Let's consider two identical jet engines operating at the same point. The first is attached on a plane's wing which is moving and the second is attached on the ground and remains still in order to be tested. Given that they both consume the same amount of energy (fuel) the total energy E transfered into the system is same in each case. In the first case, the energy E is converted into kinetic energy of the gas exiting the engine (Kgas) and kinetic energy of the plane (Kplane). In the second case, while the engine is attached, almost all E of the fuel is converted into kinetic energy of the gas. This means that the gas' velocity exiting the engine in this case is greater than in the case of the plane moving. If this is true, even though the two engines are at the same operation point, they have different gas exit velocities and therefore they generate different amounts of thrust. This is strange because in calculating gas velocities through an engine we don't consider factors like the engine's speed but only thermodynamics.
-
$\begingroup$ It's not the same operating point. And how is the speed of the air going into the engine not important? Since when does thermodynamics allow you to neglect kinetic energy (of a part) of the working fluid? $\endgroup$– jjackCommented Dec 8, 2017 at 22:10
-
$\begingroup$ If the engine can't move forward, but it "wants to", doesn't that create (more) (internal) heat? Like internal friction maybe? $\endgroup$– ŘídícíCommented Dec 8, 2017 at 22:18
-
$\begingroup$ @Keepthesemind You can't compare the two operating conditions. In one, the engine moves at a speed relative to the air. In the other, it doesn't. $\endgroup$– jjackCommented Dec 8, 2017 at 22:50
-
$\begingroup$ @jjack Of course you can. In scenario 1 the plane is starting to take off in windless conditions, and in scenario 2 the engine is fixed to the ground but there is maybe a small head-wind. Easy. $\endgroup$– ŘídícíCommented Dec 8, 2017 at 22:54
-
$\begingroup$ @Keepthesemind The question states that there is motion in one case and no motion in the other. But you're asking me to take the limit as one goes to the other. $\endgroup$– jjackCommented Dec 8, 2017 at 22:57
1 Answer
Consider a jet flying at a steady speed. Its kinetic energy is neither rising nor falling, so none of the fuel energy being burned is being converted into change in the airplane's kinetic energy. This jet engine could be compared to a jet engine on the ground mounted static, as long as the static jet engine is tested with a wind flow into the engine at the same rate as the flying jet has, a windspeed equal to the speed of the jet. These would be two equivalent situations and would have the same conversion of energy in the fuel to kinetic and thermal energy in the flowing air.
The real question is how does the thrust produced by the jet depend on the air speed of the wind blowing into the jet engine? The thrust measured on the ground would be the same as the thrust it takes to keep the airplane from moving at a constant speed, and not slowing down due to air friction, as it would do once the jet was turned off.
-
$\begingroup$ try posting this on the aviation stack exchange page. $\endgroup$ Commented Dec 9, 2017 at 4:55