Efficient EV Mode Driving Techniques

[larryh]

I thought I would share some of the knowledge I have gained by analyzing the data I have been gathering from my car that could prove useful in improving EV mode mileage.  But in this first post, I wish to establish a context to discuss efficient driving techniques that I plan to discuss in subsequent posts.

 

There are three basic forms of energy that we need to be concerned with when driving the car: 

1. Electrical energy from the HVB,
2. Kinetic energy from the car’s motion, and
3. Potential energy associated with the car’s elevation.

There are also several frictional forces that come into play:

1. Aerodynamic drag,
2. Tire rolling resistance,
3. Various internal frictions associated with the transmission, wheel bearings, gears, etc.
4. Slippage associated with transmitting power from the wheels to the road, etc.

The efficiency of the conversion between these different forms of energy along with the frictional forces acting on the car determines the mileage we get when driving the car.  The latter two types of energy, Kinetic and Potential, are referred to as Mechanical energy vs. Electrical energy.

 

The conversion between Kinetic and Potential Energy (in the absence of frictional forces) is 100% efficient.  When coasting, the car slows down when going up a hill, converting Kinetic to Potential Energy.  The car speeds up when going downhill converting Potential to Kinetic Energy. 

 

The motor converts Electrical to Mechanical energy.  In normal driving, the motor efficiency ranges from about 50% to 90%, depending on the power output of the motor.  The motor is more efficient (up to a limit) when outputting more power.  The motor also converts Mechanical to Electrical Energy via regen.  The efficiency of this conversion is very good, i.e. greater than 95%.  It is slightly more efficient with increasing power. 

 

In the absence of frictional forces, the total amount of Mechanical Energy output by the motor for any trip that returned to the starting location would be zero.   The Mechanical Energy output by the motor to accelerate the car would be returned back to the motor during deceleration.  Similarly, the Mechanical Energy output by the motor to go up a hill would be returned back to the motor when coming back down.  If all energy conversions were 100% efficient, the total amount of Electrical Energy consumed for the trip would also be zero. 

 

The reason that energy is consumed from the HVB is that frictional forces act on the car and the conversion between Electrical and Mechanical energy is not 100% efficient.  Frictional forces increase with the square of the speed of the car.  Going 60 mph vs 30 mph requires a lot more energy to go the same distance (more than double).  The only thing that can be done to reduce the impact of frictional forces on mileage is to drive slower.

 

To improve mileage, we also want to minimize the conversion between Electrical and Mechanical energy.  Obviously, we must convert some Electrical to Mechanical energy or the car won’t go anywhere.  However, there are many ways we can avoid unnecessary conversions. 

 

Trying to maintain a constant speed is one technique that can be used to avoid unnecessary conversion between Electrical and Mechanical Energy.  If you speed up, gaining 0.01 kWh of Mechanical energy, the conversion from Electrical to Mechanical energy might be about 80% efficient and would require 0.01/0.8 = 0.0125 kWh of Electrical energy.  If you immediately slow back down again via regen, the efficiency of the conversion from Mechanical to Electrical energy would be about 95% and you would capture 0.01*0.95 = 0.0095 kWh of Electrical energy.  Versus going at a constant speed, you will have consumed an additional 0.0125 – 0.0095 = 0.003 kWh of Electrical energy.  Actually, you will have consumed even more since frictional forces are greater at higher speeds.  This is one reason why driving in Low is inefficient, if you are not careful, you will cause unnecessary and wasteful conversions between Electrical and Mechanical energy.

 

There are several additional ways to avoid unnecessary conversions between Electrical and Mechanical energy.  If you need to stop for a stop light, it is better to slow down sooner than later.  If you wish to accelerate, it is better to accelerate slower than faster.  When going up a hill, use up some of the Kinetic Energy to go up the hill by allowing the car to slow down a little.  When descending the hill, use up some of the Potential Energy allowing the car to regain speed.  Conversions between Kinetic and Potential energy are 100% efficient.   Converting Electrical to Mechanical Energy to accelerate up the hill and then use regen to convert some of the Potential Energy back to Electrical energy is wasteful.

Edited September 24, 2014 at 01:12 AM by larryh

[larryh]

The following plot shows the mechanical energy output of the motor required to travel a mile at a given speed on four different days.  The legend indicates the outside temperature.  This is the mechanical energy required to overcome the frictional forces at a given speed.  Since the conversion from electrical to mechanical energy is from 50% to 80% efficient, the electrical energy consumed from the HVB to overcome frictional forces will range from 1/50% = 2 times greater (at slow speeds) to 1/80% = 1.25 times greater (at higher speeds) than the mechanical energy shown in the plot. 

 

The mechanical energy required to overcome frictional forces at 60 mph is about 2.5 times that required at 30 mph.  The colder the outside temperature, the greater the frictional forces.  At slow speeds, internal friction and tire rolling resistance dominate.  At higher speeds, aerodynamic drag dominates.  Colder temperatures and higher pressure mean denser air and consequently more aerodynamic drag.  All of these resistance forces increase with lower temperatures.  Although I took the measurements when the wind was calm, a small amount of wind is still present.   The wind direction and speed will impact the results.  It takes more energy to overcome headwinds than tail winds. 

