me 478 introduction to finite element...
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Heat TransferDerivation of differential equations for heat transfer conduction without convection. By conservation of energy we have:
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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.
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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.
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Similar to
The heat flux can be stated as:
Expanding this using a two term Taylor series
gives us:
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Substituting the previous equations into
Gives us the 1D heat conduction equation.
For steady‐state this becomes. or or
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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
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Expansion to 2D Conduction no Convection.
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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.
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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
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Units for variables in heat transfer.
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Heat conduction coefficients
Heat transfer coefficients
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Shape (interpolation) functions
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The total potential energy is given by.
Minimization gives you.
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where
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{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
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So.
The convection part becomes.
Integrating.
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Total element stiffness matrix becomes.
The force terms are as follows.
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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.
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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).
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Conduction terms:
Element 4 has a convection from heat loss from the flat surface at the right end.
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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.
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Apply boundary condition F1 = 100.
Solve for temps.
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Apply boundary condition F1 = 100.
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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.
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Element stiffness matrices
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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
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The known nodal temperature boundary condition of t1=200 F.
(200)4 = 800 , ‐0.5(200)4 = ‐400 ,
200o F
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Solve for the temperatures.
Determine the heat flux through element 1.
Determine the rate of heat flow by multiplying above by cross‐sectional area.