I recall hearing somewhere that the Space Shuttle's power was similar to that of the US electrical grid at some point in the past - but that might have been a myth.

If anyone thinks that they know how to do the calculation for Superheavy's power output (with reasonable accuracy) I'd be interested in learning the answer.

For answer-to-answer consistency, let's assume the theoretical maximum ISP for a sea-level MethylOx engine is 299 s, which I found through a link on the Specific Impulse Wiki Page in this article on ROCKET PROPELLANTS.

ROCKET PROPELLANT PERFORMANCE enter image description here

Let's also assume that the Raptors are 98% efficient (this is a SWAG).

In case this is helpful, I also fit a curve to Starship's orbital flight test 2 data (lift-off to MECO):

Speed(kph) = 6.2e-4t^3 + 1.55e-2t^2 + 1.75e1*t (0<=t<=160)

(Note: I am not affiliated with any of the links in this post.)

Update: I have subsequently learned that the suggested "98% efficient" assumption is in the original question was way off. There's combustion efficiency and nozzle efficiency. Combustion efficiency might be 98%. Nozzle efficiency needs to be considered too. Not all of the combustion energy is converted into forward kinetic energy. Some of it is lost because it ends up being used to spin the exhaust molecules, jiggle the intermolecular bonds of the exhaust molecules, heat up the nozzles, pre-heat the propellants, pump propellants through the engine, and cause molecules of exhaust gas to travel sideways rather than straight aft. An answer below suggests that the overall efficiency of a rocket engine is more like ~70%.


2 Answers 2


Let's focus on the energy expended by burning the fuel.

1 mol(16g) of methane has a energy output of 810kJ when burnt in excess oxygen. One watt is one Joule per second. 1kg methane therefore releases 50,625kJ or 50.625MJ of energy. Therefore, burning 134 kg/s results in 6783 MJ per second or 6.78 GW of energy output! For one engine. So this times 33 (the currently planned final engine count for Booster) results in 223.9 GW of energy output! That's quite a bit.

But how does it stack up to the US Energy Grid? From the Energy Information Administration

At the end of 2022, the United States had 1,160,169 MW(1160GW)—or about 1.16 billion kW—of total utility-scale electricity-generation capacity and about 39,486 MW—or nearly 0.04 billion kW—of small-scale solar photovoltaic electricity-generation capacity.

So the 33 Raptor Engines combined produce about a fifth of the wattage of the US National Grid. Pretty impressive if you ask me!

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    $\begingroup$ That would be the theoretical maximum based on the assumption that the Raptors are 100% efficient in converting the combustion energy of methane into power. In reality a huge amount of the energy goes to waste heat. And there are efficiency losses in the nozzle. And there is also unburnt methane that makes it into the exhaust. What is really needed is to covert the Raptor thrust into watts, but since force and power are not the same thing that's not a straightforward calculation. $\endgroup$ Dec 20, 2023 at 12:47
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    $\begingroup$ Just to be clear, unburned methane doesn't "make it into" the exhaust; the rocket intentionally runs fuel-rich (as do most rockets) for a variety of technical reasons. I don't think we can look at the methane usage and derive power output by just assuming it all burns. We know it doesn't. $\endgroup$ Dec 20, 2023 at 14:50
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    $\begingroup$ @DarthPseudonym - thanks for adding, I was going to mention the fuel rich ratio but decided not to for brevity since that's not the only reason not all of the fuel burns, even at stoichiometric not all of the fuel combusts because combustion chambers aren't perfect. But yes I would think the fuel rich mixture is the biggest contributor to unburnt fuel. $\endgroup$ Dec 20, 2023 at 16:02
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    $\begingroup$ It does depend on what you compare to what. Rocket brute output or effective thrust v total installed electrical capacity or actual operational output at any time. It more like a Germany than a USA... $\endgroup$
    – Slarty
    Dec 21, 2023 at 14:23
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    $\begingroup$ @uhoh - possibly complicating the calculation is that early in the launch some of the unburnt fuel combusts as it comes in contact with atmospheric oxygen. But as the rocket climbs this effect decreases, which is why oftentimes the exhaust can be seen turning darker as the rocket gets into the upper atmosphere. After it reaches near vacuum any fuel that doesn't combust in the presence of onboard oxidizer probably won't combust, although having been heated it probably will still radiate some energy but not all of its potential energy. $\endgroup$ 4 hours ago

The power output of the 33 Raptors at launch is nearly 60% of the electrical power of the entire US energy grid

US electrical power

The US total electrical energy generation by source is 4,178 billion kWh. A kWh is 3600 kJ since there are 3600 seconds in an hour. Multiplying we get about 15e15 kJ (15 EJ, Exa-Joules) of electrical energy in year. A year has 365x24x3600 = 35.54 million seconds. Dividing 15e15 kJ/35.54e6 seconds ~ 477e6 kW or 477 GW. That's of electrical power.

Note that the efficiency of thermal power plants is ~33% so the power released (the "heat rate") in burning the fuels is 3 times as much.

SuperHeavy Power

The power of a Raptor engine is found by Power = g x Isp x F, where g = the gravitational constant (9.81 m/s^2), F = the thrust of the Raptor engine (2,400 kN), and Isp is the specific impulse of a Raptor engine at sea level (350 s).

g*Isp is the effective exhaust velocity, which is ~3,400 m/s.

