marketing planning phase two the situation review part 1

...  u1, j, n + + λ u , j, n + + λ 1 u 1, j − 1, n + + λ 1 u 1, j + 1, n + = Ke   = − u1, j, n ⋅ K1 , e = 1, , Ke (70) For j = , when writing the initial conditions for the boundary y = , the ... and [B1 ] = [C ] = [0]) : [A ] ⋅ {U } = {F1 } ( 91) The components of {F1} are computed using the relation:  r∞ K ⋅ dτ ⋅ ϕ ( x i ,0, τ ) , i ≤ i d  ui , 1, n + k K1 ⋅ n fi ,1 =    ui , 1, n ... In time, successions the phases the object suffers while irradiate by the power laser beam are the following: phase 1, for ≤ t < t top ; - phase 2, for t top ≤ t < t vap ; phase 3, for t ≥ t vap...

Ngày tải lên: 19/06/2014, 12:20

... 12 0.3 12 0.3 12 0.3 71. 6 71. 6 44.3 21. 1 1. 0 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 68.4 42 .1 20 .1 1.0 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 64.9 40.0 19 .1 1.0 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 61. 7 38.0 18 .2 1. 0 12 0.3 12 0.3 ... 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 45.0 21. 4 1. 0 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 71. 6 45.0 21. 4 1. 0 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 71. 7 44.8 21. 3 1. 0 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 71. 6 44.7 21. 3 1. 0 12 0.3 12 0.3 ... 12 0.3 12 0.3 12 0.3 71. 6 59.0 35.4 17 .4 1. 0 10 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 56.8 35.0 16 .8 1. 0 11 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 54.9 33.9 16 .3 1. 0 12 12 0.3 12 0.3 12 0.3 12 0.3 71. 6 53.4 33.0 15 .8 1. 0 13 ...

Ngày tải lên: 19/06/2014, 12:20

... 32.7 11 25 .1 75.0 0 .18 99.3 2.26 67 .1 72.9 0.23 38.8 12 25 .1 75 .1 0 .18 99.2 2.24 71. 1 72.5 0.24 38 .1 13 24.9 75.2 0 .16 10 0.5 2.22 64.5 72 .1 0 .19 33.6 14 24.9 75.4 0 .16 10 2.9 2.23 61. 0 72.2 0 .19 ... 0 .18 10 8.5 2. 71 78.7 81. 4 0.25 38.3 24 35 .1 85 .1 0 .19 10 9.5 2.70 78.5 81. 0 0.27 40.7 25 35 .1 85 .1 0 .17 10 9.0 2.69 79.0 81. 0 0.25 37.0 26 39.9 75.2 0.22 10 4.3 2.26 63.5 72.9 0 .19 33 .1 27 40 .1 ... 72.2 0 .19 33.9 15 25 .1 75.4 0 .18 99.3 2.25 71. 0 72.8 0.24 38.3 16 25.2 78.3 0 .16 10 4.0 2.34 68.4 74.5 0. 21 35.5 17 25 .1 78.8 0 .18 10 3.7 2.43 73 .1 76.2 0.25 40.4 18 25.0 80.0 0 .19 10 4 .1 2.47 73.8...

Ngày tải lên: 19/06/2014, 12:20

... for the anti-gravity position Tsf = 40 °C 1. 5 1. 4 Thermosyphon position 1. 3 Horizontal position Anti-gravity position Rth (K/W) 1. 2 1. 1 1. 0 0.9 0.8 0.7 0.6 10 20 30 40 50 60 Q (W) Fig 10 FMHP thermal ... W 14 0 12 0 10 0 T (°C) Evaporator 60 z (mm) (a) T (°C) z (mm) 11 0 10 0 90 80 70 60 50 40 30 20 10 Q =10 Q=20 Q=30 Q=40 Q=50 Q=60 Condenser 80 60 40 20 0 10 20 30 40 50 z (mm) (c) 60 70 80 90 10 0 10 ... work, R2/R1=2 and L/R1 =1 and the number of cells are 10 0 10 0 The working gas was Argon, characterized by a specific haet ratio γ = / Considering as a Hard-Sphere gas the d = 4 .17 × 10 10 m and...

