Heat ransfer Convection
Previous lectures conduction: heat transfer without fluid motion oday (textbook nearly 00 pages) Convection: heat transfer with fluid motion Research methods different Natural Convection fluid is in natural flow, such as buoyancy effect Forced Convection fluid is forced to flow, such as wind, pump, fan Forced convection is generally more efficient than natural convection due to faster velocity of the fluid
Convection Heat transfer due to MOVEMEN of fluids conv q h( ) at the interface of surface-fluid boundary layer s q q q q convection conduction convection conduction h( hl k k( ) y s y0 ) h k L Nu hl k Nusselt Number L y 0
Convection Aim: Obtain Nusselt number Nu or h It s ratio of convection to conduction Nu hl Basic ideas: study fluid s boundary layer geometries empirical k c
Boundary Layer & Geometries Ideas: use dimensionless numbers to observe sensitive factors flow divided into laminar and turbulent geometries into external and internal external flow (free surface): flat plate, across cylinder, across sphere internal flow (inner surface): tubes Aim: Obtain Nusselt number Nu or h Illustrated by forced convection
Boundary Layers Velocity boundary layers drag/friction force hermal boundary layers heat transfer
close to the surface, velocity slow down due to friction (viscous shearing force caused by viscosity) (resistance to flow) on surface y=0, velocity is zero. It increases with y thickness of boundary layer δisdefinedas0.99v Velocity Boundary Layer y V 0.99V V Boundary Layer x
Development of Velocity Boundary Layer transition turbulent laminar buffer layer x cr turbulent described by Reynolds number Re
Reynolds Number Re V Lc VLc / Inertial force Viscous force V upstream velocity L c characteristic length (=x distance from leading edge for flat plate =D for cylinder and sphere) dynamic viscosity kg/m.s kinematic viscosity m /s Re varies with position x or r
Flow over Flat Plates Characteristic length: L Re L VL Critical Re cr = 510 5 fluid properties evaluated at FILM EMPERSURE f =( s + )/
hermal Boundary Layer Likewise, fluid with specific temperature flows over solid surface. hermal boundary layer develops s 0.99( s y ) t s x
Velocity vs hermal Boundary Layers relative thickness of velocity and thermal boundary layers is described by dimensionless parameter Prandtl number Pr Momentum diffusivity hermal diffusivity k / / 0.01 for liquid metals (small) (more conductive) 1 for gases, 0.7 for air 100,000 for heavy oil (large) (more convective) C p C k p
Now, we are ready for Nu--h general equation Nu f (Re, Pr) Based on fluid (laminar or turbulent ) & geometries
1.38 hl k Laminar: C 1/ 1/ 3 urbulent: Nu Combined: f hl k Re C f C f 1/ L 0.074 Re 1/5 L Nu Pr 0.6 4/5 0.037 ReL 0.074 Re 1/ 5 L 174 Re L Pr 0.664 ReL 10 7 Re L 510 5 1/3 Pr 60Pr 0.6 10 7 Re L 510 5 Nu hl k 4/5 L ( 0.037 Re 871) Pr Flow over Flat Plates 1/3 60Pr 0.6
Example: 60 C oil flows over a 5m long 0 C flat plate with a speed of m/s. Is the flow laminar? Determine the average friction coefficient and the average heat transfer coefficient. Determine x cr that the flow turns to turbulent. OIL =60 C V =m/s s =0 C Determine Geometry + fluid L =5m Known: Use film temperature f =( s + )/= 40 C then refer to property able we have =876 kg/m 3, Pr = 870, k = 0.144 W/m C, =410-6 m /s.
