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A Short Course on Heat Transfer

A Short Course on Heat Transfer
A Short Course on Heat Transfer Intended as a repetition from previous courses by Björn Palm, Royal Institute of Technology, Stockholm Course Contents, based on Holman’s book Heat Transfer Chapter 1: Introduction Chapter 2: SteadyState Conduction One Dimension Chapter 3: SteadyState Conduction Multiple Dimensions Chapter 4: UnsteadyState Conduction Chapter 5: Principles of Convection Chapter 6: Empirical and Practical Relations for ForcedConvection Heat Transfer Chapter 7: Natural Convection Systems Chapter 8: Radiation Heat Transfer Chapter 9: Condensation and Boiling Heat Transfer Chapter 10: Heat Exchangers (Chapter 11: Mass Transfer) 2 Part 1 Introduction 3 Introduction What is heat Heat is energy transfer caused by temperature difference 4 The four laws of thermodynamics: • Zeroth law: If two bodies both are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other, and they then are said to have the same temperature. 5 • First law: (Energy principle) Energy cannot be generated or destroyed, only converted to different forms. (”Energy consumption” is the transfer of ”prime” energy to thermal energy in the surrounding) 6 • Second law: Heat cannot by itself pass from one body to another body with higher temperature. (Entropy disorder strives to a maximum in a closed system. Shows the direction of time.) 7 • 8 Third law: The entropy of a pure, crystalline material takes its lowest value at absolute zero temperature, where it is 0. (There is a lowest limit to the temperature, 0K = 273.15°C. World record: 0.00000017 K = 170 nanokelvin) 9 Three modes of heat transfer: • Conduction Through solid bodies and ”still” fluids hot cold • Convection Through moving fluids (also boiling and condensation) • Radiation Between surfaces, through gas or vacuum hot cold 10 What do we know about conduction 11 Conduction, thermal conductivity, Fourier´s law: Fourier’s law: q = k⋅A⋅δT/δx where, q = heat flow (W) 2 A = area perpendicular to heat flow (m ) δT/δx = temperature gradient in the direction of heat flow(°C/m) k = thermal conductivity (W/(m °C)) Fourier´s law is the defining equation for the thermal conductivity. 12 Conduction, thermal conductivity, Fourier´s law: For onedimensional heat transfer (a plane wall,) with constant thermal conductivity, Fourier´s law is simplified to q = k⋅A⋅ΔT/δ where ΔT = temperature difference (°C) δ = distance or thickness (m). 13 fig. 1 Temp δT/δx0 x Hot Cold q0 14 Fig. 2 Heat transfer through a plane wall T1 ΔT T2 δ 15 Table 1 Material Thermal conductivity (20°), (W/m⋅°C) Diamond, type IIa 2600 Copper 386 Iron, wrought, 0.5 C 60 Stainless steel, 18/8 16.3 Brick 0.69 Water 0.6 Pine wood, 0.15/0.33 (cross/along fibres) Cork 0.045 Glass wool 0.038 Mineral wool 0.04 Polyurethane 0.02 Air 0.026 Argon gas 0.018 16 Example to solve: Conduction: Calculate the heat flow per square meter (heat flux) through a mineral wool insulation, 5 cm thick, if the temperatures on the two surfaces are 30 and 200°C, respectively. 17 Heat transfer by convection, Newton’s law of cooling Convection is a general term for heat transfer through a moving fluid. 18 Fig 3, Different types of convection heat transfer Boiling Forced convection Diffusion, mass transfer Condensation 19 What do we know about convection 20 All types of convection are governed by Newton’s law of cooling: q = h⋅A⋅ΔT where A = surface area where convection takes place 2 (m ) ΔT = temperature difference (°C) 2 h = heat transfer coefficient (W/(m⋅°C)) Newton’s law of cooling is the defining equation for the heat transfer coefficient h. 21 Table 2 Type of flow Approximate heat transfer 2 coefficient (W/(m⋅°C)) Turbulent flow in tubes, (diameter ≈ 50 25 mm) Water (0.5 5 m/s) 1500 20000 Air (1 10 m/s) 10 50 Laminar flow in tubes, (diameter ≈ 50 10 mm) Water 50 250 Air 2 15 Air flow past plates (1 10 m/s) 10 50 22 Type of flow Approximate heat transfer 2 coefficient (W/(m⋅°C)) Natural convection Water 200 1000 Air 2 10 Condensation Water 5000 15000 Refrigerants 1000 5000 Boiling Water 1000 40000 Refrigerants 200 5000 23 Example to solve: Convection: What is the approximate temperature difference between a hot plate and the surrounding air if the heat flux from the plate is 2 800 W/m Assume that the air is flowing past the surface with a velocity of 5 m/s giving a heat transfer coefficient of 2 20 W/(m K). 24 Radiation • All bodies send out energy in the form of electromagnetic radiation. • The wavelength and intensity is dependent on the temperature of the surface. • Radiation may be transferred through vacuum, but also through air. 25 • fig. 4, Heat transfer by radiation • ΔT = T T r 1 2 T 2 T 3 T 1 26 Heat transfer by radiation between a small body and an isothermal environment may be calculated by Newton’s law of cooling, if we define a radiation heat transfer coefficient: q = h⋅A⋅ΔT r r r where q = heat flow due to radiation (W) r h = radiation heat transfer coefficient r 2 (W/(m⋅°C)) 2 A = surface area of the small body (m ) ΔT = temp. difference between surfaces (°C). r 27 • h is a function of the geometry, the emissivity of the r surfaces and on the temperatures of the surfaces. • For radiation heat transfer between nonpolished surfaces with temperatures between 0 and 100°C, the radiation heat 2 transfer coefficient is usually between 4.5 and 12 W/(m °C) (≈5 at room temperature, ≈25 at 200°C) 28 Summing convection and radiation modes of heat transfer • If the temperature differences for convection and radiation are equal then the hvalues may be added to a total heat transfer coefficient. h + h = h c r tot • If the temperature differences are not the same, the two modes have to be treated separately. 