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The article shows that to suppress the rotational motion of the air, form a funnel with a diameter of 200 m and the velocity of rotational movement of its walls equal to V = 100 m/s, it is required to pour 80 tons of liquid nitrogen into the funnel.

The region of the rotational motion of the air, which is also observed in the vertical movement of air masses, is called a tornado [

The rotational movement is realized in a vortex under the action of the centripetal force generated by the difference of static pressures (P_{out} and P_{in}) multiplied by the corresponding area of the tornado walls:

We take the diameter of the tornado funnel at the earth surface equal to d_{funnel} = 200 m, and let the height of the funnel be equal to H = 1 km [_{funnel} is as follows:

In [

The vertical axis represents the rotational velocity of the wall of a tornado, which we denote by V_{t}. It can be

seen that the tornadoes have V_{t} within a narrow range of velocities: 60 m/s < V_{t} < 160 m/s.

The horizontal axis shows the value of _{w}―density of the wall of the funnel, _{in}, the volume Vol_{t} of the funnel is_{0} = 1.3 kg/m^{3} is the density of air under normal conditions.

According to the ideas presented in [

Since the relation_{t} can be written as follows:

where the second term shows how many times the mass of the walls of the tornado exceeds the mass of the air pressed out by the tornado. This formula shows that K_{t} is always greater than 1.

Indeed, if K_{t} were less than 1, the tornado would not have fallen onto the earth but sailed in the air. The volume of the walls of a typical tornado is about 5 times less than the volume of the funnel of_{w}/ρ_{0} ≈ 10 - 20, so that a typical value of K_{t} = 3 - 5.

The density of the air and water vapor in the tornado walls ρ_{w}_{1} in this case is only by 3 times higher than the air density under normal conditions ρ_{w}_{1} = 4 g/cm^{2}. Why it happens so is not clear. The dynamical pressure calculated on the Bernoulli formula is equal to the following:

i.e.

The dynamic pressure of the rain drops in the walls of a tornado is equal to the following:

At this pressure the tornado walls cut trees approximately like a rotating blade of the electric razor cuts the hair.

The water in the walls of a tornado is not involved in the creation of the pressure difference_{w}_{1} = 4 g/cm^{3}.

In nature there is a vertical pressure gradient associated with the gravitational field [

where R_{0} = 287 J/(kg∙degree)―the universal gas constant, g―acceleration of the free fall, γ―adiabatic coefficient. For the diatomic gas, i.e.―the air, γ = 1.4, therefore, dT/dx ≈ 10 degree/km.

We call this temperature gradient “static”.

Let the air have a vertical temperature gradient equal to ΔT = 10 degree at a height H = 1 km. We call this temperature gradient “dynamic”.

This temperature gradient will cause a vertical pressure gradient equal to

where P_{0} = 10^{5} Pa―the normal atmospheric pressure, ΔT = 10 degree―the temperature gradient, T_{0} = 300 K is the room temperature. Then from (6) it follows that

This vertical dynamic pressure gradient will result in vertical movement of the air. The velocity of this vertical air movement can be determined from the following ratio:

where ^{3}, H = 1 km, d_{funnel} = 200 m. Thus, the vertical velocity is equal to V_{vert} = 17 (m/s).

It is not difficult to show that the vertical movement of the air with a velocity V_{vert} = 17 m/s is able to rise up a water balloon with a diameter of 1 cm.

Indeed, the dynamic pressure in this case is equal to the following:

When the cross-section of the ball S_{tr} ≈ 1 cm^{2}, the lift force for this ball will be equal to 1.5 × 10^{3} dn. In the units of mg it is equal to 1.5 gram forces. That is more than the force of gravity acting on a water balloon with a diameter of 1 cm.

The lighter warm air rises up inside the tornado, lifting up the items which happen to be inside the tornado.

An obvious way to disrupt the rotational motion in a tornado is to create the pressure difference P_{in-out} equal to the magnitude of the difference between the static pressures. Then the centripetal force will disappear―it holds the water in the walls of the tornado on circular orbits and then the mode of the rotational motion of rain in the walls of the tornado will fail.

Let us calculate the quantity of liquid nitrogen to be poured inside the tornado to cease the rotational movement of the tornado walls.

The calculations will be carried out for the vertical tube section with a height of H_{1} = 50 m. We will find the required conditions to break the tornado “trunk” and lift it up over the ground to a height of H_{1} = 50 m. In this case this area will be filled with non-rotating air and the pressure difference P_{out} − P_{in}, which causes the rotational motion, will disappear.

Moreover, it is not necessary to reduce the pressure difference to zero inside and outside of the tornado funnel. According to the graph shown in

Assuming that the density of nitrogen in the gaseous state is equal to [

This means that in order to fill the funnel having a volume of 1.5 × 10^{6} m^{3} with nitrogen and create the extra pressure P = 0.1 atm, it is necessary to implant 1.5 × 10^{6} × 1.25 × 0.1 ≈ 200 ton of liquid nitrogen into the funnel. In the liquid state this volume of nitrogen will be equal to

The heat capacity of nitrogen is [^{10} J of energy.

The heat of the phase transition from liquid―gas for nitrogen is equal to [^{10} J. Thus, the total energy expenses to evaporate nitrogen and heat it to the temperature of 25˚C, is equal to 9 × 10^{10} J.

The heat capacity of air is 1 kJ/(kg∙degree), and the air inside the segment having the length of 50 meters, has a mass of 1.5 × 10^{6} m^{3} × 1.3 kg/m^{3} = 2 × 10^{6} kg. So, to cool this air mass by one degree, it will be required to spend 2 × 10^{9} J/(1 degree) of energy.

Evaporation of liquid nitrogen and its heating will result in cooling of the total mass of the air inside the funnel: 9 × 10^{10} J/[2 × 10^{9} J/(1degree)] by 45 degrees of Celsius.

This will increase the pressure inside the funnel: _{vert}_{1} ≈ 10^{5} × (45/300) = 1.5 × 10^{4} Pa. This pressure is higher than the pressure of 10^{4} Pa, which is created by the extra mass of nitrogen. So, to break the rotational motion of the air and water in the walls of a tornado, it will be required to use a less mass of nitrogen―about 80 ton.

Taking into account all the above, it is clear that the use of liquid nitrogen can result in breaking of the rotational movement of air inside the tornado. Perhaps, the height of H_{1} = 50 m, to which, as we consider, the tornado “jumps” is too high. The height of 10 meters may be enough for the tornado “to jump” and it will require a smaller amount of liquid nitrogen than we have calculated.

Sergey Niikolayevich Dolya, (2015) An Opportunity of Failure of the Air Rotational Motion. Open Access Library Journal,02,1-4. doi: 10.4236/oalib.1101403