Estudo sobre Convecção Natural e Mista em Placas Verticais e Horizontais

Convecção Livre ou Convecção Natural Placas Verticais ou Horizontais Figure 39 Streamline patterns visualised with an electrolytical precipitation method t 300s General Considerations Princípios Relacionada com forças de empuxo densidade e gravidade General Considerations cont Fluxos de fluido Plumas e jatos de conveção natural General Considerations cont Parametros Adimensionais Número de Grashof 3 2 Buoyancy Force Viscous Force s L g T T L Gr Número de Rayleigh 3 Pr s L L g T T L Ra Gr characteristic length of surface L a ther ther mody mal expa namic pr nsion coefficien operty of the fl d t ui 1 p T Liquids Tables A5 A6 Perfect Gas 1 T K General Considerations cont Convecção Mista Se Re2 1 L L Gr o 2 Free convection Re 1 L L Gr 2 Forced convection Re 1 L L Gr Correlações para convecção Mista n n n FC NC Nu Nu Nu assisting and transverse flows opposing flows 3 n Vertical Plates Placas Verticais How do conditions differ from those associated with forced convection How do conditions differ for a cooled plate sT T Vertical Plates cont xMomentum Equação para escoamento Laminar 2 2 u u u u g T T x y y Net Momentum Fluxes Inertia Forces Buoyancy Force Viscous Force Solução do campo de temperatura requer o conhecimento de u xy que deve ser obtido acoplado com a equação da energia para T xy 2 2 T T T u x y y g y u x u x p y v u x u u 2 2 2 2 g g g g g x p T T T T ou T T T T p 1 1 1 Aproximação de Boussinesq Vertical Plates cont Solução por similaridade 1 4 4 y Grx x Transformando as equações para 2 3 2 0 f ff f T 3Pr 0 T fT 1 2 2 x s T T df x f Gr u T d T T Vertical Plates cont Soluções numéricas para and f T Camada limite 5 for Pr 06 1 4 1 4 1 4 Pr 06 5 707 4 x x Gr x x x Gr Perfil de velocidade Perfil de temperatura Vertical Plates cont Número de Nusselt and L Nux Nu 1 4 1 4 0 Pr 4 4 x x x Gr Gr hx dT Nu g k d 1 2 1 4 1 2 075 Pr Pr 0 Pr 0609 1221 Pr 1238 Pr g 1 4 3 L L L o h hdx Nu Nu L Transição para a Turbulência Transitção ocorre para o Número de Rayleigh crítico 3 9 Pr 10 s x c x c g T T x Ra Gr Vertical Plates cont Correlações Empíricas Escoamento Laminar 109 L Ra 1 4 4 9 916 0670 068 1 0492Pr L L Ra Nu Qualquer condição laminar e turbulento 2 16 8 27 916 0387 0825 1 0492 Pr L L Ra Nu Horizontal Plates Placas Horizontais Superfície quente na parte inferior ou Superfície fria na parte superior sT T sT T 1 4 4 7 054 10 10 L L L Nu Ra Ra 1 3 7 11 015 10 10 L L L Nu Ra Ra P A L S Horizontal Plates cont invertendo com saídas laterais sT T sT T 1 4 5 10 027 10 10 L L L Nu Ra Ra P A L S Convecção Natural em Cilindros Esferas e Cavidades Cylinders Cilindro Horizontal Longo Número de Nusselt médio 2 1 6 12 8 27 916 0387 060 10 1 0559Pr D D D Ra Nu Ra How do conditions change for a cooled cylinder Spheres Esferas Número de Nusselt médio 1 4 4 9 916 0589 2 1 0469Pr D D Ra Nu Enclosures Confinamentos Cavidades retangulares 3 1 2 L g T T L Ra 1 2 q h T T Cavidade Horizontal 0 180deg Cavidade Vertical 90 deg Enclosures cont Cavidade Horizontal Aquecimento inferior 0 1708 L L c Ra Ra Fluid layer is thermally stable 1 L hL Nu k 4 L 1708 Ra 5 10 Thermal instability yields a regular convection pattern in the form of roll cells 5 9 3 10 7 10 RaL Buoyancy driven flow is turbulent 1 3 0074 0069 Pr L L Nu Ra Enclosures cont Aquecimento superior 180 deg Sem movimento 1 L Nu Cavidade Vertical 3 10 L Ra 3 10 L Ra A primary cellular flow is established as the core becomes progressively more quiescent and secondary corner cells develop with increasing L Ra Correlations for Eqs 950 953 L Nu 1 L Nu Enclosures cont Cavidade inclinada Relevant to flat plate solar collectors Heat transfer depends on the magnitude of relative to a critical angle whose value depends on HL Table 94 Heat transfer also depends on the magnitude of relative to a critical Rayleigh number of L Ra 1708cos RaL c Heat transfer correlations Eqs 954 957 Enclosures cont Cavidades