Ngày tải lên: 19/06/2014, 12:20

... > 12 50 6.240 0. 21 1.6 12 3.4 0 .18 5 (medium carbon) < 10 00 14 1 .1 0.36 3 .1 211 .3 0 .18 5 (medium carbon) 10 00 -12 50 1. 825 0.37 2.5 14 4.3 0 .18 5 (medium carbon) > 12 50 1. 342 0.25 1. 5 10 2.5 Table Data ... %Cu %Ni %Cr %V %Al Tliq(°C) Tsol(°C) 0 .10 0 0.25 1. 20 0.025 0. 010 0.28 0.30 0 .10 0.05 0.04 15 16 14 97 0 .16 5 0.25 1. 20 0.025 0. 010 0.28 0.30 0 .10 0.05 0.04 15 11 1484 Table Steel chemical analyses examined ... (28) C = 0.30 91 + 0.2090 ⋅ (%C ) + 0 .17 73 ⋅ (%C )2 (29) n = 6.365 − 4.5 21 ⋅ 10 −3 ⋅ T + 1. 439 ⋅ 10 −6 ⋅ T (30) m = 1. 362 + 5.7 61 ⋅ 10 −4 ⋅ T + 1. 982 ⋅ 10 −8 ⋅ T ( 31) where, QC,K = 17 160 and: So,...

Ngày tải lên: 19/06/2014, 12:20

... to: r1b ( r1 ) y ( r1 ) −  + r1 a ( r1 )    dy ( r1 ) = r1 f ( r1 ) dr (16 ) The condition at the boundary of the confined plasma region, rn , corresponds normally to a prescribed value of the ... arbitrarily small by decreasing r1 In the range ≤ r ≤ r1 one can use the Taylor’s expansion: y ( ≤ r ≤ r1 ) = y ( r1 ) + ( r − r1 ) dy ( r − r1 ) d y r ( r1 ) + ( 1) dr dr 2 and the requirement dy/dr (r ... by the relations following from the boundary conditions (16 ) and (17 ) where the approximations dy/dr (r1) ≈ (y2 – y1)/(r2 – r1) and dy/dr (rn) ≈ (yn – yn -1) /(rn – rn -1) are applied With y1,…,n...

Ngày tải lên: 19/06/2014, 12:20

... Impinging 18 7 15 200 Re 7E3 9E3 11 E3 13 E3 15 E3 17 E3 18 .3E3 18 0 16 0 14 0 Nu 12 0 10 0 80 60 40 20 0 r 10 15 Fig 10 As in Fig 8, but for H/D = 10 70 Re 7E3 9E3 11 E3 13 E3 15 E3 17 E3 18 E3 60 50 Nu 40 30 20 10 ... to an Impinging 18 5 13 300 Re 7E3 9E3 11 E3 13 E3 15 E3 17 E3 18 .3E3 250 Nu 200 15 0 10 0 50 0 10 r 15 Fig Nu evolution for H/D = and the Re indicated in the legend 1. 4 Re 7E3 9E3 1. 2 11 E3 W 0.8 0.6 ... Impinging Wall to an Impinging 17 9 0.7 7E3 9E3 11 E3 13 E3 15 E3 17 E3 18 .3E3 0.6 0.5 V 0.4 0.3 0.2 0 .1 0 0.5 r 1. 5 (a) 1. 5 7E3 9E3 11 E3 13 E3 15 E3 17 E3 18 .3E3 W 0.5 0 0.5 r 1. 5 (b) Fig Mean dimensionless...

