Heat transfer lectures 2 (convection)

Ch 4: Convection of Heat Transfer

3rd Year College of Technical

Chapter Four Convection Heat Transfer 4.1 Introduction

Convection Heat Transfer

Free Convection (by Bouncy Force)

Force Convection (by Pump or Fan)

Convection with Change Phase (by Boiling & Condensation )

External

Internal

A Plane Wall

The entry region Fully developed

A Cylinder A Sphere

‫اﻟﺤﻤﻞ اﻟﻘﺴﺮي ﻟﻠﺠﺮﻳﺎن اﻟﺨﺎرﺟﻲ‬

4.2 Force Convection for External Flow

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

From energy balance of fluid to determined Heat transfer coefficient: qconv = qcond

h x A (T s - T ∞ ) = − kA

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dT(x) dx

y = 0

4.1

66


h x

=


− k f

dT ( x )

dx T s − T ∞

y = 0

4.2

Where k f : a film thermal conductivity for fluid Local heat transfer coefficient ( h x) 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 (C p , µ , ρ , k) Table (4.1) Ranges of Heat transfer coefficients Values

w/m2 k

Type of Convection

h

Free Convection

5 - 25 25 - 250 50 - 20000 2500 - 100000

gases liquids

Force Convection Boiling & Condensation

dq = h x dA s (T s − T ∞ ) dq = h A s (T s − T ∞ ) h : Average convection coefficient h x : Local convection coefficient and A= w x dA =w dx Equal last equations and substitution substitution A and dA x

hwx (T s

− T ∞ ) = w∫ h x dx(T s − T ∞ )

4.3

0

h=

1 x

x

∫ h dx x

4.4

0

4.3 The Velocity and Thermal Boundary Layers Mechanism of heat transfer by convection: 1. Random molecular motion 2. Bulk motion

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

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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

Figure 4.3 Heat transfer coefficient and velocity distribution. 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 x c. This location is determined by a dimensionless grouping of variables called the Reynolds number νρ X Re x = µ where x is the distance from the leading edge. The critical Reynolds number is the value of ( Re Rec =5×105 ) for which transition begins, and for flow over a flat plate

The velocity boundary layer du dv

dx

+

dy

=0

Continuously equation

∂ 2u v + u = µ 2 dx dy ∂ y

Momentum direction x

∂ 2T +u = α 2 v ∂ y dx y

Energy equation

du

dv

dT

δ =

dT

5 x

Local boundary layer thickness

Re x

Assume a velocity distribution in boundary layer in one dimension flow is u=a+by+cy 2+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 2 2 B.C. 4 d u/dy =0 at y=0 A result is ‫ﻻ‬  ً ‫ﻳﻄﺒﻖ اﻟﺸﺮط اﻟﺤﺪي اﻻول واﻟﺮاﺑﻊ او‬

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68


u

3 y

=

u∞


2 δ x

1 ⎛ y ⎞

3

− ⎜⎜ ⎟⎟ 2 ⎝ δ x ⎠

4.5

Assume a temperature distribution in boundary layer in one dimension is T=a+by+cy 2+dy3 2 3 θ =a+by+cy θ =T-T =a+by+cy +dy =T-T s B.C. 1 θ =0 at y=0 B.C. 2 θ = θ ∞ at y=δ B.C. 3 θ /dy=0 d θ at y=δ 2 2 B.C. 4 d θ /dy =0 at y=0 A result is θ

T − T S

=

θ ∞

− T S

T ∞

=

3 y 2 δ t

1 ⎛ y ⎞

3

− ⎜⎜ ⎟⎟ 2 ⎝ δ t ⎠

4.6

The ratio of the velocity to thermal boundary layer thickness is 1

δt

= 1.026 Pr 3

δ

4.4 Dimensionless Parameters: Table (4.2) Dimensionless Groups of Importance for Heat Transfer and Fluid Flow Group Definition Interpretation h L Biot number Ratio of internal thermal resistance of a solid body to its Bi = surface thermal resistance k s

