Chapter 5

Principles of Force Convection

5.1 Introduction

The preceding chapters have considered the mechanism and calculation of conduction heat transfer. Convection was considered only insofar as it related to the boundary conditions imposed on a conduction problem. We now wish to examine the methods of calculating convection heat transfer and, in particular, the ways of predicting the value of the convection heat-transfer coefficient h.

Our development in this chapter is primarily analytical in character and is concerned only with forced-convection flow systems. Subsequent chapters will present empirical relations for calculating forced-convection heat transfer and will also treat the subjects of natural convection.

5.2 Viscous Flow 5.2.1 Flow on a flat plate (external flow)

Consider the flow over a flat plate as shown in Figures 5.1 and 5.2. Beginning at the leading edge of the plate, a region develops where the influence of viscous forces is felt. These viscous forces are described in terms of a shear stress between the fluid layers. If this stress is assumed to be proportional to the normal velocity휏 gradient, we have the defining equation for the viscosity,

= 5.1 푑푢 The휏 휇region푑푦 of flow that develops from the leading edge of the plate in which the effects of viscosity are observed is called the boundary layer. Some arbitrary point is used to designate the y position where the boundary layer ends; this point is usually chosen as the y coordinate where the velocity becomes 99 percent of the free-stream value.

Initially, the boundary-layer development is laminar, but at some critical distance from the leading edge, depending on the flow field and fluid properties, small disturbances in the flow begin to become amplified, and a transition process takes place until the flow becomes turbulent. The turbulent-flow region may be pictured as a random churning action with chunks of fluid moving to and fro in all directions. The transition from laminar to turbulent flow occurs when

= = > 5 10 (transiant) 푢∞푥 휌푢∞푥 5 휐 5 휇10 푅푒(laminar)∗ 5 Where푅푒 ≤ ∗ = free stream velocity, m/s.

푢 ∞= distance from leading edge, m. 2 푥 = =kinematic viscosity, m /s. 휇 휐 휌 = 5.2 푢∞푥 푅푒 푥 휐

Figure 5.1: Sketch showing different boundary-layer flow regimes on a flat plate.

Figure 5.2: Laminar velocity profile on a flat plate.

5.3 Laminar Boundary Layer on a Flat Plate

Consider the elemental control volume shown in Figure 5.3. We derive the equation of motion for the boundary layer by making a force-and-momentum balance on this element. To simplify the analysis we assume: 1. The fluid is incompressible and the flow is steady. 2. There are no pressure variations in the direction perpendicular to the plate. 3. The viscosity is constant. 4. Viscous-shear forces in the y direction are negligible.

Figure 5.4 Elemental control volume for force balance on laminar boundary layer.

The mass continuity equation for the boundary layer.

+ = 0 5.3 T휕푢he momentum휕푣 equation of the laminar boundary layer with constant properties. 휕푥 휕푦 + = 2 5.4 휕푢 휕푢 휕 푢 휕푝 2 휌 �푢 휕푥 푣 휕푦� 휇 휕푦 − 휕푥

Figure 5.5: Elemental control volume for integral momentum analysis of laminar boundary layer.

The mass flow through plane 1 in Figure 5.5 is

= 퐻 푚Voṅ Kármán∫0 휌푢 approximate푑푦 solution of equations 5.3 and 5.4 gives

. = 4 64푥 1 �2 훿 푅푒푥 The exact solution of the boundary-layer equations

= 5.5 5푥 1 �2 훿 푅푒푥 The velosity profile of the stream in x-direction within the baundary layer is given by:

= 5.6 푢 3 푦 1 푦 3 ∞ 푢And the2 훿 energy− 2 �훿 �equation of the laminar boundary layer is:

+ = + 2 5.7 휕푇 휕푇 휕 푇 휇 휕푢 2 2 푢 휕푥 푣 휕푦 훼 휕푦 휌푐푝 �휕푦� For low velocity incompressible flow, we have

+ = 2 5.8 휕푇 휕푇 휕 푇 2 There푢 휕푥 is푣 a휕푦 striking훼 휕푦 similarity between equation 5.8 and the momentum equation for costant pressure,

+ = 2 5.9 휕푢 휕푢 휕 푢 2 푢 휕푥 푣 휕푦 훼 휕푦

Example 5.1: Mass Flow and Boundary-Layer Thickness -3 Air at 27◦C and 1 atm flows over a flat plate at a speed of 2 m/s. Calculate the boundary-layer thickness at distances of 20 cm and 40 cm from the leading edge of the plate. Calculate the mass flow that enters the boundary layer between = 20 cm −5 and = 40 cm. The viscosity of air at 27◦C is 1.85×10 kg/m· s. Assume푥 unit depth in the푥 z direction. Solution The density of air is calculated from . = = = 1.177 kg/m3 ( )( 5 ) 푝 1 0132∗10 휌 =푅푇 287 27+273 푢∞푥 . . 푅푒푥 = 20휐 = = = = 25,448 At : . × 푢∞푥 휌푢∞푥 1 177∗2∗0 2 푥 . . 푥 = 40 푐푚 푅푒 = 휐 = 휇 1 85=1050−5,896 At : . × 휌푢∞푥 1 177∗2∗0 4 The푥 boundary푐푚 layer푅푒푥 thickness휇 is1 calculated85 10−5 from:

. = 4 64푥 1 �2 훿 푅푒푥 . . At = 20 , = = 0.00582 m. ( ) 4 64∗0 2 1 �2 푥 푐푚 훿 25488. . At = 40 , = = 0.00823 m. ( ) 4 64∗0 4 1� 푥 푐푚 훿 50896 2 To calculate the mass flow that enters the boundary layer from the free stream between = 20 cm and = 40 cm, we simply take the difference between the mass flow in the푥 boundary layer푥 at these two x positions. At any x position the mass flow in the boundary layer is given by the integral

= 훿 0 푚̇ = ∫ 휌푢 푑푦 3 푦 1 푦 3 ∞ 2 훿 2 훿 푢 푢 − 3� � 1 3 5 = = 2 2 8 훿 푦 푦 ∞ ∞ 0 푢 훿 − 훿 푢 훿 푚̇ =∫ 휌 ( � )�=푑푦(1.177휌 2(0.00823 0.00582) = 0.00354 kg/s 5 5 ∞ 40 20 ∆푚̇ 8 휌푢 훿 − 훿 8 ∗ −

5.4 Thermal boundary layer

Just as the hydrodynamic boundary layer was defined as that region of the flow where viscous forces are felt, a thermal boundary layer may be defined as that region where temperature gradients are present in the flow. These temperature gradients would result from a heat-exchange process between the fluid and the wall.

