# ANNA UNIVERSITY CHENNAI

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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

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

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

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 FLOW OVER FLAT PLATE Re < 2300 FLOW THROUGH PIPE Convection

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

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

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

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

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

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

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

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

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 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

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

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

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 e T 0 = -d t 0 = -c -3d -1 Convection

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

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

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

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

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

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

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

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

Dimensional Analysis FORCED CONVECTION FREE CONVECTION Convection

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

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

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

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

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

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

Continuity Equation – Laminar – Flat Plate
Mass flow Convection

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

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

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

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

Momentum Equation – Laminar – Flat Plate
Momentum flux Convection

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Heat Transfer Coefficient
Combining these equations, Convection

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

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

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

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

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

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

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

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

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

Fluid Friction and Heat Transfer
Reynolds-Colburn Analogy Convection

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

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 = x 105 R = 287 T = 300 K ρ = kg/m3 u = 2 ms-1 µ = 1.85x10-5 Convection

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

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

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

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

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

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

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

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

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

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

QUESTIONS FOR THIS SESSION
1 Flat Plate X = 0.61 m y Leading Edge AIR T = 37.8 °C u = m/s ρ = 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 °C y X = 0.61 m AIR T = 65.6 °C u = m/s ν = 0.223x10-4 m2/s k = 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 = m L = 1m JP.H/300/6.10 Convection

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

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

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

INTERNAL FLOW Where, for CIRCULAR DUCT Convection

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

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

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

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

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

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

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

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

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 = m JP.H/300/6.10 Convection

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 = m Convection

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

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 = x 10-6 m2/s α = 7.71 x 10-2 m2/h k = W/mK Convection

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

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 = 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

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

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

Free Convection X Momentum Equation GH / 220 Convection

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

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

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

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

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

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

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

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

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

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

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

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

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

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

? 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

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

Example 2 – Horizontal Duct
Heat gained from vertical wall (sides) Laminar SD/341/8.6 β = K ρ = kg/m3 ν = x 10-6 m2/s k = 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 = K ρ = kg/m3 ν = x 10-6 m2/s k = 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

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

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

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

Combined Free & Forced Convection with Air
k = W/m°C µw = x 10-5 kg/m.s µ = x 10-5 kg/m.s Gz = 15.33 Gr = x d = m Convection

Questions 1 Panel : m x 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