Dave does basic back-of-the-envelope thermodynamic calculations to prove the Fontus Airo & Ryde self filling water bottle campaign on Indiegogo is full of hot air instead of water.
It's 100% baloney!
1,140% of $30,000 goal, oh jeez, yet more mugs out there.....
1,140% of $30,000 goal, oh jeez, yet more mugs out there.....
There is a sucker born every minute. And some of those suckers are probably from the campaigners social circle, to (*cough*) kickstart the campaign so to speak.
Also, tl;dr for those without time to watch the video: Indiegogo, duh...
Also, tl;dr for those without time to watch the video: Indiegogo, duh...
It's almost like that now.
Indiegogo = instant impractical ideas
Kickstarter = probably practical, will more likely fail on financials or production.
Also, tl;dr for those without time to watch the video: Indiegogo, duh...
It's almost like that now.
Indiegogo = instant impractical ideas
Kickstarter = probably practical, will more likely fail on financials or production.
I sometimes wonder whether there will still be anyone with a sincere and diligent project proposal in the future choosing Indiegogo as crowdfunding platform. "You are known by the company you keep" -- it's hard to argue against that...
Shit, this crap is even from my home country it appears.
Oh boy. I can say you one thing, we students here at the technical universities are laughing our ass of about this.
The device truly works, you can even observe the results, the reason why you are skeptical, is because you do not take the proper initial steps prior to viewing the product.
The first step requires you to fill the entire bottle with LSD, and then chug it.
The LSD then helps you to see the machine working as intended.
Without the LSD, logic takes over and thinks look disappointing.
I'm afraid the back of the envolope calculations are wrong. Atleast the 1L/h statement. Whatever time it takes, you always need a minimum of 2264 KJ of energy to produce 1 liter of water. So that's 2264000/60/150= 250 watts for 150 minuts or 2.5 hours. To produce 1l/h you need 630 watt. So 2.5 times worse as you stated. 2264000/3600.
You devide by 2.5 twice for some reason. Very nice video anyway. Always good to see these scams busted. Keep it up !
Thanks Dave for the video. I particularly like your enraged super E letter used in system
Who knows, maybe the solar cell is 100% efficient O0
Where can I find the link Dave mentioned at 14:52?
Sorry Dave, at 4:50 you made an error of units in your calculation, mixing Energy and Power, or missing the evaluated time:
905kJ/l = 251Wh/l (not Watt/l).
So in 150 min = 2.5h, you would 'only' need a 100W panel.
This energy argument also is not directly convincing, if you think of physical 'heat pumps'.. (e.g. inversely running Sterling motor)
These machines have an efficiency, which is by definition >> 1, so you may pump these 905kJ out of the vapor by a fraction of equivalent in electrical energy, theoretically.
A practical machine of that sort (~ 40..60°C, 100% humidity) is the tumble dryer, which needs about 1.6kWh to condense several liters water out of the laundry.. if you have a modern device (EU energy rating A+++) with heat exchanger.
So the most obvious practical argument would be, that this device does not consist of the necessary miniaturized parts, like heat pump, condenser, heat exchanger, and so on.
Frank
I lost it when it became clear Dave had put a stamp + name and address on the back of envelope calculations so he could mail it.
Don't be too harsh on those people. They just forgot to mention the capsule with LIQUID NITROGEN to cool the hot side of the peltier-element.
Happens even to the best
Sorry Dave, at 4:50 you made an error of units in your calculation, mixing Energy and Power, or missing the evaluated time:
905kJ/l = 251Wh/l (not Watt/l).
So in 150 min = 2.5h, you would 'only' need a 100W panel.
frank
If you keep on deviding by 2.5 at some point this becomes a viable device.
Dave, in order to produce water from steam you have to take energy out of steam, not to bring it. You have to take 250 W from the freezer in the bottle - it's a different case.
Lat's do the math.
dew point for 40C 90% is Tc = 37 C = 310.15 K
ambient temperature Ta = 40C = 313.15 K
Diameter of a hot radiator d= 6cm= 0.06m
Speed of a cyclist = 15 km/h = 4.16 m/s
Isobaric volumetric heat capacity of air C=0.0013 J*cm^-3*K^-1 = 1300 J*m^-1*K^-1
So for Carnot cycle (ideal case) we have
W = Wh*(1-Tc/Th)
where W = power of the heat engine, Wh - Power of the heat source, Tc - temeperature of the cooler, Th - temperature of the heater.
Engine produces power W taking power Wh from the heat source and bringing part of it Wc to cooler.
As long as Carnot engine is reversable we can bring power to it taking power Wc from the cooler and moving Wh=Wc+W to the heater. We know that we have to move Wc=250W from the freezer. Let's find how much power W we need for that.
We know that Tc=310.15 K because it's a dew point we need to get water.
Let's find Th.
All Wt has to be blown out by air, moving through the radiator of the system.
Every second through the cylinder d=0.06 moving with a speed 4.16 m/s passes V=0.01 m^3/s of air.
If heat exchange is perfect air comes out with temperature Th and moves away
(Th-Ta)*C*V of power
Wh = (Th-Ta)*C*V
on the other hand Wh = W/(1-Tc/Th) = (Wh-Wc)/(1-Tc/Th)
Wh=Wc*Th/Tc
(Th-Ta)*C*V = Wc*Th/Tc
Th = V*C*Ta/(V*C - Wc/Tc) = 333.8 K
W=Wc*(Th/Tc-1) = 19 W (to move 250 W from the bottle)
What's the size of a 19 W panel?
Of course, if they want to use Peltier cooler with a 10% of efficiency of the Carnot cycle, they have to find 190 W of power. And because of that additional power we have to dissipate, our hot radiator will be much hotter and it leads to even less efficient cooling which leads to even more power consumption and so on.
And what about the water in the bottle continuously evaporating and recondensing on the peltier? Lots of wasted energy.
In ideal case in thermally isolated bottle we have a water with temperature of the dew point and 100% humidity. But in real life it will take some energy of course.
And then there's this paper from the "International Journal of Water Resources and Arid Environments" that draws a litre of water an hour using the same basic technology at 120w of solar power. Given they are likely at high temp and humidity..
But they use multiple peltier devices and they run the air first through the hot side heatsink so it can gain more moisture, and then run it through cold side heatsink.
But still published results of a liter per hour at just 120w solar.
http://www.psipw.org/attachments/article/273/IJWRAE_1%282%29142-145.pdf?
Could it be made even more efficient with better heat transfer technology in larger surface area heatsinks?
Yep. If we use cooled and dehumidified air for cooling of the hot radiator we will get Th=330
And W= 16.5 W
Could it be made even more efficient with better heat transfer technology in larger surface area heatsinks?
From equation I got follows that yes.
But Israel uses reverse osmosis for water harvesting. I think that they did the math too and it's more efficient.
Basically water is not a problem in wet environment. It's usually a problem in dry climate. And you cannot harvest water from dry air. Dew point for 40C 20% is 13C. For 40C 10% is 3C !!!
Yeah there is a lot of reverse osmosis water harvesting in that whole region. Seems to be the dominant technology, I assume that is for several reasons. Sailboats use R.O. pretty exclusively as well they call it a "watermaker", and you can even buy a hand pump R.O. system for your emergency lift raft or whatever.
In a survival situation I could likely use a couple of plastic garbage bags to harvest enough water to survive on using condensation, probably more than the indiegogo device can make.
I am curious though if that liter per hour at 120w they got in the paper could be refined and reduced to maybe 70w?, their heat sinks appeared to be off the shelf hack together stuff. How much room for improvement is there in improving thermal flow and surface area?? 10%?, 20%? 50%?