 

 

Edited September 24, 2014 at 09:32 AM by larryh

[Hybridbear]

Trying to maintain a constant speed is one technique that can be used to avoid unnecessary conversion between Electrical and Mechanical Energy.  If you speed up, gaining 0.01 kWh of Mechanical energy, the conversion from Electrical to Mechanical energy might be about 80% efficient and would require 0.01/0.8 = 0.0125 kWh of Electrical energy.  If you immediately slow back down again via regen, the efficiency of the conversion from Mechanical to Electrical energy would be about 95% and you would capture 0.01*0.95 = 0.0095 kWh of Electrical energy.  Versus going at a constant speed, you will have consumed an additional 0.0125 – 0.0095 = 0.003 kWh of Electrical energy.  Actually, you will have consumed even more since frictional forces are greater at higher speeds.  This is one reason why driving in Low is inefficient, if you are not careful, you will cause unnecessary and wasteful conversions between Electrical and Mechanical energy.

 

There are several additional ways to avoid unnecessary conversions between Electrical and Mechanical energy.  If you need to stop for a stop light, it is better to slow down sooner than later.  If you wish to accelerate, it is better to accelerate slower than faster.  When going up a hill, use up some of the Kinetic Energy to go up the hill by allowing the car to slow down a little.  When descending the hill, use up some of the Potential Energy allowing the car to regain speed.  Conversions between Kinetic and Potential energy are 100% efficient.   Converting Electrical to Mechanical Energy to accelerate up the hill and then use regen to convert some of the Potential Energy back to Electrical energy is wasteful.

I would add that there's a balance with this though. If you're coming up on a light that's red that you expect to turn green before you arrive then you're better off trying to plan so that you do the minimal amount of braking and subsequent acceleration. I've had plenty of times where I start slowing down early for a red light, as mentioned above, only to have the light turn green. In those scenarios I would have been better off maintaining my cruising speed the entire time since braking wasn't required. However, the opposite can also happen. Sometimes I'm approaching a red light that I expect to turn green so I don't slow down but then it doesn't turn green so I have to brake harder (still a 100% brake score though) and slow down to a lower speed without stopping before the light turns green versus if I had just slowed down gradually from farther back.

 

One of the Fusion Hybrid owners has said it best: "Drive smarter, not faster." Sometimes fast acceleration is best when it allows you to make the next light green versus accelerating slowly and having to stop again quickly.

[Hybridbear]

My question for efficient city driving is when you're driving from stop sign to stop sign which is very common for us. In those situations, is it best to accelerate up to speed and then coast in neutral until needing to brake or accelerate up to speed and then maintain a light pedal pressure to continue discharging the HVB until beginning to coast & brake. I need to do some experiments.

[larryh]

The following plot shows the mechanical energy recovered by different methods of stopping on a level road from 35 mph.  (Since the road is level, there is no change in potential energy.)  For the solid lines (slow stop), I immediately lift my foot off the accelerator, coast in Drive until getting closer to the stop light, and then apply the brakes to complete the stop.  You get approximately 5 kW of regen when coasting in Drive to slow the car down.  For the dashed lines (fast stop), I maintain speed until I get closer to the stop light, then I shift into Low for max regen to come to a stop at the stop light.  In both cases, the total distance traveled is the same. 

 

The green lines show the energy loss due to frictional forces vs. distance.  Since, on average, I am going faster during the fast stop (I maintain speed until the last minute), the fast stop will experience greater losses.  The total loss for the fast stop is about -0.023 kWh.  For the slow stop, it is about -0.020 kWh.  That's energy that you cannot recover through regenerative braking.  So from a frictional loss perspective, the slow stop is better, since you waste less energy.

 

The blue lines show the net amount of mechanical energy supplied by/to the motor.  I am assuming that all changes to mechanical energy (kinetic energy in this example) of the car appear at the motor.  That means that the transmission and tires are 100% efficient in transmitting the power from the motor to propel the car.  The actual efficiency is probably something like 98% efficient.  But in any case, I am lumping the transmission losses in with the frictional losses, so I can assume that 100% of the change in kinetic energy appears at the motor.  That means the net energy applied to the motor after coming to a complete stop will equal the initial kinetic energy of the car less any frictional losses.  It doesn't matter how I stop, as long as I get a 100% brake score and do not use the friction brakes.  The initial kinetic energy of the car was 0.063 kWh.  So after coming to a complete stop, the net amount of energy supplied to the motor will be -0.063 - (-0.020) = -0.043 kWh for the slow stop, and -0.063 - (-0.023) = -0.040 kWh for the fast stop.  The amount is negative because energy is being supplied to the motor for regenerative braking.  The difference in energy applied to the motor for regen for the fast vs. slow stop only differs by the frictional loss differences for the two stops. 

