ISP of Coke

[Mmmmyum]

What would the exhaust velocity of a 2L bottle of Coke with a mentos thrown into it be?

Mostly because I want to calculate the delta-v for it really.

If for some reason you don't know what I'm talking about, this: http://en.wikipedia.org/wiki/Diet_Coke_and_Mentos_eruption

[Z-Man]

In the picture on the wiki, the fountain of the diet coke bottle seems to be rising to about 3 m, round that up to 5 m, assume it is purely on a ballistic trajectory and you get a very optimistic ejection velocity of 10 m/s. And really low thrust because it's foam that is getting ejected.

[NovaSilisko]

Working with a few assumptions...

1. The mass of a full bottle of coke is 2.05 kg

2. The mass of an empty bottle of coke is 0.05 kg

3. There is a properly shaped de laval nozzle fastened over the neck of the bottle

4. The exhaust velocity of a diet coke and mentos reaction is a generous 15 m/s (probably doable with a proper nozzle)

This gives a delta-v of:

~55.7 m/s

Enough to give somebody a large whack on the head, maybe even deorbit itself. Roughly five times the delta-v needed to reach escape velocity from the surface of Phobos, so there's that too.

Edited January 14, 2014 by NovaSilisko

[Ralathon]

What would the exhaust velocity of a 2L bottle of Coke with a mentos thrown into it be?

Mostly because I want to calculate the delta-v for it really.

If for some reason you don't know what I'm talking about, this: http://en.wikipedia.org/wiki/Diet_Coke_and_Mentos_eruption

The easiest way to figure this out is to measure how high it sprays.

gh=0.5v^2 ==> v_exhaust = sqrt(2gh)

I estimate that the diet coke bottle on the second picture of the wiki sprays about 2 meters high. So this gives us an exhaust velocity of about 6.2 m/s. assuming a coke bottle weighs about 2kg full, 0.05kg empty and you expell all the liquid at constant pressure this gives you about 22m/s of dV. Comparable to a fire extinguisher or a Manned Maneuvering Unit.

However, you could significantly improve this by narrowing the nozzle. That way your chamber pressure can build to much higher pressures and therefore higher exhaust velocities. You could probably squeeze a bit more out of it that way. You can also use different agents to fuel the reaction. IIRC rock salt preforms very good. But if you modify it that much you might as well just build a dedicated bottle rocket. These can reach some impressive heights.

[Mmmmyum]

Ah, I was wondering about something like 100m/s or so.

I was mostly talking bull with a friend about taking a bottle up to the ISS and putting a mentos in it while on EVA and seeing it shoot away. So nothing like what would be needed to deorbit from the ISS's altitude (irc 1km/s or so)

[toric5]

shire it takes a km per second from that altitude? maybe in 1 orbit, but the isa even need occasional boosts to keep its orbit high. (still some atmosphere.) i think it would only take around 100-200 m/s to enter a 1 day deorbit tagectory. (assuming it only takes about 10-20 m/s to get from LKO to a 3 orbit reentry in ksp.)

[K^2]

Depends on how you build your nozzle. Mythbusters managed to get the fountain to about 29 feet. That is about 13m/s. Or I_SP of 1.34s.

The total delta-V is probably going to be significantly lower than what NovaSilisko estimates, however, since you'll probably not end up with just a mass of the empty bottle in the end, and you certainly won't have that high of an I_SP through the entire run. I wouldn't expect anything more than 20m/s of total delta-V.

So yeah, no de-orbiting from ISS, but it'd still be some sight to see it launched.

[Z-Man]

Oh, VACUUM I_SP? That is going to be much, much higher. You can assume the stored carbon dioxide releases all of its thermal energy into kinetic energy for the ejected liquid, with maybe 50-75% losses (thermodynamics and other inefficiencies). Casual research says that for 1 volume unit of coke, there must be about 4 volume units of CO₂ at standard pressure.

Dimensional analysis time!

Standard pressure p is measured in Pa = N/m² or kg/(m s²).

Density of water rho is in kg/m³.

Those are the only dimensional parameters here, the dimensionless alpha = 1/4 factor obviously needs to be multiplied with the density prior to calculations (well, in the limit that the mass of the gas is much lower than the mass of the liquid) as half the amount of liquid of double the density should yield the same result; it is the mass that matters.

Only way to combine those to a velocity is

I_SP = sqrt(p / alpha * rho)

Plugging in values of p = 10⁵ Pa, rho = 10³ kg/m³ yields I_SP = 20 m/s...

