Oberth effect and propulsive efficiency of rocket engines

[Arkalius]

So I've been reading a lot of things about the various aspects of the physics of spaceflight. I've recently stumbled on something regarding the propulsive efficiency in rocket engines in regards to energy utilization and the Oberth effect. I understand how the effect works. The thing that confuses me, though, is how propulsive efficiency is capped. The equation I've seen is:

With c as exhaust speed, and v is speed of the spacecraft.

With this equation, efficiency caps out (at 1.0) when v = c, and then starts decreasing, but this doesn't make sense to me.

In a static firing of the engine, where the craft doesn't move, all of the (non-wasted) energy from combustion is imparted on the exhaust shooting out the back.

When the craft is moving at half the exhaust velocity, all of the energy is given to the craft. As an example, say exhaust velocity is 100m/s. The craft is moving 50 m/s. The reaction mass changes from moving 50m/s with the craft to 50m/s away in the opposite direction, thus its kinetic energy hasn't changed, so all of the produced energy is with the craft. Why this turns out to be 0.8 all the time, I'm not sure... I'd think the value would depend on the specific chemical energy of the propellant being used, but anyway that's not as important.

When the craft is moving at the same speed as the exhaust velocity, not only is it getting all the energy of combustion, but it's also stealing all of the kinetic energy of the exhaust, which will have a velocity of 0. With the 100m/s example, that's an extra gain of 5kJ/kg.

However, it seems to me the efficiency should continue to climb as the craft gets faster. Say it is moving 200 m/s. The exhaust will be moving 100m/s in the same direction after leaving the craft. It's specific kinetic energy would go from 20 kJ/kg down to 5 kJ/kg, a difference of 15 kJ/kg, 3 times more than in the above situation. So why doesn't this translate into even greater propulsive efficiency for the spacecraft? According to that equation, the 2:1 velocity to exhaust velocity ratio has the same 0.8 efficiency as the 1:2 ratio (with 1:1 being 1.0 efficiency). Anyone able to explain this?

[Ralathon]

That equation describes how much energy of the propellant ends up in the vehicle. It does not take into account the Oberth effect because this equation is usually used for low velocity vehicles where Oberth doesn't play a major role (airliners etc).

Here's the math for the Oberth effect, which indeed tends towards infinity as velocity increases.

[Yasmy]

Incorrect, Ralathon. The Oberth effect is merely the consequence of power equals force times velocity, P = Fv.

Any mechanics where a force is applied independent of v automatically includes the Oberth effect.

In fact, eta_p = (F v) / (F v + 1/2 (F/c) (v^2-c^2)), which follows directly from the Oberth effect.

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This term is just the mechanical (kinetic) part.

Initially, the rocket kinetic energy, excluding the soon to be exhaust propellant is 1/2 m v^2

Initially, the rocket propellant (soon to be exhaust) kinetic energy is KE_p = 1/2 dm v^2

Thus the total initial KE is 1/2 (m+dm) v^2.

Finally, the rocket kinetic energy is KE_r = 1/2 m (v+dv)^2

Finally, exhaust kinetic energy is KE_e = 1/2 dm (v-c)^2.

By conservation of momentum, (m + dm) v = m (v+dv) + dm (v-c), so m dv = dm c.

eta_p = percentage of "usable" propellant kinetic energy gained by the rocket

eta_p = (KE gained by rocket) / (kinetic energy gained by rocket + kinetic energy remaining in fuel)

KE gained by rocket = 1/2 m (v+dv)^2 - 1/2 m v^2 ~= m dv v = dm c v.

eta_p = (dm c v) / (dm c v + 1/2 dm (v-c)^2) = (c v) / (c v + 1/2 v^2 - c v + 1/2 c^2) = (2v/c) / ((v/c)^2 + 1)

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Now let's go back to your last two examples:

Case 1: v = 100 m/s, c = 100 m/s. eta_p = 2/(1 + 1) = 1

Case 2: v = 200 m/s, c = 100 m/s. eta_p = 4/(1 + 4) = 0.8

While it is true that in case 2 the propellant loses more kinetic energy than in case 1, the rocket gains even more kinetic energy.

Compare the final propellant kinetic energy (1/2 dm (v-c)^2) to the final rocket kinetic energy gain (dm c v):

The ratio is (v-c)^2 / (2 c v)

Case 1: (100 - 100)^2 / (200) = 0

Case 2: (200 - 100)^2 / (400) = 1/4

In Case 1, the exhaust takes none of the mechanical energy of the propellant.

In Case 2, the exhaust takes one quarter as much of the mechanical energy of the propellant as the rocket gets, or 1/5 of the initial propellant KE.

eta_p is not the efficiency of kinetic energy gain (which is the Oberth effect: KE gained by rocket = dm c v, i.e., proportional to v.)

It is the efficiency of using up the propellant's kinetic energy.

(And eta = eta_p * eta_c is the efficiency of using up all of the propellant energy, where eta_c describes the efficiency of combustion and conversion into thrust.)

Finally, eta_p is capped because it is impossible for the propellant to lose more kinetic energy than it has. Kinetic energy can not go negative. eta = eta_p * eta_c is capped at 1 by energy conservation, and in practice capped to something less than 1 by thermodynamics.

Edited June 17, 2015 by Yasmy
Who can type that much without a few errors?

[Arkalius]

It's funny, after posting this, I went to bed, and then figured it out while lying there thinking about it. I was originally envisioning it as what fraction of the chemical potential energy in the propellant only. But now I realize it is a combination of that and the kinetic energy. Clearly when moving faster than the exhaust velocity, some of the kinetic energy will be left behind in the exhaust, with no way to recover it, thus the reduced efficiency. I also realized why the value is 0.8 at both 1/2 and 2x exhaust velocity. At 1/2 speed, the propellant has 4x more chemical potential energy than kinetic energy, and you can obtain all of the chemical energy on the craft, but none of the kinetic. At 2x speed, you can get 3/4 of the kinetic (which is 4x the chemical) energy and all of the chemical, so in both cases you get 80% of the total energy involved.

I didn't read through your equations in detail, but I'm assuming they more precisely explained the concepts I just summarized. I almost feel silly for posting a question the answer to which dawned on me merely minutes later, but at least others can learn from it