Nusselt number

Nu =

hc L

ratio of convection heat transfer to conduction in a fluid layer of thickness L

k f

Peclet number

Pe = Re L Pr

Prandtl number

Pr =

Cpµ

Re L

=

Reynolds number Stanton number

St =

=

ν

k α νρ L

Product of Reynolds and Prandtl numbers Ratio of molecular momentum diffusivity to thermal diffusivity Ratio of inertia to viscous forces

µ

Nu L

Dimensionless heat transfer coefficient

Re L Pr

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:

h x

=

k f dT L dx

y =0

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 h L Nu = c 4.8 k f

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The Local Nusselt Number depends only on x, Re, and Pr:

= f(x , Re ,P r )

Nu x

4.9

Once this functional relation is i s 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 Nu l

Nu = f (Re, Pr)

4.10

Nu l = C Re m Pr n

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 Nu x

Nu x h=

=

h x X

=

h x X

1 x

k f

k f

1

1

= 0.332 Re 2 Pr 3

Pr > 0.6, Re x < 5x105

4.12

Pr < 0.1, Re x < 5x105

4.13

1

= 0.565(Re Pr) 2

x

∫ h dx

4.4

x

0

Substitution Nux equation in equation 4.3 1

k f

1 x

ρν 2 3 h= 0.332( ) Pr x −0.5 dx µ X 0

∫

h = 2h x Nu L

=

h L X k f

= 0.664 Re 2 Pr 3

Pr >0.6 , Rex < 5x105

4.14

0.5 5x105

4.15

Rec > 5 ×105

B Turbulent Flow Nu x

1

1

=

h x X k f

1

= 0.0288 Re

0.8

Pr 3

C Laminar and Turbulent Flow For mixed boundary layer conditions:

h L

=

1

L

xc

∫

L

∫

( h La min ar dx + hTurbulent dx) 0

xc

With transition at Re c = 5 x105

Nu L

=

h L X

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k f

= 0.036 Pr 0.33 (Re 0.8 − 32200)

5

0.55x10

4.16

70


Nu L

=

h L X k f


= 0.036 Pr 0.33 Re 0.8

0.5 5x105

4.17

Physical Properties of fluid evaluated at the mean film Temperature T f : T f

=

T s

+ T ∞

4.18

2

4.5.2 Empirical correlations for cylinder in cross flow Edge of boundary layer

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

103 < ReD < 105 (d) Small turbulent wake Turbulent boundary layer

Laminar boundary layer

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

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

Ө =140

Ө =80

Figure 4.7 Effect turbulent flow on separation angle.

A correlation for cylinder at uniform temperature T s in cross flow of fluid has been purpose by Whitaker : Nu D

Mr. Amjed Ahmed

=

h C D k f

1

2

= 0.4 Re D 2 + 0.06 Re D 3 Pr

0.4

(

µ µ S

1

)4

4.18

72



0.67< Pr <300

10< Re D<100000

Physical Properties of fluid evaluated at T ∞ except

Re D

Where:

=

0.25< (

µ S

) <5.2

at T s :

S

νρ D µ

by Zukauskas

Nu D

=

h C D k f

1

= C Re Pr ( Pr / Pr S ) 4 m

n

4.19

0.710 Where all fluid properties are evaluated at T ∞ and Pr s at T s Table 4.3 Constants of Equation 4.11 for external flow Re D C 1-40 0.75 40-1000 0.51 3 5 10 - 2×10 0.26 5 6 2×10 - 10 0.075

m 0.4 0.5 0.6 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

Nu D

=

h C D k f

= C Re m Pr 1/3

(4.20)

In Table 4.2. In Eq. (4.26) all properties are to be evaluated at T f . Table 4.4 Constant for forced convection perpendicular to noncircular tubes Re D

Geometry

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From

To

m

C

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4.5.3 Empirical Correlations for Sphere

