Asking gas spring model

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sungyeol kim
sungyeol kim on 28 Aug 2022
Commented: Vijay on 8 Sep 2022
Hello all,
I'm newb here.
I have no idea how to make the fomula of Gas Spring model.
Hope you guys can give me ideas.
Gas spring model (sqrt = square root, ddt = double dot, dt = single dot)
△T =0, T1=T2
Costant Numbers
K=0.5229sqrt(ºR) /sec
γ=1.4 (ratio of specific heat)
R=661.9 in/ºR
Pcr = 0.5283 (critical pressure)
Cd = 0.7 (drag coefficient)
ρ = P/RT
ºR=460+ºF
ºF = 32+0.18 ºC
Variable Numbers
P0 : initial pressure
x0 : Original piston position
Fμ : Frition of piston
A0 : orifice setion-area
Pi,Vi,Ai : Pressure, Volume, Section-area of piston compartment
F : Force on piston load.
Equation of Motion (ddt = double dot, dt = single dot)
Mxddt = (P2A2-P1A1) - Fμ -F
where, P1dt = (-m12dt + ρA1xdt)*γRT1/V1
P2dt = (m12dt - ρA1xdt)*γRT1/V1
if P1≥P2, m12dt = (Cd*A0*K*P1*N12)/sqrt(T1)
N12 = [((P2/P1)^2/γ - (P2/P1)^γ+1/γ) / (((γ-1)/2)*(2/(γ+1))^(γ+1)/(γ-1))]^1/2 for P2/P1 > Pcr
N12 =1 for P2/P1 ≤ Pcr
if P1≤P2, m12dt = -(Cd*A0*K*P1*N12)/sqrt(T1)
N12 = [((P1/P2)^2/γ - (P1/P2)^γ+1/γ) / (((γ-1)/2)*(2/(γ+1))^(γ+1)/(γ-1))]^1/2 for P1/P2 > Pcr
N12 =1 for P1/P2 ≤ Pcr
V1 = A1(L-x-x0-pt)
V2 = A2*x
*Foe adiabatic process : ρ2/ρ1 = (P2/P1)^(1/γ), T2/T1 = (P2/P1)^((γ-1)/γ)
initial temperature : T0
Temp. of chamber 1 : T1 = T0(P1/P0)^((γ-1)/γ)
Temp. of chamber 2 : T2 = T0(P2/P0)^((γ-1)/γ)
ρ1 = P1/RT1 = P1/(R*T0*(P1/P0)^((γ-1)/γ) = (P0/(RT0)) * (P1/P0)^((γ-1)/γ)
ρ1 = P2/RT2 = P2/(R*T0*(P2/P0)^((γ-1)/γ) = (P0/(RT0)) * (P2/P0)^((γ-1)/γ)
continuity equation:
P1dt = (-m12dt + ρA1xdt)*γRT1/V1
P2dt = (m12dt - ρA1xdt)*γRT1/V1

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