1.00 Lambda = 14.7 for E0
1.00 Lambda = 14.124 for E10
1.00 Lambda = 9.760 for E85

Both, it will read in Lambda or AFR. The gauge reads 16.5:1 gas AFR so actual E85 AFR is around 11:1 . I've run it at 17.5:1 but its hard to tell if the mpg is any better.
Sorry, yeah 16.5:1 E85 AFR would be a bit lean, LOL.

 

I look at it as: Lambda = 1.0 for all of them. 1.0 is stoich for all of them also.

cheap wide-bands are programmed that 1.0 lambda = 14.7. We know that isn't correct when E is present.
if you are cuising on E85 and the AFR reads 14.7 on the gauge and Lambda = 1.0 then the AFR is 9.7 and the gauge is wrong

 

I just went into my editor, highlighted the 62mph and the 75 mph lines and multiplied by .9 across the board. Is this the right thing to do for mpg? If you can't tell, tuning is still quite a mystery to me.

 

I would Data-Log a vehicle over an flat strech of highway and maintain a given speed. Then I tune the truck and go back out over the same strech and maintain the same speed. I would compare the airflow and throttle angle. More MPG would show less airflow and less throttle angle.

Ok none of us travel on a flat raod at 60 mph from start to finish. After you have dialed in the criuse speed start working on a drive cycle. Say you have a hill you go up everyday. That would ba a good place to start. Do the same thing only on this hill.

 

I had a 2006 Express van in today that had an intake gasket leak.
After resealing the intake the idle airflow was close to the target airflow. It had been 4 g/sec lower.
After I was done my fuel trims were +15% to +17%
I adjusted the target AFR and guess what they were after?

 

Is this a riddle, or are you going to tell us?


My MPG's seem to be un-change-able...

 

The only way to increase MPG is to decrease airflow. Tuning the engine is the best way to do that with increasing Brake Mean Effective Pressure (BMEP).
The definition of BMEP is: the average (mean) pressure which, if imposed on the pistons uniformly from the top to the bottom of each power stroke, would produce the measured (brake) power output.

If I don’t remember all of this correctly I apologies now

The optimal Compression ratio for gas is about 15 to 1. Now this is 100% gas and I am not sure what the octane rating is. This is from a class I took

Stoich for 100% gas is 14.7 to 1, but this can be leaned out at cruise safely. 100% gas is getting rare to find

If you are leaving the engine alone then tuning is your option. This leaves timing and EGR as the primary adjustments.

You would increase the timing close to Mean Best Timing (MBT) and hope you had good gas on every fill up

The EGR system hurts MPG by reducing combustion chamber temperature by reducing combustion chamber pressure by adding an inert gas that lowers the combustion chamber pressure.

Knocking/Pinging is abnormal combustion in which the air-fuel mixture ignites prematurely due to exposure to high temperature and pressure, creating an unwanted high-frequency noise. When the compression ratio is increased, the temperature at compression top dead center (TDC) also rises, increasing the probability of knocking. Pinging reduces engine torque.

In order to lower the temperature at compression TDC, reducing the amount of hot exhaust gas remaining inside the combustion chamber is effective instead of adding hot EGR gas. For example, with a compression ratio of 10:1, a residual gas temperature of 750 deg. C, and an intake air temperature of 25 deg. C, if 10% of the exhaust gas remains, the temperature inside the cylinder before compression increases by roughly 70 deg. C, and the temperature at compression TDC is calculated to increase by roughly 160 deg. C. Therefore, it can be easily inferred that the amount of residual gas has an major impact on knocking.
One option to significantly reduce residual gas is the adoption of a 4-2-1 exhaust system. One option to significantly reduce residual gas is the adoption of a 4-2-1 exhaust system. When the exhaust manifold is short, the high pressure wave from the gas emerging immediately after cylinder No. 3’s exhaust valves open, for example, arrives at cylinder No.1 as it finishes its exhaust stroke and enters its intake stroke. As a result, exhaust gas which has just moved out of the cylinder is forced back inside the combustion chamber, increasing the amount of hot residual gas. With a short exhaust manifold, the high pressure wave arrives at the next cylinder within a short amount of time, causing this adverse effect to continue from low to high engine speeds. However, with a long 4-2-1 exhaust system, since it takes time for the high pressure wave to reach the next cylinder, the effect mentioned is limited to extra-low engine speeds, making the reduction of residual gas at almost all engine speeds possible.