Solution: Reynolds number at the end of the plate Re L V L ReL 4.1310 4 (m / s)(5m) 6 410 m / s 5 510 OIL =60 C V =m/s Yes, laminar flow L =5m s =0 C Now friction and Nu and x cr based on the table Re Nu h L 4.1310 hl k k Nu L 0.664 Re 4 1/ L C f Pr 1/ 3 1.38 / Re L 1.38 (4.1310 1 4 1/ 0.664 (4.1310 4 ) 1/ 0.144W / mc 1918 55.[ W / m. ] 5m C 5 V xcr (m / s) xcr Recr 510 6 410 m / s 870 ) 1/ 3 1918 x cr 60. 5m 0.00653
Flow across Cylinder & Sphere Characteristic length: external diameter D Re V D Critical Re cr = 10 5 diameter D for cylinder & sphere
Characteristic length in general hydraulic diameter D h =4A c /p for noncircular tube (A c : cross-section area p: perimeter) empirical correlations (skip)
Internal Flow external flow : boundary layer free to grow internal flow : boundary layer has limit to grow depending on pressure drop fluid properties evaluated at averaged temp m =1/( i + e ), however, both changes in flow direction i =mean temperature at the inlet e =mean temperature at the outlet
How to calculate Q? q const s Q e q s A i mc q s A mc p( e i p ) (A = perimeter*l) s const m C p d m h( ) da ( da pdx) s m e s ( s i )exp( ha mc p ) dimensionless parameter NU (number of transfer unit) Q ha ha( ) ave here are ways to express ave s m ave
ways to calculate average temp difference ln ha Q i e i e i s e s i e ln ln ln i s i e s e More accurate logarithmic mean temp difference s const am e i s ave arithmetic mean temp difference
Laminar and urbulent flows Characteristic length: diameter D Laminar: ransition: urbulent: 300 > Re Re 10000(4000) Re 300 Re> 10000(4000) V m D 10000 refers to the new edition of Heat ransfer textbook
Velocity Boundary Layer velocity boundary layer hydrodynamically developed region hydrodynamic entry length L h hydrodynamic entry length laminar: L h,laminar 0.05 Re D turbulent: L h,turbulent 10D entry region f x or h x
hermal Boundary Layer thermal developed region thermal entry length L t thermal entry length laminar: L t,laminar 0.05 Re Pr D turbulent: L t,turbulent 10D entry region f x or h x
Velocity vs hermal Boundary Layers fully developed region both hydrodynamic & thermal regions are developed->constant f and h often assume fully developed region for entire tube for simplicity!!!! otherwise, see references f x or h x Now, we are ready for Nu (h)!!!! entry region
Developed Velocity profile (parabolic) Nu = 3.66 for s = const Nu = 4.36 for Nu Re Pr D 1.86( ) L Circular Pipes friction factor f =64/Re (laminar) q s const 1/3 ( bulk surface ) r V ( r) V m (1 ) R (laminar) (laminar) 0.14 (laminar,developing) friction factor f =0.184Re -0. (turbulent,developed) Nu =0.03 Re 0.8 Pr n (turbulent,developed) 160 Pr 0.7,Re>10000 n =0.4 (heating) 0.3(cooling)
Pressure Drop inside ubes P f L D V m also P total 0.5 V in f Lin D f L D L f D out V pressure must decrease after the obstruction Bernoulli s Equation ρgh for potential if locations are different out f L D : 0. 0.5 for inlet e.g. : for channel only : 1 for outlet inside V
Example: 0 C oil flows inside 00m long 0 C surface temp pipeline with a speed of m/s. D =30cm. Determine (a) average heat transfer coefficient (b) exit oil temperature (c) rate of heat transfer (4) pressure drop inside pipe. OIL 0 C m/s Determine Geometry + fluid s =0 C L =00m Known: Since we don t know the exit oil temp so can t calculate mean temp. We evaluate the oil properties at the inlet temp 0 C then refer to able we have =888 kg/m 3, Pr = 10400, k = 0.145 W/m C,=90110-6 m /s b(0c) =0.8 kg/m.s (bulk), s(0c) =3.85 kg/m.s (surface), C p = 1880J/kg C
s =0 C OIL 0 C m/s Solution: laminar or turbulent? L =00m V ( / )(0.3 ) Re m m s m D 90110 m / s 666 6 Laminar! L t,laminar 0.05 Re Pr D =0.05666 10400 0.3m =104000m Very long, developing entry region! 300 Nu hd k Re Pr D 1.86( ) L 1/ 3 b ( ) s 0.14 666 1.86( 10400 0.3m ) 00m 1/ 3 ( 0.8 3.85 ) 0.14 3.6 h k D Nu 0.145W / mc (3.6) 15.8[ W / m ] 0.3m C (a) ok!
