Power split device and electrical/mechanical power

Introduction

Many resources and threads exist that describe the Prius power split device. They correctly describe how RPMs can be calculated and how torque is split. But I did not find any that would correctly describe how the power is split. The solution seems easy at first: Power is RPM multiplied by torque.

Many correctly state that the engine torque is split about 28% to MG1 and 72% to the fixed gears of the final drive and MG2. They conclude that power is also split at the same ratio. I question that. Why? Because MG1/MG2/ICE almost never have identical RPMs. If you multiply the split torque with the correct RPM numbers you will see that power is almost never split 28/72%.

Let me first list some of the great resources I found on the power split device and HSD in general. Then I will give a compact summary of the necessary formulas to understand what I stated above. After that I have some detailed examples.

Resources
Toyota Prius - Power Split Device
prius.ecrostech.com/original/Understanding/PowerSplitDevice.htm
ICE to wheel transmission efficiency value at 60 mph | PriusChat
Introduction to Prius Power Flow | PriusChat
Two MG1 Questions | PriusChat
ecee.colorado.edu/~ecen5017/notes/OakRidge_2004Prius.pdf (Backup Link: www.osti.gov/scitech/biblio/890029)
info.ornl.gov/sites/publications/files/Pub26762.pdf
www.ae.pwr.wroc.pl/filez/20110606092430_HEV_Toyota.pdf

Compact summary of the formulas

In the Prius Gen 2 the two motors/generators and the engine are connected to the planetary gearset (power split device, PSD) as follows:

• sun gear = MG1

• carrier = engine (ICE)

• ring gear = MG2 and chain to the fixed gears of the final drive

RPM

The gear ratios between sun gear, carrier and ring gear are defined by to number of teeth of the sun gear (30 teeth) and ring gear (78 teeth). The number of teeth on the planetary gears have no effect on this.

If you know the RPMs of two components the RPM of the third component can be calculated:

• MG1_RPM = 3.6 * ICE_RPM - 2.6 * MG2_RPM

• MG2_RPM = (3.6 * ICE_RPM - MG1_RPM) / 2.6

• ICE_RPM = (MG1_RPM + 2.6 * MG2_RPM) / 3.6

The constants 2.6 and 3.6 result from the number of teeth:
2.6 = 78 / 30
3.6 = (78 + 30) / 30

Torque

The torque of MG1 can be calculated from the engine torque (as long as RPMs are constant):
MG1_TORQUE = ICE_TORQUE / -3.6 = ICE_TORQUE * -27.778 %

Notice the negative sign. This is not the torque transmitted by the engine, but the torque that MG1 has to generate as counterforce to keep the RPM constant. Toyota uses the same negative sign in the torque values transmitted over the CAN bus (see e. g. ScanGauge or Torque Pro App). So I keep consistent with that.

Power

Power is RPM multiplied by torque. I'm metric, so RPM is given in 1/min and torque as Nm. The power in kW can then be calculated as:

POWER = RPM * TORQUE / 9549.3

Examples

Now I'm going to calculate some examples with those formulas. Numbers are made up, but the scale is realistic. For each example I will calculate how much power of the engine is transmitted to MG1. The rest of the engine power of course goes to MG2 and the chain to the fixed gears of the final drive. To keep things simple I do not include mechanical loss or power consumption for auxiliaries.

Example 1
Prius is stationary and in D. The battery is being charged using the engine.

Given:
MG2_RPM = 0/min
ICE_RPM = 1600/min
ICE_TORQUE = 72 Nm

Result
MG1_RPM = 3.6 * 1600/min - 2.6 * 0/min = 5760/min
MG1_TORQUE = 72 Nm / -3.6 = -20 Nm
ICE_POWER = 1600/min * 72 Nm / 9549.3 = 12.06 kW
MG1_POWER = 5760/min * -20 Nm / 9549.3 = -12.06 kW
Portion of engine power transmitted to MG1: 100%

Discussion
Yes, this is a special case. But no special formulas are required. And yes, about 72% of the engine torque is going to the final drive chain (MG2 is not doing anything). The brakes keep the final drive from rotating, therefore the power going over that path is 0 kW, even if torque is going that path.
Only 1/3.6 (~28%) of the engine torque is transmitted to MG1. But at the same time the RPM of MG1 is 3.6 times the RPM of the engine. Therefore all the engine power is transmitted to MG1. Where else should it go?

Example 2
Prius is going uphill on a highway at constant speed.

Given:
MG2_RPM = 3300/min (~94 km/h, 58 MPH)
ICE_RPM = 2500/min
ICE_TORQUE = 90 Nm

Result
MG1_RPM = 3.6 * 2500/min - 2.6 * 3300/min = 420/min
MG1_TORQUE = 90 Nm / -3.6 = -25 Nm
ICE_POWER = 2500/min * 90 Nm / 9549.3 = 23.56 kW
MG1_POWER = 420/min * -25 Nm / 9549.3 = -1.10 kW
Portion of engine power transmitted to MG1: 4.7%

Discussion
This is a very nice pace in the Gen2. Engine at best efficiency and most power (95.3%) is transmitted over the mechanical path. The remaining 4.7% can, reduced by generator and electrical loss, either charge the battery or support MG2 propelling the vehicle. At this low RPM I have the impression that MG1 is not generating any power at all. It's just a counterforce for the engine. In theory you could get MG1_RPM to 0/min, but in reality ScanGauge and Torqe Pro show this rarely happening. But low RPMs of a few 100/min are possible.

