L L LT T ( ) L LT α - University of Arizona · ∆= ∆LL T α. 2. Add reaction ... Determine...

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J. H. Burge University of Arizona 1 Thermal Distortions Thermal Expansion Materials expand or contract with changing temperature. Change the temperature ( ) 2 1 2 1 L L LT T L LT L T L 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 T 1 T 2 L 2 L 1 Isotropic materials, temperature changes cause ALL dimensions to scale proportionally: Inside dimensions scale the same as outside dimensions A A’ B B’ A A T B B T α α = = Combining materials α i , L i = i L L T L T L L e i i = = α α L L i i e = α α Define equivalent CTE α e :

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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

    1

    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

    =

    =

    =

    thermal strain

    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

    = =

    Combining materials i, Li= iLL

    TLTLL

    e

    ii

    =

    =

    LLii

    e= Define equivalent CTE e:

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    2

    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)

  • J. H. Burge University of Arizona

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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

    = + = + = +

    = +

    To athermalize over a distance L, use two materials so

    1 1 2 2

    1 2

    0EL L LL L L + = =

    + = Using materials with > 0, requires L < 0

    L1, 1 L2, 2

    L1, 1

    L2, 2 L

    1 1 2 2

    1 2

    0L LL L L =

    =

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    Thermal stress Use superposition Calculate stress due to temperature change

    1. Determine expansion, as if unconstrained

    L L T =

    2. Add reaction force that provides constraint by pushing back L

    Solve more general problems the same way.

    F FLLEA

    =

    F E TA

    = =

  • J. H. Burge University of Arizona

    5

    J. H. Burge 30

    Control of thermally induced stresses

    Design preload element to take thermal strain preload stiffness K is

  • J. H. Burge University of Arizona

    6

    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 T = By geometry: or More generally, Integrate for deflection

    T1

    T2 h

    L

    1

    2

    1 2

    1

    L Lh

    L Th

    TL R y

    =

    =

    = =

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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 h

    =

    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 = By analogy : and Generally:

    1

    2 h

    L

    1

    2

    ( )1

    L Th

    L Ty

    TL R y

    =

    =

    = =

  • J. H. Burge University of Arizona

    8

    Integrate to get deflections y(z) Use boundary conditions: For cantilever: at z=0, =0 and y=0 So 0=0 and C=0 For simple support at the ends: y=0 at z=0 and at z=L

    So 0= ( )

    2TL

    y

    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

    =

    =

    = = +

    =

    = = + +

    ( ) 21( )2y

    TL L

    y

    =

    ( ) ( )2 21 12 2 2 8y

    T TL L Ly y

    = =

  • J. H. Burge University of Arizona

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    Heat flow causes thermal gradients For steady state: Qin = Qout

    Define thermal conductivity : ( )1 2T TQ

    L

    =

    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

    L

    = = 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

    = = = =

    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 T For a constant heat source, with power H, the thermal gradient is

    proportional to HT

    So the distortion will be proportional to

    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

    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 governs the rate at which the transient response decays exponentially (1-e-t/) 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 T, so the transient distortion will be proportional to

    This provides a merit function for transient thermal stability

    P

    DC

    =

    D

    2aCD

    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

    D

    D Better stability

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    Thermal time constant: For BK7 glass: = 1.1 W/(m K) = 0.011 J/s /(cm K) = 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.

    ()/0=0.37 (2)/0=0.14 (3)/0=0.05 (4)/0=0.02 (5)/0=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

    http://en.wikipedia.org/wiki/Image:Early_stages_of_Hubble_construction.jpg

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    Use different materials to compensate

    L1, 1

    L2, 2 L

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    Make everything out of the same material (including the mirrors)

    If all optical surfaces and spacing stay in proportion, then the system will still work! Spitzer Telescope with mirrors and mechanics made of beryllium

    T