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

Transcription

1 A5: High Heat Load Design 2/22/00 Qun Shen (CHESS) Beam line components: X-ray optics: Crotch absorbers Primary beam stops Vacuum windows Apertures and slits Mirrors Crystal monochromators Multilayers Karl Smolenski (CHESS) x-ray mirror beam stops exit-port crotch monochromator beryllium windows x-ray beam position monitor Topics: Heat load concerns Typical procedures Design principles & guidelines Commonly used methods Design examples 1 Shen 2/14/00

2 Table of Contents Why heat load concerns.4 Incident power load calculations.. 6 SR power absorption in materials.. 9 Calculation examples 12 Heat transfer basics Three modes of heat transfer.. 15 Concept of thermal resistance.. 17 General case of heat conduction Finite-element analysis Heat transfer by forced convection Basic fluid dynamics Thermal considerations How to improve convective heat transfer.. 27 Thermal deformation and stress. 32 Strain and stress basics Thermal stress Thermal strain in x-ray optics.. 35 Designs for high-heat-load monochromators 37 Homework assignment Shen 2/14/00

3 Reference Articles F.M. Anthony, High heat load optics: an historical overview, Optical Engineering 34, 313 (1995). K.J. Kim, Optical and power characteristics of synchrotron radiation Optical 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 third generation synchrotron sources: cryogenic thin-crystal monochromators, Rev. Sci. Instrum. 65, 2792 (1994). W.K. Lee, D.M. Mills, et al., High heat load monochromator development at the Advanced Photon Source, Optical Engineering 34, 418 (1995). A.K. Freund, Diamond single crystals: the ultimate monochromator materials for high-power x-ray beams, Optical Engineering 34, 432 (1995). D.H. Bilderback, The potential of cryogenic silicon and bermanium x-ray monochromators for user with large synchrotron heat loads, Nucl. Instrum. Methods A 246, 434 (1986). D.H. Bilderback, A.K. Freund, G.S. Knapp, D.M. Mills, The Advanced Photon Source Compton Award, October R.K. Smither, et al., Liquid gallium cooling of silicon crystals in high intensity photon beams, Rev. Sci. Instrum. 60, 1486 (1989). J. Arthur, Experience with microchannel and pin-post water cooling of silicon monochromator crystals, Optical Engineering 34, 441 (1995). K.W. Smolenski, Q. Shen, P. Doing, Improved internally water-cooled monochromators 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/berylliumwindow for CHESS wiggler beam lines, Proc. SPIE 3151, 116 (1997). K.W. Smolenski, et al., Design and initial testing of a synchrotron radiation absorber for the CESR B-factory, Proc. SPIE 1739, 186 (1992). R.D. Watson, High heat flux issues for plasma facing components in fusion reactors, Proc. SPIE 1739, 306 (1992). A. Bar-Cohen, Thermal management of high-power microelectronic components: 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). 3 Shen 2/14/00

4 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 beam of λ=0.91 Å is 3x10 13 photons/sec, a typical value for CHESS F1 100 ma, the power in this beam is P= 13600eV x 1.6x10 19 Joules/eV x 3x10 13 /sec = Watts. Tiny! However, this beam is selected from a white wiggler spectrum by a monochromator Si (111) which has a bandwidth of E/E ~ X-ray Flux (a.u.) E/E ~ X-ray Energy (kev) The wiggler spectrum is characterized by a critical energy of E c =22 kev. The effective bandwidth may be viewed as E/E ~ 2, then the total power incident onto the Si monochromator has to be on the order of P 0 ~ W x 22/13.6 x 2/ = 2.1 kw!! 4 Shen 2/14/00

