Calculating fuel use?

[Arugela]

I'm trying to backwards calculate fuel use:

411.6 Dv/velocity to get to mach 1.2.

I use:

411.6/9.81/ISP example: 411.6/9.81/455=0.09221359680075276406 (l1 aerospike figures for isp?)

then I use a calculator to find the approprate multiplier. ln(1.x) to equal the above. (speedcrunch allows inputs while seeing the answer as you change inputs.)

ln(1.0392)=0.0384511863742527303

 

I then divide my starting weight by the 1.0392, or to simplify 1.04.

 

Say I have 2000 tons. That is then 2000/1.04=1924 I round this to 1920. 2000 - 1920 ln(1.1)=0.095. 2000/1.1=1818.181818. 2000-1818.18= 180 tons of fuel approximately.

 

Is this correct up to this point?

 

Secondly. If this is using a fuel+lox combo of hydrogen. If I used a fuel mix of 5:1 would I divide this by 6 to find 30 tons of hydrogen and 150 tons of lox? Or is it 180 tons of hydrogen plus 900 tons of lox?

 

Is any of this even remotely correct?

 

Edited January 31, 2022 by Arugela

[SOXBLOX]

Here's an equation for the amount of fuel burnt in a certain time of burn.

M_pb = (F * T_b) / (g₀ * I_sp)

Where Mpb is the mass of propellant burnt

g0 is 9.81 m/s^2

Tb is the time of burn.

There are other useful equations here on Atomic Rockets.

Edited January 30, 2022 by SOXBLOX

[Arugela]

I've seen that, but I don't have the information(and can't find it) to fill it in and use it.

I have no idea what the burn time is or other things.

Mpb=(F?*Tb?)/(9.81*3600or450)

That is all I have.

In the flight I'm looking at I need to get to mach 1.2. I have a plane that is 2000 or 2160 tons. It has 168 tons of hydrogen gas(ignore the amount for arguments sake.) on board and is using 4-8xsabre engines with 440,000 lbs of thrust each at sea level. I'm not sure how that fills in.

This is why I asked about how the tonnage works. If I end up with 180 tons of fuel over 411.6 delta v, but supply the air externally is it in effect only 30 tons with a fuel mix ratio of 5:1? I'm not sure how it works. Does the energy or fuel usage only count the hydrogen part of does it also involve the oxygen.

If I were to use an aerospike and store hydrogen and lox on board and used 180 tons of fuel. Is that 180 tons hydrogen or is that the mix of 30tons hydrogen and 150tons oxygen within that mass? I'm assuming the fuel mass includes both the fuel and the oxidizer.

 

Edited January 30, 2022 by Arugela

[K^2]

Always keep track of the units. Make sure they are consistent. The reason you multiply I_SP by g₀ is that in the west it's typical to use I_SP in the units of seconds. Then the number would be on the order of a few hundred. If you're seeing something like 350s, you need to multiply it by g₀ which is in units of m/s². In that case, the product I_SP * g₀ is in units of m/s, which is what you're looking for. Some references will give I_SP already in units of m/s. The numbers will then be on the order of a few thousand. You might see something like 3,200m/s. In that case, you do not need to multiply by g₀. So that should resolve some confusion over which number to use where.

You can verify that this works with the formula that SOXBLOX gave you. (F * T) / (I_SP * g₀) => (N * s) / (m/s). Substitute N = kg*m/s² and you get (kg*m/s)/(m/s) = kg. Which is the units you want for mass. It's always a good check to run.

This is also the analysis you should be running to see if you got the right formula. See what are the units of available information and whether they are useful. In your case, your input is final velocity. One relevant formula is rocket equation.

dV = ln(M_i/M_f) * g₀ * I_SP

Again, you can quickly verify that you end up with m/s on the left side. M_i/M_f is the mass ratio between the mass of your craft fueled and mass after the fuel burned off. So M_f - M_i is the quantity you seek. However, this assumes no aerodynamic losses. Your dV is actually a lot higher than 411m/s if you are traveling through atmosphere. How much higher? Well, that depends on how long it takes you to accelerate, and that ultimately depends on the thrust setting, in which case, you'd go back to the formula SOXBLOX provided.

