1 Shen 2/14/00

A5: High Heat Load Design 2/22/00

Qun Shen (CHESS)

Beam line components: Crotch absorbersPrimary beam stopsVacuum windowsApertures and slits

X-ray optics: MirrorsCrystal monochromatorsMultilayers

Topics: ⇒⇒ Heat load concerns⇒⇒ Typical procedures⇒⇒ Design principles & guidelines⇒⇒ Commonly used methods⇒⇒ Design examples

exit-port crotchbeam stopsx-ray mirror

beryllium windows x-ray beamposition monitormonochromator

Karl Smolenski (CHESS)

2 Shen 2/14/00

Table of Contents

Why heat load concerns ………………………………….4

6

Incident power load calculations …………………………….. 6SR power absorption in materials ………………………….. 9Calculation examples ………………………………………… 12

Heat transfer basics ………...…………………………… 15

Three modes of heat transfer ………………………..……… 15Concept of thermal resistance ………………………..………17General case of heat conduction …………………..……….. 19Finite-element analysis ……………………………..……….. 21

Heat transfer by forced convection ………....…………. 22

Basic fluid dynamics ………………………………..………... 22Thermal considerations ………………………..…………...… 26How to improve convective heat transfer ………………….. 27

Thermal deformation and stress ………………………. 32

Strain and stress basics …………………………..………... 32Thermal stress …………..……………………...…………...…34Thermal strain in x-ray optics ……………………………….. 35Designs for high-heat-load monochromators ……………… 37

Homework assignment ……………...……………….….43

3 Shen 2/14/00

Reference Articles

F.M. Anthony, “High heat load optics: an historical overview”, Optical Engi-neering 34, 313 (1995).

K.J. Kim, “Optical and power characteristics of synchrotron radiationOptical Engineering 34, 343 (1995).

F.P. Incropera, D.P. DeWitt, Introduction to Heat Transfer (John Wiley &Sons, New York, 1985).

G.S. Knapp, et al., “Solution to the high heat loads from undulators at thirdgeneration synchrotron sources: cryogenic thin-crystal monochroma-tors”, Rev. Sci. Instrum. 65, 2792 (1994).

W.K. Lee, D.M. Mills, et al., “High heat load monochromator developmentat the Advanced Photon Source”, Optical Engineering 34, 418 (1995).

A.K. Freund, “Diamond single crystals: the ultimate monochromator materi-als for high-power x-ray beams”, Optical Engineering 34, 432 (1995).

D.H. Bilderback, “The potential of cryogenic silicon and bermanium x-raymonochromators for user with large synchrotron heat loads”, Nucl. In-strum. Methods A 246, 434 (1986).

D.H. Bilderback, A.K. Freund, G.S. Knapp, D.M. Mills, The Advanced Pho-ton Source Compton Award , October 1998.

R.K. Smither, et al., “Liquid gallium cooling of silicon crystals in high inten-sity photon beams”, Rev. Sci. Instrum. 60, 1486 (1989).

J. Arthur, “Experience with microchannel and pin-post water cooling of sili-con monochromator crystals”, Optical Engineering 34, 441 (1995).

K.W. Smolenski, Q. Shen, P. Doing, “Improved internally water-cooledmonochromators for a high-power wiggler beam line at CHESS”, Proc.SPIE 3151, 181 (1997).

Q. Shen, K.W. Smolenski, E. Fontes, “Design of a graphite-filter/beryllium-window for CHESS wiggler beam lines”, Proc. SPIE 3151, 116 (1997).

K.W. Smolenski, et al., “Design and initial testing of a synchrotron radiationabsorber for the CESR B-factory”, Proc. SPIE 1739, 186 (1992).

R.D. Watson, “High heat flux issues for plasma facing components in fusionreactors”, Proc. SPIE 1739, 306 (1992).

A. Bar-Cohen, “Thermal management of high-power microelectronic com-ponents: state-of-the-art and future challenges”, Proc. SPIE 1739, 321(1992).

D.B. Tuckerman, “Heat-transfer microstructures for intergrated circuits”,Ph.D. Thesis, Stanford University (1984); D.B. Tucherman and R.F.W.Pease, “High-performance heat sinking for VLSI”, IEEE Elec. Dev. Lett.EDL-2, 126 (1981).

