Bad mpg?

[blanchard7684]

But,,, unless the engine is at wide open throttle, neither engine is full of air. While driving around a manifold vacuum of 15 in-hg is pretty normal. 15in Hg converts over to about 7.3 psia (pounds per square inch absolute).

At sea level atmospheric pressure it typically around 14.7 psia, this pressure is also referred to as 0 psig (pounds per square inch gauge) or one atmosphere.

So driving around at that 7.3 psia in the manifold, each cylinder is only getting filled with about half of the air it could hold.

Now assuming you believe Boyles law (unlike someone else here) we can easily see what pressure another displacement engine would be at if it was ingesting that same amount of air.

Lets use this example,,,say you're driving down the road at steady throttle cruising with a manifold vacuum of 15 in-hg or 7.3 psia in a 6.2L Tahoe, burning however much fuel that amount of air can burn at stochiometric ratio (14.1 for 10% ethanol gasoline or 14.7 for pure gasoline). What manifold pressure (which will still be less that atmospheric pressure) would be required to fit that same amount of air in a 5.3L?

Using Boyles law and rearranging the equation to solve for P2, that makes the pressure in a 5.3's cylinders (that's holding the exact same amount of air as the 6.2 cylinder is at 7.3 psia), come out to 8.5 psia, or about 17.3 in-hg vacuum.

So for the 5.3 ingest the exact same amount of air, it would need to open its throttle body a little further than the 6.2 and allow 8.5 psia (of the 14.7 psia available in the atmosphere) to come in. Its still running at vacuum, but it's at higher pressure (closer to atmospheric) than the 6.2, but its holding the exact same amount of air.

Now that both engines have the same amount of air in them (just one at a slightly higher pressure at bottom dead center to hold it), if you inject the same amount of fuel, you'll end up with the same afr for both engines, even though they have different displacements. And if you burn the same amount of fuel, at the same afr, in two engines with the same efficiencies,,, then they make the same power at those manifold vacuum points ( 7.3 psia and 8.5 psia for the 6.2 and 5.3 respectively in this example).

This is the whole argument above about different cylinder pressures for displacements. All the pressures talked about so far in this thread are in vacuum/ below atmospheric pressure, just some are higher (closer to atmospheric) than others.

What I described above can happen for all power levels below whatever the power level is when the smaller engine reaches wide open throttle (355hp in the case of a 5.3).



...

Error 1. Overall you are using an overly simplistic equation of state to describe a system that operates via heat and mass transfer.

Error 2. Pressure is a measure of restriction in a flowing system. You are conflating mass flow of air to a measure of upstream pressure (in the intake manifold).

Error 3. You are assuming manifold pressure is equivalent to pressure at BDC of intake stroke. If it was the same, then how is anything flowing if there is no additional motive force from pressure drop?

Error 4.

"Rearranging and solving for P2 makes the pressure in 5.3's cylinders (that is holding the exact same amount of air as 6.2 at 7.3 psia) come out to 8.5 psia"

The natural and logical inference from this is that, for every increment of displacement I remove, I gain cylinder pressure at BDC. The higher pressure at BDC means the smaller engine maintains the same power as a larger displacement engine.

Error 5. You are taking a specific operating condition for a 6.2 and substituting it into a 5.3 and assuming the only variable that changes is manifold pressure. If throttle opens up a bit, so does engine speed.

Error 6. This example says the entire induction system upstream of the short block, is absolutely meaningless.

Error 7. The difference in performance between a 5.3 and 6.2 are only reflected at wot, and not at any other rpm range.

 

[swathdiver]

I find it interesting that, after decades of Corporate Average Fuel Economy testing,
leading from 3-speed to 10-speed transmissions and myriad technologies for improving MpGs,
'they' have never bothered to state a road speed range where a given vehicle gets near its best MpGs.
Whatever the 'city' MpG' cited by the Monroney sticker, it ain't easy to achieve that number.
It's a bit easier on the highway.
I'd prefer they say something like,
'if you want to get at least 2xMpG, keep it under 65MpH on the highway.'

It's just about all spelled out beginning here:

Detailed Test Information

EPA tests vehicles by running them through a series of driving routines, or schedules, that specify vehicle speed for each point in time during the laboratory tests.

fueleconomy.gov

Using this data I've made a very accurate spreadsheet to calculate city versus highway usage for a tank of fuel in both miles per gallon and more accurately, gallons per hour.

I haven't been here since Christmas and see all this bantering going on. Did the OPs gas mileage go up yet? Does he understand what affects it?

 

[B-train]

It's just about all spelled out beginning here:

Detailed Test Information

EPA tests vehicles by running them through a series of driving routines, or schedules, that specify vehicle speed for each point in time during the laboratory tests.

fueleconomy.gov

Using this data I've made a very accurate spreadsheet to calculate city versus highway usage for a tank of fuel in both miles per gallon and more accurately, gallons per hour.

