ME_446_-_Lec_02-Math._Models_of_Systems

# ME_446_-_Lec_02-Math._Models_of_Systems - Mathematical...

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Unformatted text preview: Mathematical Models of Systems - I 446 - 2 Prof. Neil A. Duffie University of Wisconsin-Madison Wisconsin- Neil A. Duffie, 1996 All rights reserved. Need for Mathematical Models Must have a quantitative mathematical model in order to understand and control complex systems. Must have fundamental method for modelling many physical systems: - Mechanical - Electrical - Hydraulic - Biological Steps in Modelling Understand the physical system and its components Make appropriate simplifying assumptions Use basic principles to formulate the mathematical model Write differential and algebraic equations describing the model Check the model for validity What will the model be used for? Solutions of the differential and algebraic equations allows system response and performance to analyzed and designed. The Laplace transformation will be applied to the model to allow convenient manipulation and dynamic analysis. Input-output relationships for systems Inputand components will be obtained. Controller models will be designed that can be implemented in hardware. Mechanical Spring-Mass System Spring Consider a mass M on a frictionless surface connected to a rigid wall by a spring with stiffness k: mass M [kg] spring k [N/m] Need a model for the position of the mass as a function of time. Analyze Physical System Choose a sign convention for the position variable y(t): -0+ (Note that sign y(t), [m] convention for velocity and acceleration mass are same as M [kg] position!) spring k [N/m] Use fundamental physical principles to model system: Newton's Law Draw Free-Body Diagram The spring force is the only force acting on the mass: y(t) M ky(t) Spring exerts a force proportional to and in opposition to movement of the mass: Newtons Law: F = Ma dy 2(t) ky(t) = M dt 2 Obtain Differential Equation ky(t) = M M dy2(t) dt2 dy2(t) + ky(t) = 0 dt2 2 M dy (t) + y(t) = 0 k dt2 2 1 dy (t) + y(t) = 0 2 n dt 2 where n = k M y(t) ky(t) M n is the (undamped) natural frequency n is an abstract term with generic meaning Second-order process System Characteristics 2 1 dy (t) + y(t) = 0 2 n dt 2 y(t) ky(t) M The system has no input. No external force acts on the mass. In the d.e., this is indicated by the zero on the right-hand side. right- System Characteristics 2 1 dy (t) + y(t) = 0 2 n dt 2 y(t) ky(t) M The system has no damping. There is no energy dissipation in the system. In the d.e., this is indicated by the absence of a first derivative term. Spring-Mass-Damper Spring- Massy(t) [m] x(t) [m] mass M [kg] damper spring c [N-s/m] k [N/m] [N Mass is perturbed by manipulating end position x(t) of spring with stiffness k. Dashpot with damping coefficient c resists motion in proportion to velocity Spring-Mass-Damper y(t) [m] Spring- MassOutput x(t) [m] mass Input M [kg] damper spring c [N-s/m] k [N/m] F s = k x(t) - y(t) y(t) [m] M Fd [N] F = Fs - Fd Fd = c dy(t) dt 2 Fs [N] M dy (t) = dt2 Spring-Mass-Damper y(t) [m] Spring- MassOutput x(t) [m] mass Input M [kg] damper spring c [N-s/m] k [N/m] 2 dy (t) dy(t) M = k x(t) - y(t) - c dt dt2 2 2 dy(t) 1 dy (t) + + y(t) = x(t) n dt n2 dt2 k [r/s] Second-order process n = M c = 2 kM Second-Order Process Response Second2 = 0.2 y(t) [m] = 1.0 0 -1 0 t [s] 5 Step input: x(t) = 1 m n = 2 r/s q(t) cm 3 /s Input Hydraulic Cylinder y(t) cm Output A cm2 flow in flow out Example piston moves to right q(t) cm 3 /s Input Hydraulic Cylinder y(t) cm Output A cm2 A dy(t) = q(t) dt cm3 s cm s cm-2 Integration process t dy(t) = Kq(t) dt K = 1 = gain A y(t) = 0 Kq(t)dt Integration Process Response 2 A = 2 cm y(t) [m] 2 A = 4 cm 0 -1 2 0 t [s] 3 5 Step input: q(t) = 1 cm /s R RC Circuit C [F] Input i(t) x(t) [amps] [volts] e = 0 Output y(t) [volts] volts volts x(t) - Ri(t) - y(t) = 0 i(t) = C dy(t) dt amps R RC Circuit C [F] Input i(t) x(t) [volts] [amps] x(t) - RC Output y(t) [volts] dy(t) - y (t) = 0 volts dt First-order dy(t) + y (t) = x (t) volts process dt Time constant: = RC seconds Gain: 1 volt/volt First-Order Process Response First1.2 y(t) [volts] 0 -1 0 t [s] 5 Step input: x(t) = 1 volt = 1s Mixing Valve and Pipe valve position x(t) [%] concentration c(t) [%] Input water mixture Output concentrated length d [m] solution flow rate q [m/s] [s] [%] Delay process D = d q c(t) = x(t - D) Common Process Types Delay: c(t) = Km(t - D) Integration: First-order: dc(t) = Km(t) dt dc(t) + c(t) = Km(t) dt Second-order: 2 1 dc (t) + 2 dc(t) + c(t) = Km(t) 2 n dt n dt2 Component Combination Example valve voltage v(t) [v] + flow control valve mixture valve position y(t) [cm] pipe length flow rate dp [m] q p [m/s] cylinder flow q(t) [cm3/s] water concentrated solution hydraulic supply concentration return c(t) [%] Mixing valve and pipe (delay) Hydraulic cylinder (integration) Flow control valve (2nd-order) Component Combination Example flow control valve 2 1 dq (t) + 2 dq(t) + q(t) = K''m(t) n dt n2 dt2 cylinder dy(t) = Kq(t) dt mixing valve x(t) = K'y(t) pipe c(t) = x(t - D) input output Component Combination Example Result: third-order with delay 3 1 dc (t) + 2 dc2(t) + dc(t) n dt2 dt n2 dt3 = K'''m(t - D) Verify Simplify? ...
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## This note was uploaded on 04/27/2010 for the course ME 446 taught by Professor Nd during the Spring '10 term at Universität für Bodenkultur Wien.

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