Using this equationQuotepumpPower = Δpressure*mDot/(density*efficiency)If we ignore the inlet pressure to the pump and use 80% for the efficiency:methanePumpPower = (22,000,000Pa * 69kg/s) / (422.8kg/m³ * 80%) = 4488kWloxPumpPower = (22,000,000Pa * 247kg/s) / (1141kg/m³ * 80%) = 5953kWTotal power = 10,441kWand knowing that a single battery system can output about 400kW max, we can back calculate the flow rate for 22MPamDot = pumpPower*density*efficiency/ΔpressuremDot = 400kW * 1141kg/m³ * 60% / 22,000,000Pa = 12.4 kg/sec (LOX)Which would be 3.4 kg/sec of Methane (at slightly less power)F = mdot * v = 42.2 kg/sec * 2800m/sec = 118kN.So each full size Tesla Battery on board can pump methalox up enough to get 118kN of thrust. Assuming we are running 4+1 redundancy we have 236kN of thrust. At 1.5kW/kg that's 1.3t of batteries.for a 250t Starship, that's 1 m/sec of acceleration. Not enough for the Moon, let alone Mars.Plenty for maneuvering thrusters.
pumpPower = Δpressure*mDot/(density*efficiency)If we ignore the inlet pressure to the pump and use 80% for the efficiency:methanePumpPower = (22,000,000Pa * 69kg/s) / (422.8kg/m³ * 80%) = 4488kWloxPumpPower = (22,000,000Pa * 247kg/s) / (1141kg/m³ * 80%) = 5953kWTotal power = 10,441kW
Quote from: InterestedEngineer on 04/27/2025 02:34 amFirst, the 270Wh/kg is for the 100% to 0% range, nobody runs lithium batteries that way. They degrade quickly when you charge them past 80% and you want the low side to be 20% so there's some spare.Practically speaking the energy density of Lithium-ion is half of their rated spec (60% if you want to be picky).While what you say is true for an EV battery where you'd like to minimize the capacity loss over 10 years and 200,000 miles of driving, it doesn't really apply to spacecraft. Maintaining the battery at 50% between uses, charging to 100% when needed, and discharging down to 5% (leaving a small safety margin) would work fine for long-lived orbital craft.The bigger issue is that 270 Wh/kg is for bleeding-edge consumer products, not space rated, derated batteries. Taking that into account, I might buy into your 60% derating. But this IS SpaceX, and they might very well use lightly modified consumer batteries for non-human-spaceflight purposes simply because it's easier and cheaper. Using 95% of the battery (100->5%, leaving a small safety margin) would work fine for space use, especially if you left the batteries at 50% for any long-duration quiescent periods, and only fully charged when necessary.
First, the 270Wh/kg is for the 100% to 0% range, nobody runs lithium batteries that way. They degrade quickly when you charge them past 80% and you want the low side to be 20% so there's some spare.Practically speaking the energy density of Lithium-ion is half of their rated spec (60% if you want to be picky).
