chapter 2 convection dr. r. velraj, professor anna university chennai heat transfer

CHAPTER 2

CONVECTION

Dr. R. VELRAJ, PROFESSORANNA UNIVERSITY CHENNAI

HEAT TRANSFER

1. Introduction to Convection

2. Boundary Layer Concepts

CHAPTER 2 (CONVECTION) – SESSION 1

IN THIS SESSION

Newton’s Law of Cooling

Q = h A (Tw – T∞)

Convection 1

GOVERNING LAW

h – convective heat transfer coefficient

A – surface area over which convection occurs

(Tw – T∞) – temperature potential difference

Flow Regimes on a flat plate

FLAT PLATE

LAMINARREGION

TRANSITION TURBULENTREGION

uu∞

u∞

u

x

y

Convection 2

CONCEPT OF BOUNDARY LAYER

u = 0 at y = 0 u = u∞ at y = δ

Laminar Region (Re < 5 x 105)FLOW REGIMES ON A FLAT PLATE

FLAT PLATE

x

y

u∞

u

LAMINAR BOUNDARY LAYER

dydu

τ - Shear stressµ - Dynamic viscosity (proportionality constant)

Reynolds’ no.

xuRe

Convection 3

Laminar Region REYNOLDS’ NUMBER

ρ Density, kg / m3

u∞ Free Stream Velocity, m / sx Distance from leading edge, mµ Dynamic viscosity, kg / m-s

Re < 5 x 105 FLOW OVER FLAT PLATERe < 2300 FLOW THROUGH PIPE

xuRe

Convection 4

Transition RegionFLOW REGIMES ON A FLAT PLATE

5 x 105 < Re < 106 FLOW OVER FLAT PLATE2000 < Re < 4000 FLOW THROUGH PIPE

FLAT PLATE

TRANSITION

Convection 5

Turbulent RegionFLOW REGIMES ON A FLAT PLATE

FLAT PLATELAMINAR SUB LAYER

BUFFER ZONE

TURBULENT CORETURBULENT BOUNDARY LAYER

x

y

Convection 6

u∞

u

Flow DevelopmentFLOW THROUGH TUBE

x

y

Convection 7

UNIFORM INLET FLOW

BOUNDARY LAYER

FULLY DEVELOPEDFLOW

STARTING LENGTH

x

y

Convection 8

THERMAL BOUNDARY LAYER

FLAT PLATE

T∞

δtTW

TEMPERATURE PROFILE

Convection 9

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.

Number of independent dimensionless groups that can be formed from a set of ‘m’ variables

having ‘n’ basic dimensions is (m – n)

Convection 10

Buckingham’s Pi theorem

End of Session

QUESTIONS FOR THIS SESSION

1. What is Newton’s Law of Cooling ?2. Draw the boundary layer for a flow over a flat

plate indicating the velocity distribution in the laminar and turbulent flow region.

3. Draw the boundary layer for flow over through tube.

4. Define Buckingham’s π theorem

Convection 9


• 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.

Number of independent dimensionless groups that can be formed from a set of ‘m’ variables

having ‘n’ basic dimensions is (m – n)

Convection 10

Buckingham’s Pi theorem

Convection 11

Dimensional Analysis for Forced ConvectionConsider a case of fluid flowing across a heated tubeS No. Variable Symbol Dimension

1 Tube Diameter D L2 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 12

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 13

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 Uπ2 = 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 14


Starting with π1

Equating the sum of exponents of each basic dimension to 0, we get equations for:M 0 = b + c + dL 0 = a – 3b + d + 1 + eT 0 = -dt 0 = -c -3d -1

tL

TtML

LtM

LM

LtTLMdcb

a33

0000 )(1

Convection 15


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

Similarly for π2

number) (Reynolds Re

UD1

TtL

TtML

LtM

LM

LtTLMigf

e3

2

330000 )(1

Convection 16


Equating the sum of exponentsM 0 = f + g + IL 0 = e – 3f – g + i + 2T 0 = -i – 1t 0 = -g – 3i -2Solving, we get e = 0, f = 0, g = 1, i = 1 giving

number) (Prandtl PrkCp

2

Convection 17


By following a similar procedure, we can obtain

The relationship between dimensionless groups can be expressed as F(π1, π2, π3) = 0. Thus,

number) (Nusselt Nuk

hD3

Pr)(Re, Nu

Convection 18


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.

p.U.Ch

Re.PrNu

St

Convection 19

Dimensional Analysis for Free Convection

TS (SURFACE)

T∞ (FLUID)g

LFLUID PROPERTIES

ρ,µ, CP, k, βg

Free Convection on a Vertical Plate

Convection 20

Dimensional Analysis for Free ConvectionFree Convection on a Vertical Plate

In free convection, the variable U is replaced by the variables ΔT, β and g.

Pertinent Variables in Free ConvectionS.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 21


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 T8 Significant length L L9 Heat Transfer Coefficient h M t-2 T-1

Convection 22


Choosing L, ρ, µ and k as the core (arbitrarily), the groups formed is represented as:

π1 = La ρb µc kd ΔTπ2 = Le ρf µi kj βgπ3 = Ll ρm µn ko Cp

π4 = Lp ρq µr ks h

Convection 23


Following the procedure outlined in last section,we get:

π1 = (L2 ρ2 k ΔT) / µ2

π2 = (Lµβg) / kπ3 = (µCp) / k = Pr (Prandtl number)π4 = (hL) / k = Nu (Nusselt number)

Grashof Number2

32

21TgL

.