Consider the system shown in Figure 5.6. The temperature of the wall is , the temperature of the fluid outside the thermal boundary layer is , and the thickness푇푤 of the thermal boundary layer is designated as . At the wall, the푇∞ velocity is zero, and the heat transfer into the fluid takes place by훿 conduction.푡 Thus the local heat flux per unit area is:

= 5.10 푞 휕푇 퐴 휕푦 From− Newton’s푘 �푤푎푙푙 law of cooling

= ( ) 5.11 푞 푤 ∞ where퐴 ℎ h 푇is the− 푇convection heat-transfer coefficient. Combining these equations, we have

= 휕푇 ( ) 5.12 −푘휕푦�푤푎푙푙 ℎ 푇푤−푇∞

Figure 5.6: Temperature profile in the thermal boundary layer.

Then we need only find the temperature gradiant at the wall to evaluate . Therefore, the temperature distribution is: ℎ = = 5.13 휃 푇−푇푤 3 푦 1 푦 3 휃∞ 푇∞−푇푤 2 훿푡 − 2 �훿푡�

The thermal boundary layer can calculated from the equation below:

= 1 1 5.14 . 3 �3 훿푡 1 −1 푥표 �4 �3 훿 1 026 푃푟 � − � 푥 � � When the plate is heated over the entire length, = 0, and

표 = 푥 . 5.15 훿푡 1 −1 �3 Where훿 1 026 the푃푟 Prandtl number is dimensionless when a consistent set of units is used:

= = = 5.16 휇 휐 �휌 푐푝휇 푘 푃푟 훼 �휌푐푝 푘 The local convective heat transfer coefficient is calculated from the equation as below:

= 0.332 1 −1 5.17 1 3 �3 1 푢∞ �2 푥표 �4 �3 ℎ푥 푘 푃푟 � 휐푥 � � − � 푥 � � The equation may be nondimensionalized by multiplying both sides by , producing the dimensionless group on the left side, 푥 푘 = 5.18 ℎ푥푥 푥 called푁푢 the Nusselt푘 number afterWilhelm Nusselt, who made significant contributions to the theory of convection heat transfer. Finally,

= 0.332 1 −1 5.19 3 �3 1 1 푥표 �4 �3 �2 푁푢푥 푃푟 푅푒푥 � − � 푥 � � or, for the plate heated over its entire length, = 0 and

표 = 0.332 푥 0.6 < > 50 5.20 a 1 1 �3 �2 푁푢푥 푃푟 푅푒푥 푃푟 Equation (5.20 a) is applicable to fluids having Prandtl numbers between about 0.6 and 50. It would not apply to fluids with very low Prandtl numbers like liquid metals or to high- Prandtl-number fluids like heavy oils or silicones. For a very wide range of Prandtl numbers, Churchill and Ozoe have correlated a large amount of data to give the following relation for laminar flow on an isothermal flat plate:

. = 1 1 > 100 5.20 b �2 �3 0 3387.푅푒푥 푃푟 1 푥 2 �4 푥 푁푢 0 0468 �3 푅푒 푃푟 �1+� 푃푟 � � Equations (5.17), (5.19), and (5.20 a) express the local values of the heat-transfer coefficient in terms of the distance from the leading edge of the plate and the fluid properties. For the case where = 0 the average heat-transfer coefficient and

Nusselt number may be obtained by푥표 integrating over the length of the plate:

= 퐿 = 2 5.21 ∫0 ℎ푥푑푥 � 퐿 푥=퐿 assumingℎ ∫0 푑푥 the heatedℎ section is at the constant temperature . For the plate heated

over the entire length, 푇푤 = = 2 5.22 ℎ�퐿 푁푢Or���� 퐿 푘 푁푢푥=퐿

= = 0.664 5.23 ℎ�퐿 1� 1 2 �3 퐿 퐿 Where:푁푢���� 푘 푅푒 푃푟

= 5.24 휌푢∞퐿 퐿 For푅푒 a plate휇 where heating starts at = , it can be shown that the average heat

transfer coefficient can be expressed as푥 푥표

= 2 3� 푥표 4 1−� �퐿� 표 ℎ �푥 −퐿 ℎ푥=퐿 � 퐿 퐿−푥표 � In this case, the total heat transfer for the plate would be

= ( )( )

표 The푞푡표푡푎푙 foregoingℎ�푥 −퐿 analysis퐿 − 푥표 was푇푤 based− 푇∞ on the assumption that the fluid properties were constant throughout the flow. When there is an appreciable variation between wall and free-stream conditions, it is recommended that the properties be evaluated at the so-called film temperature , defined as the arithmetic mean between the wall and free-stream temperature, 푇푓 = 5.25 푇푤+푇∞ 푇푓 2 Constant Heat Flux

The above analysis has considered the laminar heat transfer from an isothermal surface. In many practical problems the surface heat flux is essentially constant, and the objective is to find the distribution of the plate-surface temperature for given fluid-flow conditions. For the constant-heat-flux case it can be shown that the local Nusselt number is given by

= 0.453 5.26 1 1 �2 �3 푁푢푥 푅푒푥 푃푟 which may be expressed in terms of the wall heat flux and temperature difference as

= ( ) 5.27 푞푤푥 푁푢푥 푘 푇푤−푇∞ Where: : heat flux, W/m2

푞Note푤 that the heat flux = is assumed constant over the entire plate surface. 푞 푞푤 퐴 = ( ) = 1 퐿 1 퐿 푞푤푥

