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larryh

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Everything posted by larryh

  1. larryh
    Transmission Efficiency I computed transmission efficiency for several more accelerations. The following plot shows the percentage of energy output from the motor that is converted into kinetic energy to accelerate the car, i.e. transmission plus tire slippage efficiency. It looks like efficiency is about 96%. The accuracy of the measurements is roughly +/- 2%.
  2. larryh
    Collecting Accurate Measurement Data The plot below shows some of the measurements I collected when determining the output power of the motor required to maintain a constant speed of 20 mph. I collected the measurements on a relatively level East-West road when the wind was calm. I traveled in both directions to verify that I get the same results going in either direction. I collect GPS, power, and speed measurements from the car about every half a second. The x-axis is GPS longitude along the road. The total distance in the plot is about 1,600 feet. Roads aren’t level for very long distances. The green line shows the elevation of the road from Lidar data. You need to add approximately 900 meters to what is shown. The elevation varies by about 0.4 meters or about 1.3 ft. The Lidar data is accurate to within a few inches. The purple line shows speed when traveling East and the light blue line shows speed while traveling West. You need to add 19 mph to what is shown. When going East, I start the measurements at -92.748 Longitude and end the measurements at -92.745 Longitude. The speed varies by just under 0.2 mph. The output power of the motor while traveling East is shown by the blue line. The red line shows the power measurements when traveling West. You will notice that the red line is higher than the blue line. That makes sense since there is a slight downward slope to the road going in the Easterly direction—gravity is assisting in propelling the car so the motor doesn’t have to work so hard. Without compensating for the speed or elevation changes, the average power measured going East was 0.99 kW. When going west it was 1.23 kW. This is a difference of 0.24 kW, or a 24% error (I should get the same results in either direction). If I adjust for the elevation change in the road using PE = mgh (where PE is potential energy, m is the car’s mass, g is the Earth’s gravitational constant, and h is the elevation) and for the speed changes of the car during the trip using KE = 0.5mv^2 (where KE is kinetic energy, m is the car’s mass, and v is the car’s speed), I get 1.19 kW for the Eastbound direction and 1.24 kW for the Westbound direction. The difference is now 0.05 kW for an error of 4%--much better. Note that the 1.3 foot elevation change has 10 times more impact on the results than the 0.2 mph speed change.
  3. larryh
    The capacity of the HVB is 7.6 kWh. The maximum that the car will charge the HVB to is 7.2 kWh. However, normally it charges to about 98% of 7.2 kWh or 7.06 kWh. That is 93% of capacity. The max cell voltage is approximately 4.08 V. The car will not allow the HVB energy to fall below 1.0 kWh, or 13% of capacity.
  4. larryh
    The car will attempt to fully charge the battery before the next Go Time. It will first make sure that it charges during hours with the lowest rates. However, it necessary to fully the charge the car before the next Go time, it will charge using higher rates. If you don't set a Go Time, it will fully charge the car within 24 hours. You cannot program the car to automatically charge to a predetermined state of charge.
  5. larryh
    Efficiency of Motor in Provide Kinetic Energy The following plot shows the percentage of energy from the motor that is converted into kinetic energy from five different accelerations. I am not sure how accurate my calculations are. There is a lot of uncertainty involved. But it looks like the slower you accelerate, the greater the percentage of energy transmitted from the motor to provide the kinetic energy required to accelerate the car, i.e. slow acceleration is more efficient than fast acceleration. I would have to collect a lot more data to provide conclusive results.
  6. larryh
    Transmission Efficiency We can calculate the predicted power output and actual power output of the motor during acceleration in the previous post by computing the slope of the red and blue lines between 6 and 24 seconds. Doing this, I compute the actual power output of the motor to be 28.8 kW. The predicted power output is 26.4 kW. Efficiency is 26.4 / 28.8 = 92%. 8% of the power output of the motor is being lost by the transmission and tires. But this not really the calculation that is of interest. Of greater interest is how much additional energy the motor must output to provide the kinetic energy to accelerate the car. To do that, we must first subtract the energy used to maintain speed (overcome aerodynamic drag, etc.) from the previous plot. When I do that, I calculate the actual power output of the motor, above the power required to maintain speed, to be 27.7 kW. This is the portion of the motor power output used exclusively to accelerate the car (beyond that required to overcome frictional forces). The predicted power required to provide the kinetic energy is 22.8kW. Only 22.8 / 27.7 = 82% of the output power of the motor is actually converted into kinetic energy to accelerate the car. The rest is lost. If we assume that the motor is 88% efficient converting electricity to mechanical energy, then 72% of the energy from the HVB used to speed of the car is actually transformed into kinetic energy. It requires 0.16 kWh of kinetic energy to accelerate to 55 mph. That means that we will consume about 0.16 / 0.72 = 0.22 kWh of energy from the HVB to provide that kinetic energy. The remaining 0.06 kWh of energy is lost forever and cannot be recaptured by regenerative braking.
