3.1 INTRODUCTION Convection is one of the three basic mechanisms of heat

transfer. It is a heat transfer that takes place between moving fluid and solid surface. Like conduction heat transfer, convection heat transfer requires material medium. Heat transfer by convection combines heat transfer by conduction and bulk flow of fluid.

Convection heat transfer depends on several parameters like: Fluid properties

Dynamic viscosity, μ. Thermal conductivity, k. Density, ρ. Specific heat capacity, cp.

Fluid velocity, v. Geometry of solid surface. Roughness of solid surface. Type of fluid flow (turbulent or laminar).

3.1 INTRODUCTION… The dependence of convection heat transfer on

so many variables makes it very complex and it is usually determined experimentally.

The convection heat transfer rate is given by Newton’s law of cooling,

(3.1)Where

h=convection heat transfer coefficient,A=heat transfer surface area,Ts=surface temperature,T∞=fluid temperature far away from surface.

)(.

.

TThAQ sconv

3.1 INTRODUCTION…Non dimensional convection coefficientThe convection heat transfer coefficient, h, is

nondimensionalized to obtain the Nusselt number, Nu.

(3.2)

Where k = thermal conductivity, = characteristic length.

khNu

3.1 INTRODUCTION…The Nusselt number can also be given as the

ratio of convection heat transfer rate to conduction heat transfer rate.

(3.3)

ratetransferheatConductionratetransferheatConvectionNu

.

..

.

cond

conv

Q

QNu

3.2 VELOCITY BOUNDARY LAYER

Fig. 3.1 Velocity boundary layer development over flat plate

uu 99.0 is the value of y for which

3.2 VELOCITY BOUNDARY LAYER… For external flows the velocity boundary layer concept

provides the basis for determining the local friction coefficient. (3.4)

Where the surface shear stress for Newtonian fluid is obtained from the relation, (3.5)

The drag or friction force over the surface is determined by(3.6)

2

21

u

C sf

0

y

s yu

2

2

uACF fD

3.3 THERMAL BOUNDARY LAYER

Fig. 3.2 Thermal boundary layer development over isothermal flat plate

t is the value of y for which the ratio 99.0)/()( TTTT ss

3.3 THERMAL BOUNDARY LAYER… At any distance x from the leading edge, the local heat flux may be obtained

by applying Fourier’s law to the fluid at y= 0. That is,(3.7)

By combining equation (3.7) with Newton’s law of cooling we then obtain

(3.8)

The relative thickness of the velocity and thermal boundary layers is described by Prandtl number,

 

The hydrodynamic and thermal boundary layers are related in terms of Prandtl number as (3.9)

0

y

fs yTkq

TT

yTk

hs

yf

0

kC

ydiffusivitHeatydiffusivitMomentum P

Pr

1Pr026.11

t

3.3 THERMAL BOUNDARY LAYER…

Fig. 3.3 Relative thickness of the velocity and thermal boundary layers

3.4 LAMINAR AND TURBULENT FLOW

Fig.3.4 Laminar, transition and turbulent flow regions

3.4 LAMINAR AND TURBULENT FLOW…The location where the flow turns to turbulent

is determined by a dimensionless grouping of variables called the Reynolds number.(3.10)

Where u∞=free stream velocity=characteristic length of the geometry=kinematic velocity of the fluid

The critical Reynolds number at which flow turns to turbulent is about 5x105.

u

forcesVicousforcesInertiaRe

3.5 FLOW OVER FLAT PLATES The friction and heat transfer coefficients for a flat plate

can be determined by solving the conservation of mass, momentum and energy equations. The average Nusselt number can be expressed as(3.11)

Where C, m and n are constants and L is length of the plate.