Gr

FORCED CONVECTION

FREE CONVECTION

Convection 24


(Gr.Pr) Nu F

Pr)(Re, Nu

PRANDTL NUMBER

Convection 25

kpc

Pr

Multiplying with ρ in the numerator and denominator,

ydiffusivit Thermalydiffusivit Molecular

kpc

Pr

Prair = 0.7 Prwater = 4.5 Prliquid Na = 0.011

PRANDTL NUMBER

Convection 26

Pr = 1

δt

δh

Pr >> 1Pr << 1

δt = δh

δt

δh

δh = Hydrodynamic thicknessδt = Thermal Boundary layer thickness

End of Session


1. What are the dimensionless numbers involved in forced convection and free convection ?

2. Define Prandtl number.3. List the advantages of using liquid metal as

heat transfer fluid.4. Draw the hydrodynamic and thermal

boundary layer (in the same plane) for Pr << 1, Pr >> 1 & Pr = 1.

1. Continuity Equation

2. Momentum Equation

3. Energy Equation

Convection 27

What is …

Convection 28

Laminar – Momentum Equation –Flat Plate

FLAT PLATE

dydx

u∞ x

y

Momentum EquationAssumptions1. Fluid is incompressible2. Flow is steady3. No pressure variations in the direction

perpendicular to the plate4. Viscosity is constant5. Viscous-shear forces in ‘y’ direction are

negligible.

Convection 29

Laminar Boundary Layer on a Flat Plate

Convection 30

Continuity Equation

x

y

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

Velocitydy

yv

v

dxxu

u

v

u

Convection 31

Continuity Equation – Laminar – Flat Plate

Mass flowdxdy

xv

v

vdx

udy dydx

xu

u

Convection 32

Continuity Equation

Mass balanceMass balance on the element yields:

OrMass Continuity Equation

0yv

xu

dxdyyv

vdydxxu

uvdxudy

Convection 33

Momentum Equation – Laminar – Flat Plate

x

y

Pressure Forces

p - Pressure

dydxxp

p

dy.p

Convection 34


x

y

dy

yu

yyu

dx.Shear

Stresses

yu

.dx.

µ - Dynamic viscosityu - Velocity in x direction v - Velocity in y direction

Convection 35


Newton’s 2nd Law

dmVd

F xx

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 36


Momentum flux

dydxxu

u2

. uudy

dxdyyu

udyyv

v

. uvdx

Convection 37


Momentum and Force AnalysisNet pressure force

Net Viscous-Shear force

dxdyxp

dxdyyu2

2

Convection 38


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:

xp

yu

yv

vxu

u 2

2

Convection 39

Energy Equation – Assumptions1. Incompressible steady flow2. Constant viscosity, thermal conductivity and

specific heat.3. Negligible heat conduction in the direction of flow

(x direction).

FLAT PLATE

dydx

u∞

x

y

Convection 40

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 41


x

y


Energy Convected

dydxxT

Tdxxu

ucp

uTdycp

vTdxcp

dxdyyT

Tdyyv

vcp

Convection 42


x

y


Heat Conducted

yT

kdx

dy

yT

yyT

kdx

dy

yu

.dxyu

Net Viscous Work

Convection 43



Net Viscous Work

dy

yu

.dxyu

dxdyyu 2

Convection 44


Writing energy balance corresponding to the quantities shown in figure, assuming unit depth in the z direction, and neglecting second-order differentials:

dxdyyv

xu

TyT

vxT

uc p

dxdyyu

dxdyy

Tk

2

2

2

Convection 45


Using the continuity relation and dividing the whole equation by ρcp

for Low Velocity incompressible flow

2

p2

2

yu

cyT

yT

vxT

u

2

2

yT

yT

vxT

u

Convection 46

Energy Equation & Momentum Equation

2

2

yT

yT

vxT

u

2

2

yu

yu

vxu

u

Energy Equation

Momentum Equation(constant pressure)

The solution to the two equations will have exactly the same form when α = ν

End of Session


1. What is the momentum equation for the laminar boundary layer on a flat plate?

2. What are the assumptions involved in derivation of momentum equation?

3. Write the energy equation for laminar boundary layer on a flat plate

4. Explain the analogy between momentum and energy equation.

Convection 47

Integral form of Momentum Equation

Integral form of Momentum equation can be obtained using Von Kármán method:

(for constant pressure condition)

0yw0 y

uudy)uu(

dxd

Convection 48


Polynomial equation for velocity

Boundary Conditions

34

2321 yCyCyCCu

0 yat 0u yat uu

yat 0yu

00yu2

2

yat

Convection 49


Applying the boundary conditions, we get

Substituting,

Velocity Equation3y

21y

23

uu

u

23

C20C1 0C3 34u

21

C

Convection 50

Integral form of Momentum EquationUsing expression for velocity in integral equation,