If your target speed wasn't supersonic, there would be a reasonably simple way to get an estimate. You can use the quadratic drag model, estimate drag coefficient, which you can actually get from the glide ratio, decide what the TWR of the engines are, and determine how long and under what thrust you can get to the target speed. Then you have the F and T and can figure out the rest. But This isn't going to cut it for supersonic flight. Drag changes drastically between subsonic and supersonic speeds, and the transonic region is particularly bad, because you pick up wave drag. Estimating the fuel use on getting to a Mach 1.2 is extremely complex procedure that will depend on the types of engines used, whether they are capable of afterburn or similar boost, configuration of the aircraft, and a number of other parameters that are way, way beyond information you have available.

[Arugela]

I've been reversing the rocket equation with a calculator that let me input the formula and see the result as I change values(linux's "speedcrunch" calculator.). I derive the number for Mi/Mj then divide the starting mass by that value to get the fuel usage. I wasn't sure if that was a viable way to do it.

I started with mach 1.2 = 411.6 velocity/Dv.

I take the velocity target and divide by the known ISP and g0.

411.6/9.81/450=0.0932381923207611281

I then put in ln(x) into the calculator and change the value manually until my calculator gets the same value as the above calculation:

ln(1.1)=0.09531017980432486004

I then divide the starting mass by 1.1.

2000/1.1 = 1818.18181818181818181818

2000-1818.18181818181818181818 = 181.81818181818181818182

I round this to 180 tons of fuel.

That is where my questions start. Is this realistic? And two does this mass represent the full mass of fuel. If I use a simplified 5:1 oxidizer to fuel is this the correct mass?

180/6 = 30 tons.

30x5=150 tons.

So, did I correctly derive the fuel usage? At least in the simplified sense.

30 tons hydrogen

150 tons oxidizer? Lox/Lair or whatever it is in that mix ratio.

I assumed aerodynamic losses had to be added. I was actually hoping they could be nullified by an air frame. Although I did use 9400 delta v overall instead of 7600-7800 delta v to try to compensate. Although I'm assuming that is for rockets and not spaceplane I'm trying to figure out. So, I'm assuming it might be more unless it's very efficient. I haven't gotten to drag and aerodynamics yet fully. That sort of thing is my next step if I can figure any of it out.

Interestingly a ration of 6:1 from a starting point of 2160 tons gets you:

2160/1.1=196.x

196/7=28

28*6=168.

So, if you are providing the oxidizer this requires only 28 tons of hydrogen. Assuming this is correct.

Edited January 31, 2022 by Arugela

[mikegarrison]

Yeah, it's going to be a log function. It's a result of having to carry fuel to burn to carry fuel. Airplanes end up with the same issues.

I didn't check your math, but if you are using logs and the numbers seem about right, it's probably OK.

[Arugela]

I needed delta V of 514.5(411.6(mach1.2x1.25) for aerodynamic inefficiencies.) I found that 630 second was about right.

I ended up using lbs since I'm converting to lbs and tons:

F = 8xsabre engines 440,000x8 = 3,520,000 lbs of thrust

Tb = 630 seconds

ISP= 3600

Mbp = (3,520,000x630)/(9.81x3600) = 62793.06829765545361875637 lbs of fuel. /2000 = 31.39 tons of fuel.

2160-32 = 2128  (<-tons)

2160/2128 = 1.015038

ln(1.015038)*9.81*3600 = 527.11426305211844473706 Dv.

Now my question is. Is that 32 tons all hydrogen or is it, at a mix of 8:1, 28 tons lox/air and 4 tons hydrogen?!

