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Check out the AMAZING airlift video! I have checked (with a manometer) and about 0.2 psi has pumped water to over 6 ft high. (It worked all night using 3/16 inch tubing). The whole pallet planter thing works great after a few modifications to the original idea! As you work on them you discover things and adapt. Originally, it was going to be just one pallet standing up. So now they are 2 pallets joined together (to get a decent 7 inch gap) and the pallets have legs to stand on and the legs collect water in their "boots" to be recycled back up top again. Also, the floors of each section now slope back. The concrete on the bottoms of the "boots" is to prevent the weight of the pallets breaking the plastic buckets.

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This step looks hard but it really isn't. If the plastic of the base of the bucket takes all the weight of the planter and moves, it will break and leak. So you need to make strong concrete round the base. In this case, I made the concrete using wet masonry sand as a sand mold and a flower pot that is a bit wider to make the hole in the mold., I used a pallet with plywood under the sand and then a 3 to 1 mix of coarse sand and cement for the concrete of the foot. So lots of pictures and a few image notes to explain the steps. The pictures are screenshots from a camera movie so not wonderful quality but its better than watching my youtube movies! As I write, they are curing, it is cold so it will be a while. BUT, I have already made one pallet planter to a slightly different pattern and that first one does not have the feet at all. I haven't used a sand mold in years but it works great. (Your sand has to be just the right firmness and wetness. If it sumps too easily you may have to add a portion of clay. (if you have clay!)

Simply install a swing check valve to stop the back flow. Buy my manual $50 at www.olomanagardens.com Aquaponic Manual.

No check valve is needed, the answer is in the next step. Note that the momentum of the water in the horizontal pipe causes the flow to be in only one direction and the constriction prevents a wasteful back and forth "vibrating" as it flows.

Hey, I"m having difficulties visualizing this. Could you explain where these tubes "Wide tube", "air delivery tube" etc are being connected. I attached a picture with how I see this step, hoping you could shed some light on it.

Hey thanks for the photo. Makes sense, I don't know why I was confused in the first place :S. Do you find the lifts work better with 1/4OD or 3/16OD Tubing? I Tried 1/2OD and only air was travelling through it, I even turned the air flow way down.

Do you have any algae developing in the clear tubing?

I've been doing small tests running something similar but with a flapper valve where the water goes in. In this case, imagine a flapper valve at the end of the 10" section at the bottom of the bucket. This is also an old experiment, since then I've developed something new.

Here are my results:

Static Head: 7.25", Lift water to: 65", Water Flow rate 0.22ml/s With air flow rate 17.24 ml/s. Efficiency, ~1.3%

Static Head: 5.25", Lift water to: 65", Water flow rate 0.11ml/s with air flow rate 4.76 ml/s. Efficiency ~1.8%

Static Head: 4.25", Lift water to: 65", Water flow rate 0.083 ml/s with air flow rate 2.94 ml/s. Efficiency ~2.83%

Static Head: 3.5", Lift water to 20", Water flow rate 0.31 ml/s with air flow rate 1ml/s. Efficiency 24.33%

Now, there are a lot more trials with varying performance and efficiency. I've just shown the results where I dialed back the air flow to get some water flowing with less and less static head (submergence). In the last trial I cut back the pumping height to 20".

I'm wondering how to increase the efficiency without increasing submergence. Any thoughts?

Hi Robert, thanks for doing all that experimenting and for sharing your results. Have you used the constriction at all or just your flapper? How are you working out efficiency? Are you going off the watts used by your air pump or are you just doing Mass by g (acceleration due to gravity) by height? (The potential energy formula) I suspect some error in the last number 24.33% because static head X air flow is 3.5 while water pumped by height is 6.2 which is over unity. (which is impossible). If you measured water flow correctly, however it is a really good sign. You used low submergence but I suspect you probably got useful work from higher pressure through the airlift pump. That is is great because ordinarily, higher pressure would cause blow back and even greater loss of efficiency. What is the ideal submergence for you? And the perfect pumping height for you? Have you tried splitting up the flow and using 2 airlift pumps from one source and pumping to different heights? Better I stop this post now before we both get lost!