29 Example to solve: Convection + radiation: A cold bottle of beer (+5°C) is placed in a room where the temperature of the air and of the walls is 25°C. Calculate the approximate heat flux caused by radiation and by natural convection 30 Overall heat transfer coefficient We define the overall heat transfer coefficient by the equation q = U⋅A⋅ΔT tot 2 where U = the overall heat transfer coefficient (W/(m⋅°C)) 2 A = surface area on either side of the wall (m ) ΔT = difference between the fluid temperatures tot sufficiently far from the wall. 31 Relation between U, h and k • U can be related to the h , h in the fluids and to k and δ of 1 2 the wall through the temperature differences: • ΔT on either side of a wall can be written, according to Newton’s law of cooling as, respectively: ΔT = q/(h⋅A ) 1 1 1 ΔT = q/(h⋅A ) 2 2 2 • ΔT in the wall is, according to Fourier´s law: ΔT = q⋅δ/(k⋅A ) W W 32 • The total temperature difference is ΔT = q/(U⋅A) Tot • But ΔT = ΔT + ΔT + ΔT tot 1 2 W • q (at steady state) must be equal at the inside, the outside and in the wall, thus • 1/(U⋅A)= 1/(h⋅A )+ δ/(k⋅A )+ 1/(h⋅A ) 1 1 W 2 2 33 • For the case of a plane wall, the areas are also equal, and the relation is even simpler. • If the wall has two or more layers of different materials, additional terms of δ/(k⋅A ) have to be added. W • For a curved surface, the surfaces are not the same. The overall heat transfer coefficient may be referred to any of the surfaces, but when specifying the Uvalue, it must always be stated to which area it is connected. 34 Fig. 5 ΔT 1 q = U⋅A⋅ΔT q tot T ΔT tot ΔT w k δ ΔT 2 35 Heat transfer resistance • Thermal resistances, are defined analogous to electric resistances. • The thermal correspondence to Ohms law shows how the thermal resistance has to be defined: Ohms law: Voltage = Current ⋅ Resistance Thermal analogy gives: ΔT = q ⋅ R th 36 • Thermal resistance in the fluid R = 1/(h⋅A ) 1 • Thermal resistance in the wall R = δ/(k⋅A) • Total thermal resistance R =1/(U⋅A) tot • The total thermal resistance is thus the sum of the resistances R = R + R + R tot 1 W 2 37 Example to solve: Overall heat transfer coefficient: 2 The heat flux through a plane wall is 5000 W/m and the 2 overall heat transfer coefficient is 1000 W/(m⋅°C). Calculate the temperature difference 38 Part 2 More about convection 39 Definition of dimensionless parameters • A large number of parameters are needed to describe heat transfer. • Parameters may be grouped together to form a small number of dimensionless similarity parameters. These give simpler and more general equations by which heat transfer coefficients may be calculated. 40 Reynolds number The Reynolds number is defined as Re = u⋅x/ν where u = velocity of fluid (m/s) x = characteristic length (m). (For a tube, x = d). 2 ν = kinematic viscosity of fluid (m /s) • Reynolds number determines whether the fluid flow is laminar or turbulent. • Reynolds number determines the ratio of the inertia and viscous forces in the flow. 41 Reynolds number Glasrör Bläckstråle Laminärt strömningssätt Turbulent strömningssätt 42 Nusselt number The Nusselt number is defined as Nu = h⋅x/k 2 where h = surface heat transfer coefficient (W/(m⋅°C)) x = characteristic length (m). k = thermal conductivity (W/(m⋅°C)) • The Nusselt number: a dimensionless temperature gradient at the surface. • From Nu, the heat transfer coefficient can easily be calculated. 43 Example to solve: Water (k = 0.608 W/(m⋅K)) is flowing through a tube with inner diameter 15 mm. The Nusselt number is found to be 70. What is the heat transfer coefficient 44 Prandtl number The definition of the Prandtl number is Pr = ν/α = c⋅μ/k p 2 where ν = kinematic viscosity of fluid (m /s) 2 α = thermal diffusivity (m /s) c = specific heat (J/kg⋅°C) p 2 μ = dynamic viscosity (N⋅s/m ) k = thermal conductivity (W/(m⋅°C)) • The Prandtl number is a thermodynamic property of the fluid. 45 Grashof number The Grashof number is defined as 3 2 Gr = g⋅β⋅ΔT⋅x /ν 2 where g = acceleration of gravity (m/s ) β = volumetric thermal expansion coeff. (1/°C) For (ideal) gases, (β = 1/T ). abs ΔT = temperature difference between surface and fluid (°C) x = characteristic length (m). 2 ν = kinematic viscosity of fluid (m /s) • The Grashof number indicates, in free convection, whether the flow is laminar or turbulent. • It is the ratio between buoyancy and viscous forces. 46 Graetz number The Graetz number is defined as Gz = Re ⋅ Pr ⋅ d/x where d = (hydraulic) diameter of channel (m) x = distance from entrance of channel (m) • Gz is used when calculating heat transfer in laminar tube flow. Can be interpreted as a dimensionless, inverted length (distance from entrance). 47 Rayleigh number The Rayleigh number is defined as Ra = Gr ⋅ Pr The Rayleigh number often appears in equations for free convection. 48 Laminar and turbulent flow, velocity boundary layer Fig. 6 49 • The part of the flow where the velocity is influenced by the surface is called the boundary layer. • Near the front edge of the plate the thickness of the boundary layer is thin and it then grows successively thicker. • As long as the boundary layer is thin, there is no mixing between layers at different distances from the plate. The flow is then said to be laminar. • At some distance from the leading edge, the laminar layer will become unstable, and eddies will develop, mixing the different layers. The flow is then becoming turbulent. 50 • Because of the mixing, the difference in velocity between layers is much smaller in turbulent flow than in laminar, and the velocity profile thus much flatter. • Even in the turbulent region there is a laminar sublayer closest to the surface. In this sublayer, the temperature profile is nearly linear. • The type of flow, laminar or turbulent, can be determined from the Reynolds number calculated with the distance from the leading edge as the characteristic length. Transition will 5 occur at Re ≈ 5⋅10 . 51 Fig. 7 52 • In tube flow, at some distance from the entrance, the boundary layers from opposite sides will meet. At this point the flow is fully developed. • The fully developed flow may be turbulent or laminar. • The Reynolds number is calculated using the tube diameter as the characteristic length. Transition takes place at approximately Re ≈ 2300. • The heat transfer coefficients are generally higher in turbulent than in laminar flow. 53 Example to solve: 6 2 Water (ν = 0.86⋅10 m /s) flows through a tube with the diameter 12 mm at a velocity of 2 m/s. Determine if the flow is laminar or turbulent 54 Thermal boundary layer • At heated (or cooled) surfaces, a thermal boundary layer will form in which the temperature change from the wall temperature to the temperature of the undisturbed fluid. • In laminar flow the velocity and the thermal boundary layers will be similar, but the thicknesses will not necessarily be the same. • The relative thickness of the thermal and velocity boundary layers is in laminar flow related to the Prandtl number by n δ /δ ≈ Pr where n is a positive exponent (≈1/3). v th 55 • For gases, the Prandtl number is usually between 0.7 and 1, and in laminar flow the thermal and velocity boundary layer thicknesses are thus approximately equal. • For liquid metals, the Pr1 and the thermal boundary layer (in laminar flow) is considerably thicker than the velocity boundary layer, while for oils, Pr1 and the velocity boundary layer is the thickest. (For water, Pr range from 13.4 at 0°C to 1.75 at 100°C). • For turbulent boundary layers, the mixing within the layer will result in more or less equal thicknesses of the velocity and turbulent layers. 