Unrammed air increases fuel milage?

[spork]

Doesn't make a bit of sense to me but it's true. I'm flying from BIKF to LOWI, but I can get well over 2000nm range flying FL190 at 2600rpm so I'm thinking of going farther. Why does this happen? Shouldn't the hotter air (if only marginally) have a negative effect on mileage?

[mer8771]

Un-rammed air should be colder not warmer. Or have missed the point? Anyway I'll go on with what little I know in hopes it helps you. A ram or scope will force air into a charger of any type. The higher density air will then be compressed higher by the charger. Thus bringing the air pressure up going into the cylinders. Higher pressure means you can not lean it out as much. Un-rammed air but using the charger will have the effect of allowing you to lean it out further. If it is un-rammed and un charged then you can lean it back even further. The air temp will rise the more you compress the air, this is true on the air frame as well. To a point compressed air can be cool until it reaches a point that the compression surpasses the heat being exchanged by the moving air. Now as far as the millage, the slower you go the less drag you will have on the aircraft. However there is a point at witch the speed being to low will introduce more drag by the angel of attack. So if you lean it out un-rammed and uncompressed and find the speed at which you start to re introduce drag by the AOA should in theory give you the best millage. Compressing the air to critical altitude then leaning to 5% of PTT then slowing to best glide angel will in theory give you the best range. After all that I thank planing to stop for a refuel would be safer and easier albeit ever so slightly longer on time.

Please for the love of god tell me where I am wrong and provide references so that we (I) can learn. I am not trying to be a smart a** this is my understanding and I don't want to tell other people wrong things.

Bernoulli's principle is one formula that will give some of the answers.

The equation is PV = nRT
P = pressure
V = volume
T = temperature
n = the amount of gas, if you don't ad or remove any you can ignore this.
R = a constant to get all this in the correct units.

So basicaly PV/T is the same before and after and as long as you use the same units it all works.
Remember to do this in terms of changes in temperature. If you want this in absolute temperature you need to convert to Kelvin (google absolute temperature)

In Thermodynamics there are different fundamental THERMODYNAMIC PROCESSES namely Isobaric, Isthermal, Isentropic, and Polytropic and Isochoirc processes.

Each of these processes has its own unique method of heating or cooling of the air/gas depending on the boundary conditions or type of surrounding or keeping some fundamental parameters like pressure, temperature, volume or entropy a constant.

Depending on the type of process chosen the values and relations between the pressure, temperature ( as mentioned in your case) will vary.

The basic equation is the Ideal Gas Equation PV = mRT which remains applicable in all cases in any process.

But there are another set of equations derived from the above ie PV^n = C, where n is called the polytropic index, and all the above processes have a particular value of n. And depending on this value of n for each process the amount of heat or pressure required to heat up or compress air to a particular value will depend on or vary.

It all depends on your choice of process and the variables will change according to the particular equation.

QUESTION

What compression ratio Vmax / Vmin will raise the air temperature from 20C to 1000C in an adiabatic process?

the gas is diatomic

ANSWER

For an adiabatic process, with Q = 0, the first law of thermodynamics is:

Change in thermal energy = Work. Compressing a gas adiabatically (W > 0) increases the thermal energy. So an adiabatic compression raises the temperature of a gas.

Pf = Pi(Vi/Vf)^gamma

gamma = 1.40 for a diatomic gas (like air)

Pf = Pi(Vmax/Vmin)^1.40


Wait a minute:

(Tf) Vf^(gamma - 1) = (Ti) Vi^(gamma - 1)

(1,273K)Vf^(gamma-1) = (293K)Vi^(gamma -1)

(1,273K) Vf^0.4 = (293K) Vi^0.4

1,273K / 293K = Vmax^0.4/Vmin^0.4

4.3447 = (Vmax^0.4 / Vmin^0.4)

(4.3447)^2.5 = (Vmin^0.4 / Vmax^0.4)^2.5


Vmax/Vmin = 39.346


So you would need to increase pressure to 2,513 psi if you started from atmospheric pressure.

P.S Thank you for making my brain hurt.