Calculating total fuel mass from fuel ratio and volume.

[Zatnikitelman]

I seem to be having a major brain fart tonight and am trying to figure out what my total fuel mass would be (for Orbiter) given an oxidizer:fuel ratio of 6:1 using LOX and LH2, in a 2.5m radius cylinder 6.15m in height. LOX and LH2 have densities of 1141 kg/m^3 and 70kg/m^3 respectively. For simplicity, I'm ignoring stuff like tank thickness, common bulkhead thickness etc.
I've got lots of equations scribbled down here on my notepad, but none that will fit together later, I'm just completely stuck at this point. Can anyone help guide me toward how to solve this problem?
Thanks,
Zatnikitelman-Matt

 

[Notebook]

HAd a go, and this is my thinking so far.

http://i89.photobucket.com/albums/k207/Notebook_04/totalfuelmass.jpg

Double check, as it is 6:50 here...

N.

 

[jedidia]

I'm getting a total mass of... 119.306 tons? Now I'm surprised by how much you can stuff into that cylinder... you'll probably best take it with a grain of salt.

I simply calculated (LOX-density *6 + LH2 density) / 7 * volume of tank.

 

[Notebook]

I'm having second thoughts on my method. The 6:1 ratio would be LOX/LH2 by mass. Not sure you can just divide the volumes the way I have, without taking the densities into account?

Too early in the morning, have a try later, after coffee and breakfast.

Edit: After re-reading jedidia's post, I think thats what I needed to do

N.

 

[jedidia]

Ah no. I just noticed that the MASS ratio is 6:1, not the volume, so forgett the above.

erm... this is more complicated than I thought at first.

I ran a calculation that provided much more realistic results, but couldn't be accurate because the mass ratio in the end didn't quite add up (maybe some typo or rounding error somewhere along the line). I'd suggest to first calculate the volume ratio within 1 m^3 and then derive the mass for the whole tank filling. I don't have the time to re-run it currently...

 

[Notebook]

This reminds me of a reply I got in this thread:

http://www.orbiter-forum.com/showthread.php?t=6705

From Miner1, #8.

I'm wondering if its a similar problem?

N.

 

[jedidia]

Lunchbreak, let's give this another shot!

in a volume of 2 m^3, we have 1141 kg of LOX and 70 kg of LH2. That's a mass ratio of 16.3:1 on a volume ratio of 1:1. A mass ratio of 6:1 would ergo result in a volume ratio of 0.3681:1.

Let's check this: it means that 100% of the volume is 1.3681, resulting in the LH2 taking up 73.09% of the volume, leaving 26.91% for LOX. In the volume of one m^3, that means 307.043 kg of LOX and 51.163 kg of LH2. 51.163 * 6 is 306.978, which would be accurate if I would use some less radical rounding. The density of the mix is ergo a rough 358.16 kg/m^3, times 120.755 m^3 tank volume results in a total mass of 43249.6 kg. Looks a LOT more realistic than what came out in my first attempt, and the mass ratios check out too, so I'd say this is it.

re-calculate yourself to get a more accurate result if needed.

 

[Notebook]

You may be right jedidia, and as maths wimp on Chrismas Eve, I'll be right behind behind you!
Either way, we shall sort this out, and I wish you and yours all the best.

N.

 

[Zatnikitelman]

Lunchbreak, let's give this another shot!

in a volume of 2 m^3, we have 1141 kg of LOX and 70 kg of LH2. That's a mass ratio of 16.3:1 on a volume ratio of 1:1. A mass ratio of 6:1 would ergo result in a volume ratio of 0.3681:1.

Let's check this: it means that 100% of the volume is 1.3681, resulting in the LH2 taking up 73.09% of the volume, leaving 26.91% for LOX. In the volume of one m^3, that means 307.043 kg of LOX and 51.163 kg of LH2. 51.163 * 6 is 306.978, which would be accurate if I would use some less radical rounding. The density of the mix is ergo a rough 358.16 kg/m^3, times 120.755 m^3 tank volume results in a total mass of 43249.6 kg. Looks a LOT more realistic than what came out in my first attempt, and the mass ratios check out too, so I'd say this is it.

re-calculate yourself to get a more accurate result if needed.

First of all, thanks to both of you for all the help.
I'm having a little trouble however figuring how you got the volume ratio of 0.3681:1. The numbers you get from it look right however. To get the 0.3681, that would mean dividing the LOX mass by 6 right? That gives me 2.716, the inverse of which is 0.3681. You've helped me a great deal, but I'm just a little confused here on how you initially derived the numbers.
Thanks,
Zatman

 

[jedidia]

you have a mass ratio of 16.3:1 with a volume ratio of 1:1, since 1141 kg = 70kg * 16.3

What we want is a mass ratio of 6:1, so we have to divide the LOX-mass by x, which is 16.3 / 6 = 2.716666666... (that's where the rounding errors come from, by the way...). So if we divide the volume fraction of the LOX by this, we get 1/2.766666aso = 0.3681. So the volume ratio of the gases at a mass ratio of 6:1 is 0.3681:1.

If you have further qquestions, just ask. If this were a math test, I'd probably not get much points for it, since I didn't really describe the derivation...

 

[Notebook]

For my input #2 above, the 6:1 mass ratio should have realted to the fact that the oxidiser is the 6 times factor. I multiplied the LH2.
Ignore all I've done above, its wrong, and I had only Weetabix.

Have another go after the hols.

N.

