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Old 01-23-2012, 10:41 AM   #40 (permalink)
mort
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Quote:
Originally Posted by drmiller100 View Post
What makes one engine more efficient then the next?
Hello drmiller100,
Well I don't know what make any give design choice better and under which conditions - more swirl, hemi head, indexed spark plugs etc. - but in so far as which engine gets better mpg based on being right-sized I have an opinion.
I'm going to describe an imaginary Otto cycle gasoline engine but other cycles behave similarly.
You emphasized pumping losses in your post, so I will look at that first. From the mechanical point of view, the exhaust pumping losses, and impact on efficiency, are straight forward. Any back pressure is subtracted from the average expansion stroke pressure to find the effective power stroke pressure. So if the average expansion pressure is 200 psi and the back pressure is 5 psi then the engine produces 2.5% less power than it could. Intake pumping losses are approximated the same way, with a couple of constraints. Intake partial vacuum is also subtracted from average expansion pressure. So for an engine with a closed throttle (about 10 psi), but still high in the power band - expansion pressure of 200 psi - the engine produces about 5% less power. The actual pumping power is easy to compute. It's mass flow times suction force. For an average engine, throttle pumping power is always less than about 5% of shaft power. The constraints on intake pumping are that the pressure can't go below a complete vacuum. In the case of exhaust back pressure, that can go as high as the average expansion pressure and the engine still chug along, if barely. The exhaust gasses are compressed to the point that the tail pipe will be shrieking with leaks, but almost all of the exhaust mass will be ejected. The intake gets rarefied until there is nothing there. And the maximum pressure across the throttle is 1 atm, about 14.7 psi.

However, the most important figure determining efficiency is compression ratio. For the Otto cycle the expansion ratio (ER) equals the compression ratio (CR), except that the fuel does take time to burn. In the Diesel cycle the effective ER depends on how far down the expansion stroke fuel continues to be injected. Modern Diesel engines must meet emission limits that limit maximum power, but also allow for high efficiency at high power output.

In the Atkinson cycle, the ideal expansion ratio is just enough more than the compression ratio by the amount of heat added by the burning fuel. Conventional designs allow for about 8.5:1 effective CR and about 11:1 ER, based the fuel burn adding about 25% more than the heat of compression.

Compression ratio determines efficiency because the way a heat engine works is to take your working fluid, raise its temperature, add some more heat by burning fuel and then allow the whole blob to cool off through a machine that turns the heat in to motion. The amount of heat you can turn into usable work is some portion of that flow from the highest temperature to the temperature you end at. The end temperature can't be below the room temp. where your engine is. In fact, for conventional materials, the ejection temperature must be well above the boiling point of water.

Imagine an engine with a CR of 10:1, at some reasonable rpm the compression temperature will be about 650 K. Assume, and this is purely fictional to make the numbers easy, that burning the fuel adds 350 K so the peak temp. is 1000K. The ejection temp is 300K. The maximum thermal efficiency is is the highest temp. minus the lowest temp. all divided by the highest temp. So (1000-300)/1000 x 100 = 70%
Now imagine the CR is reduced to 5:1. Now the compression temp. is about 500K, if you add the same amount of heat the temperature rises by the same proportion. The 350K we added in the 10:1 engine is a rise of 350/650 about 54%. The same rise in the 5:1 engine is about 270K. So the maximum efficiency is (770-300)/770 X 100 = 61%

Why does intake throttling reduce efficiency? It's like reducing the compression ratio. If you reduce the amount of mixture going into the cylinder, that reduces the compression pressure and temperature. Consider reducing the mixture (and amount of fuel) to 1/5th. Now the compression temp. is about 360K the amount of fuel added is 1/5 and the amount of heat added is 360 * .54 /5 = 38. So the max. efficiency is (398 - 300)/398 X 100 = 25%

Here's a crazy exaggeration. If your engine has a very wide efficiency band you could get 1/5 the maximum power by either running at 1/5 the maximum rpm. Say your maximum power occurs at 5000 rpm, you can gear down to run at 1/5th power at 1000 rpm and use about 1/5th as much gas as full power. Or you can throttle down to 1/5th power at 5000 rpm and use about half as much fuel as at full power. Or put another way if you could get 30 mpg at cruising 1000 rpm, you'd get about 10 mpg throttled down but in first gear at 5000 rpm but at the same cruise speed.

Ideally you run your engine at the highest compression pressures you can for the amount of power required. While many gas savers will tell you to limit your use of brakes. It's clear that the big efficiency killer is allowing the throttle to close. Pedal to the metal at all times!

-mort
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adam86 (03-14-2012), drmiller100 (01-23-2012), Ken Fry (01-23-2012), t vago (04-23-2012)