Carrying out integration leads to

0y0

332

yu

dyy

21y

23

1y

21y

23

udxd

u

23

u28039

dxd 2

Convection 51

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

dxu13

140dx

u13140

d

constu

x13

1402

2

Convection 52

Integral form of Momentum EquationAt x=0, δ=0; so

Writing in terms of Reynolds number

BL thickness in terms of Reynolds number

Exact solution of BL equation

2/1xRe64.4

x

u

x64.4

x

2/1xRe0.5

x

Convection 53

Integral form of Energy Equation

FLAT PLATE

T∞

δtTW

TEMPERATURE PROFILE

wallyT

kAq

Convection 54


Polynomial equation for temperature

Boundary Conditions

34

2321 yCyCyCC

0 yat wTT tTT yat

t0yT

yat 00y

T2

2

yat

Convection 55


3

tt

y21y

23

Applying boundary conditions

Integral form of Energy Equation is given by:

0y

H0 y

T udy)TT(

dxd

Convection 56

Integral form of Energy EquationIntegral form of Energy Equation is given by:

Writing in terms of θ,

0y

H0 y

T udy)TT(

dxd

0y

H0 y

udy)(dxd

Convection 57

Integral form of Energy EquationWhere,

Using temperature & velocity profile equation in LHS

dyy

21y

23y

21y

23

1(dxd

u 3

H0

3

tt

t0y 23

yT

Convection 58

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

23

2803

203

dxd

u 42

3/1t Pr026.11

Convection 59

Heat Transfer Coefficient

wall

"

yT

kqAq

)TT(hq w"

Combining these equations,

TT

)y/T(kh

w

wall

Convection 60

Making an energy balance at the surface,

solving,

Local Nusselt Number

k23k

23

TT)y/T(k

htw

w

2/1x

3/1x RePr332.0Nu

End of Session


1. What is the assumption made by Von Karmen to solve the integral momentum equation ?

2. Write the velocity profile and the temperature profile equation used by Von Karmen in solving the momentum and energy equation

3. Write the equation to determine hydrodynamic & thermal boundary layer thicknesses

Part 14

CONVECTION

FORCED CONVECTION(FLOW OVER A FLAT PLATE)

CORRELATIONS

Convection 61


wall

"

yT

kqAq

)TT(hq w"

Combining these equations,

TT

)y/T(kh

w

wall

Convection 62

Nusselt NumberMaking an energy balance at the surface,

Using expression for δT

Introducing Nusselt no. Local Nusselt Number

tw

w k23

TT)y/T(k

h

2/1x

3/1x RePr332.0Nu

2/13/1

x xu

Prk332.0h

kxh

Nu xx

Convection 63

Nusselt Number

Ratio of temperature gradients by conduction and convection at the surface

khL Nu Number, Nusselt

Nusselt Number is an indicative of temperature gradient at the wall in the normal direction

Convection 64

Nusselt NumberAverage Nusselt number is obtained from

LxL0

L0 x h2

dx

dxhh

Lxx Nu2kLh

Nu

Average Nusselt Number

2/1L

3/1x RePr664.0

kLh

Nu

Convection 65

Use of Correlations1. External Flow

Flow over a Flat PlateFlow across cylinderFlow across sphereFlow across bank of tubes

2. Internal FlowFlow through tubes & ducts

Convection 66

Use of Correlations

Separate correlations are available for • Laminar• Constant temperature surfaces• Constant heat flux boundary condition

• Turbulent• Constant temperature surfaces• Constant heat flux boundary condition

• Combined laminar & turbulent conditions

Special correlations are available for liquid metals

Convection 67

Fluid Friction and Heat Transfer

Shear stress at the wall may be expressed in terms of friction coefficient Cf :

Also,

Using velocity distribution equation,

2u

C2

fw

ww y

u

u

23

w

Convection 68

Fluid Friction and Heat TransferMaking use of relation for boundary layer thickness:

Combining equations,

2/1

w xu

64.4u

23

2/1x2

2/1fx Re332.0

u1

xu

64.4u

23

2C

Convection 69


The equation may be rewritten as:

Where,

2/1x

3/2

p

x

x

x RePr332.0uc

hPrRe

Nu

2/1x

3/2x Re332.0PrSt

uch

Stp

xx

2/1x

3/1x RePr332.0Nu

Convection 70


Reynolds-Colburn Analogy2

CPrSt fx3/2

x

2/1x

fx Re332.02

C

End of Session


1. What is the significance of Nusselt Number 2. What is the relationship between local and average

Nusselt number for a flow over a flat plate in the laminar region ?

3. What is drag coefficient ?4. Why separate correlations are available for liquid

metals ?5. What is Reynolds-Colburn analogy ?

Part 15

CONVECTION

FORCED CONVECTION(FLOW OVER A FLAT PLATE)

PROBLEMS

Convection 71

Example – Mass flow and BL thickness

CalculateBoundary Layer Thickness at x = 20 cm & 40 cmMass flow which enters the boundary layer between x=20 cm and x = 40 cm.Assume unit depth in z direction.

FLAT PLATE

AIR2 m/s, 27 °C, 1 atm

1.85x10-5 kg/m.sx

y

Convection 72

Example – Mass flow and BL thickness

Density of Air

Reynolds number

When x = 20 cm, Re = 27,580When x = 40 cm, Re = 55,160

3m/kg77.1RTp

p = 1.0132 x 105 R = 287 T = 300 Kρ = 1.177 kg/m3 u = 2 ms-1 µ = 1.85x10-5

ux

Re

Convection 73

Example – Mass flow and BL thicknessBoundary Layer Thickness

When x = 20 cm, δ = 0.00559 mWhen x = 40 cm, δ = 0.0079 m

Re = 27,580 when x = 20 cm (calculated)Re = 55,160 when x = 40 cm (calculated)

2/1xRe64.4

x

Convection 74

Example – Mass flow and BL thicknessMass flow entering the Boundary Layer

Velocity, u is given by

Evaluating the integral with this velocity distribution,

0

dy.u.

