Heat transfer lectures 2 (convection)

Ch 4: Convection of Heat Transfer 3rd Year College of Technical

Mr. Amjed Ahmed 66

Chapter Four

Convection Heat Transfer

4.1 Introduction

4.2 Force Convection for External Flow الحمل القسري للجريان الخارجي From energy balance of fluid to determined Heat transfer coefficient:

qconv = qcond

0=−= )∞ ydx

dT(x)kA- TA (Th sx 4.1

Figure 4.1 Developments of the Velocity and Thermal Boundary Layers in Flow Over a Flat Surface of Arbitrary Shape.

Free Convection (by Bouncy Force)

Force Convection (by Pump or Fan)

Convection with Change Phase (by Boiling & Condensation )

Internal The entry region Fully developed

ExternalA Plane Wall A Cylinder A Sphere

Convection Heat Transfer


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Figure 4.2 Laminar and Turbulent flow Over a Flat Surface plate.

( )

∞

=

−

−=

TTdx

xdTkh

s

yf

x

0

4.2

Where kf: a film thermal conductivity for fluid Local heat transfer coefficient (hx) depends on:

1. Geometry of surface 2. Surface and fluid Temperature 3. Velocity and type of fluid flow (Laminar or Turbulent ) 4. Physical properties fluid (Cp, µ , ρ, k)

Table (4.1) Ranges of Heat transfer coefficients Values

Type of Convection h w/m2 k Free Convection 5 - 25

Force Convection gases 25 - 250 liquids 50 - 20000

Boiling & Condensation 2500 - 100000

)(

)(

∞

∞

−=

−=

TTAhdq

TTdAhdq

ss

ssx

h : Average convection coefficient xh : Local convection coefficient

A= w x and dA =w dx Equal last equations and substitution A and dA

)()(0

∞∞ −=− ∫ TTdxhwTTwxh s

x

xs 4.3

∫=x

xdxhx

h0

1 4.4

4.3 The Velocity and Thermal Boundary Layers

Mechanism of heat transfer by convection: 1. Random molecular motion 2. Bulk motion


Mr. Amjed Ahmed 68

The velocity boundary layer development on a flat plate. The boundary layer is initially laminar ,but some distance from the leading edge, small disturbance are amplified and transition region, and the boundary layer eventually becomes completely turbulent

Three different regions may be delineated in Turbulent Boundary layer is 1. A laminar sublayer in which transport is dominated by diffusion and the velocity

profile is nearly linear. 2. A buffer layer in which diffusion and turbulent mixing are comparable 3. A turbulent zone in which transport is dominated by turbulent mixing.

In calculating boundary layer behavior it is frequently reasonable to assume that transition begins at some location xc. This location is determined by a dimensionless grouping of variables called the Reynolds number

µνρX

x =Re

where x is the distance from the leading edge. The critical Reynolds number is the value of (Rec =5×105) for which transition begins, and for flow over a flat plate

The velocity boundary layer

0=+dydv

dxdu Continuously equation

2

2

yu

dydvu

dxduv

∂∂

=+ µ Momentum direction x

2

2

yT

ydTu

dxdTv

∂∂

=+ α Energy equation

x

xRe5

=δ Local boundary layer thickness

Assume a velocity distribution in boundary layer in one dimension flow is u=a+by+cy2+dy3

B.C. 1 u=0 at y=0 B.C. 2 u=u∞ at y=δ B.C. 3 du/dy=0 at y=δ B.C. 4 d2u/dy2=0 at y=0 A result is يطبق الشرط الحدي االول والرابع اوًال

Figure 4.3 Heat transfer coefficient and velocity distribution.