4 Shen 2/14/00

Why do we have to be concerned with heat loads?

⇒⇒ Demands for intense hard x-ray beams:

An estimate:

Suppose the intensity of a monochromatic x-ray beamof λ=0.91 Å is 3x1013 photons/sec, a typical value forCHESS F1 station @ 100 mA, the power in this beam is

∆P= 13600eV x 1.6x10−19Joules/eV x 3x1013/sec = 0.065 Watts.

Tiny! However, this beam is selected from a white wig-gler spectrum by a monochromator Si (111) which has abandwidth of ∆E/E ~ 10−4.

The wiggler spectrum is characterized by a critical en-ergy of Ec=22 keV. The effective bandwidth may be viewedas ∆E/E ~ 2, then the total power incident onto the Si mono-chromator has to be on the order of

P0 ~ 0.065 W x 22/13.6 x 2/0.0001 = 2.1 kW !!

1 10 100

0.01

0.1

1

∆E/E ~ 10-4

X-r

ay F

lux

(a.u

.)

X-ray Energy (keV)

5 Shen 2/14/00

⇒⇒ Beam line integrity and x-ray beam stability:

q Personnel and equipment safety – beam stops, aper-tures, masks, …

q Storage-ring operation – windows, vapor pressures of

q Thermal distortions and drifts of optics – demands distor-tions less than a couple of arc-seconds.

⇒⇒ Possible failures and concerns:

All components:

q Melting of uncooled components

q Evaporation due to high vapor pressure at hi-temp

q Breaking due to thermal stress from nonuniform heating

q Thermal cycling fatigue

Optical components:

q Loss of x-ray flux due to thermal deformation

q Degradation of storage ring brilliance or brightness

q Beam position drifts due to temperature change

q Change in x-ray energy due to d-spacing change

6 Shen 2/14/00

SR Power and Power Absorption

1.1. Incident power load calculations:

(1) Bending magnet radiation:

Bending radius ][/][3.3][ TBGeVEm =ρ

Critical energy ][][665.0][ 2 TBGeVEkeVc =ε

][/][2.2 3 mGeVE ρ=

Vertically integrated power

][/][][14]/[ 4 mAIGeVEmrWP ρ=

Vertical distribution of power:Gaussian-like, HWHM = 0.64/γ

Example:

CHESS hard-bend radiation: m32=ρ , GeV3.5=E

=> W/mr17332/5.03.514 4 =××=P @ 500 mA

At 10 m from the source, the average power density

2W/mm5.13mm101.064.02mm10

W173 =××××

=dAdP .

7 Shen 2/14/00

(2) Wiggler radiation:

Sinusoidal magnetic field )/2sin(0 wzBB λπ=

Deflection parameter ][][943.0 0 TBcmK wλ= , 1>>K

Maximum deflection γδ /K=

Critical energy ][][665.0][ 02max TBGeVEkeVc =ε

2max )/(1)( δφεφε −= cc

Total integrated power][][][][633.0][ 2

02 AImLTBGeVEkWP =

Vertical distribution of power:Gaussian-like, HWHM = 0.64/γ

Horizontal distribution of power:

half circle: 2)/(1)0,( δφφ −=∞f

Example:

F-line wiggler: T2.10 =B , GeV3.5=E , m25.2=L , 22=K

ð kW8.285.025.22.13.5633.0 22 =××××=P @ 500 mA

ð Horizontal angular span: mr2.4/22 == γδ K

At 10 m: 2W/mm536mm101.064.02mm42

kW8.28 =××××

=dAdP .

φδ−δ

Ec

8 Shen 2/14/00

(3) Undulator radiation:

Deflection parameter 1≤K , or maximum γδ /1≤ ,creating peaks in SR spectrum.

Total average power is the same as that for wiggler.

Angular distribution of power: ),( ψφKf

Vertical ψ:

Horizontal φ:

K-J. Kim, Optical Eng.34, 342-352 (1995).

