Total Energy Efficiency of a Rocket Engine

[shynung]

Whenever I ask about rocket engine efficiency, I often found the answer in the form of specific impulse (I_SP), given in seconds, representing impulse per unit of propellant weight.

However, there is a problem. Some rockets, namely electrically-powered ones, have high I_SP values, but also have considerable power requirements, often necessitating heavy power generators, which could nullify mass loss from needing less propellant mass.

In short, I ask this: How to calculate rocket efficiency from energy usage perspective, including energy generated from sources other than the propellant?

Thanks in advance.

[K^2]

Energy efficiency is a bit different from what you are asking, but also a very interesting subject for rocketry. Some day, we should have a topic about it. What you are asking about is still ultimatley about mass efficiency, but including a lot of dead weight you have to carry around.

One way to sort of work around this issue is define effective I_SP. The whole point of specific impulse is impulse per mass of propellant. But it doesn't have to be just propellant. You can include any additional dead weight. So for example, normally you say that your rocket's mass started out at some m₀ and decreased to m₁. Rearranging the delta-V formula, we can get I_SP as the following.

I_SP = ÃŽâ€V / (g₀ ln(m₀/m₁))

What you can do is instead of taking m₁ to be the total mass of the rocket after it has exhausted the fuel to be the just the mass of the useful load. In other words, you'd subtract mass of any batteries, generators, solar pannels, and anything else you don't actually need to reach that delta-V. The resulting effective I_SP will be significantly lower than actual, because it reflects mass efficiency of your propulsion system.

Naturally, the above is going to depend very much on how much delta-V you actually plan to achieve. If your power plant has a fixed weight, the more delta-V you achieve, the less of an impact the dead weight makes. Which is one of the reasons why ion drives make more sense on long journeys than short ones. But this can apply to more conventional engines as well. For example, we are happy with low I_SP of the monoprop for the RCS, because we don't need much dleta-V out of it, and a full bi-prop engine is just too much dead weight to carry around. There are some other advantages there as well, but it's a big part of it.

P.S. Just for future reference if anyone's searching for energy efficiency. For conventional rocket following Tsiolkovsky Rocket Equation, the specific work done on propellant reducing mass of the rocket from m₀ to m₁ to achieve ÃŽâ€V is given by the following equation.

ÃŽâ€E = (m₀/m₁ - 1)ÃŽâ€V² / (2 ln(m₀/m₁))

This has units of energy/kg. The above achieves absolute minimum, regardless of ÃŽâ€V, at the following mass ratio.

m₀/m₁ = -2/W(2/ln(2))

Where W(z) is Lambert W Function, also known as product logarithm, defined as solution of z = W(x)Exp(W(z)). Numerically, this evaluates to approximately m₀/m₁ = 4.29155.

Given this optimal mass ratio, efficiency of conventional rocket, not counting thermal or other losses of the engine itself, is approximately 64.76%, which is way more than you might have thought. Especially, when you consider that base efficiency of an ion drive can be over 90%.

This does lead to an interesting conclusion. Since mass ratio here is fixed for any delta-V, the optimal I_SP depends on desired delta-V. (One can use the equation for I_SP mentioned above.) More specifically, exhaust velocity should exit at about 62.75% of delta-V, which gives I_SP of about 0.064s for every 1m/s of delta-V desired.

Of course, all of this assumes that the limiting factor is energy, rather than mass. However, with conventional rockets, amount of energy itself is limited by chemical energy in the available mass of the fuel, and that tends to fall short of the mark. For 9km/s of delta-V, which is what you'd typically want to enter LEO, energy-optimal I_SP would have been about 576s, with real chemical fuels making it as high as 450s. On the other hand, with conventional rocket, we really don't care about energy use, so much as we care about the mass. So if we had fuels with I_SP well over the 576s above, we'd probably use them anyways despite it being a bit wasteful because it would make for smaller, lighter rockets.

This theory does, however, directly apply to ion drives. Not anything that would use on-board battery, but anything that would use sollar or RTG power, certainly. Assuming a long duration mission, minimizing energy requirement for mission delta-V minimizes the size of the generator or solar array. Whether that's more important than reducing total payload is a separate question, since for energy-optimal operation the probe would have to cary more than four times it's own weight in propellant.

Edited February 17, 2014 by K^2

[shynung]

So, you're trying to tell me that, in order to get effective I_SP, I must assume all power-generating parts as propellant mass? The reverse of payload mass fraction?

