efficient.  Suppose so far the motor has produced 0.75 kWh of mechanical energy from 1 kWh of electricity supplied by the HVB.  Efficiency is then currently 75%.

 

Now you stop and regen provides 0.1 kWh of energy to the HVB.  Since regen is perfect, the motor converts 0.1 kWh of mechanical energy to 0.1 kWh of electrical energy.

 

Now the motor has output 0.75 kWh - 0.1 kWh = 0.65 kWh of energy and the HVB has output 0.9 kWh of energy.  Efficiency has now gone down to 72.2%. 

 

Assume you now accelerate and the motor is 80% efficient in converting electricity to mechanical energy.  The motor will now have to convert 0.5 kWh of electricity to 0.4 kWh of mechanical energy to get back to 75% efficiency.  The motor will have output a total of 0.65 kWh + 0.4 kWh = 1.05 kWh of mechanical energy and the HVB will have output a total of 0.9 kWh + 0.5 kWh = 1.4 kWh of electrical energy.   You now have to accelerate such that you provide 4 times the KE that you recovered from stopping.  Basically, you have to accelerate to twice the speed from which you stopped.  You can't keep doing this without breaking a few speed limits (and exceeding the Energi's maximum speed).   In general, you don't want any unnecessary regen. 

 

Note that with P&G you get very inefficient regen during the glide.  The internal resistance of about -1.3 kW is present the entire time during the glide and no electricity is generated.  Fortunately, the energy loss associated with the internal resistance is orders of magnitude smaller compared to the energy you need to supply during the pulse.  If you are not going to get anything from the 1.3 kW power loss of the motor, it may be more advantageous to regen say maybe 5 kWh at 95% efficiency instead.  One would have to do the math to determine what is best (actually I have done the math and regen is more efficient).

 

Edited September 7, 2014 at 11:20 PM by larryh

[larryh]

If I am calculating things correctly, it looks like P&G at 45 mph will be more efficient if I leave the car in drive during the glide rather than shifting to neutral.  The motor will consume about -5 kW of power for regen.  Regen efficiency is estimated to be around 90%.   But most likely, the motor will have to produce more power for the extra regen during the glide, negating any increase in efficiency. 

 

Estimated efficiency for CC is 66%.

Estimated efficiency for P&G shifting into neutral is 72%.

Estimated efficiency for P&G leaving car in drive is 77%.

Optimal P&G efficiency (-3.5 kW of regen) is 77.5%.

Edited September 9, 2014 at 08:12 PM by larryh

[Hybridbear]

I wish Torque Pro would work with the Focus Electric. I'd guess that the more simple transmission setup of the BEV would result in a much lower internal resistance in neutral.

[larryh]

I doubt that the Focus Electric motor has lower internal resistance than the Fusion Energi.  I suspect that's just the way EV motors are.  However, the Focus Electric does not have the overhead of the planetary gear system (PGS) dragging it down.  For the same speed, the Focus Electric should require less power to propel the car since it does not have to also turn the planetary gear system.  I'm not sure what the power loss would be associated with the PGS. 

 

Are their any scanners, other than the prototype Scan Gauge II, that work with the Focus Electric? 

Edited September 8, 2014 at 08:36 PM by larryh

[Hybridbear]

However, the Focus Electric does not have the overhead of the planetary gear system (PGS) dragging it down.  For the same speed, the Focus Electric should require less power to propel the car since it does not have to also turn the planetary gear system.  I'm not sure what the power loss would be associated with the PGS.

My thought exactly.

 

Are their any scanners, other than the prototype Scan Gauge II, that work with the Focus Electric? 

FORScan.

[larryh]

The following plot shows the efficiency of the electric motor for my commute home from work.  The commute is 8 miles.  The blue line shows efficiency as a percentage, i.e. the ratio of the total energy output by the motor divided by the total energy output by the HVB.  The red line shows speed in mph.  You can see that approximately 70% of the energy from the HVB is converted to mechanical energy by the electric motor.  The greater the efficiency, the higher the MPGe. 

