Escoamento em Dutos: Capítulo 8 - Condições de Entrada e Escoamento Desenvolvido

Escoamento em dutos Capítulo 8 Entrance Conditions Região de Entrada e Região com Escoamento Completamente Desenvolvido Velocidade em x muda na região de entrada Por quê 10 D x 60 D x turbulento 10 Escoamento 0 05Re D x laminar Escoamento 2 300 Re Re fdh h fd h fd e normalment D u C Entrance Conditions cont Problema Térmico Assumir esc laminar com temp uniforme na região de entrada com temperatura const na superfície do tubo com ou fluxo de calor imposto 0 i T r T s i T T sq Forma adimensional do perfil de temperatura Para tornase independente de x for s and s T q Condições térmicamente desenvolvida Mean Quantities O Conceito de Velocidade Média u r m u r x u ro Velocidade média Hence Ac c m c u r x dA u A or Ac c m u r x dA Para escoamento incompressível em um duto circular de raio ro 2 0 2 or m o u u r x r dr r m ACum r u r x ro rdr r dx dr xp dx xp Para escoamento desenvolvido 0 0 x u v 0 2 2 2 2 2 2 0 rdr dx dx p d rdr p rdr p rdx dr dr d rdx rdx F r r r Sx Simplificando os termos 1 dx dp dr r d r r cte dx dp dr r d r r 1 sendo dr du r cte dx dp dr dr r du d r dr du dr r d r 1 integrando 2 1 2 1 1 2 ln 4 1 2 1 2 1 c r c r dx dp r u r c r dx dp dr du c r dx dp dr du r dx dr dp r dr dr dr r du d Com condições de contorno 0 0 4 0 1 0 2 2 c r u x p r c r u r o o Desta forma 2 2 1 4 1 o o r r dx r dp u r Integrando e calculando a velocidade média dx dp r u o m 8 2 Esc laminar Esc laminar u r Escoamento turbulento Fully Developed Flow Condição completamente desenvolvida Assumindo propriedades constantes o perfil de velocidades permaneceinalterado na região completamente desenvolvida fator de atrito 2 2 m dp dx D f u Escoamento laminar em um duto circular 64 ReD f Escoamento turbulento em um tubo circular com paredes lisas 2 0790 1n Re D 164 f Fully Developed Flow cont Escoamento turbulento com parede rugosa diagrama de Moody Queda de pressão 2 1 2 2 1 2 um p p p f x x D Potência de bombeamento pm P p Entry Lengths Região de Entrada Hidrodinâmica e Térmica Número de Reynolds Re m h D u D Diametro hidraulico 4 c h A D P ou 4 Re m h D u D m P Para um tubo circular 4 Re m D u D m D 10 D x 60 D x turbulento 10 Escoamento 0 05Re D x laminar Escoamento 2 300 Re Re fdh h fd h fd e normalment D u C 10 D x turbulento Escoamento 0 D x 100 Pr 0 05Re Pr D x laminar Escoamento t fd h fd t fd p Temperatura média r x u r x T r x Tm x Mean Quantities c ont 07102010 Energia média energia da mistura c t A p c p m E uc T dA mc T Hence Ac p c m p uc T dA T mc Para escoamento incompressível propriedades constante em um duto circular 0 2 2 or m m o T u x r T x r r dr u r Lei de Newton para o resfriamento s s m q h T T What is the essential difference between use of for internal flow and for external flow Tm T Fully Developed Flow cont CONDIÇÃO PLENAMENTE DESENVOLVIDO CPD para temperatura TS isotérmica 0 s s m fd t T x T r x x T x T x considerando que o o r r s s m s m r r T r T T f x r T T T T Assumindo propriedades constantes s s m q k h f x T T k h f x Variação de h na região de entrada e RCP Mean Temperature Determinação da variação da temperatura de mistura ao longo do tubo Determinação de Tm x he hs pv pvdpv c dT dh c dT du pv u h p v Aplicando conservação de energia 0 0 dx dx pv m d c T pv m c T pv m c T dq h m h dq m v m v m v conv s e conv desta forma Integrando da entrada a saída Análise Integral 1 conv p m o m i q mc T T pv md c T dq m v conv HIPÓTESES 1 Gás ideal 2 Líquidos incompressíveis logo R c R ou c c c RT pv p v v p m m v v p d