5 Beam line integrity and x-ray beam stability: Personnel and equipment safety beam stops, apertures, masks, Storage-ring operation windows, vapor pressures of Thermal distortions and drifts of optics demands distortions less than a couple of arc-seconds. Possible failures and concerns: All components: Melting of uncooled components Evaporation due to high vapor pressure at hi-temp Breaking due to thermal stress from nonuniform heating Thermal cycling fatigue Optical components: Loss of x-ray flux due to thermal deformation Degradation of storage ring brilliance or brightness Beam position drifts due to temperature change Change in x-ray energy due to d-spacing change 5 Shen 2/14/00

6 SR Power and Power Absorption 1. Incident power load calculations: (1) Bending magnet radiation: Bending radius ρ[ m ] = 3.3E[ GeV]/ B[ T ] Critical energy ε [ ] c kev = E [ GeV] B[ T ] Vertically integrated power P[ W / mr] = 14E 4[ GeV ] I[ A]/ ρ[ m] = 2.2 E3[ GeV ]/ ρ[ m] Vertical distribution of power: Gaussian-like, HWHM = 0.64/γ Example: CHESS hard-bend radiation: ρ = 32 m, E = 5.3 GeV => P = /32 = ma At 10 m from the source, the average power density dp da = 173 W 10 mm mm = 13.5 W/mm 2. 6 Shen 2/14/00

7 (2) Wiggler radiation: Sinusoidal magnetic field B = B sin(2πz / λ ) 0 w Deflection parameter K = λ [ cm ] B 0[ T ], K >> 1 Maximum deflection δ = K / γ max[ 2 0 max c ( φ ) = ε c 1 ( φ / δ ) 2 Critical energy ε c kev] = 0.665E [ GeV] B [ T ] ε w E c Total integrated power P kw] = 0.633E 2 [ GeV] B [ T] L[ m] I[ A ] [ 2 0 δ δ φ Example: Vertical distribution of power: Gaussian-like, HWHM = 0.64/γ Horizontal distribution of power: half circle: ( φ, 0) = 1 ( φ / δ ) 2 f F-line wiggler: B 0 =1.2 T, E = 5.3 GeV, L = 2.25 m, K = = P = ma Horizontal angular span: 2 δ = 2K / γ = 4.2 mr dp 28.8 kw At 10 m: = = 536 W/mm 2. da 42 mm mm 7 Shen 2/14/00

8 (3) Undulator radiation: Deflection parameter K 1, or maximum δ 1/ γ, creating peaks in SR spectrum. Total average power is the same as that for wiggler. Angular distribution of power: ( φ, ψ ) f K Vertical ψ: Horizontal φ: K-J. Kim, Optical Eng. 34, (1995). 8 Shen 2/14/00

9 2. SR power absorption in materials: For heat transfer analysis of a abs (ψ,φ,z) P beam line component for a given incident ψ synchrotron radiation beam, it is obviously important to know z how much power is z absorbed in the material and how the power is spatially distributed. This is not easy as one might think, for the following reasons: (1) SR spectrum is white and contains x-rays in a wide range of energies: P trans [ 1 exp( µ ( ε) z ] dε Pabs ( ψ, φ, z) = Ptrans ( ψ, φ, z, ε) e ) ; (2) Beam line components have different functions, e.g. a beryllium window is designed to pass most x-rays while absorbing some power. Thus one cannot use the incident power as the absorbed power in these filter-type components; (3) For absorbers such as beam stops, even though one can assume that the whole SR power is deposited entirely on its incident surface (as a worst case scenario), often a more accurate account of power absorption as a function of depth is necessary, especially if low-z materials are used; (4) SR spectrum can be modified by previous beam line components such as windows and filters, as well as x-ray optics such as mirrors. 9 Shen 2/14/00

10 For these reasons, this step of the analysis is often done by a numerical calculation, in which one can specify a series of absorbing materials, and integrate absorbed power over 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). all bend-magnet & wiggler calculations; composite materials; 3-d volumetric absorption; account for wiggler ε c variations in PHOTON2; 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). complete absorber calculations; composite materials; wigglers and undulators; mirrors and filters; crystal optics; multilayers, etc; interface to other programs, such as XAFS etc. 10 Shen 2/14/00