56 Forced and free convection • Fluid flow may be caused by force (by a fan or a pump or any other means external to the fluid itself). free convection (the movement is caused by temperature induced density differences in the fluid). • In forced convection, the type of flow (turbulent or laminar) is determined from the Reynolds number • In free convection, the type of flow is determined by the Grashof number 57 • The dimensionless equations are different for the two cases: In forced convection: Nu = f(Re, Pr) In free convection: Nu = f(Gr, Pr) 58 Methods for calculating heat transfer coefficients (in one phase flow) Forced convection 59 Turbulent flow in tubes and channels The heat transfer coefficient is generally referred to the logarithmic mean temperature difference (LMTD) defined as: ϑ = (T T ) (T T ) / ln(T T ) / (T T ) ln h2 c2 h1 c1 h2 c2 h1 c1 ΔTT−Δ 21 ΔT 2 ln = ΔT 1 where ΔT and ΔT are the temperature differences at each 1 2 end of the heated section (inlet and outlet) 60 Turbulent flow in tubes and channels Fully developed turbulent flow in tubes, the Dittus Boelter equation: 0.8 n Nu = 0.023⋅Re ⋅ Pr where n = 0.4 for heating of the fluid n = 0.3 for cooling of the fluid Valid in smooth tubes for Re 10000 0.6 Pr 100. Fluid properties at bulk temperature (≈average of inlet and outlet temperatures). May, for fluids with low viscosity (μ 2⋅μ ), be used when Re 2300, that is for the H2O whole turbulent region. 61 Entrance region, turbulent flow For the entrance region and for short tubes (10L/d400): 0.8 1/3 0.055 Nu = 0.036⋅Re ⋅ Pr ⋅ (d/L) where d = diameter of tube (m) L = length of tube (m) Thermal properties at bulk temperature. 62 Noncircular crosssections, hydraulic diameter Calculate the diameters as hydraulic diameters, D , defined H as D = 4⋅A/P H 2 where A = cross sectional area of the flow (m ) P = wetted perimeter (m) Example: 2 A = 4 cm ⋅ 8 cm =32 cm Cross 4 cm P = 4 + 4 +8 + 8 = 24 cm section ⇒ D = 4 ⋅ 32 / 24 = 5.33 cm H 8 cm 63 Laminar flow in tubes and channels When the temperature field is fully developed, the Nusselt number is constant The value of the constant depends on the boundary conditions and on the shape of the cross section The temperature field is considered to be fully developed for Gz 10. 64 Nu number in fully developed laminar flow in tubes and channels: Geometry of cross Nu Nu section (Constant wall temp) (Constant heat flux) Triangular (equilat.) 2.47 1.89 Square 2.98 3.09 Circular 3.66 4.36 Two infinite plates 7.54 8.24 65 Entrance region in laminar flow For the entrance region in laminar flow (Gz10), (average Nusselt number) 1/3 0.14 Nu = 1.86⋅Gz ⋅ (μ / μ ) w where μ = the dynamic viscosity of the fluid at the mean bulk temperature μ = the dynamic viscosity of the fluid w at the wall temperature 66 Fig. 9 (Fig. 65 in Holman) 67 Flow across a plate, laminar and turbulent flow Plate heated (or cooled) right from the leading edge, 5 laminar flow (Re 5⋅10 , 0.6Pr60, isothermal surface): x 1/2 1/3 Nu = 0.332 ⋅ Re⋅ Pr Local Nu number at x. x x 1/2 1/3 Nu = 0.664 ⋅ Re⋅ Pr Average Nu number for L L distance 0 to L from edge. Thermodynamic properties at film temperature (average of wall and free stream temperatures). 68 Turbulent flow across plates 5 7 For (5⋅10 Re 10 ) 0.8 1/3 Nu = 0.0296 ⋅Re ⋅Pr Local Nusselt number at x x x Average Nu number for both the laminar part of the flow 5 7 closest to the edge and the turbulent part (5⋅10 Re 10 ) 0.8 1/3 Nu = (0.037⋅Re 871) ⋅ Pr L L 69 Forced convection across single cylinders Average Nusselt number for the circumference: n 1/3 Nu = C⋅ Re ⋅ Pr Reynolds number (x=d) C n 0.4 4 0.989 0.330 4 40 0.911 0.385 40 4000 0.683 0.466 4000 40’000 0.193 0.618 40’000 400’000 0.0266 0.805 Fluid properties at film temperature. 70 Forced convection across tube banks Average Nusselt number: n 1/3 Nu = C⋅ Re ⋅ Pr Constant C and the exponent n are dependent on the ratios of the tube spacings (normal and perpendicular to flow) and the tube diameter (See table 64 in Holman). The Reynolds number should be calculated using the maximum velocity occurring in the tube bank. 71 Example to solve: Find the average Nusselt no as air flows past a 0.3 m heated 6 2 plate at 0.5 m/s. For the air, ν =15.7 ⋅ 10 m /s, Pr = 0.72. 72 Free convection • In free convection the fluid flow is induced by density differences caused by temperature differences. • The flow is caused by buoyancy forces, is dependent on gravitation to appear. • The thermal and velocity boundary layers in free convection are quite different from those in forced convection. 73 • The critical Grashof number, where the transition occurs from laminar to turbulent flow depends on the geometry. 74 Free convection boundary layer on a vertical flat plate 75 General correlation for free convection In many different geometries, the average Nusselt number may be calculated by as m Nu = C⋅(Gr⋅ Pr) The constant C and the exponent m depend on the geometry and on the size of (Gr⋅Pr). Thermodynamic properties at film temperature. 76 Constant C and exponent m in general corr. for free conv. Geometry C m Gr⋅ Pr 4 9 Vertical plates 10 10 0.59 1/4 9 13 and cylinders 10 10 0.10 1/3 4 9 Horizontal cylinders 10 10 0.53 1/4 9 12 10 10 0.13 1/3 4 6 Upper surface of heated 0.54 1/4 2⋅10 8⋅10 plates, or 6 11 lower surface of cooled plate 0.15 1/3 8⋅10 10 5 11 Lower surface of heated 10 10 0.27 1/4 plates, or upper surface of cooled plates 77 3 Free convection Gr⋅Pr/(ΔT⋅L ) tabulated • The group 3 2 Gr⋅Pr/(ΔT⋅L ) = g⋅β/ν is only dependent on material properties and may be found in the Collection of formulas and tables for different fluids. 