Radial Momentum
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The Radial Momentum Approach

Radial Momentum induces low pressure.

Radial Momentum

Radial Momentum is the arithmetic sum of the individual momenta of all the particles of a system, taken from the center of mass.

Like linear momentum, Radial Momentum obeys conservation laws. So the fragments of an explosion, or the molecules of a radially expanding fluid, once set in radial motion, tend to stay in motion and maintain the total Radial Momentum, until some force acts on the individual elements. Once fluid molecules start expanding radially, Radial Momentum keeps these molecules expanding into larger and larger volumes ... and the result is that the density and the pressure both fall and induce lift.

I propose to demonstrate, with simple experiments and models, that Radial Momentum, and not fluid velocity, accounts for lift or pressure drop in many systems.

 

Radial Momentum, the Basic Math

Say a stationary 1 kg bomb explodes into 6 pieces. The fragments speed out, radially, in all directions. Its center of mass does not move and so it still has no net linear momentum.  Its fragments, however, all have momentum relative to the center of mass. Each of the fragments travels at 1-m/sec, relative to the center of mass and has momentum, relative to that center of mass of 1/6 kg-m/sec. The total Radial Momentum of the system, then = 1 kg-m/sec.

The basic math is straight forward and proceeds from Newtonian Physics. Two derivations follow, one using a cube for a control volume and the other using a sphere. Both lead to the result, P = 1/3 [MR2  / mV].  The pressure of a fluid, expanding into a vacuum equals one third of the square of the radial momentum, over the product of the mass and the volume.

 

Basic Derivations for Particles Expanding in a Cube

Eqn Operation Formula Units
1 Impulse from particle hitting inelastic wall and reversing direction. Y = 2 * m/6 * v kg-m / sec
2 Frequency of impact of walls of a cube of side, s f =  v / s 1 / sec
3 Force on one wall F = Y * f kg-m/sec2
4 Force, on one wall, from (1,2,3) F = 1/3 * m * v2 /  s kg-m / sec2
5 Pressure on Wall of Area A P = F / A = F s / V kg / m-sec2
6 P, from (4, 5), for volume V P = 1/3 * (m v2) / V kg / m-sec2
7 Radial Momentum = sum(mv) MR =  (m * v) kg-m / sec
8 squaring MR2 = (mv)2 kg2-m2 / sec2
9 rearranging (mv)2 = MR2 kg2-m2 / sec2
10 divide by mass m * v2 = MR2 / m kg2-m / sec2
11 Pressure, from (6, 10) P = 1/3 * [MR2 / mV] kg / m2-sec2
12 Pressure, from 11 P = 1/3 [MR2  / mV] kg / m2-sec2
13 Rearranging MR2  = 3 P m V kg2-m2 / sec2
14 taking sqrt MR = sqrt ( 3 P m V ) kg-m / sec
15 basic physics, Energy E = 1/2 mv2 kg-m2/sec2
16 Energy, from (10, 15) E = 1/2 MR2 / m kg-m2/sec2
17 Rearranging MR = sqrt(2Em) kg-m / sec
18 Pressure, from (12, 16) P = 2/3 [E/V] kg / m-sec2
19 Energy E = 3/2 P*V kg-m2/sec2

 

Basic Derivations for Particles Expanding in a Sphere

20 Frequency of impact, sphere of radius, r f = v / 2r 1 / sec
21 Force on wall, from (1, 3, 20) F =  m * v2 / r kg-m/sec2
22 Area of sphere of radius r A = 4 pi-r2 m2
23 Volume of sphere of radius r V = 4/3 pi-r3 m3
24 Pressure = F / A, from (21, 22) P = 1/4 m v2 / pi-r3 kg / m2-sec2
25 Pressure, from (23, 24) P = 1/3 (m v2) / V kg / m2-sec2
26 definition MR = mv kg-m / sec
27 from 25, 26 (same as 12 above) P = 1/3 [MR2  / mV] kg / m2-sec2

 

 

Radial Momentum, Quantitative Exercises

 

Exercise 1: Expanding Ring

A 1-kg device rests between two plates that are 10 cm apart and which have a vacuum between them. At t = 0, the device explodes and releases energy of 2-joules, all of which carries tiny fragmented particles of the device out from the center of mass, in an expanding ring 10 cm high.  After one second, one minute, and one hour, what is the radius of the ring and what is the particle density of the ring.  