푇��푤���−���푇��∞� 퐿 ∫0 푇푤 − 푇∞ 푑푥 퐿 ∫0 푘푁푢푥 푑푥 = 5.28 . 푤퐿 푞 �푘 1� 1 푤 ∞ 2 �3 푇�����−���푇��� 0 6795푅푒퐿 푃푟 = ( ) 3 푤 푥=퐿 푞 2 ℎ 푇푤 − 푇∞ For constant heat flux case and the properties evaluated at the film temperature:

. = 1 1 > 100 5.29 �2 �3 0 4637.푅푒푥 푃푟 1 푥 2 �4 푥 푁푢 0 0207 �3 푅푒 푃푟 �1+� 푃푟 � � Example 5.1: -3 Air at 27◦C and 1 atm flows over a flat plate at a speed of 2 m/s. Calculate the boundary-layer thickness at distances of 20 cm and 40 cm from the leading edge of the plate. Calculate the mass flow that enters the boundary layer between = 20 cm −5 and = 40 cm. The viscosity of air at 27◦C is 1.85×10 kg/m· s. Assume푥 unit depth in the푥 z direction. Example 5.2:Isothermal Flat Plate Heated Over Entire Length XAMPLE 5-4 For the flow system in Example 5.1 assume that the plate is heated over its entire length to a temperature of 60◦C. Calculate the heat transferred in (a) the first 20 cm of the plate and (b) the first 40 cm of the plate. Solution

= , = 27 , = 2 m/s, =60˚C 푇푤+푇∞ 푇푓 2 푇∞ ˚퐶 푢∞ 푇푤 = = 43.5 + 273 = 316.5 K 60+27 푇We푓 find2 the properties of air at film temperature. = 17.36 10 m2/s, = 0.7 , = 0.02749 W/m.˚C. −6 휐 = 1.006∗ Kj/kg. ˚C. 푃푟 푘 푐At푝 = . = = = 23041 풙 ퟐퟎ 풄풎 . 푢∞푥 2∗0 2 푥 −6 푅푒 = 휐 17 36∗10 ℎ푥푥 푥 푁푢 = 0푘.332 = 0.332 (0.7) (23041) = 44.74 1 1 1 1 �3 �2 �3 �2 푥 푥 . 푁푢= =푃푟44.74푅푒 = 6.15 2 ∗ . W/m . ˚C 푘 0 02749 푥 푥 Theℎ average푁푢 푥 value of the∗ heat0 2 -transfer coefficient is twice this value, or

= 2 = 2 6.15 = 12.30

ℎ� = ℎ푥( ∗ ) = 12.3 (0.2)(60 27) = 81.18 W

At푞 ℎ�=퐴 푇푤 − 푇 ∞ ∗ − . = = = 46082 풙 ퟒퟎ 풄풎 . 푢∞푥 2∗0 4 푥 −6 푅푒 = 휐 17 36∗10 ℎ푥푥 푥 푁푢 = 0푘.332 = 0.332 (0.7) (46082) = 63.28 1 1 1 1 �3 �2 �3 �2 푥 푥 . 푁푢= =푃푟63.28푅푒 = 4.349 2∗ . W/m . ˚C 푘 0 02749 ℎ푥 푁푢푥 푥 ∗ 0 4 The average value of the heat-transfer coefficient is twice this value, or

= 2 = 2 4.349 = 8.698

ℎ� = ℎ푥( ∗ ) = 8.698 (0.4)(60 27) = 114.8 W

푞 ℎ�퐴 푇푤 − 푇∞ ∗ −

EXAMPLE 5.3: Flat Plate with Constant Heat Flux A 1.0-kW heater is constructed of a glass plate with an electrically conducting film that produces a constant heat flux. The plate is 60 cm by 60 cm and placed in an airstream at 27◦C, 1 atm with =5 m/s. Calculate the average temperature

difference along the plate. 푢∞ Solution Properties should be evaluated at the film temperature, but we do not know the plate temperature. for an initial calculation, we take the properties at the free-stream conditions of At = 27 ˚C we find the properties of the fluid 2 =푇∞15.96 10 m /s, = 0.708 , = 0.02624 W/m.˚C. −6 . = = = 1.88 10 휐 ∗ . 푃푟 푘therefore the flow is laminar 푢∞퐿 5∗0 6 5 −6 푅푒퐿 휐 15 96∗10 ∗ = . 푤퐿 푞 �푘 1� 1 푤 ∞ 2 �3 푇�����−���푇��� 0 6795푅푒퐿 푃푟 . . . = 1000 = 241.85 ˚C . ( . 0 6 ) ( . ) � 0 62 �� �0 02624� 푤 ∞ 5 1� 1� 푇�����−���푇��� 0 6795 1 88∗10 2 0 708 3 = 241.85 + 27 = 268.85 ˚C

푇Now푤 we find the properties at the film temperature

. = = = (147.927 + 273) = 421 푇푤+푇∞ 268 85+27 푓 푇 = 28.222 10 2m2/s, = 0.687 , = 0.035 W/m.˚C.퐾 . −6 = = 1.06 10 휐 . ∗ 푃푟 푘 5∗0 6 5 퐿 −6 푅푒 28 22∗10 . ∗ . . = 1000 = 243.6 ˚C . ( . 0 6) ( . ) � 0 62 �� �0 035 � 푤 ∞ 5 1� 1� 푇�����−���푇��� 0 6795 1 06∗10 2 0 687 3 = ( ) 푞푤푥 푁푢푥 푘 푇푤−푇∞ = 푞푤푥 푇푤 − 푇∞ 푘푁푢푥 = 푞푤푥 푇푤 − 푇∞ 푘푁푢푥 = 0.453 1 1 �2 � 푥 푥 3 푁푢 = 푅푒 푃푟 ( ) . 푞푤푥 푤 ∞ 1� 1� 푇 − 푇 푘 0 453 푅푒푥 2푃푟 3 At =

푥 퐿= ( ) . 푞푤퐿 1� 1� 푤 ∞ 퐿 2 3 푇 − 푇 푘 0 453 푅푒 . 푃푟 . . = 1000 = 365.9 ˚C . ( . 0 6) ( . ) � 0 62 �� �0 035 � 푤 ∞ 5 1� 1� 푇 − 푇 0 453 1 06∗10 2 0 687 3