Or can you not use lbs of thrust and get the correct result. I was using lbs of thrust and lbs as a base unit. 2000 tons per conversion is straight forward.

That seems like way too little overall unless the oxidizer has to be stacked on stop of it. Although it's not too bad if you consider pure hydrogen from previous calculations.

 

 

Edited January 31, 2022 by Arugela

[K^2]

12 hours ago, Arugela said:

I then put in ln(x) into the calculator and change the value manually until my calculator gets the same value as the above calculation:

ln(1.1)=0.09531017980432486004

This works, but FYI, the inverse operation of natural logarithm is exponentiation, so you can do exp(0.093) = 1.097

Also, for small values of x, exp(x) is approximately 1 + x. As you can see, it wasn't that far off. As x gets larger, this approximation gets worse, though, so if you have access to calculator, better to just run the number. Still, a neat trick if you ever have to do this math on a napkin somewhere.

12 hours ago, Arugela said:

I round this to 180 tons of fuel.

That is where my questions start. Is this realistic? And two does this mass represent the full mass of fuel. If I use a simplified 5:1 oxidizer to fuel is this the correct mass?

180/6 = 30 tons.

30x5=150 tons.

Keeping in mind that we're still dealing with no drag, yeah, that seems reasonable. The 5:1 split is for kerlox. If you're running LH2/LOX, the ratio is actually 8:1, but that just changes the split, not the total.

12 hours ago, Arugela said:

I assumed aerodynamic losses had to be added. I was actually hoping they could be nullified by an air frame.

For a space plane doing horizontal takeoff, it's probably going to be significant. At the very least, lifting surface has to generate drag to generate lift. If we assume good aerodynamics, it could be as low as 10% of the weight - this corresponds to a glide ratio of 10. This value can be way lower for performance gliders, but it's also higher for something like a Space Shuttle. I doubt it will ever be better than 10% for any space plane, so it's a good starting point.

So for a very rough estimate, ignoring transonic and supersonic effects, the changes in weight as the aircraft flies, etc, you can just take 10% of the plane's weight and use it as the thrust your engines have to generate and plug it into the formula SOXBLOX provided.

M_fuel = (0.1 * g₀ * M_craft * T) / (I_SP * g₀) = 0.1 * M_craft * T / I_SP

So we still need to know time T it takes to get to the target speed. To get a good estimate for that, you need to know the TWR, climb profile, and how much parasitic drag you get on top of lift drag. We can often discount parasitic drag for a good airframe until you hit about Mach 0.8 or so because of wave drag. Since you go above that, we really ought to be taking it into account, but lets pretend it's not there to start with. In that case, we keep drag at constant 10% of the aircraft weight, leaving you with TWR of 1.9 for acceleration. We'll also assume that you aren't climbing at all, maintaining constant altitude.

T = 411.6m/s / (F_net/M) = 411.6m/s / (g₀ * 1.9) = 411.6m/s / (9.8m/s² * 1.9) = 22.1s

So now we can plug that back into the equation above.

M_fuel = 0.1 * 2000kg * 22.1s / 450s = 9.8kg

This isn't a lot, but keep in mind that this involves absolutely zero climb. If we throw in a generous climb angle, say 30°, then the available TWR will go down by half of the weight, so the equation for T will have 1.4 instead of 1.9, increasing time to 30s. You will climb to about 3km in these 30s and the fuel penalty goes up to about 13kg.

Finally, there isn't a great way with dealing with parasitic drag without knowing exact configuration, but typically, jet fighters that aren't capable of TWR of at least 1 can't punch through the sound barrier without afterburners. So we can get a rough estimate of what impact this will have if we simply drop TWR by 1 at about Mach 0.8. Again, lets keep the climb at 30°. We split the computation into two parts. Before Mach 0.8 and after.

T₁ = 299m/s / (g₀ * 1.4) = 21.8s

T₂ = 112m/s / (g₀ * 0.4) = 28.6s

And we have to add that 1 * weight to drag. So the mass also has to be computed in two parts.

