First Order Simultaneous ODE_Example

# First Order Simultaneous ODE_Example - Â dTb/dt =(Tw â€“...

This preview shows pages 1–9. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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)
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
4th order Runge Kutta n Define the vector T = [Tb Tw]T n Then, dT = [dTb dTw]T

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Â¡ dTb/dt = (Tw â€“ Tb) (hA/mbcb) Â¡ dTw/dt = (Tb â€“ Tw) (hA/mwcw) 4th order Runge Kutta (Contâ€™d) n The Formulation: Â¡ k1 = x*dT/dt|T = T(t) Â¡ k2 = x*dT/dt|T = T(t) + 0.5 k1 Â¡ k3 = x*dT/dt|T = T(t) + 0.5 k2 Â¡ k4 = x*dT/dt|T = T(t) + k3 Â¡ T(t+x) = T(t) + (1/6)k1 + (1/3)k2 + (1/3)k3 + (1/6)k4 Â¡ 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 Â¡ Temperature of the bath = 294 K Â¡ Temperature of the bar = 509 K n After 30 seconds Â¡ Temperature of the bath = 295.7 K Â¡ 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...
View Full Document

## This note was uploaded on 06/23/2011 for the course ME 218 taught by Professor Unknown during the Fall '08 term at University of Texas.

### Page1 / 9

First Order Simultaneous ODE_Example - Â dTb/dt =(Tw â€“...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online