I'm taking a look at what you can actually get from a fully fueled Starship in final tanking orbit. It seems that you could either throw a 150-ton payload entirely out of the Solar System (with hyperbolic excess velocity of 467 m/s) or put it into an elliptical orbit with perihelion at just 0.2 AU. In other words, it could send twice the mass of Gateway anywhere in the Solar System. Not because you'd want to use Starship to do that, of course, but because it speaks to the power available from this approach. (If anyone wants to check my math, that would be great.)I'm using TheRadicalModerate's numbers for the following: Starship dry mass is 120 metric tons (mt), full load of propellants is 1208 mt, and payload is 150 mt. Depot dry mass is 88.5 mt and full load of propellant is 1588 mt.We start with a fully-fueled Starship and a fully-fueled depot in circular LEO at 281 km with velocity 7.7 kps. This accords with SpaceX's FAA filing. The total weight of the two, taken together, is 3154 mt. After the burn to the final fueling orbit (FFO), the total weight must be 1566.5 mt. That is, the dry masses and payload mass don't change, and--between the two vehicles--there's just enough propellant to fully fuel the Starship. This is a mass ratio very close to 0.5. Using the ISP of 3.728 kps from Wikipedia (for Raptor V3 vacuum), Δv = 3.728 * ln(2) or about 2.6 kps for a total perigee velocity of 10.3 kps for the FFO. This gives the FFO an apogee of 50,000 km (altitude), which is a bit higher than the 34,534 km in the FAA filing, but well within the (huge) tolerances they cited: (+116,000 to -24,000)! This orbit has a 15-hour period, so plenty of time for the depot to transfer the remaining propellant before the Starship reaches perigee and needs to do it's big burn.Fully fueled, and with that 150 mt payload, our Starship weighs 147.8 mt. After the final burn, mass it just 270 mt, for a mass ratio of 0.183. Δv = -3.728 * ln(0.183) = 6.34 kps. Perigee velocity is 16.7 kps, which has hyperbolic excess (with respect to Earth) of 12.6 kps.If we add that to Earth's velocity around the sun, that's 47 kps which is just over solar escape velocity. Using full precision calculations, I get 467 m/s hyperbolic excess.If we subtract, we get velocity of 25.1 kps, which points to an orbit with perihelion at 0.2 AU--well inside the orbit of Mercury.Again, that with 150 mt of payload plus the 120 mt dry mass of Starship itself. So, unless I've made an error, this is vastly more mass than anyone has ever sent outside of LEO--even to Mars.Obviously, Starship isn't even the right vehicle for this; you'd want something with very little dry mass compared to the payload. I just like the idea of saying that the refueling scheme makes it powerful enough to throw things entirely out of the Solar System! (Assuming that's really true, of course.)
cross checking with some reasonably well known Raptor numbersQuotepumpPower = Δpressure*mDot/(density*efficiency)If we ignore the inlet pressure to the pump and use 80% for the efficiency:methanePumpPower = (22,000,000Pa * 69kg/s) / (422.8kg/m³ * 80%) = 4488kWloxPumpPower = (22,000,000Pa * 247kg/s) / (1141kg/m³ * 80%) = 5953kWTotal power = 10,441kWΔpressure = 600 barmDot = 500kg/secdensity = 422.8kg/m³efficiency = 70%pumpPower (LOX) = 600 *100e3 * 500 / (423 * 0.7) = 100MWF = mdot * velocity = 750kg/sec * 3600m/sec = 2.7MN (VacuumRaptor)Yep, checks out.It's all linearly proportional. If you want 270kN you need 10MW. If you want 27kN you need 1MW, which will be the upper limit for electric pumps.
Quote from: InterestedEngineer on 04/28/2025 02:50 amcross checking with some reasonably well known Raptor numbersQuotepumpPower = Δpressure*mDot/(density*efficiency)If we ignore the inlet pressure to the pump and use 80% for the efficiency:methanePumpPower = (22,000,000Pa * 69kg/s) / (422.8kg/m³ * 80%) = 4488kWloxPumpPower = (22,000,000Pa * 247kg/s) / (1141kg/m³ * 80%) = 5953kWTotal power = 10,441kWΔpressure = 600 barmDot = 500kg/secdensity = 422.8kg/m³efficiency = 70%pumpPower (LOX) = 600 *100e3 * 500 / (423 * 0.7) = 100MWF = mdot * velocity = 750kg/sec * 3600m/sec = 2.7MN (VacuumRaptor)Yep, checks out.It's all linearly proportional. If you want 270kN you need 10MW. If you want 27kN you need 1MW, which will be the upper limit for electric pumps.All the energy numbers were based on a 15s thruster burn. The number was a bad guess. Here's a better (but less favorable) one:Let's assume we want the Raptors off some particular height above the surface, with the Starship descending velocity specified. Time to whip out the weird kinematic equation that doesn't depend on time:v² = u² + 2ahWe want v=0 at h=0, and the acceleration is going to be negative with respect to the velocity. So:u² = 2ah, anda = u²/(2h)Let's look at switching over to the thrusters 100m above the surface, with a 5m/s downward velocity. That gives a = 0.125m/s². (Note that the actual acceleration has to be gLocal + 0.125m/s², to counteract gravity.) But that means it takes 50s to do the last 100m, if you allocate 25% extra time as margin. You can reduce this, but only at the cost of increasing the thrust even further, which makes the pump power higher.I looked at supercapacitor solutions a bit, and they seem to be well below the kind of power we'd need, especially for the longer discharge time. So, at least for the landing thrusters, I'm ready to write off the electric pump idea. It may still be useful for ullage thrusters, however. That leaves us with either a real turbopump or a pressure-fed solution.