9 Shen 2/14/00

2.2. SR power absorption in materials:

For heat trans-fer analysis of abeam line componentfor a given incidentsynchrotron radiationbeam, it is obviouslyimportant to knowhow much power isabsorbed in the ma-terial and how the power is spatially distributed. This is noteasy as one might think, for the following reasons:

(1) SR spectrum is white and contains x-rays in awide range of energies:

[ ] ε∆εµεφψφψ dzzPzP etransabs ))(exp(1),,,(),,( −−∫= ;

(2) Beam line components have different functions,e.g. a beryllium window is designed to pass mostx-rays while absorbing some power. Thus onecannot use the incident power as the absorbedpower in these filter-type components;

(3) For absorbers such as beam stops, even thoughone can assume that the whole SR power is de-posited entirely on its incident surface (as a worstcase scenario), often a more accurate account ofpower absorption as a function of depth is neces-sary, especially if low-Z materials are used;

(4) SR spectrum can be modified by previous beamline components such as windows and filters, aswell as x-ray optics such as mirrors.

z∆z

ψPtrans

Pabs(ψ,φ,z)

10 Shen 2/14/00

For these reasons, this step of the analysis is oftendone by a numerical calculation, in which one can specify aseries of absorbing materials, and integrate absorbed powerover SR spectrum.

Two program packages can be used for this purpose:

(1) PHOTON or PHOTON2:

Chapman et al., Nucl. Instrum. Meth. A 266, 191 (1988).Dejus, PHOTON2 program, APS/ANL (1992).Dejus, et al., Nucl. Instrum. Meth. A 319, 207 (1992).

q all bend-magnet & wiggler calculations;q composite materials;q 3-d volumetric absorption;q account for wiggler εc variations in PHOTON2;q does not include optics, or undulators.

(2) XOP:

Sanchez del Rio & Dejus, Proc. SPIE 3448, 340 (1998).Sanchez del Rio & Dejus, Proc. SPIE 3152, 148 (1997).Sanchez del Rio et al., Proc. SPIE 3152, 312 (1997).

q complete absorber calculations;q composite materials;q wigglers and undulators;q mirrors and filters;q crystal optics;q multilayers, etc;q interface to other programs, such as XAFS etc.

11 Shen 2/14/00

12 Shen 2/14/00

3.3. Calculation examples:

Example #1: Calculated power absorption using PHOTON invarious materials for a proposed Cornell B-factory [Shen & Bilder-back, Proc. SPIE 1739, 191 (1992)]:

13 Shen 2/14/00

Example #2: Calculated power distribution and absorption in aberyllium window for wiggler B at the Advanced Photon Source[Dejus, et al., Nucl. Instrum. Meth. A 319, 207 (1992)]:

14 Shen 2/14/00

Example #3: Calculated SR spectrum using XOP package, for a49-pole wiggler at CHESS G-line, with graphite filter and berylliumwindows plus a Rh-coated white-beam mirror at an incident angleof 0.44 mrad:

0 10 20 30 40 50 60 70 800

1

2

3

4

5

CHESS G-line wiggler

3.17kW

9.85kW

13

.5ke

V

49pole Ec=14.9keV

w/ 0.5mm C-filter w/ 0.5mm-C&0.5mm-Be Rh-coated mir @0.44mr

Flu

x (x

1015

ph/

s/0.

1%/5

00m

A/1

.6m

r)

Photon Energy (keV)

15 Shen 2/14/00

Heat Transfer Basics

1.1. Three modes of heat transfer: APq /"= = heat flux

−− transfer of energy or power due to temperature differences.

Conduction: L

TTkAP

q)(

" 0−==

k = thermal conductivity (material property,temperature dependent, maybe anisotropic)

Convection: )(" bs TThq −⋅=h = film coefficient (empirical, depends on

coolant properties, flow rate, type of flow,and surface geometry)

Tb = bulk fluid temperature. Temperature rise∆Tb at outlet depends on mass flow rate m&and fluid heat capacity cp: bp TcmP ∆&= .

Radiation: 4" sTq σε=σ = Stefan-Boltzmann constant

= 5.67x10-8 W/m2-K4

ε = emissivity (ε ≈ 0.01-0.1 for polished met-als, and 0.3-0.9 for others)

Note: σTs4 ~ 5.67 W/cm2 @ 1000K.

