OBD II Data for ICE

[larryh]

I made the following posts in the Fusion Hybrid forum which explain the operation of hybrid mode during city and highway driving:

 

http://fordfusionhybridforum.com/topic/8723-efficiency-on-longer-highway-trips/?p=84199

http://fordfusionhybridforum.com/topic/8723-efficiency-on-longer-highway-trips/?p=84289

http://fordfusionhybridforum.com/topic/8723-efficiency-on-longer-highway-trips/?p=84307

[larryh]

Many people have wondered whether it is more efficient for the ICE to cycle on and off, alternating with EV mode, or to simply let the ICE power the wheels directly (without charging the HVB) when driving at high speeds.  The following is my answer to that question. 

 

At speeds of 60 mph or more, driving on a level grade, you want to avoid running in EV mode or charging the HVB.   You do not want the ICE to cycle on and off, alternating with EV mode, at high speeds.  You only want this to occur at slower speeds. 

 

Power Flow Paths from the ICE to the Wheels

 

There are two main paths for power to flow from the ICE to the wheels, the direct path, where the ICE transmits power directly to the wheels, and the indirect path, where the ICE charges the HVB, which then later in EV mode, powers the motor, which in turn powers the wheels.  These paths are illustrated below:

 

Direct Path:  ICE => Planetary Gearset => Countershaft => Wheels

Indirect Path:  ICE => Planetary Gearset => Countershaft => Motor/Generator => HVB => Motor => Countershaft => Wheels

 

The black arrows indicate flow of mechanical power.  The red arrows indicate flow of electrical power.  The indirect path is far less efficient than the direct path for transmitting power from the ICE to the wheels.  My best guess is about 80% of the power from the ICE is transmitted to the wheels via the direct path, and at most 70% for the indirect path.  This would suggest that we would never want to use the indirect path at all, i.e. EV mode, since it requires more energy, and consequently more gas, than the direct path.  But there are additional factors that come into play.

 

ICE Combustion Efficiency

 

The ICE is more efficient when producing between 15 – 35 kW of power than when producing less than 15 kW of power.  My best guess is that the ICE is about 37% efficient when producing 15 – 35 kW of power, i.e. 37% of the energy released from the combustion of gas is converted to mechanical power by the ICE.  At lower power, the ICE may only be 25% efficient or less.  So we would like to operate the ICE where it is most efficient, between 15 – 35 kW of power.  And that is basically what the car does.  You will rarely see the ICE producing less than 15 kW of power.

 

Overall Efficiency of Power Flow Paths

 

For the indirect path, the ICE will always be operating at power levels of 15 kW of power or more when charging the HVB, even at slower speeds.  So for this path, about 37%*70% = 26% of the energy released from the combustion of gas will make it to the wheels.

 

For the direct path, the ICE will be operating at power levels of 15 kW or more for speeds greater than about 55 mph, and about 37%*80% = 30% of the energy released by the combustion of gas will make it to the wheels.  For slow speeds, if the ICE is not also charging the HVB and the ICE power is less than 15 kW, only about 25%*80% = 20% of the energy makes it to the wheels.  If the ICE is also charging the HVB, then the power from the ICE will exceed 15 kW, and again the efficiency of the direct path will be 30%.  When charging the HVB, we get a bonus.  The ICE is now at a more efficient operating point and the portion of the gas being consumed to power the wheels is less than what it would be if we did not charge the HVB. 

 

Low Speed Efficiency

 

At low speeds, we have:

 

Efficiency of Direct Path when charging HVB: 30%

Efficiency of Indirect Path: 26%

Efficiency of Direct Path when not charging HVB:  20%

 

For best mileage at low speeds, we would never want to use the direct path without also charging the HVB (the least efficient path).  That means we want to alternate between the direct path and charging the HVB, and the indirect path to use up the accumulated energy stored in the HVB.   We want the ICE to cycle on and off, charging the HVB, alternating with EV mode.    The overall efficiency will then be between 30% and 26%, but it will be higher than the 20% associated with the direct path when not also charging the HVB. 

 

High Speed Efficiency

 

At high speeds, we have:

 

Efficiency of Direct Path when not charging HVB:  30%

Efficiency of Direct Path when charging HVB: 30%

Efficiency of Indirect Path: 26%

 

For best mileage at high speeds, we would never want to use the indirect path (the least efficient path).  We also don’t want to use the direct path and charge the HVB.  We would have to use up the accumulated energy stored in the HVB at a later time using the indirect path.  We certainly don’t want to do at higher speeds.  We would have been better off simply not charging the HVB to begin with.   So we would have to use up the accumulated energy stored in the HVB when driving at slower speeds.  But that means spending more time using the less efficient indirect path when driving at slower speeds to use up the accumulated energy in the HVB.  The conclusion is that we only want to use the direct path without charging the HVB at higher speeds.  About 30% of the energy released from the combustion of gas will make it to the wheels at high speed via the direct path without charging the HVB.

