Thinking about energy/speed -- and confused

[SageBrush]

The esteemed Wayne Brown has this table in one his website pages.
I'm unclear why the energy delta/speed delta is not higher for the numbers I checked. E.g., speed doubles from 40 to 80 mph, but the table shows energy consumption increasing 288/153 equal to 1.88.

I cannot think of any system that has an efficiency gain as the speed increases over that interval other than the ICE, and I find it hard to believe that offsets the geometric increase in air resistance occurring.

[fuzzy1]

That is not an efficiency gain as speed increases, it is a loss.

Are you confusing energy per mile (watt hours per mile) with power (watts)?

[darelldd]

Oops. Never mind.

[SageBrush]

Sorry I missed your post Darrell

Breaking down the 40 mph row,
air is 45.5 Wh/mile, other is 107.3 Wh/mile
At 80 mph, other tends to increase linearly, while air resistance is the speed change squared. Since we have doubled the speed, other should be 214.6 and air 182 for a total energy consumption of 396.6 Wh/mile.

I figure there is a 99.99% chance I am wrong in my reasoning above -- please help

[bwilson4web]

Let's start with total drag formula:

At 45 mph, the NHW11 takes about 10 hp to overcome the rolling and aerodynamic drag (see red line for power required as a function of velocity: )

7,450 Wh= 10 hp * (745Wh/hp)
165.6 Wh/mile = 7,450 (Wh)/45 (mi/h)

Compare that to Wayne Brown's numbers for the NHW20:

152.8 Wh/mile @40 mph
165.6 Wh/mile @45 mph (NHW11 from Toyota formula)
181.3 Wh/mile @45 mph
190.8 Wh/mile @50 mph

It turns out that at 42 mph, there is a distinct change in the control laws however Wayne's data suggests a discontinuity in drag between 40 and 45 mph. IMHO, our numbers are closer together and Wayne's data reflects the control law change that requires the engine to run all of the time versus cycling the engine as needed but implemented in the drag relationship versus ICE engine efficiency.

There is a control law discontinuity that occurs around 42 mph. The engine BSFC suffers a hit the closer the vehicle travels at 42 mph due to an increase in START/STOP cycles. As we approach 42 mph from lower speed, the energy cost of START/STOP begins to make a measurable impact on engine efficiency. Approaching 42 mph from a higher speed, puts the engine in low power regions that also suffer ICE efficiency problems due to partial power operation:


I suspect Wayne's model builds in the engine discontinuity into the drag relationship. In mine, I treat drag, rolling and aerodynamic as a continuous function.

I believe the discontinuity around 42 mph is a function of the engine thermodynamic efficiency. It hasn't been until recently that we've had enough engine BSFC data to make an accurate engine performance chart. So far, no one has worked out how to calculate the START/STOP loss but I have some ideas.

Measuring START/STOP energy loss is a real technical challenge. My high resolution data suggests engine START/STOP events occur in ~0.25 seconds. Even the excellent Graham miniscanner has difficulty recording down to this level of detail. Also there is a curious 'dead zone' in engine efficiency in the 1,700-1,800 rpm range. Something is going on but I don't have a good explanation. Of course the ZVW30 will makes all of our NHW11/NHW20 data moot.

Bob Wilson

[Frayadjacent]

I had always heard the rule of thumb is from about 50mph and up, 'wind resistance' doubles every 10mph of increased velocity.

[richard schumacher]

Sage, "W-hr/mile" is energy *per mile*. But at 80 MPH each mile takes half the time as at 40 MPH. For an efficiency comparison we most commonly use energy per *unit time*, not distance. So just multiply the "MPH" and the "W-hr/mile" values given above and get:

40 x 153 = 6 kW (six thousand Joules per second)
80 x 288 = 23 kW (23 thousand Joules per second)

It needs nearly 4x the power at 80 MPH as at 40 MPH, pretty nearly proportional to the square of the speed difference, which is as expected when neglecting fixed losses.

