(*Calculate the conductance of Rb table's differential pumping tube*)

(*transmission probability for tube of length h and diameter d
see Santeler, Journal of Vac. Sci. and Tech., 1985 Vol 4 issue 4 p 338*)
(*       α  =  (1/(1+(6/8)(((1+(1/(3+(6/7)(h/d)))))/(d/h))))      *)

(*addition of multiple components of differing diamters
(Subscript[α, i]) and (Subscript[A, i]) are transmission probability and cross sectional area through ith element *)
(*   (1/(Subscript[A, 1]))((1/(Subscript[α, total]))-1)=(∑_(i=1)^n)((1/(Subscript[A, i]))((1/(Subscript[α, i]))-1))+ (∑_(i=1)^n)((1/(Subscript[A, i+1]))-(1/(Subscript[A, i])))  *)

(*Units and stuffs *)
J=kg*m^2/s^2
GasConstant = 8.3144621 * (J/mol/Kelvin)
Temperature = (20 + 273.15) Kelvin
MolarMass =0.0854678 kg/mol
m = 1000*mm
mm = (liter/1000000)^(1/3)
inch = 25.4*mm

(*Define parameters of differential pumping tube, starting at small end *)diameter = {.125,.156,.188,.234,.266,.297,.344,.375,.406,.438,.484,.516}
length = 0*diameter + 0.5
length[[1]]=0.79

(*if you want to reverse to check conductance in the other direction*)
diameter = Reverse[diameter]
length=Reverse[length]

(*do the calculation*)

A = Pi*(diameter/2)^2//N

α = (1/(1+(6/8)(((1+(1/(3+(6/7)(length/diameter)))))/(diameter/length))))//N

transprob =x /.N[Solve[(1/(A[[1]]))((1/x)-1)==(∑_(i=1)^(Length[diameter]))((1/(A[[i]]))((1/(α[[i]]))-1))+(∑_(i=1)^(Length[diameter]-1))((1/(A[[i+1]]))-(1/(A[[i]])))*If[A[[i+1]]< A[[i]],1,0],x]]

(*multiply by input aperture conductance to find total conductance*)
conductance = Simplify[A[[1]]*inch^2*(Sqrt[((GasConstant*Temperature)/(2*Pi*MolarMass))])*transprob,s>0]

(*notice that overall conductance it doesn't depend on direction!*)