# How much pressure do I need to power a keg full of coffee

1. Oct 31, 2016

### dcbrwn

Hey everyone,
I came across physicsforum.com while trying to answer a puzzling question I'm facing. I'm building a portable nitro coffee dispenser and need to know how much PSI I need to put in a small nitrogen tank to adequately dispense the beverage.
Here's the facts;
• I have a 64oz keg that needs to maintain a pressure of 35 psi throughout the duration of serving. The average size of a pour will be 8 oz. Temperature of the fluid and keg will be 10 degrees Celsius.
• The 64oz keg can be pressurized to 35 PSI prior to hooking up the 4 oz tank (via a 5 LBS tank.)
• I have a 4oz tank (co2 tank that will instead run pure nitrogen. Limited to 1800 psi.) I have a regulator that will run at the desired 35 PSI. Length of tubing from tank to keg is 7". Temperature should be around 18-19 degrees Celsius.
• From what I've read the faucet provides about 2-3 PSI resistance. I'm unsure how much PSI is lost during the pour but based on what I've seen testing it I'd say there's a 5 PSI loss per 8 oz.
• The 4oz tank will be filled from a 5 LBS tank that runs at 2200 PSI but the regulator can only handle 500 PSI. After running the numbers I believe I can only ever achieve a PSI of 450-470 PSI in the 4 oz tank.
• Do I need to buy a new regulator that can output a higher PSI and how much more pressure do I need? (I've found an 800 PSI output regulator that I might be able to squeeze 900 PSI out of. I don't want to buy it if it's not enough PSI to serve a 64oz keg.)
I've played around with the numbers but I really don't know how to calculate the PSI in the keg as the volume of liquid decreases.
I do know from my *flawed* math and experimentation that 400 PSI in a 4 oz tank isn't enough to serve the entire 64oz while maintaining 35 PSI. Can anybody show me what I'm doing wrong? I had thought 350-400 PSI would have been enough to dispense the 64 oz at atleast 30 PSI but my tests aren't confirming my hypothesis thus far (to be fair I did have a leaky fitting I've fixed. I ran out of N2 so I haven't had a chance to test it again.)
I appreciate any insight sharper minds can provide!

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2. Oct 31, 2016

### Merlin3189

If you start with the 64oz* tank absolutely full of liquid, then all the gas is in the 4oz container and the connectors (whose volume we will need to know, but for now I'll ignore.**)
At the end there is no liquid and the gas fills the 64 oz tank and the 4oz tank and connectors.
So your gas expands from 4oz+ at unknown pressure to 68oz+ at 35psi.
Now if we were to assume no temperature change for the gas, it would be easy. But you seem to have gas initially at 18-19C going into a tank which is at 10C, with considerable thermal capacity (the metal and the liquid). So you could try assuming the gas changed from 4oz+, at 18.5c to 68oz+ at 10C and 35 psi. That seems like a worst case, so the result you get would be good enough.

* I'm assuming that oz are some USA units of volume. If not we need to convert to volumes first. (Edit: if the 64oz is weight of water and the 4oz is weight of CO2 gas, then the calculation becomes more complex, but your problem eases because the ratio of volumes becomes much smaller.)
** rereading your post, I see the connectors can be precharged to 35psi. Since they also end at 35psi there is no net change in pressure. So you need only allow for the cooling from 292K to 283K, which may be small enough to ignore.

3. Oct 31, 2016

### dcbrwn

Sorry about that. My mind slipped when writing that. The keg volume is 1.893 liters and the nitrogen tank is .1182 liters. Tonight I tried a real-world experiment to see what would happen (plus I wanted some coffee.) I hooked the mini keg up to my large N2 tank and set the regulator to 30 psi. Then I cut off the N2 flow from the main tank and poured myself a .2366 liter glass (8 ounces) cup of coffee (this had been chilled to 10C.) After checking the keg pressure gauge I noticed it had dropped about 10 PSI. This was a much bigger drop than I had expected but explains why I was using so much N2. I *think* I've got the math worked out; however, it could all be rubbish.
• For every .0295 liters poured I need 1.25 PSI so I'll need ~80 PSI to empty the keg.
• (80P * 1.893L) / 14.7P = 10.30L of N2
• The gas tank volume is .1182 liters, so if I ran it at 1300 PSI;
• (1300P * .1182L) / 14.7P= 10.45L of N2
So assuming the approach and math is correct I should fill the tank to 1300 PSI? Sorry for the "oz" goof-up. Since I've been brewing coffee for the tests I've been doing many calculations in S.I. ounces. I do appreciate the help! I'll post some photos of my rig soon (I'd send you some nitro coffee if it were possible.)

4. Nov 1, 2016

### jack action

I'll do the math @Merlin3189 explained.

The equation is:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
Where:

$P$ is the gas pressure;
$V$ is the gas volume;
$T$ is the temperature;
$1$ represents state 1 of the gas;
$2$ represents state 2 of the gas.

