me 478 introduction to finite element...

Heat TransferDerivation of differential equations for heat transfer conduction without convection.  By conservation of energy we have:

Where

Ein is the energy entering the control volume, in units of joules (J) or kW *h or Btu.U is the change in stored energy, in units of kW *h (kWh) or Btu.qx is the heat conducted (heat flux) into the control volume at surfaceedge x, in units of kW/m2 or Btu/(h-ft2).qx+dx is the heat conducted out of the control volume at the surface edge x + dx.t is time, in h or s (in U.S. customary units) or s (in SI units).Q is the internal heat source (heat generated per unit time per unit volumeis positive), in kW/m3 or Btu/(h-ft3) (a heat sink, heat drawn out of thevolume, is negative).A is the cross-sectional area perpendicular to heat flow q, in m2 or ft2.

Fourier’s law of heat conduction gives us.

Kxx is the thermal conductivity in the x direction, in kW/(m * C) or Btu/(h‐ft‐F).T is the temperature, in C or F.dT=dx is the temperature gradient, in C/m or F/ft.This equation states that the heat flux in the x direction is proportional to the gradient of temperature in the x direction. The minus sign in the above equation  states heat flow is positive in the direction opposite the direction of temperature increase.

Similar to

The heat flux can be stated as:

Expanding this using a two term Taylor series

gives us:

Substituting the previous equations into

Gives us the 1D heat conduction equation.

For steady‐state this becomes.                                               or  or

where TB represents a known boundary temperature and S1 is a surface where the temperature is known, and

On an insulated boundary, qx = 0.

Boundary Conditions

Expansion to 2D Conduction no Convection.

2D Conduction with Convection

For a given control volume we get the Following:

Newton’s law of cooling gives us.

P in above denotes the perimeter around the constant cross-sectional area A.

Divide by Adx/dt, and simplifying, we obtain the equation for 1D heat conduction with convection as:

Equating the heat flow in the solid wall to the heat flow in the fluid at the solid/fluid interface, we have

Units for variables in heat transfer.

Heat conduction coefficients

Heat transfer coefficients

Shape (interpolation) functions

The total potential energy is given by.

Minimization gives you.

where

{fQ}  is a heat source (positive, sink negative) is analogous to a body‐force, and {fq} is heat flux, (positive into the surface) and {fh) is heat transfer or convection) are similar to surface tractions (distributed loading).

[k] can be given by

So.

The convection part becomes.

Integrating.

Total  element stiffness matrix becomes.

The force terms are as follows.

The convection at the free end of an element gives us.

Or

But S3 (the surface over which convection occurs) now equal tothe cross‐sectional area A of the rod. 

Direct assembly of globa K matrix is the same as for structural problems.

The global force matrix is given by. kW or Btu/h                    kW or Btu/h

The global equation is 

Solve for the Nodal Temperatures. 

Then solve for the element temperature gradients and heat fluxes.

kW/oC or Btu/(h‐oF).

Conduction terms:

Element 4 has a convection from heat loss from the flat surface at the right end.

Collect global force terms.

In this example, there is no heat source (Q = 0)  or heat flux (q = 0) and the only convection is at the right end.

On the element level.

Apply boundary condition F1 = 100.

Solve for temps.

Apply boundary condition F1 = 100.

Next Example. one‐dimensional rod, determine the temperatures at 3‐in. increments along the length of the rod and the rate of heat flow through element 1. Let Kxx = 3 Btu/(h-in.-F), h = 1.0 Btu/(h-in2-F),  and . 

The temperature at the left end of the rod is constant at 200 F.

Element stiffness matrices

Element 3 has an additional (convection) term owing to heat loss from the exposed surface at its right end.

Loading Q = 0, q = 0, and 

The known nodal temperature boundary condition of t1=200 F.

(200)4 = 800 , ‐0.5(200)4 = ‐400 ,

200o F

Solve for the temperatures.

Determine the heat flux through element 1.

Determine the rate of heat flow  by multiplying above by cross‐sectional area.