Chinese scientists demonstrated anti-gravity pump, which uses capillary effect and super-hydrophobic screen to flow water upward by 1cm using internal energy of water.

How much water will be cooled when lifted to 10m OR how high water need to flow up to cool by 1K?

How much of that energy can be extracted back using power generator?

How much of energy can be extracted by cooling water from 20C to 10C?

  • $\begingroup$ What Chinese scientists? Do you have a link to a published article or a news article about it? $\endgroup$
    – Kyle Kanos
    Jun 18 '15 at 12:48
  • 1
    $\begingroup$ @KyleKanos: I've added a link $\endgroup$ Jun 18 '15 at 15:21
  • $\begingroup$ The thermodynamics on this problem says that the efficiency of this process is so low that it is a no-go. With a 20-25 C temperature difference (somewhat bigger than your 10 deg C temperature difference), the electricity costs approximately $0.25/kw-hr, and it only gets worse as the temperature difference goes down. See en.wikipedia.org/wiki/Ocean_thermal_energy_conversion $\endgroup$ Jan 24 '17 at 2:32

It doesn't work by using the internal energy of the water - it works by using the surface tension of the water.

If we take a water droplet of radius $r$ the pressure inside the droplet is:

$$ P = \frac{2\gamma}{r} $$

where $\gamma$ is the surface tension of the water (about 0.073 N/m at room temp). Similarly if we take a column of water of height $h$ the pressure at the bottom of the column is:

$$ P = \rho g h $$

where $\rho$ is the density of water and $g$ is the gravitational contact. So to find the height at which the pressure is the same as in the droplet we just equate the two pressures:

$$ \rho g h = \frac{2\gamma}{r} $$


$$ h = \frac{2\gamma}{\rho g r} $$

For example, if we take a water droplet with a radius of 1 mm we find it has the same pressure as a water column with a height of about 1.5 cm.

And this is how the pump works. The superhydrophobic grid just acts as a base for the water column because the water won't wet it:


Suppose a droplet of water of radius $r$ touches the grid. If the pressure is higher than at the base of the column, i.e. if $h \lt 2\gamma/(\rho g r)$ then the droplet will merge with the base of the column and increase the pressure, which pushes the column up. We can keep adding droplets and the column will keep rising until $h = 2\gamma/(\rho g r)$.

We should note there is not perpetual motion effect in this pump. It takes energy to create the water droplets because we have to do work against the surface tension. This energy then goes into increasing the gravitational potential energy of the water column. In the video in the linked article a pump is being used to create a stream of small droplets, and it's the work done by the pump that ultimately goes into pudhing the water up.

  • $\begingroup$ But, can we restart this process at higher level and stack it to be 10m high? $\endgroup$ Jun 18 '15 at 21:33

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