 

Note that every time I slow down, and regenerative braking occurs, efficiency drops.  At the beginning efficiency jumps to about 75% (during acceleration) and then at the first stop light, it plummets to a negative value, i.e. the battery did work but the motor didn't.  In fact, the motor received more energy from the wheels (during a short downhill) than it output to the wheels.   Not enough regen occurred to make up for the energy consumed from the HVB during the trip prior to the stop light.  Efficiency varies much more in the beginning and begins to level out at the end.  Since I haven't used much energy yet at the beginning, any changes in current efficiency have a much greater impact on the cumulative efficiency. 

 

Regen can never increase efficiency, it can only decrease efficiency.  However, efficiency in city driving, where lots of regen occurs, is significantly higher than a drive at constant speed taking the same length of time.  This seems like a contradiction.  However, in city driving, there is also a lot of acceleration from the stops.  The motor operates more efficiently during acceleration than driving at constant speed.  So although all the regen lowers overall efficiency, all the acceleration more than makes up for it. 

 

I averaged about 33 mph for the commute.  If I drove at a constant 33 mph to work, efficiency would be about 60% rather than the 68% shown on the chart at the end.  So that means I should get higher MPGe in city driving than driving constant speed at 33 mph since it is more efficient?  Right?  Unfortunately, it doesn't work that way.  Rolling resistance, aerodynamic drag, and internal frictions increase significantly at higher speeds.  If I drove at a constant 33 mph, the motor would have to output about 0.68 kWh of energy.   For the actual speeds traveled during the commute, I estimate that the amount of energy required for my commute to overcome this resistance was 1.01 kWh.  I spent much of the commute driving significantly faster than 33 mph where much more power is required to overcome the resistance.  The actual amount of energy output by the motor was 1.11 kWh.  So I am missing 1.11 - 1.01 = 0.09 kWh.  The remaining difference is due to an elevation change.  The elevation at home is about 23 meters more than at work. 

 

 

Edited September 15, 2014 at 08:43 PM by larryh

[larryh]

I have also measured the efficiency achieved when going up and down a 9% grade at 30 mph.  The length of the hill was 0.6 miles in each direction.  The following table shows the energy output of the HVB and Motor:

 

Direction  HVB        Motor      Efficiency

Down       -0.23 kWh  -0.24 kWh  97%

Up          0.47 kWh   0.38 kWh  81%

Combined    0.24 kWh   0.15 kWh  61%

 

So 97% downhill regen efficiency plus 81% uphill acceleration efficiency yields 61% overall efficiency.  This is the same efficiency one would expect traveling at a constant speed of 30 mph.  The MPGe for the trip was 171.  Not very good for 30 mph. 

 

If there were no aerodynamic drag, rolling resistance, and internal friction, the mechanical energy applied to the motor going downhilll should equal the energy required to go back uphill.  But since it exists, we see a 0.15 kWh difference.  That means a total of 0.15 kWh of energy was lost due to these sources in the 1.2 miles traveled.

 

Had I simply gone at constant speed for those 1.2 miles, the motor would have output exactly the same amount of energy from the HVB.  And at 61% efficiency, the HVB would have output exactly the same amount of energy and MPGe would be exactly the same.  Normally I would expect at least 220 MPGe at 30 mph.  I think energy is being lost somewhere--not all of the energy from the motor is making it to the road.  On a level road at constant 30 mph, the motor should have had to output less than 0.15 kWh of energy. 

Edited September 9, 2014 at 12:31 AM by larryh

[larryh]

At least the Energi has regen.  Without regen, as is the case with conventional cars, efficiency would be down to 31% and mileage down to 86 MPGe.

[larryh]

I tried a simple experiment this morning.  Starting from the same stopping point going about 55 mph before the stop sign, I tried two different methods to stop. 

 

- Lifting my foot off the accelerator with the car in Drive about 0.4 miles before the stop sign then using the brake as I approached the stop sign.   Regen occurred slowly over a long period of time. 

 

- Shifting the car into neutral until I got closer to the stop sign (to stop consuming/adding energy from/to the HVB), then shifting into Low to brake, and then shifting back to neutral, and eventually back to drive and applying the brake.  Regen occurred rapidly over a short period of time.

 

With the slow regen method, the motor/generator produced 0.0761 kWh of electricity.  With the fast regen method, the motor/generator produced 0.0896 kWh of electricity.   For fast regen method, the motor/generator produced 18% more electricity.  This takes into account the power used by the car's accessories.