78 Free convection The exponent m • The exponent m is often 1/3 (turbulent flow) or ¼ (laminar flow). • With m = 1/3, h is independent of L as it cancels out of the equation: 1/3 Nu = C⋅(Gr ⋅ Pr) 3 2 1/3 h⋅L/k = C ⋅(g⋅β⋅ΔT⋅L /ν ) ⋅ Pr ⇒ h independent of L 79 Free convection Vertical plates, vertical cylinders • Isothermal vertical plates: L = height of plate. • Vertical cylinders: treat as vertical plates, if 1/4 D/L ≥ 35/Gr 4 • For Gr 10 , use graphical solution (fig. 12) 80 Free convection, simplified correlations for air For a given fluid, the dimensionless equation 34 may be rewritten to give the heat transfer coefficient directly. Two cases have to be discerned, laminar and turbulent: For laminar flow: ¼ 4 8 h = K ⋅ (ΔT/H) for 10 Gr⋅Pr 10 l For turbulent flow 1/3 8 12 h = K ⋅ ΔT for 10 Gr⋅Pr 10 t Where K and K from Table 6 for air (assuming SIunits). l t 81 Free convection, simplified correlations for air Table 3: Constants K and K for air l t 3 T K K (Gr⋅Pr)/(ΔT⋅H ) film l t 7 50 1.57 1.88 34.8⋅10 7 0 1.49 1.66 14.5⋅10 7 50 1.41 1.48 6.75⋅10 7 100 1.35 1.33 3.47⋅10 7 200 1.27 1.14 1.18⋅10 6 300 1.21 1.01 5.1⋅10 6 400 1.15 0.91 2.54⋅10 6 600 1.06 0.76 0.85⋅10 82 Vertical Isothermal Plates, Nu vs GrPr 83 Free convection from horizontal cylinders • Constants C and m from table. • Tube diameter is characteristic length. 84 Free convection from horizontal surfaces Two cases: 1. Top side of a heated plate and bottom side of a cooled plate. (Gravitation will force fluid away from surface) 2. Bottom side of a heated surface and top side of a cooled plate. (A stable layer of fluid will form, decreasing the heat transfer). • C and m from table. C is exactly twice as high in case 1. • Characteristic length: L = A/P where A is the area and P the perimeter of the surface. 85 Free convection in between vertical plates Correlation for the average Nusselt number: −12 / ⎡⎤ C C 1 2 Nu=+ ⎢⎥ 2 12 / as (/ Ra⋅ s L) (Ra⋅ s/ L) ss ⎣⎦ where s = distance between plates (m) L = plate height (m) Ra = (Gr ⋅ Pr), with s as characteristic length. s Nu = average Nusselt no, with s as the a,s characteristic length. • ΔT in Gr is the temperature difference between the fin surface and the undisturbed air. 86 Free convection in between vertical plates Constants C and C 1 2 Boundary condition C C 1 2 Symmetric isothermal plates 576 2.87 Symmetric isoflux plates 48 2.51 Isothermal/adiabatic plates 144 2.87 Isoflux/adiabatic plates 24 2.51 87 Free convection in between vertical isothermal plates 7 6 5 s = 15 mm Plate s = 10 mm 4 s = 8 mm 3 s = 6 mm 2 s = 5 mm 1 s = 4 mm 0 0 50 100 150 200 250 300 Dt/H Dt/H (air 60°C) 88 h (W/(m2 K))Free convection in between vertical plates, different boundary conditions, (plate distance 8 mm, air 60°C) 7 6 5 4 3 2 h, Isothermal h, Isoflux 1 h, Isothermal/adiabatic h, Isoflux/adiabatic 0 0 50 100 150 200 250 300 Dt/H 89 h (W/(m2 K))Flow in an enclosed space 90 Free convection in enclosed spaces (fig 15) Define a heat transfer coefficient using the surface temperatures between which heat is transferred: h = q/(A⋅(T T )) 1 2 Use the distance δ between the surfaces as the characteristic length The Nusselt number is expressed as the ratio between an apparent, or effective thermal conductivity k and the normal e conductivity k. Nu = k / k e 91 We get: q/A = h ⋅ (T T ) = (Nu ⋅ k/δ )⋅ (T T ) = k /δ ⋅ (T T ) 1 2 1 2 e 1 2 with Nu ⋅ k = k e 92 For vertical enclosures and constant heat flux 1/4 0.012 0.30 Nu = 0.42 ⋅ (Gr ⋅ Pr) ⋅ Pr ⋅ (L/δ) 4 7 Conditions: 10 Gr ⋅ Pr 10 1 Pr 20000 10 L/δ 40 where δ = the distance between the plates (m) L = the height of the enclosure (m) 93 Vertical enclosures Larger values of Gr⋅Pr 1/3 Nu = 0.046 ⋅ (Gr ⋅ Pr) 6 9 Conditions 10 Gr ⋅ Pr 10 1 Pr 20 1 L/δ 40 Use δ, as characteristic length in Nu and Gr, ΔT = difference between the surfaces. 94 Vertical enclosures Low Gr ⋅ Pr 3 At Gr ⋅ Pr 10 there is no free convection and heat is transferred by conduction only (Nu = 1). 95 Vertical enclosures, comparison of eqs. 41 and 42. 100 Nu eq38 Nu eq. 41 Nu eq. 42 Nu eq39 Conduction 10 Nu=1 1 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09 1.E+10 Gr Pr 96 NuHorizontal enclosures Two cases: 1.If the top plate is the hotter, a stable situation will occur and heat is transferred by pure conduction, and thus Nu = 1 (or k /k =1). e 2.If the bottom plate is the hotter, pure conduction will occur at Gr 1700, while at higher values, convection cells will occur, increasing heat transfer. See table 73 in Holman. 97 Combined free and forced convection To determine which type of convection that is dominant, the following inequality may be used. 2 Gr / Re 10 If this criterion is fulfilled, free convection is dominant. 98 Example to solve: Find the Nusselt no at a heated vertical plate, if the Grashof 8 no is known to be 1⋅10 and Pr is 1. 99 Part 3 Radiation 100 Radiation Thermal radiation.: ”That electromagnetic radiation emitted by a body as a result of its temperature”. Thermal radiation is restricted to a limited range of the electromagnetic spectrum. 101 Radiation 102 103 Blackbody radiation A blackbody is a perfect radiator. Three characteristics: • It absorbs all incident radiation • It radiates more energy than any real surface at the same temperature • The emitted radiation is independent of direction Also: Blackbody radiation obey certain simple laws 104 StefanBoltzmann’s law The total power radiated from a blackbody is calculated from StefanBoltzmann´s law: 4 E = σ⋅T b where E = total power radiated per unit area from b 2 a blackbody (W/m ) 8 2 4 σ = 5.669 ⋅ 10 W/(m⋅K ). (StefanBoltzmann constant) T = absolute temperature (K) 105 Planck distribution law The wavelength distribution of emitted blackbody radiation is determined from the Planck distribution law (fig. 19): C 1 ET (,λ )= λ ,b 5 λλ ⋅⋅ exp(CT / ( ))− 1 2 2 8 4 2 where C = 2π⋅h⋅c = 3.742⋅10 W⋅μm / m 1 o 4 C = (h⋅c / k) =1.439 ⋅ 10 μm K 2 o λ = wavelength (μm) T = absolute temperature (T) h = Planck constant 106 k = Boltzmann constant c = speed of light in vacuum 0 107 Ù Result of increasing temperature on radiation • Higher intensity • Shorter wavelength higher frequency 108 Wien’s displacement law The wavelength of maximum emissive power is determined by Wien’s displacement law: λ ⋅ T = C = 2897.8 μm K max 3 109 Radiation from real surfaces Real surfaces: • emit