Generally, properties of the fluid vary with temperature. To make the heat transfer analysis simple, properties are evaluated at film temperature given by (3.12)

In heat transfer analysis, we are usually interested in the heat transfer and drag force on the entire surface of the plate which are determined using the average heat transfer and friction coefficients.

nmLC

khLNu PrRe

2

TTT s

f

3.5 FLOW OVER FLAT PLATES…The local and average Nusselt number and

friction coefficient are determined for laminar and turbulent flows separately,

Laminar flowThe boundary layer thickness

(3.13)

Local friction coefficient(3.14)

21

Re

5

x

x

212,

Re

664.02/1

x

xxf u

C

3.5 FLOW OVER FLAT PLATES…Combining equations (3.9) and (3.13), the

thermal boundary layer thickness is given by(3.15)

 Local Nusselt number

(3.16)Average friction coefficient

(3.17)

)(Pr)(Re026.15Pr

026.11

2/11

xt

x

)6.0(PrRe332.0 31

21

prkxhNu x

xx

21

Re

328.1

L

fC

3.5 FLOW OVER FLAT PLATES…Average Nusselt number 

(3.18)The critical Reynolds number

)6.0(PrRe664.0 31

21

prkhLNu L

5105Re

Cr

Crxu

3.5 FLOW OVER FLAT PLATES…Turbulent flow

Local friction coefficient (3.19)Local Nusselt number

(3.20)Average friction coefficient

(3.21)Average Nusselt number

(3.22)

75

51, 10Re105

Re

0592.0 x

x

xfC

75

31

54

10Re105

606.0PrRe0296.0

x

xx

x

prkxh

Nu

75

51 10Re105

Re

074.0 L

L

fC

753

15

4

10Re105

606.0PrRe037.0

L

Lpr

khLNu

3.5 FLOW OVER FLAT PLATES…For combined laminar and turbulent flow

(3.23)

(3.24)The heat transfer rate is obtained from

(3.25)Where

The drag force is calculated as (3.26)

75

51 10Re105

Re1742

Re

074.0 L

LL

fC

753

15

4

10Re105

606.0Pr871Re037.0

L

Lpr

khLNu

)(.

.

TThAQ sconv

L

kNuh

2

2 uACF fD

3.5 FLOW OVER FLAT PLATES…Example 3.1

An electric air heater consists of a horizontal array of thin metal strips that are each 10mm long in the direction of an airstream that is in parallel flows over the top of the strips. Each strip is 0.2m wide, and 25 strips are arranged side by side, forming a continuous and smooth surface over which the air flows at 2m/s. During operation each strip is maintained at 5000C and the air is at 250C.

a. What is the rate of convection heat transfer from the first strip? The fifth strip? The tenth strip? All the strips?

b. For air velocities at 3, 5, and 10 m/s, determine the convection heat rates for all the locations of part (a). Represent your results in tabular or bar graph form.

3.5 FLOW OVER FLAT PLATES…Solution

3.5 FLOW OVER FLAT PLATES…Properties of air (Tf=535K, 1atm): =43.54x10-

6m2/s, k=0.0429W/m.K, Pr=0.683.a)The location of transition is determined from

Since xc >>L=0.25m, the air flow is laminar over the entire heater. For the first strip,

where h1 is obtained from

Then

mu

xc 9.102

1054.431051056

55

TTwLhQ s11

.

KmwLkh x

231

21

1 /8.53PrRe664.0

wQ 1.511

.

3.5 FLOW OVER FLAT PLATES…for the fifth strip,

In similar manner for the tenth strip and entire 25 strips,

40

.

505

.

QQQ TTwLhTTwLhQ ss 45 40505

.

KmwL

kh

andKmwL

kh

x

x

231

21

04.040

231

21

05.050

/9.26PrRe664.04

/1.24PrRe664.05

wQ 2.125

.

wTTwLhTTwLhQ ss 3.8910 9010010

.

wTTwLhQ s 3.255250250

.

3.5 FLOW OVER FLAT PLATES…b)

velocity, m/s

heat transfer rate, w

first strip fifth strip tenth strip entire strip2 51.08 12.06 8.29 255.395 80.76 19.06 13.11 403.8010 114.20 26.96 18.53 571.06

3.6 FLOW ACROSS CYLINDERS AND SPHERESThe average Nusselt number for flow over a

cylinder is given empirically as proposed by Churchill and Bernstein:

(3.27)

Equation (3.26) is applicable for conditions where Re.Pr > 0.2. The fluid properties are evaluated at film temperature .