 

The red line shows the net electrical energy from the HVB vs. distance for the two stops.  This is where the efficiency of the conversion between electrical and mechanical energy comes into play.  For the slow stop, approximately 95% of the mechanical power applied to the motor is converted into electricity for the entire stop.  For the fast stop, about 70% of the electrical power is converted to mechanical power to maintain speed, and then during regenerative braking, about 97% of the mechanical power is converted to electrical power.  The total electrical energy recovered during the slow stop is -0.041 kWh.  The electrical energy recovered during the fast stop is -0.031 kWh.  You get 0.01 kWh less electricity recovered with the fast stop.  The slow stop is more efficient. 

 

The problem with the fast stop, is that the motor is rather inefficient in converting electrical to mechanical energy going 35 mph.  You can see this via the difference between the red and blue lines.  For the fast stop, the separation between the red and blue dashed lines increases while providing power to the motor to maintain constant speed for the first 0.12 miles. This difference represents a loss during the conversion of electrical to mechanical power by the motor that you can never recover.  For the remainder of the stop, during regenerative braking, the conversion from mechanical to electrical energy is very efficient, and very little additional separation occurs between the lines.  But unfortunately, the damage is already done. 

 

For the slow stop, the regen efficiency remains high throughout the stop, so very little separation occurs between the red and blue solid lines.  Note that after about 0.16 miles, I start applying the brakes and regen increases.

 

So there are two main reasons why slow braking was more efficient than fast braking:

1.  The higher average speed during fast braking means more frictional losses.

2.  The motor converts mechanical to electrical energy (via regen) far more efficiently than it converts electrical to mechanical energy (to propel the car).  You don't want to supply any more electrical power to the motor than is necessary during the stop.  If the motor converted electrical to mechanical power as efficiently as regen, then the only difference in the energy recovered during regen would be due to frictional losses.

 

 

 

Edited September 24, 2014 at 11:06 PM by larryh

[larryh]

My question for efficient city driving is when you're driving from stop sign to stop sign which is very common for us. In those situations, is it best to accelerate up to speed and then coast in neutral until needing to brake or accelerate up to speed and then maintain a light pedal pressure to continue discharging the HVB until beginning to coast & brake. I need to do some experiments.

 

Shifting into neutral and applying light pressure to the accelerator do the same thing.  You should notice no difference in efficiency between the two, provided you are able to control the accelerator so that the motor neither consumes no regens electricity.  But that is hard to do.  The less energy consumed from the HVB to accelerate up to speed, the better.  Ideally, you would like to totally avoid regen.  You would slowly increase speed, then shift into neutral and glide to a stop without using the brakes.   But that is not really practical.  If you have regen, you could have supplied that much less energy to the motor to begin with and avoided the 20-30% loss associated with converting electrical to mechanical power (energy that you can never recover). 

Edited September 24, 2014 at 11:50 PM by larryh

[larryh]

I would add that there's a balance with this though. If you're coming up on a light that's red that you expect to turn green before you arrive then you're better off trying to plan so that you do the minimal amount of braking and subsequent acceleration. I've had plenty of times where I start slowing down early for a red light, as mentioned above, only to have the light turn green. In those scenarios I would have been better off maintaining my cruising speed the entire time since braking wasn't required. However, the opposite can also happen. Sometimes I'm approaching a red light that I expect to turn green so I don't slow down but then it doesn't turn green so I have to brake harder (still a 100% brake score though) and slow down to a lower speed without stopping before the light turns green versus if I had just slowed down gradually from farther back.

 

One of the Fusion Hybrid owners has said it best: "Drive smarter, not faster." Sometimes fast acceleration is best when it allows you to make the next light green versus accelerating slowly and having to stop again quickly.

 

You definitely want to avoid regen and then immediately having to accelerate back up to speed.  You will have captured most of the change in kinetic energy in the HVB when slowing down.  But, when accelerating to resume your original speed, the conversion from electrical to mechanical energy will be around 80% efficient.  You have to supply 25% more energy than was captured through regen to get back up to speed.   You now have less energy remaining in the HVB than had you maintained constant speed. 

 

However, note that if you slow down by coasting in neutral or applying slight pressure to the accelerator to avoid regen, and then accelerate back up to speed, efficiency is actually increased over maintaining constant speed.  That is the principle of pulse and glide.  The key difference between pulse and glide and the previous paragraph is pulse and glide avoids regen.   As long as you avoid regen during speed changes, you don't loose much efficiency.  You may even gain efficiency.  You can think of regen as magnifying the inefficiency of the motor converting electrical energy to mechanical energy when making unnecessary speed changes.  The motor was inefficient getting up to speed the first time, you now slow down and capture regen, and then make the motor do the inefficient conversion a second time to get back up to speed. 

