Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution

Two dimensional model steady state analysis

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Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse/Rockbed 4 Conservation of Energy gives m 1 C 1 du 1 dt = - h 1 A 1 ( u 1 - T a ) - h 2 A 2 ( u 1 - u 2 ) m 2 C 2 du 2 dt = - h 2 A 2 ( u 2 - u 1 ) Can write system du 1 dt = - ( k 1 + k 2 ) u 1 + k 2 u 2 + k 1 T a du 2 dt = εk 2 u 1 - εk 2 u 2 with k 1 = h 1 A 1 m 1 C 1 k 2 = h 2 A 2 m 1 C 1 ε = m 1 C 1 m 2 C 2 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (7/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse/Rockbed 5 Model Design Allows simulation to choose the size of rock bed and amount of airflow based on size of greenhouse Varying quantities and material changes coefficients Coefficients are known based on thermal properties of gases and building materials Given initial conditions u 1 (0) = u 10 and u 2 (0) = u 20 can easily simulate Analysis allows optimal design Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (8/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Greenhouse/Rockbed 6 Model: Actual determining the values of the kinetic parameters for a particular greenhouse/rockbed configuration can be a very difficult problem This is the most important problem in design Suppose that we have k 1 = 7 8 k 2 = 3 4 ε = 1 3 T a = 16 C Then du 1 dt = - 13 8 u 1 + 3 4 u 2 + 14 du 2 dt = 1 4 u 1 - 1 4 u 2 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (9/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Model Analysis - Matrix Form 1 Model in Matrix Form ( Note : We define du 1 ( t ) dt = ˙ u 1 .) ˙ u 1 ˙ u 2 = - 13 8 3 4 1 4 - 1 4 ! u 1 u 2 + 14 0 which has the form ˙ u = Ku + b with initial condition u (0) = u 0 = u 10 u 20 Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (10/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Model Analysis - Expectations 2 Qualitative Model Expectations The only energy input into the system is the environment at 16 C With this constant environmental temperature, expect lim t →∞ u 1 ( t ) u 2 ( t ) = lim t →∞ u ( t ) = 16 16 = u e Model uses Newton’s Law of Cooling , so expect an exponential decay toward u e Joseph M. Mahaffy, h [email protected] i Lecture Notes – Systems of Two First Order Eq — (11/32)
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Introduction Greenhouse/Rockbed Example Direction Fields and Phase Portraits Two Dimensional Model Steady State Analysis Eigenvalue Analysis Model Solution Model Analysis - Steady State 3 Model Analysis - Steady State: At steady state , ˙u = 0 Need to solve Ku + b = 0 or Ku = - b This solves the linear system - 13 8 3 4 1 4 - 1 4 !
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