lecture 1 - ME 344 — What is a dynamic system? ° The...

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Unformatted text preview: ME 344 — What is a dynamic system? ° The system whose behavior as a function of time is important — How to understand a dynamic system? - Model _ - Linear — Non-linear — How to control a dynamic system? - Classic ° Modern ME 344 - Example 1: - Example 2: "V ‘H/KTVC +VL:0 NW - WWW, laws : l/L : A“ /\:L1; MK, .5? I‘qu Lew Va =——g; w(/ + YKRT%—+L :9 Cab/“(77140? CK : {L = , ‘ "1% 00 i 7”: W ,e i“ W 7W Hwé‘é-iféi ban! 6 (Co/71mm W i W. +9 32% +21%; ’5 ,, wezéeéangfiéZ-‘ii W" E‘LKHL 5%??? +z’—%6w * TTT f PM” ' “rm 0W , * (fig/“0%” WL‘YWW FED: we wxwéx—ebkrwr : M Mzow'XkaX) ;> we kr ’522: mkam WM gfivzcfiirgl “Hum 551 r; Mme «m :> m? +b>‘<+%e><= birbklxlf (x 7ng +5 >'< »~‘ >'< «we -' 0&7? : ‘ A‘ A” v’ ‘ “A. A It X A ' z 3 ‘ I AA “'er ‘F‘ v _—:> mr%>‘<z=m>‘<>‘<‘»l<xr>é+tx>é >0? ‘ . ‘ Q EX *5 :MX -' Xr+l< 950%?) = mk‘*%CXr~X u ‘T +197?le “HZXV’ FUNDAMENTAL DESCRIPTION OF SYSTEMS IN TERMS OF ENERGY Energy: ability to do work (joules—nm) 2 E = JFdx => Work done by a force, F, in going from point 1 to point 2 1 Power: rate of change of energy _aE p___ at We define systems in terms of energy and power using conservation laws. Recall First Law of Themodynamics: Q=U+W—static behavior Qzenergy supplied to system U=change in internal energy of the system W=work done by the system Differentiating this expression with respect to time, we obtain the analogous expression for power. ipini :a—E+ipouti 1 ' at ‘ l => Net power into system 2 Time rate of change of energy in system + Net power out of system Alternatively : ipini'ipomiza—E 1 ' 1 ' at Point: it is possible to determine the dynamic (static) responses of a system by accounting for the power (energy) flow in the system. Energy Storage and Dissipation Elements - . O Conant.“ Electrical System gs‘ . Mk” E = jvidt = Jv(idt) = Jvdq Potential Energy Capacitance q=q(v) q=Cy E = Jvidt = Ji(vdt) = Jidk Kinetic Energy Inductance >\=)\(i) )\=Li E = Jvidt Dissipation Resistance =v(i) v=Ri Translation System B = J‘det = {P(vdt) = IFdx Potential Energy Spring x=x(F) x=F/k E = Jdet = Jv(th) = Jvdp Kinetic Energy Mass p=p(v) p=Mv E 2 Idet Dissipation Damper =F(v) F=bv Rotation System B = erdt = [1((DClt) = [1116 Potential Energy Rotary Spring E = Irwdt = Imfidt) = Icodh Kinetic Energy Mass E: [det 0= 6 (T) =h(w) 0= T/k h= Jco Dissipation Damper T = ’1’ (co) _ T =Bw Fluid System B = jPth = [P(th) = deU Potential Energy Fluid Cap. U= U(P) U=CP E = {mm = jQ(Pdt) = der Kinetic Energy Fluid Inertance r: P(Q) I‘=IQ E = [13th Dissipation Fluid Resistance P = P(Q) P =RQ Note: k is a measure of spring stiffness (force need for a given displacement) C is a measure of spring compliance (displacement resulting from a given force) Gmliswis t’Olém‘ f‘r €5.11on WNW/h FUNDAMENTALS OF BOND GRAPH‘MODELING METHOD Conservation of Energy is key (note power and energy are scalar quantities): 6E ZPinJ = a. + zpouLj Net power into system=rate of change of energy in systemLi-net power out of system Define Power Conjugates variables: Translation F, force V, velocity Rotation T, torque Q , angular velocity ' Fluid/Hydraulic P, pressure Q, flow rate Electrical _ e, voltage i, current H Thermal T, temperature £15 , time-rate of change of entropy, 5 dt , . Flow-Effort variables (developed by James C. Maxwell) ' Effort variables “e”: ‘pushing’: F, T, P, e, T » . .1) y_ t ' ’- ' 11:9— ' Flow variables . f. motlon .V,9, (2:1: dt ) Power, P=ef I. Energy, E= IPdt = Jefdt Define energy variables: Generalized momentum: p = jedt Generalized displacement: q = det Energy of a system can be rewritten as: E = dep “kinetic” energy E = Jedq “potential” energy ...
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lecture 1 - ME 344 — What is a dynamic system? ° The...

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