Heat Transfer Project

Heat Transfer (ENGR 4123) Department of Engineering and Physics University of Central Oklahoma Introduction In simple physics and chemistry, heat is described as a form of thermal energy. Heat can be transferred in three ways. They are conduction, conviction, and radiation. Conduction is heat transfer by means of molecular agitation within a material without any motion of the material as a whole. That’s mean the high temperature point on the material will transfer the higher energy to the lower energy point which is the collar point. In this project we will be talking about the conduction heat transfer special in two dimatonal direction. Finite-Difference Equations is best way to develop the finite-difference equations by an alternative method called the energy balance method. Problem Statement. A common arrangement for heating a large surface area is to move warm air through rectangular ducts below the surface. The ducts are square and located midway between the top and bottom surfaces that are exposed to room air and insulated, respectively. For the condition when the floor and duct temperatures are 30 and 80 C, respectively, and the thermal conductivity of concrete is 1.4 W/m K, calculate the heat rate from each duct, per unit length of duct. Use a grid spacing with x= 2 y, where y= 0.125L and L =150mm. literature review: Model geometry: Equations and Boundary: We will use the energy balance method. The best way to develop the finite-difference equations by an alternative method called the energy balance method. For two-dimensional conditions: energy exchange is influenced by conduction between (m, n) and its four adjoining nodes. Ein + Eg = 0 …Eq.1 q’(U,D,R,L) + q’g= 0 …Eq.2 Where U is upper side of the node, D is down side of the node, R is right side of the node, and L is left side of the node. Here the energy transfer from the node by conduction on the left side of the node … Eq.3 where ( y 1) is the heat transfer area, and the (Tm–1,n Tm,n)/ x is the finite- difference approximation to the temperature gradient at the boundary between the two nodes. Here the equation for the rigth side of the node: …Eq.4 Here the equation for the upper side of the node …Eq.5 Here the equation for the down side of the node …Eq.6 Here we consider there is no energy generation. So we solve the equation to find the heat rate q. The will be 32 equations. We solve them by using Formulation as a Matrix Equation. The matrix inversion begins by show the equations as …Eq.7 Where a and c are known. Also will include coefficients and constant such as K and T. the equation will be expressed as [A][T]=[C] …Eq.8 Where …Eq.9 Analytical solution for problem 4.67 Given: T1=80C, T2=30C, L=150mm, k=1.4 W/mK FIND: q′? W/m ASSUMPTIONS: (1) S.S, (2) 2-D, (3) Constant properties, (3) No internal volumetric generation. Analysis: y = 0.125L = 18.75 mm x = 2 y = 37.50 mm By using all the equations above we get the temperature for T1, T2, T3, T4, and T5. Note: the equation in the appendix and solve them by the computer. T1=42 C, T2= 44.3 C, T3= 54 C, T4= 54.9 C, T5=55 C The heat loss per unit length Reference: • Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera, David P. Dewitt “Fundamentals of Heat and Mass Transfer”, 7th edition ,March 21, 2011