3y21y

23

uu

u85

dyy

21y

23

u0

3

Convection 75

Example – Mass flow and BL thicknessMass flow entering the Boundary Layer

2040u85

m

ρ = 1.177 kg/m3 u∞=2 m/sδ40 = 0.0079 m δ20 = 0.00559 m

kg/s 310x399.3m

Convection 76

Example – Isothermal flat plate (heated)

Plate is heated over its entire length to 60 °CCalculateHeat Transferred (a) at the first 20 cm of the plate(b) at the first 40 cm of the plate

Flat Plate, T = 60 °C

AIR2 m/s, 27 °C, 1 atm

µ = 1.85x10-5 kg/m.sx

y

Convection 77

Example – Isothermal flat plate (heated)Formulae Used

Heat Flow

Nusselt No.

Reynolds No.

All properties are evaluated at film temperature

2/1x

3/1xx RePr332.0

kxh

Nu

xu

Re

)TT.(A.hq w

Convection 78

Example – Isothermal flat plate (heated)Film Temperature

ν=17.36x10-6 m2/s Pr = 0.7k=0.02749 W/m°C cp=1.006 kJ/kg K

K5.316C5.432

6027Tf

Properties of air at Film Temperature:

Convection 79

Example – Isothermal flat plate (heated)At x = 20 cmReynolds No.

Nusselt No.


u∞= 2 m/s Tf = 316.5 Kν = 17.36x10-6 m2/s Pr = 0.7k = 0.02749 W/m°C cp=1.006 kJ/kg K

Substituted Values

041,23xu

Re

74.44RePr332.0Nu 2/1x

3/1x

Cm/W15.6xk

Nuh 2xx

Convection 80

Example – Isothermal flat plate (heated)At x = 20 cm

Heat Flow

h = 6.15 W/m2 K Tw = 60 °CA = 0.2 m2 T∞ = 27 °C

Substituted Values

)TT.(A.hq w

W 81.18q

K.m/W3.12h.2h 2L

Convection 81

Example – Isothermal flat plate (heated)At x = 40 cmReynolds No.

Nusselt No.


u∞= 2 m/s Tf = 316.5 Kν = 17.36x10-6 m2/s Pr = 0.7k = 0.02749 W/m°C cp=1.006 kJ/kg K

Substituted Values

082,46xu

Re

28.63RePr332.0Nu 2/1x

3/1x

Cm/W349.4xk

Nuh 2xx

Convection 82

Example – Isothermal flat plate (heated)At x = 40 cm

Heat Flow

h = 4.349 W/m2 K Tw = 60 °CA = 0.4 m2 T∞ = 27 °C

Substituted Values

)TT.(A.hq w

W 114.8q

K.m/W698.8h.2h 2L

End of Session


Calculate:Boundary Layer Thickness & Drag Coefficient at a distance of 0.61 m from leading edge of plate

AIRT = 37.8 °Cu = 0.915 m/sρ = 1.126 kg/m3

ν = 0.167x10-4 m2/sFlat Plate

X = 0.61 m

yLeading

Edge

1

End of Session


Calculate:Local heat transfer coefficient and the heat transfer for 0.61 m length taking width of plate as 1 m

2AIR

T = 65.6 °Cu = 0.915 m/sν = 0.223x10-4 m2/sk = 0.0313 W/mK

Flat Plate at 121.1 °C

y

X = 0.61 m

Part 21

CONVECTION

FORCED CONVECTION(EXTERNAL FLOW)

CORRELATIONS & PROBLEMS

Convection 83

CORRELATIONS – EXTERNAL FLOW

FLAT PLATE

Laminar Flow

333.00

333.05.0xx ]75.0)x/x(1.[PrRe332.0Nu

xL Nu2Nu

Flat PlateLeading

Edge

X0

δt

δh

FLAT PLATETurbulent Flow (Fully turbulent from leading edge)

Combined Laminar and Turbulent Flow

Convection 84


33.08.0xx PrRe0296.0Nu

33.08.0L PrRe037.0Nu

333.08.0LL PrARe037.0Nu

5.0cr

8.0cr Re664.0Re037.0A

CYLINDERGeneralised EquationNuD – Nusselt number based on diameterAll properties to be taken at film temperature

Convection 85


333.0mDD PrReCNu

Re D C m0.4 – 4 0.989 0.3304.0 – 40 0.911 0.385

40 – 4000 0.683 0.4664000 – 40000 0.193 0.618

TUBE BANKS

Convection 86


INLINE

St

STAGGERED

SL

St

D

SL

D

SL

TUBE BANKS

Convection 87


333.0n PrRec13.1Nu For N ≥ 10

10N1NucNu 1 ≤ N ≤ 10Re to be calculated based on max. fluid velocity Vmax

INLINE

STAGGERED

where

u.)DS/(SV TTmax

u)DS(2/SV DLmax

5.02T

2LD )2/S(SS

TUBE BANKS (INLINE)