 

[Notebook]

Did a plot of the tank, with a moving bulkhead seperating the propellants:

Imagine the bulkhead moves from bottom to top of the tank, with LOX below the bulkhead, and LH2 above.

http://i89.photobucket.com/albums/k207/Notebook_04/TotalMass_7_.jpg

Plot shows the LOX increasing in mass, the LH2 decreasing, as you would expect.
Mathcad note: Had to multiply the tank height 6.15m by 100 to give 615. This is used as the Bulkhead_Position variable. As this is used as a subscript for other variables, it has to be an integer. Its divided by 100 and given metre units in the calculations.

So, just have to find a value for Bulkhead_Position where the LOX mass is 6 times the LH2 mass.
I can think of one way to do that.
Use y=ax+b for a straight line graph, find for each propellant, and treat as simultaneous equations?
Might be another way of running through the two column matrices and comparing values till you get close? Sounds like programming.

Any other methods welcome.

N.

 

[Notebook]

Had a quick "cheat" and put a 6*LH2 mass trace on the plot:

http://i89.photobucket.com/albums/k207/Notebook_04/TotalMass_8_.jpg

Then used Mathcad trace tool to get the Bulkhead_Position where it crossed the LOX trace.

Then worked out volumes and masses for that position, 1.65m:

http://i89.photobucket.com/albums/k207/Notebook_04/TotalMass_9_.jpg

Looks close, just need maths method to check?

N.

 

[Fizyk]

Let's try it like this.

You want to have [math]V_H[/math] volume of LH2 and [math]V_O[/math] volume of LOX. Their sum should be the total volume of the tank:

[math]V_H + V_O = V[/math]

Mass ratio should be 6:1, so:

[math]\frac{\rh V_O}{\rho_H V_H} = 6[/math]

where [math]\rho_H[/math] and [math]\rh[/math] are the densities of LH2 and LOX respectively.

Time to solve. From the second equation we get:

[math]V_O = 6 \frac{\rho_H}{\rh}V_H[/math]

Substituting into the first equation:

[math]V_H \left( 1 + 6\frac{\rho_H}{\rh} \right) = V[/math]

[math]V_H = \frac{V}{1+6\frac{\rho_H}{\rh}}[/math]

Putting in the numbers:

[math]\frac{\rho_H}{\rh} = 0.061[/math]

[math]V = \pi R^2 h = 120.755 m^3[/math]

[math]V_H = 88.4 m^3[/math]

[math]V_O = 32.355 m^3[/math]

Total mass:

[math]M = \rho_H V_H + \rh V_O = 43105.055 kg[/math]

Pretty much the same result that jedidia got, difference is probably due to rounding.

Hope that helped

 

[Notebook]

Thanks Fizyk, thats nicely done.

I'll have a go at putting that into Mathcad land

EDIT:
For completeness here is the Mathcad version.
http://i89.photobucket.com/albums/k207/Notebook_04/TotalMass_Fizyk2_.jpg


Many thanks, N.

 

[perseus]

Hello, other methods to Calculations of size LH2 and LO2
If we used O2 and H2 by the high push and stop ISP within the chemical rockets, this is the dimensions that must have the tanks.
Chemical reaction of the motor (nothing polluting, by the way could be obtained O2 and H2 in the Moon by electrolysis)
O2 + 2 H2 ---> 2 H2 O
Calculating the molecular weights:
O2 = 32; H2 = 2; H2Or =18 u.m.a. (P. Atomic O =16 P. Atomic. H =1)
Using the law of conservation of mass and calculating I number of masses based on the total mass of fuel (The initial total mass can be calculated with the leaf excell the mass of fuel it will be approximately the difference between initial mass - final mass )
nº of masses = nº grams/p.m.
X = I number of grams of H2
(MX) =numero of grams of O2
M = I number of =gramos total grams of fuel that formed of H2O
O2 + 2 H2 => 2 H2 Or
(MX) /32 + 2 X/2 = 2 M/18
Solving equation
grams of I oxygenate = 0,9145 * M (M in grams)
grams of hydrogen = 0,0855 * M (M in grams)
(Almost all the weight is of I oxygenate. But the volumes depend on the density.)
d = m/V => V = md d (density)
d(O2) =1.14 gr/cm3 (I eliminate - 183º C Temperature of boiling - 182.97º C.)
d(H2) =0.07 gr/cm3 (I eliminate - 252.76º C Temperature of boiling - 252.76º C.)
The necessary volumes based on the total mass of fuel:
VO2= 914.5* M dO2 (if we put M in kg)
VH2= 85.5 * M dH2 (if we put M in kg) volumes in cm3
the height of the Hydrogen tank based on (diameter in meters)
Supposed cylindrical V It rolled = PI (D/2)^2 h (D = diameter h = height PI=3.14159.. ) =>
hO2= 4* VO2/(D^2*PI)
hH2= 4*VH2 /(D^2*PI )
In order to calculate the height to apply it formulates empiricist for motors of LO2 and LH2 (I oxygenate and hydrogen liquid.)
hO2=0.00102 *M/D^2 M = total initial Mass of fuel (Kg)
h H2= 0,00155 *M/D^2 D = diameter (meters)
hH2/ hO2= 1.52 LH2 length is 1,52 times the one of LO2
If they are in series the overall height it is: ho2 = height tank of O2 (meters)
h total = hO2 + hH2 ho2 = height tank of O2 (meters)
h total = 0.00257* M/D^2 h total = height total if they are in series
For different fuel and forms from the tanks this you formulate change but it is come analogous for I calculate of dimensions of the tanks