L L LT T ( ) L LT α - University of Arizona · ∆= ∆LL T α. 2. Add reaction ... Determine...
Transcript of L L LT T ( ) L LT α - University of Arizona · ∆= ∆LL T α. 2. Add reaction ... Determine...
J. H. Burge University of Arizona
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Thermal Distortions Thermal Expansion
Materials expand or contract with changing temperature. Change the temperature
( )2 1 2 1L L L T TL L T
L TL
α
α
ε α
− = −
∆ = ∆
∆≡ = ∆
thermal strain
α is the Coefficient of Thermal Expansion CTE Aluminum ~23 ppm/°C Optical Glass ~3 – 10 ppm/°C
Temp T1
T2
L2
L1
Isotropic materials, temperature changes cause ALL dimensions to scale proportionally: Inside dimensions scale the same as outside dimensions
A
A’
B
B’
A A TB B T
αα
∆ = ∆∆ = ∆
Combining materials αi, Li
∑= iLL
TLTLL
e
ii
∆=
∆=∆ ∑α
α
LLii
e∑=
ααDefine equivalent CTE αe:
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Low CTE materials: Borosilicate glass (Pyrex) ~3 ppm/°C Fused silica 0.6 ppm/°C Invar ~1 ppm/°C Super Invar ~0.3 ppm/°C Zerodur (Schott) 0 ULE (Corning) 0 CFRP (can be tuned to 0)
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Athermalizing -- Combining two materials:
( )
1 2
1 2
1 1 2 2
1 1 2 2
L L LL L L
L T L TL L T
α αα α
= +∆ = ∆ + ∆= ∆ + ∆
= + ∆
To athermalize over a distance L, use two materials so
1 1 2 2
1 2
0EL L LL L Lα α α+ = =
+ = Using materials with α > 0, requires L < 0
L1, α1 L2, α2
L1, α1
L2, α2 L
1 1 2 2
1 2
0L LL L Lα α− =
− =
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Thermal stress Use superposition Calculate stress due to temperature change
1. Determine expansion, as if unconstrained
L L Tα∆ = ∆
2. Add reaction force that provides constraint by pushing back ∆L
Solve more general problems the same way.
F FLLEA
∆ =
F E TA
σ α= = ∆
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J. H. Burge 30
Control of thermally induced stresses
• Design preload element to take thermal strain – preload stiffness K is << stiffness of constraint– Simplified model, use Li, αi, LM, αM
FCTLFCTLLLLL
LL
MMee
MFMTGFGT
MG
−∆=+∆∆+∆=∆+∆
∆=∆
αα
L2 L3L1
LM
∑ ∆=∆=∆ TLTLL iieGT αα
optic
Rigid constraint
Compliant preload
L1 L2 L3
Path G∑== iM LLL
LLM =
2. Determine relationship between force and displacementUse equivalent compliance Ce
Path M
TLL MMT ∆=∆ α
• Use superposition to determine change in stress1. Allow unconstrained expansion, use equivalent CTE
F
F
∑==∆ FCFCL ieGF
FCL MMF −=∆( )
Me
eM
CCTLF
+∆−
=αα
3. Solve for the force required to hold the thing together
Add this to the preload force
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Temperature gradients cause distortions
Gradient = 1 2T TT
y h−∂
=∂
This will cause the beam to bend in an arc, in the same way that the applied moment did. The arc length along surface 1 is longer than the arc length along surface 2 by the amount
( )1 2 1 2L L L T Tα− = − By geometry: or More generally, Integrate for deflection
T1
T2 h
L
∆θ
1
2
1 2
1
L Lh
L Th
TL R y
θ
αθ
θ α
−∆ =
∆∆ =
∆ ∂= =
∆ ∂
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Materials with inhomogeneous CTE, coupled with a bulk temperate change behave the same way as above:
CTE Gradient = 1 2
y hα αα −∂
=∂
This will cause the beam to bend in an arc, in the same way that the thermal gradient did. If the temperature is changed uniformly, the arc length along surface 1 is longer than the arc length along surface 2 by the amount
( )1 2 1 2L L L T α α− = ∆ − By analogy : and Generally:
α1
α2 h
L
∆θ 1
2
( )1
L Th
L Ty
TL R y
αθ
αθ
αθ
∆∆ = ∆
∂∆ = ∆
∂
∂∆ = =∆ ∂
