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To break the vortex into bits, we create a very simple flow system - the building block of a vortex, or its discrete element.

The building block of a vortex

Consider a duct with compressed gas tank mounted to the inlet:

duct1

We shall move from simple to more complex concepts. For this reason, let's first study the behavior of the above duct system when the system is stationary. Then we can move on to the rectilinear motion of this duct system. Last, we can study its rotational motion.

The duct has adiabatic walls - no heat comes in or goes out through them. The same applies to the tank. The gas in the tank is stored at some high pressure and has room (static) temperature.

How does the duct system behave when not moving?

Let the system be stationary in the stationary frame of reference F. At some moment, gas is allowed to flow through the duct and leave through the outlet. This type of gas flow in adiabatic duct is known as "Fanno flow". The following takes place:
1.at inlet, gas velocity is lowest; at the outlet, gas velocity is highest
2.at inlet, gas pressure is highest; at the outlet, gas pressure is lowest (pressure difference drives the flow)
3.at inlet, gas density is highest; at the outlet, gas density is lowest
4.at inlet, the static temperature of the gas is highest; at the outlet, the static temperature of the gas is lowest
5.at any position in the duct, the mass flux is the same (continuity condition, mass conservation)
6.at any position in the duct, the total temperature is the same (adiabatic condition, enthalpy conservation)

We are specifically interested in the temperatures. At inlet, gas velocity is close to zero; at outlet, let the gas velocity be c. Then at inlet, the total temperature is equal to the static temperature of the tank. Denote this temperature with

T_inf

Important!!!

The static temperature of the tank gas is denoted with "T-infinity" to show that it is the same as the ambient temperature (the temperature at infinity). This is a substantial part of the explanation of the vortex tube effect. Vortex tubes work with compressed, ambient-temperature air. But this air has first been compressed to some high pressure with a compressor. In the process, not just the pressure, but also the static temperature of the compressed air rises. Then the tank is left to cool to ambient temperature, which also leads to some decrease of its pressure. In this way, some invested energy is lost and this air is in essence, "pre-cooled" - its mere expansion to atmospheric pressure will already result in cold air. The vortex tube adds resistance to this expansion, thus cooling it even more, as we shall see below.

The total temperature at any cross-section of the duct is thus equal to this same value "T-infinity", including at outlet. At inlet,

eq1

At outlet,

eq2

Then

eq3

Then the observer in the stationary frame of reference F will see total temperature difference

eq4

and static temperature difference

eq5

This means that if the gas at the exit is stopped adiabatically, it will show the same thermodynamic temperature as the tank (all friction is neglected).  Therefore, no cooling is observed yet.

(We neglect any cooling due to expansion before the gas reaches the inlet of the duct. At inlet, the gas is assumed to have exactly tank temperature, which, if the gas is injected leads to certain expansion and thus some cooling in real, non-ideal systems; in these, the static temperature difference for this reason can be slightly bigger)

What is the rectilinear motion of this duct system like?
 

We set the system in uniform rectilinear motion; the tank is at the leading end. Let the velocity of the system be equal to c. The inertial laboratory frame of the moving duct we denote with F'. The stationary frame we denote by F. Since the moving system F' is inertial, there is no acceleration. For the observer in F' the flow through the duct takes place just as before when the system was not moving. That is, the observer in F' sees Fanno flow in the duct. See all details above for the non-moving duct.
 

Does the moving observer see cooling?

No.
The observer in the moving frame of reference F' will see total temperature difference

eq6

and static temperature difference

eq5

This means that if the gas at the exit is stopped adiabatically by the observer in F', it will show the same thermodynamic temperature as the tank (all friction is neglected). Therefore, no cooling is observed F'.

Does the stationary observer see cooling?

Yes!!! Here is why.
First, let's look at the ejected gas. In the stationary frame, the duct system moves with velocity c, while in the moving frame, the gas leaves the duct with velocity c (in direction, opposite to the motion). This means, that the ejected gas has velocity of zero in the stationary frame.