A

L

PT T0

Tb

Tsq″forced convection

q″Ts

16 Shen 2/14/00

Thermophysical properties of selected solids (@ 300K):

Tmelt

(K)ρ

(g/cm3)cp

(J/g-K)k

(W/cm-K)α

(10-6/K)

Aluminum 933 2.70 0.90 2.37 23.2Beryllium 1550 1.85 1.82 2.00 11.5Copper 1358 8.93 0.38 4.01 17.0Glidcop 8.90 3.65 16.6Tungsten 3660 19.3 0.13 1.74 4.5S Steel (304) 1670 7.90 0.48 0.15 16

Germanium 1211 5.36 0.32 0.60 5.8Silicon 1685 2.33 0.71 1.48 2.6Diamond 2a NA 3.5 0.51 23 1.1

F.P. Incropera, D.P. DeWitt, Introduction to Heat Transfer (John Wiley &Sons, New York, 1985).

Y.S. Youloukian, in A Physicist’s Desk Reference, 2nd ed., edited by H.L.Anderson (Am. Inst. Phys., New York, 1989).

W.K. Lee et al., Optical Engineering 34, 419 (1995).

17 Shen 2/14/00

2.2. Concept of thermal resistance:

−− for steady-state, 1-D, linear conduction and convection.

Analogy to electric circuit:RVI =

V ó ∆TI ó P

Thermal resistance:kALRcond = , for conduction;

hARconv

1= , for convection;

vpflow cvcmR

&&11 == , for fluid flow.

The concept of thermal resistance is useful in estimatingoverall temperature rises in heat transfer designs.

These simple equations also tell us the ways to improve heattransfer.

Conduction: -- Increasing area by inclination: q” => q”sinθ.

-- Choose a material with larger k.

-- Make L shorter (heat sink closer).

θq”

18 Shen 2/14/00

Example of thermal resistance in series:

Total thermal resistance vchAkAtR

ptotal &ρ

11 ++=

ð totalRPTT ⋅+= 0max

C118

)0077.022.0(403012.45.009.63

15.01

15.04

4.04030

o=

++×+=

××+

×+

××+=

P

water flow rate v&

t k

T0=30oC

TmaxA

h

P = 40 W, A = 0.5 cm2, t = 4 mmCu k = 4 W/cm-Kfilm coefficient h = 1 W/cm2-Kvolume flow rate v& = 0.5 gpmspecific heat cv=ρcp

= 4.12 J/cm3-K(1 gpm = 63.09 cm3/sec)

19 Shen 2/14/00

3.3. General case of heat conduction:

The most general equation that governs heat conductionin a complex system is derived by energy conservation,which is given by

tTcqTk

∂∂=+∇⋅∇ p)( ρ& ,

or, in Cartesian coordinates:

tTcq

zTk

zyTk

yxTk

x ∂∂=+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂

p)()()( ρ& ,

where ),,,( tzyxTT = is the time-dependent temperature dis-tribution that one tries to solve, k is the thermal conductivity,q& is the internal heat generation rate per unit volume, ρ isthe density, and cp is the heat capacity.

In practice, one often employs a finite element analysis(FEA) to solve this complicated equation. Nowadays thereare powerful commercial FEA software packages (e.g. AN-SYS) that are used for this purpose. The best part is that apackage such as ANSYS can handle complicated geome-tries, and can now even take mo dels from AutoCAD!

q&Initial and boundary conditions:e.g. T =T0 at back surface, andconst. heat flux at front surface.

20 Shen 2/14/00

One still needs to identify the correct boundary conditionsand heat generations. There are in general three types ofboundary conditions:

F.P. Incropera, D.P. DeWitt, Introduction to Heat Transfer (John Wiley &Sons, New York, 1985).

21 Shen 2/14/00

4.4. Finite-elements analysis:

q Type of analysisq Material propertiesq Geometry definitionq Heat loads and boundary conditionsq Calculationq Review of results

ANSYS Example: A beryllium window at CHESS A-line with amis-steered beam.

22 Shen 2/14/00

Heat Transfer by Forced Convection

−− lots of empirical formulas and parameters, and be carefulwith definitions and units!

1.1. Basic fluid dynamics:

Reynolds number : represents the flow rate

µρ uDRe ≡ ,

where ρ is the fluid density, µ dynamic viscosity, u linearvelocity, and D the hydraulic diameter.