[donee]

Hi Sage...,

Think about the car in a vacumm. The rolling resistance is approximately constant and represents a constant force against the car. The faster you go, against the same force, the more power is needed. Double speed, doubles power.

At 40 mph some force is is used to overcome air drag. Wayne says 29.8 % of energy is used to overcome air drag, which is the same as saying that 29.8% of power is against air drag. That means 70.2 % is against rolling resistance.

Double the 40 mph rolling resistance power gives 140.4 % of the 40 mph total power is needed to overcome rolling at 80 mph. Wayne says the % of energy needed to overcome air drag at 80 mph is 63.4%. So, this 140.4 % of the 40 mph power is 36.6 percent of the car's overall power. Which means the overall power at 80 mph is 383.6 % of the 40 mph power.

Do these numbers seem better to you ?

But, your question is why is the energy consumption only 1.88 times as much then. Because of engine efficiency at partial power. Cruising the Prius at 40 mph only needs 8 hp on Bob's chart. This is about as slow as the engine can go - 1100 rpm at the pretty much fixed engine torque the car operates on. At 80 mph the power needed is 3.836 times 8 , or 30.7 hp according to the numbers above. Oh, look, Bob's chart says 35.0 hp - and the Gen 1 Prius has a .29 Cd, versus the .26 Gen 2. 30.7 * .29/.26 = 34.2 - double check of math and varying sources of performance data. So, that is good.

So now we can compute how much worse in efficiency the engine/drivetrain of the Prius is (mostly engine) at 8 HP versus 30.7 HP. If its running at 3.836 times the energy output per unit time (definition of power) but only consuming 1.88 times as much energy input per unit time, then its .49 as efficient at 8 HP , then 30.7 HP. Which seems reasonable. If the engine is near 30 % efficient at 30.7 hp, then its near 15 % efficient at 8 HP. Argonne says the engine is 25 % efficient at 12.5 hp. So, its that dropping down from 12.5 to 8 hp that results in the biggest engine efficiency hit.

I imagine somebody will pull up the BSFC chart for the Prius engine and get the same ratio.

This illustrates that energy consumption and engine efficiency are not the same thing in cars. Because, even with a 1.5 liter engine in a Prius, the engine is too large to be efficient at allot of the drving regimes. Which is why we need to have hybrid cars. Because, in the Hybrid, the engine can run at 30 hp for a bit, then turn off. Rather than at 8 hp continuously.

Another way to think about this is the Car System requires operation of the engine at inefficient power levels.

There is a flaw in these numbers however. Which is why I think my analysis of them above is correct, versus what other people have said (control law variations). Because the Prius control law results in like 70 mpg at 53 mph and 1280 RPM. Below about 50 mph fuel economy (WH/mile) gets worse. Fuel economy does not get better as one goes slower until one can get into a sub-40 mph pulse and glide situation. And you do not see that in the chart. So, I think the chart does not take into account control law stuff at all. Its just based on steady state power requirements, and the engine efficiency at that requirement.

What practical conclusions can we draw from this? The 2010 Prius has a bigger engine. But what if they had done a 1.0 liter 3 cylinder engine? With that engine, the 53 mph power requirement of 15 hp would be up into the 25 % engine efficiency range, from the approximately 15 % efficiency at 15 hp. That means that SHM mode would yeild 70 * 25/15 or 116 mpg performance.

The highway EPA improvement in the 2010 Prius is due to the highway EPA now including 65 mph cruising speeds, AND the engine apparently has a variable speed oil pump, and no belt losses , I think.

[caln8iv]

What does wh stand for -- literally -- looks like it should be ??? per hour.
Could some explain in the simplest terms the concept of wh?

[fuzzy1]

Originally Posted by caln8iv

What does wh stand for -- literally -- looks like it should be ??? per hour.
Could some explain in the simplest terms the concept of wh?

Watt-hours, which is (watts)*(hours).

One thousand watt-hours is a kilowatt hour, the same unit of energy that your electric meter reads.

One watt-hour is 3600 watt-seconds, a.k.a. 3600 joules.