Advanced Design for Mechanical System Pract10 1 Thermal analysis There is a resemblance between the...
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Transcript of Advanced Design for Mechanical System Pract10 1 Thermal analysis There is a resemblance between the...
Advanced Design for Mechanical System Pract10
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Thermal analysis
•There is a resemblance between the thermal problem and the stress analysis.
•The same element types, even the same FE mesh, can be used for both analysis
•Thermal analysis means primarily calculation of temperature within a solid body
•A by-product of the temperature calculation is information about the magnitude and direction of heat flow in the body.
•The results of thermal analysis, nodal temperature, can be used to evaluate the distribution of stresses due to the presence of a thermal gradient within the body
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Pipe with thermal gradient along the thickness
T =
25 °
C
T =
100°
C
• Thermal conductivity 20 W/m°C• Coefficient of thermal expansion 12 10-6 /°C
Internal diameter = 40mm
Thickness = 10mm
Length =600mm
Flux= 0
Flux= 0
ra
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Gradient along the thickness
iie
eii r
r
rr
TTTrT ln
)/ln()(
2W/m dr
dTkq
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The thermal problem
Heat moves within the body by conduction
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Fourier equation of heat flow
fx, heat flux for unit area (W/m2)
k thermal conductivity (W/m°C)
Negative sign means that heat flows is in a direction opposite to the direction of temperature increase
The temperature represents the potential for the heat flow.
Consider an isotropic material and image that exist a temperature gradient in the x direction
x
Tkf x
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zT
yT
xT
f
f
f
x
Tkf
z
y
x
x κ
fx, fy, fz heat flux for unit area (W/m2)
matrix of thermal conductivity (W/m°C)
Boundary condition on surface with normal n
s
SnS
fn
TkTT
f
Hp: non isotropic material
the direction of the heat flow in general is not parallel to the direction of the temperature increase
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Conductivity
•The matrix of conductivity is in general a full 3 by 3 matrix
•If x,y,z are principal axes of the material is a diagonal matrix
•In the special case of isotropy 11=22=33=
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t
Tcq
f
f
f
v
z
y
x
zyx-
By considering a differential element of volume dV and writing the balance equation (rate in – rate out)= (rate of increase within), it is obtained:
where:
qv is the rate of internal heat generation for unit volume (W/m3)
c is the specific heat (J/kg °C)
is the density (kg/m3)
t is the time (s)
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If the problem becomes steady-state. For material isotropic the equation becomes
vqTk
vv qy
T
y
T
x
TkqTk
2
2
2
2
2
22 or
0/ tT
Where is the gradient operator.
If, in addition, k is independent of the coordinates results
Solution: T=T(x,y,z) that also meets prescribed boundary conditions
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• The heat flow trough the surface of the body is analogous to surface load in the stress analysis.
• A distributed internal source is analogous to body forces in stress analysis.
• Prescribed temperature are analogous to prescribed displacements• For a steady state problem, the mathematics of all this leads to the
global FE equation
QTKT
where:KT depends on the conductivity of the material ([W/°C])T is the vector of the nodal temperature ([°C])Q is the vector of the thermal loads ([W])
•Convection and radiation boundary conditions, if present, contribute term to both KT and Q.
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Hypothesis for FE in thermal analysis
• In solid body the heat transfer occurs by conduction
• In fluid the transmission by mean of convection must be take into account
• In many structural problems the thermal problem and the mechanical problem are not coupled
• This hypothesis is no longer valid when the deformation can generate heat and cause a variation of the mechanical properties
• Thermal conductivity and other properties of the material must be considered dependent on the temperature
• The conductivity matrix can be generated by a direct method for very simple elements, otherwise a formal procedure is required
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Thermal bar element
The rate of heat flow, q=Af, is constant and direct axially.
T1>0, T2=0 T2>0, T1=0
Nodal heat flow is positive when direct within the element
An actual bar can be modeled by several of these elements if temperature at several locations between its ends were required
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2
1
2
1
matrixty conductivielement the
//
//
q
q
T
T
k
lAklAk
LAkLAk
T
In matrix format, these results are:
Temperature within the element are obtained by interpolating nodal temperatures
e
n
n T
T
T
T
NNNT NT
...
... 2
1
21
•The form of the interpolation determines the complexity of the temperature field that an element can represent
•The shape function can be exactly those used also to interpolate a displacement field
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NB BTT
y
x
T
T
T
yNyNyN
xNxNxN
yT
xT
n
n
n
/
/ where
.../...//
/...//
/
/ 2
1
21
21
volumeelement
dVk TT κBB
In Cartesian coordinates, temperature gradients in a plane element are
For solid a third row expressing T/z is added
The expression for an element conductivity matrix can be shown to be
The array of thermal conductivities becomes the scalar k for bar elements.
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Remarks•The thermal problem is a scalar field problem
•A FE temperature field is continuous within elements and across interelement boundaries
•Temperature gradient, like strains in stress analysis, are typically not interelement continuous
•If radiation is not present, the rate of heat flow is proportional to temperature differences
•Upon assembly of elements, nodal rates of heat flow qi from separate elements are combined at shared nodes.
•Thus the net flow into node i of the structure:
except at nodes where Ti is prescribed, nodes on a structure boundary across which heat is transferred, or at internal nodes where Qi arise from qv
0 ii qQ
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Where
f is the flux normal to the surface and positive inward
h is the heat transfer coefficient
Tf e T temperature of the fluid and of the surface of the body
For the increment, dS, of the element surface subjected to convection formal procedure gives
The matrix combine with the element conductivity matrix and hence contribute to KT the vector contribute to Q
Convection boundary conditions
)( TThf f
dShThdS fTT NNN :tor vec:matrix
If the material conductivity is dependent on temperature a nonlinearity is present
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Radiation boundary conditions
44
44
fluxheat radiated
4
fluxheat received
4
1)/1()/1(TTf
TTTTf
rr
SB
rSBSBrSB
Consider two infinite parallel planes. Let the planes have temperature Tr and T , and each is a perfect absorber and perfect radiator (black body).
