- Find The Solution Of The One Dimensioanal Heat Equation Given Using The Boundary Conditions Provided 1 (59.78 KiB) Viewed 23 times
USING THE BOUNDARY CONDITIONS PROVIDED
Problem 2 – 1D heat conduction The ODE that models heat conduction across a 1D rod just like the one depicted in problem 1 is given by: d dx ka(v) ) + s(x) = 0 X dT + dx 0<x<L The rod’s temperature is fixed on the left end and loses heat through air convection on the left end. This ODE is a restatement of energy balance. Here: • T is temperature along the x-axis • k is the thermal conductivity o Typical values: 100-500 W/m/C A(x) is the area of cross-section • This can be constant, piece-wise constant, a linear or a quadratic function with the min and max areas of cross section being 0.05 m² and 0.1 m², respectively s(x) is heat generated/withdrawn at x per unit length o This can be constant, linear, or quadratic with the min and max values being 0 and 10 W/m • L is the length of the bar o Assume this is 10 m The boundary conditions are: T(x = 0) = 300°C dT kdx= (TL = 1 = -h(T(L) – Tair) Here, h is the convection coefficient (typical values for air are 2-20 W/mº/C and Tair is the air temperature (assume 20° C).