Let's say state $1$ is when the keg is full of coffee and state $2$ is when the keg is empty. You are looking for $P_1$, so rewriting the previous equation:
$$P_1 = P_2 \frac{V_2}{V_1}\frac{T_1}{T_2}$$
• When the keg is empty, $V_2$ is the sum of the volume of the keg, the bottle and the hose between both of them. If we ignore the hose for now, that is 68 oz;
• When the keg is full, $V_1$ is the sum of the volume of the bottle and the hose between both of them. If we ignore the hose for now, that is 4 oz;
• If we assume the temperature of the gas is equal to the temperature of the coffee when the keg is full, then $T_1$ = 283 K (i.e. 10°C);
• If we assume the temperature of the gas is equal to the room temperature when the keg is empty, then $T_2$ = 291.5 K (i.e 18.5°C);
• Of course, the desired empty keg pressure is $P_2$ = 35 psi.
So:
$$P_1 = 35 \frac{68}{4}\frac{283}{291.5}$$
Or $P_1$ = 578 psi. If you add the hose volume, this number should go down, so you can assume it is a worst case scenario. If we assume the temperature stays the same, the value goes up to 595 psi, an absolute maximum value.

Just for checking, with your little experiment, you had a full keg ($V_1$) and a partially emptied keg ($V_2$ = $V_1$ + 8 oz). Let's assume the temperature is the same, then:
$$30 = 20 \frac{V_1 + 8}{V_1}$$
For this to be true, it would mean that $V_1$ = 16 oz, or that your hose has 12 oz of volume, which seems unusually large. If it was, then comparing a full and empty keg would give:
$$P_1 = 35 \frac{64+16}{16}\frac{283}{291.5}$$
Or $P_1$ = 170 psi, which is obviously not true from your first experiment.

That can mean only one thing: you are loosing gas somehow every time you pour a glass. Is the gas mix up with the coffee and goes out with it? Or do you have a leak? How much gas goes out with every pour is probably dependent on the pressure when you pour that glass, so it becomes a more complex problem to solve.

Sorry if I raise more questions than I answer.

5. Nov 2, 2016

### 256bits

Maybe not.
When the 4 oz cylinder is hiked up to the keg, the pressure in the hose and the dead space in the keg is at atmospheric.
The 4 oz cylinder has to charge that space up to 35 psi.

By the way, using absolute pressure I get a dead space ( hose and keg ) of 28 oz.

If I assume dead space is at 0 psi, then the pressure in the cylinder comes out to 1004 psi absolute ( 990 guage approx. )
( since the dead space has already a 14.7 psi, in real terms the 990 guage for the cyl should drop a bit.

You mean up.

6. Nov 2, 2016

### 256bits

Fluid oz is a volume measurement, and in context it is usually abbreviated to just oz. , so you should be OK with that unit.

7. Nov 2, 2016

### jack action

There it is, I forgot to convert gauge pressure to absolute pressure. So you need to add the atmospheric pressure (14.7 psi) to all pressure readings:
$$P_1 = (35 + 14.7) \frac{68}{4}\frac{283}{291.5} - 14.7 = 805\ psi$$
I also confirm that doing the same with the second experiment gives 28 oz.

But I thought of something: I assumed that the keg was initially full in the second experiment, but that wasn't specified. Maybe that wasn't the case?
No, I mean down. If you increase $V_1$, then $\frac{V_1 + \Delta V}{V_1}$ will decrease. If $V_1 = 0$, the ratio is infinite, if $V_1 = \infty$, then the ratio is 1.

8. Nov 2, 2016

### dcbrwn

Thanks for the help guys! I'm starting to see where I went wrong with my original calculations (I'll admit I was a C student in high school physics, though I do enjoy trying to understand how fusion reactors work.)

Yes the keg does lose some N2 when coffee is dispensed. The dissolved N2 bubbles is what gives it a smooth creamy texture (mouth-feel) and seems to sweeten the taste slightly. If there's a way I can measure the amount of dissolved N2 in a solution please let me know and I'll try my best to provide accurate measures. Perhaps I could weigh the keg full of coffee before adding nitrogen then weight it with nitrogen after it's sat overnight? Understanding how the N2 dissolves in the solution would be helpful since I'm trying to become a guru of nitro coffee. There's also a small restriction plate in the faucet that reacts with the pressurized fluid to create the head. I'd hazard a guess from what I've read that N2 absorption is related to temperature of the fluid and surface area. Brewers use a diffusion stone at the bottom of the keg to force the gas into tiny (.5 micron) bubbles which definitely speeds up the process.

The line from the tank to the keg is about 15.24cm long and only 3.75mm in diameter. But there is also the regulator body; I have no idea what the volume of that is since I can't seem to get the little thing taken apart. I doubt there's very much volume in it as it's a very small regulator.

Also, for the real-world experiment the keg was not full. I'll be making more coffee tonight and this time I'll fill it completely. I'll also measure the "drip tube" (the tube that feeds the fluid from the bottom of the keg to the faucet) since it will initially need to be filled when the first cup is drawn through the faucet.

I'm actually relieved to see that the PSI is more around the ~1000 PSI area as I've bought a new fill-regulator that has a minimum output of 1100 PSI. With the new regulator I should have more than enough pressure to empty a keg. I know from my first fills of the 4oz tank that I can't get it 100% filled (unless I were to draw it down to a very deep vacuum first.) I suspect I can get 80% full if I freeze the 4oz tank first and purge it the best I can prior to filling. Luckily these CO2 tanks are regularly filled to 1800PSI and are rated for 3000PSI so it looks like I'll have a good safety margin if I can get it to ~1400-1500PSI.

Once I get the new regulator I'll post some pictures of my dispenser. You guys are welcome to come try some nitro coffee if your in the Southwest Michigan area! If you're a fan of coffee and science I suggest trying to find a local coffee house that has nitro coffee. It's a delicious way to serve coffee. From what I've learned thus far nitrogen is quickly becoming my favorite gas. (Oxygen is great and all but I love how inert nitrogen is - it gives the coffee a very long shelf-life.)