EXAMPLE 5.4:Plate with Unheated Starting Length Air at 1 atm and 300 K flows across a 20-cm-square plate at a free-stream velocity of 20 m/s. The last half of the plate is heated to a constant temperature of 350 K. Calculate the heat lost by the plate. Solution

First we evaluate the air properties at the film temperature

= = = 325 푇푤+푇∞ 350+300 푓 푇 = 18.232 10 2 m2/s, = 0퐾.02814 W/m.˚C, = 0.7 −6 휐At = ∗ 푘 푃푟

. 푥= 퐿 = = 2.194 10 . 푢∞퐿 20∗0 2 5 퐿 −6 푅푒 < 5휐 10 18 23∗10 ∗ 5 푅푒Therefore퐿 ∗ the flow is laminar

= 0.332 1 −1 1 3 �3 1 푢∞ �2 푥표 �4 �3 ℎ푥 푘 푃푟 � 휐푥 � � − � 푥 � � . = 0.332 0.02814( 0.7) 1 −1� . . 1� . 3� 3 1� 20 2 0 1 4 3 −6 ℎ퐿 ∗ �18 23∗10 ∗0 2� � − �0 2� � = 26.253 W/m2.˚C.

ℎ퐿 = 2 3� 푥표 4 1−� �퐿� 표 ℎ�푥 −퐿 ℎ푥=퐿 � 퐿 퐿−푥표 � . . = 26.253 2(0.2) 3� 0. 1 . 4 1−� �0 2� 표 ℎ�푥 −퐿 � 0 2−0 1 � = 42.566 W/m2.˚C

푥표−퐿 ℎ� = ( ) ( )

푥표−퐿 표 푤 ∞ 푞 = 42ℎ� .566퐿(0−.2푥 0∗.1푤) 푇0.2−(350푇 300) = 42.566 W

EXAMPLE푞 5.5:− Oil Flow∗ Over Heated− Flat Plate Engine oil at 20◦C is forced over a 20-cm-square plate at a velocity of 1.2 m/s. The plate is heated to a uniform temperature of 60◦C. Calculate the heat lost by the plate. Solution We first evaluate the film temperature:

= = = 40 + 273 = 313 K 푇푤+푇∞ 60+20 푓 푇 = 0.000242 m2/s,2 = 0.144 W/m.˚C, = 2870, = 876 kg/m3

. . 휐 = = 푘 = 1000 푃푟 휌 . 푢∞푥 1 2∗0 2 푥 휐 0 00024 푅푒 . . ( ) ( ) = 1 1 = 1 1 = 152.2 �2 �3 �2 �3 0 3387.푅푒푥 푃푟 0 3387 1000. 2870 1 1 푥 2 �4 2 �4 푁푢 0 0468 �3 0 0468 �3 �1+� 푃푟 � � �1+� 2870 � � . = = 152.2 = 109.6 2 . W/m . ˚C 푘 0 144 푥 푥 ℎ = 2푁푢=푥2 109.6∗= 02192 .2 W/m2. ˚C

ℎ� = ℎ푥( ∗ ) = 219.2 (0.2) (60 20) = 350.6 W 2 푤 ∞ 5.5푞 Theℎ�퐴 푇 Relation− 푇 Between∗ Fluid Friction− and Heat Transfer

We have already seen that the temperature and flow fields are related. Now we seek an expression whereby the frictional resistance may be directly related to heat transfer.

The shear stress at the wall may be expressed in terms of a friction coefficient :

퐶푓 = 2 5.30 휌푢∞ 휏푤 퐶푓 2 The exact solution of the boundary-layer equations yields

= 0.332 5.31 퐶푓푥 −1 �2 푥 2 = 0.332푅푒 5.20 a 1 1 �3 �2 푁푢Equation푥 (5.20 a)푃푟 may be푅푒 rewritten푥 in the following form:

= = 0.332 푁푢푥 ℎ푥 −2 −1 �3 �2 푅푒푥푃푟 휌푐푝푢∞ 푥 The group on the left is called푃푟 the Stanton푅푒 number,

= ℎ푥 푆푡푥 휌푐푝푢∞ So that

= 0.332 5.32 2 1 �3 �2 Upon푆푡푥푃푟 comparing Equations푅푒푥 (5.31) and (5.32), we note that the right sides are alike except for a difference of about 3 percent in the constant, which is the result of the approximate nature of the integral boundary-layer analysis. We recognize this approximation And write

= 0.6 < < 60 5.33 푓푥 2� 퐶 푥 3 Equation푆푡 푃푟 (5.323), called the Reynolds-Colburn analogy,푝푟 expresses the relation between fluid friction and heat transfer for laminar flow on a flat plate. The heat-transfer coefficient thus could be determined by making measurements of the frictional drag on a plate under conditions in which no heat transfer is involved. It turns out that Equation (5.33) can also be applied to turbulent flow over a flat plate and in a modified way to turbulent flow in a tube. It does not apply to laminar tube flow.

EXAMPLE 5.3: A 1.0-kW heater is constructed of a glass plate with an electrically conducting film that produces a constant heat flux. The plate is 60 cm by 60 cm and placed in an airstream at 27◦C, 1 atm with =5 m/s. Calculate the average temperature difference along the plate. 푢∞ EXAMPLE 5.6 Drag Force on a Flat Plate For the flow system in Example 5.2 compute the drag force exerted on the first 40 cm of the plate using the analogy between fluid friction and heat transfer. Solution We use Equation (5.33) to compute the friction coefficient and then calculate the drag force. An average friction coefficient is desired, so

= 2 퐶푓푥 �3 푆푡푥푃푟 2 From example 5.2

= , = 27 , = 2 m/s, =60˚C, = 1.006 Kj/kg. ˚C. 푇푤+푇∞ 푇 푓 2 푇∞ ˚퐶 푢∞ 푇푤 푐푝

= = 43.5 + 273 = 316.5 K 60+27 푓 . 푇 = =2 = 1.115 kg/m3 ( )( . 5) 푝 1 0132∗10 For휌 40푅푇 cm 287length316 5 From example 5.2

