Demostylus,
Is the kinda’ efficiency curve you plotted the reason why much less powerful engines in the same chassis get would get better gas mileage, or is that an effect of engine weight/gearing?
Demostylus,
Is the kinda’ efficiency curve you plotted the reason why much less powerful engines in the same chassis get would get better gas mileage, or is that an effect of engine weight/gearing?
That’s another pretty complicated question, Jonathon.
The efficiency gain in the curve is entirely illusory. The engine drag is a kind of “overhead” that shouldn’t really be figured into efficiency the way that Gordon Taylor tried to in the graph zut linked to.
A smaller engine has a smaller “overhead”. It takes much less power to keep a 3 litre engine hot and spinning than it does a 6 litre engine in the same car. The 3 litre engine also pumps out less unburnt fuel than the 6 litre.
The “overhead” difference above would account for most of the difference in fuel consumption seen in a standardised test, where both vehicles were loaded on a dyno through the same cycle.
In the real world, the results might not turn out like that. Some drivers drive according to throttle position, e.g. accelerate away from the lights at about half throttle. The 6 litre engine will deliver much worse fuel consumption for such a driver. Some drivers, OTOH, might try to compensate for the smaller 3 litre engine by running it at greater throttle position and higher rpm. That could conceivable cause the 3 litre engine to use more fuel.
Like I said, it’s complicated.
Is the engine drag static and determined by engine type, or is it proportional to RPMS, or is it a multi-factorial output that is the sum of a static element combined with a (frictional?) element that scales with RPM?
For friction this is true, but you’ve neglected the pumping losses from sucking air through a partially closed throttle. Pumping losses fall with throttle opening.
Page 12 of the following link makes the assertion:
**"Throttling loss increases from zero at wide-open throttle (WOT) to about half of all fuel usage at idle (other half is friction loss)
At typical highway cruise condition (≈ 1/3 of BMEP at WOT), about 15% loss due to throttling" **
So for your table, you should have another pumping loss term equal to the engine drag at 0% throttle opening and decreasing progressively to zero at 100% throttle opening. How this effects the overall argument I’m not sure - too many variables, not enough data. It’s worth noting that in fuel-economy competitions, engines are alternately run at open throttle to accelerate and then turned off to coast, suggesting that there is a real efficiency gain to be had from running at open throttle.
WARNING! 3 megabyte Powerpoint link!
http://carambola.usc.edu/AME436S05/Lecture7%20files/AME436-S05-lecture7.ppt
Much smaller Google html conversion of the same link:
http://64.233.183.104/search?q=cache:nipcPvmLObMJ:carambola.usc.edu/AME436S05/Lecture7%20files/AME436-S05-lecture7.ppt+otto+cycle+pumping+losses&hl=en
I disagree. You’re changing the definition of the problem. Xema’s original post that you quoted states: “You drive uphill at a speed that corresponds to an efficient engine RPM, then shut off the engine and coast downhill.” In your hypothetical example, you’re ignoring that. Moreover, even if you did keep the engine on during the coast downhill (not a bad idea for safety reasons), you’d put the car in neutral and let the engine idle. If you do that, the engine is running at a much lower speed, and is much more efficient.
Let’s run through some numbers, using the real engine map on this page. To make it easy, let me alter your scenario a bit. Let’s assume 50min at 10kW/ 4000rpm (flat ground) versus a 10min hill at 50kW/ 4000rpm. In addition, let’s assume 5kW of power to drive the power steering/power brakes and so forth.
Case 1: Flat ground: 50 min at 4000rpm, 15 kW. The overall efficiency here is about 22%, so the total energy used is (15)/(.22)*50 = 3410kW-min.
Case 2: Uphill: 10 min at 4000rpm, 55 kW. The overall efficiency here is about 35%, so the energy used is (55)/(.35)*10 = 1570kW-min. In addition, we’ll idle downhill: 40 min at 1000rpm, 5kW. The overall efficiency here is about 28%, so the energy used is (5)/(.28)*40 = 714kW-min. Total energy usage in this case is 2284kW-min, 33% less than in case 1.
Note that case 2 would have a greater energy savings yet if we turned the engine off, as Xema first suggested. As you say, this assumes something of an optimally-shaped hill, so a real-life scenario is unlikely to be anywhere near so favorable.
“Illusory”? Why would you say that? Lower frictional losses mean higher efficiency. There’s nothing illusory about it. Yes, real world conditions are more complex than just looking at an equation or a chart, but in general – and particularly in the specific case we’re looking at, constant highway speeds – a smaller engine of similar design will be more efficient.
Just for the record, I believe that coasting downhill with the engine off is illegal in some states. I was told this by a California Highway Patrol officer during a casual chat at a gas station. I have not looked it up in the state codes either in California or here in Montana, but he ought to know.
Note that we have been talking about coasting down a moderate hill, where routine braking is not needed. I have done this many times and not yet found it to be hazardous. The steering is stiffer, but for the moderate inputs needed this is scarecely noticeable, let alone difficult. Brake boost nearly always uses a vaccuum reservoir, which retains its boost capability long after the engine is shut off, provided it isn’t used much.
So, based on a fair amount of experience, I simply see no problem here.
I further note that there are plenty of vehicles out there with neither power-assisted steering nor brakes.
Noted, but not especially relevant to a thread concerned with vehicle efficiency.
I’m still trying to figure out where you guys are finding hills that take 10 minutes to climb and 40 minutes to decend. Are you all driving to Death Valley, or will it all even out in the end when you have to drive home, uphill for 40 minutes, and downhill for 10?
As noted above, we are postulating a case where you can climb a rather steep slope and then coast down a rather gradual one - no net gain or loss of altitude.
Coasting without engine braking is dangerous. As a matter of fact, it was one of the first vehicle safety laws ever passed. Cars were once allowed to have “freewheeling” as a feature. When you were going downhill, the car would basically disconnect the engine from the driveline. This was banned by federal law in the thirties. Driving with the engine turned off is just as dangerous, since you wouldn’t have power brakes or power steering, and if you’ve ever driven a car that has power steering that failed, it’s not like you’re driving a car with manual steering. It’s much much harder to steer since you have to overcome the inertia of all that fluid. Cars with power brakes are not made to be braked without power, they are much more difficult to brake that way.
I guess you missed post #27 above.
To eliminate this red herring, and get back to the issue of efficiency, let’s postulate a vehicle with unboosted steering and brakes.
If you postulate a car with unassisted braking and steering, you STILL shouldn’t do it. You will probably overheat the brakes, and since your theoretical car(find me a car made today that has neither) probably was made in the sixties when cars were made with all drum brakes, you need engine braking or else you’ll end up with a car that won’t stop period. I guess if you were on a straight downhill road, without any curves, or traffic so you wouldn’t have to touch your brakes, then it would make some sense. I have no idea where such a road might exist. I don’t get the fishy references here. I thought a red herring was a clue in a mystery novel that pointed to the wrong person as being the murderer.
Vehicle doesn’t need to be a car to illustrate the point - a motorcycle would work fine.
I’ve several times noted that we are talking about coasting down a shallow slope where little to no braking is required.
Agreed.
The road need not be straight, since (as noted) steering works just fine. Long & gradual slopes are not hard to find.
More generally, it’s something that tends to distract from the true matter at hand.
My 1950 Studebaker freewheeled when you engaged overdrive. I believe this was available into the sixties.