 

With the slow regen method, the car claims that 0.062 kWh of energy was added to the HVB.  With the fast regen method, the car claims that 0.078 kWh of energy was added.  However, the slow regen method took much longer than the fast regen method.  The car's accessories used twice as much energy during the slow regen method than the fast regen method.  Taking that into account, the fast regen method added 19% more energy to the HVB. 

 

So it appears to make a difference in the amount of regen for slow vs. faster braking.  You can get about 18-19% more regen with faster braking if you do it right.  However, if you continue to press the accelerator to maintain the car's momentum until the last possible moment, and then brake at the maximum charge limit, you probably won't realize much, if any, of the 18-19% additional regen.  It may be wiped out by the additional time you were using energy by applying the accelerator. 

 

Note that you don't want to apply the brakes when the car is in neutral.  You will get no regen from the brakes driving in neutral.

 

One would have to experiment a lot more to determine the optimal braking strategy for various situations.

 

It appears the better strategy for stopping for a stop light is to lift your foot off the accelerator as as soon as possible and coast to the stop light, applying the brake as required when you get close to the stop light.  You get more regen when stopping faster, but to maintain speed until closer to the stop light, you consume more energy than you gain in regen.  I am assuming that you leave the car in drive.

 

The following table shows the amount of kinetic energy at various speeds stored in the HVB with the two different methods of stopping, i.e. maintain constant speed until you get close to the stop light and then apply maximum regen braking vs. coasting in drive and then applying the brake when close to the stop light.  This assumes you have plenty of time to stop and coast most of the way to the stop light.

 

Initial Speed      Max Regen     Coast

(mph)

30                 59%           69%

40                 52%           62%

50                 46%           57%

 

If you need to stop quickly (no time to coast), then the amount of kinetic energy stored in the HVB (max regen) is approximately as shown in the following table:

 

Initial Speed      Max Regen

(mph)

30                 81%

40                 79%

50                 77%

 

Note that the motor converts approximately 95% of the mechanical energy to electrical energy even when coasting.  However, with coasting, you slow down much more slowly and more energy is lost due to aerodynamic drag, rolling resistance, and internal frictions, which is then not available to the motor to convert to electricity. 

Edited September 12, 2014 at 11:23 PM by larryh

[Hybridbear]

What about coasting in neutral and then braking near the limit versus braking gently over time?

[larryh]

Coasting in neutral will most likely yield slightly worse results than coasting in drive.  Since you are not slowing down as fast in neutral, more kinetic energy will be lost due to aerodynamic drag, rolling resistance, and internal frictions.  You are already getting around 95% regen efficiency with coasting in drive.  You are not going to get much more efficiency by slowing down at a faster rate.  So you are just losing more energy due to resistance and not getting much gain in return from improved regen efficiency. 

Edited September 12, 2014 at 11:25 PM by larryh

[larryh]

When coasting in neutral, I slow from 50 mph to around 40 mph and then shift into L to complete the stop.  The total distance to stop is the same as in post #120.  I lose 0.05 kWh of kinetic energy due to resistance while in neutral.  Of the remaining kinetic energy, I capture 0.064 kWh.  I have captured a total of 51% of the available kinetic energy.  If I leave it in drive, coast closer to the stop sign, and then brake until I come to a complete stop, I capture 57% of the available kinetic energy.  So coasting in neutral is not optimal. 

Edited September 13, 2014 at 06:59 PM by larryh

[Hybridbear]

One item that I don't think has been covered in this thread is that the HVB fan appears to do more than just cool the HVB. It also cools the DCDC converter. On a recent 200+ mile drive to Wisconsin I observed that the HVB temp started out at 82F. The DCDC converter temp also started out around 80F. After a short time the DCDC converter temp rose to 165-175F. It stabilized at about 174F. It stayed constant at that temperature until the HVB fan came on. The HVB fan ran until the HVB temp dropped from 84F to 66F. While the HVB  fan was running the DCDC converter temp dropped and stabilized at about 80F. Within just a few minutes after the HVB fan turned off the DCDC converter temp had risen again to about 174F.

 

Outside temperatures were 39-42F during this trip.

[larryh]

I wonder where the sensors are that measure temperature.   If the sensors were buried deep within the HVB and converter, I would think it would take quite some time for the temperature to rise or fall.