with its adjacent nodes. The designation, i, j, implies a general location in a two-dimensional system where i is a general index in the x direction and j is they index. Adjacent node indices are shown in Figure 17.14. The grid is set up with constant node width, Dx; and constant height, Dy: It may be convenient to make the grid ‘‘square,’’ that is, Dx ¼ Dy; but for now we will allow these dimensions to be different. A direct application of equation (6-10) to node i, j yields dQ dt ¼ @ @t ZZZ c:v: er dV (17-58) The heat input term, dQ/dt, may be evaluated allowing for conduction into node i, j from the adjacent nodes and by energy generation within the medium. Evaluating dQ/dt in this manner, we obtain dQ dt ¼ k Dy Dx (TiÀ1,j À Ti, j) þ k Dy Dx (Tiþ1,j À Ti,j) þ k Dx Dy (Ti,jÀ1 À Ti,j) þ k Dx Dy (Ti,jþ1 À Ti,j) þ _ q Dx Dy ð17-59Þ The first two terms in this expression relate conduction in the x direction, the third and fourth express y-directional conduction, and the last is the generation term. All of these terms are positive; heat transfer is assumed positive. The rate of energy increase within node i, j may be written simply as @ @t ZZZ c:v: er dV ¼ rcTj tþDt À rcTj t Dt ! Dx Dy (17-60) Equation (17-58) indicates that the expressions given by equations (17-59) and (17-60) may be equated. Setting these expressions equal to each other and simplifying, we have k Dy Dx ½TiÀ1,j þ Tiþ1,j À 2Ti,j þ k Dx Dy ½Ti,jÀ1 þ Ti,jþ1 À 2Ti,j þ _ q Dx Dy ¼ rcTi,j j tþDt À rcTi,j j t Dt ! Dx Dy ð17-61Þ This expression has been considered in a more complete form in the next chapter. For the present we will not consider time-variant terms; moreover, we will consider the nodes to be square, that is, Dx ¼ Dy: With these simplifications equation (17-61) becomes TiÀ1,j þ Tiþ1,j þ Ti,jÀ1 þ Ti,jþ1 À 4Ti,j þ _ q Dx2 k ¼ 0 (17-62) In the absence of internal generation, equation, (17-62) may be solved for Tij to yield Ti,j ¼ TiÀ1,j þ Tiþ1,j þ Ti,jÀ1 þ Ti,jþ1 4 (17-63) or, the temperature of node i, j is the arithmetic mean of the temperatures of its adjacent nodes. A simple example showing the use of equation (17-63) in solving a two-dimensional heat conduction problem follows. EXAMPLE 4 A hollow square duct of the configuration shown (left) has its surfaces maintained at 200 and 100 K, respectively. Determine the steady-state heat transfer rate between the hot and cold surfaces of this duct. The wall material has a thermal conductivity of 1:21 W/mÁK: We may take advantage of the eightfold symmetry of this figure to lay out the simple square grid shown below (right). 244 Chapter 17 Steady-State Conduction