Anulares Cilindros Concentricos 2 1n eff i o o i k q T T r r 1 4 1 4 0 386 0 861 eff c k Pr Ra k Pr or keffk 1 if the value calculated above is less than unity Enclosures cont The length scale in Rac is given by Concentric Spheres 4 3 5 3 3 5 3 5 2 o i c i o ln r r L r r 4 1 1 eff i o i o k T T q r r 1 4 1 4 0 74 0 861 eff s k Pr Ra k Pr or keffk 1 if the value calculated above is less than unity The length scale in Ras is given by 4 3 5 3 1 3 7 5 7 5 1 1 2 i o s i o r r L r r Convecção natural Transferência de Massa Mass Transfer Transferência de Massa A analogia entre transferência de calor e massa pode ser utilizada se as variações da densidade com a temperatura são muito inferiores as variações oriundas dos gradientes de concentração Nestes casos m L L AB h L Sh f Gr Sc D 3 2 s L g L Gr s s A s B A B 2 s Mass Transfer cont Placas Verticais analogous to Eq 924 n L L Sh C Gr Sc Placas Horizontais at lower surface or at upper surface s s 1 4 L 027 L Sh GR Sc 4 9 9 13 laminar 0 59 1 4 10 10 turbulent 0 10 1 3 10 10 L L C n Ra C n Ra 5 10 10 10 L Ra Analoga a Eq 932 Problem Solar Receiver Problem 931 Convection and radiation losses from the surface of a central solar receiver T 300 K oo q conv q rad q 10 S 5 Wm2 KNOWN Dimensions and emissivity of cylindrical solar receiver Incident solar flux Temperature of ambient air FIND a Heat loss and collection efficiency for a prescribed receiver temperature b Effect of receiver temperature on heat losses and collector efficiency ASSUMPTIONS 1 Steadystate 2 Ambient air is quiescent 3 Incident solar flux is uniformly distributed over receiver surface 4 All of the incident solar flux is absorbed by the receiver 5 Negligible irradiation from the surroundings 6 Uniform receiver surface temperature 7 Curvature of cylinder has a negligible effect on boundary layer development 8 Constant properties Problem Solar Receiver cont PROPERTIES Table A4 air Tf 550 K k 00439 WmK 456 106 m2s 667 106 m2s Pr 0683 182 103 K1 ANALYSIS a The total heat loss is 4 rad conv s s s s q q q A T hA T T With RaL g Ts TL3 98 ms2 182 103 K1 500K 12m3456 667 1012 m4s2 507 1012 the Churchill and Chu correlation yields 2 1 6 2 2 L 8 27 9 16 0387 Ra 00439 W m K k h 0825 0825 424 683W m K L 12m 1 0492 Pr Hence with As DL 264 m2 4 2 8 2 4 2 2 q 264m 02 567 10 W m K 800K 264m 683W m K 500K 6 5 6 rad conv q q q 123 10 W 901 10 W 213 10 W Problem Solar Receiver cont With 7 s A qs 264 10 W the collector efficiency is 7 6 s s 7 s s 264 10 213 10 W A q q 100 100 919 A q 264 10 W b As shown below because of its dependence on temperature to the fourth power qrad increases more significantly with increasing Ts than does qconv and the effect on the efficiency is pronounced 600 700 800 900 1000 Receiver temperature K 0 1E6 2E6 3E6 4E6 5E6 Heat rate W Convection Radiation Total 600 700 800 900 1000 Receiver temperature K 75 80 85 90 95 100 Collector efficiency COMMENTS The collector efficiency is also reduced by the inability to have a perfectly absorbing receiver Partial reflection of the incident solar flux will reduce the efficiency by at least several percent Problem Batch Reactor Problem 974 Use of saturated steam to heat a pharmaceutical in a batch reactor KNOWN Volume thermophysical properties and initial and final temperatures of a pharmaceutical Diameter and length of submerged tubing Pressure of saturated steam flowing through the tubing FIND a Initial rate of heat transfer to the pharmaceutical b Time required to heat the pharmaceutical to 70C and the amount of steam condensed during the process Problem Batch Reactor cont SCHEMATIC Pharmaceutical 1100 kgm3 c 2000 JkgK 40x10 s Pr 10 0002 K1 k 0250 WmK 6 m2 D 15 mm L 15 m Tubing Tt T C s o 127 psat 2455 bars T C V 200 L i 25o T 70 C f o Saturated steam ASSUMPTIONS 1 Pharmaceutical may be approximated as an infinite quiescent fluid of uniform but timevarying temperature 2 Free convection heat transfer from the coil may be approximated as that