Damn. Not much higher after all, and that is before the unknown loss and dimensional-analysis-is-so-wrong factor. Still, this is assuming all liquid gets ejected, whereas the earthbound experiments probably only spill a part.

[PakledHostage]

So yeah, no de-orbiting from ISS, but it'd still be some sight to see it launched.

Just for reference, I worked out the rough amount of delta-V that it would take to lower an object's PE from the ISS's orbital altitude. NORAD posts the current orbital elements of the ISS as:

ISS (ZARYA)

1 25544U 98067A 14014.57553442 .00012953 00000-0 23115-3 0 4603

2 25544 51.6498 140.9654 0005920 28.6094 82.6983 15.50409075867495

This means that the ISS is currently orbiting at roughly 420 km altitude with an eccentricity of 0.0005920, or about ±2 km. Assuming a circular orbit about the Earth at 420 km altitude:

- Lowering an objects PE to 300 km would take about 35 m/s delta-V.

- Lowering an object's PE to 200 km would take about 63 m/s.

- A 100 km PE would require about 94 m/s delta-V.

- A 0 km PE would require about 123 m/s delta-V.

[Mmmmyum]

Just for reference, I worked out the rough amount of delta-V that it would take to lower an object's PE from the ISS's orbital altitude. NORAD posts the current orbital elements of the ISS as:

This means that the ISS is currently orbiting at roughly 420 km altitude with an eccentricity of 0.0005920, or about ±2 km. Assuming a circular orbit about the Earth at 420 km altitude:

- Lowering an objects PE to 300 km would take about 35 m/s delta-V.

- Lowering an object's PE to 200 km would take about 63 m/s.

- A 100 km PE would require about 94 m/s delta-V.

- A 0 km PE would require about 123 m/s delta-V.

Where do you get your energy requirements from? I'm using KSP with RSS for my estimate. Also from a 300km orbit, not a 420km one.

[ZetaX]

KSP is not reality...

[PakledHostage]

Where do you get your energy requirements from?

The equations on my MathCAD spreadsheet.

[K^2]

Where do you get your energy requirements from? I'm using KSP with RSS for my estimate. Also from a 300km orbit, not a 420km one.

It's not about energy requirements. It's a simple Hohmann transfer equation. His numbers check out.

So a single bottle, no dice. But something with 2-3 stages of coke and metos might do the trick. Of course, if the top stage is a single bottle, you'd need at least 6 for a stage before, which is entirely feasible, but a 3-stager is probably out of the question. At least, as far as launching something from ISS.

[PakledHostage]

I don't have time to work it out right now (I am supposed to be working), but I am interested in Z-man's point about vacuum Isp vs. the Isp that K^2 calculated for the bottle at something close to 1 atmosphere pressure. I think it is a relevant point. Also, maybe we could get a better estimate of the total delta-V if we can find a video that shows how empty the bottle is when it is finished expelling liquid.

[Mars90000000]

Not enough, that's the answer.

[K^2]

I don't have time to work it out right now (I am supposed to be working), but I am interested in Z-man's point about vacuum Isp vs. the Isp that K^2 calculated for the bottle at something close to 1 atmosphere pressure. I think it is a relevant point. Also, maybe we could get a better estimate of the total delta-V if we can find a video that shows how empty the bottle is when it is finished expelling liquid.

It's relevant, but not going to make a big enough difference, unfortunately. Mostly, because there is no way to ensure that all of the energy is given up before fluid exits the bottle, and that's your biggest limiting factor even on Earth. Even if it boosts you to 20m/s as per Z-Man's estimate, which is definitely high end, then best you can count on from a bottle is a 30-something range on delta-V. With two stages of that you might get into dense enough atmo to start de-orbiting rapidly.

[Ralathon]

You have to consider that in a vacuum the mentos reaction isn't going to do much. It'll generate a bubble, but that bubble doesn't move away from the mentos, so the reaction stops pretty quickly and you end with a mentos suspended in a bubble of foamy coke.

On the other hand. When launching this in vacuum the water in the bottle will start to boil. Furthermore, the acceleration will push all the liquid towards the nozzle. So you've effectively built a bottle rocket that has a constant internal pressure equal to the vapor pressure. Assuming this all goes relatively fast you can also assume that no liquid will be left inside the bottle. It doesn't have time to form ice and it gets pushed towards the nozzle by the acceleration.

