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**ANNA UNIVERSITY CHENNAI**

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

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**1. Introduction to Convection 2. Boundary Layer Concepts**

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

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**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

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**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

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**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

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**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

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**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

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**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

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**Flow Development FLOW THROUGH TUBE Convection 7 y x BOUNDARY LAYER**

UNIFORM INLET FLOW FULLY DEVELOPED FLOW STARTING LENGTH Convection

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**THERMAL BOUNDARY LAYER**

x y FLAT PLATE T∞ δt TW TEMPERATURE PROFILE Convection

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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

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**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

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**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

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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

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**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

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**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

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**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

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**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

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**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

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**Dimensional Analysis for Forced Convection**

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

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**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

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**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

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**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

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**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

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**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 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

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**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

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**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

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**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

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Dimensional Analysis FORCED CONVECTION FREE CONVECTION Convection

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**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

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**PRANDTL NUMBER Pr << 1 Pr >> 1 δt δh δt δh Pr = 1 δt = δh**

δh = Hydrodynamic thickness δt = Thermal Boundary layer thickness Convection

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**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

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What is … Continuity Equation Momentum Equation Energy Equation Convection

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**Laminar – Momentum Equation –Flat Plate**

x y u∞ dy dx FLAT PLATE Convection

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**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

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**Continuity Equation Convection 30 Velocity**

x y u - Velocity in x direction v - Velocity in y direction Convection

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**Continuity Equation – Laminar – Flat Plate**

Mass flow Convection

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**Continuity Equation Convection 32 Equation Mass balance**

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

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**Momentum Equation – Laminar – Flat Plate**

Pressure Forces x y p - Pressure Convection

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**Momentum Equation – Laminar – Flat Plate**

Shear Stresses x y µ - Dynamic viscosity u - Velocity in x direction v - Velocity in y direction Convection

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**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

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**Momentum Equation – Laminar – Flat Plate**

Momentum flux Convection

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**Momentum Equation – Laminar – Flat Plate**

Momentum and Force Analysis Net pressure force Net Viscous-Shear force Convection

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**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

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**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

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**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

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**Energy Equation – Laminar – Flat Plate**

x y Energy Convected u - Velocity in x direction v - Velocity in y direction Convection

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**Energy Equation – Laminar – Flat Plate**

Heat Conducted x y Net Viscous Work u - Velocity in x direction v - Velocity in y direction Convection

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**Energy Equation – Laminar – Flat Plate**

Net Viscous Work u - Velocity in x direction v - Velocity in y direction Convection

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**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

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**Energy Equation – Laminar – Flat Plate**

Using the continuity relation and dividing the whole equation by ρcp for Low Velocity incompressible flow Convection

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**Energy Equation & Momentum Equation**

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

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**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

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**Integral form of Momentum Equation**

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

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**Integral form of Momentum Equation**

Polynomial equation for velocity Boundary Conditions Convection

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**Integral form of Momentum Equation**

Applying the boundary conditions, we get Substituting, Velocity Equation Convection

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**Integral form of Momentum Equation**

Using expression for velocity in integral equation, Carrying out integration leads to Convection

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**Integral form of Momentum Equation**

Since ρ and u∞ are constants, the variables may be separated to give Convection

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**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

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**Integral form of Energy Equation**

FLAT PLATE T∞ δt TW TEMPERATURE PROFILE Convection

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**Integral form of Energy Equation**

Polynomial equation for temperature Boundary Conditions Convection

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**Integral form of Energy Equation**

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

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**Integral form of Energy Equation**

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

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**Integral form of Energy Equation**

Where, Using temperature & velocity profile equation in LHS Convection

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**Integral form of Energy Equation**

Performing algebraic manipulation and making the substitution ζ (zeta) = δt / δ Convection

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**Heat Transfer Coefficient**

Combining these equations, Convection

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

Nusselt Number Convection

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**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

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**(FLOW OVER A FLAT PLATE)**

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

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**Heat Transfer Coefficient**

Combining these equations, Convection

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**Nusselt Number Convection 62 Making an energy balance at the surface,**