54

85

41

32

31

21

28200Re1

Pr4.01

PrRe62.03.0

khDNu

2

TTT s

f

3.6 FLOW ACROSS CYLINDERS AND SPHERES…The average Nusselt number for flow over a

cylinder can be expressed in compact form as

(3.28)Where the constants C and m are obtained from table 3.1.

Heat transfer to or from a bank (or bundle) of tubes in cross flow is relevant to numerous industrial applications, such as steam generation in a boiler or air cooling in the coil of an air conditioner.

31

PrRemCkhDNu

3.6 FLOW ACROSS CYLINDERS AND SPHERES…Table 3.1 Constants used in equation (3.28)

3.6 FLOW ACROSS CYLINDERS AND SPHERES…The tube rows of a bank are either staggered or

aligned in the direction of the fluid velocity V (Fig. 3.6)

Fig. 3.5 Tube bank in cross flow

3.6 FLOW ACROSS CYLINDERS AND SPHERES…

Fig. 3.6 Aligned and staggered tube arrangement

3.6 FLOW ACROSS CYLINDERS AND SPHERES…For airflow across tube bundles composed of 10 or

more rows (NL10), the average Nusselt number can be obtained by the Grimison correlation,

(3.29)

The constants C1 and m are obtained from table 3.2.

The maximum Reynolds number is given by(3.30)

7.0Pr

000,40Re200010

Re max,max,1 D

LmD

NC

khDNu

DV

Dmax

max,Re

3.6 FLOW ACROSS CYLINDERS AND SPHERES…For the aligned arrangement the maximum velocity,

Vmax occurs at the transverse plane A1 of Fig. 3.6a and is given for an incompressible fluid as (3.31)

For the staggered configuration, the maximum velocity may occur at either the transverse plane A1

or the diagonal plane A2 . It will occur at A2 if the rows are spaced such that

Where SD is given by

VDS

SVT

T

max

2)()(2 DSSorDSDS T

DTD

21

22

2

T

LDSSS

3.6 FLOW ACROSS CYLINDERS AND SPHERES…And

(3.32)If SD>(ST+D)/2, the maximum velocity occurs at A1

and is given by equation (3.31).For flow of fluids other than air equation (3.28) is

modified by inserting 1.13Pr1/3.

(3.33)

All the properties in these equations are evaluated at film temperature. If NL<10 a correction factor given in table 3.3 is used as(3.34)

VDSSVD

T

2max

7.0Pr

000,40Re200010

PrRe13.1 max,3

1

max,1 D

LmD

NC

khDNu

10210 LL NN NuCNu

3.6 FLOW ACROSS CYLINDERS AND SPHERES…Table 3.2 Constants used in equations (3.29) and (3.33)

3.6 FLOW ACROSS CYLINDERS AND SPHERES…Table 3.3 Constant C2 used in equation (3.34)

3.6 FLOW ACROSS CYLINDERS AND SPHERES…The heat transfer rate could be more predicted by using

the log mean temperature difference instead of using T=Ts-T∞.

(3.35)

Where Ti and To are temperatures of the fluid as it enters and leaves the bank, respectively.

The outlet temperature, which is needed to determine Tlm may be estimated from

(3.36)

Where N is the total number of tubes in the bank , NT is the number of tubes in the transverse plane and V is the speed at inlet.

osis

osislm TTTT

TTTTT

/ln

PTTis

os

cSVNDNh

TTTT

exp

3.6 FLOW ACROSS CYLINDERS AND SPHERES…Once Tlm is known, the heat transfer rate may be

computed from(3.37)

For flows over a sphere, the average Nusselt number can be obtained by the Whitaker correlation, (3.38)

Equation (3.37) is valid for 3.5≤Re≤ 80,000 and 0.7≤Pr≤380. The fluid properties are evaluated at film temperature except s which is evaluated at the surface temperature, Ts.

lmTDLhNQ .

