I clearly didn't understand how this works, but the explanation and claim that it reduce the temperature of room by 5°c made me wonder, here I am trying to understand the Venturi-effect on temperature

This is something I found on web, Using old soda bottle and board they made some e-cooler

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Note:- This system doesn't work, an enthusiastic professor did an experiment to prove it doesn't work. but is it possible reduce the temperature of room or a house using Venturi-effect ?


2 Answers 2


I don't think it will drop the temperature at all.

What it might accomplish is increase the velocity of the air in the room. That will in turn affect the convective coefficient

The convective heat transfer coefficient for air flow can be approximated to (see engineering toolbox link)

$$h_c = 10.45 - v + 10 v^{1/2}$$


  • $h_c$ = heat transfer coefficient ($\frac{kCal}{m^2h°C}$)

  • $v$ = relative speed between object surface and air (m/s)

Because the coefficient will increase the exchange with the bodies in the room will change also.

The Total rate of exchanged heat is:

$$\dot{Q} = h_c A \Delta T$$


  • $h_c$ = heat transfer coefficient ($\frac{kCal}{m^2h°C}$)
  • A is the total exchange surface
  • $\Delta T$ the temperature difference

The $\dot{Q} $ is important because human beings do not understand Temperature directly/ Rather we understand temperature by the exchange heat energy.

So, increasing the velocity of the air in the room, will probably increase the convective transfer coefficient, and (might) make you feel a bit cooler. However, it should not drop the temperature.


It's correct to say that the temperature of the air in the jet coming from the neck of the bottle will be less than the temperature of the air in the bottle. The temperature of the air in the bottle will also be a little higher than the temperature of the outside air. The question is, will both the temperature and relative humidity of the jet air be sufficiently low to result in a net "comfort zone" in the room.

The "comfort zone" is a region of ambient air temperature and relative humidity that induces a comfortable feeling, and it has been mapped out by ASHRAE (American Society of Heating Refrigeration and Air Conditioning Engineers). Such data can be found for instance at https://www.dartmouth.edu/~cushman/courses/engs44/comfort.pdf

I assume this device is meant only for Summer months, when outdoor humidity and temperatures are very uncomfortable. In order to work then, the device must deliver air to the room that will result in the temperature and humidity space according to the ASHRAE standard.

It's clear that the temperature of the jet air exiting the bottles will be a little lower than the temperature of the air in the bottle. That's simply because the expansion process through the nozzle is roughly isentropic. It's very likely that, for normal outside atmospheric conditions, the air pressure in the bottle will exceed room pressure only slightly. That's because the pressure in the bottle can be greater than the air pressure in the room only when outside wind collides with the bottles. Such a collision converts the kinetic energy of the wind velocity to a stagnation pressure. For normal wind speeds, however, this increase in pressure is small.

The velocity of the air jet into the room will thus be small, and the net temperature difference between outside air and air jet velocity will be small. Considering also that when the air jet mixes with room air, it's kinetic energy is dissipated, being converted to heat. The net effect is that the final temperature of the jet air after it mixes with room air will be slightly elevated above its temperature when it exits the bottle. The conclusion here is that there can be only a minimal sensible cooling effect from this device. Most importantly, sensible cooling is only part of the air conditioning process, and the real question is what's the story with relative humidity.

Often the most important process of air conditioning is the removal of water vapor from the air. The ASHRAE graphs show that the relative humidity necessary for comfort is often much less than the relative humidity present in Summer outside air. What happens to the water vapor (absolute humidity) of the air that's pushed through these bottles? Unfortunately nothing much.

In the vast majority of air conditioning processes, water vapor must be removed from the outside air before it can become comfortable for humans. Water removal is usually the most intensive process in air condition, not the reduction in (sensible) temperature. This device offers little mechanism for the removal of water vapor. The very slight increase in bottle pressure and the relatively low pressure drop encountered by the air in moving through the bottles ensures that the water vapor in the outside air remains in the air pushed into the room. If the jet air temperature would be lowered to a value below the dew point of the outside air, condensation would remove some of this water vapor. It's very unlikely that such a device could accomplish such low temperatures, for the reasons already explained. The net result is that this device pushes more water vapor into the room - not a very good means for conditioning outside air for comfort.

Bottom line is that for much of the typical Summer months in areas that require air conditioning, this device will not be able to produce a comfort zone, as defined by ASHRAE standards.

  • 1
    $\begingroup$ There is no pump. It is passive wind powered. See Step 5 of the instructions. $\endgroup$
    – Transistor
    Commented Sep 25, 2020 at 18:35
  • $\begingroup$ Transistor, thanks. I missed that point. I'll modify my answer accordingly. $\endgroup$
    – ttonon
    Commented Sep 26, 2020 at 21:36

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