Ngày tải lên: 19/06/2014, 12:20

... 300 10 00 11 00 12 00 13 00 Distance [nm] 14 00 15 00 16 00 Fig Spatial temperature evolution resulting from a particular random throwing 15 ADNS are present, A/ρC = 10 13 K.s 1 and τ = ns are used The ... illustrate the main principle of the model, Fig plots the evolution of the temperature (10 ) as a function of the 1D spatial coordinate in a case for which n ADNS = 15 , A/ρC = 10 13 K.s 1 and τ = ... micro-balloon At the end of the laser chain, the final optic assembly is in charge for the frequency conversion of the laser beam from the 10 53 nm (1 ) to 3 51 nm (3ω) before its focusing on the target...

Ngày tải lên: 19/06/2014, 12:20

... evolution on the Au coated metal assemblies Parameters G0 (10 16 J m-3 s -1 K -1) Ce0(J m-3 K-2) ke0(J m -1 s -1 K -1) Cl (10 6 J m-3 K -1) A (10 7 s -1 K-2) B (10 11 s -1 K -1) Au 2 .1 68 318 2.5 1. 18 1. 25 Ag 3 .1 63 ... as (ω) (3ω ) (1 ) Q abs I( ω ) = Q abs (3ω, 1 ) I3ω + Q abs (3ω, 1 ) I1ω (3ω ) (19 ) (1 ) Where Q abs (3ω, 1 ) and Q abs (3ω, 1 ) are the absorption efficiencies at 3ω and 1 It is noteworthy ... into account the presence of two wavelengths at the same time: here the 3ω and 1 For this configuration, a particular attention has been paid to the influence of the wavelength on the defects energy...

Ngày tải lên: 19/06/2014, 12:20

... ∆Pgrav 12 .82 17 . 31% 12 .82 7.96% 12 .82 10 .96% 12 .82 14 .62% ∆Pacc 10 .24 13 .84% 10 .24 6.36% 10 .24 8.76% 10 .24 11 .68% ∆Pin 15 .35 20.74% 15 .35 9.54% 15 .35 13 .12 % 15 .35 17 . 51% ∆Pfrict ,1 0.96 1. 29% ... [kPa] 19 0 18 0 17 0 16 0 15 0 14 0 13 0 12 0 11 0 10 0 10 20 30 t [s] 40 50 60 Fig Experimental recording of total pressure drop oscillation showing “shark-fin” shape (SIET labs) 272 Two Phase Flow, Phase ... 0.82% 0.96 1. 09% ∆Pfrict,2φ 10 . 61 14.33% 39.84 24.75% 23.54 20 .12 % 14 .97 17 .07% ∆Pex 24.06 32.50% 81. 73 50.79% 54.07 46.22% 33.36 38.04% ∆Ptot 74.03 10 0% 16 0.94 10 0% 11 6.97 10 0% 87.69 10 0% Table...

Ngày tải lên: 19/06/2014, 12:20

... distribution No 10 11 Ave Exp.(°C) 19 .9 19 .8 20.3 19 .1 20 .1 20.0 19 .7 20.0 19 .5 20.0 19 .9 19 .8 Num.(°C) 19 .6 19 .4 19 .3 19 .2 20.2 19 .9 19 .3 19 .2 20.0 19 .9 19 .3 19 .6 Dev (°C) -0.3 -0.4 -1. 0 0 .1 0 .1 -0 .1 -0.4 ... at the surface temperature of 99 °C, the majority of the heat flux removed has been credited to the forced convection by the droplet impingement for the single phase spray cooling In the two phase ... out So, this part will be covered by the slip flow phenomenon and the convective heat transfer related to the slip flow on the hydrophobic surface In the two -phase flow, various two -phase heat...

