Introduction to Prius Power Flow

Not quite correct with your statement there. The ICE is connected to the road wheels, and 72% of it's torque takes that rout. The other 28% goes to MG1, and the output from that can be supplied to MG2 to return this torque minus any losses to the wheels. MG1's output could alternatively be returned to the battery. If MG1 is allowed to spin freely (no current is taken from it) no torque is sent to the wheels from the ICE and the only drive path is electric only from the battery through MG2.

John

 

While it is absolutely correct that the ICE is not directly mechanically connected to the drive wheels, the ICE still contributes mechanical energy to drive the wheels (and not solely by providing electricity) by means of spinning MG2 which IS directly connected to the drive wheels. When going in reverse, you are correct, the wheels are entirely driven by electric power.

 

Reverse in the Prius is rather an unusual arrangement. If the ICE starts while you are in reverse it actually reduces the amount of torque to the wheels. This is because the ICE can only run in one direction and 72% of it's torque goes to the wheels always in a forwards direction. Taking power from MG1 splits the output 28% through MG2 in reverse and 72% forwards from the ICE. Extra power is supplied from the battery to help turn MG2.

John

 

Bob, I know you are a legend on Priuschat. As I said before I learned a lot from your posts and I know this thread well since years. Yet I hope a student is allowed to question the words of its master. I agree to almost everything in this thread, except some power/torque confusion, which has important consequences:

Replace "engine power" with "engine torque" and I agree.

28% to 72% is the ratio the torque is split, not the power. For the rest I agree.

I do not agree with this efficiency calculation. For the total efficiency we have to multiply the efficiency of the electrial path (.84) respectively of the mechanical path (.98) with the ratio that the power is split. Not with the ratio that the torque is split. I showed in Power split device and electrical/mechanical power | PriusChat that the ratio the power is split is not fixed but varies with the speeds of engine, MG1 and MG2.

I agree that ~25 mph is the lowest possible speed for energy recirculate mode. To the audience: Check the beloved Toyota Prius - Power Split Device you will see that at 25 mph MG1 RPM can't be negative without the engine RPM falling below ~1100 RPM.
I do not agree to the term "engine power" that should be "engine torque". And I might be getting you wrong, but it sounds like you suggest that energy recirculate mode is always used above 25 mph. Which might be true at constant speed on flat ground, but not in general. Think about hard acceleration and steep hills.
You can test that again with the simulator or by driving with gauges: Positive MG1 RPM = normal mode, negative MG1 RPM = recirculate mode (at least when engine is running and all RPMs are stable, transitions between RPMs are a topic on its own).

 

Thanks for explaining your way to calculate the split power. I once had the exact same thought. It seems to be the most logical and only way to calculate it. But the Prius has another surprise hidden there. And it fully complies to Newton's Laws, no worries about that. Yes, the reference frame is important. In the case of the Prius the reference frame is the housing of the transaxle. RPMs are relative to that.

Do you find any component in the PSD rotating with engine RPM "n" and 28% of the engine torque? With engine RPM and 72% of the engine torque? I would assume not. But you will find MG1 rotating at the speed dictated by the gears of the PSD. MG1 generates a counter torque of -28% of the engine torque. You can see MG1 torque on the CAN bus. Multiplying MG1 RPM with -28% engine torque gives us the mechanical power of MG1.

Let me explain planetary gears with a different approach. Suppose we have two toothed racks facing each other and a bar between them:

Suppose that the blue rack at the top is static, it's bolted to the ceiling. That is also our reference frame. We pull the center of the red bar to the right. We apply the force F_p and move with speed v_p.

Questions:

1. Which counterforce F_s do we have to apply to the green rack at the bottom to stop further acceleration once the center of the bar reaches speed v_p?
2. Which final speed v_s has the green rack?
3. How much power P_p is applied at the center of the red bar?
4. How much braking power P_s is required at the green rack to stop further acceleration?

Answers:

1. F_s = -F_p / 2
2. v_s = v_p * 2
3. P_p = F_p * v_p
4. P_s = F_s * v_s = (-F_p / 2) * v_p * 2 = -F_p * v_p = -P_p

The green rack moves at double the speed but only half the counterforce is required to keep the speed constant (= the forces in balance). All the power applied to the center of the red bar is transfered to the green rack and an equal amount of negative braking power is required to keep the balance.

Now to the blue rack at the top:

Questions:

1. Which counterforce F_r do the bolts, that keep the blue rack in place, have to generate?
2. Is there a braking power P_r required?

Answers:

1. It's the same counterforce as for the green rack: F_r = F_s = -F_p / 2
2. The speed of the blue rack is 0 m/s, therefore the braking power is: P_r = 0 m/s * F_r = 0 W

Ok, we see that we are able to split a force 50% / 50% to the racks. But the power is transfered 100% to the green rack which is able to move relative to the reference frame.

Locking at the second drawing you see that the bar will soon lose its contact to the teeth of the racks. We should introduce some more bars, like eight of them in a circle:

Nice, a pinion. Now we can move until we run out of tooth rack.

Running out of toothed rack we can solve by extending the toothed racks and bending them into circles. Let's say we make a small circle out of the green rack and a large circle out of the blue rack. Just so that the red pinion fits between them. We get: A planetary gear set! The sun gear is green and the ring gear is blue. You only need to add some more red pinions and connect them to a carrier. If I find time I'll draw that also and transform the force from the above formulas to torque. Splitting the force 50%/50% from the red pinion to the green and blue teeth will still be the same, but since sun (green) and ring (blue) gear have different diameters the resulting torque is not the same for sun and ring gear.

About your bulldozer example: The power leaving the engines is fully converted to heat and noise and maybe some earth moved under the tracks. Most of the power is leaving the tracks (btw. they are moving), the tracks have a friction on the ground. So we have a speed (moving tracks) and a force (friction). Multiply them and you get the the power leaving the tracks.

Thanks for reminding me that forces need a counterforce, but not each counterforce needs power. In my example the blue toothed rack transmits a counterforce but is not moving. Therefore all the power is going to the green rack. If the blue rack would also move then it would need some braking power. The green rack would move slower and require less braking power. The sum of the braking power would be the same. And all forces would also be the same.