
Ring Physics The Geometry and Algorithms of the Control Volume To the Tune of Salty Dog © 1999 by Ed Seykota
Ring to Ring Flow Computation Algorithm The goal is to update the mass [M], pressure [P], temperature [T] and momentum, [X] states of the control volume in response to the flows occurring in and out during a small time interval [dT]. Rather than using the standard approach (balancing the work done as the gas flows in and out of the control volume) consider the flows in and out are the result of Radial Momentum. As such, assume the gas arriving and leaving carries the of the control volume from which it originates. Assume the gases will wind up occupying volumes proportional to their masses. So, first perform and adiabatic expansion (or compression) of both gasses into these volumes. Then combine the two gasses to find the new total. Find the starting volumes from the gas velocities and their control volumes: Volume_old_1 = Volume_Total * (1  Velocity_old * dT / dR) Volume_new_1 = Volume_Total[r1] * (Velocity[r1] * dT/ dR) 
Find the masses of the gasses from their volumes and densities:
Mass_old = Volume_old * density_old
Mass_new = Volume_new * density_new
Find the new volumes by partitioning the control according to mass:
Volume_old_2 = (O  I) * Face * M_old / (M_old + M_new)
Volume_new_2 = (O  I) * Face * M_new / (M_old + M_new)
Expand (or compress both gasses adiabiatically) into their new volumes:
Pressure_old = Pressure_old * (Volume_old_1/ Volume_old_2)^{1.4}
Temperature_old = T * (P/P)
Pressure_new = Pressure_new * (Volume_new_1/ Volume_new_2)^{1.4}
Temperature_new = = T * (P/P)
Mix the Gasses:
Mass_total = Mass_old _ Mass_new
Pressure_total = (Pressure_new * Mass_new + Pressure_old * Mass_old) / Mass_total
Temperature_total = (Temperature_new * Mass_new + Temperature_old * Mass_old) / Mass_total
Momentum is the sum of the momenta:
Momentum_total = Momentum_old + Momentum_new
Internal Ring Activity Computation Algorithm
Within a ring, the states change through drag and pressure acceleration.
Drag is the resistive pressure of the air against the table and the disk. At low Reynolds this is generally a viscosity effect based on the nonslip property. At high Reynolds number friction is generally a kinetic roughness effect. See Friction Studies for more details. Drag is proportional to the contact surface area (the sum of the top and bottom areas). The pressure gradient across the ring width [dR] times the face normal to the flow gives the resistive force. The resulting acceleration is the drag from the top and bottom faces divided by the mass in the ring. This acceleration decrements the forward velocity over the next time interval [dT]. The drag also generates internal thermal power equal to the acceleration times the momentum. Units: m/s^{2} * kgm/s = kgm^{2}/s^{3}. This thermal power increments the thermal energy of the gas and raises the temperature.
Drag_Force = Drag_Pressure * Face_top
Drag_Acceleration = Drag_Force / Mass
Velocity = Velocity  Drag_Acceleration * dT
Drag_Power = Drag_Acceleration * Momentum
Thermal_Energy = 3/2 * Mass * k * Temperature + Drag_Power * dT
Temperature = 2/3 * Thermal_Energy / (Mass * k)
Pressure acceleration is the effect of the pressure gradient across the width of a ring to change the velocity of the gas. As with drag, the force equals the pressure times the face area normal the flow. In this case, however, the force is 1/2 the pressure drop since the average gas molecule is just half way between the inner and outer faces. The resultant acceleration increments or decrements the velocity. In addition, the acceleration attends a power loss that this model carries to debit the very pressure that induced it.
Pressure_Acceleration = Pressure_Drop / Mass / 2.0
Velocity = Velocity + Pressure_Acceleration * dT
Pressure_Power_Loss = Pressure_Acceleration * Momentum
Pressure_Energy = Pressure * Volume  Pressure_Power_Loss * dT
Pressure = Pressure_Energy / Volume
A Palimpsest of Levitator Equations
Variable 
Equation 
Notes 
Face Area (m^{2}) 
Circumference times height 

Frequency (1/s) 
Velocity divided by length 

Mass (kg) 
Accumulate mass flux. 

Mass Flux (kg/s) 
Mass times frequency 

Momentum (kgm/s) 
Accumulate momentum flux. 

Momentum Flux (kgm/s^{2}) 
Momentum times frequency 

Pressure (pa) 
Energy divided by the 1.4 power of volume 

Thermal Energy (J) 
Accumulate thermal energy flux 

Thermal Energy Flux (J/s) 
Thermal energy times frequency 

Turnover (fraction of mass) 
Frequency times delta time 

Top Area (m^{2}) 
Circumference times width 

Velocity (m/s) 

Momentum divided by mass. 
Volume (m^{3}) 

Circumference times width times height 
Gas Physics^{1}  A Review of Basic Principles of Gas Physics.
The Energy and pressure of a molecule are a function of velocity. A molecule bouncing around inside a cube has kinetic energy, E = mv^{2} and pressure, P = 1/3 mv^{2}/V. 
Consider a cube of side L containing a molecule of a gas with momentum mv_{x}. Each time it hits and recoils from a wall, it imparts an impulse, i = 2 * mv_{x}. It repeats this with a frequency, f = v_{x}/(2*L) so the force on the wall, F = f * i = mv_{x}^{2}/L. For a number [N] of molecules, the force is F = Nmv_{x}^{2}/L. Now since the molecule may move in any one of three directions the average velocity [v^{2}] = 3 * v_{x}^{2}. Thus, the pressure on any one wall [P] = F/A = 1/3 Nmv^{2}/V and PV = 1/3 Nmv^{2}.
Since the kinetic energy [Ek] = 1/2 Nmv^{2}, PV = 2/3 Ek and Ek = 3/2 PV.
And since PV = nRT = NkT, E = 3/2 NkT and T = 2/3 E / (Nk)
Note: R = 8.31 J/molK and k = R/6*10^{23} = 1.38 * 10^{23} J/moleculeK
Isothermal Expansion
For isothermal (constant temperature) processes, consider PV = NkT. Since T is constant, P = NkT / V, and pressure is directly inverse to volume.
Adiabatic Expansion
For adiabatic (no heat flows in or out to the control volume) expansion, temperature is free to vary. Temperature generally falls as volume rises and as pressure falls. In particular, PV^{g} = K where g = cp/cv. For diatomic gasses such as nitrogen and oxygen g = 1.4. Therefore P = P_{0}*(V_{0} / V)^{1.4} and pressure drops a little faster than it would if the gas were free to absorb heat and maintain constant temperature.

References: 1. Teaching Introductory Physics, Swartz, 1998, AIP Press, pages 236 and 239.