The physics is,

Power = Force * Velocity = 2,400 kN * 3,400 m/s = 8,160,000 kW = 8.16 GW for each Raptor engine.

For 33 Raptor engines that's ~ 270 GW.

Again the thermal energy (heat rate) of the rocket fuels will be greater. Rocket engines are typically 70% efficient.

SuperHeavy vs US Electrical Power

So 270 GW/477 GW ~ 56% i.e. more than half the US electricity output.

Note: Lots of technical detail below

See: FINAL NOTE (below) for final update

After comments from @Erin Anne & @phil1008 (below)

Something is amiss. To check @hi-bye125 's calculations I used the fuel flow rates on the side-bar (CH4 140 kg/s, O2 510 kg/s) and the reaction stoichiometry and enthalpy i.e -890 kJ/mol (exothermic) CH4 for the reaction: CH4 + 2O2 -> CO2 + 2H2O

For CH4 the molar flow rate is 140,000/16 = 8750 mol/s For 2O2 the molar flow rate is 510,000/(2*32) = 7970 mol/s That's ~ 10% fuel rich. Fuel-rich is normal in liquid-fueled rocket engines. This gives a 3.65:1 oxidiser to fuel ratio. The text gives a 3.8:1 ratio, while 4:1 is stoichiometric.

Taking the smaller 7970 mol/s * 890 kJ/mol = 7.09 GW/engine, or 234 GW for 33 Raptor engines. Notice this is close to @hi-bye125 's 223.9 GW, but using a different method.

My result is 270 GW/234 GW = 115% efficient, which is impossible.

AFAIK the Space Shuttle Main Engines (RS-25) were the most efficient engines ever made. I can't find the numbers, but I remember they were about 75% efficient. I'd be surprised if Raptors reached 70%.

Some solutions:

1. Sidebar note 12 says fuel flows are calculated at 2.23 MN thrust & Isp = 350 s. Power = Force x Velocity = Thrust x Isp x g = 2,230 x 350 x 9.81 = 7,660 MW (7.66 MW) Or ~ 253 GE total That's still 253 GW/234 GW = 108% efficient

2. The Raptor Wiki page statistics are from Elon tweets! Notably here and here. These are hardly technical specifications released by the manufacturer.

3 It's possible some numbers, such as thrust &/or Isp's - are from test conditions, while others - such as fuel flows - are from routine operations.

4 We're trusting links to (Elon's) tweets, unreferenced notes on Wikipedia pages, and references that link to people's web pages. These aren't manufacturers data sheets or technical data in the public domain - such as the RS-25 Shuttle engines.

FINAL NOTE I found this presentation from SpaceX.

See page 23 LHS for sea level calculations.

Isp = 334 s ~ 3,275 m/s

Thrust = 3,050kN

(But the same document on page 20, says Raptor 3 thrust is 590,000 lb at sea level or ~2,630 kN!)

Power = 3,275 * 3,050 = 9,988,750 kW ~ 10 GW

For 33 Raptor engines that's 330 GW

Fuel flows - this is where things are different from the Wikipedia page

LOX flow rate = 737.2 kg/s

LCH4 flow rate = 194 kg/s

Oxidiser/Fuel ration = 3.8 (which is still fuel rich)

Based on full O2 burn, 737,200/64 = 11,520 moles CH4 complete combustion.

Enthalpy = 11,520 mol/s * 890 kJ/mol = 10,250,000 kW = 10.25 GW

That's still a claim of 9,988,750/10,250,000 = 97% efficiency! (And more in a vacuum)

This is a better result, but I'm not completely happy.

Using Elon's own numbers of ~2,250 kN and 350 sec

Power = 7,875,000 kW (7.875 GW) per Raptor engine at sea level and the fuel flows from the SpaceX presentation

Efficiency = 7.875/10.25 = 77%, which is exceptional if true.

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    $\begingroup$ I think it's important to note that this answer is about the average power generated and the previous upvoted answer is about the grid generation capacity $\endgroup$
    – Erin Anne
    Apr 12 at 19:17
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    $\begingroup$ Great answer. I'm not convinced that the ISP of the Raptor from Elon's Tweet is accurate since it appears to exceed the theoretical maximum of 299 s (see table in answer), so your estimate may be a little bit high, but other than that, the math looks good to me. $\endgroup$
    – phil1008
    Apr 12 at 21:17
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    $\begingroup$ But, there appears to be something wrong with one or both of the answers. The first answer says that the energy of combustion should be 223.9 GW. The second answer says that the energy measured in terms of work done is 270 GW. This implies that the engines are 270/223.9 = 121% efficient. Have I understood this correctly? If yes, then I think that both answers might need to be reviewed and the calculations tweaked so that they arrive at values that are more reasonable. Anyone with expertise in liquid fuel's rocket engines ... please help! $\endgroup$
    – phil1008
    Apr 12 at 21:45
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    $\begingroup$ @phil1008 the table you refer to uses a reference combustion chamber pressure of 1000psi; Raptor's chamber pressure is reputed to be 4400 psi for Raptor 1 and 5100 psi for Raptor 2 (see sidebar) $\endgroup$
    – Erin Anne
    Apr 12 at 23:43
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    $\begingroup$ @ErinAnne Good point! I did some math (from this answer) and worked out that the increased chamber pressures should increase the theoretical max ISP from the table's value of 299 s to 328.3 s (for 4400 psi) and 330.7 s (for 5100 psi). $\endgroup$
    – phil1008
    Apr 13 at 0:59

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