Second, let's look at the moving gas tank. It moves with velocity c and the gas in it has static temperature

eq7

Then the stationary observer concludes, that the gas in the tank and at the inlet has total temperature

eq8

As we saw above, the static temperature of the gas at the outlet of the duct is

eq9

because

eq10

But this static temperature is the same in all frames of reference. This means that the stationary observer sees the ejected gas to have total temperature
 

eq11

Therefore, the observer in the stationary frame of reference F will see total temperature difference

total_T_gradient

eq12

(let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 108.25K separation in total temperature) and static temperature difference

static_T_gradient

eq5


(let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 54.1K maximum cooling of air, not counting any expansion effects when the gas may be injected from the tank into the duct inlet, similarly to the injection done in vortex tubes, where the maximum cooling may be somewhat larger due to nozzle expansion)

This means since the gas at the exit has been stopped adiabatically, it exhibits a drop in thermodynamic temperature compared to the tank (all friction is neglected).  Therefore cooling is observed F. Not just cooling, but "temperature separation" is also observed. The leading end of the duct system appears hot, while its trailing end appears cold. This conclusion comes from the values of the total/static temperatures at the two ends of the system.
 

Where did the energy go?

The observer in the stationary frame F sees a gas parcel with high initial kinetic energy and room (static) temperature. Upon exiting the moving system, not only has the parcel stopped, it has also been cooled! This goes against intuition. Usually, when a moving parcel is forced to stop adiabatically, its static temperature goes up. The conclusion is:

The energy of the gas parcel was delivered as propulsion to the moving system!


Discretization of Vortex Tube Flow : Rotational Motion

The discrete vortex element (the duct) rotates about an axis with uniform angular velocity:

rot_duct_general1

The rotation axis is perpendicular to the duct; the rotation axis is at the duct outlet. Gas from the tank flows towards the center of rotation. As before, the static temperature of the compressed gas in the tank is

T_inf

Important!!!

The static temperature of the tank gas is denoted with "T-infinity" to show that it is the same as the ambient temperature (the temperature at infinity). This is a substantial part of the explanation of the vortex tube effect. Vortex tubes work with compressed, ambient-temperature air. But this air has first been compressed to some high pressure with a compressor. In the process, not just the pressure, but also the static temperature of the compressed air rises. Then the tank is left to cool to ambient temperature, which also leads to some decrease of its pressure. In this way, some invested energy is lost and this air is in essence, "pre-cooled" - its mere expansion to atmospheric pressure will already result in cold air. The vortex tube adds resistance to this expansion, thus cooling it even more, as we shall see below.

The rotating reference frame is F', the stationary frame is F.

What happens in the rotating frame?

The rotating observer notices that the compressed gas in the tank has ambient static temperature. The gas in the duct moves against the centrifugal force, as if "climbing up" a gravitational well. In doing so, it does work. This work can come only from the pressure energy contained in the gas. For this reason, as the gas moves towards the center, it expands, loses internal energy and becomes cold.

What happens to the total temperature T of the gas? In the "primed" frame F', this temperature is denoted by T' and is known as "relative total temperature", meaning "the total temperature as perceived by the rotating observer". Because the rotating frame is not inertial, the laws of physics here have different mathematical form. Remember the definition of "relative total temperature" in moving inertial frame? It was

T_rel

where v' is the "relative velocity", or the gas velocity relative to the moving observer. In a moving, inertial frame T' is conserved throughout the duct. This meant, that the relative total temperature is the same at any cross-section of uniformly moving adiabatic duct. But in the non-inertial, rotating frame this is no longer true! In rotating frame, there is another temperature that is conserved. It is called "rotary total temperature" or "rotary stagnation temperature". Total temperature is related to the total enthalpy of the gas; Rotary total temperature is related to the rotational total enthalpy of the gas, also known as rothalpy. Therefore, the temperature that is conserved in rotating frames is

T_rot

the rotary total temperature; v' is the relative velocity of the gas, omega is the angular velocity of the frame and r is the radial position. We omit the derivation of Trot; it can be derived from energy conservation considerations. At this time, it will suffice to say that in a rotating reference frame, at any cross-section of the adiabatic duct, the rothalpy of the gas is conserved, e.g.

rothalpy_cons

 

What happens in the stationary frame?

The stationary observer monitors the total and static temperatures of the rotating gas. To derive an expression for the total temperature in F, use the velocity addition formula

velocity-addition

where V is the velocity of the gas in the stationary frame. Now express v' in the velocity addition formula,

velocity-addition1

substitute it into the conservation of rothalpy statement

deriv1

deriv2

cancel the squared cross product and get

deriv3

in other words

deriv4

Now notice the sum of the first 2 terms is the definition of total temperature in the stationary frame; also cancel the factors of "2" in the 3rd fraction and thus

deriv5

This is the vectorial form of Euler’s turbine equation, as shown in the previous section. This equation shows a quantity seen in F that is conserved at any cross-section of the rotating duct! Now, all ejected gas begins its motion at the same radial position R: the duct inlet. Denote the linear speed of the inlet (the linear speed of the rotating inlet, as seen in the stationary frame F) with c,

omega_r

Look again at the conservation condition above. At the outlet, r=0; at inlet, v=c; then

deriv6

which gives

eq12

The observer in the stationary frame sees this total temperature difference between inlet and outlet! (let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 108.25K separation in total temperature). What is the static temperature difference? At inlet,

deriv7

At outlet,

deriv8

Express the static temperatures at inlet/outlet through the total temperatures at inlet/outlet and take the difference:

eq5

precisely as in the case of rectilinear motion!