s =0 C e A e s OIL 0 C m/s ( s ha i )exp( mc p L =00m ) pl DL ( 0.3m)(00m) 188.5m 3 0.3 m Across sectionv m (888kg/ m )( ( ) s ( s )exp( i ha mc 15.8W / m C188.5m 0 (0 0) exp( ) 19.75 15.5kg/ s1880j / kgc. p ) )(m / C s) 15.5kg/ (b) ok! s
s =0 C OIL 0 C m/s Use logarithmic mean temperature L =00m ln 19.75 0 e i 19. 875C 0 19.75 s e ln ln 0 0 s i Q ha (15.8W / m C)(188.5m )(19.875C) 59. 19kW ln (c) ok!
s =0 C OIL 0 C m/s L =00m P In tube laminar we have f =64/Re =64/666=0.0961 P f L D f L V D m V m 00m 0.0961 0.3m 3 (888kg/ m )(m / s) 113.78N / m (d) ok!
Summary Select the right empirical correlation Determine: geometry (easy), fluid (tricky, use average temp) Calculate Re and check the flow regime (laminar or turbulent) Calculate hydrodynamic/thermal entrance length (fully developed flow vs. developing) Nusselt number can be obtained from an appropriate correlation: Nu = f(re, Pr) We need to determine some properties and plug them into the correlation!!! (film temperature, inlet/exit temperature, special cases of our interest, etc) End up with h, Q
Natural Ventilation temperature difference density difference buoyancy force Buoyancy will induce a flow current due to the gravitational field and the variation in the density field. cold friction r r
Natural Convection h? conv q h( ) s Basic idea: similar as forced convection Study fluid (air) boundary layers Study geometries Aim: Obtain Nusselt number Nu or h!!!!
Boundary Layers & Geometries boundary layers laminar turbulent geometries external flow (over surface): flat plate, across cylinder, across sphere internal flow (enclosures): rectangular enclosure, cylinder, sphere Remember the aim: to find Nu or h
Boundary Layers turbulent laminar stationary fluid V= 0 s > analysis of boundary-layer follows entirely that of forced convection
Grashof Number Gr Buoyancyforce g ( s ) Lc Gr Viscousforce g gravitational acceleration m/s 1 volume expansion coefficient 1/K s surface temperature C temperature of fluid far from the surface C L c characteristic length m υ kinematic viscosity m /s he Grashof number similar as Reynolds number for natural convection General equation Nu f (Gr, Pr) Remember Re 3
Laminar or urbulent critical Gr at10 9 for vertical plate Gr 10 9 laminar Gr > 10 9 turbulent Ready for Nu for different flows & geometries fluid properties evaluated at FILM EMPERSURE f =( s + )/ hl k c n n Nu C( Gr.Pr) C( Ra) g ( s Ra Gr. Pr ) L 3 c Pr Rayleigh number
Inside Enclosures e.g. double-pane window L c : distance between hot and cold surfaces Fluid property: Nu=1,conduction Cold Down Hot Up vertical horizontal (a) hot top (b) hot bottom
Inside Enclosures continued Hot up no motion,conduction Horizontal heavy fluid heavy fluid stays (Nu=1) Ra<1708 no convection current (Nu=1) 3x10 5 >Ra>1708 Bénard Cells Ra>3x10 5 turbulent flow (celles broken) Hot bottom
Inside Enclosures Nu f ( Ra ) L many empirical formula fluid properties evaluated at avg =( 1 + )/ Various correlations horizontal rectangular enclosures inclined rectangular enclosures vertical rectangular enclosures concentric cylinders concentric spheres
Example: Double pane window maintained at temps 63 o C(336K) and 7 o C(80 K). Estimate its natural convective heat loss As an exercise!!!
Forced or natural convection Gr Re presents the relative importance of natural relative to forced convection < 0.1 then natural convection negligible >10 then forced convection negligible (0.1 10) neither is negligible Nu combined = (Nu forcedn Nu natural n ) 1/n n is between 3 (vertical surface) and 4 (horizontal surface)
Simplification in Building Physics Rough surfaces -> higher h Orientation actual fluid temp and not just the temp difference h vs Δ for floor, wall
Forced convection h vs air speed
Commonly used forced convection for exterior surfaces like windows walls etc