Backyard Aquaponics
A place to discuss aquaponics

Hi Frank

Ran a small test today. Baby pump iwaki md 6z, capacitor run, no detectable amp varn with head varn. Couldn't get it higher than 2m cause of the splashing. Only time it dropped was with no water in it.
Would have done it with a bigger pump but it would have made more of a mess of the laundry

Post it! I've been pondering the subject myself.

One nice thing here is that people seem to try to couch their disagreements gently and as if we are all reasonable people (even jerks get a civil reception unless they persist) rather than jumping on heretics and fools with knives drawn.

I use water pump for aeration as well, if that is what you topic was

Amp draw is more at higher head pressures. The motor works harder to pump the water higher.

Sleepe has tested this and like I suspected that has shown not to be true: the motor does not work harder to pump the water higher because then the pump will automatically pump less water, which compensates.



that was another issue people refused to accept: that airlift pumps are very inefficient at pumping at any head higher than the water level and that the aeration advantage does NOT compensate this in the least.

another question is aeration by diffusers: in order for that to be energy efficient, you must produce as small bubbles as possible as deep as possible so as to expose the largest possible surface the longest time possible.

That means very large, high tech (expensive) diffusers which are prone to clogging.

As you need a pump anyway to cope with the head higher than the water level a good design with long and wide gutters will make sure the water gets aerated quite enough. An ebb and flow system will increase aeration thanks to the oxygen clinging to media and roots.

No, the item I was bashed on was the question I asked (not statement I made) if oxygenation by electrolysis, i.e the electric splitting of water into oxygen and hydrogen, which takes extremely little power to do might be a solution.

No sooner than I posted it I was accused of spamming and suspected of being sponsored, all in very abusive language.

I know that there are a heap of scams about hydrogen production systems, but that was not the issue and very clearly stated so: the issue is oxygen production.

So what do you all think?

Frank

will have to test this for myself as soon as i fit my valves to the pump outlet.

i'd suspect that the amps would go higher. just listen to a pump that had had its output severly restricted........you will hear the RPM drop, hence higher load hence higher amps...........but the proof will be in the test.

from my previous post:



but please do your test

it is better to have two confirmations

Frank

Frank,

Let's see:
It releases (240*2)k=480k joules to make 16g of O2 (plus the required H2) into water, so assume a 50% efficiency for the reverse process and we get about 10^6j or 275watt*hr. Tilapia supposedly need .3g of O2 per kg of fish, so you would need .3/16*275watt*hr=5watts/kg of fish. Looks workable from a power standpoint!

I'm not sure about impact of electrolysis on water quality: hydrogen and electrolysis byproducts (like chlorine from NaCl) *might* be a problem. Another deal-killer might be CO2 buildup: a bubbler normally gets rid of that as well as bringing in O2. Perhaps it would be removed in the growbeds. This might also be a solution for other odd electrolysis chemistry: run it through the beds and let them first clean it up rather than sending it first to the less-adaptable biology of the fish.

I like the idea of not needing so much surface exposed in cool weather and the inefficiency would add more heat to the water for warm water fish.

DISCLAIMER: The 50% number is based on the most efficient systems running 66%, according to my quick research. I have no idea what actual performance would be.

sorry, but I'm a little slow

I'm reasonably good at pumps (and still learning)
but I still have a lot (if not most) to learn on all other issues,

so have mercy


where did you get the equations?

what releases 480 kjoules? shouldn't tat be "it takes"?
why assume 50% efficiency?
what reverse process?
shouldn't that be 275 watt/hr (twice) instead of 275 watt*hr?

Tilapia need .3g of O2 per kg of fish per what? per hour, per day?
Isn't it better to start calculations not from fish weight but from feed amount per day?
This way O2 needs of all uneaten food is also calculated, whether you overfeed or not.

I'm sure I could start a search to find all this out.
but your help will be quicker.

electrolysis byproducts are an issue: it is strictly discouraged when any salt is present

on CO2:

so does a (or several) waterfall(s) without extra energy input.
you already have invested this energy.


the inefficiency of what? of a centrifugal pump?
though that is true (lost energy is transformed mostly in heat), warming up water with a centrifugal pump seems inefficient too.
There are other, better methods.

greets

frank

It may not be your fault: I've been accused of skipping steps. I've also been accused of too much detail....



I looked up the energy released when water is formed from H2 and O: about 286kjoule/mol (I either got it from another source that gave it as 240kj or misread it. Then I doubled it because of another equation I found involving the atomic bonds, but shouldn't have. *grrrr*!), so that 286,000joules is the energy given off by formation of 1 mol of water (containing 16g of O) http://en.wikipedia.org/wiki/Standard_e ... _formation.