41

4.032

21

PrRe06.0Re4.02

skhDNu

3.6 FLOW ACROSS CYLINDERS AND SPHERES…Example 3.2A preheater involves the use of condensing

steam at 1000C on the inside of bank of tubes to heat air that enters at 1atm and 250C. The air moves at 5m/s in cross flow over the tubes. Each tube is 1m long and has an outside diameter of 10mm. The bank consists of 196 tube in a square, aligned array for which ST=SL=15mm. What is the total rate of heat transfer to the air’?

3.6 FLOW ACROSS CYLINDERS AND SPHERES…Solution

3.6 FLOW ACROSS CYLINDERS AND SPHERES…Atmospheric air (T∞=298K):=15.8x10-6m2/s,

k=0.0263W/mK, Pr=0.707, cp=1007J/kgK, =1.17kg/m3; (Ts=373K): Pr=0.695

The total heat transfer rate

From tables 3.2 and 3.3, C = 0.27, m = 0.63 and C2=0.99.

lm

osis

osis TDLNhTTTTTTTTDLNhQ

/ln

.

9494108.1501.015Re,/155

515

6max,max

D

T

T smVDS

SV

3.6 FLOW ACROSS CYLINDERS AND SPHERES…From the Zhukauskas correlation

Hence,

C

cSVNhDNTTTT

KmWDkNuh

Nu

pTTisos

D

D

0

2

4/136.063.0

7.271007015.014517.1

20019601.0exp75

exp

./20001.0/0263.09.75/

9.75695.0/707.0707.0949427.099.0

kWTDLNhQ lm 5.587.27/75ln

7.2775101.0196200.

3.7 FLOW IN TUBES

Fig. 3.7 Hydrodynamic boundary layer development in tube

3.7 FLOW IN TUBES…

Fig. 3.8 Thermal boundary layer development in tube

3.7 FLOW IN TUBES…The Reynolds number for flow in a circular

tube is defined as(3.39)

Where ρ is the fluid densityD is tube diameter is fluid dynamic viscosityum is the average fluid velocity given by

(3.40)

Where is mass flow rate of fluid.

Dum

D Re

Amum

.

.m

3.7 FLOW IN TUBES…In a fully developed flow the critical Reynolds

number is 2300. The pressure drop during the flow is given by

(3.41)Where f is the friction factor and L is tube length

For laminar and turbulent flows the hydrodynamic entry lengths may be obtained from (3.42) (3.43)

2

2mu

DLfP

Dx Dlamhfd Re05.0,

Dxturbhfd 10,

3.7 FLOW IN TUBES…For laminar and turbulent flows the thermal entry

lengths may be obtained from (3.44) (3.45)

Laminar flowFor hydrodynamically developed laminar flow in a

tube, the velocity profile is profile is parabolic and given by (3.46)

Dx Dlamtfd PrRe05.0,

Dxturbtfd 10,

)1(2)( 2

2

om r

ruru

3.7 FLOW IN TUBES…And the surface shear stress is (3.47)

The surface shear stress can also be written as(3.48)

It follows from equations (3.46) and (3.47) that the friction coefficient Cf can be given as (3.49)

Du

drdu m

rrs

o

82

2

2m

fsuC

DfC

Re16

3.7 FLOW IN TUBES…The friction factor f used in the pressure drop

calculation in laminar flow is given by

(3.50)The average Nusselt number for the

hydrodynamically or thermally developed laminar flow is given by Sider and Tate

(3.51)All the properties are evaluated at bulk mean

fluid temperature, except for s, which is evaluated at the surface temperature.

D

fRe64

14.03

1PrRe86.1

s

b

LD

khDNu

3.7 FLOW IN TUBES…Turbulent flowThe friction factor for fully developed turbulent flow in

a smooth tube is given by(3.52)

For flows in smooth or rough tubes the friction factor is obtained from Moody diagram of Fig. 3.9.