Mr. Amjed Ahmed 69

3

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

∞ xx

yyuu

δδ 4.5

Assume a temperature distribution in boundary layer in one dimension is T=a+by+cy2+dy3

θ=a+by+cy2+dy3 θ=T-Ts

B.C. 1 θ =0 at y=0 B.C. 2 θ = θ ∞ at y=δ B.C. 3 dθ /dy=0 at y=δ B.C. 4 d2θ /dy2=0 at y=0 A result is

3

21

23

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−−

=∞∞ ttS

S yyTTTT

δδθθ 4.6

The ratio of the velocity to thermal boundary layer thickness is

31

Pr0261.δδt =

4.4 Dimensionless Parameters:

Table (4.2) Dimensionless Groups of Importance for Heat Transfer and Fluid Flow

4.5 Heat transfer coefficient

To calculate heat transfer coefficient by several methods: 1. Solve the boundary layer equation(Exact solution). 2. Using experimental data.(empirical correlation)

In convection heat transfer the key unknown is the heat transfer coefficient. From Eq. (4.1) we obtain the following equation in terms of the dimensionless Parameters:

0== yf

x dxdT

Lk

h 4.7

Inspection of this equation suggests that the appropriate dimensionless form of the heat transfer coefficient is the so-called Nusselt number after Wilhelm Nussult, Nu, defined by

f

c

kLhNu = 4.8

Group Definition Interpretation Biot number

skLh

Bi = Ratio of internal thermal resistance of a solid body to its surface thermal resistance

Nusselt number f

c

kLhNu = ratio of convection heat transfer to conduction in a fluid

layer of thickness L

Peclet number PrRe LPe = Product of Reynolds and Prandtl numbers Prandtl number

ανµ

==k

CpPr Ratio of molecular momentum diffusivity to thermal diffusivity

Reynolds number µ

νρLL =Re Ratio of inertia to viscous forces

Stanton number PrReL

LNuSt = Dimensionless heat transfer coefficient


Mr. Amjed Ahmed 70

The Local Nusselt Number depends only on x, Re, and Pr: )Re r,P,f(xNux = 4.9

Once this functional relation is known, either from an analysis or from experiments with a particular fluid. Moreover, from the local value of Nu, we can first obtain the local value of h and then an average value of the heat transfer coefficient h and an average Nusselt number lNu

Pr)(Re,fNu = 4.10 nm

l CNu PrRe= 4.11 To determine parameters (C, m and n) from experimental data to give Empirical

Correlations

4.5.1 Empirical Correlations for Flat plate in Parallel flow A Laminar Flow Rec < 5 ×105

31

21

PrRe332.0==f

xx k

XhNu Pr > 0.6, Rex < 5x105 4.12

21

Pr)(Re565.0==f

xx k

XhNu Pr < 0.1, Rex < 5x105 4.13

∫=x

xdxhx

h0

1 4.4

Substitution Nux equation in equation 4.3

x

xf

hh

dxxXk

h

2

Pr)(332.00

5.031

21

=

= ∫ −

µρν

31

21

PrRe664.0==f

LL k

XhNu Pr >0.6 , Rex < 5x105 4.14

B Turbulent Flow Rec > 5 ×105 31

8.0 PrRe0288.0==f

xx k

XhNu 0.5<Pr <60 & Rex > 5x105 4.15

C Laminar and Turbulent Flow For mixed boundary layer conditions:

)(1

0min ∫∫ +=

L

xcTurbulent

xc

arLaL dxhdxhL

h

With transition at Rec = 5 x105

)32200(RePr036.0 8.033.0 −==f

LL k

XhNu 0.5<Pr<60 & Rex>5x105 4.16


Mr. Amjed Ahmed 71

8.033.0 RePr036.0==f

LL k

XhNu 0.5<Pr <60 & Rex > 5x105 4.17

Physical Properties of fluid evaluated at the mean film Temperature fT :

2∞+

=TTT s

f 4.18

4.5.2 Empirical correlations for cylinder in cross flow

Figure4.4 Schematic Sketch of the Boundary Layer on a Circular Cylinder Near the Separation Point.

Edge of boundary layer

103 < ReD < 105 (d)

Small turbulent wake

Turbulent boundary layer

Laminar boundary layer

ReD > 105 (e)Figure 4.5 Flow Patterns for Cross-Flow over a Circular Cylinder at Various

Reynolds Numbers.


Mr. Amjed Ahmed 72

A correlation for cylinder at uniform temperature Ts in cross flow of fluid has been

purpose by Whitaker:

41

4.032

21

)(PrRe06.0Re4.0S

DDf

CD k

DhNuµµ

+== 4.18

Figure 4.7 Effect turbulent flow on separation angle.