Convection 88


For 10 ROWS or MORE

ST / D

SL / D

1.25 1.5 2.0 3.0

C n C n C n C n

1.25 0.35 0.59 0.28 0.608 0.1 0.704 0.063 0.751.5 0.37 0.586 0.25 0.62 0.1 0.702 0.068 0.742 0.42 0.57 0.29 0.60 0.23 0.632 0.198 0.653 0.29 0.60 0.357 0.584 0.37 0.581 0.286 0.61

TUBE BANKS (STAGGERED)

Convection 89


For 10 ROWS or MORE

ST / D

SL / D

1.25 1.5 2.0 3.0

C n C n C n C n

0.6 - - - - - - .213 .6361 - - .497 .558 - - - -

1.5 .451 .568 .46 .562 .452 .568 .488 .5683 .31 .592 .356 .58 .44 .562 .421 .574

TUBE BANKS ( C1 values )

Convection 90


For LESS than 10 ROWS

ST – STAGGEREDIN – INLINE

N 1 2 3 4 5 6 7 8 9 10ST .68 .75 .83 .89 .92 .95 .97 .98 .99 1IN .64 .8 .87 .9 .92 .94 .96 .98 .99 1

10N1NucNu

CALCULATEHeat Transfer Coefficient for full length of plateRate of Energy Dissipation from the plateConvection 91

Example – Heated Flat Plate

Flat Plate at 90 °C

y

XAIR

T = 0 °Cu = 75 m/s

45 cm LONG, 60 cm WIDEAssume transition takes place at Re X, C = 5 x 105

Properties of air at Film Temperature

Critical Length (distance at which transition takes place)

Convection 92

Film Temperature

u∞ = 75 m/s ν=17.45x10-6 m2/sk=2.8 x 10-2 W/m°C Pr = 0.698

K318C452

090Tf


5cc 105

xuRe

cm 11.6xc

Convection 93


u∞= 75 m/s L = 0.45 mν = 17.45x10-6 m2/s Pr = 0.698k = 2.8 x 10-2 W/m°C

Substituted Values

6L 1093.1

LuRe

2732Pr870Re037.0Nu 3/15/4LL

Cm/W170Lk

Nuh 2LL


Convection 94

RATE OF ENERGY DISSIPATION FROM THE PLATE

hL = 170 W/m2 K A = 0.45 x 0.6 m2

TS = 90 °C T∞ = 0 °CSubstituted

Values

)TT(Ah2Q SL

kW 8.262Q


End of Session

QUESTIONS FOR THIS SESSION1. 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.

2. 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.

Part 22

CONVECTION

FORCED CONVECTION


CALCULATEHeat lost while standing in the wind

Convection 95

Example – Flow over Cylinder

D = 30 cmAIRT = 10 °Cu = 36 km/h

H = 1.7 mTS = 30 °C

Assume a man (represented as a cylinder) standing in the direction of wind


Reynolds Number

Convection 96

Film Temperature

ν = 15x10-6 m2/s k = 2.59 x 10-2 W/m°CPr = 0.707

K393C202

1030Tf

5D 102

DuRe

u∞ = 10

m/sD = 0.3 m


Convection 97

Rate of Heat Lost

ReD= 2 x 105 Pr = 0.707k = 2.59 x 10-2 W/m°CTS = 30 °C T∞ = 10 °C

Substituted Values

7.444PrRe027.0Nu 333.0805.0DD

Cm/W39.38Dk

Nuh 2DD

kW2.1)TT(AhQ SD


CALCULATETotal heat transfer per unit length for tube bank and the exit air temperatureConvection 98

Example – Flow through Tube Banks

AIR1 atm, 10 °C

u = 7 m/s

15 ROWS HIGH5 ROWS DEEPSL = ST = 3.81 cm

Heating of air with in-line tube bank

SL Tsurface = 65°C

2.54 cm

ST


Constants for use ( C & n ) from table

Convection 99

Film Temperature

µ = 1.894 x 10-5 kg/ms ρ = 1.137 kg/m3

k = 0.027 W/m°C Pr = 0.706

K5.310C5.372

1065Tf

5.154.281.3

DSL C =

0.25n = 0.62

5.154.281.3

DST


Convection 100

Maximum Velocity

D = 0.0254 m ST = 3.81 u∞ = 7 m/s n = 0.62µ = 1.894 x 10-5 kg/ms ρ = 1.137 kg/m3

c = 0.25 k = 0.027 W/m°C

s/m21u.)DS/(SV TTmax

020,32DV

Re max

35.155RecNu n

Cm/W14.165D/)k.Nu(h 2D


Convection 101

Heat Transferred

m/m985.5DLNA 2

)TT(hAQ w

Correction Factor ( C1 ) = 0.92 (from table) Total heat transfer surface area (assuming unit length)

N = 15 D = 0.0254 m L = 1m


Convection 102

)TT(mc2

TTThAQ 1,2,p

2,1,w

Subscripts 1 & 2 denote entrance & exit temperatures

Substituting

s/kg99.4S)15(um L

3m/kg246.1)RT/(p


C08.19T 2,

Convection 103

Heat Transferred

m/W6.45)TT(mcQ 1,2,p


Convection 104

Mixing Cup Temperature / Bulk Mean Temperatureis the temperature, the fluid would assume if placed in a

mixing chamber and allowed to come to equilibrium.