We can figure out the exhaust velocity via Bernoulli's principle. sqrt(2p/rho) = v_exhaust. Vapour pressure for water at room temperature is about 2kpa if memory serves. So that gives an exhaust velocity of about 2m/s. This gives 7.4 m/s of dV to a simple bottle of water on this principle alone. However, for carbonated drinks the internal pressure is going to be waaaay higher thanks to the dissolved CO2. The average can of coca cola is has a staggering vapor pressure of 200kPa at room temperature (That's the point it reaches equilibrium). If we apply the same method using this pressure you get a massive 74dV. If you heat up the cola bottle to a nice warm temperature I imagine it has enough dV to deorbit itself.

[Mmmmyum]

Also correction about deorbit burn, I was wrong. This is a bit more complicated than I thought it would be lol

[K^2]

You have to consider that in a vacuum the mentos reaction isn't going to do much. It'll generate a bubble, but that bubble doesn't move away from the mentos, so the reaction stops pretty quickly and you end with a mentos suspended in a bubble of foamy coke.

[...]

Furthermore, the acceleration will push all the liquid towards the nozzle.

I'll let you figure out how these two relate.

On the other hand. When launching this in vacuum the water in the bottle will start to boil. Furthermore, the acceleration will push all the liquid towards the nozzle. So you've effectively built a bottle rocket that has a constant internal pressure equal to the vapor pressure. Assuming this all goes relatively fast you can also assume that no liquid will be left inside the bottle. It doesn't have time to form ice and it gets pushed towards the nozzle by the acceleration.

We can figure out the exhaust velocity via Bernoulli's principle. sqrt(2p/rho) = v_exhaust. Vapour pressure for water at room temperature is about 2kpa if memory serves. So that gives an exhaust velocity of about 2m/s. This gives 7.4 m/s of dV to a simple bottle of water on this principle alone. However, for carbonated drinks the internal pressure is going to be waaaay higher thanks to the dissolved CO2. The average can of coca cola is has a staggering vapor pressure of 200kPa at room temperature (That's the point it reaches equilibrium). If we apply the same method using this pressure you get a massive 74dV. If you heat up the cola bottle to a nice warm temperature I imagine it has enough dV to deorbit itself.

Where do I even start with all of this.

First, you aren't propelling yourself with gas. It's going to be coke forced out of the nozzle. So you can throw the Bernoulli Equation for Ideal Gas out. You are way over-estimating available energy this way.

Second, main purpose of mentos in a bottle is the same as that of a boiling chip. You need it to speed up carbon exiting liquid. As soon as you open up that valve, pressure in the bottle will drop. And it's the rate at which CO2 can be released compared to rate at which fluid is forced out that's going to determine pressure inside. Good luck estimating that.

Finally, as soon as water begins to boil, temperature will drop, and that will mean both vapor pressure and CO2 pressure will drop as well. Care to enter that into your estimates? I did not think so.

As I've said earlier, the main limiting factor is the fraction of energy stored in Coke that's going to be used up on its way out. And that just isn't going to change a hell of a lot in vacuum. You will get a boost to I_SP, but nowhere near that much. There is no way a bottle will deorbit itself in a single stage.

[Ralathon]

I'll let you figure out how these two relate.

Mentos sink in cola. There is a hole in the bottom. I'll let you figure out how these two relate

Where do I even start with all of this.

First, you aren't propelling yourself with gas. It's going to be coke forced out of the nozzle. So you can throw the Bernoulli Equation for Ideal Gas out. You are way over-estimating available energy this way.

Where did I ever say the gas was the reaction mass? The reaction mass is of course the coke, but the gas provides the pressure. It's exactly like one of those secondary school bottle rockets except the pressure is provided via vapor pressure instead of some external pump. Last I checked Bernoulli applies just as well to liquids as it does for gasses provided that you trow viscosity out of the window, a pretty reasonable assumption considering the size and viscosity of water.

Second, main purpose of mentos in a bottle is the same as that of a boiling chip. You need it to speed up carbon exiting liquid. As soon as you open up that valve, pressure in the bottle will drop. And it's the rate at which CO2 can be released compared to rate at which fluid is forced out that's going to determine pressure inside. Good luck estimating that.

The mentos merely provides nucleation cores for the CO2 to release. A lower pressure will cause nucleation just as well if not better. The lower pressures in the turbulent flows are the whole reason that a cola bottle starts to foam when you open it after shaking. Considering that a mentos provides 1 point while the vacuum affects the whole bottle the mentos is pretty irrelevant. Admittedly, the chamber pressure will be significantly lower than the equilibrium pressure. But considering that CO2 can escape fast enough to create geisers in 1 atmosphere I wouldn't put it that much below equilibrium.