and absorb less than blackbodies • reflect radiation • emit and absorb differently depending on angle and wavelength • do not obey the simple laws 110 Blackbody and real surface emissions 111 Spectral emissivity and total emissivity To account for ”real surface” behavior we introduce the spectral emissivity, defined by E (λ,T) = ε (λ,T) ⋅ E (λ,T) λλλ,black and the total emissivity defined by 4 E = ε ⋅ E = ε ⋅ σ ⋅ T b (ε = integrated average) 112 Gray diffuse body To simplify matters it is common to assume the emissivity to be independent on wavelength and direction. Such a surface is called a gray diffuse body, for which E (λ,T) = ε ⋅ E (λ,T) λλλ,black ε = constant = ε (0 ε 1) λ 113 Absorptivity, reflectivity, transmittivity Incident radiation may be absorbed, reflected or transmitted We define Absorptivity α: Fraction of incident radiation absorbed Reflectivity ρ: Fraction of incident radiation reflected Transmittivity τ: Fraction of incident radiation transmitted thus α + ρ + τ = 1 ρ α 114 τ Kirchhoff’s identity The (total) emissivity and the (total) absorptivity of a surface are equal at equal temperatures (wavelengths): ε = α 115 Radiation exchange between blackbodies To calculate radiation exchange we must take into account • surface areas • surface geometries • position in relation to each other This is done by the shape factor, F 12 F = fraction of radiation leaving surface 1 intercepted 12 by surface 2. 2 1 116 Net exchange of radiation, law of reciprocity Net exchange of radiation between blackbodies: 4 4 q = F ⋅ A ⋅ (E E ) = F ⋅ A ⋅ σ ⋅ (T T ) 12 12 1 b1 b2 12 1 1 2 Law of reciprocity F ⋅ A ⋅ = F ⋅ A 12 1 21 2 117 Ö Shape factors Shape factors are often difficult to calculate. See diagrams and formulas in Holman, Figs 8.128.16 and CFT pp. 4547 Important special case: Small (convex) surface (1) surrounded by other surface (2): F = 1. 12 118 ÖÖ Radiation exchange between real surfaces, simple case For the special case of F = 1 the exchange is calculated as 12 4 4 q = ε ⋅ A ⋅ (E E ) = ε ⋅ A ⋅ σ ⋅ (T T ) 12 1 1 b1 b2 1 1 1 2 Comparing to Newton’s law of cooling: q = h ⋅ A ⋅ (T T ) r 1 1 2 4 4 h = ε ⋅ σ ⋅ (T T ) / (T T ) = ε ⋅ h r 1 1 2 1 2 1 r, black h = f(T , T ), from table r, black 1 2 F = 1 12 A 1 119 Table 8 (p.30 in CFT) 120 Total emissivities of selected materials Material Temp (°C) Emissivity ε Aluminum, commercial sheet 100 0.09 Copper, polished 100 0.052 Iron, darkgray surface 100 0.31 Glass, smooth 22 0.94 Snowwhite enamel varnish 23 0.906 Black shiny lacquer 24 0.875 Roofing paper 21 0.91 Porcelain, glazed 22 0.92 Red brick 23 0.93 Alpaints 100 0.270.67 121 Radiation exchange calculated by resistance networks Consider the blackbody emissive power as the driving potential 4 4 q = (E E ) / R = σ ⋅ (T T ) / R 12 b1 b2 rad 1 2 rad Consider the radiation resistance as the sum of surface and space resistances R = R + R + R rad surface1 space surface2 J 1 J 2 E E b1 b2 R R R surface1 space surface2 122 Radiation exchange by resistance networks, assumptions Assume: • All surfaces are gray, • All surfaces are uniform in temperature. • Reflective and emissive properties are constant over the surfaces. Define: • Irradiation, G = total radiation / (unit time, unit area) • Radiosity, J = total radiation leaving /(unit time, unit area) (including reflected radiation) Assume these properties are uniform over each surface. 123 Surface resistance: Radiosity = emitted radiation + radiation reflected: J = ε ⋅ E + ρ ⋅ G b where ε = emissivity ρ = reflectivity of surface Assume surfaces opaque (τ=0) = ρ = 1 α = 1 ε (as α = ε). = J = ε⋅ E + (1 ε) ⋅ G b or G = (J ε ⋅ E ) / (1 ε) b 124 The net energy leaving the surface per unit area: q / A = J G = ε⋅ E + (1 ε) ⋅ G G = b =ε ⋅ E ε ⋅ G = ε ⋅ (E G) b b () EJ− ε⋅ A b q= ⋅−() EJ= b = 11−εε()−⋅/(ε A) Consider (E J) as the driving potential. b = R = (1ε)/(ε ⋅ A) surface 125 Space resistance: Radiation from 1 to 2 (per unit time): J⋅A⋅F . 1 1 12 Radiation from 2 to 1 (per unit time): J⋅A⋅F 2 2 21. Net exchange of radiation: q = J⋅A⋅F J⋅A⋅F = (J J )⋅A⋅F = (J J )⋅A⋅F 12 1 1 12 2 2 21 1 2 1 12 1 2 2 21 (as A⋅F = A⋅F ). 1 12 2 21 Consider (J J ) as the driving potential, 1 2 = R = 1/(A⋅F ) space 1 12 126 Heat exchange due to radiation, two bodies 4 4 EE− EE−σ⋅−() TT bb 12 bb 12 1 2 q=== 1−ε 1−ε 1 R RR++R 1 2 rad surface12 space surface ++ ε⋅AA⋅Fε⋅A 11 1 12 22 127 Special case, two parallel infinite plates 4 4 qσ⋅−() TT 1 2 = 11 A +− 1 F = 1 and A = A 12 1 2 εε 12 Special case, two concentric cylinders or spheres 4 4 qσ⋅−() TT 1 2 = 11 A A 1 1 +⋅()−1 F = 1 12 εε A 1 22 4 4 If A A q /A = ε ⋅ σ ⋅ (T T ) 1 2 1 1 1 2 128 Three body problem • Calculate all resistances. • For the nodes J , J and J , the sum of the energy flows into 1 2 3 each of the nodes must be zero. Example, node J 1 J R J 2 space12 1 EJ− JJ− JJ− E b11 21 31 b E b ++= 0 R AF⋅ AF⋅ ⎛⎞ 1−ε surface 112 113 R 1 surface ⎜⎟ ε⋅ A ⎝⎠ 11⋅ R space13 R space23 • Solve J J 1 3 J • Calculate the heat exchanges 3 R surface E b 129 (q/A) 32 (q/A) 13 Radiation shields 4 4 qσ⋅−() TT 1 2 1 3 2 1 = 2 A11// εε+−1 12 No shield With shield (q/A) = (q/A) = (q/A) 13 32 4 4 4 4 σ⋅−() TTσ⋅−() TT q 1 3 3 2 == A11// εε+−11// εε+− 1 1 13 32 Everything is known except the temperature T 3. 130 Simplest case, all emissivities equal: 4 4 4 4 4 4 4 T T = T T = T = ½ ⋅ (T + T ) 1 3 3 2 3 1 2 4 4 1 ⋅⋅σ() TT − q 2 1 2 = A 11// εε+−1 = 32 Emissivities assumed equal, = heat flux reduced to half 131 Multiple shields of equal emissivities It may be shown by similar reasoning that the heat flux will be reduced to 1 (q/A) =⋅ (q / A) with shields without shields n+1 where n is the number of shields. 132 Different emissivities Two plates with equal emissivities ε 1 One shield of different emissivity ε : 2 2 −1 1 ε 1 (q/A) =⋅ (q / A) with shields 11 without shields 2 +−1 εε 12 ε ε = largest decrease in heat flux 2 1 Assume ε = 1. = 1 1 21− 1 (q/A) =⋅ (q / A) =⋅ε⋅(q / A) with shields 1 without shields 2 without shields 211 +− 2 ε 2 thus factor equal to one half times the emissivity of the shield. 133 Part 4 Conduction, Boiling, Condensation and Diffusion 134 Conduction Introduction Fourier’s law: q = k⋅A⋅δT/δx For a plane wall, the temperature gradient is constant q = k⋅A⋅ΔT/δ 135 Conduction in cylindrical shells When heat is conducted through a wall which is not plane, the temperature gradient will no longer be constant through the wall. q = k⋅A⋅δT/δx At any section, q is constant Assume k constant ⇒ (A⋅δT/δx) = constant. Thus, if the area A changes with x, as it will for a curved surface, the temperature gradient must also change. 