Perhpas my efficiency equation is wrong? All i'm doing is Output/Input (Note that my calcs are wrong ,my spread sheet was selecting the wrong values, I've corrected it now). I did another one which calculated efficiency based on work and pumping height form this website <<http://www.pumpsandsystems.com/topics/pumps/pumps/...>>

For the calc in the link, i'm selecting a trial where the pump did 0.0833ml/s of water for 2.9411ml/s of air. its pumping in 4.25" of submergance to a lifting height of 65".

Electrical power: My air pump is 3.5 Watts, good to 2.1psi. In order for my system to work I have to dial back the air flow with a restrictor so that I'm using about 1/27th the maximum air flow (79.82ml.s is max air flow), so, i'm making a guess that in a nice world I could potentially use 1/27th the electrical power. So, for this trial electrical power is 3.5w / 27 = 0.1296 w = 0001737hp = Pa.

Water power: The pumping height is 65" = 5.41'. The water flow rate is 0.0833ml/s = 0.001321 gal/min. Ps = (0.001321 * 5.41) / 3960 = 0.000001805

(Pw / Pa) * 100 = (0.000001805 / 0.0001737) * 100 = 1.038%

Now, granted that is different than a 2.33% efficiency of out/in, but I didn't want to set up my excel spreadsheet with more stuff. However, my calcs may be wrong for this particular application. Any equations for air lift pump efficiency I should be looking at?

Hi Robert, to be fair you are using a 2.1 psi pump to produce about 0.3 psi or less. So you are really using something designed for a totally different purpose to produce your air. How efficient is it at producing air at 2.1 psi? So you need to get a tube that goes 58 inches deep under water to see. 58 inches under water is 2.1 psi. Measure the volume of air coming out. The formula can just be MGH for potential energy of the air. For instance if you are pumping 60 liters per minute of air to 1 meter deep, you are displacing 60 liters per minute of water. So the potential energy calculation is mgh m is mass in kg. And watts is energy per second. 1 liter of water weighs 1 kg. g is acceleration due to gravity. 9.8 meters per second per second. (we can use 10 for handiness instead of 9.8 so 1 by 10 by (1 meter high). That makes this a 10 watt power source. But of course, you probably need a 15 watt pump to do this due to losses and inefficiencies when you compress the air. My point here is that if you can calculate the efficiency of compressing air, then just assume this same efficiency in your "ideal world pump" that makes the very low pressure you need. So I think you need to be working out how efficient it is at using the low pressure air to pump water. And that is actually pretty easy. Efficiency of that process is just (volume of water pumped multiplied by height you pump to) Divided by (volume of air by depth under water that you are pumping it to). I think you need to make sure of your depth pressure with such low pressures. I think the best way is to let your pump go into an open bottom container that is in water to the required depth. All the excess air just bubbles out, and you get the amount of air and pressure that your airlift needs. Thanks Brian

This is a helpful link. I made little vortex thingys, but my air flow was too small. I think you might only be able to measure effects in the 1/2 inch diameter tubing (1 cm to 1.2 cm diameter tubing) and maybe 5 to 15 liters per minute of air (at about 2.5 to 3 psi).

http://www.tippettsfountains.com/fluidicsbasics.php

I think there is a possibility of using higher pressure air with low submergence. There is a kind of fluidics "capacitor" called a vortex capacitor. If you put that where I have the constriction, there is a small chance that you get to transfer maybe 1/3 the air energy to it. I don't think it is efficient to produce 4 or 7 inches of air pressure, but it is probably fine if you are producing 14 or 16 inches of air pressure. I think it might be a good idea to produce something with, say, 6 inches, of submergence that can use 18 inches of air pressure. your vortex capacitor would both prevent backflow and gradually build up a powerful flow towards the joint where the air is introduced. I haven't really tried this. I made a few half hearted attempts to do it last summer. I think the vortex capacitor has to be either facing up or down, The center of the vortex enters into the horizontal tube. I had a quick look for links about vortex capacitors but didn't find any. They were used in fluid logic circuits. Fluid logic was used in the Apollo moon rockets and it is returning in nano technology applications. Brian