= 8.698

� . ℎ = = = 3.88 10 . ℎ� 8 698 −3 푆푡� 휌푐푝푢∞ 1 115∗1006∗2 ∗ = = 3.88 10 (0.7) = 3.06 10 퐶���푓� 2 2 �3 −3 �3 −3 2 푆푡� 푃푟 ∗ ∗ = = 3.06 10 1.115 (2) = 0.0136 2 2 N/m 휌푢∞ −3 2 푤 ��푓� 2 휏����= 퐶 = 0.0136 ∗0.4 = ∗5.44 N.m∗

퐷 휏���푤�퐿 ∗

5.6 Turbulent-Boundary-Layer Heat Transfer

Schlichting has surveyed experimental measurements of friction coefficients for turbulent flow on flat plates. We present the results of that survey so that they may be employed in the calculation of turbulent heat transfer with the fluid-friction–heat- transfer analogy. The local skin-friction coefficient is given by

= 0.0592 1 5 10 < < 10 5.34 −5 . 5 7 퐶푓푥 = 0.370(log푅푒푥 ) ∗ 10 푅푒<푥 < 10 5.35 −2 584 7 9 퐶The푓푥 average-friction푅푒푥 coefficient for a flat푅푒 plate푥 with a laminar boundary layer up to and turbulent thereafter can be calculated from

crit 푅푒 . = < 10 ( ) . for laminar and turbulent 5.36 0 455 퐴 9 2 584 where퐶푓̅ logthe푅푒 constant퐿 − 푅푒A 퐿depends 푅푒on퐿 in accordance with Table 5.1. A somewhat

simpler formula can be obtained for푅푒 lowercrit Reynolds numbers as

. = < 10 5.37 0 074 퐴 7 푓 1 퐿 ̅ 5 푅푒퐿 퐶 푅푒퐿 − 푅푒

Applying the fluid-friction analogy

/ = we obtain2 3 the퐶푓 local turbulent heat transfer as: 푆푡 푃푟 2

/ = 0.0296 5 10 < < 10 1 5.38 2 3 −5 5 7 푆푡Or푥 푃푟 푅푒푥 ∗ 푅푒푥 / = 0.185(log ) . 10 < < 10 5.39 The average2 3 heat transfer over the entire−2 584 laminar-turbulent7 boundary layer9 is 푆푡푥 푃푟 푅푒푥 푅푒푥

/ = 5.40 2 3 퐶푓 푆푡 � 푃푟 2

For = 5 × 10 and < 10 , Equation (5.37) can be used to obtain 5 7 푐푟푖푡 퐿 푅푒 / = 0.037 푅푒 871 5.41 1 2 3 − �5 −1 푆푡� =푃푟 푅푒퐿 − 푅푒퐿 ( ) 푁푢���� � 푅푒퐿Pr 푆푡 = = (0.037 . 871) 1 5.42 ℎ퐿 3 0 8 푁푢�For��� 퐿higher푘 Reynolds푃푟 numbers푅푒 the퐿 friction− coefficient from Equation (5.36) may be used, so that For 10 < < 10 and = 5 × 10 7 9 5 = = [0퐿 .228 (log 푐푟푖푡) . 871] 푅푒 푅푒 1 5.43 � ℎ퐿 −2 584 3 푁푢 퐿 푘 푅푒퐿 푅푒퐿 − 푃푟 If differs from 5 × 10 , An alternative equation is suggested by Whitaker that 5 may푅푒 give푐푟푖푡 better results with some liquids because of the viscosity-ratio term:

. . = 0.036 ( 9200) 1 5.44 4 0 43 0 8 휇∞ 푁푢0��.�7�퐿< < 380푃푟 푅푒퐿 − �휇푤�

2 10푃푟< < 5.5 10 5 6 0.26∗ < 푅푒< 퐿3.5 ∗ 휇∞ : the휇 viscosity푤 evaluated at .

휇∞ : the viscosity evaluated at 푇∞. 휇For푤 the gases the viscosity ratio푇 is푤 dropped and the properties are evaluated at . 푇푓 Costant Heat Flux For constant-wall-heat flux in turbulent flow that the local Nusselt number is only about 4 percent higher than for the isothermal surface; that is,

= 1.04 ] 5.45

Example푁푢푥 5.7:Turbulent푁푢푥 푇푤=푐표푛푠푡푎푛푡 Heat Transfer from Isothermal Flat Plate Air at 20◦C and 1 atm flows over a flat plate at 35 m/s. The plate is 75 cm long and is maintained at 60◦C. Assuming unit depth in the z direction, calculate the heat transfer from the plate. Solution We evaluate properties at the film temperature: = = = 40 + 273 = 313 K 푇푤+푇∞ 60+20 푓 . 푇 = =2 2 = 1.128 kg/m3. ( )( 5) 푝 1 0132∗10 휌 = 푅푇1.906 28710313 kg/m.s, = 0.02723 W/m.˚C, = 0.7, = 1.007 kJ/kg.˚C. −5 . . = = = = 1.553 10 > 5 10 푝 휇 ∗ . 푘 푃푟 퐶 푢∞퐿 휌푢∞퐿 1 128∗35∗0 75 6 5 −5 then푅푒퐿 the 휐flow is 휇turbulent1 906 ∗10 ∗ ∗ therefore we use equation 5.42

= = (0.037 . 871) 1 � ℎ퐿 3 0 8 퐿 퐿 푁푢���� = (푘0.7)푃푟[(0.037)(푅푒1.553− 10 ) . 871] = 2180 1 3( )( . ) 6 0 8 ���=�퐿 = = 79.1 2 푁푢 . ∗ W/m−.˚C. 푘 2180 0 02723 ℎ� = 푁푢����(퐿 퐿 ) 0=75(79.1)(0.75)(60 20) = 2373 W.