c T pv d c c m p conv mc dT dq Mean Temperature cont Análise diferencial para obter Tm x conv s s m dq q Pdx h T T Pdx 2 m s s m p p dT q P P h T T dx mc mc Fluxo de calor constante m s p dT q P f x dx mc s m m i p q P T x T x mc Taxa transferida de calor conv s q q PL mcpdTm Mean Temperature cont Temperatura da superfície constante From Eq 2 with s m m p T T T d T dT P h T dx dx mc Integrando de x0 até uma posição qualquer x exp s m x s m i p T T x Px h T T mc 2 m s s m p p dT q P P h T T dx mc mc 0 x T T T T x p x p x m S m i S mc dx Ph T T d mc dx Ph T T d 0 1 x x x h h dx x com Entre a entrada e a saída exp exp s m o o s i s m i p p T T T hA PL h T T T mc mc exp s m x s m i p T T x Px h T T mc dA x T h T dq q da taxa transferida de calor transferido ao longo do tubo cálculo m S x conv conv 0 Diferença de temperatura Média logarítmica conv s m q hA T 3 1n o i m o i T T T T T exp exp s m o o s i s m i p p T T T hA PL h T T T mc mc L i o p L p i o h T T ln PL mc h mc PL T T ln Mean Temperature cont Fluido externo com temperatura uniforme 1 exp exp m o o s i m i p p tot T T T UA T T T mc mc R m s m tot T q UA T R Eq 3 with replaced by m s T T T Note Replacement of by Tso if outer surface temperature is uniform T 3 1n o i m o i T T T T T Fully Developed Flow Escoamento completamente desenvolvido Escoamento laminar em um Tubo Circular O número de Nusselt local é constante na região completamente desenvolvida mas este valor depende da condição térmica da superfície Equação 2 1 2 0 0 R r u r u x u e v com r r r T r r v T x T u m Fluxo de calor uniforme na superfície sq dx dT R r u r r r T r dx dT x T m m t fd m t fd 2 1 1 2 1 Sendo que Tmx varia linearmente em x e c cte dx dT u m m 2 2 4 2 2 1 2 4 2 2 3 4 1 16 1 16 3 2 ln 16 4 R r R r dx dT u R x T r x T c r c R r c r r x T dr R r c r r T r m m S sendo dx dT u R T x x T uTrdr u R T m m s m R m m 2 0 2 48 11 2 Para Fluxo de calor uniforme na superfície 4 2 D u D m P c m P q dx dT m p s m 36 4 11 48 f D f m s s k hD Nu ou D k h T h T q Temperatura Uniforme na Surperfíci e por solução numérica sT 366 D hD Nu k Entry Region Região de Entrada Laminar flow in a circular tube GzNúmero de Graetz Perfil de velocidades estabelecido Perfil de velocidades desenvolvendo junto com o perfil de temperatura Entry Region cont Número de Nusselt para escoamento Laminar num duto circular com temperatura uniforme na superfície Solução combinada 1 3 014 Re Pr 2 D s L D 014 186 Re Pr 1 3 D D s Nu L D 1 3 014 Re Pr 2 D s L D Comprimento de entrada térmico 2 3 00668 Re Pr 366 1 004 Re Pr D D D D L Nu D L Escoamento Turbulento em um Tubo circular Para superfície lisa e condições de escoamento turbulento a Equação de Dittus Boelter pode ser usada Re 10000 D 0023Re4 5 Prn D D Nu 1 2 2 3 8 Re 1000 Pr 1 127 8 Pr 1 D D f Nu f Os efeitos da rugosidade da parede e o escoamento de transição pode ser cosiderada utilizando a correlação de Gnielinski Re 3000 D 03 04 s m s m n T T n T T Fully Developed cont Superfície lisa 2 07901nRe D 164 f Superfície rugosa 0 e Figure 83 f Tubos Nãocirculares diametro hidraulico 4 c h A D P Escoamento Laminar O número de Nusselt local é constante Table 81 e depende da Condição de contorno térmica da superfície e da razão de aspecto do duto s or s T q Escoamento Turbulento DittusBoelter ou Gnielinski podem ser usadas com O diametro hidraulico independente da condição térmica da superfície Entry Region cont Average Nusselt Number for Turbulent Flow in a Circular Tube Effects of entry and surface thermal conditions are less pronounced for turbulent flow and can