M_fuel = 0.1 * 2000kg * 21.8s / 450s + 1.1 * 2000kg * 28.6s / 450s = 150kg

This is starting to be more believable. Also, due to a longer track at 30°, you will climb to a bit over 5km with this profile.

As you can see, transonic region is the part that makes this whole thing so much more costly. Unfortunately, it's also where the estimate is the roughest. I can easily see this number being wrong by half in either direction. You can alleviate a lot of that penalty by going to an even higher TWR. If your engines can handle something like TWR of 3, this fuel cost can be cut significantly. That will also reduce both the relative and absolute errors in this estimate, so you should be able to get a more reliable number for higher TWR. But basically, without doing ALL of the engineering on the craft, this is about as detailed as you're going to get.

[Arugela]

Is ISP always neutral and in seconds. Or is it fundamentally m/s and needs to be converted if I went with a full non metric system calculation?

I'm working with things like 460 ISP in space and low end sabre engines haveing ISP of 3600 at best. Not sure if that is in metrics or is neutral.

Here is a calculation sheet showing it:

[mikegarrison]

5 hours ago, Arugela said:

Is ISP always neutral and in seconds. Or is it fundamentally m/s and needs to be converted if I went with a full non metric system calculation?

I'm working with things like 460 ISP in space and low end sabre engines haveing ISP of 3600 at best. Not sure if that is in metrics or is neutral.

Here is a calculation sheet showing it:

I_SP is generally quoted in seconds. When it is given in seconds, there are no other units to convert, and it doesn't matter if you are using SI or English Customary or whatever.

Edited February 3, 2022 by mikegarrison

[K^2]

17 hours ago, Arugela said:

Is ISP always neutral and in seconds. Or is it fundamentally m/s and needs to be converted if I went with a full non metric system calculation?

I'm working with things like 460 ISP in space and low end sabre engines haveing ISP of 3600 at best. Not sure if that is in metrics or is neutral.

You need to look up what the units are. If the source you are using isn't quoting units, look for another source.

I_SP stands for specific impulse. Impulse is the change in momentum and has units of N*s or alternatively kg * m/s. (These are the same in metric, because N = kg * m/s²) And the word "specific" means the amount you get per some quality of something. More precisely, in this case, it is the amount of impulse you get from a quantity of propellant. And this is where the two definitions diverge in how you define quantity of propellant. You can define impulse per mass of propellant used, and then if you do the math, the units are m/s. Or you can define it per weight of propellant, and then you get units of seconds. Of course, weight of propellant is just mass of propellant times g₀ and that's why the two definitions are different by this exact factor.

So again, check your units.

 

For a historical note, defining I_SP in m/s is more correct. In fact, that's what you supposed to be working with in metric, and no g₀ should be showing up in a rocket equation. Unfortunately, a lot of Western literature on rocketry has been dominated by United States early on, and as we all know, American aerospace engineers worked in imperial units. What's worse, while imperial system has a unit of mass, which is called slug, engineers never used it. Instead, engineers would use pounds. Which is, of course, a unit of weight and not a unit of mass. At the same time, pound would be used as a unit of force, and so you have impulse in lb * s, the fuel quantity also in lb, and the specific impulse in lb *s / lb = seconds... And then, of course, to fix this in the rocket formula, you have to introduce the g₀ factor to everyone's annoyance.

You will never find such silliness in Soviet papers on rocketry. Unfortunately, because rocketry in USSR was primarily being developed during the cold war, a lot of the materials were being immediately classified, and very little of it was getting translated into English and other languages until much, much later. Much of US research was also getting classified, but still, significantly more of it was published in the West than there was of translated Soviet works, and so the American style of using seconds for specific impulse stuck.

But anyways, that's why there are two systems of measurements and why you should always watch your units.

