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3.2 Cryogenic Convection Heat Transfer

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3.2 Cryogenic Convection Heat Transfer 3.2 Cryogenic Convection Heat Transfer Involves process of heat transfer between q solid material and adjacent cryogenic fluid TTf = mean Classic heat transfer problem (Newton’s law) 2 m& f q(kW/m ) = h (Ts –Tf) Configurations of interest TT= Internal forced flow (single phase, f mean ) q Free convection (single phase,TT f = ∞ ) Internal two phase flow Pool boiling (two phase) Understanding is primarily empirical leading to correlations based on dimensionless numbers Issue is relevant to the design of: Liquid T Heat exchangers f Ts Cryogenic fluid storage Superconducting magnets Low temperature instrumentation USPAS Short Course Boston, MA 6/14 to 6/18/2010 1 Single phase internal flow heat transfer Forced T D m, f T Convection s & Q luid correlationsClassical f The heat transfer coefficient in a classical fluid system is generally correlated in the form where the Nusselt number, hD where, NuD ≡ and D is the characteristic length k f For laminar flow, NuD = constant ~ 4 (depending on b.c.) For turbulent flow (Re > 2000) D μ C n m Pr ≡ f p Nu(ReD = , f D Pr)= C D Reand Pr (Prandtl number) k f Dittus-Boelter Correlation for classical fluids (+/- 15%) 4 2 5 5 0Nu .D = 023 Re Pr Note that fluid properties should be TTs+ f computed at T (the “film temperature”): T ≡ f f 2 USPAS Short Course Boston, MA 6/14 to 6/18/2010 2 Johannes Correlation (1972) Improved correlation specifically for helium (+/- 8.3%) 0− . 716 4 2 ⎛ T ⎞ 0 .Nu = 02595 Re5 ⎜ s ⎟ Pr D D ⎜ ⎟ ⎝ Tf ⎠ Last factor takes care of temperature dependent properties Note that one often does not know Tf, so iteration may be necessary. Example USPAS Short Course Boston, MA 6/14 to 6/18/2010 3 Application: Cryogenic heat exchangers Common types of heat exchangers used in cryogenic systems P1 , T1 P , T Forced flow single phase fluid-fluid 2 2 E.g. counterflow heat exchanger in refrigerator/liquefier Forced single phase flow - boiling liquid (Tube in shell HX) E.g. LN precooler in a cooling circuit 2 m Static boiling liquid-liquid E.g. Liquid subcooler in a magnet system USPAS Short Course Boston, MA 6/14 to 6/18/2010 4 Simple 1-D heat exchanger Ts D m& , f T Differential equation describing the temperature of the fluid in the tube: dTf m& C +hP()f T − S T0 = dx For constant Ts, the solution of this eq ialtponenx e ans itionua Ts -Tf . ⎛ hP ⎞ Increasing m TTTT−s( f = s ) − f0 exp⎜− x⎟ ⎝ m& C ⎠ x USPAS Short Course Boston, MA 6/14 to 6/18/2010 5 He (Ti = 300 K) He (Tf = 80 K) Liquid nitrogen precooler Ts D & ,Tm f Assumptions & givens Ts is a constant @ 77 K (NBP of LN2) LN2 (Ts = 77 K) -6 3 Helium gas (Cp = 5.2 kJ/kg K; μ = 15 x 10 Pa s; ρ = 