INTERNAL FLOWUNIFORM INLET FLOW

BOUNDARY LAYER

FULLY DEVELOPED FLOW

STARTING LENGTH

A

pzmpm TdAcvTc)Av(

INTERNAL FLOW

A

zm dAvA1

v

Az

Az

m dAv

TdAvT

0

0

r0 z

r0 z

mdr.r2.v

dr.r2.TvT

Where, A

zm

m TdAvAv

1T

A

pzmpm TdAcvTc)Av(

forCIRCULAR

DUCT

R0 z

m2m dr.r.T.vuR2

T

Convection 105

Convection 106

INTERNAL FLOW

2

2

rru

ru

vxu

u

Momentum Equation

(constant pressure)

Energy Equation

rT

r1

rT

ck

xT

u 2

2

p

For Slug flow…

rT

r1

rT

xT

u 2

2

CALCULATEAir side heat transfer coefficient across the tube bundleConvection


AIRT∞ = 15 °C u∞ = 6 m/s

7 ROWS in direction of flow

SL = ST = 20.5 mm

Water passing through Staggered tube bankTsurface = 70°C

1.64 cm

SL

1

End of Session


2. What is ‘bulk mean temperature or mixing cup temperature’ ?

3. What is slug flow ?4. Write the momentum and energy

equation for the flow through a tube.

Part 23

CONVECTION

FORCED CONVECTION(INTERNAL FLOW)


Convection 107

CORRELATIONS – INTERNAL FLOWProperties to be evaluated at Bulk Mean Temperature

Tm = (Tmi + Tmo) / 2Tmi – Mean Temperature at inletTmo – Mean Temperature at outlet

LAMINAR FLOWFully developed Thermal Layer

Constant Wall Temperature

Constant Heat Flux

D L ,66.3Nu0.6Pr ,36.4Nu

Convection 108

CORRELATIONS – INTERNAL FLOWLAMINAR FLOW (contd.)Entry region (Hydrodynamic layer fully developed, thermal layer developing)

Simultaneous development of hydrodynamic & thermal layers

67.0D

D

Pr]Re)L/D[(04.01PrRe)L/D(0668.0

66.3Nu

8.0D

D

]x/D.Pr[Re16.01)x/D.Pr(Re104.0

66.3Nu

0.7Pr

0.6Pr

Convection 109

CORRELATIONS – INTERNAL FLOWTURBULENT FLOWFully Developed flow (Dittus-Boelter equation)

n = 0.4 for heating of fluids / n = 0.3 for cooling of fluids0.6 < 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

n8.0D PrRe023.0Nu

14.0wm

n8.0D /PrRe027.0Nu

CALCULATE1. Reynolds number2. Heat Transfer Coefficient3. Difference between wall temperature and bulk (mean)

temperature.

Convection 110

Example 1

Water flowing through pipe with constant wall heat flux

Douter = 2 cm

WATERT = 25 °C

m = 0.01 kg/s

Constant Wall Heat Fluxqs = 1 kW/m2

Properties of water at 25 °C

< 2300. Flow is LAMINAR

For Constant Heat Flux

Convection 111

µ = 8.96 x 10-4 kg/ms k = 0.6109 W/m°C

364.4NuD

Example 1

709uD

Re

skg/m 28.31)01.0(

01.0Am

u 2

Cm/W3.133D/)k.Nu(h 2DD

D = 0.02 m

Difference between Wall Temperature and Bulk (mean) Temperature

Convection 112

Example 1

)TT(hq ms''s

C5.73.133

1000hq

TT''s

ms

CALCULATEAverage heat transfer coefficient by using Sieder-Tate equationConvection 113

Example 2

Water flowing through Copper Tube with constant wall temperature

Douter = 2.2 cm

WATERTinitial = 15 °CTfinal = 60 °Cu = 2 m/s

Constant Wall TemperatureTs = 95 °C

Properties of water

Convection 114

Bulk (mean) Temperature

µ = 0.69 x 10-3 N.s/m2 ρ = 990 kg/m3

k = 0.63 W/m°C cp = 4160 J/kg.KD = 0.022 m

K5.310C5.372

6015Tf

Example 2

63213uD

Re

56.4kc

Pr p

Convection 115

µ = 0.69 x 10-3 N.s/m2 ρ = 990 kg/m3

k = 0.63 W/m°C cp = 4160 J/kg.Kµs = 0.3 x 10-3 N.s/m2 D = 0.022 m

Example 2

84.346(Pr))(Re027.0Nu14.0

s

3/15/4DD

Cm/W9932D/)k.Nu(h 2DD

Estimate the heat leakage per meter length per unit temperature difference.