Edited September 25, 2014 at 12:21 AM by larryh

[larryh]

The following plot shows the mechanical energy recovered by different methods of stopping on a level road from 35 mph.  (Since the road is level, there is no change in potential energy.)  For the solid lines (slow stop), I immediately lift my foot off the accelerator, coast in Drive until getting closer to the stop light, and then apply the brakes to complete the stop.  You get approximately 5 kW of regen when coasting in Drive to slow the car down.  For the dashed lines (fast stop), I maintain speed until I get closer to the stop light, then I shift into Low for max regen to come to a stop at the stop light.  In both cases, the total distance traveled is the same. 

 

 

Another method of stopping is to shift into neutral until you come closer to the stop sign and then shift into L to complete the stop.  However, you will get less regen with this method than with slow braking.  You will coast for about 0.14 miles and speed will have dropped to 29 mph.  The total amount of energy lost due to frictional forces is -0.024 kWh.  This is slightly more than with the fast braking method, even though your average speed will be slower when shifting into neutral.  This is because of internal friction in the motor itself is now slowing the car down in addition to the other frictional forces.  So now -0.063 - (-0.024) = -0.039 kWh of kinetic energy remain for regen.  Regen will recover about 97% of that or -0.039 * 0.97 = -0.038 kWh of electrical energy.  This is better than the fast stop (-0.031 kWh), but less than the slow stop (-0.043 kWh).  The main reason for less regen than the slow stop is because the average speed is higher when shifting into neutral so more energy is lost due to frictional forces. 

Edited September 26, 2014 at 09:45 AM by larryh

[larryh]

This chart illustrates the energy consumed during acceleration from a stop to 40 mph over a distance of 0.25 miles. For the dashed lines (fast accel), I accelerated over 4 bars on the Empower screen for a distance of 0.05 miles to 40 mph and then maintained that speed for the remainder of the 0.25 miles. For the solid lines (slow accel), I accelerated at less than 2 bars for the entire 0.25 miles at which point I reached 40 mph. Slow acceleration consumed less energy from the HVB than fast acceleration. Mainly, because the energy to overcome friction is much greater at higher speeds and the speed over the entire trip was higher for fast acceleration vs. slow acceleration. In addition, motor efficiency is higher at higher power levels. The power output of the motor remained high longer for slow acceleration vs. fast acceleration.

 

The green lines show the power loss from friction.  Since the car is going faster throughout the entire trip for fast acceleration vs. slow acceleration, there is more energy loss from friction with fast acceleration, i.e. the dashed green line is above the solid green line.  In this case, the loss for fast acceleration was 0.042 kWh, and for slow acceleration the loss was 0.034 kWh.  0.042 - 0.034 = 0.008 kWh of additional energy was required for fast acceleration.

 

The blue lines show the mechanical energy supplied by the motor.  The motor has to supply the kinetic energy for 40 mph plus the energy losses from friction.  So for fast acceleration, the motor had to supply the 0.008 kWh more energy to overcome the additional losses from friction. 

 

The red lines show the electrical energy supplied by the HVB.  With fast acceleration, the motor was operating at high power levels only during acceleration.  For the remainder of the 0.25 miles, much lower power was required to overcome friction and maintain constant speed.  With slow acceleration, the motor was operating at high power levels throughout the entire 0.25 miles.  Since the motor was operating at higher power levels for a longer period of time with slow acceleration, the motor operated slightly more efficiently for slow acceleration, and hence the motor required slightly less electrical energy from the HVB to produce the same amount of mechanical power.  For fast acceleration 0.150 kWh of electrical energy was consumed, and for slow 0.140 kWh of electrical energy was consumed. 

 

Similar to fast braking, there are two main reasons fast acceleration is less efficient:

1.  Being at higher speeds throughout the trip with fast vs. slow acceleration means more energy loss due to friction which must be supplied by the motor and HVB.

2.  Being at lower power levels for much of the trip while maintaining constant speed with fast vs. slow acceleration means the motor operates less efficiently. 

 

 

Edited September 26, 2014 at 04:44 PM by larryh

[Fat Fusion]

Larry would it be fair to say that speed is the overriding factor in all of this?  All the micromanaging in the world can be negated by the wind resistance if you go "too fast?"  Nice analysis by the way.

[larryh]

Yes, the majority of energy consumed from the HVB is used to overcome frictional losses, which greatly increase with speed.  If you encounter a 20 mph head wind going 40 mph, the aerodynamic drag on the car will almost double, meaning now the trip will require almost twice as much energy.   If you are going 60 mph, the difference won't be quite as dramatic, but will still be significant.  Hopefully you will not experience the full force of a 20 mph head wind during the entire trip. 

 

To a lesser extent, motor efficiency determines the remaining energy consumption from the HVB.  The less change in Potential and Kinetic energy (the fewer stops and hills), the less the impact. 

Edited September 26, 2014 at 04:58 PM by larryh

[Fat Fusion]

I agree with everything you're saying except, "This is one reason why driving in Low is inefficient,"

 

IMO it's more efficient if you're careful about your driving since you're less likely to use the friction brakes to assist the regen.  I would say, "Driving in L can be slightly less efficient."