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12 3. Calculation examples: Example #1: Calculated power absorption using PHOTON in various materials for a proposed Cornell B-factory [Shen & Bilderback, Proc. SPIE 1739, 191 (1992)]: 12 Shen 2/14/00

13 Example #2: Calculated power distribution and absorption in a beryllium window for wiggler B at the Advanced Photon Source [Dejus, et al., Nucl. Instrum. Meth. A 319, 207 (1992)]: 13 Shen 2/14/00

14 Example #3: Calculated SR spectrum using XOP package, for a 49-pole wiggler at CHESS G-line, with graphite filter and beryllium windows plus a Rh-coated white-beam mirror at an incident angle of 0.44 mrad: 5 CHESS G-line wiggler Flux (x10 15 ph/s/0.1%/500ma/1.6mr) keV 3.17kW 9.85kW 49pole E c =14.9keV w/ 0.5mm C-filter w/ 0.5mm-C&0.5mm-Be Rh-coated Photon Energy (kev) 14 Shen 2/14/00

15 Heat Transfer Basics 1. Three modes of heat transfer: q P / A "= = heat flux transfer of energy or power due to temperature differences. Conduction: P k ( T T0 ) q" = = A L k = thermal conductivity (material property, temperature dependent, maybe anisotropic) Convection: q" = h ( T s Tb ) h = film coefficient (empirical, depends on coolant properties, flow rate, type of flow, and surface geometry) T b = bulk fluid temperature. Temperature rise T b at outlet depends on mass flow rate m& and fluid heat capacity c p : P = mc & p T. P T T 0 A L b forced convection T b q T s Radiation: q" =ε σ T 4 s σ = Stefan-Boltzmann constant = 5.67x10-8 W/m 2 -K 4 ε = emissivity (ε for polished metals, and for others) Note: σt 4 s ~ 5.67 W/cm 1000K. T s q 15 Shen 2/14/00

16 Thermophysical properties of selected solids 300K): T melt (K) ρ (g/cm 3 ) c p (J/g-K) k (W/cm-K) α (10-6 /K) Aluminum Beryllium Copper Glidcop Tungsten S Steel (304) Germanium Silicon Diamond 2a NA 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, 2 nd ed., edited by H.L. Anderson (Am. Inst. Phys., New York, 1989). W.K. Lee et al., Optical Engineering 34, 419 (1995). 16 Shen 2/14/00

17 2. Concept of thermal resistance: for steady-state, 1-D, linear conduction and convection. Analogy to electric circuit: Thermal resistance: I = V R V T I P L R cond =, for conduction; ka 1 R conv =, for convection; ha 1 1 R flow = =, for fluid flow. mc & v& c The concept of thermal resistance is useful in estimating overall temperature rises in heat transfer designs. These simple equations also tell us the ways to improve heat transfer. p v Conduction: -- Increasing area by inclination: q => q sinθ. q θ -- Choose a material with larger k. -- Make L shorter (heat sink closer). 17 Shen 2/14/00

18 Example of thermal resistance in series: T 0 =30 o C P A t k h T max water flow rate v& P = 40 W, A = 0.5 cm 2, t = 4 mm Cu k = 4 W/cm-K film coefficient h = 1 W/cm 2 -K volume flow rate v& = 0.5 gpm specific heat c v =ρc p = 4.12 J/cm 3 -K (1 gpm = cm 3 /sec) Total thermal resistance T max = 118 C R total = T 0 + P Rtotal = = ( ) o = t ka ha 1 ρ c p v& Shen 2/14/00