136 Conduction in cylindrical shells The simplified version of Fourier´s law may still be used if we use an average area (or average diameter) For a cylindrical wall the area is calculated as the logarithmic mean of the inside and outside surface. A = (A A )/ln(A /A ) = ln 2 1 2 1 π ⋅ L ⋅ d = π ⋅ L ⋅ (d d ) / ln(d / d ) ln 2 1 2 1 137 Conduction in cylindrical shells 138 Conduction in fins Flänsarnas area=A f Area mellan flänsarna =A 2 t t t f 1 t ϑ f rot ϑ mf t 2 Fig 11.5 i termo 139 Conduction in fins Definition of fin efficiency Because of the lower ΔT, lower heat flux from fin ⇒ The fin surface is less efficient than the base surface, We define a fin efficiency η as the ratio between the actual f heat transferred and the heat which would be transferred if the whole fin had the base temperature. η ≡ ΔT / ΔT f m, fin base (Assuming equal h on fin and base) 140 Conduction in fins For straight fins with constant cross section: η ≡ tanh(m ⋅ L) / (m ⋅ L) f where L = the length of the fin (m) 1/2 m = (h ⋅ P)/(k ⋅ A) where P = perimeter of the fin (m) A = cross section area perpendicular 2 to heat flow (m ) For many types of fins, the fin efficiency may be estimated from the following diagram. 141 142 Overall heat transfer coefficient For a finned surface, the area is calculated as A = A + η ⋅ A tot base f fin The Uvalue of a finned tube is calculated as: 1/(U⋅A) = 1/(h⋅A ) + δ/(k⋅A ) + 1/h⋅(A + η⋅A ) 1 1 ln 2 base f fin 2 where A = inside surface area (m ) 1 2 A = outside surface area in between fins (m ) base 2 A = logarithmic mean area of tube (m ) ln 2 A = fin area (m ) fin 143 Electric analogy for solving 2d conduction problems Make electric model of 2d conduction problem in: • Electrolyte tray • Thin conducting foil • Resistance mesh Boundary conditions: Isothermal or adiabatic Inside of wall, T 1 Line of symmetry Line of symmetry Line of symmetry Points of equal Rail, T 2 potential Outside of wall, T 2 144 Ö Electric analogy models… Electrolyte tray or foil continuous models Resistance mesh: discreet model Measurements show : Potential in node = mean of the four surrounding nodes Measurements not necessary Calculate R = 50 Ohm 145 Excel model of wall with rail, horizontal cut AB C D E F G H I J K L 1 200 200 200 200 200 200 200 200 200 200 200 200 2 188.7026 188.6737 188.5876 188.447 188.2564 188.0241 187.7626 187.49 187.2305 187.013 186.8668 186.8151 3 177.463 177.4044 177.2299 176.9438 176.5547 176.0775 175.5362 174.9668 174.419 173.9545 173.6391 173.5268 4 166.3407 166.2511 165.9838 165.5437 164.941 164.1949 163.3379 162.4221 161.5243 160.747 160.2085 160.0137 5 155.3975 155.2755 154.9105 154.3061 153.4708 152.4232 151.1985 149.8595 148.5091 147.3007 146.434 146.1111 6 144.6982 144.5429 144.0767 143.2993 142.2131 =(D5+E6+C6+D7)/4 140.8286 139.1733 137.3 082 135.3518 133.5128 132.1159 131.5626 7 134.3096 134.1213 133.5542 132.6013 131.2538 129.5046 127.3581 124.8484 122.077 119.283 116.9541 115.9075 8 124.2975 124.0787 123.4173 122.298 120.6961 118.578 115.906 112.6502 108.8248 104.5881 100.51 98.15913 9 114.723 114.4787 113.7382 112.4775 110.6545 108.2053 105.0376 101.0218 95.98379 89.73458 82.33865 75.70911 10 105.637 105.375 104.5793 103.2192 101.2393 98.55121 95.01738 90.41565 84.35401 76.02783 63.40091 40 11 97.07489 96.8051 95.98494 94.58083 92.53218 89.74285 86.06506 81.26942 74.98878 66.6218 55.23719 40 12 89.05243 88.78556 87.97451 86.58699 84.56579 81.82295 78.2306 73.60818 67.7099 60.23341 50.92604 40 13 81.56375 81.31023 80.54056 79.22686 77.32104 74.75256 71.42625 67.22281 62.00923 55.67591 48.23357 40 =(A13+B14+B14+A15)/4 14 74.58213 74.35108 73.65065 72.45886 70.73898 68.44003 65.49901 61.84761 57.42829 52.22745 46.33231 40 15 68.06261 67.86134 67.25212 66.21895 64.736 62.76959 60.28217 57.24033 53.62889 49.47328 44.86824 40 16 61.94566 61.77957 61.27757 60.42882 59.21651 57.62016 55.61977 53.20265 50.37367 47.16856 43.66736 40 17 56.1609 56.03372 55.64977 55.00226 54.08106 52.87477 51.3741 49.57686 47.49456 45.15991 42.63266 40 18 50.63054 50.54463 50.28553 49.84939 49.23071 48.42378 47.42502 46.23611 44.86782 43.34387 41.70337 40 19 45.27201 45.22874 45.09834 44.87908 44.5686 44.16462 43.66609 43.07473 42.39673 41.64439 40.83694 40 20 40 40 40 40 40 40 40 40 40 40 40 40 146 Tools/Alternatives/Calculate/Iteration 147 Ù Transient heat transfer Lumped capacitance method Use when temperature of the body is approx. uniform. I.E. if R R h k / L th, surface th, internal We define the Biot number (Bi) as Bi = h⋅L / k where h = heat transfer coefficient k = thermal conductivity of solid L = characteristic length of solid body = V/A =volume of body /exterior surface area 148 Transient heat transfer Lumped capacitance method… Method can be used when Bi 0.1. After a sudden change in ambient temperature: ΔT = ΔT ⋅ exp(Bi ⋅ Fo) 0 where ΔT = temperature difference at time τ. ΔT = temperature difference at τ =0. 0 2 Fo = k⋅τ/(ρ⋅c⋅L ) = Fourier number (dim. less time) p τ = time after stepchange in ambient temperature k = thermal conductivity of solid body ρ = density of solid body c = specific heat of solid body p 149 Transient heat transfer Lumped capacitance method… Product of the Biot and Fourier: hV⋅(/A) k⋅τ hA⋅ Bi⋅= Fo⋅=⋅τ 2 kρρ ⋅⋅cV(/A) ⋅⋅ cV pp Numerator: = 1/R th, convection Denominator: C Heat capacity of the solid th R = 1/(h⋅A) and C = m⋅c = ρ ⋅ V ⋅ c th th p p = Bi ⋅ Fo = τ /(R⋅C) th 150 Transient heat transfer Lumped capacitance method… τ −⋅ ΔT RC⋅ th th = e = ΔT 0 Compare equation for discharge of a condenser through a resistor Lumped capacitance method handy for quick estimates 151 Transient heat transfer Heisler charts For simple geometries, use diagrams (Heisler charts) where ΔT/ΔT = f (Fo, Bi) 0 Diagrams are found in: • Collection of formulas and tables, pp. 4243, • book by Holman, fig. 47 413. Note that the definition of the characteristic length used in Bi and Fo in the charts are not the same as used above Charts valid for Fo 0.2. 