 

The only time I don't drive in L is right after getting a full charge.  Once I drop down to 98% or so, then I put it in L every single time.  Much better "drive-ability" in L, IMO.

[larryh]

If you drive all the time in L, it is very easy to cause unnecessary regen.  You have to be very careful and not allow regen to slow down the car when you don't need to slow down.  When cruising, you will have to maintain constant pressure on the accelerator to prevent regen.  If you allow regen to slow the car down and then immediately accelerate back up to speed, you have wasted energy.  If you are using L deliberately to slow down the car so you can avoid using the friction brakes, then that is fine. But as I have indicated in my previous posts, if possible, it is more efficient to coast to a stop in D than it is in L and avoid using the brakes at all until the very end of the stop.  To achieve the same effect in L, you would have to slow down using regen well before the stop light and then continue until you reach the stop light at a greatly reduced speed.

Edited September 26, 2014 at 09:57 PM by larryh

[larryh]

The following plot, which I have posted in another thread, shows the efficiency of the electric motor for a given mechanical power output observed during some of my trips.  The red line is a rough estimate of the efficiency.  In actuality, efficiency peaks somewhere around 32 kW and then begins to drop off again.  Efficiency probably peaks in the low 90's.  I don't have enough data at high power output to make an accurate estimate. 

 

The following table provides a rough idea of the efficiency of the motor traveling at constant speed on a level paved road with no wind:

 

Speed   Mechanical   Efficiency

(mph)   Power (kW)

  1      0.05        55%

 10      0.6         57%

 20      1.3         61%

 30      2.6         65%

 40      4.6         68%

 50      7.6         71%

 60     11.7         73%

 

 

 

Edited September 26, 2014 at 08:17 PM by larryh

[larryh]

You can estimate the amount of HVB electrical energy, E, consumed for a trip using the following equation:

 

E = M/e = (dPE + dKE - F - B)/e,

 

where:

 

M is the mechanical energy output by the motor,

 

e, is the efficiency of the motor.  See post 14.  For regen efficiency r, the equation is E = M*r.

 

dPE is the change in potential energy for the trip, i.e. the PE of the car at the end minus the PE of the car at the start of the trip.  If the elevation is the same at the start and end of the trip, then this quantity is 0.

 

dKE is the change in kinetic energy for the trip, i.e. the KE of the car at the end minus the KE of the car at the start.  If the car is at rest at the start and end of the trip, then this quantity is 0.

 

F is the energy loss from friction.  This quantity is defined to be negative.  This depends on the speeds of the car during the trip. See post 2.

 

B is the energy lost when the friction brakes are applied.  This quantity is defined to be negative. Hopefully you get a 100% brake score, in which case this quantity is 0.

 

As an example of using this, suppose we again consider the braking example in post 5.  We start at 40 mph and immediately brake to slow down to 20 mph.  We then continue at 20 mph until closer to the stop light. Finally, we brake again to come to a complete stop at the light.

 

If you go through the math, you can determine that if the regen power of the motor is -20 kW during the initial braking, it will take about 6 seconds to reduce speed from 40 mph to 20 mph.  Since we assume that the road is level, the change in PE is 0.  The change in kinetic energy from 40 mph to 20 mph is -0.043 kWh.  The energy loss from friction is -0.006 kWh (I used Microsoft Excel to compute this).  Assuming regen is 97% efficient, we then get:

 

E = M * 0.97 = (-0.043 + 0.006) * 0.97 = -0.036 kWh of energy supplied to the HVB.

 

For the next 22 seconds, we travel at constant speed of 20 mph.  The change in potential energy and kinetic energy during this segment of the trip is 0.  The energy loss from friction is -0.011 kWh.  The motor has to supply this much energy to maintain constant speed at 20 mph.  Assuming the motor is 61% efficient at 20 mph, we get:

 

E = M / 0.61 = (0.011) / 0.61 = 0.018 kWh of energy supplied by the HVB.

 

For the final 8 seconds, we again apply the brakes to come to a stop.  The change in potential energy when reducing speed from 20 mph to 0 mph is -0.021 kWh.  The energy loss from friction is -0.001 kWh.  Assuming 95% regen efficiency,

 

E = M * 0.95 = (-0.021 + 0.001) * 0.95 = -0.019 kWh of energy is supplied to the HVB.

