FEM for Heat Transfer Analysis, Linear Steady-State heat transfer. A student team is tasked to analyze a case of linear steady-state heat transfer in a research lab. The team uses a finite element model consisting of just one two-dimensional isoparametric element. The element shape in the x-y space and r-s space are shown below. The origin of the element coordinate system x-y is at node 3. The thickness is t = 1 cm (perpendicular to the x-y plane). Thickness t = 1 cm Conductivity = k 2 cm 1 4 cm 2 4 cm Convective boundary condition with constant h The thermal conductivity k of this material is 1 4 <2- kx = ky = k = k = 400 W mC W Watt denotes ' mC meter C Convective heat transfer boundary condition The surface area that includes Node 1 and Node 4 is facing an air flow with environmental temperature . It has a heat flow input qs from convection heat transfer: qs = h (0e - 05) in which the temperature of the environment is 0 = 90 C and the convective heat transfer coefficient is h = 500 In general, the FE matrix equation for steady-state heat transfer is W mC (K* + KC)+10 +AQ + 1+AQ In this case, there is no heat generation, no surface conduction heat flow input, and no concentrated nodal heat flow input, so the first term in the RHS is a zero vector, t+At Q = 0 Nodal temperature boundary condition The surface that includes Node 2 and Node 3 is kept at a temperature of 0 C. 0C 02 = 0 C and 3 = 0C Apply the isoparametric finite element method to: (1)Determine the temperature interpolation matrix H(r, s) of the element. (2) Determine the Jocobian matrix J of the element. (3) Determine the B(r, s) matrix of the element.