152 Two or three dimensional problems: multiply the solutions for the corresponding onedimensional cases. Examples: The temperature in the center of a short cylinder: (ΔT/ΔT ) = (T/ΔT ) ⋅ 0 center, short cylinder 0 center, infinite cylinder ⋅(ΔT/ΔT ) 0 center, infinite wall Temperature in the center of a cube: 3 (ΔT/ΔT ) = (ΔT/ΔT ) 0 center, cube 0 center, infinite wall 153 Boiling Two types of boiling: pool boiling, (as on the outside of tubes) and flow boiling (boiling inside tubes) 154 Pool Boiling The boiling curve 155 Pool Boiling A simple correlation for nucleate pool boiling: (0.12 0.2 ⋅ log10 Rp) 0.55 0.5 0.67 h = C ⋅ 55 ⋅ p ⋅ (log p ) ⋅ M ⋅ (q/A) nb r 10 r where C = 1 for horizontal, plane surface C = 1.7 for horizontal copper cylinders p = p/p = reduced pressure r critical R = Surface roughness (μm) (about 1 for many p technically smooth surfaces) M = molecular weight of the fluid (kg/kmol) 2 q/A = heat flux (W/m ) 156 Flow boiling Flow regimes in flow boiling 157 Flow boiling Flow boiling of refrigerants in horizontal tubes: 2 2 0.4 Complete evaporation: Nu = 1.0 ⋅ 10 ⋅ (Re ⋅ K ) m f 3 0.5 Incomplete evaporation: Nu = 1.1 ⋅ 10 ⋅ Re ⋅ K m f where Re = 4 ⋅ m’ /(π⋅d⋅μ ) l where m’ = mass flow (kg/s) d = tube diameter (m) 2 μ = dynamic viscosity of liquid (Ns/m ) l 158 K = Δi/(L ⋅ g) f where Δi = specific enthalpy difference across tube (J/kg) L = tube length (m) 2 g = acceleration of gravity (m/s ) The equations give average Nusselt number for the tube. 159 Condensation Two types of condensation: Film condensation (when the liquid wets the surface) Drop condensation (opposite) Drop condensation best but difficult to achieve. Calculate for film condensation Heat transfer resistance in condensation is entirely due to conduction of heat through the liquid. 160 Condensation ⇒ Heat transfer coefficient may be calculated from the film thickness as h = k /δ where k = thermal resistance of liquid (W/(m⋅°C)) δ = film thickness (m) Film thickness in laminar flow at a distance x from the top 1/4 δ = 4⋅ μ⋅ k ⋅x⋅ ΔT / g⋅h ⋅ρ⋅(ρ ρ ) fg v 2 where μ = dynamic viscosity of liquid (Ns/m ) 161 x = distance from top of surface (m) ΔT = temperature difference between surface and vapour (°C) 2 g = acceleration of gravity (m/s ) h = heat of vaporisation (J/kg) fg 3 ρ = density of liquid (kg/m ) 3 ρ = density of vapour (kg/m ) v 162 Condensation This gives the local heat transfer coefficient 3 1/4 h = g⋅h ⋅ρ⋅(ρ ρ )⋅ k / 4⋅ μ ⋅x⋅ ΔT fg v and by integrating, the average heat transfer coefficient h =4/3⋅h = av L 3 1/4 = 4/3⋅g⋅h ⋅ρ⋅(ρ ρ )⋅ k / 4⋅ μ ⋅L⋅ ΔT = fg v 3 1/4 = 0.943⋅g⋅h ⋅ρ⋅(ρ ρ )⋅ k / μ ⋅L⋅ ΔT fg v Fluid properties at film temperature. 163 Condensation Alternative eq if q/A rather than ΔT is known: 1/3 h = 0.924⋅k⋅g⋅h ⋅ρ⋅(ρ ρ ) / μ ⋅L⋅ q/A av fg v For single horizontal tubes: exchange L for d, and the constant 0.943 is changed to 0.725. (For n horizontal tubes placed on top of each other, the diameter d should be multiplied by n). 164 For condensation inside horizontal tubes, The liquid will more or less fill the tube. It has been suggested that could be used in this case too, but with the constant changed from 0.943 to 0.555. As a rule of thumb, condensation heat transfer coefficients of 2 refrigerants (not NH ) is about 2000 W/(m⋅K). 3 165 Condensation In turbulent flow (Re1800), plane vertical surfaces 1/2 Nu = 0.0030 ⋅ G ⋅ C for Re 1800 v where Nu = h ⋅ L / k 3 2 G = g⋅L /ν 3 C = g⋅ρ⋅L /( k⋅ν ⋅ΔT) v Re = 4⋅m’/(P⋅μ) where m’ = mass flow (kg/s) P = wetted perimeter (= width for plane wall, = π⋅d for vertical tube) 166 m m Diffusion d d Luft air Våt yta Wet surface p p′′ p p′′ p p å åv å åv å å Water or frost collects on cold surfaces, and wet surfaces dry up: In these processes water vapor is transferred by diffusion and convection to or from the surface. Heat of vaporization is transferred to or from the surface. 167 Diffusion The heat flow is calculated as: q = m´ ⋅ h d d fg where m´ = mass of water changing phase per unit time We define a diffusion heat transfer coefficient h as d q = h ⋅ A ⋅ ΔT d d 2 where A = wet area (m ) ΔT = temp diff between wet surface and air 168 Diffusion The ratio between the diffusion heat transfer coefficient h d and the convection heat transfer coefficient h may be c calculated by the equation h / h ≈ C ⋅ Δp/ΔT d c where C = 1520 above a wet surface and C = 1750 above a frozen surface 169 ΔT = T” T , temperature difference between vw v wet (frozen) surface and air (°C) Δp = p” p = difference in partial pressure of vw v water vapour between the surface and the free air (in bar) p” = saturation pressure of water at the vw temperature of the surface (from steam table). p = ϕ ⋅ p” v v ϕ = relative humidity () p” = saturation pressure of water at the v temperature of the free air (from steam table). 170 Diffusion Note that h / h may be positive or negative, indicating heat d c flow in opposite directions. The ratio may also be attained directly from the T, p diagram v (fig 28) To find the heat transfer by diffusion, we first calculate the convection heat transfer coefficient according to Lesson 2, then calculate the ratio h / h . d c Then we can calculate q and q as h = h + h = h (1 + d tot tot c d c h /h ) d c 171 h t t h t h t, p diagram för å fuktig luft Diffusion p x å p å (α /α ) (α /α ) d kw frost d kw t 172 ϕPart 5 Heat exchangers 173 Heat exchangers Three main types of heat exchangers: In recuperative heat exchangers, the heat is transferred from one fluid to the other through a dividing wall. In regenerative heat exchangers, heat is transferred via a moving solid part, (a rotating wheel with a large number of narrow channels). In evaporative heat exchangers, hot water is cooled by partly evaporating the water. 