19 3. General case of heat conduction: q& Initial and boundary conditions: e.g. T =T 0 at back surface, and const. heat flux at front surface. The most general equation that governs heat conduction in a complex system is derived by energy conservation, which is given by T ( k T) + q& = ρ cp, t or, in Cartesian coordinates: T T T T ( k ) + ( k ) + ( k ) + q& = ρ cp, x x y y z z t where T = T ( x, y, z, t) is the time-dependent temperature distribution that one tries to solve, k is the thermal conductivity, q& is the internal heat generation rate per unit volume, ρ is the density, and c p is the heat capacity. In practice, one often employs a finite element analysis (FEA) to solve this complicated equation. Nowadays there are powerful commercial FEA software packages (e.g. AN- SYS) that are used for this purpose. The best part is that a package such as ANSYS can handle complicated geometries, and can now even take mo dels from AutoCAD! 19 Shen 2/14/00

20 One still needs to identify the correct boundary conditions and heat generations. There are in general three types of boundary conditions: F.P. Incropera, D.P. DeWitt, Introduction to Heat Transfer (John Wiley & Sons, New York, 1985). 20 Shen 2/14/00

21 4. Finite-elements analysis: Type of analysis Material properties Geometry definition Heat loads and boundary conditions Calculation Review of results ANSYS Example: A beryllium window at CHESS A-line with a mis-steered beam. 21 Shen 2/14/00

22 Heat Transfer by Forced Convection lots of empirical formulas and parameters, and be careful with definitions and units! 1. Basic fluid dynamics: Reynolds number : represents the flow rate ρ ud Re, µ where ρ is the fluid density, µ dynamic viscosity, u linear velocity, and D the hydraulic diameter. Example: For 1 gpm water flow in a ¼ ID pipe, viscosity at room temperature for water is 0.9 cp (centipoise = 0.01g/cm-sec), Reynolds number is: Re 3 1.0g/cm 63.09cm / sec cm = [ π ( / 2) 2 cm 2] 0.009g/cm - sec 3 = Shen 2/14/00

23 Hydraulic diameter : represents characteristic size of the flow channel. It is defined as four times the cross-section area divided by the perimeter of the flow channel: D 4Ac. p D Circular: D = diameter b Rectangular: D ab = 2 a + b a b >> a: Square (b = a): D 2a D = a Prandtl number : represents fluid properties µ c p Pr, k where µ is the dynamic or absolute viscosity that relates to kinematic viscosity ν via µ = ρν [AIP Handbook, 3 rd ed., pp (McGraw-Hill, New York, 1972)]. Example: For water, k= W/cm-K, µ =0.009 g/cm-sec, c p =4.2 J/g-K, thus: Pr = 6.2 at room temperature. 23 Shen 2/14/00

24 Flow conditions : laminar vs. turbulent Critical Reynolds number: Re 2300 Re 2300: laminar flow Re 2300: turbulent flow begins Re 4000: fully turbulent flow. 24 Shen 2/14/00

25 Friction factor : pressure drop or gradient P/ x along flow channel is related to the work needed to maintain the fluid s kinetic energy ρu 2 /2: P x = f D ρ u Laminar flow: f = 64 Re Turbulent flow: f = Re 1/ 4, Re (smooth pipes) f = Re 1/ 5, Re Shen 2/14/00

26 2. Thermal considerations: Nusselt number: represents heat transfer coefficient h h D Nu. k The whole aspect of heat transfer by forced convection is reduced to finding the proper Nu. Laminar flow: exact solutions for circular channels Nu = 4.36, for constant heat flux q s Nu = 3.66, for constant temperature T s Example: For laminar water flow in ¼ tubing to transfer a constant heat flux, the film coefficient is h = Nu k D = = 0.042W/cm 2 -K. Turbulent flow: empirical correlations Dittus-Boelter equation: Nu = 0.023Re0.8Pr 0. 3 Example: For fully developed turbulent water flow in ¼ tubing with a flow rate of 1 gpm, the convection film coefficient is h = Nu k D = [ ] = 0.79 W/cm 2 -K. 26 Shen 2/14/00