0.3 kg/m , k = 0.1 W/m K) Allowed pressure drop, Δp= 10 kPa Properties are average values Helium mass flow rate = 1 g/s between 300 K and 80 K Find the length and diameter of the HX (copper tubing) Total heat transfer: = 1 g/s x 5.2 J/g K x 220 K Q = m& C p (Tin −Tout ) = 1144 W Δ Log mean T: ΔT x = 0 − ΔT x = L f ( ) f ( ) (300-77) –(80-77) K ΔTlm = = ⎛ΔTf ()x = 0 ⎞ ln⎜ ⎟ ln [(300-77)/(80-77)] ⎝ ΔTf ()x = L ⎠ = 51 K USPAS Short Course Boston, MA 6/14 to 6/18/2010 6 Liquid nitrogen precooler (continued) Q UA= hπ DL = = 1144 W/51 K = 22.4 W/K ΔTlm The heat transfer coefficient is a function of ReD and Pr = 0.67 Assuming the flow is turbulent and fully developed, use the Dittus Boelter correlation hD 0 . 8 0 . 3 4m& NuD0= . 023 = D Reand Pr ReD = k f πD μ Substituting the ReD and solving for h 8.0 0.0247 x 0.1 W/m K x (10-3 kg/s)0.8 k f ⎛ 4m& ⎞ 3.0 0h = . 023 ⎜ ⎟ Pr = D π⎝ D μ⎠ (15 x 10-6 Pa s)0.8 x D1.8 = 0.07/D(m)1.8 hπDL = 0.224 x (L/D0.8) = 22.4 W/K 0.8 0.2 or L/D = 100 m 1 equation for two unknowns USPAS Short Course Boston, MA 6/14 to 6/18/2010 7 Liquid nitrogen precooler (continued) Pressure drop equation provides the other equation for L & D 2 L ⎛ m& ⎞ Δp = f ⎜ ⎟ ; A = πD2/4 and f ~ 0.02 (guess) 2ρD ⎜ A ⎟ flow ⎝ flow ⎠ 4m Re = & D πD μ 4 x 0.001 kg/s -6 Substituting for ReD and f π x 0.0062 m x 15 x 10 Pa s 2 ~ 13,700 m L -3 2 L Δ0p . ≈ 016& 0.016 x (10 kg/s) ⎛ ⎞ 5 = ⎜ 5 ⎟ ρ D 0.3 kg/m3 ⎝ D ⎠ ⎛ L ⎞ 2nd equation for Δp= 5.33 x 10-8 ⎜ ⎟ Substitute: ⎝ D5 ⎠ two unknowns Eq. 1: L = 100 D0.8 Δp= 5.33 x 10-6/D4.2 with Δp = 10,000 Pa D = [5.33 x 10-6/Δp]1/4.2 = 6.2 mm and L = 100 x (0.0062 m)0.8 = 1.7 m USPAS Short Course Boston, MA 6/14 to 6/18/2010 8 Single phase free convection heat transfer g g Q L or Q Q D Compressible fluid effect: Heat transfer warms the fluid near the heated surface, reducing density and generating convetive flow.c Free convection heat transfer is correlated in terms of the Rayleigh number, 3 n gβΔ TL NuL = f(), Pr Gr CRa ~ L whereRaL≡ GrPr = Dthν where g is the acceleration of gravity, β is the bulk expansivity, ν is the kinematic viscosity (μ/ρ) and D is the thermal diffusivity (k/ρC). th L (or D) are the scale length of the problem (in the direction of g) USPAS Short Course Boston, MA 6/14 to 6/18/2010 9 Free convection correlations Free convection correlation for low For very low Rayleigh temperature helium number, Nu = 1 corresponding to pure conduction heat transfer For Ra < 109, the boundary layer flow is laminar (conv. fluids) 0 . 25 NuL0≈ . 59RaL For Ra > 109, the boundary layer is turbulent 0 . 33 NuL ≈0 . 