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).

For this explanation, assume we are discussing a piston engine operating on gasoline for fuel. (This analysis works for Gasoline, Methanol, Diesel fuel, Jet fuel, Whale Oil, Whatever. Each fuel has it's own weight and energy content.)

Gasoline, according to P&W Aircraft 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 779 foot-pounds. By arithmetic, it can be shown that one horsepower (550 ft-lbs per second) is the equivalent of 42.4 BTU's per minute or 2542.4 BTU's per hour.

How is that useful?  Here is an example. We have tested a reasonably good 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 2542.4 (the number of BTU's-per-hour in one HP), you discover, to your surprise, that it is 1062 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 / 1062 HP = 28.3 % (which is fairly good by piston engine standards).

The calculation for Thermal Efficiency (TE) is:

Going back to our 300 HP example, TE =  0.1338 x 300 HP /  142 PPH = 0.283 (28.3 %)

(Note that when using % in a calculation, you must divide the percentage number by 100. That is how 28.3 % becomes 0.283.)

If you prefer gallons per hour, the Thermal Efficiency calculation is:

HP = TE x FUEL FLOW (GPH) x 5.92 (# per gallon) x 19,000 / 2542 (BTU per HP per Hour)
which reduces to:
                  HP = TE x FUEL FLOW (GPH) x 44.25
solving for THERMAL EFFICIENCY:       TE = 0.0226 x HP / FUEL FLOW (GPH)
solving for FUEL FLOW:                        FUEL FLOW (GPH) = 0.0226 x HP / TE

The value of this Thermal Efficiency relationship is that, by assuming a reasonable TE value (27% - 29%), you can estimate the amount of fuel required to produce a given amount of power. (That will lead to an even more valuable equation a bit later.)

Here is an example. Suppose you need to produce 300 HP. What will be the required fuel flow assuming 28.3% TE ?

            Simple. FUEL FLOW = 0.1338 x 300 HP / 0.283    (28.3 %),  or
                         FUEL FLOW = 142 PPH or 24 GPH.

BRAKE SPECIFIC FUEL CONSUMPTION (BSFC)

A more commonly used yardstick for expressing thermal efficiency is known as Brake Specific Fuel Consumption (BSFC). It is simply fuel flow (in pounds-per-hour) divided by measured HP, and is expressed in Pounds-per-Hour-per-HP.

BSFC = Fuel Flow (PPH) ÷ Horsepower
or    BSFC = 5.92 x Fuel Flow (GPH) ÷ Horsepower
(equation 2)

This tool is also an important yardstick for comparing the performance of one engine to another and for evaluating the reasonableness of performance claims.

An excellent BSFC for a well-developed, naturally-aspirated, high-performance liquid-cooled engine at 100% power is in the neighborhood of 0.44 – 0.45. Claims of gasoline engine BSFC values less than 0.42 at max power tend to be suspect. At reduced power settings (in the region of 70% and below) BSFC values of 0.38 have been achieved, but they are not commonplace.

The operator manual for a 300 HP Lycoming IO-540-K, L, or M series engine shows a full power fuel flow of 24 GPH which is a BSFC of 0.474 ( 24 * 5.92 ÷ 300 ) and a TE of 28.3% (explained above). Those numbers aren't too bad for an air cooled engine which meets the FAR-required detonation margins. However, the turbocharged TIO-540-V2AD requires a MINIMUM of 39.2 GPH at 350 HP for a BSFC of 0.663 and a TE of 20.4%.

So if someone tells you that they have a piston engine which, at max power, makes 300 HP on 20 GPH of gasoline, you quickly calculate a BSFC of 0.39 and a Thermal Efficiency of 34.4%. You should be highly suspicious of such a claim.

In a four-stroke naturally aspirated engine, the theoretical maximum amount of air that each cylinder can ingest during the intake cycle is equal to the swept volume of that cylinder (0.7854 x bore x bore x stroke).

Since each cylinder has one intake stroke every two revolutions of the crankshaft, then the theoretical maximum volume of air it can ingest during each rotation of the crankshaft is equal to one-half its displacement. The actual amount of air the engine ingests compared to the theoretical maximum is called volumetric efficiency (VE). An engine operating at 100% VE is ingesting its' total displacement every two crankshaft revolutions.

There are many factors which determine the torque an engine can produce and the RPM at which the maximum torque occurs, but the fundamental determinant is the mass of air the engine can ingest into the cylinders, and there is a nearly-linear relationship between VE and maximum torque.