Convection 116

Example 3

Heat Leakage from an air conditioning duct

400 X 800 mm

AIRT = 20 °Cu = 7 m/s

Properties of air

Equivalent or Hydraulic Diameter

Assuming pipe wall temperature to be higher than air temperature, then Nusselt number is given by:

Convection 117

ν = 15.06 x 10-6 m2/s α = 7.71 x 10-2 m2/hk = 0.0259 W/mK

703.0/Pr

Example 3

4107.23uD

Re

m571.0)8.04.0(2)8.04.0(4

PA4

Dh

38.398PrRe023.0Nu 4.05/4D

Heat Leakage per unit length per unit temperature difference:

Convection 118

NuD = 398.38 k = 0.0259 W/mKD = 0.571 m

Example 3

Cm/W07.18D/)k.Nu(h 2DD

CW/m 37.432.1207.18hPQ

CALCULATE1. Heat transfer coefficient2. Total amount of heat transferred

Convection

Questions

Water flowing through a heated tube

Douter = 1.5 cmL = 3 m

WATERTinitial = 50 °CTfinal = 64 °Cu = 1 m/s

Constant Wall TemperatureTs = 90 °C

1

CALCULATEHeat transfer coefficient of air

Convection

Questions

Air flowing through annulus

ID = 3.125 cmOD = 5 cm

AIRTinitial = 16 °CTfinal = 32 °Cu = 30 m/s

Tsurface (of inner tube) = 50 °C

2

Part 24

CONVECTION

FREE CONVECTION

Free Convection Boundary LayerHeated Vertical Plate

Convection 119

TsTy

T∞

u(y)

δδt

VELOCITY BOUNDARY

LAYER

THERMAL BOUNDARY

LAYER

y, v

x, U

T∞, ρ∞,

g

Free Convection – Governing Equations

Continuity Equation

X Momentum Eqn.

Energy Equation

0yv

xu

Convection 120

2

2

yu

gxp1

yu

xu

u

2

2

p yT

ck

yT

xT

u

Free ConvectionX Momentum Equation

u0, ρρ∞ (density outside boundary layer)

Convection 121

2

2

yu

gxp1

yu

xu

u

gxp

2

2

yu

)(g

yu

xu

u

Free ConvectionX Momentum Equation

Convection 122

2

2

yu

)(g

yu

xu

u

pTp1

)TT(

2

2

yu

)TT(gyu

xu

u

T

Free ConvectionVolumetric Coefficient of thermal expansion, β

Convection 123

pT1

T1

RTp1

2

2p RT

pT1

TRp

T

RT/p

Free Convection

Convection 124

2

2

yu

)TT(gyu

xu

u

0yv

xu

2

2

yT

yT

xT

u

Summarizing the governing equations,

Free Convection

Convection 125

2*

*2

L

*20

s*

**

*

**

y

uRe

1T

uL)TT(g

yu

xu

u

Lx

x*

2*

*2

L*

**

*

**

y

TPrRe

1yT

xT

u

Identification of Dimensionless Groups

Ly

y* *0

*

uu

u

TTTT

T*

*

0yv

xu

*

*

*

*

Free Convection

2L

L2

0

2s

3

20

s

ReGr

)/Lu(/)TT(Lg

uL)TT(g

Rearranging

Where,

2s

3 )TT(Lg

Gr number,Grashof L

“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 FlowsVertical SurfacesLaminar (Gr.Pr < 109)Constant Wall Temperature

Constant Heat Flux

Turbulent (Gr.Pr > 109)

25.0x

25.05.0x Gr.Pr)952.0(Pr 508.0Nu

Convection 127

11xx

52.0xxx 10NuGr10forPr).Nu.Gr(6.0Nu

333.0x Pr)Gr( 1.0Nu

Free Convection in External FlowsHorizontal SurfacesCharacteristic Length

Constant Wall Temperature

Constant Heat Flux

plate theof Perimeterplate theof area Surface

L

Convection 128

6425.0 108PrGr102forPr)Gr(54.0Nu 1163/1 10PrGr108forPr)Gr(15.0Nu

8x

333.0 102PrGrforPr)Gr(13.0Nu 1162.0 10PrGr10forPr)Gr(16.0Nu

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

)negligible convection (free convection Forced1)Re/Gr( 2

Convection 129

)negligible convection (forced convection Free1)Re/Gr( 2

forced)and (free convection Mixed1)Re/Gr( 2

Combined Free & Forced ConvectionExternal Flow

Internal Flow (LAMINAR)

Graetz numberConvection 130

A)Re/(GrPrRe332.0Nu 2xx

3/12/1x if

A)Re/(GrGr)Pr952.0(Pr508.0Nu 2xx

4/1x

4/1-2/1 if

333.0333.1333.014.0

w)Gr.Gz(012.0Gz75.1Nu

)L/D.(Pr.ReGz D

Combined Free & Forced ConvectionInternal Flow (TURBULENT)

Applicable for ReD > 2000 and RaD (D/L) < 5000OrReD > 800 and RaD (D/L) > 2x 104

Convection 131

36.007.0D

21.027.0D (D/L)GrPrRe69.4Nu

Minimum spacing (L) to avoid interference of free convection boundary layers

Convection 132

Example – Convection between Vertical Plates

TSurface = 80 °C

?