You are correct, it would "take" that to reverse the process and break water into H2 and O2 in a perfect system. According to my research the best electrolysis is 66% efficient, so I said 50% to be reasonable (we aren't going to be working with as many special electrolytes, etc). So, double the 240kj and you get roughly 500,000 joules of energy. A watt is 1 joule per second (and, conversely, a joule is 1 watt*second) so you get 500,000watt*sec=140watt.hr for 16 grams of O2 (Always happy to go into any level of detail....more?)


Per hour. Can't you read what I didn't write?!


Yep, feed/day would be better, but I was just trying some BOE (back of envelope) quick calcs to see if the idea is feasible from an energy use standpoint.


Well, maybe. But pumping higher takes energy. Why use more power than needed?
I suppose if one is looking for ways to improve aeration rather than reduce heat loss this would be great, but I think if one has enough waterfall to outgas CO2 one would have enough to absorb O2. Two sides of the same coin. If only there were a way to preferentially electrolyze the CO2 back into O2 and, maybe, solid C. I wonder....


Nope: the electrolysis process. I used a dish of brine as a child as a power control for an electric arc I built. It quickly turned into a steaming, bubbling batch of nasty chemicals. (why is there no mad-scientist smiley?)

I am right now watching the movie "Troy" on television starring Brad Pitt as Achilles.

Chemic is my Achilles heel, I admit (I'm impressed by your knowledge).

And you seem to have a talent for explaining. Thanks. I am learning.
(just bear with me, English is not my mother language, neither that of my father, more that of my cousins)

Trying to follow your line of thought in a simplified version.

If mixing H2 and O releases energy, it is quite logical that splitting it requires energy. I now think I have an idea of how much.

So if I am right it would take 140 watts to produce 16 g of O in one hour by electrolysis?
And that would keep 16/.3= 53 kgs of Tilapia happy for that hour?

I have no idea if that is a good energy balance.


Why would you need to pump water higher?
You need to pump the water from the sump to the rim of the top tank in any case.
The rim of the tank is not the water level.
From the top tank it flows to the grow beds.
The top water level in the growbeds is not the level in the top tank.
then it flows to the sump from which it is pumped up again.
Top water level in the sump is not the level in your growbeds.
That should give you three level differences which with good design allow for waterfalls or other aeration enhancements.

As for heat: if mixing H2 and O releases heat, wouldn't splitting it also absorb heat? Isn't that where the energy losses go to? wouldn't the water get colder?

what am I missing?

frank

The pump will use more power pumping to a higher head. End of story.

http://www.drymine.com.au/part1.htm

Its basic mechanics. Shut the discharge valve off (create high head) and the pump will try harder to push the water. Usually making a not very nice noise.

According to http://aqua.ucdavis.edu/dbweb/outreach/aqua/282FS.PDF, yes.


Not sure what you mean by that....


I have been trying to minimize height and pumping energy. The system I've set up would have fish in sump, pumped to growbed, dropping directly into sump. Pumping height of about 18" so the drop from growbed bottom to fish tank surface is only about 8" (say 20cm. This distance is a maximum and will actually vary about 10cm from start to end, so might only be 10cm at start.


Reality has friction: nothing works perfectly.

2 H2+O2--->2 H2O + 286,000 joules
You can run this in a fuel cell and get about 60% as electricity and 40% as heat (because of imperfections in reality you can't get 100% of it as electricity *grin*)

2 H2O + 286,000 joules--->2 H2+O2
286kj of electricity has to actually get into the water molecules and stay there as chemical energy, but this is only about 66% efficient and the rest goes as waste heat, again because of that pesky reality.

DOH! I meant a 20" (50cm) not 18".

I tested this intuitive truth today and was amazed at the result: it is wrong: the higher the head the lower the amps (lower flow as well, of course).

height
meters feet amps
.5 1.5 1.59 3/4" tube
1.5 5 1.32 3/4" tube
1.8 6 1.09 1" tube
3.6? 12 1.03
I didn't bother noting watts, but they did decrease similarly

So, if one is testing the efficiency of a pump one will not only need to measure the head and flow, but the watts as well. Note that watts is not the same as amps x volts because motors are inductive loads (in layman terms, a motor draws more amperage than it seems like it should. Power companies typically don't charge you for this excess, at least in the USA, but it does cost them.)