from a heated horizontal cylinder 3 Negligible thermal resistance of condensing steam and tube wall 4 Negligible heat transfer from tank to surroundings 5 Constant properties PROPERTIES Table A4 Saturated water 2455 bars Tsat 400K 127C hfg 2183 106 Jkg Pharmaceutical See schematic ANALYSIS a The initial rate of heat transfer is s s i q hA T T where As DL 0707 m2 and h is obtained from Eq 934 Problem Batch Reactor cont With Pr 40 107 m2s and RaD g Ts Ti D3 98 ms2 0002 K1 102K 0015m316 1013 m4s2 422 106 D 2 2 6 1 6 1 6 D 8 27 8 27 9 16 9 16 0387 422 10 0387Ra Nu 060 060 277 1 0559 Pr 1 055910 Hence 2 D h Nu k D 277 0250W m K 0015m 462W m K and 2 2 s s i q hA T T 462W m K 0707m 102 C 33300W b Performing an energy balance at an instant of time for a control surface about the liquid s s d cT q t h t A T T t dt where the Rayleigh number and hence h changes with time due to the change in the temperature of the liquid Problem Batch Reactor cont Integrating the foregoing equation numerically the following results are obtained for the variation of T and h with t 0 100 200 300 400 500 600 700 800 900 Time ts 25 35 45 55 65 75 Temperature C 0 100 200 300 400 500 600 700 800 900 Time ts 370 390 410 430 450 470 Convection coefficient hbar Wm2K The time at which the liquid reaches 70C is ft 855s The rate at which T increases decreases with increasing time due to the corresponding reduction in Ts T and hence reductions in RaD h and q The Rayleigh number decreases from 422 106 to 216 106 while the heat rate decreases from 33300 to 14000 W The convection coefficient decreases approximately as Ts T13 while q Ts T43 Problem Batch Reactor cont The latent energy released by the condensed steam corresponds to the increase in thermal energy of the pharmaceutical Hence c fg m h f i c T T and 3 3 f i c 6 fg c T T 1100kg m 02m 2000J kg K 45 C m 907kg h 2183 10 J kg COMMENTS 1 Over such a large temperature range the fluid properties are likely to vary significantly particularly and Pr A more accurate solution could therefore be performed if the temperature dependence of the properties were known 2 Condensation of the steam is a significant process expense which is linked to the equipment capital and energy operating costs associated with steam production Problem Garment Drying Problem 9113 Determination of drying rate per unit width of a garment hanging in dry air KNOWN Wet garment at 25C hanging in a room with still dry air at 40C FIND Drying rate per unit width of garment ASSUMPTIONS 1 Analogy between heat and mass transfer applies 2 Water vapor at garment surface is saturated at Ts 3 Perfect gas behavior of vapor and air PROPERTIES Table A4 Air Tf Ts T2 305K 1 atm 1639 106 m2s Table A6 Water vapor Ts 298K 1 atm pAs 00317 bar As 1vf 002660 kgm3 Table A8 Airwater vapor 305 K DAB 027 104 m2s Sc DAB 0607 Problem Garment Drying cont ANALYSIS The drying rate per unit width of the garment is A m As A m h L where m h is the mass transfer coefficient associated with a vertical surface that models the garment From the heat and mass transfer analogy Eq 924 yields L n L Sh C Gr Sc where GrL gL32 and s Since the still air is dry B pBRB T where RB MB 8314 102 m3barkmolK29 kgkmol 000287 m3barkgK With pB 1 atm 10133 bar 3 3 10133 bar 11280 kg m 000287 m bar kg K 313 K The density of the airvapor mixture at the surface is s As Bs With pBs 1 atm pAs 10133 bar 00317 bar 09816 bar Bs 3 Bs 3 B s p 09816 bar 11477 kg m R T 000287m barkg K 298 K Problem Garment Drying cont Hence s 00266 11477 kgm3 11743 kgm3 and s 2 1512 kgm3 The Grashof number is then 3 2 3 9 L 2 3 6 2 98 m s 11743 11280 kg m 1 m Gr 1467 10 11512 kg m 1639 10 m s and GrL Sc 8905 108 Hence from Section 961 C and n are 059 and 1 4 and the convection coefficient is then L 4 2 8 1 4 AB m D 027 10 m s h Sh 059 8905 10 000275 ms L 1 m The drying rate is then 3 3 5 mA 2750 10 ms 10m 00226 0 kg m 621 10 kgs m COMMENTS Since s the buoyancy driven flow descends along the garment