I think I'll make an example, just to be clear.

I'll assume an ion-drive satellite in polar orbit around Kerbin, carrying a gravity meter and an antenna. The orbital plane is perpendicular to the Sun's direction, so it has sunlight coverage all around. It has a radial xenon tank, a Gigantor XL panel, and a PB-ION thruster (the bare minimum for it to work).

Stats straight out of the Wiki:

A QBE probe(80 kg) + GRAVMAX sensor(5 kg) + Comm16 antenna(5 kg) totals for 90 kg of payload mass.

A Gigantor XL panel(350 kg) + PB-X-50R tank(70 kg full, 30 kg dry) + PB-ION thruster(250 kg) totals 630 kg of engine+generator mass, and 40 kg of propellant mass.

Using standard Tsiolkovsky equation, I get 4200*9.81*ln(760/720) = 2227.67 m/s² of ÃŽâ€V.

However, to get this ÃŽâ€V, I still have to lug around 630 kg of parts, which I assume is completely useless once all propellant has been used (immovable sats are dead sats).

Using the above ÃŽâ€V and the modified equation, the effective I_SP of the entire propulsion system is 2227.67/(9.81*ln(760/90)) = 106.44 seconds, if I'm correct.

As a comparison, the same probe is now powered by a chemical rocket (LV-909), carrying fuel in an FL-T100 tank. It also carries the lightest solar panel (OX-STAT), just to keep it alive.

The LV-909(500 kg) + FL-T100(562.5 kg full, 62.5 kg dry) totals 562.5 kg of engine mass, and 500 kg of propellant mass. The OX-STAT add only a measly 5 kg to the payload mass.

Again, using the standard equation: ÃŽâ€V = 390*9.81*ln(1157.5/657.5) = 2163.83 m/s², losing to the ion drive by 63.84 m/s².

Plugging it to the modified equation yields: 2163.83/(9.81*ln(1157.5/95)) = 88.22 seconds of effective I_SP.

The difference of effective I_SP between the two systems are just 18.22 seconds.

That's kinda disappointing...

[K^2]

Keep in mind that this is a measure of performance, but not necessarily the only one to consider. Also, even a small difference in I_SP adds up over a long mission to considerable savings, thanks to that log function in the rocket equation.

Ultimately, of course, if you are sending an interplanetary probe, what you are trying to minimize is the weight to LEO/LKO that gets the mission done. But all else being the same, one with higher effective I_SP will be the ligher probe. This is why I recommend it as a measure of efficiency. As for how much it adds up, look at your own example. You have total of 760kg to LKO or 1,158kg to LKO to accomplish the same mission. The difference in specific impulse isn't large, but it's a big enough fraction of the total impulse your probe builds up, so it makes a considerable impact.

[Alchemist]

That is very strongly dependent on payload/dry mass relation and/or how much delta v you try to get out of this. If you have your propulsion stage dry mass so much larger than your payload you'll get terrible efficiency. In both cases you are approaching the delta v of the propulsion stage without payload which means terrible efficiency. Just compare to what happens if you double the payload - delta v will change minimally, and you will almost double the efficient ISP.

Maybe try plotting delta v or efficient ISP against payload mass for both propulsion systems

[shynung]

Thanks a lot for the explanation, I really appreciate it.

[cryogen]

P.S. Just for future reference if anyone's searching for energy efficiency. For conventional rocket following Tsiolkovsky Rocket Equation, the specific work done on propellant reducing mass of the rocket from m0 to m1 to achieve ÃŽâ€V is given by the following equation.

ÃŽâ€E = (m0/m1 - 1)ÃŽâ€V² / (2 ln(m0/m1))

This has units of energy/kg. The above achieves absolute minimum, regardless of ÃŽâ€V, at the following mass ratio.

m0/m1 = -2/W(2/ln(2))

Where W(z) is Lambert W Function, also known as product logarithm, defined as solution of z = W(x)Exp(W(z)). Numerically, this evaluates to approximately m0/m1 = 4.29155.

Given this optimal mass ratio, efficiency of conventional rocket, not counting thermal or other losses of the engine itself, is approximately 64.76%, which is way more than you might have thought. Especially, when you consider that base efficiency of an ion drive can be over 90%.