푞 ℎ�퐴 푇푤 − 푇∞ − 5.7 Turbulent-Boundary-Layer Thickness The turbulent boundary layer thichness is calculated from the equation below: 1. The boundary layer is fully turbulent from the leading edge of the plate. . = 5.46 0 381푥 1 훿 5 2. 푅푒The푥 boundary layer follows a laminor growth pattern up to = 5 10 and a turbulent growth thereafter. 5 푐푟푖푡푖푐푎푙 = 0푅푒.381 1 10,256∗ 5 10 < < 10 5.47 − 5 −1 5 7 푥 푥 푥 Example훿 � 5.8:푅푒 − 푅푒 � 푥 ∗ 푅푒 Turbulent-Boundary-Layer Thickness Calculate the turbulent-boundary-layer thickness at the end of the plate for Example 5.7, assuming that it develops (a) from the leading edge of the plate and (b) from the transition point at = 5 10 . 5 Solution 푅푒푐푟푖푡 ∗ Since we have already calculated the Reynolds number as = 1.553 × 10 , it is a 6 simple matter 푅푒퐿 to insert this value in Equations (5.46) and (5.47) along with = = 0.75 m to give . (0.381)(0.75) (a) = = 1 = 0.0165 m= 16.5 mm 푥 퐿 0 381푥 (1.553 106)5 1 5 ∗ 훿 푅푒푥 (b) = 0.381 1 10,256 − 5 −1 훿 � 푅푒푥 − 푅푒푥 � 푥 = (0.381)(1.553 10 ) (10,256)(1.553 10 ) 0.75 = 9.9 1 mm 6 −5 6 −1 훿 � ∗ − ∗ � ∗

5.8 Internal Forced Convection

5.8.1 Heat Transfer in Laminar Tube Flow

Consider the flow in a tube as shown in Figure 5.7. A boundary layer develops at the entrance, as shown. Eventually the boundary layer fills the entire tube, and the flow is said to be fully developed. If the flow is laminar, a parabolic velocity profile is experienced, as shown in Figure 5.7a. When the flow is turbulent, a somewhat blunter profile is observed, as in Figure 5.7b. In a tube, the Reynolds number is again used as a criterion for laminar and turbulent flow. For

= > 2300 5.48 푢푚퐷ℎ ℎ 푅푒Where퐷 푣 = 4퐴푐 퐷ℎ 푃

the flow is usually observed to be turbulent is the tube diameter.

Again, a range of Reynolds numbers for transition퐷 may be observed, depending on the pipe roughness and smoothness of the flow. The generally accepted range for transition is 2000 < < 4000

푅푒퐷

Figure 5.7 Velocity profile for (a) laminar flow in a tube and (b) turbulent tube flow. Mean velocity The value of the mean velocity in a tube is determined from

푢푚 = = ( , ) 5.49

( , ) 푚̇ 휌푢푚퐴푐 ∫ 휌 푢 푟 푥 푑퐴푐 = 푟표 = ( , ) 5.50 ∫0 휌푢 푟 푥 2휋푟푑푟 2 푟표 2 2 푢푚 휌휋푟표 푟표 ∫0 푢 푟 푥 푟푑푟

The velocity distribution for the fully developed flow in tube may be written

= 1 5.51 2 푢 푟 2 푢The표 temperature− 푟표 distribution in the tube can calculated from the equation below:

= 5.52 2 1 휕푇 푢표푟표 푟 2 1 푟 4 Where:푇 − 푇푐 훼 휕푥 4 ��푟표� − 4 �푟표� � : center temperature.

푇푐 The Bulk Temperature

In tube flow the convection heat-transfer coefficient is usually defined by

= ( ) (local heat flux) 5.53 푞 W퐴 hereℎ 푇 푤 − 푇푏 :is the wall temperature and

푇푤 :is the so-called bulk temperature, or energy-average fluid temperature across the 푇tube,푏 which may be calculated from

= = 푟0 5.54 ∫0 휌2휋푟푑푟푢푐푝푇 푏 푟0 푇 푇� ∫0 휌2휋푟푑푟푢푐푝 = + 2 5.55 7 푢표푟표 휕푇 푇and푏 for푇 the푐 wall96 temperature훼 휕푥

= + 2 5.56 3 푢표푟표 휕푇 푤 푐 The푇 heat푇 -transfer16 훼 coefficient휕푥 is calculated from

= ( ) = 5.57 휕푇 푤 푏 푞 ℎ퐴 푇 − 푇 푘퐴 �휕푟�푟=푟표

= 휕푇 ( ) 5.58 푘�휕푟�푟=푟표 푤 푏 Theℎ temperature푇 −푇 gradient is given by

= = 3 5.59 휕푇 푢표 휕푇 푟 푟 푢표푟표 휕푇 2 표 휕푟�푟=푟표 훼 휕푥 �2 − 4푟 � 표 4훼 휕푥 Substituting Equations (5푟.55=푟), (5.56), and (5.59) in Equation (5.58) gives

= = 5.60 24 푘 48 푘 표 Expressedℎ 11 푟 in terms11 퐷 of the Nusselt number, the result is

= = 4.364 5.61 ℎ푑 푁푢퐷 푘 = 4.364 푘 5ℎ.8.2 Turbulent∗ 푑 Flow in a Tube

The developed velocity profile for turbulent flow in a tube will appear as shown in Figure 5.8.

Figure 5.8 Velocity profile in turbulent tube flow.

= = = 1 5.62 ℎ 푁푢퐷 푓 푝 푚 퐷 푆푡 휌.푐 푢 푅푒 푃푟 8 푃푟 ≈ = for 4 10 < 2 10 and 5.63 0 316 3 5 1 퐷 푓 4 ∗ 푅푒 ≤ ∗ 푅푒퐷 1 = 0.0395 4 푁푢퐷 − 푅푒퐷푃푟 푅푒3퐷 = 0.0395 4 for 4 10 < 2 10 and 1 5.64 3 5 푁푢퐷 퐷 퐷 푃푟 = 푅푒 ∗ 1푅푒 ≤ ∗ ≈ 2 5.65 3 푓 푆푡푃푟 8 3 푃푟 ≠ 4 = 0.0395 1 for 4 10 < 2 10 and 1 5.66 3 3 5 퐷 퐷 5.8.3푁푢 Empirical푅푒퐷 푃푟Relation for∗ Flow푅푒 In ≤Tube∗ 푃푟 ≠

Thus, for the tube flow depicted in Figure 5.9 the total energy added can be expressed in terms of abulk-temperature difference by = ( ) = ( ) 5.67

푝 푏2 푏1 푤 푏 푎푣 푞 푚̇ 푐+ 푇 − 푇 ℎ퐴 푇 − 푇 = 1 2 푇푏 푇푏 푇푏 2

Figure 5.9 Total heat transfer in terms of bulk-temperature difference.