Example: For 1 gpm water flow in a ¼” ID pipe, viscosityat room temperature for water is 0.9 cp (centipoise =0.01g/cm-sec), Reynolds number is:

14056sec-g/cm009.0]cm)2/54.225.0([

cm54.225.0sec/cm09.63g/cm0.122

33=

×××××

=π

Re .

23 Shen 2/14/00

Hydraulic diameter : represents characteristic size of theflow channel. It is defined as four timesthe cross-section area divided by theperimeter of the flow channel:

pA

D c4≡ .

Circular: D = diameter

Rectangular: ba

abD+

= 2

b >> a: aD 2≈

Square (b = a): aD =

Prandtl number : represents fluid properties

k

cPr pµ

≡ ,

where µ is the dynamic or absolute viscos-ity that relates to kinematic viscosity ν via

νρµ = [AIP Handbook, 3rd ed., pp.232-248(McGraw-Hill, New York, 1972)].

Example: For water, k=0.0061 W/cm-K,µ =0.009 g/cm-sec, cp=4.2 J/g-K, thus:Pr = 6.2 at room temperature.

D

b

a

24 Shen 2/14/00

Flow conditions : laminar vs. turbulent

Critical Reynolds number: 2300≈Re

2300≤Re : laminar flow2300≥Re : turbulent flow begins4000≥Re : fully turbulent flow.

25 Shen 2/14/00

Friction factor : pressure drop or gradient ∆P/∆x along flowchannel is related to the work needed to maintain the fluid’skinetic energy ρu2/2:

221 u

Df

xP ρ

∆∆

−=

Laminar flow:Re

f 64=

Turbulent flow: 4/1316.0 −= Ref , 20000≤Re(smooth pipes) 5/1184.0 −= Ref , 20000≥Re

26 Shen 2/14/00

2.2. Thermal considerations:

Nusselt number: represents heat transfer coefficient h

kDhNu ≡ .

The whole aspect of heat transfer by forced convection isreduced to finding the proper Nu.

Laminar flow: exact solutions for circular channels

36.4=Nu , for constant heat flux qs”66.3=Nu , for constant temperature Ts

Example: For laminar water flow in ¼” tubing totransfer a constant heat flux, the film coefficient is

042.054.225.0

0061.036.4 =×

×=⋅=D

kNuh W/cm2-K.

Turbulent flow: empirical correlations

Dittus-Boelter equation: 3.08.0023.0 PrReNu =

Example: For fully developed turbulent water flow in¼” tubing with a flow rate of 1 gpm, the convectionfilm coefficient is

54.225.00061.0]2.614056023.0[ 3.08.0

××××=⋅=

DkNuh

= 0.79 W/cm2-K.

27 Shen 2/14/00

3.3. How to improve convective heat transfer:

TAhq ∆⋅⋅=

Improving film-coefficient h:D

kNuh ⋅=

−− key is to bring fast-flowing fluid closer to surface.

q High-flow-rate turbulent flow: 8.08.0 ~ vmh &&∝

3 gpm water flow through ¼” tube: h = 2.0 W/cm2-K

q Narrower channels: Dh 1∝ for laminar8.11 Dh ∝ for turbulent

micro-channel laminar flow: D ~ 50 µm,

=×=⋅=005.0

0061.036.4D

kNuh 5.3 W/cm2-K

−− D.B. Tuckerman, Ph.D. Thesis, Stanford University(1984); D.B. Tucherman and R.F.W. Pease, IEEEElec. Dev. Lett. EDL-2, 126 (1981).

−− J. Arthur, et al., Rev. Sci. Instrum. 63, 433 (1992).

q Better heat conducting coolant: e.g. liquid metals

−− R.K. Smither, W. Lee, A. Macrander, D. Mills, and S.Rogers, Rev. Sci. Instrum. 63, 1746 (1992).

R.K. Smither, Proc. SPIE 1739, 116 (1992).

q Jet impingement cooling:

−− S. Sharma, L.E. Berman, J.B. Hastings, and M. Hart,Proc. SPIE 1739, 116 (1992).

q Channel inserts to increase turbulence.

28 Shen 2/14/00

Increasing surface area A:

q Use of fins:

q Rectangular coolant channels with high aspect ratiocan be viewed as equivalent to fins:

29 Shen 2/14/00

q Fin effectiveness fε : defined as the ratio of the fin

heat transfer rate to the heat transfer rate that wouldexist without the fin.