 

Thus we have for the three segments of the trip:

 

Segment    Motor           HVB

           Energy (kWh)    Energy (kWh)

1          -0.037          -0.036

2           0.011           0.018

3          -0.020          -0.019

Total      -0.045          -0.036

 

So this method of stopping results in less regen (-0.036 kWh) than slow braking (-0.041 kWh).  The energy loss from friction for slow braking was -0.020 kWh and for this method it is -0.018 kWh.  Thus this method has more energy applied to the motor for regen (-0.045) vs. slow braking (-0.043 kWh), the entire difference being explained by the difference in energy losses from friction for the two methods.  With this method, you will be traveling at slower speeds most of the time compared to slow braking, and hence less friction.  However, even though more mechanical energy is applied to the motor with this method, less electrical energy is supplied to the HVB.  The motor is relatively inefficient when maintaining constant speed at 20 mph and this severely impacts the overall efficiency for the trip (80% regen efficiency for this method vs. 95% regen efficiency for slow braking).  I think the best strategy for stopping is simply to let up on the accelerator while in D well before the stop sign, and then when you get closer, apply the brakes to complete the stop. 

Edited September 27, 2014 at 02:38 PM by larryh

[larryh]

Using analysis similar to the previous post, one can determine the impact on efficiency of slowing down and then immediately speeding back up again to resume the original speed.   It appears that doing this may slightly improve efficiency. 

 

Suppose you are going 40 mph, slow down to 35 mph, and then immediately speed back up again to 40 mph.  Since the average speed will be less after slowing down, the energy loss from friction is going to be less than if you maintained constant speed.  So less mechanical energy will be required for the trip.  At the same time, the efficiency of the trip does not seem to be impacted.  As a result, you may improve efficiency slightly by doing this, simply because frictional losses are reduced by the temporary lower speed. 

Consider five different ways of slowing down:

1. Reducing the power output of the motor to half what is required to maintain 40 mph.
2. Reducing the power output of the motor to about 0.5 kW, far less than required to maintain 40 mph.

3. Coasting in Neutral.

4. Coasting in Drive

5. Coasting in Low.

 

The latter two result in regen. 

 

For each of these ways of slowing down, the energy consumption is as follows:

 

Method        Motor Mechanical     HVB Electrical

              Energy (kWh)         Energy (kWh)

1              0.019                0.030            62%

2              0.003                0.010            27%

3             -0.004                0.000             0%

4             -0.011               -0.010            95%

5             -0.015               -0.015            97%

 

For the first two methods, energy is consumed from the HVB to produce the motor mechanical power output and the efficiency is as shown.  For coasting in Neutral, there is internal friction from the motor that slows the car down in additional to the other sources of friction.  The electrical connection between the motor and HVB is disconnected.   The last two result in regen.  The regen efficiency is as shown.

 

When accelerating back up to speed, the energy consumed is the same, regardless of the method used to slow down:

 

Motor Mechanical     HVB Electrical   Efficiency

Energy (kWh)         Energy (kWh)

0.023                0.027            84%

 

Summing these two results, we can calculate the overall efficiency of the portion of the trip where the car slowed down and sped back up again:

 

Method        Motor Mechanical     HVB Electrical    Efficiency

              Energy (kWh)         Energy (kWh)

1              0.041                0.057            72%

2              0.025                0.037            69%

3              0.018                0.027            71%

4              0.012                0.016            61%

5              0.007                0.012            67%

 

The motor efficiency for constant 40 mph is around 67%, similar to the efficiency of the portion of the trip where the car slowed down and sped back up.  It appears that slowing down from 40 mph to 35 mph and speeding back up again will have little impact on overall efficiency of the entire trip.  The difference in efficiency of the different methods is not significant, probably just noise. 

 

So if you drive in L, efficiency is probably not impacted much by unnecessary regen.  I will have to actually try it out and see if this prediction holds true or there is a flaw in this analysis. 

Edited September 27, 2014 at 10:30 PM by larryh

[larryh]

I went out and did 5 cycles of slowing down and resuming speed in Low, one right after the other.  The efficiency that I measured from the actual data was 64%.  I also measured the efficiency over the same route for constant 40 mph to be 68%.    It is very difficult to determine how accurate the data is when subjecting the car to extremes of acceleration and deceleration.  In any case, no one would drive in such an extreme manner.  So as far as I can tell, unnecessary acceleration/decelerations when driving in L will not significantly impact efficiency.

 

However, the data I collected shows that the motor output 25% more energy when slowing down and speeding up vs. constant speed.  Thus the HVB had to output 25% more energy for the former.  I don't know why--it could just be noisy data.  Until I understand this better, my conclusion about the impact slowing down and speeding up is indeterminate. 

Edited September 28, 2014 at 10:48 AM by larryh

[larryh]

I have discovered the source of the missing energy in the previous post.  Acceleration and deceleration cause additional energy losses above and beyond the normal frictional energy losses that occur at constant speed.  If you accelerate quickly, you lose a significant amount of energy induced by the acceleration itself, extra energy that the motor has to provide.  You don't want to accelerate quickly and you don't want to speed up and slow down unnecessarily.

 

The following plots show this additional loss of energy for high, moderate, and slow acceleration/deceleration rates.  The red lines show speed vs. distance (speed has been divided by 1000).  The blue lines show the accumulated additional energy loss induced by acceleration/deceleration vs. distance.  The loss for fast acceleration was four times the loss for slow acceleration.  When drawing the blue lines, I have subtracted out the usual constant speed frictional losses as well as the energy associated with the change in the car's kinetic energy (speed) from energy output by the motor.  What remains is solely a function of acceleration and deceleration.

