A long steel beam with rectangular cross section of 40 cm by 60 cm is mounted on an insulating wall as shown in the following sketch. The rod is heated by radiant heaters that maintain the top and bottom surfaces at 300°C. A stream of air at 130°C cools the exposed face through a heat transfer coefficient of 20 W/(m2K). Using a node spacing of 1 cm, determine the temperature distribution in the rod and the rate of heat input to the rod. The thermal conductivity of the steel is 40 W/(m K).

Question

GIVEN

Rectangular steel beam mounted on an insulating surface, heated top and bottom, with exposed face cooled by an air flow

FIND

(a) Temperature distribution in the rod (b) Rate of heat input to the rod

SKETCH

Answer 1

Since the rod is long, we can consider a two-dimensional solution. By symmetry, the rod can be

divided along its horizontal midplane. The sketch below shows the resulting geometry along with the

node and control volume locations.



Define the top surface temperature as Ttop = 300°C, and the ambient temperature as T? = 130°C.

We now need to determine an energy balance for each control volume.

Along the top edge we have



For the central nodes we have from



Along the left edge, an energy balance on the two nodes at j = 2 and 3 gives



Along the bottom edge, an energy balance on the three nodes at i = 2, 3, and 4 gives



For the node at i = 1, j = 1, an energy balance gives



For the node at i = 5, j = 1, an energy balance gives



Finally, along the right edge, an energy balance on the two nodes at j = 2 and 3 gives



This set of difference equations can be written as a matrix equation as follows



where the coefficient matrix A is given by



and the temperature vector T and constant vector C are given by

T (1,1)



Inverting the matrix A with a spreadsheet program and multiplying the constant vector C by the

inverted matrix, we get the vector of nodal temperatures



To determine the heat flow to the rod, consider the surface of the exposed control volumes. The rate of

convective heat transfer from these surfaces must equal the rate of heat input to the rod. Remembering

to double the value of account for the symmetry, we have



Inserting the nodal temperatures from the solution vector T given above, we find