First Order Simultaneous ODE_Example

First Order Simultaneous ODE_Example - dTb/dt = (Tw Tb)...

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Click to edit Master subtitle style Quenching: Example of Numerical Solution to a System of 1st Order ODEs Prof. Dragan Djurdjanovic ME 218 Fall 2008
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The Problem n In the quenching operation, used to harden a metal bar, the bar is heated to a high temperature Tb and then quickly immersed in a water bath at temperature Tw. Bar (Tb) Water (Tw)
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Thermal Balance Equations n The consideration of thermal balance of the bar and water bath lead to the following equations ¡ (mbcb/hA)(dTb/dt) + Tb = Tw ¡ (mwcw/hA)(dTw/dt) + Tw = Tb
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Given Data n h = 500 W/m2-K (Convection heat transfer coefficient) n A = 0.01 m2 (Surface area of metal bar) n mb = 0.2 kg (Mass of metal bar) n mw = 5 kgs (Mass of water in bath) n cb = 225 J/kg-K (Specific heat of bar) n cw = 1000 J/kg-K (Specific heat of water) n Initial conditions: ¡ Tw(0) = 290 K ¡ Tb(0) = 950 K
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4th order Runge Kutta n Define the vector T = [Tb Tw]T n Then, dT = [dTb dTw]T
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Unformatted text preview: dTb/dt = (Tw Tb) (hA/mbcb) dTw/dt = (Tb Tw) (hA/mwcw) 4th order Runge Kutta (Contd) n The Formulation: k1 = x*dT/dt|T = T(t) k2 = x*dT/dt|T = T(t) + 0.5 k1 k3 = x*dT/dt|T = T(t) + 0.5 k2 k4 = x*dT/dt|T = T(t) + k3 T(t+x) = T(t) + (1/6)k1 + (1/3)k2 + (1/3)k3 + (1/6)k4 x = time step Solution with MATLAB n Need the temperature of the bar after 10 & 30 seconds n Using a step size of 0.1 seconds n After 10 seconds Temperature of the bath = 294 K Temperature of the bar = 509 K n After 30 seconds Temperature of the bath = 295.7 K Temperature of the bar = 318.5 K Plot 5 10 15 20 25 30 35 200 300 400 500 600 700 800 900 1000 T b T w Te m pe rat ur e (K ) Time (seconds) Conclusion n With time, the temperature of the bar and the water bath approach the same value, leading to thermal equilibrium...
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First Order Simultaneous ODE_Example - dTb/dt = (Tw Tb)...

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