I think we might be talking about two different things.I'm talking about peak power output (joules/second), units nominally kilowatts.
Don't worry folks, this is still on topic... kind of. Electric pumped methalox thrusters for 2-10kN of constant thrust for settling fuel for refueling is entirely doable, and electric pumps make total sense for that application, as starting them and stopping them is far easier. Elon put so many batteries on Starship it's almost rounding error for the total power involved.
Quote from: InterestedEngineer on 04/28/2025 06:57 pmI think we might be talking about two different things.I'm talking about peak power output (joules/second), units nominally kilowatts.Understood. But if the power pulse is short enough, you could maybe play games with supercapacitors, which have better specific power. However, if the pulse width is 50 seconds, that's not possible.QuoteDon't worry folks, this is still on topic... kind of. Electric pumped methalox thrusters for 2-10kN of constant thrust for settling fuel for refueling is entirely doable, and electric pumps make total sense for that application, as starting them and stopping them is far easier. Elon put so many batteries on Starship it's almost rounding error for the total power involved.The case for electrically-pumped ullage thrusters is a lot less compelling if they can't share the same prop distribution system as everything else. The other alternative is a pressure-fed system, which consists of two kinds of tanks:1) LCH4 and LOX tanks, suitably sized which segregate and hold liquid prop at very high pressures.2) GCH4 and GOX tanks, which provide the pressurant for the liquids.This is a replenishable system (always a nice property for Starship), because you can always take the pressure off, pump liquid into the small tanks from the mains, and then seal them. Then the pressurant tanks are heated to supercriticality, and continue to be warmed to keep them there. This is especially nice for attitude control, which you need the thrusters are hot standby for long periods of time, but with very little actual duty cycle during that period.The downside is that it's an insanely complex system, requiring lots and lots of COPVs, and about a zillion valves.
Why not just have the small tanks that have liquid "pumped from the mains"?, and instead of pressure fed, electrically pumped?
The big circles are clearly the CH4 and LOX fill/drain. The medium-size circles are likely vent and low-point drain (?) lines. In the four outside corners look to be alignment and/or latching features. At the top center the two rectangular connections are probably power/data.Interesting how there's a "DMZ" separation break in the adapter plane between the CH4 and LOX sections, presumably for safety.
Quote from: Twark_Main on 05/08/2025 03:00 amThe big circles are clearly the CH4 and LOX fill/drain. The medium-size circles are likely vent and low-point drain (?) lines. In the four outside corners look to be alignment and/or latching features. At the top center the two rectangular connections are probably power/data.Interesting how there's a "DMZ" separation break in the adapter plane between the CH4 and LOX sections, presumably for safety.That seems right, but I don't see what it buys them. The two plates have vertical symmetry (well, sorta vertical symmetry, at least for the big lines), but they can't be made androgynous. And it's a terrible arrangement for the GSE.The photos show a two-ring segment. Maybe it's a truncated forward barrel, and the regular QD is still present in the tail skirt assembly? But if that's true, what's this going to connect to? And why aren't they trying to make the QD do double duty for both GSE and refueling? Is every client Starship going to have to have two QDs?