• Empirical Relation In Laminar Flow In Tube

Constant Heat Flux

Fully developed laminar flow in circular tube

= = 4.36 5.68 ℎ퐷 퐷 Constant푁푢 푘 Surface Temperature

Fully developed laminar flow in circular tube

= = 3.66 5.69 ℎ퐷 퐷 푁푢Laminar푘 flow in the Entrance Region

. = = 3.66 + 퐷 5.70 ℎ퐷 0 065. �퐿�푅푒 푃푟 퐷 2 푁푢���� 푘 퐷 3 1+0 04��퐿�푅푒 푃푟�

When the difference between the surface and the fluid temperatures is large, it may be necessary to account for the variation of viscosity with temperature. The average Nusselt number for developing laminar flow in a circular tube in that case can be determined from [Sieder and Tate (1936),

. = 1.86 1 > 10 5.71 푅푒퐷 Pr 퐷 3 휇푏 0 14 푅푒퐷 Pr 퐷 All푁푢���� 퐷properties are� evaluated퐿 � � at휇 푤the� bulk mean fluid temperature,퐿 except for , which is evaluated at the . 휇푤 푇푤 Entry Length

, 0.05

퐿푡 푙푎푚푖푛푎푟 ≈ 푅푒퐷푃푟퐷 page 439 in the book.

• Imperical Relations for Turbulent Flow in Tube

For fully developed, smooth surface and moderate temperature differences:

= 0.023 . 0.6 < < 100 5.72 0 8 푛 푁푢퐷 푅푒퐷 푃푟 푃푟 The properties in this equation are evaluated at the average fluid bulk temperature, and the exponent has the following values: 0.4푛 for heating of the fluid = 0.3 for cooling of the fluid 푛 � More recent information by Gnielinski suggests that better results for turbulent flow in smooth tubes may be obtained from the following: 0.5 < < 1.5 = 0.0214( . 100) . 10 < < 5 10 5.73 0 8 0 4 푃푟 푁푢퐷 푅푒퐷 − 푃푟 � 4 6 Or 푅푒퐷 ∗ 1.5 < < 500 = 0.012( . 280) . 10 < < 10 5.74 0 87 0 4 푃푟 푁푢 퐷 푅푒퐷 − 푃푟 � 3 6 푅푒퐷 To take into account the property variations, Sieder and Tate recommend the following relation: . = 0.027 . 1 5.75 푏 0 14 0 8 3 휇 For푁푢퐷 entrance 푅푒region퐷 푃푟 �휇푤�

The above equations are apply to fully developed turbulent flow in tubes. In the entrance region the flow is not developed, and Nusselt recommended the following equation: . = 0.036 . For 10 < < 400 5.76 1 0 055 0 8 3 퐷 퐿 푁푢퐷 푅푒퐷 푃푟 �퐿� 퐷 Entry Length

, 10

퐿 푡 푡푢푟푏푢푙푒푛푡 ≈ 퐷 The above equations offer simplicity in computation, but uncertainties on the order of ±25 percent are not uncommon. Petukhov has developed a more accurate, although more complicated, expression for fully developed turbulent flow in smooth tubes:

= 푓 . 5.77 . �.8�푅푒퐷푃푟 휇푏 푛 2 퐷 푓 0 5 3 휇푤 푁푢 1 07+12 7�8� �푃푟 −1� � � Where: = 0.11 for > ,

푛= 0.25 for 푇푤 < 푇푏, and 푛 = 0 for constant푇푤 푇푏 heat flux or for gases. 푛 All properties are evaluated at ( ) = except for and . 푇푤 +푇푏 푇The푓 friction2 factor may be obtained휇푏 either휇푤 from the following for smooth tubes: = (1.82 log 1.64 ) 5.78 −2 10 퐷 Equation푓 (5.77) is푅푒 applicable− for the following ranges: 0.5 < < 200 for 6 percent accuracy

0.5 < 푃푟 < 2000 for 10 percent accuracy 10 > 푃푟 > 5 10 4 6 0.8 < 푅푒/퐷 <∗40 푏 푤 All above휇 휇equations are for smooth pipes. For rough pipe we use the equation below:

2 = 3 푓 푏 푆푡 푃푟푓 2 8 3 = 8 푁푢퐷 푓 푅푒퐷푃푟 푓 The friction푃푟 coefficient f is defined by

= 2 퐿 푢푚 An∆푝 empirical푓 퐷 휌 relation2 for the friction factor for rough tubes is given as

. = 1 325 . . . 2 푓 휀 5 74 �ln�3 7퐷�+ 0 9� 푅푒퐷 For 10 < < 10 , and 5000 < < 10 −6 휀 −3 8 퐷 • Isothermal퐷 Parallel Plates 푅푒 The average Nusselt number for the thermal entrance region of flow between isothermal parallel plates of length L is expressed as (Edwards et al., 1979)

. = = 7.45 + 퐷ℎ entry region, laminar flow 2800 ℎ 0 03� �푅푒퐷 푃푟 ℎ퐷 . 퐿 ℎ 퐷ℎ 2 푁푢 푘 퐷ℎ 3 푅푒 ≤ = 2 1+0 016�� 퐿 �푅푒퐷ℎ 푃푟�

Where퐷ℎ 푆 : is the space between the two plates.