=

Nu D

h C D

1

2

= 2 + 0.4 Re D 2 + 0.06 Re D 3 Pr

k f

0.4

(

µ µ S

1

)4

4.21

0.71< Pr <380 3.5< ReD<76000 1< (

/

S

) < 3.2

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

h=

h=

1 x

1

x

∫ h dx

4.4

x

0

x

∫ ax

−0.1

x 0

h=

dx

h = 1.1ax −0.1

a x 0.9 x 0.9

= 1.1h( x )

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 x10 5 2 m /s. At a velocity of 2 m/s the Reynolds number is

=

D νρ

=

h C D

uD

(2 m/s)(2.5 x 10 -6 m)

= = = 3.18 v µ 1.57 ×10 −5 The Reynolds number range of interest is therefore 1 to 40, so the correlation equation from Eq. (4.77) and Table 3.1 is Re D

Nu D

1

= C Re

k f

m

n

Pr (Pr/ Pr S ) 4

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

h C = 799 U ∞0.4

2

W/m °C

q = h A s (T s − T ∞ ) = (799 U ∞0.4 )π (25x10 (25x10 -6 )(6x10 -3 )(230-20) 0.4

q= 0.079 U ∞

Mr. Amjed Ahmed

W

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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.

A=0.6x0.2+2(0.1x0.6)+2(0.1x0.2) A= 0.28 m

2

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

Re L

=

νρ L µ

=

1.092 × 30 × 0.6 19.123 ×10

−6

= 1.03 ×10 6

From Eq. (4.15) the average Nusselt number is

Nu L

=

h L X k f

= 0.036 Pr 0.33 Re 0.8 = 0.036(0.71)0.33 (1.03 x 10 6)0.8

=2075 the average convection heat transfer coefficient becomes

h L

=

Nu L k f

=

2075 × 0.0265

= 91.6 W/m 2 K

0 .6 The surface area that dissipates heat is 0.28 m 2 and the rate of heat loss from the crankcase is therefore

L

q = h A s (T s − T ∞ ) = (91.6 W/m2 K)(0.28) (350 - 276)(K) =1898W

Mr. Amjed Ahmed

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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). 3 Physical properties of air at the film temperature (310K): (Density:1.177 kg/m , Viscosity: 1.846 x10-5 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/m 3, Viscosity: 2.075x10-5 Nm/s2, thermal conductivity: 0.03 W/m oC and Specific heat: 1.009 x 3 o 10 J /Kg C) 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/m 3, Viscosity: 2.075x10-5 Nm/s 2, thermal conductivity: 0.03003 W/m oC and Specific heat: 1.009 x 103 J /Kg oC)

Mr. Amjed Ahmed

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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 Figure 7.9 ). The configuration is characterized by the tube diameter D D and by the transverse pitch S T T and longitudinal pitch S L 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.

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A Grimison has obtained a correlation to know the average heat transfer coefficient is m

Nu D = 1.13C 1 Re D ,max Pr 1/ 3

4.22

⎡ N ≥ 10 ⎤ ⎢ ⎥ ⎢2000 < Re D ,max < 40000 ⎥ ⎢ Pr ≥ 0.7 ⎥ ⎣ ⎦ where C 1 and m are listed in Table 4.5 Table 4.5 Constant for fluid over tube bank (N ≥ 10). 10).

All properties appearing in the above equations are evaluated at the film temperature T s. If N N < 10, a correction factor may be applied such that 4.23

where C 2 is given in Table 4.6. Table 4.6 Correction factorC 2 of Equation 4.23 for N<10

B Zhukauskas has obtained a correlation to know the average heat transfer coefficient is 1/ 4

Nu D = C Re

m D ,max

⎛ Pr ⎞ ⎟⎟ Pr ⎜⎜ Pr ⎝ S ⎠ 0.36

4.24 ⎡ N ≥ 20 ⎤ ⎢ 6⎥ ⎢1000 < Re D ,max < 2 × 10 ⎥ ⎢0.7 < Pr < 500 ⎥ ⎣ ⎦

where all properties except Pr s are evaluated at T f and Pr s are evaluated at T s

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

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

Nu D

N < 20

= C 3 Nu D

4.25

N ≥ 20

where C 3 is given in Table 4.8. 3 Table 4.8 Correction factor C C 3 of Equation Equation 4.25 for N<20 and Re D > 10 .