Finally, as soon as water begins to boil, temperature will drop, and that will mean both vapor pressure and CO2 pressure will drop as well. Care to enter that into your estimates? I did not think so.

Number of moles in 2 liters of steam at vapor pressure and room temperature: 0.024 mol

Enthalpy of vaporization for water: 40.68 kJ/mol

Corresponding temperature drop in 2L of water: 0.11 degrees. Yea, this is going to be very relevant... The entire reaction simply goes way too fast for thermal effects to really kick in. Even if you go with a conservative vapor pressure of 1kPa you're left with an empty bottle in just 7 seconds.

As I've said earlier, the main limiting factor is the fraction of energy stored in Coke that's going to be used up on its way out. And that just isn't going to change a hell of a lot in vacuum. You will get a boost to I_SP, but nowhere near that much. There is no way a bottle will deorbit itself in a single stage.

I never came to that conclusion. A room temperature bottle wont deorbit itself. But if you heat it up I can see it working. Steam rockets are a thing after all.

[sal_vager]

On the subject of the mentos leaving the bottle, you could put something in the way that coke can pass through but not the mentos

[DisarmingBaton5]

On the subject of the mentos leaving the bottle, you could put something in the way that coke can pass through but not the mentos

Or would a grille be too heavy?

[Kinglet]

Working with a few assumptions...

1. The mass of a full bottle of coke is 2.05 kg

2. The mass of an empty bottle of coke is 0.05 kg

3. There is a properly shaped de laval nozzle fastened over the neck of the bottle

4. The exhaust velocity of a diet coke and mentos reaction is a generous 15 m/s (probably doable with a proper nozzle)

This gives a delta-v of:

~55.7 m/s

Enough to give somebody a large whack on the head, maybe even deorbit itself. Roughly five times the delta-v needed to reach escape velocity from the surface of Phobos, so there's that too.

If there is not enough funding for a proper Phobos mission we will use coke as a propellant.

[K^2]

Mentos sink in cola. There is a hole in the bottom. I'll let you figure out how these two relate

You aren't seriously calling something that can be fixed with a pair of matchsticks through the throat a problem, do you?

Where did I ever say the gas was the reaction mass? The reaction mass is of course the coke, but the gas provides the pressure. It's exactly like one of those secondary school bottle rockets except the pressure is provided via vapor pressure instead of some external pump. Last I checked Bernoulli applies just as well to liquids as it does for gasses provided that you trow viscosity out of the window, a pretty reasonable assumption considering the size and viscosity of water.

Sorry, I assumed your numbers were wacky because you used equation for compressible flow. But if not, you just did your math wrong. 200kPa @ 1g/cm³ yields 20m/s. Likewise, 2kPa gives you 2m/s. I don't know where you picked up an extra factor of 3.7. Even if you did 200kPa + 1bar, it's still only 24.5m/s.

If we are talking about 200kPa driving the stream, by the way, it gives us a good estimate on losses. Since the above suggests 20m/s exit velocity, while we see an only 13m/s exit velocity on Earth.

Edit: Unless that's meant to be 200kPa total, working against atmospheric? In that case, Mythbusters managed something very close to theoretical maximum.

The mentos merely provides nucleation cores for the CO2 to release.

Isn't that what I said? But pressure drop without nucleation sites is not going to do anything. It's like having superheated water. It won't boil if there aren't sufficiently large bubbles for vapor to expand into. You still want metos in there. Or some other source of nucleation sites.

Number of moles in 2 liters of steam at vapor pressure and room temperature: 0.024 mol

Enthalpy of vaporization for water: 40.68 kJ/mol

Corresponding temperature drop in 2L of water: 0.11 degrees. Yea, this is going to be very relevant... The entire reaction simply goes way too fast for thermal effects to really kick in. Even if you go with a conservative vapor pressure of 1kPa you're left with an empty bottle in just 7 seconds.

Totally fair. Objection on temperature withdrawn.

I never came to that conclusion. A room temperature bottle wont deorbit itself. But if you heat it up I can see it working. Steam rockets are a thing after all.

Bottle ruptures at around 500kPa. That gives you a little over 30m/s with the Bernoulli method, and since we know that it overestimates actual I_SP and that it only holds while the pressure does, even if you want to turn it into a steam rocket, you'll need much better casing than a coke bottle.

Edited January 16, 2014 by K^2