 

Not trying to contradict you, but why is it that when some people disable their EGR they claim their mileage drops? I was under the assumption that the EGR adds inert gas and is basically like decreasing the displacement of the engine, therefore reducing fuel consumption?

 

If they turn off the EGR system they will need to increase timing because the EGR adder table is now turned off.

A 6.0L with EGR will have less timing on the main spark tables than a 6.0L without EGR. A OE 9.4 to 1 LQ4 has less timing than a 10.08 to 1 LQ9

I think a L33 5.3L 9.9 to 1 has more timing than a LM7 5.3L 9.5 to 1 engine, but I don't have the file to look at with me.

To get the most MPG from the engine you need to get the most power from the energy stored in the fuel. Adding EGR will not do that. When an engine is running at MBT you will get the best MPG. Think of MBT as when a drag race car is running and the exhaust is crisp and kind of popping. They are getting the most power from engine. If you retard the timing it won't sound crisp and you would be wasting energy in the fuel

A "combustion engine" is a device which converts the chemical energy stored in a fuel into heat energy, and then converts a portion of that heat energy into mechanical work. Any combustion engine can be effectively visualized using what is commonly known as the "Black Box" model. The energy source for an engine is the chemical energy stored in the fuel. That energy is released by the oxidization of the fuel (combustion) by an oxidizing medium, which in most cases is the oxygen which makes up about 19% of the air we breathe. Variations on that theme include the use of oxidizing additives (Nitrous Oxide, for example) and high-energy fuels which contain a substantial supply of oxidizer in their makeup (Nitromethane, for example).

Gasoline, according to Pratt & Whitney data sheets, has a specific gravity of 0.71, and therefore a weight of about 5.92 pounds per gallon, and releases approximately 19,000 BTU of energy per pound of fuel burned.

What is a BTU? A "British Thermal Unit" is defined as the heat energy required to raise the temperature of one pound of pure water by one degree F, and is equivalent to 778 foot-pounds of work / energy. By arithmetic, it can be shown that one horsepower (33,000 ft-lbs per minute) is the equivalent of 42.4 BTU's per minute or 2545 BTU's per hour (33,000 ÷ 778 = 42.4165 ≈ 42.4; 42.4165 × 60 = 2544.98).

How is that useful? Here is an example. Testing a reasonably good 4-stroke piston engine which converts approximately 24 gallons of gasoline per hour ( 142 pounds of fuel per hour ) into 300 measured horsepower.

So how much of the total fuel energy does this engine convert into horsepower? If you burn 24 gallons of gasoline (142 pounds) over the course of one hour, you release 2,699,520 BTU's of energy (19,000 x 142). If you divide the 2,699,520 BTU's by 2545 (the number of BTU's-per-hour in one HP), you discover, to your surprise, that it is 1061 HP. But the engine is only making 300 HP. Where is all the rest of that energy going?

It is a known fact that a piston engine does a rather inefficient job of converting fuel energy into power. The rule of thumb approximation is that nearly 1/3 of the fuel energy goes out the exhaust pipe as lost heat, approximately 1/3 of the fuel energy is lost to the cooling system (coolant, oil and surrounding airflow), leaving roughly 1/3 of the energy (best case) available for power output. Some of that power is lost to making the pistons go up and down, driving accessories (oil pump, coolant pump, alternator, vacuum pump, hydraulic pump, etc.), losses from pumping air through the engine, thrashing the oil in the crankcase, and friction in various forms.

The difference between the energy content of the fuel consumed and the useful power extracted from the engine is known as Thermal Efficiency (TE). So in our 300-HP engine example, the TE is 300 HP / 1061 HP = 28.3 % (which is fairly good by contemporary standards for 4-stroke production piston engines).

 

Thanks for the explanation. While a lot of this is way over my head I enjoy reading it and continue to learn.