δδ

L

3.5 cm

Twater = 20 °C

WATER

PLATE

Properties of water at Film Temperature

< 1 x 109 (LAMINAR)

Convection 133

Let, δ be the boundary layer thickness at trailing edgeMinimum spacing required = L = 2δFilm temperature = t∞ = (80 + 20) / 2 = 50 °C

Pr = 3.54 β = 0.48x10-3 K-1 ν = 0.567 x 10-6 m2/s

7

2

3

1076.3

Tgx

Grx

910894.0Pr.Gr


Convection 134

Boundary layer thickness (δ)

Minimum Space to avoid interference

039.0.Pr)952.0(Pr93.3 25.025.05.0

x

x Grx


Pr = 3.54 x = 0.035 m Gr = 0.2526x109

mm36.1035.0038.0

mm72.22

Part 24

Questions

1. Draw the free convection boundary layer on a heated vertical plate.

2. Write the governing equations for free convection3. What is the significance of Grashof number ?4. Explain the situations under which combined free

and forced convection should be considered.

Part 25

CONVECTION

FREE CONVECTIONProblems

Heat lost by pipe / metre length

Convection 135

Example 1 – Vertical PipeDouter = 10 cm

AIR (ambient)T = 20 °C

TSurface = 100 °C

Vertical pipe kept in a room

L = 30 cm

?


Convection 136

Film Temperature

Pr = 0.696 β = 0.003003 K-1

ν = 18.97 x 10-6 m2/s k = 0.02896 W/m°C

K293C602

20100Tf

Example 1 – Vertical Pipe

103

s3

LL 1025.12Pr)TT(Lg

Pr.GrRa

L = 3 m T∞ = 100°C TS = 20°C

Checking

Then,

Convection 137

4/1L )Gr(35

LD

Example 1 – Vertical Pipe

RaL = 12.25 x 1010 k = 0.02896 W/m°CL = 3 m T∞ = 100°C TS = 20°C

true. is 4/110)106.17(35

1.03

488Ra1.0Nu 3/1LL

K71.4L/)k.Nu(h L2W/m

m/W37.118)TT).(D.(hQ s

Heat gained by duct / metre length

Convection 138

Example 2 – Horizontal Duct60 cm

Horizontal un-insulated Air Conditioning Duct

AIR (ambient)T = 25 °C

TSurface = 15 °C 30 cm

?


Rate of Heat Gained per unit length of duct

Convection 139

Film Temperature

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

K293C202

2515Tf

bottomtopside QQQ2Q

Example 2 – Horizontal Duct

)TT(L2)hh()Lh2(Q scharbottomtopcharside

392

3

1001.0Pr)(

Pr. LTTLg

GrRa sLL

Heat gained from vertical wall (sides)

Laminar 45.13Ra59.0Nu 4/1

LL

539 107.23.0 1001.0 LRa


K162.1L/)k.Nu(h L2W/m

m/W97.6)TT(Lh2Q svv

β = 0.00341 K-1 ρ = 1.205 kg/m3

ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°CT∞ = 25°C TS = 15°C Pr = 0.705

Heat gained from top & bottom surfacesCharacteristic Length

Laminar

Similarly for bottom surface,

31.12Ra54.0Nu 4/1LL

539 107.23.0 1001.0 LRa


K063.1L/)k.Nu(h Lt2W/m

m/W726.9)TT(L2)hh(Q sbtbt

β = 0.00341 K-1 ρ = 1.205 kg/m3

ν = 15.06 x 10-6 m2/s k = 0.02593 W/m°CT∞ = 25°C TS = 15°C Pr = 0.705

m3.02/wP/AL

TOP

SURF

ACE

4/1Lb )Ra(27.0)L/k(h

Convection 142

Rate of Heat GainedExample 2 – Horizontal Duct

bottomtopside QQQ2Q

m/W7.16Q

Qside = 6.97 W/m Q top + bottom = 9.73 W/m

Calculate the heat transfer coefficient

Convection 143

Combined Free & Forced Convection with Air

TSurface = 140 °C

0.4m

25 mmAIR

Tair = 27 °Cu = 30 cm/s

TUBEAIR

Air flowing through a horizontal tube3

Convection 144

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°Cµw = 2.337 x 10-5 kg/m.s

K5.356C5.832

27140Tf


53.3uD

Re

D = 0.025 mu = 0.3 m/s

Convection 145

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°Cµw = 2.337 x 10-5 kg/m.s


353Re m

uD

5

2

310007.1

TgxGr

4677Ld

.Pr.Gr 33.15Ld

.Pr.ReGz

Convection 146

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 = 15.33Gr = 1.007 x 109 d = 0.025 m


333.0333.1333.014.0

w)Gr.Gz(012.0Gz75.1Nu

7.7Nu

K4.9d/)k.Nu(h 2W/m

Panel : 0.915 m x 0.915 mOne side insulated, other side at 65.6 °CAmbient is at 10 °C

Questions

1

INSU

LATED

SURFA

CE

HOT

SURFA

CE

INSULATEDSURFACE

HOTSURFACE

HOTSURFACE

INSULATEDSURFACE

Calculate the mean heat transfer coefficient due to free convection

Estimate the heat gained by the duct.

Convection

Questions

30 X 20 cmDuct Surface at 5 °CAIR

T = 25 °C

2

Air flow through Rectangular Duct

Calculate the heat transferred considering combined free and forced convection

Convection

Questions

Air flowing through a tube

D = 20 mmL = 1 m

AIRT = 27 °C

u = 30 cm/s

Horizontal TubeTsurface = 127 °C

3

chapter 2 convection dr. r. velraj, professor anna university chennai heat transfer

Documents

convection session

pipe convection

t w t convection

u u slide

g free convection

forced convection influence

session slide

tube x y convection