  7. larryh
    Transmission Losses During Acceleration In Post 10, I provided a graph showing the actual and predicted energy output of the motor during slow acceleration to 20 mph. In that analysis, I assumed that the transmission was 100% efficient, i.e. all of the power output by the motor is used to propel the car. A real transmission has losses. A portion of the power output by the motor is consumed to overcome internal friction within the transmission and converted to heat causing the transmission to warm up. In addition, the tires do not grip the pavement perfectly. The tires slip some on the pavement which causes the tires and pavement to warm up. The following plot is similar to the one in Post 10. This time I accelerate faster up to 50 mph in 27 seconds and then maintain that speed until a total of 38 seconds has elapsed. The blue line shows the predicted energy output of the motor if the transmission and tires were 100% efficient in transmitting power to propel the car. It includes the kinetic energy needed to accelerate the car to 50 mph plus the energy needed to overcome friction (aerodynamic drag, tire rolling resistance, and internal friction within the car), i.e. the power required to maintain speed. The red line shows the actual energy output of the motor. There is a large difference in the predicted and actual energy output of the motor. Energy is being lost somewhere. I am assuming that is being lost by the transmission and by tire slippage on the pavement.
  8. larryh
    The EBH consumes about 400 watts of power. If you run it for 3 hours, that's about 1.2 kWh of electricity. I estimate for my 8 mile commute to work, the additional power consumed due to a cold transmission is about 0.5 kWh of energy. The commute requires about 2.4 kWh when it is below 0 F and about 1.7 kWh in the summer. The remaining 0.2 kWh is mainly due to the denser air. However, the EBH does allow me to drive to work entirely in EV mode when it is below 0 F. Without it, the ICE would come on. The ICE would consume at least 0.07 gallons of gas. That's about 2.4 kWh of energy. So in this case, I am saving energy using the EBH.
  9. larryh
    Motor Input for Trip With Acceleration It’s not really the Motor Output that we are concerned with. Instead, we are interested in the amount of electrical energy the HVB must supply to the motor. We can use the model of the electric motor in post 8 to predict the amount of energy that the HVB must supply for the trip in the previous post. The results are shown in the plot below. This time I am plotting the energy required from the HVB to propel the car.
  10. larryh
    Motor Output for Trip With Acceleration So far, I have only considered EV dynamics for constant speed and coast downs. Now let’s see what happens when acceleration is involved. In the trip under consideration, I start out at rest and accelerate slowly to 20 mph (actually I overshoot 20 mph a bit) and then maintain constant speed. I can predict the power output of the motor as follows. The motor is going to have to supply the kinetic energy required to accelerate the car to 20 mph. Plus it is going to have to supply the power required to overcome friction (which is a function of speed). For convenience, I will assume that 100% of the output power of the motor is transmitted to the road to propel the car. Of course, in reality, there are some losses in transferring power to the road, but they should be small. The kinetic energy of the car is given by 0.5 * mass * speed^2. The power required to overcome friction at a given speed is given by the chart in post 2. Adding the energy for these two sources together, I plot the results in the chart below. The entire trip takes just over 60 seconds. The blue line shows the predicted accumulated energy output of the motor vs time. The red line shows the actual. The two lines match quite closely. They are not going to coincide perfectly. The road is not perfectly level, there may be slight wind gusts, the transmission is not 100% efficient, etc. They coincide well within measurement error.
  11. larryh
    The engine block heater does very little to warm the transmission fluid temperature in the winter. The ICE needs to get a lot hotter than 100 F to warm the transmission fluid. As an example, when the outside temperature was 3 F, the EBH heat the ICE to 72 F. The TFT was still at 12.1 F.
  12. larryh
    Actual Motor Efficiency The actual motor efficiency, including internal friction, is then Po / Pi. I have plotted this ratio vs. input power using the experimental results below. Efficiency increases with increasing power. Input power of 10 kW corresponds to 50 mph. Input power of 3.7 kW corresponds to 30 mph.