 

Normally, when you drive at constant speed, the motor must supply the energy needed to overcome frictional losses associated with aerodynamic drag, tire rolling resistance, internal frictions inside the transmission, etc.  However, when you accelerate/decelerate, additional energy losses occur above and beyond these constant speed frictional losses.  From the plots below, it appears that most of the loss occurs during acceleration and the transition between acceleration/deceleration.  There doesn't seem to be much loss during deceleration.

 

In the fast acceleration plot, you can see spikes in the curve whenever the car transitions between accelerating and decelerating.  This is probably due, in part, to the rotational inertia of the motor, generator, and planetary gear system.   During acceleration, the rotational speed was increasing, and during deceleration, it is now decreasing.  The greater the difference between the acceleration and deceleration rates, i.e. the more abrupt the change, the greater the energy spikes. 

 

It appears that you want to avoid unnecessarily slowing down and speeding up.  The acceleration/deceleration specific energy losses will require the motor to output extra energy to make up these additional losses.  The faster the acceleration, the greater the loss.  Abrupt changes between acceleration/deceleration only add to the loss.  These energy losses are significant.  They can be 25% of the normal constant speed frictional losses. 

 

So if you drive in Low, you need to be wary of these acceleration/deceleration energy losses.  You don't want to change speed abruptly or unnecessarily.  When doing so, these additional energy losses will quickly add up. 

 

This provides another reason why slow acceleration is better than fast acceleration.  There is more mechanical energy loss associated with fast acceleration.

 

 

Edited September 28, 2014 at 04:09 PM by larryh

[larryh]

The following plot shows the mechanical energy loss when transforming the mechanical energy output from the motor to kinetic energy associated with the motion of the car during acceleration.  It is energy loss above the normal losses associated with aerodynamic drag, tire rolling resistance, and internal frictions of the transmission, i.e. the normal frictional losses when maintaining constant speed. The blue line is for max acceleration. The red line is for acceleration at about 1.8 bars on the Empower screen. Acceleration begins from a stop to 40 mph. 40 mph was achieved after about 0.05 miles with max acceleration and after about 0.2 miles with 1.8 bar acceleration. About 10% of the mechanical energy from the motor is lost and not transformed into kinetic energy during acceleration. More is lost the faster you accelerate.

 

When you accelerate, not all the power is transmitted to the wheels to increase the speed of the car.  I'm not sure what the reason for it is, but it probably has to do with the rotational inertia of the wheels, planetary gear system, motor, and other gears in the transmission.  So you want to avoid making any unnecessary changes in speed, i.e. constant speed will minimize this loss.  You certainly don't want to accelerate quickly.  So there are three main factors that make fast acceleration less efficient than slow acceleration:

 

1.  The average speed is higher with fast acceleration meaning there is more energy loss from the constant speed frictional losses.

2.  Motor efficiency is less.  You are spending less time accelerating when motor efficiency is the greatest (transforming electrical to mechanical energy). 

3.  More energy is lost when transforming the mechanical energy output from the motor to kinetic energy associated with the motion of the car.

 

You should heed the driving coach to obtain high scores for acceleration. There seems to be a similar loss (but perhaps smaller) during deceleration and regenerative braking. 

 

Edited October 5, 2014 at 05:11 PM by larryh

[larryh]

When the car is cold, the internal friction is significantly higher than when warm.  The following table shows the motor power output to propel the car at 20 mph vs. transmission fluid temperature (TFT) in degrees Fahrenheit during a short trip.  The outside temperature was 30 F.

 

TFT (F) Motor Power (kW)

41      2.32

46      2.00

50      1.78

53      1.78

82      1.42

 

So at the start of the trip, the power required by the motor was 65% greater than at the end.

Edited October 10, 2014 at 02:45 PM by larryh

[Rexracer]

Larryh, love all the data, and appreciate you collecting/posting it. But as an Engineer, even I can't wade through it all to come to "conclusions" about the message.

Could you do a TL;DR version of how you interpret all the information? Maybe even top 3 points?

Thanks again for your extra work to help educate all of us.

[larryh]

Suppose you are driving 35 mph on a level street with a stop sign.  You want to come to a complete stop at the sign and then resume speed consuming the least amount of energy possible from the HVB.  Essentially, you want to store the maximum amount of energy possible in HVB during the stop and use the minimum amount of energy required to get back up to speed.