The friction factor can be obtained from the Colebrook equation for flows in smooth or rough tubes in transition and turbulent flows.

(3.53)

The average Nusselt number for turbulent flow in a tube is given from the Chilton-Colburn correlation as

(3.54)

2.0Re184.0 Df

fD

f Re51.2

7.3.log0.21

31

PrRe125.0 fNuD

3.7 FLOW IN TUBES…

Fig. 3.9 The Moody Diagram

3.7 FLOW IN TUBES…Material Condition ε (mm)Steel Sheet metal, new

Stainless, newCommercial, newRivetedRusted

0.050.0020.0463.02.0

Iron Cast, newWrought, newGalvanized, newAsphalted cast

0.260.0460.150.12

Brass Drawn, new 0.002Plastic Drawn tubing 0.0015Glass - 0 (smooth)Concrete Smoothed

Rough0.042.0

Rubber Smoothed 0.01Wood Stave 0.5

Table 3.4 Roughness Values for Commercial Ducts

3.7 FLOW IN TUBES…Example 3.3An engine oil cooler consists of a bundle of 25

smooth tubes, each of length L=2.5m and diameter of D=110mm.

a) If oil at total flow rate of 24kg/s is in fully developed flow through the tubes, what are the power drop and the pump power requirements?

b) Compute and plot the pressure drop and pump power requirement as a function of flow rate for 10-30kg/s.

3.7 FLOW IN TUBES…Solution

Properties of Engine oil (300 K): =884kg/m3, μ=0.486kg/s.m.

3.7 FLOW IN TUBES…a) Considering flow through a single tube

Hence, the flow is laminar and

5.251486.001.025

2444Re.

Dm

D

2545.05.251

64Re64

D

f

smDmum /8.13

4/01.088425/24

4/ 22

.

1

3.7 FLOW IN TUBES…The pressure drop is,

The pump power required is

MpamNLDu

fp m 38.5/1038.55.201.02

8.138842545.02

2622

kwmpVpP 146884241038.5 6

..

3.7 FLOW IN TUBES…b) Plot of pressure drop and pump power

3.7 FLOW IN TUBES…Example 3.4Water at 150C (=999.1kg/m3 and =1.138x10-3

kg/ms) is flowing in a 4cm diameter and 30m long horizontal pipe made of new stainless steel steadily at a rate of 5L/s. Determine (a) the pressure drop and (b) the pumping power requirement to overcome this pressure drop.

L = 30 m

D = 4 cm

Water5 L/s

3.7 FLOW IN TUBES…SolutionThe density and dynamic viscosity of water are given

to be (=999.1kg/m3 and =1.138x10-3 kg/ms, respectively. The roughness of new stainless steel is 0.002 mm (Table 3.4).

First we calculate the mean velocity and the Reynolds number to determine the flow regime:

which is greater than 10,000. Therefore, the flow is turbulent.

53

3

2

3

2

1040.1skg/m 10138.1

m) m/s)(0.04 98.3)(kg/m 1.999(Re

/m 98.34/m) (0.04

/m 0.0054/

D

ssDV

AV

m

cm

V

V

3.7 FLOW IN TUBES…The relative roughness of the pipe is

The friction factor can be determined from the

Moody chart, but to avoid the reading error, we determine it from the Colebrook equation using an equation solver (or an iterative scheme),

105m 04.0

m 102/ 56

D

fffD

f 5

5

1040.151.2

7.3105log0.21

Re51.2

7.3/log0.21

3.7 FLOW IN TUBES…It gives f = 0.0171. Then the pressure drop and the

required power input become (a)

(b)

Therefore, useful power input in the amount of 0.508 kW is needed to overcome the frictional losses in the pipe.

101.5kPakN/m1

kPa1mskg1000

kN12

m/s)(3.98kg/m(999.104.0

m300.0171

2ρV

DLfΔP

2

23

2m

m

kW 0.508

/smkPa 1

kW 1)kPa 5.101)(/m 005.0(3

3upump, sPVW