For long rectangular channels with high aspect ratio(L >> H >> wr),

rf hw

k2=ε ,

where k here is the thermal conductivity of the fin orrib material.

Example: For 1 mm by 10 mm long rectangularcooling channels in copper (k = 4 W/cm-K), film coeffi-cient h = 1 W/cm2-K, and rib width wr =1 mm, the ef-fectiveness of a single rib is

94.81.01

42 =××=fε .

The overall ratio of the effective surface area Af withthe channels to that A without the channels is thus

rc

frcf

ww

ww

A

A

++

=ε

.

For wr = wc: 0.52

1=

+= ff

A

A ε

30 Shen 2/14/00

Coolant phase change: −− change in ∆T dependence

q Nucleate boiling:

3)(" sats TTCq −≈

where satT is the saturation temperature or theboiling temperature that depends on pressure.Constant C depends on latent heat, surface ten-sion, Prandtl number, among others.

F.P. Incropera, D.P. DeWitt, Introduction to Heat Transfer (JohnWiley & Sons, New York, 1985).

31 Shen 2/14/00

q Critical heat flux max"q :is obviously an important quantity to consider in adesign involving boiling. A requirement is that eve-rywhere on the channel wall, heat flux transferredto the fluid, Thq ∆⋅=" , should be less (a lot less, tobe safe) than max"q .

Critical heat flux max"q is usually in excess of 100W/cm2, and values as high as 2600 W/cm2 havebeen reported in the literature [Morimoto, et al.,“Design of the crotch for Spring-8”, in Vacuum De-sign of Synch. Light Sources, AIP Conf. Proc. 236,110 (1991)].

q Caution: nucleate boiling results are highly empiri-cal. It should be used with considerable caution.

q Binary fluid-solid mixture as coolant:

−− P.W. Lorffen, et al. (Oxford Instrum.), Proc. SPIE3448, 88 (1998).

32 Shen 2/14/00

Thermal Deformation and Stress

1.1. Strain and stress basics:

Hooke’s law:llEE ∆εσ ==

(stress σ and strain ε are 2nd ranktensors for anisotropic or crystallinematerials)

Young’s modulus: E = tensile stress / linear strainShear modulus: G = shear stress / rotational strainPoisson’s ratio: ν = (E−2G) / 2G

Tensile strength: maximum tensile stressYield stress σy: stress level at which plastic deforma-

tion starts to develop

E (106 psi) σy (103 psi)

Be 42 60Al 10 24Cu 16 37

Si crystal 19 10-50

(1 psi = 6894.8 Pa =68948 dyne/cm2, 1 atm = 14.7 psi).

l∆l

33 Shen 2/14/00

Example: temperature dependent stress-strain curvesfor beryllium sheets (Brush-Wellman, Inc.)

34 Shen 2/14/00

2.2. Thermal stress:

Strain due to thermal expansion: Tll ∆α∆ε ⋅=≡

with α being the linear thermal expansion coefficient.

Stress due to thermal expansion: TE ∆ασ = .

Example: Suppose that a thin beryllium window at a beamline reaches a maximum temperature of 100oC at the centerwhere the SR white beam hits. Given that α = 11.5x10-6 /oCand E = 42x106 psi for beryllium, thermal stress at the centercan be estimated as

psi225,36)25100(105.111042 66 =−××××= −σ ,

which is below the yield stress of 60,000 psi for beryllium.

Caution: For a design with a complex shape and boundaryconditions, using TE ∆ασ = may not provide the accuratestress values. The only way to obtain a more accuratethermal-strain/stress distribution is to use a finite-elementsanalysis. ANSYS package allows for a subsequent thermalstress evaluation after a thermal analysis is performed.

Design criterion: yieldσσ ≤

35 Shen 2/14/00

3.3. Thermal strain in x-ray optics:

⇒ R.K. Smither, Proc. SPIE 1739, 116 (1992).

Three types of strains: −− thermal bump

−− overall bending

−− d-spacing change ∆d

36 Shen 2/14/00

For a matching double-crystal monochromator:the main effect is the loss of x-ray throughput or flux.

Symptoms:

q Flux not linear with storage-ring current:

q Rocking curve of the 2nd crystal is much wider than intrin-sic Darwin width.