 

 

First, find the particle velocity, from E = 1/2 mv2.

v = sqrt(2E/m) = sqrt(2 * 2-Joule / 1-kg) = sqrt(4) m/sec = 2 m/sec

In the above diagram, height, h = 10 cm

Radius, r = t * v = t * 2 m/sec

Volume, V = 2-pi-r * h = .6282 * r

Density, D = mass / Volume = 1-kg/V

 

Time 1 Second 1 Minute 1 Hour
Radius (m) 2 120 7200
Volume (m3) 1.256 785.384 4523.04
Density (kg/m3 ) .796 .00127 .000221

 

 

Exercise 2: Levitator

Air entering a spool of thread, and exiting against a card below the spool, lifts the card up against the spool. This curious device is an example of a levitator. Compute and display the pressure, density, velocity, mass and mass flux between the plates of a levitator. Explain how the levitator works.

 

Simple Levitator

 

Cross Section of Levitator

 

Computer Simulation of the Levitator

 

The graph shows the values of key variables versus distance in a computer simulation. The center line is at the far left of the graph and the far right is a distance of 20 millimeters, representing the outer edge of the levitator disk. The simulation begins at 1.59 mm, the radius of the hole at the base of the fluid delivery tube.

Since the pump is the flow motivator, the pressure is the highest at the center of the levitator, far left on the graph. The initial velocity inherits from this motivation. As soon as the fluid (air) begins to traverse the gap between the top and the ceiling, it continues to move, by its own momentum, through a series of larger and larger rings. 

As it enters larger rings, its density and therefore its pressure falls, as in example #1, above. The decreasing pressure ahead even further contributes toward accelerating the fluid, so the velocity rises, in a positive feedback cycle. As the fluid continues to expand, its density and pressure continue to fall, eventually falling well below the ambient pressure beneath the card.  The low pressure in this Active Region, accounts for all the lift.  Indeed, in experiments with water as the fluid, using clear plastic plates, the Active Region appears as a white cavitation ring, just outside the intake port. The low pressure draw out tiny air bubbles.

For radial expansion situations, the density and pressure both achieve a minimum in the Active Region and that the pressure drop in that region accounts for most all of the lift.

Meanwhile, at the far outer edge of the disk, the pressure is ambient so the pressure between the plates must also be ambient at the junction. Indeed, to promote the flow from the cavitation ring to the edge of the card, after the Active Effect plays out, there must be a pressure gradient to motivate the flow. This appears as a slightly downward-sloping pressure line from just past the cavitation rang to the edge of the card.

Opposing the radial expansion are two forces. First, friction converts some energy to heat. This shows up as a bifurcation of the trajectories of the pressure and density lines. Second, the back pressure from the ambient air at the edge of the disk provides a net positive pressure slope against which the fluid must climb.

Just as the initially emergent air experiences positive feedback and an attendant increase in velocity, another self-reinforcing phenomenon occurs at the end of the Active Region. As pressure starts to rise against ambient pressure, and low-momentum air ahead, it experiences additional deceleration. This, in turn, further reduces velocity and momentum. This positive-feedback induces a rapid rise of pressure, or hydraulic jump. After the jump, the pressure decreases toward the circumference, per normal gradient flow. The low-pressure within the Active Region contributes all the lift.

One to validate the model is to notice the existence of a cavitation ring in the Active Region. Another is to notice the existence of a hydraulic jump in a similar configuration in which water impacts the back of a smooth dinner plate. Again, the Active Effect is evident between the central column of fluid and the hydraulic jump at a radius of about an inch beyond there.

 

Water Stream on Plate shows Active Effect and Hydraulic Jump

 

A third way is quantitative. While direct measurement of pressure and density between the plates is difficult, in cases where the levitator plate adheres to the table, application of an additional downward force to the plate results in (1) an increase in the gap size,  (2) an increase in airflow and (3) the induction of additional upward force to balance the increased weight. By measuring the gap size, airflow and disk weight, the author found a reasonably good fit with the model.

Full formal presentation of this model, including justifying each equation, and presenting the specifics of the numerical simulation of a set of integral difference equations is out of scope for this paper. It appears here to provide additional insight into the importance of density effects to the induction of lift and to indicate that the theory of Radial Momentum may lead to numerical simulations that produce reasonably accurate results.