27 3. How to improve convective heat transfer: q = h A T Improving film-coefficient h: h = Nu k D key is to bring fast-flowing fluid closer to surface. High-flow-rate turbulent flow: h m& 0.8 ~ v& gpm water flow through ¼ tube: h = 2.0 W/cm 2 -K Narrower channels: h 1 D for laminar h 1 D 1.8 for turbulent micro-channel laminar flow: D ~ 50 µm, Nu k h = = = 5.3 W/cm 2 -K D D.B. Tuckerman, Ph.D. Thesis, Stanford University (1984); D.B. Tucherman and R.F.W. Pease, IEEE Elec. Dev. Lett. EDL-2, 126 (1981). J. Arthur, et al., Rev. Sci. Instrum. 63, 433 (1992). 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). Jet impingement cooling: S. Sharma, L.E. Berman, J.B. Hastings, and M. Hart, Proc. SPIE 1739, 116 (1992). Channel inserts to increase turbulence. 27 Shen 2/14/00

28 Increasing surface area A: Use of fins: Rectangular coolant channels with high aspect ratio can be viewed as equivalent to fins: 28 Shen 2/14/00

29 Fin effectiveness ε f : defined as the ratio of the fin heat transfer rate to the heat transfer rate that would exist without the fin. For long rectangular channels with high aspect ratio (L >> H >> w r ), 2k ε f =, hw where k here is the thermal conductivity of the fin or rib material. r Example: For 1 mm by 10 mm long rectangular cooling channels in copper (k = 4 W/cm-K), film coefficient h = 1 W/cm 2 -K, and rib width w r =1 mm, the effectiveness of a single rib is 2 4 ε f = = The overall ratio of the effective surface area A f with the channels to that A without the channels is thus A f A = w c w + w ε c r + w r f. Af 1+ ε f For w r = w c : = = 5. 0 A 2 29 Shen 2/14/00

30 Coolant phase change: change in T dependence Nucleate boiling: q " C ( T sat ) 3 s T where T sat is the saturation temperature or the boiling temperature that depends on pressure. Constant C depends on latent heat, surface tension, Prandtl number, among others. F.P. Incropera, D.P. DeWitt, Introduction to Heat Transfer (John Wiley & Sons, New York, 1985). 30 Shen 2/14/00

31 Critical heat flux q " max : is obviously an important quantity to consider in a design involving boiling. A requirement is that everywhere on the channel wall, heat flux transferred to the fluid, q" = h T, should be less (a lot less, to be safe) than q ". max Critical heat flux q" max is usually in excess of 100 W/cm 2, and values as high as 2600 W/cm 2 have been reported in the literature [Morimoto, et al., Design of the crotch for Spring-8, in Vacuum Design of Synch. Light Sources, AIP Conf. Proc. 236, 110 (1991)]. Caution: nucleate boiling results are highly empirical. It should be used with considerable caution. Binary fluid-solid mixture as coolant: P.W. Lorffen, et al. (Oxford Instrum.), Proc. SPIE 3448, 88 (1998). 31 Shen 2/14/00

32 Thermal Deformation and Stress 1. Strain and stress basics: l l Hooke s law: Young s modulus: Shear modulus: Poisson s ratio: Tensile strength: Yield stress σ y : l σ = Eε = E l (stress σ and strain ε are 2 nd rank tensors for anisotropic or crystalline materials) E = tensile stress / linear strain G = shear stress / rotational strain ν = (E 2G) / 2G maximum tensile stress stress level at which plastic deformation starts to develop E (10 6 psi) σ y (10 3 psi) Be Al Cu Si crystal (1 psi = Pa =68948 dyne/cm 2, 1 atm = 14.7 psi). 32 Shen 2/14/00