Edited July 25, 2014 at 11:25 AM by larryh

[larryh]

Power Flow Paths from the ICE to the Wheels

 

There are two main paths for power to flow from the ICE to the wheels, the direct path, where the ICE transmits power directly to the wheels, and the indirect path, where the ICE charges the HVB, which then later in EV mode, powers the motor, which in turn powers the wheels.  These paths are illustrated below:

 

Direct Path:  ICE => Planetary Gearset => Countershaft => Wheels

Indirect Path:  ICE => Planetary Gearset => Countershaft => Motor/Generator => HVB => Motor => Countershaft => Wheels

 

The indirect path is far less efficient than the direct path for transmitting power from the ICE to the wheels.  My best guess is about 80% of the power from the ICE is transmitted to the wheels via the direct path, and at most 70% for the indirect path.  This would suggest that we would never want to use the indirect path at all, i.e. EV mode, since it requires more energy, and consequently more gas, than the direct path.  But there are additional factors that come into play.

 

From the OBD II data I have logged, it looks likes the motor/generator is between 75% - 85% efficient in converting electrical to mechanical power.  It is more efficient when producing greater power.  The motor/generator is about 97% efficient in converting mechanical power to electrical power.  In the negative split mode of operation, where the ICE consumes mechanical power from the ICE and converts it to electric power, which is then recycled back to the generator and reconverted back to mechanical power (to provide the reaction torque for the ICE to power the planetary gearset), the generator outputs about 78% of the mechanical power consumed by the motor.   So if the generator is 80% efficient (converting electrical to mechanical power) and the motor is 97% efficient (converting mechanical to electrical power), this matches the 78% efficiency observation in negative split mode, i.e. 80%*97% = 78%. 

 

Also, I observe that the HVB seems to be about 97% efficient in storing energy and 97% efficient in retrieving it. 

 

That makes the efficiency of the portion of the indirect path above in red about 97%*97%*97%*80% = 73%.   Power from the ICE is converted to electricity (97% efficiency), the electricity is stored in the HVB (97% efficiency), it is later retrieved from the HVB in EV mode (97% efficiency), and that electricity is used to power the motor (80% efficiency).  That makes the indirect path between 73% - 78% the efficiency of the direct path.  There is approximately a 22% - 27% energy loss. 

 

I think about 80% of the mechanical power generated by the ICE makes it to the wheels in the direct path.   That means about 60% of the power makes it to the wheels in the indirect path. 

Edited July 27, 2014 at 10:53 PM by larryh

[larryh]

I measured the speed of the car, coasting in neutral starting from 55 mph until stopped at about two second intervals, going both directions on a relatively level road.  It took 200 seconds to stop in one direction and 150 seconds going the other way.  Each time, I plotted the force slowing the car down (due to aerodynamic drag, rolling resistance, internal frictions of the car, etc.) vs. speed.  The plots were fitted to two degree polynomials with R^2 = 0.9994, meaning a very good correlation.  Averaging the two results to account for the calm wind and the road not being perfectly level, I get:

 

F =0.1997*v^2 -14.975*v -143.02,

 

where F is the force from aerodynamic drag, rolling resistance, etc. in Newtons and v is speed of the car in meters/second.

 

The constant force at all speeds, -143.02 N, should be due mainly to the rolling resistance of the tires.  The force due to the rolling resistance of the tires is approximately F = Rmg, where F = 143.02 N is the force on the car slowing it down due to rolling resistance, R is the rolling resistance, m is the mass of the car (about 1870 kg with everything in it), and g is the gravitational constant.  Solving for R, I get:

 

143.02/(1870*9.902) = 0.0077.

 

I doubt this is a very accurate way to measure the rolling resistance of the tires since there are other factors slowing down the car.  But at least it provides an upper bound.  You can find the rolling resistance for various tires here:  http://en.wikipedia.org/wiki/Low_rolling_resistance_tire

Edited August 2, 2014 at 07:34 PM by larryh

[larryh]

The generator torque as a function of rpm is as follows, where y is the generator torque in Nm (positive) and x is the generator rpm (negative):

 

y = -1.3x/2000, -2000 <= x;

y = 2.5e-8x^2 – 0.0001x + 1, -2000 <= x <= -6000;

y = x/6000 + 3.5, -6000 <= x

 

It seems strange that torque is such a perfect piece-wise linear/parabolic function with round and non-arbitrary constants. It looks like the torque is being controlled by one of the control modules for some reason rather than spinning freely. This happens in EV mode in all gears.