  13. larryh
    Modelling an Electric Motor The above results suggest a very simple model for how an electric motor operates. The motor converts a fraction of the total electrical power received from the HVB battery to mechanical power. Before it can output any power, it must convert at least enough electrical power to overcome the power losses due to internal friction. Thus the power output of the motor is: Po = e * Pi – Pl(rpm), where Po is the mechanical output power of the motor, Pi is the electrical input power to the motor, e is the efficiency of the inverter and motor in converting electrical to mechanical power, and Pl is the mechanical power loss of the motor due to internal friction which is a function of the rpms of the motor. The following graph plots motor efficiency neglecting internal friction, e = (Po + Pl) / Pi, vs. input power using the experimental results. As you can see, e is a constant of approximately 88%.
  14. larryh
    Incoporating Motor Internal Power Loss in Experimental Results Now let’s incorporate the internal motor power loss into the experimental results. The motor converts electrical power to mechanical power. In order for the car to maintain constant speed, the motor must convert enough electrical power, Pm, to mechanical power to overcome the internal friction of the motor as well as the other sources of friction experienced by the car, Pc (aerodynamic drag, tire rolling resistance, friction internal to the car other than from the motor). During the coast downs, we measured Pm + Pc, the total internal friction from all sources. Both the internal motor friction and the other sources of friction were slowing the car down. During the constant speed experiment, we measured the power output of the motor. This is the output power of the motor after losses due to internal motor friction, i.e. Pc. These power measurements do not include the losses due to internal motor friction. The sensor at the motor output only measures the power being output by the motor. It does not measure any losses occurring inside the motor. The following is the same plot as in post 5, but with the power loss associated with the internal friction of the motor is subtracted from the power losses due to all sources of friction during the coast downs. Now the red (power output of motor for constant speed) and blue (coast down power losses from friction excluding internal motor friction) lines coincide. The motor outputs exactly the power needed to overcome friction. The laws of physics are preserved! There are some slight discrepancies near the extremes of the blue line. But those are regions were I don’t have good coast down experimental results. To compute acceleration, I take the derivative of speed vs. time from the coast down results. But at the edges, I cannot compute accurate derivatives—I need to record data with speeds lower and greater than the edge speeds to compute an accurate derivative.
  15. larryh
    Motor Internal Power Loss During the coast downs, the mechanical power output from the motor was negative. I am interpreting that to mean that there is friction internal to the motor which slows the motor, and the car, down. A spinning motor must experience friction from the various bearings inside the motor. The spinning rotor has to move lots of air. And, there is probably a long list of other phenomena that cause power losses internal to the motor. Before the motor can output any power, enough input power must be supplied to overcome the internal power losses. If less power is supplied, the motor will slow down. In additional to the various mechanical power losses mentioned previously, there are also various sources of electrical power losses. For example, the inverter is not 100% efficient in converting DC to AC. The following plot shows the estimated power loss due to internal friction of the motor vs. speed during the coast downs. At 20 mph, the loss is approximately 0.7 kW. At 40 mph, it is approximately 1.3 kW.
  16. larryh
    Combining Results of the Two Experiments Let’s now combine the results from the two experiments. One would expect that the mechanical output power of the motor should match the power loss from friction for the car to maintain constant speed. If the motor provided more power, then the additional power over friction would accelerate the car. If the motor provided less power, then the greater power loss from friction over the output power of the motor would slow the car down. The following chart shows the power loss from friction and the motor input/output power for various speeds, i.e. I combined the plots in posts 2 and 4. I am showing the power loss from friction as a positive number to compare with the motor input/output power. Something is wrong. The motor appears to be outputting less power (red line) than is required to overcome friction (blue line). According to the plot, the car should be slowing down—the motor is not providing enough power. There appears to be something wrong with this analysis. What is missing is that the motor has internal friction which we have not taken into account.
  17. larryh
    Coast Down Analysis The next step is to compute the power loss from friction in the coast down experiment. If there were no friction, the car would maintain constant speed indefinitely. But since the car is slowing down, frictional forces (aerodynamic drag, tire rolling resistance, internal frictions) must be at work. Force of friction = mass * acceleration. We have a plot of the car’s speed as a function of time. The derivative provides the acceleration of the car vs. time. The curb weight of the car is 3913 lbs. We need to add the weight of the contents of the car, which I estimate to be about 260 lbs. We also need to account for the moment of inertia of the various rotating parts of the car, including the wheels and gears. This increases the apparent mass of the car during coast down. Generally this amounts to about 3% - 4% of the mass of the car. I will use 100 lbs. The mass of the car is thus approximately 1940 kg. From Power = Force * Speed, we can compute the power loss from friction. The plot below shows the power loss from friction as a function of speed. At 40 mph, the power loss is -5.5 kW. At 20 mph, it is -1.7 kW. The power loss as a function of speed is given by a quadratic function as indicated in the plot below.