In the rotating frame of reference, the observer can see a room temperature reservoir; and all gas along the duct gradually becoming cold in both static temperature and relative total temperature (the total temperature observed in the moving frame). It is important to understand, that in the rotating frame, the entire system is COLD. The warmest spot in it is the tank at room temperature

T_inf

This means radial heat transfer is not the reason for the radial temperature separation.

The stationary observer sees

rot_duct_gradient1

in both static and total temperature! (let c = 330 m/s, just below sonic. For air, cp=1006 J/(kg.K) --> 54.1K maximum cooling of air, not counting any expansion effects when the gas may be injected from the tank into the duct inlet, similarly to the injection done in vortex tubes, where the maximum cooling may be somewhat larger due to nozzle expansion)

In total temperatures, there is HEATING at periphery and COOLING at center! Air trajectories are SPIRALS! No doubt, the physics of this system is highly relevant to the vortex tube effect.

Where did the energy go?

The observer in the stationary frame F sees a gas parcel with high initial kinetic energy and room temperature (at periphery). Upon exiting the rotating system, not only has the parcel stopped, it has also been cooled! This goes against intuition. Usually, when a moving parcel is forced to stop adiabatically, its static temperature goes up. The conclusion is:

The energy of the gas parcel was delivered as propulsion to the rotating system! This is ANGULAR PROPULSION.

How To Model the Temperature Separation With CFD

How to prepare the geometry

The geometry is simple - a straight duct. The effect does not depend on size, thus choose the duct to be as long as you wish. For example, it can be chosen to be 15 m long, having rectangular cross-section of 0.3 m x 0.4 m (width x height). One end of the duct is the inlet; the other - the outlet. In this case, the geometries were prepared with Gambit (currently available only with legacy license, soon to be phased out). Alternatively, ICEM or other meshers may be used to build the grid for the duct. Here is a segment of meshed duct:

duct-mesh

How to position the duct

Make sure to position the duct outlet face so that it goes through the origin (0,0,0) and is parallel to one of the coordinate axes, e.g. z. This will be the rotation axis. The direction of rotation (clockwise vs. counterclockwise) is unimportant. The face at (0,0,0) is outlet, the face at the other duct end is inlet.

What are the CFD simulation parameters in Ansys FLUENT?

The FLUENT solver parameters are:
•3d, double precision
•density-based solver, steady simulation (steady-state solution)
•energy equation is ON
•viscous model: k-epsilon standard or realizable; k-omega SST is best; standard wall functions
•fluid: air, ideal gas
•operating conditions: all set to zero, no gravity
•fluid zone: set as "frame motion", choose rotation axis and constant rotation rate that will give c < 330 m/s
•Solution: "second order upwind"
•start with low Courant number and URFs: below 1.
•set all monitors to 1E-19

The FLUENT boundary conditions on the duct are:
•central outlet (exit) : pressure outlet, choose your pressure value (this value does not influence the cooling)
•inlet: either mass-flow-inlet or pressure-inlet.
•inlet reference frame: relative to adjacent cell zone. This ensures zero incidence angle of the inlet flow and proper inlet velocity magnitude!
•inlet flow direction: normal to boundary. Choose inlet temperature of 300 K.
•duct walls: no-slip; moving wall, rotational motion with 0 rad/s relative to adjacent fluid cell zone.

Ansys FLUENT simulation results:

rstraight_duct_15m_1a

FLUENT should predict cold temperatures very close (within 1.5 %) to their theoretical value given by the above formula.
•In total temperatures, heating and cooling should be symmetric with respect to the inlet temperature
•the temperature separation in thermodynamic (static) temperatures is 1/2 of the total temperature separation.

Engineer_Stacked_Reversed

Horizontal_Rev

Large parts of this work were accomplished without funding. If you find the information on this site helpful, please consider donating to this project.

 

The information contained in this site is based on the following research articles written by Jeliazko G Polihronov and collaborators: