ANNA UNIVERSITY CHENNAI HEAT TRANSFER CHAPTER 2 CONVECTION Dr. R. VELRAJ, PROFESSOR ANNA UNIVERSITY CHENNAI

1. Introduction to Convection 2. Boundary Layer Concepts IN THIS SESSION 1. Introduction to Convection 2. Boundary Layer Concepts CHAPTER 2 (CONVECTION) – SESSION 1

Newton’s Law of Cooling Q = h A (Tw – T∞) GOVERNING LAW Newton’s Law of Cooling Q = h A (Tw – T∞) h – convective heat transfer coefficient A – surface area over which convection occurs (Tw – T∞) – temperature potential difference Convection 1

Flow Regimes on a flat plate CONCEPT OF BOUNDARY LAYER Flow Regimes on a flat plate x y FLAT PLATE LAMINAR REGION TRANSITION TURBULENT u u∞ u = 0 at y = 0 u = u∞ at y = δ Convection 2

Laminar Region (Re < 5 x 105) FLOW REGIMES ON A FLAT PLATE Laminar Region (Re < 5 x 105) FLAT PLATE x y u∞ u LAMINAR BOUNDARY LAYER Reynolds’ no. τ - Shear stress µ - Dynamic viscosity (proportionality constant) Convection 3

Laminar Region REYNOLDS’ NUMBER ρ Density, kg / m3 u∞ Free Stream Velocity, m / s x Distance from leading edge, m µ Dynamic viscosity, kg / m-s Re < 5 x 105 FLOW OVER FLAT PLATE Re < 2300 FLOW THROUGH PIPE Convection 4

FLOW REGIMES ON A FLAT PLATE Transition Region FLAT PLATE TRANSITION 5 x 105 < Re < 106 FLOW OVER FLAT PLATE 2000 < Re < 4000 FLOW THROUGH PIPE Convection 5

FLOW REGIMES ON A FLAT PLATE Turbulent Region x y TURBULENT BOUNDARY LAYER u∞ u LAMINAR SUB LAYER BUFFER ZONE TURBULENT CORE FLAT PLATE Convection 6

Flow Development FLOW THROUGH TUBE Convection 7 y x BOUNDARY LAYER UNIFORM INLET FLOW FULLY DEVELOPED FLOW STARTING LENGTH Convection 7

THERMAL BOUNDARY LAYER x y FLAT PLATE T∞ δt TW TEMPERATURE PROFILE Convection 8

Dimensional Analysis Reduces the number of independent variables in a problem. Experimental data can be conveniently presented in terms of dimensionless numbers. Buckingham’s Pi theorem is used a rule of thumb for determining the dimensionless groups that can be obtained from a set of variables. Convection 9

Buckingham’s Pi theorem Number of independent dimensionless groups that can be formed from a set of ‘m’ variables having ‘n’ basic dimensions is (m – n) Convection 10

QUESTIONS FOR THIS SESSION What is Newton’s Law of Cooling ? Draw the boundary layer for a flow over a flat plate indicating the velocity distribution in the laminar and turbulent flow region. Draw the boundary layer for flow over through tube. Define Buckingham’s π theorem End of Session

Dimensional Analysis Reduces the number of independent variables in a problem. Experimental data can be conveniently presented in terms of dimensionless numbers. Buckingham’s Pi theorem is used a rule of thumb for determining the dimensionless groups that can be obtained from a set of variables. Convection 9

Buckingham’s Pi theorem Number of independent dimensionless groups that can be formed from a set of ‘m’ variables having ‘n’ basic dimensions is (m – n) Convection 10

Dimensional Analysis for Forced Convection Consider a case of fluid flowing across a heated tube S No. Variable Symbol Dimension 1 Tube Diameter D L 2 Fluid Density ρ M L-3 3 Fluid Velocity U L t-1 4 Fluid Viscosity µ M L-1 t-1 5 Specific Heat Cp L2 t-2 T-1 6 Thermal Conductivity k M L t-3 T-1 7 Heat Transfer Coefficient h M t-3 T-1 Convection 11

Dimensional Analysis for Forced Convection There are 7 (m) variables and 4 (n) basic dimensions. 3 (m-n) dimensionless parameters symbolized as π1 ,π2, π3 can be formed. Each dimensionless parameter will be formed by combining a core group of ‘n’ variables with one of the remaining variables not in the core. The core group will include variables with all of the basic dimensions Convection 12

Dimensional Analysis for Forced Convection Choosing D, ρ, µ and k as the core (arbitrarily), the groups formed is represented as: π1 = Da ρb µc kd U π2 = De ρf µg kh Cp π3 = Dj ρl µm kn h Since these groups are to be dimensionless, the variables are raised to certain exponents (a, b, c,….) Convection 13

Dimensional Analysis for Forced Convection Starting with π1 Equating the sum of exponents of each basic dimension to 0, we get equations for: M 0 = b + c + d L 0 = a – 3b + d + 1 + e T 0 = -d t 0 = -c -3d -1 Convection 14

Dimensional Analysis for Forced Convection Solving these equations, we get: d = 0, c = -1, b = 1, a = 1 giving Similarly for π2 Convection 15

Dimensional Analysis for Forced Convection Equating the sum of exponents M 0 = f + g + I L 0 = e – 3f – g + i + 2 T 0 = -i – 1 t 0 = -g – 3i -2 Solving, we get e = 0, f = 0, g = 1, i = 1 giving Convection 16

Dimensional Analysis for Forced Convection By following a similar procedure, we can obtain The relationship between dimensionless groups can be expressed as F(π1, π2, π3) = 0. Thus, Convection 17

Dimensional Analysis for Forced Convection Influence of selecting the core variables Choosing different core variables leads to different dimensionless parameters. If D, ρ, µ, Cp were chosen, then the π groups obtained would be Re, Pr and St. St is Stanton number, a non dimensional form of heat transfer coefficient. Convection 18

Dimensional Analysis for Free Convection Free Convection on a Vertical Plate g T∞ (FLUID) TS (SURFACE) L FLUID PROPERTIES ρ,µ, CP, k, βg Convection 19

Dimensional Analysis for Free Convection Free Convection on a Vertical Plate In free convection, the variable U is replaced by the variables ΔT, β and g. Pertinent Variables in Free Convection S.No. Variable Symbol Dimension 1 Fluid Density ρ M L-3 2 Fluid Viscosity µ M L-1 t-1 3 Fluid Heat Capacity Cp L2 t-2 T-1 4 Fluid Thermal Conductivity k M L t-2 T-1 Convection 20

Dimensional Analysis for Free Convection Pertinent Variables in Free Convection (contd.) S.No. Variable Symbol Dimension 5 Fluid Coefficient of Thermal Expansion β T-1 6 Gravitational acceleration g L t-2 7 Temperature difference ΔT T 8 Significant length L 9 Heat Transfer Coefficient h M t-2 T-1 Convection 21

Dimensional Analysis for Free Convection Choosing L, ρ, µ and k as the core (arbitrarily), the groups formed is represented as: π1 = La ρb µc kd ΔT π2 = Le ρf µi kj βg π3 = Ll ρm µn ko Cp π4 = Lp ρq µr ks h Convection 22

Dimensional Analysis for Free Convection Following the procedure outlined in last section, we get: π1 = (L2 ρ2 k ΔT) / µ2 π2 = (Lµβg) / k π3 = (µCp) / k = Pr (Prandtl number) π4 = (hL) / k = Nu (Nusselt number) Grashof Number Convection 23

Dimensional Analysis FORCED CONVECTION FREE CONVECTION Convection 24

Prair = 0.7 Prwater = 4.5 Prliquid Na = 0.011 PRANDTL NUMBER Multiplying with ρ in the numerator and denominator, Prair = 0.7 Prwater = 4.5 Prliquid Na = 0.011 Convection 25

PRANDTL NUMBER Pr << 1 Pr >> 1 δt δh δt δh Pr = 1 δt = δh δh = Hydrodynamic thickness δt = Thermal Boundary layer thickness Convection 26

QUESTIONS FOR THIS SESSION What are the dimensionless numbers involved in forced convection and free convection ? Define Prandtl number. List the advantages of using liquid metal as heat transfer fluid. Draw the hydrodynamic and thermal boundary layer (in the same plane) for Pr << 1, Pr >> 1 & Pr = 1. End of Session

What is … Continuity Equation Momentum Equation Energy Equation Convection 27

Laminar – Momentum Equation –Flat Plate x y u∞ dy dx FLAT PLATE Convection 28

Laminar Boundary Layer on a Flat Plate Momentum Equation Assumptions Fluid is incompressible Flow is steady No pressure variations in the direction perpendicular to the plate Viscosity is constant Viscous-shear forces in ‘y’ direction are negligible. Convection 29

Continuity Equation Convection 30 Velocity x y u - Velocity in x direction v - Velocity in y direction Convection 30

Continuity Equation – Laminar – Flat Plate Mass flow Convection 31

Continuity Equation Convection 32 Equation Mass balance Mass balance on the element yields: Or Mass Continuity Equation Convection 32

Momentum Equation – Laminar – Flat Plate Pressure Forces x y p - Pressure Convection 33

Momentum Equation – Laminar – Flat Plate Shear Stresses x y µ - Dynamic viscosity u - Velocity in x direction v - Velocity in y direction Convection 34

Momentum Equation – Laminar – Flat Plate Newton’s 2nd Law Momentum flux in x direction is the product of mass flow through a particular side of control volume and x component of velocity at that point Convection 35

Momentum Equation – Laminar – Flat Plate Momentum flux Convection 36

Momentum Equation – Laminar – Flat Plate Momentum and Force Analysis Net pressure force Net Viscous-Shear force Convection 37

Momentum Equation – Laminar – Flat Plate Equating the sum of viscous-shear and pressure forces to the net momentum transfer in x direction, making use of continuity relation and neglecting second order differentials: Convection 38

Energy Equation – Assumptions Incompressible steady flow Constant viscosity, thermal conductivity and specific heat. Negligible heat conduction in the direction of flow (x direction). FLAT PLATE dy dx u∞ x y Convection 39

Energy Equation – Laminar – Flat Plate x y dy dx Energy convected in (left face + bottom face) + heat conducted in bottom face + net viscous work done on element Energy convected out in (right face + top face) + heat conducted out from top face Convection 40

Energy Equation – Laminar – Flat Plate x y Energy Convected u - Velocity in x direction v - Velocity in y direction Convection 41

Energy Equation – Laminar – Flat Plate Heat Conducted x y Net Viscous Work u - Velocity in x direction v - Velocity in y direction Convection 42

Energy Equation – Laminar – Flat Plate Net Viscous Work u - Velocity in x direction v - Velocity in y direction Convection 43

Energy Equation – Laminar – Flat Plate Writing energy balance corresponding to the quantities shown in figure, assuming unit depth in the z direction, and neglecting second-order differentials: Convection 44

Energy Equation – Laminar – Flat Plate Using the continuity relation and dividing the whole equation by ρcp for Low Velocity incompressible flow Convection 45

Energy Equation & Momentum Equation (constant pressure) The solution to the two equations will have exactly the same form when α = ν Convection 46

QUESTIONS FOR THIS SESSION What is the momentum equation for the laminar boundary layer on a flat plate? What are the assumptions involved in derivation of momentum equation? Write the energy equation for laminar boundary layer on a flat plate Explain the analogy between momentum and energy equation. End of Session

Integral form of Momentum Equation Integral form of Momentum equation can be obtained using Von Kármán method: (for constant pressure condition) Convection 47

Integral form of Momentum Equation Polynomial equation for velocity Boundary Conditions Convection 48

Integral form of Momentum Equation Applying the boundary conditions, we get Substituting, Velocity Equation Convection 49

Integral form of Momentum Equation Using expression for velocity in integral equation, Carrying out integration leads to Convection 50

Integral form of Momentum Equation Since ρ and u∞ are constants, the variables may be separated to give Convection 51

Integral form of Momentum Equation At x=0, δ=0; so Writing in terms of Reynolds number BL thickness in terms of Reynolds number Exact solution of BL equation Convection 52

Integral form of Energy Equation FLAT PLATE T∞ δt TW TEMPERATURE PROFILE Convection 53

Integral form of Energy Equation Polynomial equation for temperature Boundary Conditions Convection 54

Integral form of Energy Equation Applying boundary conditions Integral form of Energy Equation is given by: Convection 55

Integral form of Energy Equation Integral form of Energy Equation is given by: Writing in terms of θ, Convection 56

Integral form of Energy Equation Where, Using temperature & velocity profile equation in LHS Convection 57

Integral form of Energy Equation Performing algebraic manipulation and making the substitution ζ (zeta) = δt / δ Convection 58

Heat Transfer Coefficient Combining these equations, Convection 59

Convection 60 Making an energy balance at the surface, solving, Local Nusselt Number Convection 60

QUESTIONS FOR THIS SESSION What is the assumption made by Von Karmen to solve the integral momentum equation ? Write the velocity profile and the temperature profile equation used by Von Karmen in solving the momentum and energy equation Write the equation to determine hydrodynamic & thermal boundary layer thicknesses End of Session

(FLOW OVER A FLAT PLATE) CONVECTION FORCED CONVECTION (FLOW OVER A FLAT PLATE) CORRELATIONS Part 14

Heat Transfer Coefficient Combining these equations, Convection 61

Nusselt Number Convection 62 Making an energy balance at the surface, Using expression for δT Introducing Nusselt no. Local Nusselt Number Convection 62

Nusselt Number Convection 63 Ratio of temperature gradients by conduction and convection at the surface Nusselt Number is an indicative of temperature gradient at the wall in the normal direction Convection 63

Nusselt Number Convection 64 Average Nusselt number is obtained from

Use of Correlations Convection 65 External Flow Flow over a Flat Plate Flow across cylinder Flow across sphere Flow across bank of tubes Internal Flow Flow through tubes & ducts Convection 65

Use of Correlations Convection 66 Separate correlations are available for Laminar Constant temperature surfaces Constant heat flux boundary condition Turbulent Combined laminar & turbulent conditions Special correlations are available for liquid metals Convection 66

Fluid Friction and Heat Transfer Shear stress at the wall may be expressed in terms of friction coefficient Cf : Also, Using velocity distribution equation, Convection 67

Fluid Friction and Heat Transfer Making use of relation for boundary layer thickness: Combining equations, Convection 68

Fluid Friction and Heat Transfer The equation may be rewritten as: Where, Convection 69

Fluid Friction and Heat Transfer Reynolds-Colburn Analogy Convection 70

QUESTIONS FOR THIS SESSION What is the significance of Nusselt Number What is the relationship between local and average Nusselt number for a flow over a flat plate in the laminar region ? What is drag coefficient ? Why separate correlations are available for liquid metals ? What is Reynolds-Colburn analogy ? End of Session

(FLOW OVER A FLAT PLATE) CONVECTION FORCED CONVECTION (FLOW OVER A FLAT PLATE) PROBLEMS Part 15

Example – Mass flow and BL thickness AIR 2 m/s, 27 °C, 1 atm 1.85x10-5 kg/m.s x y FLAT PLATE Calculate Boundary Layer Thickness at x = 20 cm & 40 cm Mass flow which enters the boundary layer between x=20 cm and x = 40 cm. Assume unit depth in z direction. Holman, 221, Ex5.3 Convection 71

Example – Mass flow and BL thickness Density of Air Reynolds number When x = 20 cm, Re = 27,580 When x = 40 cm, Re = 55,160 p = 1.0132 x 105 R = 287 T = 300 K ρ = 1.177 kg/m3 u = 2 ms-1 µ = 1.85x10-5 Convection 72

Example – Mass flow and BL thickness Boundary Layer Thickness When x = 20 cm, δ = 0.00559 m When x = 40 cm, δ = 0.0079 m Re = 27,580 when x = 20 cm (calculated) Re = 55,160 when x = 40 cm (calculated) Convection 73

Example – Mass flow and BL thickness Mass flow entering the Boundary Layer Velocity, u is given by Evaluating the integral with this velocity distribution, Convection 74

Example – Mass flow and BL thickness Mass flow entering the Boundary Layer ρ = 1.177 kg/m3 u∞=2 m/s δ40 = 0.0079 m δ20 = 0.00559 m Convection 75

Plate is heated over its entire length to 60 °C Example – Isothermal flat plate (heated) AIR 2 m/s, 27 °C, 1 atm µ = 1.85x10-5 kg/m.s x y Flat Plate, T = 60 °C Plate is heated over its entire length to 60 °C Calculate Heat Transferred at the first 20 cm of the plate at the first 40 cm of the plate Holman 232, Ex-5.4 Convection 76

All properties are evaluated at film temperature Example – Isothermal flat plate (heated) Formulae Used Heat Flow Nusselt No. Reynolds No. All properties are evaluated at film temperature Convection 77

Example – Isothermal flat plate (heated) Film Temperature Properties of air at Film Temperature: ν=17.36x10-6 m2/s Pr = 0.7 k=0.02749 W/m°C cp=1.006 kJ/kg K Convection 78

Example – Isothermal flat plate (heated) At x = 20 cm Reynolds No. Nusselt No. Heat Transfer Coefficient u∞= 2 m/s Tf = 316.5 K ν = 17.36x10-6 m2/s Pr = 0.7 k = 0.02749 W/m°C cp=1.006 kJ/kg K Substituted Values Convection 79

Example – Isothermal flat plate (heated) At x = 20 cm Heat Flow h = 6.15 W/m2 K Tw = 60 °C A = 0.2 m2 T∞ = 27 °C Substituted Values Convection 80

Example – Isothermal flat plate (heated) At x = 40 cm Reynolds No. Nusselt No. Heat Transfer Coefficient u∞= 2 m/s Tf = 316.5 K ν = 17.36x10-6 m2/s Pr = 0.7 k = 0.02749 W/m°C cp=1.006 kJ/kg K Substituted Values Convection 81

Example – Isothermal flat plate (heated) At x = 40 cm Heat Flow h = 4.349 W/m2 K Tw = 60 °C A = 0.4 m2 T∞ = 27 °C Substituted Values Convection 82

QUESTIONS FOR THIS SESSION 1 Flat Plate X = 0.61 m y Leading Edge AIR T = 37.8 °C u = 0.915 m/s ρ = 1.126 kg/m3 ν = 0.167x10-4 m2/s Sachdeva Pg322, 7.19 Calculate: Boundary Layer Thickness & Drag Coefficient at a distance of 0.61 m from leading edge of plate End of Session

QUESTIONS FOR THIS SESSION 2 Flat Plate at 121.1 °C y X = 0.61 m AIR T = 65.6 °C u = 0.915 m/s ν = 0.223x10-4 m2/s k = 0.0313 W/mK Calculate: Local heat transfer coefficient and the heat transfer for 0.61 m length taking width of plate as 1 m End of Session

CORRELATIONS & PROBLEMS FORCED CONVECTION CONVECTION Part 21 (EXTERNAL FLOW) CORRELATIONS & PROBLEMS Part 21

CORRELATIONS – EXTERNAL FLOW FLAT PLATE Laminar Flow X0 Flat Plate Leading Edge δh δt Convection 83

CORRELATIONS – EXTERNAL FLOW FLAT PLATE Turbulent Flow (Fully turbulent from leading edge) Combined Laminar and Turbulent Flow Convection 84

CORRELATIONS – EXTERNAL FLOW CYLINDER Generalised Equation NuD – Nusselt number based on diameter All properties to be taken at film temperature Re D C m 0.4 – 4 0.989 0.330 4.0 – 40 0.911 0.385 40 – 4000 0.683 0.466 4000 – 40000 0.193 0.618 Convection 85

CORRELATIONS – EXTERNAL FLOW TUBE BANKS St SL D SL St D INLINE STAGGERED Convection 86

CORRELATIONS – EXTERNAL FLOW TUBE BANKS For N ≥ 10 1 ≤ N ≤ 10 Re to be calculated based on max. fluid velocity Vmax INLINE STAGGERED where Convection 87

CORRELATIONS – EXTERNAL FLOW TUBE BANKS (INLINE) For 10 ROWS or MORE ST / D SL / D 1.25 1.5 2.0 3.0 C n 0.35 0.59 0.28 0.608 0.1 0.704 0.063 0.75 0.37 0.586 0.25 0.62 0.702 0.068 0.74 2 0.42 0.57 0.29 0.60 0.23 0.632 0.198 0.65 3 0.357 0.584 0.581 0.286 0.61 Convection 88

CORRELATIONS – EXTERNAL FLOW TUBE BANKS (STAGGERED) For 10 ROWS or MORE ST / D SL / D 1.25 1.5 2.0 3.0 C n 0.6 - .213 .636 1 .497 .558 .451 .568 .46 .562 .452 .488 3 .31 .592 .356 .58 .44 .421 .574 Convection 89

CORRELATIONS – EXTERNAL FLOW TUBE BANKS ( C1 values ) For LESS than 10 ROWS ST – STAGGERED IN – INLINE N 1 2 3 4 5 6 7 8 9 10 ST .68 .75 .83 .89 .92 .95 .97 .98 .99 IN .64 .8 .87 .9 .94 .96 Convection 90

Example – Heated Flat Plate Flat Plate at 90 °C y X AIR T = 0 °C u = 75 m/s 45 cm LONG, 60 cm WIDE Assume transition takes place at Re X, C = 5 x 105 CALCULATE Heat Transfer Coefficient for full length of plate Rate of Energy Dissipation from the plate Sachdeva Pg. 284, Ex7.7 Convection 91

Example – Heated Flat Plate Film Temperature Properties of air at Film Temperature Critical Length (distance at which transition takes place) u∞ = 75 m/s ν=17.45x10-6 m2/s k=2.8 x 10-2 W/m°C Pr = 0.698 Convection 92

Example – Heated Flat Plate Heat Transfer Coefficient u∞= 75 m/s L = 0.45 m ν = 17.45x10-6 m2/s Pr = 0.698 k = 2.8 x 10-2 W/m°C Substituted Values Convection 93

RATE OF ENERGY DISSIPATION FROM THE PLATE Example – Heated Flat Plate RATE OF ENERGY DISSIPATION FROM THE PLATE hL = 170 W/m2 K A = 0.45 x 0.6 m2 TS = 90 °C T∞ = 0 °C Substituted Values Convection 94

QUESTIONS FOR THIS SESSION Air at 1 atm and 350C flows across 5.9 cm diameter cylinder at a velocity of 50m/s. The cylinder surface is maintained at a temperature of 1500C. Calculate the heat loss per unit length of the cylinder. A fine wire having a diameter of 3.94 X 10-5 m is placed in a 1 atm airstream at 250C having a flow velocity of 50 m/s perpendicular to the wire. An electric current is passed through the wire, raising its surface temperature to 500C. Calculate the heat loss per unit length. End of Session

CONVECTION FORCED CONVECTION CORRELATIONS & PROBLEMS Part 22

Example – Flow over Cylinder Assume a man (represented as a cylinder) standing in the direction of wind D = 30 cm AIR T = 10 °C u = 36 km/h H = 1.7 m TS = 30 °C CALCULATE Heat lost while standing in the wind Sachdeva Pg291, Ex7.8 Convection 95

Example – Flow over Cylinder Film Temperature Properties of air at Film Temperature Reynolds Number ν = 15x10-6 m2/s k = 2.59 x 10-2 W/m°C Pr = 0.707 u∞ = 10 m/s D = 0.3 m Convection 96

Example – Flow over Cylinder Rate of Heat Lost ReD= 2 x 105 Pr = 0.707 k = 2.59 x 10-2 W/m°C TS = 30 °C T∞ = 10 °C Substituted Values Convection 97

Heating of air with in-line tube bank Example – Flow through Tube Banks Heating of air with in-line tube bank ST SL Tsurface = 65°C 2.54 cm 15 ROWS HIGH 5 ROWS DEEP SL = ST = 3.81 cm AIR 1 atm, 10 °C u = 7 m/s CALCULATE Total heat transfer per unit length for tube bank and the exit air temperature JP.H/300/6.10 Convection 98

Example – Flow through Tube Banks Film Temperature Properties of air at Film Temperature Constants for use ( C & n ) from table µ = 1.894 x 10-5 kg/ms ρ = 1.137 kg/m3 k = 0.027 W/m°C Pr = 0.706 JP.H/300/6.10 C = 0.25 n = 0.62 Convection 99

Example – Flow through Tube Banks Maximum Velocity JP.H/300/6.10 D = 0.0254 m ST = 3.81 u∞ = 7 m/s n = 0.62 µ = 1.894 x 10-5 kg/ms ρ = 1.137 kg/m3 c = 0.25 k = 0.027 W/m°C Convection 100

Example – Flow through Tube Banks Correction Factor ( C1 ) = 0.92 (from table) Total heat transfer surface area (assuming unit length) Heat Transferred N = 15 D = 0.0254 m L = 1m JP.H/300/6.10 Convection 101

Example – Flow through Tube Banks Subscripts 1 & 2 denote entrance & exit temperatures Substituting JP.H/300/6.10 Convection 102

Example – Flow through Tube Banks Heat Transferred JP.H/300/6.10 Convection 103

INTERNAL FLOW Convection 104 BOUNDARY LAYER UNIFORM INLET FLOW FULLY DEVELOPED FLOW STARTING LENGTH Mixing Cup Temperature / Bulk Mean Temperature is the temperature, the fluid would assume if placed in a mixing chamber and allowed to come to equilibrium. Convection 104

INTERNAL FLOW Where, for CIRCULAR DUCT Convection 105

Momentum Equation INTERNAL FLOW (constant pressure) Energy Equation For Slug flow… Convection 106

Water passing through Staggered tube bank 7 ROWS in direction of flow Example – Flow through Tube Banks Water passing through Staggered tube bank Tsurface = 70°C 1 SL 7 ROWS in direction of flow SL = ST = 20.5 mm AIR T∞ = 15 °C u∞ = 6 m/s 1.64 cm CALCULATE Air side heat transfer coefficient across the tube bundle Sachdeva 297, Ex7.12 Convection

QUESTIONS FOR THIS SESSION What is ‘bulk mean temperature or mixing cup temperature’ ? What is slug flow ? Write the momentum and energy equation for the flow through a tube. End of Session

CORRELATIONS & PROBLEMS FORCED CONVECTION CONVECTION Part 23 (INTERNAL FLOW) CORRELATIONS & PROBLEMS Part 23

CORRELATIONS – INTERNAL FLOW Properties to be evaluated at Bulk Mean Temperature Tm = (Tmi + Tmo) / 2 Tmi – Mean Temperature at inlet Tmo – Mean Temperature at outlet LAMINAR FLOW Fully developed Thermal Layer Constant Wall Temperature Constant Heat Flux GH / 220 Convection 107

CORRELATIONS – INTERNAL FLOW LAMINAR FLOW (contd.) Entry region (Hydrodynamic layer fully developed, thermal layer developing) Simultaneous development of hydrodynamic & thermal layers GH / 220 Convection 108

CORRELATIONS – INTERNAL FLOW TURBULENT FLOW Fully Developed flow (Dittus-Boelter equation) n = 0.4 for heating of fluids / n = 0.3 for cooling of fluids 0.6 < Pr < 100, 2500 < Re < 1.25 x 106 ; L/D > 60 Fully Developed flow (Sieder-Tate equation) 0.7 < Pr < 16,700 ; ReD ≥ 10,000 ; L / D ≥ 60 GH / 220 Convection 109

Constant Wall Heat Flux Example 1 Douter = 2 cm WATER T = 25 °C m = 0.01 kg/s Constant Wall Heat Flux qs = 1 kW/m2 Water flowing through pipe with constant wall heat flux CALCULATE Reynolds number 2. Heat Transfer Coefficient 3. Difference between wall temperature and bulk (mean) temperature. GH/228/5.5 Convection 110

Example 1 Convection 111 Properties of water at 25 °C < 2300. Flow is LAMINAR For Constant Heat Flux µ = 8.96 x 10-4 kg/ms k = 0.6109 W/m°C GH/228/5.5 D = 0.02 m Convection 111

Example 1 Difference between Wall Temperature and Bulk (mean) Temperature GH/228/5.5 Convection 112

Constant Wall Temperature Example 2 Douter = 2.2 cm WATER Tinitial = 15 °C Tfinal = 60 °C u = 2 m/s Constant Wall Temperature Ts = 95 °C Water flowing through Copper Tube with constant wall temperature Sachdeva 315, 7.22 CALCULATE Average heat transfer coefficient by using Sieder-Tate equation Convection 113

Example 2 Convection 114 Bulk (mean) Temperature Properties of water µ = 0.69 x 10-3 N.s/m2 ρ = 990 kg/m3 k = 0.63 W/m°C cp = 4160 J/kg.K D = 0.022 m JP.H/300/6.10 Convection 114

Example 2 Convection 115 µ = 0.69 x 10-3 N.s/m2 ρ = 990 kg/m3 JP.H/300/6.10 µ = 0.69 x 10-3 N.s/m2 ρ = 990 kg/m3 k = 0.63 W/m°C cp = 4160 J/kg.K µs = 0.3 x 10-3 N.s/m2 D = 0.022 m Convection 115

Heat Leakage from an air conditioning duct Example 3 400 X 800 mm AIR T = 20 °C u = 7 m/s Heat Leakage from an air conditioning duct Sachdeva 3.16, Ex7.24 Estimate the heat leakage per meter length per unit temperature difference. Convection 116

Example 3 Convection 117 Properties of air Equivalent or Hydraulic Diameter Assuming pipe wall temperature to be higher than air temperature, then Nusselt number is given by: ν = 15.06 x 10-6 m2/s α = 7.71 x 10-2 m2/h k = 0.0259 W/mK Convection 117

Example 3 Heat Leakage per unit length per unit temperature difference: NuD = 398.38 k = 0.0259 W/mK D = 0.571 m Convection 118

Constant Wall Temperature Questions 1 Douter = 1.5 cm L = 3 m WATER Tinitial = 50 °C Tfinal = 64 °C u = 1 m/s Constant Wall Temperature Ts = 90 °C Water flowing through a heated tube SC/314/7.21 CALCULATE 1. Heat transfer coefficient 2. Total amount of heat transferred Convection

Tsurface (of inner tube) = 50 °C Air flowing through annulus Questions 2 ID = 3.125 cm OD = 5 cm AIR Tinitial = 16 °C Tfinal = 32 °C u = 30 m/s Tsurface (of inner tube) = 50 °C Air flowing through annulus SC/317/7.25 CALCULATE Heat transfer coefficient of air Convection

CONVECTION FREE CONVECTION Part 24

VELOCITY BOUNDARY LAYER THERMAL BOUNDARY LAYER Free Convection Boundary Layer Heated Vertical Plate Ts Ty T∞ u(y) δ δt VELOCITY BOUNDARY LAYER THERMAL BOUNDARY LAYER y, v x, U T∞, ρ∞, g GH / 220 Convection 119

Free Convection – Governing Equations Continuity Equation X Momentum Eqn. Energy Equation GH / 220 Convection 120

Free Convection Convection 121 X Momentum Equation u0, ρρ∞ (density outside boundary layer) GH / 220 Convection 121

Free Convection X Momentum Equation GH / 220 Convection 122

Free Convection Convection 123 Volumetric Coefficient of thermal expansion, β GH / 220 Convection 123

Free Convection Convection 124 Summarizing the governing equations, GH / 220 Convection 124

Free Convection Convection 125 Identification of Dimensionless Groups GH / 220 Convection 125

Free Convection Rearranging Where, GH / 220 “ratio of buoyancy force to the viscous force in fluid” This number plays similar role in free convection as does the Reynolds number in forced convection

Free Convection in External Flows Vertical Surfaces Laminar (Gr.Pr < 109) Constant Wall Temperature Constant Heat Flux Turbulent (Gr.Pr > 109) GH / 220 Convection 127

Free Convection in External Flows Horizontal Surfaces Characteristic Length Constant Wall Temperature Constant Heat Flux GH / 220 Convection 128

Combined Free & Forced Convection When air is flowing over heated surface at a low velocity, the effect of free and forced convections are equally important GH / 220 Convection 129

Combined Free & Forced Convection External Flow Internal Flow (LAMINAR) Graetz number GH / 220 Convection 130

Combined Free & Forced Convection Internal Flow (TURBULENT) Applicable for ReD > 2000 and RaD (D/L) < 5000 Or ReD > 800 and RaD (D/L) > 2x 104 GH / 220 Convection 131

Example – Convection between Vertical Plates δ δ TSurface = 80 °C PLATE 3.5 cm Twater = 20 °C WATER Yadav 475, Ex 13.4 L Minimum spacing (L) to avoid interference of free convection boundary layers ? Convection 132