Ө =80 Ө =140

Figure 4.6 Circumferential Variation of Nusselt number at High Reynolds Numbers for a Circular Cylinder in Cross-Flow.


Mr. Amjed Ahmed 73

0.67< Pr <300 10< ReD<100000 0.25< )(Sµ

µ <5.2

Physical Properties of fluid evaluated at T∞ except Sµ at Ts :

Where: µ

νρDD =Re

by Zukauskas

41

PrPrPrRe )/(Ck

DhNu Snm

f

CD == 4.19

0.7<Pr<500 1<Re<106

n= 0.37 Pr<10 n= 0.36 Pr>10

Where all fluid properties are evaluated at T∞ and Prs at Ts Table 4.3 Constants of Equation 4.11 for external flow

ReD C m 1-40 0.75 0.4

40-1000 0.51 0.5 103 - 2×105 0.26 0.6 2×105 - 106 0.075 0.7

For cylinders with non circular cross sections in gases, Jakob compiled data from two

sources and presented the coefficients of the correlation equation 1/3PrRem

f

CD C

kDhNu == (4.20)

In Table 4.2. In Eq. (4.26) all properties are to be evaluated at Tf.

Table 4.4 Constant for forced convection perpendicular to noncircular tubes

Geometry From To m C ReD


Mr. Amjed Ahmed 74

4.5.3 Empirical Correlations for Sphere

41

4.032

21

)(PrRe06.0Re4.02S

DDf

CD k

DhNuµµ

++== 4.21

0.71< Pr <380 3.5< ReD<76000 1< )/( Sµµ < 3.2

Example 4.1: Derivative equation to average heat transfer coefficient at h(x)= ax-0.1 for flat plate

Solution

∫=x

xdxhx

h0

1 4.4

∫ −=x

dxaxx

h0

1.01

9.0

9.0xxah =

)(1.11.1 1.0 xhaxh == − Example 4.2 A 25µm-diameter polished-platinum wire 6 mm long is to be used for a hot-wire

anemometer to measure the velocity of 20°C air in the range between 2 and 10 m/s (see below Fig). Its temperature is to be maintained at 230°C by adjusting the current. Calculate heat transfer rate required current as a function of air velocity.

Solution Since the wire is very thin, conduction along it can be neglected; also, the temperature

gradient in the wire at any cross section can be disregarded. At the free stream temperature, the air has a thermal conductivity of 0.0251 W/m °C and a kinematic viscosity of 1.57 x105 m2/s. At a velocity of 2 m/s the Reynolds number is

18.31057.1

m) 10 x m/s)(2.5 (2Re 5

-6

=×

=== −vuDD

D µνρ

The Reynolds number range of interest is therefore 1 to 40, so the correlation equation from Eq. (4.77) and Table 3.1 is

41

)Pr(Pr/PrRe Snm

f

CD C

kDhNu ==

C=0.75, m= 0.4 n = 0.37 Neglecting the small variation in Prandtl number from 20° to 230°C, the average

convection heat transfer coefficient as a function of velocity is Ch = 799 U∞0.4 W/m2 °C

)( ∞−= TTAhq ss = (799 U∞0.4)π(25x10-6)(6x10-3)(230-20)

q= 0.079 U∞0.4 W


Mr. Amjed Ahmed 75

Example 4.3 The crankcase of an automobile is approximately 0.6 m long, 0.2 m wide, and 0.1 m

deep. Assuming that the surface temperature of the crankcase is 350 K, estimate the rate of heat flow from the crankcase to atmospheric air at 276 K at a road speed of 30 m/s. Assume that the vibration of the engine and the chassis induce the transition from laminar to turbulent flow so near to the leading edge that, for practical purposes, the boundary layer is turbulent over the entire surface. Neglect radiation and use for the front and rear surfaces the same average convection heat transfer coefficient as for the bottom and sides.