174 Heat exchangers Regenerativ värmeväxlare fyllnings material i en rotor som långsamt roterar Evaporativ värmeväxlare (kyltorn) Rekuperativ värmeväxlare kyltorns fyllning späd vatten luft kylt vatten varmt vatten pump för cirkulation av vatten genom kyltornsfyllningen Fig 29 (Fig. 11.18 i termo) 175 Heat exchangers Two basic types of recuperative, Counterflow and parallel flow (fig 30). Motströms Medströms 2 1 1 2 t t t 1 1 Δ 1 ϑ Δ 2 1 θ ϑ 1 t 2 Δ t 2 2 ϑΔ 1 2 Area Area 0 A 0 A a) b) Counterflow Parallel flow 176 ϑθ= 2Heat exchangers The heat exchange rate from flow: q = m⋅ c⋅(T T ) = m⋅ c (T T ) h h h1 h2 c c c1 c2 where m = mass flow (kg/s) c = heat capacity (J/(kg⋅°C)) index h and c refers to the hot and cold fluid index 1 and 2 refers to the two ends. The product (m⋅c) is called the heat capacity rate and is often written as C. 177 Heat exchangers Heat exchange rate from size: q = U⋅A⋅ϑ ln where ϑ = log mean temp difference (or LMTD) ln (For parallel flow and counter flow heat exchangers) ϑ = (T T ) (T T ) / ln(T T ) / (T T ) ln h2 c2 h1 c1 h2 c2 h1 c1 = (ΔT ΔT ) / ln(ΔT / ΔT ) 2 1 2 1 178 Heat exchangers A third basic type of flow is crossflow. The temperatures of both fluids at the outlets are different from one side to the other of the flow channel. In most heat exchangers, the flow is neither purely parallel, counterflow or crossflow, but rather a mixture of these types. The logarithmic mean temperature difference may still be used, if corrected by factor F.: q = U⋅A⋅F⋅ϑ ln 179 Heat exchangers Cross flow heat exchanger, temperature profile Δ 1 1. θ Δ 2 2. 180 Heat exchangers A fourth basic type: Condensers and evaporators: the temperature of one fluid is constant The logarithmic mean temperature may be used without any correction (F=1) 181 Heat exchangers Temperature effectiveness, NTUmethod: The temperature effectiveness is defined as ε = actual heat transfer / maximum possible heat transfer The maximum possible heat transfer would result if the temperature of the fluid with the lowest heat capacity rate (C =m⋅c) at the outlet of the heat exchanger reach the inlet temperature of the other fluid. 182 Heat exchangers (This fluid is referred to in Holman as the minimum fluid. = = the fluid with largest temperature change). The definition gives ε = C ⋅ΔT / C ⋅ΔT = ΔT / ΔT min min min max min max where ΔT = temperature change of minimum fluid. min ΔT = the difference between the max inlet temperatures. 183 Heat exchangers The temperature effectiveness is a function of only two variables: UA/C and C / C min min max ε = f(C /C , UA/C ) min max min UA/C is called the number of transfer units (NTU). min See fig. 33. Equations for f are found in table 103 i Holman. 184 η= 2 0,9 0,8 0,7 0,6 0,5 0,4 10 7 5 3 2 1,5 0,3 kA = 1,0 W 1 0,75 0,2 0,1 0,50 0,25 Heat exchangers MOTSTRÖMSVÄRMEVÄXLARE 1,0 η 1 0,8 ε Δ 1 1 θ 0,6 2 Δ 2 UA/C min A 0,4 ΔΔ 1 2 η== ; η 1 2 θθ W=⋅ mc 1 1 p 0,2 1 W=⋅ mc 2 2 p 2 0,05 (ηη=W⋅ /W ) 21 1 2 0 0 0,2 0,4 0,6 0,8 1,0 W 1 W 2 fig. 11.25 i Termo C /C min max 185 Lesson 6 Methods of enhancing heat transfer 186 Enhanced heat transfer What is enhanced heat transfer ”An enhanced heat transfer surface has a special surface geometry that provides a higher h⋅A value per unit base surface area than does a plain surface.” When should it be applied 187 When should enhanced heat transfer be applied To the side of a heat exchanger where the dominant heat transfer resistance lies. It is the overall heat transfer resistance which determines the temperature difference. Example: 2 Heat exchanger with surface areas equal on both sides (1m ). Total heat transfer resistance (wall is neglected): 1/(UA) = 1/(h⋅A) + 1/(h⋅A) 1 2 188 2⋅ Assume h = 50 W/(m⋅°C) (forced flow air) 1 2⋅ and h = 1000 W/(m °C) (forced flow water) 2⋅ The overall heat transfer resistance is 1/(UA)=1/(50⋅1) + 1/(1000⋅1) = 0.02 +0.001 =0.021°C/W If we double the heat transfer coefficient on the water side from 1000 to 2000 we would get 1/(UA)=1/(50⋅1)+1/(2000⋅1)=0.02 + 0.0005=0.0205°C/W No improvement 189 If we double the heat transfer coefficient on the air side, we would get 1/(UA)=1/(100⋅1)+1/(1000⋅1) = 0.01+0.001 = 0.011°C/W A reduction to half of the original total heat transfer resistance. 190 Other aspects of enhanced heat transfer: The costs of enhanced surfaces are usually higher than for the plain surfaces. The enhanced surfaces are also usually more sensitive to fouling, which may be important in some applications. 191 What is enhanced heat transfer good for Enhanced heat transfer may be used for any of four reasons: 1. To reduce the temperature difference between the fluids at a given heat exchanger size and capacity. 2. To reduce the size of the heat exchanger at a given capacity and temperature difference. 3. To increase the capacity of a given surface area. 4. To reduce the pumping power for a given capacity and temperature difference. 192 Enhanced heat transfer in gas flow on finned surfaces Louvered fins Offset strip fins Enhancement is achieved by making the fins short in the flow direction. In this way the boundary layer is not allowed to grow 193 Singlephase flow inside tubes Surface with small protrusions, to disturb the boundary layer. (”dimples”, grooves or minute fins). Turbulator in the tube, inducing large undulations in the flow. 194 Enhancement of boiling heat transfer Pool boiling Enhanced surfaces act by facilitating the nucleation of vapour bubbles. Nucleation is facilitated by porous surface structure. Small amounts of vapour is trapped during boiling. Evaporation takes place in thin liquid films in the structure. Enhancement larger for single tubes than for tube bundles. 195 Flow boiling Nucleation is often suppressed by the convection. Evaporation takes place at the vapour liquid interface. A porous coating will induce nucleate boiling and enhance heat transfer. Most of the porous surfaces can not be made inside tubes. Enhancements generally increase the convection in the fluid. ”Microfins” (0.2mm), give hvalues 24 times smooth tube. 196 Star shaped aluminium insert: divides the tube into several channels. Area is increased, Hydraulic diameter is decreased, 197 Enhanced heat transfer in condensation Heat transfer resistance in condensation is due to the resistance in the liquid film. Enhanced heat transfer is achieved by reducing the thickness of this film. Surface structure where surface tension act to gather the liquid at certain locations while keeping other areas free of liquid. Surfaces often have fins with narrow spacing, surface tension drags liquid from tips to ”valleys” in between fins. 198 Enhanced heat transfer in condensation inside tubes Microfins have been shown to increase the heat transfer coefficients by a factor of 2 to 3. 199 Home assignments: Exercise 1: e Exercise 13 : a, b Exercise 2: e Exercise 14 : b, d Exercise 3: c Exercise 15 : a, c Exercise 4 : d Exercise 5 : c Exercise 6 : c Exercise 7 : a, b Exercise 8 : c, d Exercise 9 : a, b Exercise 10 : d Exercise 11 : c, d Exercise 12 : a, b 200
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