be neglected For long tubes 60 L D D D fd Nu Nu For short tubes 60 L D 1 D m D fd Nu C Nu L D 1 23 C m Noncircular Tubes Laminar Flow depends strongly on aspect ratio as well as entry region and surface thermal conditions See references 11 and 12 NuDh Entry Region cont When determining for any tube geometry or flow condition all properties are to be evaluated at D Nu 2 m m i m o T T T Why do solutions to internal flow problems often require iteration Turbulent Flow As a first approximation correlations for a circular tube may be used with D replaced by Dh Annulus The Concentric Tube Annulus Fluid flow through region formed by concentric tubes Convection heat transfer may be from or to inner surface of outer tube and outer surface of inner tube Surface thermal conditions may be characterized by uniform temperature or uniform heat flux s i s o T T i q qo Convection coefficients are associated with each surface where i i s i m q h T T o o s o m q h T T Annulus cont i h o h i o h D h D Nu Nu k k Fully Developed Laminar Flow Nusselt numbers depend on and surface thermal conditions Tables 82 83 i o D D Fully Developed Turbulent Flow Correlations for a circular tube may be used with replaced by D h D h o i D D D Problem Gas Turbine Vane Problem 843 For an air passage used to cool a gas turbine vane calculate the air outlet temperature and heat removed from the vane KNOWN Diameter and length of copper tubing Temperature of collector plate to which tubing is soldered Water inlet temperature and flow rate ASSUMPTIONS 1 Steadystate conditions 2 Ideal gas with negligible viscous dissipation and pressure variation PROPERTIES Table A4 Air assume m T 780 K 1 atm cp 1094 JkgK k 00563 WmK 3637 107 Nsm2 Pr 0706 Ts 650C 923 K 1 atm 4042 107 Nsm2 FIND a Outlet temperature of the air coolant for the prescribed conditions and b Compute and plot the air outlet temperature Tmo as a function of flow rate 01 m 06 kgh Compare this result with those for vanes having passage diameters of 2 and 4 mm Problem Gas Turbine Vane cont ANALYSIS a For constant wall temperature heating from Eq 841b s mo s mi p T T PLh exp T T mc 1 where P D For flow in a circular passage D 7 2 4 018kg h 1 3600s h 4m Re 584 D 0003m 3637 10 N s m 2 The flow is laminar and since LD 75 mm3 mm 25 the SiederTate correlation including combined entry length yields 014 13 D D s Re Pr hD Nu 186 k L D 3 014 13 7 2 7 00563W m K 584 0706 3637 10 h 186 875W m K 0003m 25 4042 10 Hence the air outlet temperature is 2 mo 650 T 0003m 0075m 875W m K exp 018 3600 kg s 1094J kg K 650 427 C Tmo 578 C Problem Gas Turbine Vane cont 01 02 03 04 05 06 Flow rate mdot kgh 500 550 600 650 Outlet temperature Tmo C Based upon the calculation for Tmo 578C m T 775 K which is in good agreement with our assumption to evaluate the thermophysical properties Why is Tmo independent of D From Eq 3 note that h is inversely proportional to D h D1 From Eq 1 note that on the righthand side the product Ph will be independent of D Hence Tmo will depend only on m This is of course a consequence of the laminar flow condition and will not be the same for turbulent flow b Using the IHT Correlations Tool Internal Flow for Laminar Flow with combined entry length along with the energy balance and rate equations above the outlet temperature Tmo was calculated as a function of flow rate for diameters of D 2 3 and 4 mm The plot below shows that Tmo decreases nearly linearly with increasing flow rate but is independent of passage diameter Problem Oven Exhaust Gases Problem 852 Determine effect of ambient air