I (mA)

Flux

~ 5-10 arc-sec

37 Shen 2/14/00

Designs for high-heat-load monochromators:

⇒ Micro-channel water-cooling

−− J. Arthur, et al., Rev. Sci. Instrum. 63, 433 (1992).

q Laminar water flowq High heat-transfer film coefficient

q Difficult to fabricate and be strain-freeq Glass diffusion bonds suffer radiation damage

38 Shen 2/14/00

⇒ Pin-post water-cooling

−− J. Arthur, et al., Rev. Sci. Instrum. 63, 433 (1992).−− T. Ishikawa, et al., Proc. SPIE 3448, 2 (1998).

q Turbulent water flowq High flow rate has higher cooling capacityq High heat-transfer film coefficient

q Difficult to fabricate and be strain-freeq Glass frit bonds suffer radiation damage

⇒ Mini-channel water-cooling

−− K.W. Smolenski, et al., Proc. SPIE 3151, 181 (1997).−− A.K. Freund, et al., Proc. SPIE 3151, 216 (1997).

q Turbulent water flowq High flow rate has higher cooling capacityq Decent heat-transfer film coefficientq Metal (Ag, Au, Al) diffusion or In-based solder bonds

q Fabrication strain?

39 Shen 2/14/00

CHESS Design −− K.W. Smolenski, et al., Proc. SPIE 3151, 181(1997).

40 Shen 2/14/00

⇒ Cryogenic cooling

--- D.H. Bilderback, A.K. Freund, G.S. Knapp, D.M. Mills, TheAdvanced Photon Source Compton Award, October 1998.

q Turbulent LN2 flowq High thermal conductivity at low temperaturesq Almost zero thermal expanstion coefficient

q Narrow operable temperature range limits the totalcooling capacity to ~ 2kW, for LN2. So this is so farfor undulator beams only.

q Possible liquid propane cooling???

H2O LN2 C3H8

Density (g/cm3) 1.0 0.81 0.71 Specific heat (J/g-K) 4.2 2.0 2.0 Working pressure (bar) 1 10 2

Working ∆T (K) 75 27 170Relative cooling capacity 1 0.1 0.5

Working temperature (K) 300 <140 α/k for Si (10-6cm/W) 2.0 0.05

⇒ Isotopically pure Si ?

--- W.S. Capinski, et al., “Thermal conductivity of isotopicallyenriched Si”, Appl. Phys. Lett. 71, 2109 (1997).

41 Shen 2/14/00

42 Shen 2/14/00

⇒ Use of diamond crystals −− limited to small beams

−− J. Sellschop, Proc. SPIE 3448, 40 (1998).−− A.K. Freund, et al., Proc. SPIE 3448, 53 (1998).

43 Shen 2/14/00

Homework Assignment

A white-beam x-ray mirror made of Glidcop (as shown schematically in thefigure) is designed to absorb 15 kW of a wiggler beam at a fixed incidentangle of 3.9 mrad, and is water-cooled with a total flow rate of 12 gpmthrough 22 parallel mini-channels. The mirror is located 19 m from thewiggler which has a horizontal opening angle of 4.2 mrad and is operatedin a storage ring of 5.3 GeV and 500 mA.

(a) Assuming that the absorbed power is uniformly distributed within thenatural opening angles of the wiggler, calculate the power density onthe mirror surface.

(b) Given the water channel dimensions in the figure, find out the hydraulicdiameter D, Reynolds number Re, the heat transfer film coefficient hand the pressure drop over the 1 m length (assuming smooth pipe con-dition).

(c) Using concepts of thermal resistance and fin effectiveness, and ignor-ing the heating up of water flow for now, calculate the temperature riseon the mirror surface.

(d) Calculate the water temperature difference between the inlet on oneend and the outlet on the other end of the mirror. Estimate the ‘thermaltilt’ due to the water temperature rise along the 1 m length of the mirror,by taking an effective thickness from the mirror surface to the bottom ofthe flow channels.

(e) What is the angular deviation ∆θ in the mirror-reflected beam due to this‘tilt’? Estimate the energy shift ∆Ε due to ∆θ for 12 keV x-rays from a Si(111) double-crystal monochromator located downstream of the mirror.