# Help understanding chemistry (an idea for harnessing energy)

1. Jan 20, 2010

### andrewbb

What causes H20 molecules to rise through evaporation or boiling?

My theory is the H20 molecule is separated from other H20 molecules through the introduction of heat or perhaps air flow. Then the electrical charge of the H20 molecule is attracted to the electrical charge of another atom or molecule in the air. That creates a balloon effect that causes it to rise. Can someone help me understand what is happening at that stage of a water molecule rising from the water's surface into the air?

I'd like to identify the specific charges on each atom/molecule that is creating that temporary cohesion. Once those charges are identified, a chemical engineer could make the process very efficient.

See the attached.

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2. Jan 20, 2010

### Yitzach

What exactly is the capillary action doing? Capillary action carries water upward.

Water molecules have a vibration distribution that is equal to the black body distribution of the corresponding temperature. As the temperature rises, more of the distribution is past the point of evaporation. As more of the distribution gets past the point of evaporation, the fast the water evaporates.
The point of evaporation is determined by the amount of energy required to break the "hydrogen bonds" and the London forces. Hydrogen bonds are "70% covalent bond" according to my chemistry prof. This allows the hydrogen switch from one oxygen to the next creating hydroxide and hydronium.

3. Jan 20, 2010

### andrewbb

What is happening to each water molecule as it is carried upward?

As temperature rises, the molecules become more dispersed. The molecule can then attach itself to a molecule from the air above based on its electrical charge. That creates a balloon effect and carries both molecules upward.

IOW, I'd like to understand the micro look at how capillary action works.

4. Jan 21, 2010

### Char. Limit

The average atomic mass of air is about 28.96. The atomic mass of water is about 18, and never mind whether you use grams or AMU, the ratio stays the same.

Because of this, water vapor is lighter than air, and therefore rises.

That's how I see it.

5. Jan 21, 2010

### andrewbb

That's good to know.

So how does that water molecule separate from its cohesiveness to the other water molecules? It must be attracted to another molecule in a certain way.

6. Jan 21, 2010

### Char. Limit

Vapor pressure type stuff, maybe?

I'm not sure about this, but I think water molecules are always escaping and getting recaptured, in equilibrium.

In boiling it's obvious. The molecules are becoming a gas already just from temperature. Evaporation I'm not so sure of...
The exact atomic mass of air, I just found, is... 28.96708801991 g/mol, for pedantry's sake.

7. Jan 21, 2010

### Yitzach

Water vapor rises because the ideal gas law applies. Some corrections could be in order but that is too late to think about. But as the temp of the water vapor rises, it's density decreases and it rises like alcohol on water (for different reasons).

Water condenses because the presence of other water allows the new water molecule finds a lower energy state with the hydrogen bonds and the london forces.
Even when water is boiling, there is probability of condensing back into the pot. This is an equilbrium reaction depending on the partial pressure of water vapor and the tempurture of the water. This is where dew/frost point comes from. The dew point is the temp of the water at which evaporation/condensation are at equilibruim with the partial pressure of water vapor. In the desert where the partial pressure of water vapor is extremely low, the dew point is a frost point. A bowl of water can freeze overnight in a desert, even if the temp did not get below freezing, because enough water leaves the bowl quickly taking large quantities if energy with it. That is called evaporative cooling. If you questions, ask in the morning.

8. Jan 21, 2010

### andrewbb

That helps, but what is happening at the molecule by molecule level? How does a single molecule separate from its cohesive bond and begin floating up? Something above must attract it, yes?

9. Jan 21, 2010

### Staff: Mentor

Single molecules do not float up.

They can diffuse in every direction - including up. That's not the same.

Last edited by a moderator: Aug 13, 2013
10. Jan 21, 2010

### Yitzach

A molecule separates from the bulk because is has enough thermal energy to do so. Likewise, a molecule returns to the bulk because it loses energy to do so and so is the favoured condition.

Our friend is asking about the flow of water vapour upwards when it is hot, not asking about the diffusion of water vapour because a certain area of air is drier.
pV=nRT: p=pressure, V=volume, n=number of moles, T=Temperature, R=gas constant. As the temperature rises, the volume of the gas increases, assuming constant pressure and amount of gas. As the volume increases, the density falls. As density falls, the gas rises in the atmosphere.

The most limiting part of your device is going to be the removal of water vapour from the air in the closed system. Otherwise, the only thing your going to get is a rise in pressure. In which case you would be better off putting your wheel over an open pot/source of water or pressure release valve.
I would recommend adding something that will decrease the energy holding the water down, so that the water will evaporate at a lower energy, allowing the device to be useful in more environments.
The dew point of the vessel should be well above ambient temperature so the environment can naturally remove the water from the air of the device. Or it should be used in an environment where the ambient temperature is significantly below the desired dew point of the vessel.