Solution Using the properties of air at 313 K, the Reynolds number is

66 1003.1

10123.196.030092.1Re ×=

×××

== −µνρL

L

From Eq. (4.15) the average Nusselt number is 8.033.0 RePr036.0==

f

LL k

XhNu = 0.036(0.71)0.33 (1.03 x 106)0.8

=2075 the average convection heat transfer coefficient becomes

K W/m91.66.0

0265.02075 2=×

==L

kNuh fL

L

The surface area that dissipates heat is 0.28 m2 and the rate of heat loss from the crankcase is therefore

)( ∞−= TTAhq ss = (91.6 W/m2 K)(0.28) (350 - 276)(K) =1898W

A=0.6x0.2+2(0.1x0.6)+2(0.1x0.2)

A= 0.28 m2


Mr. Amjed Ahmed 76

Example 4.5 A Wire is placed in a 1 atm air stream at 25°C having a flow velocity of 50 m/s a cross

to the wire. An electric current is passed through the wire, raising its surface temperature to 323K. Calculate the heat loss per unit length for the different geometry of wire: (a) A circular shape (diameter 2 x 10-3 m ), (b) A square shape (slat 1.5 x 10-3 m).

Physical properties of air at the film temperature (310K): (Density:1.177 kg/m3, Viscosity: 1.846x10-5 Nm/s2, thermal conductivity: 0.02704 W/m oC and Specific heat: 1.005 x 103 J /Kg oC)

Physical properties of air at the surface temperature(323 K): (Density:0.998 kg/m3, Viscosity: 2.075x10-5 Nm/s2, thermal conductivity: 0.03 W/m oC and Specific heat: 1.009 x 103 J /Kg oC)

Example 4.4 Air flow a flat plate (length 1 m and weight: 0.4 m) with temperature 34 oC and velocity

10 m/s and the plate temperature is 120 oC What is the heat transfer rate from the plate to the air? What is the heat transfer rate if the air velocity is double? What is the heat transfer coefficient if the air velocity is 25 m/s in turbulent flow only? Physical Properties of air is (Density:1.008 kg/m3, Viscosity: 2.075x10-5 Nm/s2,

thermal conductivity: 0.03003 W/m oC and Specific heat: 1.009 x 103 J /Kg oC)


Mr. Amjed Ahmed 77

4.5 External Flow Across Banks of Tubes 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

Figure 4.8 Schematic of a tube bank in cross flow. The tube rows of a bank are either Staggered or Aligned in the direction of the fluid

velocity V (Figure 7.9). The configuration is characterized by the tube diameter D and by the transverse pitch ST and longitudinal pitch SL measured between tube centers

(a) Aligned (b) Staggered.

Figure 4.9 Tube arrangements in a bank.

The heat transfer coefficient associated with a tube is determined by its position in the bank. The coefficient for a tube in the first row is approximately equal to that for a single tube in cross flow, whereas larger heat transfer coefficients are associated with tubes of the inner rows. The tubes of the first few rows act as turbulence grids, which increase the heat transfer coefficient for tubes in the following rows.


Mr. Amjed Ahmed 78

A Grimison has obtained a correlation to know the average heat transfer coefficient is 3/1

max,1 PrRe13.1m

DD CNu = 4.22

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

≥

<<

≥

7.0Pr40000Re2000

10

max,D

N

where C1 and m are listed in Table 4.5

Table 4.5 Constant for fluid over tube bank (N≥10).

All properties appearing in the above equations are evaluated at the film temperature Ts. If N < 10, a correction factor may be applied such that

4.23 where C2 is given in Table 4.6.

Table 4.6 Correction factorC2 of Equation 4.23 for N<10

B Zhukauskas has obtained a correlation to know the average heat transfer coefficient is

4/136.0

max, PrPrPrRe ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

S

mDD CNu 4.24

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

<<

×<<

≥

500Pr7.0102Re1000

206

max,D

N

where all properties except Prs are evaluated at Tf and Prs are evaluated at Ts


Mr. Amjed Ahmed 79

Table 4.7 Constants of Equation 4.24 for the tube bank in cross flow

If N < 20, a correction factor may be applied such that

20320 ≥<=

ND

ND NuCNu 4.25

where C3 is given in Table 4.8. Table 4.8 Correction factorC3 of Equation 4.25 for N<20 and ReD > 103.