If SB is the Boltzman constant, the net heat flux received by the surface at temperature T is
If the emissivity (ratio of total emissive power to that of a blackbody at the same temperature) is introduced, than
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Accounted for the fact that the surface are not parallel, often not flat, and certainly not infinite, a shape factor is introduced
F is a factor that accounts for the geometry of the radiating surface and their emissivities
44 TTFf rSB
The calculation of F is sufficiently complicated that it may be done by a separate computer program.
The flux f is an average value.
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TTTTFhTThf rrrrr 2244 dove
This equation can be written as
T is the absolute [K]
hr= hr(T) is a temperature dependent heat transfer coefficientThis makes the problem non linear and requires iterative solution
•The flux expressions have the same form for convection and radiation boundary conditions
•Accordingly, a radiation boundary condition leads to matrix and vector expressions having the same form as for convection, with h and Tf replaced by hr and Tr
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Modeling considerations
• Element types, and shapes for a thermal analysis may be dictated less by thermal analysis than by an anticipated stress analysis based on the same mesh
• The mesh demands of stress analysis are usually more severe
• Also a modest temperature gradient may create forces of constraint that produce large strain gradients, especially near holes, grooves and other stress raisers.
• Also for thermal analysis the present symmetry must be considered: heat flux trough a plane of symmetry is zero
• Heat flux must be parallel to an insulate boundary or a plane of symmetry
• Temperature contours (isotherms) should be parallel to a boundary of constant temperature and normal to plane of symmetry
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Non linearity
• If in the body there are appreciable difference of temperature, it is necessary to regard conductivity as temperature dependent
• If there is convection with a temperature dependent coefficient of heat transfer
• If is present radiation
)()(
)(
)(
TT
TQQ
TKK TT
QTKT
The problem is non linear
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• Iteration is stopped when a convergence test is satisfied.
•For example
valuefixed1
i
iii T
TTe
•Iterative solution based on the method of direct substitution:
- assume an initial temperature field T0
- generate KT and Q based on these temperature
- solve the equation for T1
- use the newly computed temperature for new values of k, h, and hr
- generate a new KT and Q
- solve for a new T )()( 121 TT QTKT
)()( 010 TT QTKT
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Thermal transient
cdV
t
assemble
NNc
cC
TT
QTCTK
T
T
matrix capacity heat element assembling built is
matrix capacity heat the and
t
where
)(
The solution must be integrated with respect to time
•When steady state conditions do not prevail, temperature change in a unit volume of material is resisted by “thermal mass” that depends on the mass density and the specific heat c.
•The solution become
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Dependence on temperature of some material properties
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Flanged Joint• Sections of a pipe are connected by a flanged joint.
• Each flange is connected to the pipe by two circumferential welds
• Bolts draw the flanges together and compress a gasket between them
Evaluate the steady state temperature field in the pipe and in the flange for use in a subsequent stress analysis
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Data of the problem
• Boundary conditions:
- fluid in the pipe has temperature 0°C
- vapor condensing on the outside of the pipe has temperature 100 °C
• Heat transfer coefficient h:
- inside the pipe is 5000 W/m2
- outside the pipe is 20000 W/m2
• The material of the pipe and of the flanges is steel with the following thermal properties
- Thermal conductivity = 20 W/m°C
- Coefficient of thermal expansion = 12 10-6 /°C
- Specific heat = 480 J/Kg °C
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Preliminary Analysis
• The maximum possible value of flux through the pipe wall is approximated by regarding the wall as plane and using the limiting surface temperature
• This is a maximum value because:
- the inside of the pipe must be warmer than 0°C
- the outside of the pipe must be cooler than 100 °C
2lim /286000
077.0084.0
010020 mW
r
Tkf
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Axis of revolution
Model
Radial
Plane of symmetry
Plane of symmetry
Axisymmetric solid of revolution
Gap
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Boundary conditions
• Between the welds, along IJ, pipe and flanges touch only at random and isolated point, if at all.
• Conductivity across this cylindrical surface is very low compared with solid metal and can be assumed equal zero
• This is a pessimistic assumption for eventual stress analysis, because it increase the thermal gradient and therefore increase stresses.
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Temperature field
• The lowest temperature is about 3 °C along the right half of the inside boundary AB
• The highest temperature is 99.99 °C along the outer surface CDEFG• Temperature contour are interelements continuous, except along IJ
where discontinuity is expected
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Thermal flux
• Flux arrows are perpendicular to the temperature contour because the material is isotropic
• Flux arrows agree with the expectation of preliminary analysis• Radial flux near AB is 170.000 W/m2 which, as expected is less that the
approximate limiting value.
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Thermomechanical analysis• The mesh used for the thermal analysis is used again, now with
computed nodal temperatures used to load the model.
• Two significant question about boundary arise:
- should nodes along BC be fixed against axial motion or not? Allowing movement gives no credit to resistance offered by the gasket and the bolts, while fixity gives too much credit.
We elect to run the stress analysis twice, first allowing motion along BC and second preventing it.
- is it proper to let nodes along IJ move independently?
If sides of the interface move apart, as the results shows, the answer is yes.
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Maximum and minimum stresses (in MPa) for different axial restraint along BC
Only node B axially restrained
All nodes on BC axially restrained
r z r z
Maximum stress 113 216 244 105 229 257
LocationI,J L-N I,J I,J M-B I,J,K
Minimum stress -68 -142 -199 -107 -225 -279
LocationG G G G G G