Ngày tải lên: 19/06/2014, 12:20

... S., & Kim, M H (2 010 ) On the Mechanism of Pool Boiling Critical Heat Flux Enhancement in Nanofluids ASME J Heat Transfer, Vol 13 2, pp 0 615 01- 10 615 01- 11 338 Two Phase Flow, Phase Change and Numerical ... When the radius of a drop on the top surface reaches the size of the cavities, two phenomena enter in a competition The drop can either (i) coalesce with the drops in the 326 Two Phase Flow, Phase ... 410 5- 411 6 Kim, H D & Kim, M H (2007) Effect of nanoparticle deposition on capillary wicking that influences the critical heat flux in nanofluids Applied Physics Letters, Vol 91, pp 014 104 -1- 014 104-3...

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... Fig The mass flow rate increases with decreasing G/H aspect ratio, due to the decreasing frictional pressure term MASS FLOW RATE ml + 1x10 G/H = 0,0 01 G/H = G/H = 10 00 1x10 1x10 1x10 1x10 1x10 10 ... numerically in the case of laminar flow at the steady state is presented in Fig MASS FLOW RATE ml 4x10 -2 B/H = 1* 10 -3 B/H = 1* 10 B/H = 1, 1 + B/H = 1* 10 B/H = 1* 10 8x10 3 0 4x10 8x10 ** MODIFIED ... fluid on the wall of a tube, Physics of Fluids, 12 (10 ), 2367-23 71 Bretherton, F P (19 61) The motion of long bubbles in tubes, Journal of Fluid Mechanics, 10 (2), 16 6 -18 8 Cooper, M G (19 69) The microlayer...

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... (9) 1/ 2β 11 n2 (0)t dn2 = β 11 n2 ⇒ n2 (t) = dt + β 11 n1 (0)t (10 ) where n1 (0) is the initial number of particles per unit volume Introducing t1/2 = 1/ β 11 n1 (0), recalling that the volume ... 2v1 and that this particle deposits as a sediment without undergoing another collisions Using relations (5), (6) and (7) we write: dn1 n1 (0) = − β 11 n2 ⇒ n1 (t) = dt + β 11 n1 (0)t (9) 1/ 2β 11 ... 2. 51 t (s) 7.8 × 10 −6 1. 1 × 10 −3 1. 7 × 10 −3 1. 7 × 10 −6 6.2 × 10 −6 Table Values of t for some common materials at RT 3.5.2.2 Measuring circuit The measurement of the temperature rise δT of the...

Ngày tải lên: 19/06/2014, 12:20

... platinum ν0 = 10 1 Hz, Λ = 335 39 81 2 71 8433 4469 ν0 = 10 0 Hz, Λ = 10 6 12 59 86 2667 14 13 ν0 = 10 1 Hz, Λ = 33 398 27 843 447 ν0 = 10 2 Hz, Λ = 10 ,6 12 6 8,6 267 14 1 Table Values of the thermal length ... distance at Re = 11 00 and 17 00 14 vol % (only DIW) 0.2 vol % 0.4 vol % 0.6 vol % 0.8 vol % 13 Nu 12 11 10 800 10 00 12 00 14 00 16 00 18 00 Re Fig Effect of Reynolds number and TiO2 nanoparticle concentrations ... around the particles [12 -14 ], the particle size effects [15 ], agglomeration of nanoparticles [16 -18 ], the microconvection mechanism due to Brownian motion of nanoparticles in the liquid [19 - 21] ,...

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... 30 (14 ): p 11 36 -11 50 [9] Nan, C.W., et al., Effective thermal conductivity of particulate composites with interfacial thermal resistance Journal of Applied Physics, 19 97 81( 10): p 6692-6699 [10 ] ... the finite differences method (Zienkievicz &Taylor, 19 91) Furthermore, using tha same method from (Zienkievicz 458 Two Phase Flow, Phase Change and Numerical Modeling &Taylor, 19 91) , if the thermal ... field with the fractal coordinates (b) 466 Two Phase Flow, Phase Change and Numerical Modeling The presence of the interface couples together the two previous Equations (60) and ( 61) in the form:...