 

When in neutral, the motor torque as a function of rpm is as follows, where y is the motor torque in Nm (negative) and x is the motor rpm (positive):

 

y = -0.0011*x, x <=2000,

y = -0.0001*x - 2.0, 2000 <= x <= 4000,

y = 0.0001*x - 2.8, 4000 <= x <= 6000,

y = 0.0006*x - 5.8, 6000 <= x <= 8000,

y = -1.0, x >= 8000

 

The torque is negative, which means that the motor is experiencing resistance, slowing the motor and the car down.  Enough electrical power must be supplied to the motor to overcome this negative torque before the motor can begin to supply power to the wheels.  I suspect that this resistance is independent of the load on the motor.

 

At 40 mph, the power associated with this resistance is about -1.2 kW.  The mechanical output power of the motor is about 4.4 kW.  This resistance alone accounts for a 27% loss of efficiency converting electrical power to mechanical power.   I estimate the total efficiency loss at 40 mph is 39%.  The remaining losses are from the inverter and other sources. 

Edited September 5, 2014 at 02:33 PM by larryh

[Hybridbear]

The Transmission Fluid Temp got hotter today than going east since the outside temp was slightly warmed. It peaked at 175F this afternoon versus 169F yesterday morning driving to Wausau. The motor coil temps peaked at 180+F today. 

[larryh]

In the summer, when the EV mode is set to Auto on the freeway going 65 mph, only the electric motor propels the car.  In the winter, the motor, the generator, and the ICE all work together to propel the car.  The motor supplies 8.0 kW of power, the generator supplies 5.4 kW power, and the ICE supplies 5.1 kW of power for a total of 18.5 kW of power required to drive at 65 mph. 

 

MPGe is 64 when all three provide traction.  When only the ICE runs in EV Later mode, MPGe is 32.5.  Although it looks appealing to be getting 64 MPGe vs 32.5 MPGe, using all three (motor/generator/ICE) to propel the car is not very efficient.  You would most likely be better off avoiding this mode of operation by using EV Later rather than EV Now on the freeway and reserve the HVB to propel the car at slower speeds.

 

When all three are powering the car, the efficiency of the motor/generator is 82%, which is quite good.  However, the efficiency of the ICE is only 28%.  Normally, ICE efficiency is around 37% on the Freeway in EV Later mode.  The ICE is not providing enough power to run efficiently.  The ICE rpms is about 1175.

Edited November 22, 2014 at 02:50 PM by larryh

[larryh]

When in neutral, the motor torque as a function of rpm is as follows, where y is the motor torque in Nm (negative) and x is the motor rpm (positive):

 

y = -0.0011*x, x <=2000,

y = -0.0001*x - 2.0, 2000 <= x <= 4000,

y = 0.0001*x - 2.8, 4000 <= x <= 6000,

y = 0.0006*x - 5.8, 6000 <= x <= 8000,

y = -1.0, x >= 8000

 

The torque is negative, which means that the motor is experiencing resistance, slowing the motor and the car down.  Enough electrical power must be supplied to the motor to overcome this negative torque before the motor can begin to supply power to the wheels.  I suspect that this resistance is independent of the load on the motor.

 

At 40 mph, the power associated with this resistance is about -1.2 kW.  The mechanical output power of the motor is about 4.4 kW.  This resistance alone accounts for a 27% loss of efficiency converting electrical power to mechanical power.   I estimate the total efficiency loss at 40 mph is 39%.  The remaining losses are from the inverter and other sources. 

 

I believe the Torque PID for the electric motor is a synthetic PID, and that is the reason I can express Torque with the above equations.  The actual name of the PID is Motor Torque from AC Source.  There probably is no actual sensor measuring torque at the output of the motor.  Instead, the above equations are used to estimate the torque loss at the motor output based on motor RPM.  The above equations apply when there is no significant regen.  When there is significant regen, no torque loss is assumed.

 

From the data I have collected, it appears that the mechanical power output (PO) of the motor is related to the electrical power input (PI) of the motor by the following relation:

 

PO = 0.87 * PI - PR, PI > -4 kW  (no significant regen)

PO = PI / 0.98, PI < -4 kW (significant regen)

 

PR is power loss of the motor due to the torque loss using the equations above (note that power is computed as 2*pi*rpm*torque/60 in watts).  If you do not supply the motor with electrical power, it is going to slow down due to friction.  The torque loss from this friction is given by the equations above. 

 

The first equation, PO = 0.87 * PI - PR, states that the inverter/motor is 87% efficient (neglecting friction), i.e. 87% of the electrical power is converted to mechanical power.  After taking into account the power loss due to friction, PR, the remaining motor output power is PO = 0.87 * PI - PR.

 

When there is significant regen, it is assumed torque loss is minimal and motor/inverter efficiency is estimated to be 98% (this is a very uncertain measurement--I would need to collect more data to get a better estimate). 

 

Similarly, the torque measurements for the generator are probably also synthesized.  The name of the PID is Generator Torque from AC Source.