The Reynolds number Re Re D ,max is based on the maximum fluid velocity occurring within the tube bank Re D, max =

ρ V max D

or

µ

Re D ,max =

V max D v

The mass conservation requirement for an incompressible fluid (a) Aligned Tube V max =

S T S T − D

4.26

V

(b) Staggered Tube V max = V max =

S T S T − D

at A1

V

S T 2( S D − D )

V

at A2

due to

A2 > A1

due to

A2 < A1 S D <

S T + D 2

(

S D = S L + ( S T / 2) 2

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4.27

2

)

1/ 2

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Start

Flow Chart Banks of Tubes

V max Aligned or Staggered Tube

Re D,max =

ρ V max D

Pr =

Cpµ

µ

k

Grimison

Yes

Method m

Nu D = 1.13C 1 Re D ,max Pr 1/ 3

No Yes

C 1 & m from Tabel 4.5

Zhukauskas Method 1/ 4

Nu D = C Re

m D ,max

⎛ Pr ⎞ ⎟⎟ Pr 0.36 ⎜⎜ Pr ⎝ S ⎠

No

C & m from Tabel 4.7

No

Stop

No

Rows No.>20

Rows No.≥10

Yes

Nu D Yes

N < 20

= C 3 Nu D

Nu D

N ≥ 20

C 3 From Table 4.8

N <10

= C 2 Nu D

N ≥10

C 2 From Table 4.6

h q=hA(T s-T ∞ )

End

Row N

N: No of Row

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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

⎛ ρ V 2 max ⎞ ⎟⎟ f ∆ P = Nx⎜⎜ 2 ⎝ ⎠

4.28

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

P L =S L /D P T T =S T T /D

Figure 4.10 Friction factor f f and correction factor x x In In Aligned tube Aligned tube bundle. P L =S L /D P T T =S T T /D

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

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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 S L= 34.3 mm and S T T = 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: 3 Air (T = 15°C): ρ = 1.217 kg/m , C p = 1007 J/kg K, v = 14.82 X 10 -6 /s, k = 0.0253 W/m K, Pr = 0.710. Air (T s = 70°C): Pr = 0.701. -6 2 Air (T f = 43°C): v = 17.4 X 10 m /s, k = 0.0274 W/m K, Pr = 0.705 ∞

2 1/2 1- Since S D = [S L2 + (S T ] = 37.7 mm is greater than (S T T /2) T + D)/2=24mm, the maximum velocity occurs on the transverse plane, A1 of Figure 4.11. Hence from f rom Equation 4.26

with

It follows from Table 4.7 that

1/ 4

Nu D = C Re

m D ,max

⎛ Pr ⎞ ⎟⎟ Pr ⎜⎜ Pr ⎝ S ⎠

C 3 from table 4.8

0.36

2. The pressure drop may be obtained from Equation 4.28

⎛ ρ V 2 max ⎞ ⎟⎟ f ∆ P = Nx⎜⎜ 2 ⎝ ⎠

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with Re D ,max = 13943 and P T = 1.91 it follows from Figure 4.11 that x= 1.04 T=(S =(S T T /D) with Re D ,max = 13943 and (P T T /P L ) = 0.91 it follows from Figure 4.10 that f = 0.35 Hence with N= 7

Mr. Amjed Ahmed

⎛ 1.217 × 12.6 ⎞ 2 ∆ P = 7 × 1.04⎜ ⎟0.35 = 246 N / m 2 ⎝ ⎠

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Heat transfer lectures 2 (convection)

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