Ngày tải lên: 19/06/2014, 12:20

... (11 9) The relative concentration nr = E(s)   1 −  + s2  K (s)    (12 0) iii The relative thermal conductivity kr = K ( s ) ( K ( s ) − E ( s ) ) (12 1) Fig 14 The fractalisation of the thermal ...  2 1+ s 1+ s  1+ s  (12 3) and for s → the quasi-autonomous regime (of soliton packet type), Q ( β , s → 1) =  β − β0  − s2 2s + sech  ;s  2 1+ s 1+ s  1+ s  (12 4) For s = the soliton (11 8a) ... through the normalized wave length (see Fig .13 b): λ= 2sK ( s ) a (11 3) and normalized phase velocity (see Fig .13 c):  E(s) 1 1  u = 4a 3 K (s) s     (11 4) In such conjecture, the followings...

Ngày tải lên: 19/06/2014, 12:20

... ρ is the density of the fluid and ρr denotes the density of the (coexisting) reacting fluid Furthermore they assumed μ(c, A1 ) = μ* (c) [1 + αtr( A 21 )]n (10 ) where n determines whether the fluid ... flow, these relationships reduce to the ones proposed by Bingham (19 22), i.e F =1 K T12 (17 ) And 0 forF <    2μD 12 =   FT12 forF ≥     (18 ) The constitutive relation given by Eqn (15 ) ... viscosities These fluids are known as the power-law models or the generalized Newtonian fluid (GNF) models, where ( T = − p1 + μ trA1 ) m A1 (7) where p is the indeterminate part of the stress due to the...

Ngày tải lên: 19/06/2014, 12:20

... SA 01- 21- 1 12 .7 5 .1 7 .18 E-02 4.5 SA 01- 31- 1 15 .4 4 .1 2.03E- 01 4.4 SA 01- 12 -1 14.7 11 .4 2.24E- 01 11. 6 SA 01- 22 -1 18.3 8.9 6.72E-02 11 .4 SA 01- 32 -1 25.3 8 .1 1.77E- 01 14.3 SA 01- 42 -1 10.9 16 .0 4 .10 E-03 12 .2 ... SA 01- 11- 2 10 .8 6.0 2.76E-02 4.5 SA 01- 21- 2 13 .0 5.0 7.83E-02 4.5 SA 01- 31- 2 15 .4 4.0 2.03E- 01 4.3 SA 01- 12-2 14 .5 11 .3 2 .16 E- 01 11. 4 SA 01- 22-2 18 .3 8.9 6.78E-02 11 .4 SA 01- 32-2 25.2 7.9 1. 85E- 01 13.9 ... by the enthalpy flow rate Initial temp [°C] Charging time [hr:min] IPF [%] 14 .9 3:45 10 SA 01- 21- 1 9.3 2:55 10 SA 01- 31- 1 15 .3 3:30 10 SA 01- 12 -1 10 .1 2:55 10 SA 01- 22 -1 13 .1 3 :15 10 SA 01- 32 -1 18.7...

Ngày tải lên: 19/06/2014, 12:20

... investigated effect 5 61 18 16 14 12 10 350 Quantity of PCM [kg] MT23 MT 21 MT19 400 50 10 0 15 0 200 250 300 Temp deviation; "A" of Fig Thermal Energy Storage Tanks Using Phase Change Material (PCM) ... 0.383760(ln τ ) (14 ) 10 ≤ τ ≤ 10 0 ηt = [0.839337 + 0.444 718 (ln τ ) (15 ) 10 00 ≤ τ Λ= ηt = Λ (1 − Λ − Λ − Λ ) 0.57722 (16 ) 0.57722 [ln(4τ ) − 1. 15444) (17 ) αrt rcao (18 ) τ= The elapsed time factor ηt ... the element and Tw is the temperature at the outer surface of the casing T fi g x i Δx xi, ρi, Ei, Pi Δqi M Heat Loss T 0(xi+Δx/2) xi +1, ρi +1, Ei +1, Pi +1 i +1 T fi +1 Fig 10 The numerical calculation...

Ngày tải lên: 19/06/2014, 12:20