For contemporary naturally-aspirated, two-valve-per-cylinder, pushrod engine technology, a VE over 95% is excellent, and 100% is achievable, but quite difficult. Only the best of the best can exceed 110%, and that is by means of extremely specialized development of the complex system comprised of the intake passages, combustion chambers, exhaust passages and valve system components.

Generally, the RPM at peak VE coincides with the RPM at the torque peak. And generally, automotive engines rarely exceed 90% VE. There is a variety of good reasons for that performance, including the design requirements for automotive engines (good low-end torque, good throttle response, high mileage, low emissions, low noise, inexpensive production costs, restrictive form factors, etc.), as well as the allowable tolerances for components in high-volume production.

For a known engine displacement and RPM, you can calculate the engine airflow at 100% VE, in sea-level-standard-day cubic feet per minute (scfm) as follows:

100% VE AIRFLOW (scfm) = DISPLACEMENT (ci) x RPM / 3456
(equation 3)

Using that equation to evaluate a 540 cubic-inch engine operating 2700 RPM reveals that, at 100% VE, the engine will flow 422 SCFM.

If I solve (equation 2) (explained back in Thermal Efficiency) how to calculate the fuel flow required for a given amount of power produced. Once you know the required fuel flow, you can calculate the airflow required for that amount of fuel, then by using that calculated airflow along with the engine displacement, the targeted operating RPM, and the achievable VE values, you can quickly determine the reasonableness of your expectations. Here's how.

Once you know the required fuel flow, you can determine the required airflow. It is generally accepted (and demonstrable) that a given engine (of reasonable design) will achieve its' best power on a mixture strength of approximately 12.5 parts of air to one part of fuel (gasoline) by weight. (Other fuels have different best-power-mixture values. Methanol, for example, is somewhere around 5.0 to 1.)

Using that best-power air-to-fuel ratio, you calculate required airflow:

AIRFLOW (pph) = 12.5 (Pounds-per-Pound) x FUEL FLOW (pph)
(equation 4)

But airflow is usually discussed in terms of volume flow (Standard Cubic Feet per Minute, SCFM). One cubic foot of air at standard atmospheric conditions (29.92 inches of HG absolute pressure, 59°F temperature) weighs 0.0765 pounds.  So the volume airflow required is:

AIRFLOW (scfm) = 12.5 (ppp) x FUEL FLOW (pph) / (60 min-per-hour x 0.0765 pounds per cubic foot)

That equation reduces to:
 REQUIRED AIRFLOW (scfm)  = 2.723 x FUEL FLOW (pph) 
(equation 5)

OK, hang on. The really useful bit is next.

When you solve (equation 2) (will explained in Thermal Efficiency) for FUEL FLOW, I get:

FUEL FLOW (pph) = HP x BSFC

Replacing "FUEL FLOW" in (equation 5) with with "HP x BSFC" from (equation 2) produces this very useful relationship:

REQUIRED AIRFLOW (scfm)  = 2.723 x HP x BSFC
(equation 6)

Now, using equation 6 you can estimate the airflow required for a given amount of horsepower, and using equation 3, you can calculate the 100% VE airflow your engine can generate at a known RPM.

Combining those two equations yields one equation which enables you to evaluate the reasonableness of any engine program by knowing just a few values: (1) required HP, (2) operating RPM, (3) engine displacement (cubic inches), and (4) an assumed reasonable BSFC.

Here it is:

REQUIRED VE = ( 9411 x HP x BSFC ) / (DISPLACEMENT x RPM)
(equation 7)

Here is an example of how useful that relationship can be. Suppose you decide that a certain 2.2 liter engine would make a great aircraft powerplant. You decide that 300 HP is a nice number, and 5200 RPM produces an acceptable mean piston speed . How reasonable is your goal?

The required VE for that engine will be:

Required VE = (9411 x 300 x .45 ) / (134 x 5200 ) = 1.82 (182 %)

Clearly that's not going to happen with a normally aspirated engine. Supercharging will be required, and you can use the 1.82 figure to calculate the approximate MAP needed (1.82 x 30" = 55" MAP) for that power level.

Here's another example. Suppose you want 300 HP from a 540 cubic inch engine at 2700 RPM, and assume a BSFC of 0.45.  Plugging the known values into equation 7 produces:

Required VE = (9411 x 300 x .45 ) / (540 x 2700) = 0.87 (87 %)

That is a very reasonable, real-world number. If you recognized the figures as those for the 300-HP Lycoming IO-540 discussed above, excellent.

Brake Mean Effective Pressure (BMEP)

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