  18. larryh
    Coast Down For the second part of the experiment, I did several coast downs with the car in neutral starting from various speeds along the same section of the road. I then spliced the results together and plotted them below. The plot shows speed as a function of time. Speed is in meters/second and time is in seconds. At time 0, the car is going 17 m/s (38 mph). After about 88 seconds, the speed has fallen to 6.9 m/s (15 mph). The speed is described by a quadratic equation of time as indicated in the chart below.
  19. larryh
    Power Required to Maintain Constant Speed For the first part of the experiment, I drove the car at various constant speeds (10, 20, 30, 40, and 50 mph) along the same section of road and measured the electrical input power and mechanical output power of the motor. The results are plotted below. Since a motor is not 100% efficient, the input power to the motor will be greater than the output power. The difference reflects losses in converting electrical to mechanical power. The car provides the rpm and torque output by the motor. The power (in watts) can be computed as 2*pi*rpm*torque/60. I suspect that the car does not actually have a sensor to measure the output torque of the motor. Instead, it estimates the torque based on the electrical input power applied to the motor. So the output power of the motor is only an estimate. Driving 30 mph requires approximately 3.7 kW of electrical power from the HVB. Driving 50 mph requires 9.8 kW. The cubic equations describing power as a function of speed are shown in the plot. The frictional force from aerodynamic drag varies with the square of speed. Power = Force * Speed. Hence power is described by a cubic function of speed.
  20. larryh
    The following posts describe an experiment I performed the other day with the car to understand the forces and their effect on the motion of an EV. For the first part of the experiment, I drove the car with cruise control set to various speeds along the same section of road and measured the electrical input power and mechanical output power of the motor required to maintain constant speed. This allowed me to plot the power required to overcome friction (aerodynamic drag, tire rolling resistance, and internal frictions) as a function of speed. Next I did a coast down on the same section of road. With the car in neutral, I recorded time vs. speed as the car slowed down. I also recorded the power output of the motor. The power output of the motor is negative during coast down. Internal friction inside the motor is slowing the car down. The outside temperature was 63 F. There was no wind. The road that I am using is level to within 1.5 feet. I drove the car for approximately 20 miles, with the ICE running, prior to the experiment. I wanted to make sure the car was thoroughly warmed up before starting. The temperature of the drivetrain components in a car have a huge impact on the internal friction within the car. When cold, the car will require significantly much more energy to maintain constant speed. As the car warms up, each time I repeated a portion of the experiment, I would get very different results.
  21. larryh
    It is four bars when you see the three dimensional battery display and have EV range left. It is two bars when you see a two dimensional battery display and are in hybrid mode after using up all your plug-in energy.
  22. larryh
    I don't know what the maximum discharge limit is for the Hybrid. Ford's web site specifies what it is for the Energi, but not the Hybrid. The above chart is actually showing the maximum charge limit for the HVB. Charging is limited to 35 kW. But when the battery is cold, the charge limit is significantly less. The discharge limit is also temperature dependent, but is not limited as much as is charging. It is usually in the upper 60's.
  23. larryh
    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.
  24. larryh
    The Energi cannot output 88 kW of power from the motor. The maximum power that can be supplied by the HVB to the electric motor is 68 kW. If the motor is 86% efficient, then the maximum output of the motor is 58 kW.
  25. larryh
    The following plot shows the Transmission Fluid Temperature (TFT) during two commutes. The blue line is in the winter when the outside temperature is -5 F. The red line is in the summer when the temperature is 68 F. In the winter, the TFT never rises above 100 F. In post 350, you can see that there is significant power loss from increased friction when TFT is below 80 F. During the winter, for more than 1/3 of the trip TFT is below 80 F. TFT must be well above 100 F to minimize the power loss. It never rises above 100 F in the winter. In the summer, for more than 85% of the trip, TFT is above 100 F. I suspect that 10% of the increase in gas consumption during the winter is from increased air density and tire rolling resistance (see previous post showing a total increase of 30%). The remaining 20% is from increased friction resulting from cold powertrain components.
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