Example – Convection between Vertical Plates Let, δ be the boundary layer thickness at trailing edge Minimum spacing required = L = 2δ Film temperature = t∞ = (80 + 20) / 2 = 50 °C Properties of water at Film Temperature < 1 x 109 (LAMINAR) Pr = 3.54 β = 0.48x10-3 K-1 ν = 0.567 x 10-6 m2/s Convection 133

Example – Convection between Vertical Plates Boundary layer thickness (δ) Minimum Space to avoid interference Pr = 3.54 x = 0.035 m Gr = 0.2526x109 Convection 134

Questions Draw the free convection boundary layer on a heated vertical plate. Write the governing equations for free convection What is the significance of Grashof number ? Explain the situations under which combined free and forced convection should be considered. Part 24

CONVECTION FREE CONVECTION Problems Part 25

Vertical pipe kept in a room Heat lost by pipe / metre length Example 1 – Vertical Pipe Douter = 10 cm TSurface = 100 °C L = 30 cm AIR (ambient) T = 20 °C SD/344/8.8 Vertical pipe kept in a room ? Heat lost by pipe / metre length Convection 135

Example 1 – Vertical Pipe Film Temperature Properties of air at Film Temperature Pr = 0.696 β = 0.003003 K-1 ν = 18.97 x 10-6 m2/s k = 0.02896 W/m°C SD/341/8.6 L = 3 m T∞ = 100°C TS = 20°C Convection 136

Example 1 – Vertical Pipe Checking Then, SD/341/8.6 RaL = 12.25 x 1010 k = 0.02896 W/m°C L = 3 m T∞ = 100°C TS = 20°C Convection 137

? Example 2 – Horizontal Duct Heat gained by duct / metre length 60 cm AIR (ambient) T = 25 °C TSurface = 15 °C 30 cm Horizontal un-insulated Air Conditioning Duct SD/341/8.6 ? Heat gained by duct / metre length Convection 138

Example 2 – Horizontal Duct Film Temperature Properties of air at Film Temperature Rate of Heat Gained per unit length of duct Pr = 0.705 β = 0.00341 K-1 ρ = 1.205 kg/m3 ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°C SD/341/8.6 Convection 139

Example 2 – Horizontal Duct Heat gained from vertical wall (sides) Laminar SD/341/8.6 β = 0.00341 K-1 ρ = 1.205 kg/m3 ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°C T∞ = 25°C TS = 15°C Pr = 0.705

Example 2 – Horizontal Duct Heat gained from top & bottom surfaces Characteristic Length Laminar Similarly for bottom surface, TOP SURFACE SD/341/8.6 β = 0.00341 K-1 ρ = 1.205 kg/m3 ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°C T∞ = 25°C TS = 15°C Pr = 0.705

Example 2 – Horizontal Duct Rate of Heat Gained SD/341/8.6 Qside = 6.97 W/m Q top + bottom = 9.73 W/m Convection 142

Calculate the heat transfer coefficient Combined Free & Forced Convection with Air Air flowing through a horizontal tube 3 TSurface = 140 °C AIR Tair = 27 °C u = 30 cm/s AIR 25 mm TUBE 0.4m SD/357/8.16 Calculate the heat transfer coefficient Convection 143

Combined Free & Forced Convection with Air Film Temperature Properties of air at Film Temperature Reynolds Number Pr = 0.695 β = 2.805x10-3 K-1 ρ = 0.99 kg/m3 µ bulk = 2.1 x 10-5 kg/m.s k = 0.0305 W/m°C µw = 2.337 x 10-5 kg/m.s D = 0.025 m u = 0.3 m/s Convection 144

Combined Free & Forced Convection with Air Pr = 0.695 β = 2.805x10-3 K-1 ρ = 0.99 kg/m3 µf = 2.1 x 10-5 kg/m.s k = 0.0305 W/m°C µw = 2.337 x 10-5 kg/m.s SD/341/8.6 Convection 145

Combined Free & Forced Convection with Air k = 0.0305 W/m°C µw = 2.337 x 10-5 kg/m.s µ = 1.8462 x 10-5 kg/m.s Gz = 15.33 Gr = 1.007 x 109 d = 0.025 m Convection 146

Questions 1 Panel : 0.915 m x 0.915 m One side insulated, other side at 65.6 °C Ambient is at 10 °C INSULATED SURFACE HOT INSULATED SURFACE HOT HOT SURFACE INSULATED Calculate the mean heat transfer coefficient due to free convection

Air flow through Rectangular Duct Questions 2 30 X 20 cm Duct Surface at 5 °C AIR T = 25 °C Air flow through Rectangular Duct Estimate the heat gained by the duct. Convection

Air flowing through a tube Questions 3 D = 20 mm L = 1 m AIR T = 27 °C u = 30 cm/s Horizontal Tube Tsurface = 127 °C Air flowing through a tube Calculate the heat transferred considering combined free and forced convection Convection