Find the temperature distribution in a rod (Figure) with internal heat generation using the finite-element method. Derive the element nodal equations using Fourier heat conduction.

q_k = -kAdT∂x

And heat conservation relationships

∑[qk + ƒ(x)] = 0

Where qk = heat flow (W), k = thermal conductivity (W/(m · °C)), A = cross-sectional area (m²), and ƒ(x) = heat source (W/cm). The rod is 50 cm long, the x - coordinate is zero at the left end, and positive to the right. The rod is also linearly tapered with a value of kA = 100 and 50 W m/°C at x = 0 and at x = 50, respectively. Divide the rod into 5 elements (6 nodes, each l0 cm long). Both ends of the rod have fixed temperatures. The heat source ƒ(x) has a constant value. Thus, the conditions are

T|_{x = 0} = 100°C             T|_{x = 50} = 50°C              ƒ(x) = 30 W/cm

The tapered areas must be treated as if they were constant over the length of an element. Therefore, average the kA values at each end of the node and take that average as a constant over the node. Develop the nodal equations that must be solved for the temperatures and temperature gradients at each of the six nodes. Assemble the equations, insert the boundary conditions, and solve the resulting set for the unknowns.

Transcribed Image Text:

KA = 100 W-m/c fe) = 30 W/ćm kA = 50 W-m/c T- 50°C 50 cm- To 100°C