Convection Heat Transfer. Introduction

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Convection Heat Transfer Reading Problems 12-1 12-8 12-40, 12-49, 12-68, 12-70, 12-87, 12-98 13-1 13-6 13-39, 13-47, 13-59 14-1 14-4 14-18, 14-24, 14-45, 14-82 Introduction Newton s Law of Cooling Controlling Factors Q conv = R conv = 1 ha T R conv = ha(t w T ) Typical Values of h (W/m 2 K) Natural gases: 3-20 Convection water: 60-900 Forced gases: 30-300 Convection oils: 60-1800 water: 100-1500 Boiling water: 3000-10 5 Condensation steam: 3000-10 5 Geometry: Flow Type: Boundary: Fluid Type: Properties: shape, size, aspect ratio and orientation forced, natural, laminar, turbulent, internal, external isothermal (T w = constant) or isoflux ( q w = constant) viscous oil, water, gases or liquid metals all properties determined at film temperature T f = (T w + T )/2 Note: ρ and ν 1/P atm see Q12-40 density: ρ ((kg/m 3 ) specific heat: C p (J/kg K) dynamic viscosity: µ, (N s/m 2 ) kinematic viscosity: ν µ/ρ (m 2 /s) thermal conductivity: k, (W/m K) thermal diffusivity: α, k/(ρ C p ) (m 2 /s) Prandtl number: P r, ν/α ( ) volumetric compressibility: β, (1/K) 1

Forced Convection The simplest forced convection configuration to consider is the flow of mass and heat near a flat plate as shown below. flow forms thin layers that can slip past one another at different velocities as Reynolds number increases the flow has a tendency to become more chaotic resulting in disordered motion known as turbulent flow transition from laminar to turbulent is called the critical Reynolds number, Re cr Re cr = U x cr ν for flow over a flat plate Re cr 500, 000 x < x cr the boundary layer is laminar; x > x cr the boundary layer is turbulent Boundary Layers Velocity Boundary Layer the region of fluid flow over the plate where viscous effects dominate is called the velocity or hydrodynamic boundary layer the velocity at the surface of the plate, y = 0, is set to zero, U @y=0 = 0 m/s because of the no slip condition at the wall 2

the velocity of the fluid progressively increases away from the wall until we reach approximately 0.99 U which is denoted as the δ, the velocity boundary layer thickness. the region beyond the velocity boundary layer is denoted as the inviscid flow region, where frictional effects are negligible and the velocity remains relatively constant at U Thermal Boundary Layer the thermal boundary layer is arbitrarily selected as the locus of points where T T w T T w = 0.99 for low Prandtl number fluids the velocity boundary layer is fully contained within the thermal boundary layer conversely, for high Prandtl number fluids the thermal boundary layer is contained within the velocity boundary layer Flow Over Plates 1. Laminar Boundary Layer Flow, Isothermal (UWT) All laminar formulations for Re < 500, 000. The local value of the Nusselt number is given as Nu x = 0.332 Re 1/2 x P r 1/3 local, laminar, UWT, P r 0.6 An average value of the heat transfer coefficient for the full extent of the plate can be obtained by using the mean value theorem. Nu L = h LL k f = 0.664 Re 1/2 L Pr 1/3 average, laminar, UWT, P r 0.6 For low Prandtl numbers, i.e. liquid metals Nu x = 0.565 Re 1/2 x P r 1/2 local, laminar, UWT, P r 0.6 2. Turbulent Boundary Layer Flow, Isothermal (UWT) All turbulent formulations for 500, 000 Re 10 7. The local Nusselt number is given as Nu x = 0.0296 Re 0.8 x P r1/3 local, turbulent, UWT, 0.6 < P r < 60 3

and the average Nusselt number is Nu L = 0.037 Re 0.8 L P r1/3 average, turbulent, UWT, 0.6 < P r < 60 3. Combined Laminar and Turbulent Boundary Layer Flow, Isothermal (UWT) When (T w T ) is constant h L = 1 L L 0 hdx = 1 L { xcr 0 } L h lam x dx + h tur x dx x cr Nu L = h LL k = (0.037 Re0.8 L 871) P r1/3 average, combined, UWT, 0.6 < P r < 60, 500, 000 Re L 10 7 4. Laminar Boundary Layer Flow, Isoflux (UWF) Nu x = 0.453 Re 1/2 x P r 1/3 local, laminar, UWF, P r 0.6 5. Turbulent Boundary Layer Flow, Isoflux (UWF) Nu x = 0.0308 Re 4/5 x P r 1/3 local, turbulent, UWF, P r 0.6 Example 6-1: Hot engine oil with a bulk temperature of 60 C flows over a horizontal, flat plate 5 m long with a wall temperature of 20 C. If the fluid has a free stream velocity of 2 m/s, determine the heat transfer rate from the oil to the plate if the plate is assumed to be of unit width. 4