temperature and wind velocity on temperature at which oven gases are discharged from a stack KNOWN Thinwalled tall stack discharging exhaust gases from an oven into the environment Problem Oven Exhaust Gases cont FIND a Outlet gas and stack surface temperatures Tmo and Tso for V5 ms and T 4 C b Effect of wind temperature and velocity on Tmo ASSUMPTIONS 1 Steadystate conditions 2 Negligible wall thermal resistance 3 Exhaust gas properties approximately those of atmospheric air 4 Negligible radiative exchange with surroundings 5 Ideal gas with negligible viscous dissipation and pressure variation 6 Fully developed flow 7 Constant properties PROPERTIES Table A4 air assume Tmo 773 K m T 823 K 1 atm cp 1104 JkgK 3764 107 Nsm2 k 00584 WmK Pr 0712 Table A4 air assume Ts 523 K T 4C 277 K Tf 400 K 1 atm 2641 106 m2s k 00338 WmK Pr 0690 Problem Oven Exhaust Gases cont The flow is turbulent and assuming fully developed conditions throughout the stack the Dittus Boelter correlation may be used to determine ih i i 4 5 03 D D h D Nu 0023Re Pr k 3 4 5 03 2 i 584 10 W m K h 0023 33827 0712 102W m K 05m ANALYSIS a From Eq 845a mo mi p PL T T T T exp U mc i o 1 1 U 1 h h Internal flow With a Reynolds number of Di 7 2 4 05kg s 4m Re 33827 D 05m 3764 10 N s m External flow Working with the ChurchillBernstein correlation and Do 6 2 5m s 05m VD Re 94660 2641 10 m s 4 5 1 2 1 3 58 D D D 1 4 2 3 062Re Pr Re Nu 03 1 205 282000 1 04 Pr i o where h and h are average coefficients for internal and external flow respectively Problem Oven Exhaust Gases cont The outlet stack surface temperature can be determined from a local surface energy balance of the form hiTmo Tso hoTso T which yields 2 i mo o so 2 i o h T h T 102 543 139 4 W m T 232 C h h 102 139 W m K Hence 2 ho 00338W m K 05m 205 139W m K The outlet gas temperature is then 2 mo 05m 6m 1 T 4 C 4 600 Cexp W m K 543 C 05kg s 1104J kg K 1 102 1 139 Problem Oven Exhaust Gases cont Due to the elevated temperatures of the gas the variation in ambient temperature has only a small effect on the gas exit temperature However the effect of the freestream velocity is more pronounced Discharge temperatures of approximately 530 and 560C would be representative of coldwindy and warmstill atmospheric conditions respectively b The effects of the air temperature and velocity are as follows 2 3 4 5 6 7 8 9 10 Freestream velocity Vms 520 530 540 550 560 Gas outlet temperature TmoC Tinf 35 C Tinf 5 C Tinf 25C COMMENTS If there are constituents in the gas discharge that condense or precipitate out at temperatures below Tso related operating conditions should be avoided Internal Flow Mass Transfer Chapter 8 Section 89 General General Considerations Convection mass transfer occurs for gas flow through a tube whose inner surface is coated with a volatile liquid evaporation or solid sublimation and characterized by a uniform concentration A s A species concentration boundary layer develops in an entry region and a fully developed condition is eventually reached 1 005Re fd c D am x Sc D The mean species concentration is c A c A A m m c u dA u A 10 60 fd c turb x D Transfer Rates Species Transfer Rates Local Species Mass Flux A m A s A m n h Longitudinal Distribution of Mean Species Concentration Applying conservation of species to a differential control volume in the tube the following expression may be obtained Problem 8112 exp A s A m m A s A m i m c x h P x u A Transfer Rates where m c m u A Total mass flow rate Total mass density Species Transfer Rate A A m o A m i m n 1 or A m s A m n h A 1 where 1n A