Using expression for δT Introducing Nusselt no. Local Nusselt Number Convection

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**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

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**Nusselt Number Convection 64 Average Nusselt number is obtained from**

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**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

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**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

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**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

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**Fluid Friction and Heat Transfer**

Making use of relation for boundary layer thickness: Combining equations, Convection

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**Fluid Friction and Heat Transfer**

The equation may be rewritten as: Where, Convection

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**Fluid Friction and Heat Transfer**

Reynolds-Colburn Analogy Convection

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**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

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**(FLOW OVER A FLAT PLATE)**

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

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**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

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**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

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**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

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**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

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**Example – Mass flow and BL thickness**

Mass flow entering the Boundary Layer ρ = kg/m3 u∞=2 m/s δ40 = m δ20 = m Convection

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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**CORRELATIONS & PROBLEMS FORCED CONVECTION CONVECTION Part 21**

(EXTERNAL FLOW) CORRELATIONS & PROBLEMS Part 21

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**CORRELATIONS – EXTERNAL FLOW**

FLAT PLATE Laminar Flow X0 Flat Plate Leading Edge δh δt Convection

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**CORRELATIONS – EXTERNAL FLOW**

FLAT PLATE Turbulent Flow (Fully turbulent from leading edge) Combined Laminar and Turbulent Flow Convection

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**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

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**CORRELATIONS – EXTERNAL FLOW**

TUBE BANKS St SL D SL St D INLINE STAGGERED Convection

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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CONVECTION FORCED CONVECTION CORRELATIONS & PROBLEMS Part 22

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**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

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**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

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**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

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**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

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**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

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**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

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**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

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**Example – Flow through Tube Banks**

Subscripts 1 & 2 denote entrance & exit temperatures Substituting JP.H/300/6.10 Convection

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**Example – Flow through Tube Banks**

Heat Transferred JP.H/300/6.10 Convection

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**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

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INTERNAL FLOW Where, for CIRCULAR DUCT Convection

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**Momentum Equation INTERNAL FLOW (constant pressure) Energy Equation**

For Slug flow… Convection

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**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

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**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

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**CORRELATIONS & PROBLEMS FORCED CONVECTION CONVECTION Part 23**

(INTERNAL FLOW) CORRELATIONS & PROBLEMS Part 23

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**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

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**CORRELATIONS – INTERNAL FLOW**

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

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**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

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**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

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**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

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

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**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

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**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

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**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

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**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

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**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

137
Example 3 Heat Leakage per unit length per unit temperature difference: NuD = k = W/mK D = m Convection

138
**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

139
**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

140
CONVECTION FREE CONVECTION Part 24

141
**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

142
**Free Convection – Governing Equations**

Continuity Equation X Momentum Eqn. Energy Equation GH / 220 Convection

143
**Free Convection Convection 121 X Momentum Equation**

u0, ρρ∞ (density outside boundary layer) GH / 220 Convection

144
Free Convection X Momentum Equation GH / 220 Convection

145
**Free Convection Convection 123**

Volumetric Coefficient of thermal expansion, β GH / 220 Convection

146
**Free Convection Convection 124 Summarizing the governing equations,**

GH / 220 Convection

147
**Free Convection Convection 125 Identification of Dimensionless Groups**

GH / 220 Convection

148
**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

149
**Free Convection in External Flows**

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

150
**Free Convection in External Flows**

Horizontal Surfaces Characteristic Length Constant Wall Temperature Constant Heat Flux GH / 220 Convection

151
**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

152
**Combined Free & Forced Convection**

External Flow Internal Flow (LAMINAR) Graetz number GH / 220 Convection

153
**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

154
**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

155
**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

156
**Example – Convection between Vertical Plates**

Boundary layer thickness (δ) Minimum Space to avoid interference Pr = x = m Gr = x109 Convection

157
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

158
CONVECTION FREE CONVECTION Problems Part 25

159
**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

160
**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

161
**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

162
**? 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

163
**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

164
**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

165
**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

166
**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

167
**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

168
**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

169
**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

170
**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

171
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

172
**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

173
**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

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