33 Example: temperature dependent stress-strain curves for beryllium sheets (Brush-Wellman, Inc.) 33 Shen 2/14/00

34 2. Thermal stress: l Strain due to thermal expansion: ε = α T l with α being the linear thermal expansion coefficient. Stress due to thermal expansion: σ = Eα T. Example: Suppose that a thin beryllium window at a beam line reaches a maximum temperature of 100 o C at the center where the SR white beam hits. Given that α = 11.5x10-6 / o C and E = 42x10 6 psi for beryllium, thermal stress at the center can be estimated as σ (100 25) = = 36,225psi, which is below the yield stress of 60,000 psi for beryllium. Caution: For a design with a complex shape and boundary conditions, using σ = Eα T may not provide the accurate stress values. The only way to obtain a more accurate thermal-strain/stress distribution is to use a finite-elements analysis. ANSYS package allows for a subsequent thermal stress evaluation after a thermal analysis is performed. Design criterion: σ σ yield 34 Shen 2/14/00

35 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 35 Shen 2/14/00

36 For a matching double-crystal monochromator: the main effect is the loss of x-ray throughput or flux. Symptoms: Flux not linear with storage-ring current: Flux I (ma) Rocking curve of the 2 nd crystal is much wider than intrinsic Darwin width. ~ 5-10 arc-sec 36 Shen 2/14/00

37 Designs for high-heat-load monochromators: Micro-channel water-cooling J. Arthur, et al., Rev. Sci. Instrum. 63, 433 (1992). Laminar water flow High heat-transfer film coefficient Difficult to fabricate and be strain-free Glass diffusion bonds suffer radiation damage 37 Shen 2/14/00

38 Pin-post water-cooling J. Arthur, et al., Rev. Sci. Instrum. 63, 433 (1992). T. Ishikawa, et al., Proc. SPIE 3448, 2 (1998). Turbulent water flow High flow rate has higher cooling capacity High heat-transfer film coefficient Difficult to fabricate and be strain-free 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). Turbulent water flow High flow rate has higher cooling capacity Decent heat-transfer film coefficient Metal (Ag, Au, Al) diffusion or In-based solder bonds Fabrication strain? 38 Shen 2/14/00

39 CHESS Design K.W. Smolenski, et al., Proc. SPIE 3151, 181 (1997). 39 Shen 2/14/00

40 Cryogenic cooling --- D.H. Bilderback, A.K. Freund, G.S. Knapp, D.M. Mills, The Advanced Photon Source Compton Award, October Turbulent LN 2 flow High thermal conductivity at low temperatures Almost zero thermal expanstion coefficient Narrow operable temperature range limits the total cooling capacity to ~ 2kW, for LN 2. So this is so far for undulator beams only. Possible liquid propane cooling??? H 2 O LN 2 C 3 H 8 Density (g/cm 3 ) Specific heat (J/g-K) Working pressure (bar) Working T (K) Relative cooling capacity Working temperature (K) 300 <140 α/k for Si (10-6 cm/w) Isotopically pure Si? --- W.S. Capinski, et al., Thermal conductivity of isotopically enriched Si, Appl. Phys. Lett. 71, 2109 (1997). 40 Shen 2/14/00

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42 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). 42 Shen 2/14/00

43 Homework Assignment A white-beam x-ray mirror made of Glidcop (as shown schematically in the figure) is designed to absorb 15 kw of a wiggler beam at a fixed incident angle of 3.9 mrad, and is water-cooled with a total flow rate of 12 gpm through 22 parallel mini-channels. The mirror is located 19 m from the wiggler which has a horizontal opening angle of 4.2 mrad and is operated in a storage ring of 5.3 GeV and 500 ma. (a) Assuming that the absorbed power is uniformly distributed within the natural opening angles of the wiggler, calculate the power density on the mirror surface. (b) Given the water channel dimensions in the figure, find out the hydraulic diameter D, Reynolds number Re, the heat transfer film coefficient h and the pressure drop over the 1 m length (assuming smooth pipe condition). (c) Using concepts of thermal resistance and fin effectiveness, and ignoring the heating up of water flow for now, calculate the temperature rise on the mirror surface. (d) Calculate the water temperature difference between the inlet on one end and the outlet on the other end of the mirror. Estimate the thermal tilt 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 of the 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. 43 Shen 2/14/00