This does lead to an interesting conclusion. Since mass ratio here is fixed for any delta-V, the optimal ISP depends on desired delta-V. (One can use the equation for ISP mentioned above.) More specifically, exhaust velocity should exit at about 62.75% of delta-V, which gives ISP of about 0.064s for every 1m/s of delta-V desired.

This is absolutely correct. Should clarify that this is for a rocket with a constant I_sp: it's possible to get higher energy efficiencies by varying I_sp over time (for example, different fuels in each stage, or variable ion voltage in electric thrusters).

In the limiting case, you can approach 100% efficiency, if v_e(t) = v_r(t) -- if the exhaust speed is always equal to the rocket's speed, at a given time. Then, the exhaust has zero velocity in the initial reference frame -- its velocity backwards exactly cancels the rocket's forward speed when it was expelled, so it ends up stationary. All of the kinetic energy ends up in the rocket; none in the propellant. (This is only a limiting case though: the propellant mass diverges to infinity, as v_e(t=0) approaches zero).

Here's some maths. Imagine a rocket with a continuously varying exhaust velocity v_e(ÃŽâ€v) (instead of parameterizing as a function of time, we're parameterizing in the cumulative delta-v of the rocket). And let U(ÃŽâ€v) be the cumulative energy of the propellant, that is δU = (1/2) δm ve². (Ignoring thermodynamic losses here). The kinetic energy of the rocket (payload + structure) is K_R(ÃŽâ€v) = (1/2) m_R ÃŽâ€v². We're interested in the energy conversion efficiency, η = K_R / U. Of course η ≤ 1.

A differential increase in ÃŽâ€v, by δÎâ€v, increases the propellant energy by:

U(ÃŽâ€v + δÎâ€v) = (1/2) δm v_e(ÃŽâ€v)² + U(ÃŽâ€v) (1 + δm/m_R)

The left term is the kinetic energy of a differential mass of propellant δm, expelled with velocity ve². Since we've increased the mass of propellant, we need more propellant to lift that propellant (rocket equation logic). The mass accelerated to ÃŽâ€v increases from m_R to m_R + δm, so we have everything before that multiplicatively. That's the term on the right, multiplying U(ÃŽâ€v).

We can relate δm to δÎâ€v, according to conservation of momentum: δÎâ€v = δm * v_e.

U(ÃŽâ€v + δÎâ€v) = (1/2) δÎâ€v v_e(ÃŽâ€v) + U(ÃŽâ€v) (1 + δÎâ€v/(m_R v_e(ÃŽâ€v)))

d/dÃŽâ€v{ U(ÃŽâ€v) } = (1/2) v_e(ÃŽâ€v) + U(ÃŽâ€v) / (m_R v_e(ÃŽâ€v)) [eq. 1]

This gives us some intuition. Increasing Isp gives us two competing effects: it increases the marginal energy of the propellant (left term), but it also decreases the mass of that propellant, and thus the energy that goes into to accelerating it (right term). As ÃŽâ€v increases, the right term becomes more important (rocket equation!!), so the optimal Isp goes up. (You can solve in several ways, that the functionally optimal Isp function is just v_e(ÃŽâ€v) = ÃŽâ€v. Easiest way is just to observe that that solution is 100% efficient, η=1, and you can't do better than that according to conservation of energy).

For constant-I_sp solutions, the maximum is at about v_e(ÃŽâ€v) = 0.63*ÃŽâ€v. (This is what K^2's post focused on).

By the rocket equation, if v_e(ÃŽâ€v) = α ÃŽâ€v, the mass ratio is e^(ÃŽâ€v / ve) = e^(1/α), and the propellant fraction is ζ = (e^(1/α) - 1) / e^(1/α) . So

1/η = U/K_R = (e^(1/α) - 1) v_e² / ÃŽâ€v² = (e^(1/α) - 1) α²

(Alternatively you could get this by solving eq. 1 directly, with v_e a constant). The maximum occurs where dη/dα = 0 (or d/dα [1/η] = 0):

0 = d/dα { (e^(1/α) - 1) α² }

0 = 2 α (e^(1/α) - 1) - e^(1/α)

(2 α - 1) e^(1/α) = 2 α

e^(1/α) = (2 α) / (2 α - 1)

e^(s + 2) = -2 / s | α -> 1/(s + 2)

s e^s = -2 / e²

s = W(-2 / e²)

α = 1/(W(-2 / e²) + 2) ~= 0.6275

η = 1/ { (e^(1/α) - 1) α² } ~= 64.76 %

Edited May 31, 2015 by cryogen