Flow Over Cylinders and Spheres 1. Boundary Layer Flow Over Circular Cylinders, Isothermal (UWT) The Churchill-Bernstein (1977) correlation for the average Nusselt number for long (L/D > 100) cylinders is Nu D = 0.3 + 0.62 Re1/2 D P r 1/3 [1 + (0.4/P r) 2/3 ] 1/4 ( ) 5/8 ReD 1 + 282, 000 4/5 average, UWT, Re D < 10 7, 0 P r, Re D P r > 0.2 All fluid properties are evaluated at T f = (T w + T )/2. 2. Boundary Layer Flow Over Non-Circular Cylinders, Isothermal (UWT) The empirical formulations of Zhukauskas and Jakob given in Table 12-3 are commonly used, where Nu D hd k = C Rem D P r1/3 see Table 12-3 for conditions 3. Boundary Layer Flow Over a Sphere, Isothermal (UWT) For flow over an isothermal sphere of diameter D, Whitaker recommends Nu D = 2 + [ 0.4 Re 1/2 D + 0.06 Re 2/3 ] D P r 0.4 ( ) 1/4 µ µ s average, UWT, 0.7 P r 380 3.5 < Re D < 80, 000 where the dynamic viscosity of the fluid in the bulk flow, µ is based on the free stream temperature, T and the dynamic viscosity of the fluid at the surface, µ s, is based on the surface temperature, T s. All other properties are based on T. 5

Example 6-2: An electric wire with a 1 mm diameter and a wall temperature of 325 K is cooled by air in cross flow with a free stream temperature of 275 K. Determine the air velocity required to maintain a steady heat loss per unit length of 70 W/m. Internal Flow Lets consider fluid flow in a duct bounded by a wall that is at a different temperature than the fluid. For simplicity we will examine a round tube of diameter D as shown below The Reynolds number is given as: Re D < 2300 Re D = U md. For flow in a tube: ν laminar flow 2300 < Re D < 4000 transition to turbulent flow Re D > 4000 turbulent flow 6

For engineering calculations, we typically assume that Re cr 2300, therefore { < Recr laminar Re D > Re cr turbulent Hydrodynamic (Velocity) Boundary Layer when the boundary layer grows to the tube radius, r, the boundary layers merge this flow length is called the flow entrance length, L h 0 x L h is the hydrodynamic entrance region L h < x L is the fully developed region or hydrodynamically developed region the hydrodynamic boundary layer thickness can be approximated as δ(x) 5x ( Um x ν ) 1/2 = 5x Rex the hydrodynamic entry length can be approximated as L h 0.05Re D D (laminar flow) Thermal Boundary Layer a thermal entrance region develops from 0 x L t the thermal entry length can be approximated as L t 0.05Re D P rd = P rl h (laminar flow) for turbulent flow L h L t 10D 7

Wall Boundary Conditions 1. Uniform Wall Heat Flux: The total heat transfer from the wall to the fluid stream can be determined by performing an energy balance over the tube. If we assume steady flow conditions, ṁ in = ṁ out = ṁ then the energy balance becomes Q = q w A = ṁ(h out h in ) = ṁc p (T out T in ) Since the wall flux q w is uniform, the local mean temperature is linear with x. T m,x = T m,i + q wa ṁc p The surface temperature can be determined from T w = T m + q w h 2. Isothermal Wall: Using Newton s law of cooling we can determine the average rate of heat transfer to or from a fluid flowing in a tube Q = ha (T w T m ) }{{} average T From an energy balance over a control volume in the fluid, we can determine Q = ṁc p dt m Equating the two equations above we find ṁc p dt m = ha (T w T m ) }{{} average T By isolating the temperature terms and integrating we obtain ln ( ) Tw T out T w T in = ha ṁc p Because of the exponential temperature decay within the tube, it is common to present the mean temperature from inlet to outlet as a log mean temperature difference where T ln = ln T out T in ( ) = Tw T out T w T in T out T in ln( T out / T in ) Q = ha T ln 8