[Arugela]

I think the subject I'm looking for is properly called mechanical engineering. This seems to have better organized data.

http://ronney.usc.edu/AME101/AME101-LectureNotes.pdf

Edit: Nvm, all education is dead. Nobody is making resources for people to study fully independently.

We need a takeover of this game by engineers to make it work correctly. 8)

I'm going to learn this moles and slugs thing. It's exactly what I have to put up with in my back yard. Should make it easier.  I'm assuming you control moles via location and density of slugs in an area to control their movement? If you get rid of the slugs the moles vanish or die?

Edited February 7, 2022 by Arugela

[Arugela]

On 2/4/2022 at 5:34 AM, K^2 said:

You need to look up what the units are. If the source you are using isn't quoting units, look for another source.

I_SP stands for specific impulse. Impulse is the change in momentum and has units of N*s or alternatively kg * m/s. (These are the same in metric, because N = kg * m/s²) And the word "specific" means the amount you get per some quality of something. More precisely, in this case, it is the amount of impulse you get from a quantity of propellant. And this is where the two definitions diverge in how you define quantity of propellant. You can define impulse per mass of propellant used, and then if you do the math, the units are m/s. Or you can define it per weight of propellant, and then you get units of seconds. Of course, weight of propellant is just mass of propellant times g₀ and that's why the two definitions are different by this exact factor.

So again, check your units.

 

For a historical note, defining I_SP in m/s is more correct. In fact, that's what you supposed to be working with in metric, and no g₀ should be showing up in a rocket equation. Unfortunately, a lot of Western literature on rocketry has been dominated by United States early on, and as we all know, American aerospace engineers worked in imperial units. What's worse, while imperial system has a unit of mass, which is called slug, engineers never used it. Instead, engineers would use pounds. Which is, of course, a unit of weight and not a unit of mass. At the same time, pound would be used as a unit of force, and so you have impulse in lb * s, the fuel quantity also in lb, and the specific impulse in lb *s / lb = seconds... And then, of course, to fix this in the rocket formula, you have to introduce the g₀ factor to everyone's annoyance.

You will never find such silliness in Soviet papers on rocketry. Unfortunately, because rocketry in USSR was primarily being developed during the cold war, a lot of the materials were being immediately classified, and very little of it was getting translated into English and other languages until much, much later. Much of US research was also getting classified, but still, significantly more of it was published in the West than there was of translated Soviet works, and so the American style of using seconds for specific impulse stuck.

But anyways, that's why there are two systems of measurements and why you should always watch your units.

Is there some specific place I messed my units up? I'm not sure what you are referring too specifically.

Edited February 14, 2022 by Arugela

[K^2]

7 hours ago, Arugela said:

Is there some specific place I messed my units up? I'm not sure what you are referring too specifically.

There are a few places where you're asking which values for I_SP to use, and some of these numbers are clearly meant to be in m/s and some in s. The easiest way to check which ones to use is plug it into formula and see if the units work out. If your dV didn't end up in m/s in the end, the input units were wrong, so you know to use the other convention. And the conversion factor is always 9.8m/s².

I strongly recommend keeping these in any computations as well. First of all, it's a simple way to check the results, and second, if any conversions have to be taken care of, they are plainly available.

[KerboNerd]

It looks like you're on the right track with your fuel use calculations! Your method for determining the change in velocity (Delta-v) and fuel consumption using specific impulse (ISP) is a standard approach in rocketry. The logarithmic part where you're using the natural log (ln) to calculate the fuel mass fraction is also correct.

As for your hydrogen and LOX (liquid oxygen) calculations, if you're using a 5:1 LOX-to-hydrogen mass ratio, you would divide the total mass of fuel (180 tons in your example) by 6. This gives you 30 tons of hydrogen and 150 tons of LOX. Your understanding of the fuel mix proportion is spot on. Just remember, these are quite complex calculations and can vary based on specific engine designs and mission profiles.