 

In general, the faster you stop, without using the friction brakes, the more regen you get.  Your average speed will be less when you stop faster, and hence, the energy consumed by friction (aerodynamic drag, tire rolling resistance, internal frictions, transmission losses, etc.) will be less.  Less energy consumed by friction means more energy available for regen.  But this is not what you want to do for a stop sign.  You want to begin stopping well before the stop sign by taking your foot off the accelerator, coasting in Drive until closer to the stop sign, and then applying the brakes to complete the stop.  If you attempt to maintain speed until you are closer to the stop sign and then apply the brakes to stop faster rather than coasting (in order to maximize regen), you will use additional energy from the HVB to maintain speed until closer to the stop sign.  The additional energy required to maintain speed until closer to the stop sign will more than offset the additional regen you would gain from stopping faster.  Alternatively, if you apply the brakes early (to maximize regen), and then maintain a slow speed until closer to the stop sign, again the extra energy from the HVB required to maintain the slow speed will more than offset the additional regen from faster braking.

 

When resuming speed after the stop sign, you want to accelerate slowly.  The faster you accelerate, the more energy consumed.  If you accelerate faster, your average speed will be faster, and hence more energy will be required to overcome friction.  In addition, the motor operates more efficiently during acceleration than when maintaining constant speed.  By accelerating slowly, the motor will operate more efficiently longer.

 

There is one additional consideration when accelerating and decelerating (stopping).  That is how efficiently the mechanical energy from the motor is transformed into kinetic energy associated with the momentum (speed) of the car and vice versa.  The faster you accelerate or stop, the less efficient the transformation.  About 10% of the energy from the motor is lost during acceleration and not available to increase the car's speed.  A smaller amount is lost when stopping and is not available for regen.  So again, you want to stop slowly and accelerate slowly to minimize this loss.

 

If 100% of the power from the motor was transformed into kinetic energy and vice versa, then you would use less energy if you stopped vs. maintaining speed.  Your average speed would be less if you stop and hence less energy lost to friction.  But this is not the case.  You use more energy when you stop than maintain speed.  At 35 mph, you lose 0.01 kWh of mechanical energy with each stop, i.e. about 17% of the kinetic energy if you do it efficiently.

 

When cruising, you want to maintain constant speed.  You don't want to unnecessarily speed up or slow down.   Mechanical energy from the motor is lost during acceleration and deceleration. 

 

The final thing to note is that the car operates more efficiently when it is warmed up versus when it is cold.  When you first start the car in cold weather, it will require 65% or more power to overcome friction than when it is warmed up.  So you want to keep the car warm. 

Edited October 10, 2014 at 07:09 PM by larryh

[Josh Benard]

Man larryh, this is impressive stuff. Here is another interesting question. It appears to me that the general consensus on this forum is that the energy cost of leaving windows down is a better decision then turning the AC on. For purposes of this question, consider that the car is at some "warm" temperature, and that if you turn on the AC it is at some medium level of power (not blasting but not basically off), and is traveling at a variety of speeds (30mph, 50mph, 70mph). Is there a speed point at which it takes less energy to turn on the AC then you lose to putting the windows down?

Edited October 10, 2014 at 11:32 PM by Josh Benard

[larryh]

At 60 mph, it takes about 0.3 - 0.4 kW more power from the HVB to propel the car with the windows open.  The AC takes a minimum of about 0.6 kW of power.  So at 60 mph, you are better off opening the windows.  You would probably have to exceed 70 mph before operating the AC would be more efficient.  But if you are going that fast, you are not driving efficiently to begin with.  Initially, the AC requires much more than 0.6 kW of power--up to 5 kW of power (that's more power than the central A/C for a large home consumes).  It will take quite some time for the power required by A/C to reduce to 0.6 kW.  For my normal 8 mile city commute, using the AC uses far more energy than opening the windows--there is no contest. 

Edited October 11, 2014 at 02:25 PM by larryh

[larryh]

For some reason, regen going downhill seems to be significantly less efficient than regen on level surfaces.  I have no idea why.  I have repeated the measurements many times on different hills and always come up with the same results.  Going uphill, energy losses are in line with what I expect--the same as driving on a level road.  But going downhill, I observe more than double the expected losses.

 

The hill I am considering in this post has an elevation change of 264 feet and an average grade of 8-9%.   I am using 1 meter resolution LiDAR Data to determine the elevation of the road with an elevation accuracy of better than 0.1 meters.  The Potential Energy errors in my measurements should be very small.  I use cruise control to go up the hill and grade assist to go downhill, both at 30 mph.

 

The motor mechanical power loss when going uphill is 3.07 kW (about 13%), which is what I would expect for 30 mph on a level road at 45 F due to friction.  The power loss when going down hill is 6.79 kW (about 32%), more than twice as much.

 

Considering the middle portion of the hill where speed was constant for both the down and up hill trips, the change in Potential Energy was 0.28 kWh.  During the uphill, the motor output 0.31 kWh of energy.  Thus there was a loss of 0.03 kWh.  During the downhill, the motor was supplied with 0.19 kWh of energy.  Thus there was a loss of 0.09 kWh of energy. 

 

The large loss of regen during downhills is going to make driving on a hilly road significantly less efficient than a level road if you maintain constant speed.  In this instance, it requires 35% more energy to go up and down the hill vs. driving on a level road. 

Edited October 11, 2014 at 09:24 PM by larryh