# Which portion of combustion energy is used to spin the compressor of a pure jet engine?

In level flight at constant speed, it is correct to assume that the quantity of energy created by fuel combustion (FE) is exactly extracted by the turbine section (ME) and is used solely to spin the compressor section?

In other words, the propulsive energy (E4) is equal to energy in air the aircraft encounters (E1).

If this assumption is incorrect, what is the correct relationship between energy quantities E1 to E4? Maybe two extremes could be used for illustration: No thrust (idle) and full thrust.

• If E1 would be containing E4, you would have no thrust: thrust means that you are putting energy into the airstream. you want as much as possible of the fuel energy to go in (useful) energy of the airstream. useful energy is kinetic, useless energy is thermal. (sorry, going to bed, someone else will make an answer out of this) Nov 24, 2016 at 21:11
• @Federico the compressor also produces thrust in a turbofan.. Nov 24, 2016 at 21:17
• @mins, much of the thrust still acts on the compressor in pure turbojet. Fluid applies force due to pressure always perpendicular to the surface, so thrust can only be applied to aft-facing surface and those are only the compressor blades and the diffusor. Nov 24, 2016 at 21:35
• @mins, $ME$ is actually rather unrelated to anything else, because that energy just runs in a loop inside the engine. $E4 = E1$ would be true if the engine produced no net thrust, but all engines actually produce some even at idle. Nov 24, 2016 at 21:47
• Why do you think Mechanical Energy ME is needed to spin the compressor? The diagram suggest that you think E1 + ME = E2, E2+FE = E3 and E3-ME = E4. That's a reasonable approximation, but it then follows immediately that E4 = E1 + FE. In a slightly better approximation, there's friction and some of the ME is needed to overcome friction losses in the compressor. Nov 25, 2016 at 12:44

The accelerated air coming out has more energy than the air ahead of the engine, even at idle-power.

### Force and mass flow rate:

In constant speed level flight, the forward thrust acting on the jet engine(s) equals all the drag forces acting rearward on the airplane.

The thrust acting forward equals the force the mass flow rate exerts rearward. Neither is the cause (action) nor the effect (reaction), since it's a two-body system (engine and air).

The forces are perfectly simultaneous, and are there for the same reason.

### The energy bit:

Energy (work) equals force times displacement.

A non-accelerating plane in mid-air, or a jet engine running on a fixed test stand, still has moving parts and a moving [steady] exhaust: displacement is happening.

### Since we can't create energy and it has to be conserved:

Let's say the total energy leaving the combustor (heated accelerated dense air) is 10 pirate-ninjas, then say 5 of which drive the turbine and all what's connected to it (including the air at the compressor). The other 5 leave the engine as an accelerated jet of air.

Say the compressor transfers 3 to the air, and all parts use 2 to rotate. The air comes with its 1 pirate-ninja of stored energy. The 3 gained by the air, now 4, meet the fuel in the combustor and release 10 (perfect lossless combustion, sorry). So, the energy stored in the fuel before combustion is 6.

The turbine-shaft-compressor is one body, the turbine absorbs 5, and transfers 3 of those to the air via the compressor, i.e., the engine parts absorb 2 to rotate, this is what you referred to as mechanical energy.

In other words—

The engine added 4 pirate-ninjas to the air from start to finish, giving the air added kinetic energy (mass flow rate), which translates to forward force as explained early on. In doing so, the engine spent 6 fuel, 2 of which to purely rotate the engine shaft, the other 4 went to the Δ mass flow rate.

The ratios above are not real, only illustrative. If you add more fuel, more energy will be released (excess oxygen is still available, that's how afterburners work), this energy will spin the engine faster, bringing in more air for a better efficient cool combustion.

The more efficiently the fuel energy is extracted, the more thrust we get. The properties of the air will have an impact; how dense is it, is the engine running on the ground, or is the plane transferring some of its kinetic energy into steady air resulting in ram-air energy, and so on.

So the ratios/percentages are not fixed, they depend on the operating conditions as well as the engine design (e.g., bypass ratio, compression ratio, etc.); even for the same type, different variants have different number of stages.

• @mins, well, the diagram here does take the losses into account, so E4=E1+FE-LO (LO=2). Nov 26, 2016 at 13:44