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Integrate to get deflections δy(z) Use boundary conditions: For cantilever: at z=0, θ=0 and δy=0 So θ0=0 and C=0 For simple support at the ends: δy=0 at z=0 and at z=L
So θ0= ( )
2TL
yα∂
−∂ and C=0
max deflection at L/2 is
( )
( )
0
0
20
0
( ) ( ) ( )
( )
1( ) ( )2
y
z
z
z
yz
d z z zdz
Tz z
yT
z z dz z z Cy
δ θ θ θ
αθ
αδ θ θ
=
=
= = + ∆
∂∆ =
∂
∂= ∆ = + +
∂∫
( ) 21( )2y
TL L
yα
δ∂
=∂
( ) ( )221 1
2 2 2 8y
T TL L Ly y
α αδ
∂ ∂ = = ∂ ∂
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Heat flow causes thermal gradients For steady state: Qin = Qout
Define thermal conductivity λ : ( )1 2T T
QL
λ−
=
λ Glass 1.1 W/(m K) Aluminum 170 W/(m K) Copper 390 W/(m K) Stainless steel 16 W/(m K)
Heat flow A TH Q A
Lλ∆
= × = Apply 1 W through 10 cm long bar of Al, A = 1 cm2
( )( )( )( )2
1 0.16 10
0.0001 170
W mHLT K FWA m m K
λ∆ = = = ° = °
⋅
Q = heat flux (J/m2/s) Q
L
T1 T2
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Steady state thermal distortion
Distortion due to temperature gradient T∇ is always
proportional to Tα∇ For a constant heat source, with power H, the thermal gradient is
proportional to HTλ
∇ ∝
So the distortion will be proportional to
αλ
This provides a figure of merit to compare sensitivity to steady state heat loading
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• Lines of constant thermal stability are shown • Performance improves toward upper left corner • SiC has the largest thermal stability due to high conductivity,
even though large CTE • Zerodur and ULE have very high thermal stability due to their
extremely low CTE, even though poor conductor • Titanium and CRES are very poor for thermal stability
αλ
Better stability
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Transient heating • Transient heat flux is when the temperature distribution changes with
time • Thermal diffusivity (D) is the ratio of thermal conductivity to heat
capacity Larger thermal diffusivity means quicker response to temperature changes. The thermal time constant τ governs the rate at which the transient response decays exponentially (1-e-t/τ) After a given time from a transient heat impulse, the temperature gradient will be proportional to 1/D. Again, the thermal distortion is proportional to α ∆T, so the transient distortion will be proportional to
This provides a merit function for transient thermal stability
P
DC
λρ
=⋅
Dα
2aCD
τ ≅∆T
a
C=1
a
C=4
2T D Tt
∂= ⋅∇
∂
∆T
∆T
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Thermal stability under transient heating
• Lines of constant thermal stability are shown • Performance improves toward upper left corner • SiC has the largest transient thermal stability • ULE has high stability due to its extremely low CTE • Glass, Titanium and CRES are very poor
Dα
D Better stability
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Thermal time constant: For BK7 glass: λ = 1.1 W/(m – K) = 0.011 J/s /(cm – K) ρ = 2.5 g/cm3 cp = 0.86 J/(g- K) For 1 cm thick BK7, heated from one side:
or ~ 3 min
Scaling this: 25 mm glass takes 20 min
This is the time that it takes for the heat to travel through the glass. It does not include the coupling to the outside.
∆Τ(τ)/∆Τ0=0.37 ∆Τ(2τ)/∆Τ0=0.14 ∆Τ(3τ)/∆Τ0=0.05 ∆Τ(4τ)/∆Τ0=0.02 ∆Τ(5τ)/∆Τ0=0.007
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Athermal System design
1. Control geometry Use low CTE materials Kovar ~ 5 ppm/°C Invar ~ 1 ppm/°C Super Invar ~ 0.3 ppm/°C Fused silica ~ 0.6 ppm/°C Practically zero CTE materials ULE Zerodur Athermalized Carbon Fiber Reinforced Plastic Composite truss for HST
Use of metering rods