1 L Ra Example USPAS Short Course Boston, MA 6/14 to 6/18/2010 10 Pool Boiling Heat Transfer (e.g. Helium) Factors affecting heat transfer curve Surface condition (roughness, insulators, oxidation) Orientation Channels (circulation) Time to develop steady state (transient heating) Liquid Heated Note: Other cryogenic fluids have basically Surface the same behavior, although the numerical Q values of q and ΔT are different. USPAS Short Course Boston, MA 6/14 to 6/18/2010 11 Nucleate Boiling Heat Transfer Nucleate boiling is the principal heat transfer mechanism for static liquids below the peak heat flux (q* ~ 10 kW/m2 for helium) Requirements for nucleate boiling Must have a thermal boundary layer of superheated liquid near the surface ΔTk δ = f ~ 1 to 10 μm for helium th q Must have surface imperfections that act as nucleation sites for formation of vapor bubbles. USPAS Short Course Boston, MA 6/14 to 6/18/2010 12 Critical Radius of Vapor Bubble r 2σ Pv PL PPv= L + σ is the surface tension r Critical radius: For a given T, p, the bubble radius that determines whether the bubble grows of collapses r > rc and the bubble will grow r < rc and the bubble will collapse Estimate the critical radius of a bubble using thermodynamics Clausius Clapeyron relation defines the slope of the vapor pressure line in terms of fundamental properties dp ⎞ Δs hfg hfg p ⎟ = = ≈ 2 If the gas can be approximated as ideal dT ⎠sat Δv T() v− L v RT and vv >> vL USPAS Short Course Boston, MA 6/14 to 6/18/2010 13 Critical radius calculation Integrating the Clausius Clapeyron relation between ps and ps + 2σ/r 2 2 −1 2σhfg TΔ RT 2σRTs rc = (e −1) ≈ ps hfg pΔ s T Example: helium at 4.2 K (NBP) Empirical evidence indicates that ΔT ~ 0.3 K This corresponds to rc ~ 17 nm Number of helium molecules in bubble ~ 10,000 Bubble has sufficient number of molecules to be treated as a thermodynamic system Actual nucleate boiling heat transfer involves heterogeneous nucleation of bubbles on a surface. This is more efficient than homogeneous nucleation and occurs for smaller ΔT. USPAS Short Course Boston, MA 6/14 to 6/18/2010 14 Nucleate Boiling Heat Transfer He I 2.5 Note that hnb is not constant because Q ~ ΔT USPAS Short Course Boston, MA 6/14 to 6/18/2010 15 Nucleate Boiling Heat Transfer Correlations The mechanism for bubble formation and detachment is very complex and difficult to model Engineering correlations are used for analysis Kutateladse orrelationc 0 . 6 0 . 125 1 / 2 ⎡ 1 / 2⎤ ⎡ 2 3 /⎤ 2 0 . 7 h ⎛ σ ⎞ qCplρ l⎛ σ ⎞ ⎛ ρ ⎞ ⎛ σ ⎞ ⎛ p ⎞ ⎜ ⎟3= . 25 × −4 10⎢ ⎜ ⎟ ⎥ ⎢ ⎜ l ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ × g 1 / 2 k gρ h⎢ ρ kρ g ⎥ ⎜ μ ⎟ ⎜ gρ ⎟ ⎜σg ρ ⎟ l⎝ l ⎠ ⎣ fg v⎝ l⎠ l ⎦ ⎣⎢ ⎝ l ⎠ ⎝ l ⎠ ⎦⎥ ⎝ ()l ⎠ m,omewhat simpler fornto a s ingiRearrang 2 0 . 3125 1 . 5 ⎡ ⎛ ρ ⎞ ⎤ pχ1 . 75⎛ ρ ⎞ 1 . 5 −9 l 3 ⎛ ⎞ l ⎛ C p ⎞ ⎛ k ⎞ 2 . 5 1q= . 90 × 10⎢g⎜ ⎟ χ ⎥ ⎜ ⎟ ⎜ ⎟ ×⎜ ⎟ ⎜ l ⎟TT()− ⎢ μ ⎥ ⎝ σ ⎠ ρ ⎜ ⎟ ⎜ ⎟ s b ⎣ ⎝ l ⎠ ⎦ ⎝ v ⎠ ⎝ h fg ⎠ ⎝ χ ⎠ 1 / 2 ⎛ σ ⎞ 2 2 . 5 where χ = ⎜ ⎟q( W / ) cm= 5 .
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