The Reynolds number ReD,max is based on the maximum fluid velocity occurring within

the tube bank

µρ DV

Dmax

max,Re = or v

DVD

maxmax,Re =

The mass conservation requirement for an incompressible fluid

(a) Aligned Tube V

DSSV

T

T

−=max

4.26

(b) Staggered Tube V

DSSV

T

T

−=max

at A1 due to A2 > A1

VDS

SVD

T

)(2max −= at A2 due to A2 < A1 4.27

( ) 2/122 )2/(

2

TLD

TD

SSS

DSS

+=

+<


Mr. Amjed Ahmed 80

Start

Grimison Method

Zhukauskas Method

Rows No.≥10

Rows No.>20

3/1max,1 PrRe13.1

m

DD CNu =

C1 & m from Tabel 4.5

µρ DV

Dmax

max,Re = k

Cpµ=Pr

h

q=hA(Ts-T∞)

End

10210 ≥<=

ND

ND NuCNu

C2 From Table 4.6

4/136.0

max, PrPrPrRe ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

S

mDD CNu

C & m from Tabel 4.7

20320 ≥<=

ND

ND NuCNu

C3 From Table 4.8

No

Yes No

Yes

Yes

Yes

No

No

Stop

maxV

Aligned or Staggered Tube

N: No of Row

Row N

Flow Chart Banks of Tubes


Mr. Amjed Ahmed 81

4.5.2 Pressure Drop The power required to move the fluid across the bank is often a major operating expense

and is directly proportional to the pressure drop. Zhukauskas where the pressure drop is given by

fVNxP ⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

2max

2ρ 4.28

The friction factor f and the correction factor x are plotted in Figures 4.10 and 4.11.

Figure 4.10 Friction factor f and correction factor x In Aligned tube bundle.

Figure 4.11 Friction factor f and correction factor x In Staggered tube bundle.

PL =SL /D

PT =ST /D

PL =SL /D

PT =ST /D


Mr. Amjed Ahmed 82

Example 4.5 A tube bundle in which the water is passed through the tubes, while air is passed in

cross flow over the tubes. Consider a staggered arrangement for which the tube outside diameter is 16.4 mm and the longitudinal and transverse pitches are SL= 34.3 mm and ST= 31.3 mm. There are seven rows of tubes in the airflow direction and eight tubes per row. Under typical operating conditions the cylinder surface temperature is at 70° C, while the air upstream temperature and velocity are 15°C and 6 m/s, respectively. Determine

1- the air-side convection coefficient ? 2- the air-side pressure drop?

Solution Properties: Air (T∞ = 15°C): ρ = 1.217 kg/m3, Cp = 1007 J/kg K,

v = 14.82 X 10-6/s, k = 0.0253 W/m K, Pr = 0.710. Air (Ts = 70°C): Pr = 0.701. Air (Tf= 43°C): v = 17.4 X 10-6 m2/s, k = 0.0274 W/m K, Pr = 0.705

1- Since SD = [SL2 + (ST/2)2]1/2 = 37.7 mm is greater than (ST + D)/2=24mm, the maximum

velocity occurs on the transverse plane, A1 of Figure 4.11. Hence from Equation 4.26 with

It follows from Table 4.7 that

4/1

36.0max, Pr

PrPrRe ⎟⎟⎠

⎞⎜⎜⎝

⎛=

S

mDD CNu

2. The pressure drop may be obtained from Equation 4.28

fVNxP ⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

2max

2ρ

C3 from table 4.8


Mr. Amjed Ahmed 83

with 13943Re max, =D and PT=(ST/D) = 1.91 it follows from Figure 4.11 that x= 1.04

with 13943Re max, =D and (PT/PL) = 0.91 it follows from Figure 4.10 that f = 0.35

Hence with N= 7 2/24635.02

6.12217.104.17 mNP =⎟⎠⎞

⎜⎝⎛ ×

×=∆

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Heat transfer lectures 2 (convection)