1. Laminar Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF) For laminar flow where Re D 2300 Nu D = 3.66 fully developed, laminar, UWT, L > L t & L h Nu D = 4.36 fully developed, laminar, UWF, L > L t & L h Nu D = 1.86 ( ) ReD P rd 1/3 ( ) 0.14 µb L µ s developing laminar flow, UWT, P r > 0.5 L < L h or L < L t For non-circular tubes the hydraulic diameter, D h = 4A c /P can be used in conjunction with Table 13-1 to determine the Reynolds number and in turn the Nusselt number. 2. Turbulent Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF) For turbulent flow where Re D 2300 the Dittus-Boelter equation (Eq. 13-68) can be used Nu D = 0.023 Re 0.8 D P rn turbulent flow, UWT or UWF, 0.7 P r 160 Re D > 2, 300 n = 0.4 heating n = 0.3 cooling For non-circular tubes, again we can use the hydraulic diameter, D h = 4A c /P to determine both the Reynolds and the Nusselt numbers. In all cases the fluid properties are evaluated at the mean fluid temperature given as T mean = 1 2 (T m,in + T m,out ) except for µ s which is evaluated at the wall temperature, T s. 9

Natural Convection What Drives Natural Convection? fluids tend to expand when heated and contract when cooled at constant pressure therefore a fluid layer adjacent to a surface will become lighter if heated and heavier if cooled by the surface a lighter fluid will flow upward and a cooler fluid will flow downward as the fluid sweeps the wall, heat transfer will occur in a similar manner to boundary layer flow however in this case the bulk fluid is stationary as opposed to moving at a constant velocity in the case of forced convection In natural convection, the Grashof number is analogous to the Reynolds number. Gr = buouancy force viscous force = gβ(t w T )L 3 ν 2 Natural Convection Over Surfaces natural convection heat transfer depends on geometry and orientation note that unlike forced convection, the velocity at the edge of the boundary layer goes to zero the velocity and temperature profiles within a boundary layer formed on a vertical plate in a stationary fluid looks as follows: 10

Natural Convection Heat Transfer Correlations The general form of the Nusselt number for natural convection is as follows: Nu = f(gr, P r) CGr m P r n where Ra = Gr P r C depends on geometry, orientation, type of flow, boundary conditions and choice of characteristic length. m depends on type of flow (laminar or turbulent) n depends on the type of fluid and type of flow Table 14-1 should be used to find Nusselt number for various combinations of geometry and boundary conditions for ideal gases β = 1/T f, (1/K) all fluid properties are evaluated at the film temperature, T f = (T w + T )/2. 1. Laminar Flow Over a Vertical Plate, Isothermal (UWT) The general form of the Nusselt number is given as Nu L = hl k f 1/4 gβ(t w T )L 3 ν = C } ν{{ 2 }}{{} α Gr P r 1/4 = C Gr 1/4 L P r 1/4 }{{} Ra 1/4 where Ra L = Gr L P r = gβ(t w T )L 3 αν 2. Laminar Flow Over a Long Horizontal Circular Cylinder, Isothermal (UWT) The general boundary layer correlation is Nu D = hd k f 1/4 gβ(t w T )D 3 ν = C } ν{{ 2 }}{{} α Gr P r 1/4 = C Gr 1/4 D P r 1/4 }{{} Ra 1/4 D where Ra D = Gr D P r = gβ(t w T )L 3 αν 11

Natural Convection From Plate Fin Heat Sinks The average Nusselt number for an isothermal plate fin heat sink with natural convection can be determined using Nu S = hs [ 576 = k f (Ra S S/L) + 2.873 2 (Ra S S/L) 0.5 ] 0.5 Two factors must be considered in the selection of the number of fins more fins results in added surface area and reduced boundary layer resistance, R = 1 ha more fins leads to a decrease fin spacing and a decrease in the heat transfer coefficient R = 1 h A A basic optimization of the fin spacing can be obtained as follows: Q = ha(t w T ) where the fins are assumed to be isothermal and the surface area is 2nHL, with the area of the fin edges ignored. For isothermal fins with t < S with S opt = 2.714 ( L Ra 1/4 L ) Ra L = gβ(t w T )L 3 P r ν 2 The corresponding value of the heat transfer coefficient is h = 1.307k f /S opt All fluid properties are evaluated at the film temperature. 12

Example 6-3: Find the optimum fin spacing, S opt and the rate of heat transfer, Q for the following plate fin heat sink cooled by natural convection. Given: W = 120 mm H = 24 mm L = 18 mm t = 1 mm T w = 80 C T = 25 C P = 1 atm fluid = air Find: S opt and the corresponding heat transfer, Q 13