Wikipedia has an article on capillary action. The height of a liquid is given by:
$$h=\frac{2\gamma\cos\theta}{gr\rho}$$
$$\gamma$$ is the surface tension in J/m^2
$$\theta$$ is the contact angle which the angle from the side of the tube under the water to the top of the water. Contact angle also has an article on Wikipedia.
g is gravity in m/s^2
r is radius of tube in m
$$\rho$$ is density of the water in kg/m^3
$$\gamma, \theta, and\ \rho$$ will be different for your device if you followed my advice and made the water evaporate at a lower temperature.

11. Jan 22, 2010

### andrewbb

While the vibration (heat) of the water molecule is important, I think it is the surface area of the molecule itself that determines if it disconnects from the body of water and connects to a molecule or atom in the air. That creates a balloon effect and both molecules rise. Since H20 is a rather lightweight molecule it continues this process until encountering enough H20 to recondense and drop back into rain.

The two Hydrogen atoms continually slide around the Oxygen atom. So if the only part touching the body of water is O, then the two H + part of the O can touch a molecule in the air and attach through cohesion (likely electromagnetic).

Does that sound reasonable?

12. Jan 22, 2010

### Char. Limit

Except for the fact that there is almost no polar molecules in the air to create a dipole-dipole force with a water molecule...

13. Jan 22, 2010

### andrewbb

I'm not familiary with chemistry, but perhaps it's not a dipole force?

How to explain evaporation? It's not heat so much as contact with air. What overcomes the cohesion of water (which is fairly strong) to make it go up?

14. Jan 23, 2010

### Char. Limit

Excellent idea except for one thing: as the amount of air above the water decreases, the evaporation rate increases. The relationship isn't direct; it's
inverse.

In short, air doesn't help the water up, air keeps the water down.

15. Jan 23, 2010

### andrewbb

That does make sense if you have a vacuum. The water would disperse rapidly.

But, if you were to increase the surface area of the water, it will evaporate more quickly.

Say you have two containers of water. One in a cookie sheet, the other in a pot. Same amount of liquid. The cookie sheet will evaporate faster. Same air. So, it seems it's the surface area of the liquid with the air that is key. Which leads to the idea that the H20 molecules are temporarily binding/attracted to molecules in the air. What do you think?

16. Jan 23, 2010

### Char. Limit

Right idea, wrong conclusion. All you can conclude from that is that if volume is constant, evaporation rate increases as surface area increases. I could, and will, just as easily claim that the reason this is so is that more molecules are at the surface, and, thus, able to escape.

17. Jan 23, 2010

### andrewbb

Escape from the cohesion force of the water? Why would an individual molecule escape from that cohesion force?

IE. What would cause that individual molecule to escape that cohesion force?

18. Jan 23, 2010

### Char. Limit

The kinetic energy of the molecule, defined using the equation $$E_k=\frac{3RT}{2N_A}$$.

19. Jan 23, 2010

### Staff: Mentor

Water molecules escape the water surface in vacuum without problems, they don't need anything to attract them.

20. Jan 26, 2010

### andrewbb

Another design based on the same principles...

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21. Jan 26, 2010

### Staff: Mentor

That is exactly what a hydroelectric dam does. Your apparatus would need to be enormous to generate a meaningful amount of power.

22. Jan 26, 2010

### Char. Limit

Not exactly (I'm from an area with one of the biggest hydroelectric dams in America, and I've studied it), but very close. It would need to be enormous.

23. Jan 27, 2010

### andrewbb

Some initial calculations show that you could build that on the side of a house (2 stories) and generate enough power for the whole house. Total cost of material is rather inexpensive <\$10k so it would have an ROI <3 years. I don't know much about chemistry and the above designs are assuming water/air, but a good chemist should be able to choose more efficient fluids/materials. The key is maximizing the temperature differential between the top and bottom.

24. Jan 27, 2010

### Char. Limit

Could you show us the calculations that give you this expression, please?

25. Jan 27, 2010

### andrewbb

Theoretically at 100 degrees F, 1 sq meter of water evaporates at 1.1 kg/second.

So assuming that can be condensed at the top, that's 1.1 liters/second of flow per square meter of water surface. If the water is 6 meters overhead, potential energy extracted is:

PE = (1.1)(10)(6) = 66 joules/second = 66 watts

At 80% hydroelectric efficiency, that's ~52 watts/sq meter of water.

That's 1250 watt hours per day per liter.

A house needs ~45,000 watts per day. So you'd need 36 sq meters of surface area of water to power the entire house (45,000/1250). You could do that with a 6x6x6 meter cube. Or you could increase the surface area of the water by placing vertical strips of cloth in the chamber to maximize evaporation rate.

It's an engineering challenge to get the condensation rate to match the evaporation rate. That's a function of maximizing temperature differential between the water in the chamber and the condensation rods at the top. Choice of fluid/gas in the chamber and the pressure are important as well. Keeping the water in the chamber hot by burying it or by conducting sunlight to it is more than feasible. Cool rods extending through the top will dissipate heat and increase the condensation rate.

I know the #'s are theoretical and are highly dependent on the condensation rate and diffusion rate of the gas inside (eg. nitrogen), but that seems an engineering challenge.

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