I haven't been here since Christmas and see all this bantering going on. Did the OPs gas mileage go up yet? Does he understand what affects it?

Good question @swathdiver . I'm about done with all the engineering classes here overcomplicating things. We have everything from a simplistic volume and pressure approach to algorithms with pressure vs vacuum.......I'll just let everyone be correct in their own way.

I hope OP can glean something from this mess and either change approach to driving style, or just put on some more miles and do a recheck after the first oil change. Either way, his initial MPG with cruise control at a highway speed of about 80 mph doesn't seem awful to me although I would expect a RWD model to get slightly better than 4WD as time goes on.

 

[blanchard7684]

What you said above is just not correct.

Before you start typing out your angry reply, take a few moments to read below and think about it for just a second.

You most likely think the AFR can't be same when mixing 10cc of fuel into two different size engines because you assume each of those engine is full of air , two different quantities of air (5.3 and 6.2 L respectively), and as such, it could never make the same ratio.

But,,, unless the engine is at wide open throttle, neither engine is full of air. While driving around a manifold vacuum of 15 in-hg is pretty normal. 15in Hg converts over to about 7.3 psia (pounds per square inch absolute).

At sea level atmospheric pressure it typically around 14.7 psia, this pressure is also referred to as 0 psig (pounds per square inch gauge) or one atmosphere.

So driving around at that 7.3 psia in the manifold, each cylinder is only getting filled with about half of the air it could hold.

Now assuming you believe Boyles law (unlike someone else here) we can easily see what pressure another displacement engine would be at if it was ingesting that same amount of air.

Lets use this example,,,say you're driving down the road at steady throttle cruising with a manifold vacuum of 15 in-hg or 7.3 psia in a 6.2L Tahoe, burning however much fuel that amount of air can burn at stochiometric ratio (14.1 for 10% ethanol gasoline or 14.7 for pure gasoline). What manifold pressure (which will still be less that atmospheric pressure) would be required to fit that same amount of air in a 5.3L?

Using Boyles law and rearranging the equation to solve for P2, that makes the pressure in a 5.3's cylinders (that's holding the exact same amount of air as the 6.2 cylinder is at 7.3 psia), come out to 8.5 psia, or about 17.3 in-hg vacuum.

So for the 5.3 ingest the exact same amount of air, it would need to open its throttle body a little further than the 6.2 and allow 8.5 psia (of the 14.7 psia available in the atmosphere) to come in. Its still running at vacuum, but it's at higher pressure (closer to atmospheric) than the 6.2, but its holding the exact same amount of air.

Now that both engines have the same amount of air in them (just one at a slightly higher pressure at bottom dead center to hold it), if you inject the same amount of fuel, you'll end up with the same afr for both engines, even though they have different displacements. And if you burn the same amount of fuel, at the same afr, in two engines with the same efficiencies,,, then they make the same power at those manifold vacuum points ( 7.3 psia and 8.5 psia for the 6.2 and 5.3 respectively in this example).

This is the whole argument above about different cylinder pressures for displacements. All the pressures talked about so far in this thread are in vacuum/ below atmospheric pressure, just some are higher (closer to atmospheric) than others.

What I described above can happen for all power levels below whatever the power level is when the smaller engine reaches wide open throttle (355hp in the case of a 5.3).

We can also find the manifold vacuum the 6.2 would be at with the 5.3 at wide open throttle using the same method. I'll skip stepping through the work because its described above, the answer is 12.56 psia.

So when the 5.3 is tapped out at wide open making its max horsepower, the 6.2 is still throttling back and only using 12.56 psia of the 14.7 psia available to it, at that same engine RPM. So it can open its throttle body a little more (because its still running at a manifold vacuum), take in more air, inject more fuel ,and make more power.

Maybe the whole psia thing and one pressure being higher, but still a vacuum / below atmospheric pressure is what confused the last guy. Hadn't thought to explain that a higher pressure in a naturally aspirated engine just meant less vacuum/ closer to atmospheric and not a positive pressure.

Long post, but its hard to get across the entire point/ explain everything, without leaving out details that could cause arguments, and still keep the post lengths short at the same time.

...

Put a 6.2 on an engine dyno.

See what power it makes at 14.7 afr at 1200 rpm. Record mass air flow and mass rate of fuel flow.

Put 5.3 on engine dyno. See what power it makes at 1200 rpm and 14.7 afr. Record the mass flow of air and fuel.

You will notice the 5.3 is making less power at same afr, same engine speed.

Power is Torque x rpm. Incorporate this into the power equation.



Torque is proportional to SQUARE of bore size.

The 5.3 is making less torque at a given rpm.

lesson concluded.