o A i A m A o A i and A A s A m Convection Coefficients Convection Mass Transfer Coefficients Obtained by analogy to heat transfer correlations for internal flow m D D AB h D Sh Nu D Pr AB v Sc D Problem Air Flow through Wetted Tube Problem 8113 Evaporation rate and outlet vapor density for airflow through wetted circular tube KNOWN Flow rate and temperature of air Tube diameter and length Presence of water film on tube inner surface FIND a Vapor density at tube outlet b Evaporation rate ASSUMPTIONS 1 Steady incompressible flow 2 Constant flow rate 3 Isothermal system water film maintained at 25C 4 Fully developed flow PROPERTIES Table A4 Air 1 atm 298K 11707 kgm3 1836 107 Nsm2 1571 106 m2s Table A6 Water vapor 298K Asat 1vg 14425 m3kg 00226 kgm3 Table A8 Airvapor 298K DAB 26 106 m2s Sc DAB 060 ANALYSIS a From Equation 884 Amo As As Ami m DL exp h m Problem Air Flow through Wetted Tube cont 4 D 7 2 4 m 4 3 10 kgs Re 2080 D 001 m 1836 10 N sm Flow is laminar and from the mass transfer analogy to Eq 857 1 3 1 3 D D Re Sc 2080 060 Sh 186 186 431 LD 100 6 2 D AB m Sh D 431 26 10 m s h 00112 ms D 001 m 3 Amo 00226 kgm 3 3 3 4 001 m 1 m 117 kgm 00112 ms 00226 kgm exp 00169 kgm 3 10 kgs b The evaporation rate is 4 6 A m c Amo Ami Amo 3 3 m 3 10 kgs kg n u A 00169 433 10 kgs 11707 kgm m Problem Air Flow through Wetted Tube cont COMMENTS With 3 3 Ao Ai m 4 DL 001 1 117 m exp h 00226 exp 00112 573 10 kgm m s 3 10 kgs the evaporation rate is 3 Ao Ai 6 A m Ao Ai 000573 00226 kgm m n h DL 00112 001 m 1m 433 10 kgs s n 00057300226 n which agrees with the result of part b Problem Water Flow Through Pipe in Furnace Problem 817 Estimate temperature of water emerging from a thinwalled tube heated by walls and air of a furnace Inner and outer convection coefficients are known D 025 m L 8 m 1 Tmo T 300 K mi Water Air T 700 K oo Tfur 700 K qcvo qrad Tt m 5 kgs T oo T oo Tt Rcvi Rrad Rcvo Tfur h 50 Wm K o 2 KNOWN Water at prescribed temperature and flow rate enters a 025 m diameter black thinwalled tube of 8m length which passes through a large furnace whose walls and air are at a temperature of Tfur T 700 K The convection coefficients for the internal water flow and external furnace air are 300 Wm2K and 50 Wm2K respectively FIND The outlet temperature of the water Tmo Problem Water Flow Through Pipe in Furnace cont D 025 m L 8 m 1 Tmo T 300 K mi Water Air T 700 K oo Tfur 700 K qcvo qrad Tt m 5 kgs T oo T oo Tt Rcvi Rrad Rcvo Tfur h 50 Wm K o 2 SCHEMATIC ASSUMPTIONS 1 Steadystate conditions 2 Tube is small object with large isothermal surroundings 3 Furnace air and walls are at the same temperature 4 Tube is thinwalled with black surface and 5 Incompressible liquid with negligible viscous dissipation PROPERTIES Table A6 Water cp 4180 JkgK ANALYSIS The linearized radiation coefficient may be estimated from Eq 19 with 1 2 2 rad t fur t fur h T T T T where tT represents the average tube wall surface temperature which can be estimated from an energy balance on the tube As represented by the thermal circuit the energy balance may be expressed as m t t fur cvi cvo rad T T T T R 1 R 1 R The thermal resistances with As PL DL are cvi i s cvo o s rad rad R 1 h A R 1 h A R 1 h Problem Water Flow Through Pipe in Furnace cont and the mean temperature of the water is approximated as m mi mo T T T 2 The outlet temperature can be calculated from Eq 845b with Tfur T T T 1 exp T T mc R mo mi p tot where tot cvi cvo rad 1 R R 1 R 1 R with 5 4 4 cvi cvo rad R 6631 10 K W R 3978 10 K W R 4724 10 K W it follows that m t T 331 K T 418 K Tmo 362 K