Example 3
Prius is going up very steep hill at lower, constant speed.

Given
MG2_RPM = 2000/min (~57 km/h, 35 MPH)
ICE_RPM = 3500/min
ICE_TORQUE = 97.2 Nm

Result
MG1_RPM = 3.6 * 3500/min - 2.6 * 2000/min = 7400/min
MG1_TORQUE = 97.2 Nm / -3.6 = -27 Nm
ICE_POWER = 3500/min * 97.2 Nm / 9549.3 = 35.63 kW
MG1_POWER = 7400/min * -27 Nm / 9549.3 = -20.92 kW
Portion of engine power transmitted to MG1: 58.7%

Discussion
Compared to example 2 much more power is transmitted over the electrical path. Only 35.63 - 20.92 = 14.71 kW are transmitted to the ring gear on the mechanical path. Let's assume that the battery is not being charged or discharged. Then the electric power generated by MG1 is consumed by MG2. Let's assume an efficiency of 90% for the path MG1 -> inverter -> MG2. Then 14.71 + 20.92 * 0,9 = 33,54 kW are transmitted to the wheels.

Example 4
Prius is going at constant highway speed on flat ground.

Given
MG2_RPM = 3300/min (~94 km/h, 58 MPH)
ICE_RPM = 1600/min
ICE_TORQUE = 72 Nm

Result
MG1_RPM = 3.6 * 1600/min - 2.6 * 3300/min = -2820/min
MG1_TORQUE = 72 Nm / -3.6 = -20 Nm
ICE_POWER = 1600/min * 72 Nm / 9549.3 = 12.06 kW
MG1_POWER = -2820/min * -20 Nm / 9549.3 = 5.91 kW
Portion of engine power transmitted to MG1: 0% (see discussion for electrical path)

Discussion
For the experts: Yes, this is heretical mode. MG1 is still receiving ~28% of the torque of the engine. At the same time it is not consuming mechanical power (generator) but outputing mechanical power (motor). Where is all this power going? MG1 transmits its power over the sun gear to the planetary gears of the carrier and the engine transmits its power to the the carrier. The planetary gears transmit the joined power to the ring gear.

Let's assume that the battery is not being charged or discharged. Then the electric power for MG1 has to be provided by MG2. MG2 is in generator mode and converts some of the power of the ring gear into electricity for MG1. Let's assume an efficiency of 90% for the path MG2 -> inverter -> MG1. MG2 has to consume 5.91 kW / 0.9 = 6.57 kW mechanical power from the ring gear. The rest of the power (12.06 + 5.91 - 6.57 = 11.4 kW) is transmitted to the wheels (as said above, mechanical loss not taken into account). So, at which ratio does the PSD split the engine power in this example? I would say: 12.06 - 6.57 = 5.49 kW go over the mechanical path and 6.57 kW go over the electrical path. Ratio electrical/mechanical: 54/46%

Summary
This theory is backed by measurements I did with ScanGauge, Torque Pro and my own app. Some hints how you can verify it yourself:

Check MG1 RPM and torque, MG2 torque and battery amps. You will see that MG1_TORQUE does not vary much when the engine is on. The typical range range is -20 to -27 Nm, as in the examples. Go on a highway and drive at constant speed/load. After some time the battery will be at SOC 55-65% and battery amps will be close to 0. Now you can try different MG1_RPM by varying ICE_RPM (watch traffic and speed limit, hills might help):

• MG1_RPM lower than +/-800/min: MG2_TORQUE will also be low (< +/-10 Nm)
• MG1_RPM high (positive or negative) results in high MG2_TORQUE (positive or negative)
  
• When MG1_RPM is positive then MG2_TORQUE is also positive, when MG1_RPM is negative then MG2_TORQUE is negative (assuming battery amp is close to 0 and values of RPM/TORQUE are not close to 0).

I could generate many more examples. E. g. engine off, B mode, engine starting, stopping.

I could show results from test drives.

I could show the Matlab model I created and the dozens of charts I plotted with it. That model is much more refined than what I introduced here. It takes into account the engines BSFC chart and 2D efficiency charts (torque, rpm) for motors and inverter.

I could talk about the Android app I wrote that records 800 PIDs per second (but is not public and still very unfinished).

Hope you liked it.

 

I apologize for the length and complexity of the post. I found no way to simplify it. I checked all the calculations again and found no mistakes in the numbers. I added some missing letters and words, corrected the link to [OakRidge_2004Prius.pdf] and added a warning about the detailed examples. Any hints about mistakes are welcome!

Thanks. I checked, but think my formula is correct. The constant is 60 * 1000 / 2π. See Torque and power or Power &amp; Torque | Engine resources | UL260i | Engines Sales

Plausibility check: [OakRidge_2004Prius.pdf] lists the following performance data for the Gen 2 engine:

-> Torque at maximum output is 57 kW * 9549.3 / 5000/min = 108.86 Nm
-> Power at max torque is 115 Nm * 4200/min / 9549.3 = 50.58 kW

Btw. conversion to freedom units:
lb-ft = Nm * 0.73756
Nm = lb-ft * 1.35582