푆

Example 5.9: Turbulent Heat Transfer in a Tube Air at 2 atm and 200◦C is heated as it flows through a tube with a diameter of 1 in (2.54 cm) at a velocity of 10 m/s. Calculate the heat transfer per unit length of tube if a constant-heat-flux condition is maintained at the wall and the wall temperature is 20◦C above the air temperature, all along the length of the tube. How much would the bulk temperature increase over a 3-m length of the tube? Solution We first calculate the Reynolds number to determine if the flow is laminar or turbulent, and then select the appropriate empirical correlation to calculate the heat transfer. The properties of air at a bulk temperature of 200◦C are

( )( . ) = = = 1.493 kg/m3 ( )( ) 5 푝 2 1 0132∗10 휌 =푅푇0.681 ,287 =4732.57 10 kg/m.s, = 0.0386 W/m.˚C, = 1.025 kj/kg. ˚C. ( . )( )( . −5 ) = = = 14756 푝 푃푟 휇 . ∗ 푘 푐 휌푢푚퐷 1 493 10 0 0254 −5 S푅푒o 퐷that the휇 flow is turbulent.2 57∗10 Check the entry length

, 10 = 10(0.0254) = 0.254 m

퐿Its푡 푡푢푟푏푢푙푒푛푡too shorter≈ than퐷 the length of the tube. Therefore the flow assumed fully developed. We therefore use Equation (5.72) to calculate the heat-transfer coefficient.

= 0.023 . 0 8 푛 푁푢For퐷 heating the푅푒 fluid퐷 푃푟 n=0.4 = 0.023(14756) . (0.681) . = 42.67 ( . )( . 0 8) 0 4 =퐷 = = 64.85 2 푁푢 . W/m .˚C. 푁푢퐷푘 42 67 0 0386 ℎ = 퐷 ( 0 0254) = (64.85)( )(0.0254)(20) = 103.5 W/m. 푞 푤 푏 퐿 = ℎ휋퐷 푇 −=푇 휋 푞 푝 푏 ( . ) 푞 =푚̇ 푐 ∆푇 =퐿(�1퐿.493� )(10)( ) = 7.565 10 2 2 kg/s. 휋퐷 0 0254 −3 푚̇ 휌푢푚 4 휋 4 ∗ ( )( . ) = = = 40.04 푞 ( . )( ) ˚C. 퐿�퐿� 3 103 5 −3 ∆푇푏 푚̇ 푐푝 7 565∗10 1025 EXAMPLE 5.10: Heating ofWater in Laminar Tube Flow Water at 60◦C enters a tube of 1-in (2.54-cm) diameter at a mean flow velocity of 2 cm/s. Calculate the exit water temperature if the tube is 3.0 m long and the wall temperature is constant at 80◦C. Solution We first evaluate the Reynolds number at the inlet bulk temperature to determine the flow regime. The properties of water at 60◦C are = 985 kg/m3

휌 = 3.02 , = 4.71 10 kg/m.s, = 0.651 W/m.˚C, = 4.18 kj/kg. ˚C. ( )( . )( .−4 ) = = = 1062 푝 푃푟 휇 . ∗ 푘 푐 휌푢푚퐷 985 0 02 0 0254 −4 푅푒 퐷 휇 4 71∗10 So the flow is laminar. Check for enrty length

, 0.05 = (0.05)(1062)(3.02)(0.0254) = 4 m

퐿Therefore푡 푙푎푚푖푛푎푟 the≈ flow푅푒 not퐷푃푟퐷 fully developed (i.e the flow in the interence region) Calculating the additional parameter, we have ( )( . )( . ) = = 27.15 > 10 푅푒퐷 Pr 퐷 1062 3 02 0 0254 So 퐿Equation (5.713) or Equation (5.70) is applicable. We do not yet know the mean bulk temperature to evaluate properties so we first make the calculation on the basis

of 60◦C,

At the wall temperature of 80◦C we have = 3.55 10 kg/m.s. −4 휇푤 ∗ . = 1.86 1 푅푒퐷 Pr 퐷 3 휇푏 0 14 퐷 퐿 휇푤 푁푢���� �( )( �. )�( . � ) . . = 1.86 1 = 5.816 . 1062 3 02 0 0254 3 4 41 0 14 ( . )( . ) 푁푢���=�퐷 = � 3 = 149�.1 �3 55�2 . W/m .˚C. 푁푢퐷푘 5 816 0 651 ( . ) ℎ = 퐷 =0(0254985)(0.02)( ) = 9.982 10 2 2 kg/s. 휋퐷 0 0254 −3 푚̇ 휌푢푚 4 휋 4 ∗ = = ( ) 푇푏1+푇푏2 푤 푝 푏2 푏1 (푞149ℎ.1휋퐷퐿)( )�(푇0.0254− )2(3) �80 푚̇ 푐 푇 −=푇(9.982 10 )(4180)( 60) 60+푇푏2 −3 = 71휋.98 ˚C � − 2 � ∗ 푇푏2 − . 푏2 푇 , = = = 66 ˚C 푇푏1+푇푏2 60+71 98 3 푇푏=푎푣푔982 kg/m2 2

휌 = 2.78 , = 4.36 10 kg/m.s, = 0.656 W/m.˚C, = 4.185 kj/kg. ˚C. ( )( . )( −. 4 ) = = = 1147 푝 푃푟 휇 . ∗ 푘 푐 휌푢푚퐷 982 0 02 0 0254 퐷 ( )( . )( . −)4 푅푒 =휇 4 63∗10 = 27.00 > 10 푅푒퐷 Pr 퐷 1147 2 78 0 0254 퐿 3 . = 1.86 1 푅푒퐷 Pr 퐷 3 휇푏 0 14 푁푢����퐷 � 퐿 .� �휇. 푤� = 1.86(27) = 5.743 1 . 0 14 3 4 36 ( . )( . ) 푁푢���=�퐷 = �3 55� = 148.3 2 . W/m .˚C. 푁푢퐷푘 5 743 0 656 (ℎ148.3퐷)( )(0.02540 0254)(3) 80 = (9.982 10 )(4185)( 60) 60+푇푏2 −3 = 71휋.88 ˚C � − 2 � ∗ 푇푏2 −

푇5.9푏2 Flow Across Cylinders And Spheres As the flow progresses along the front side of the cylinder, the pressure would decrease and then increase along the back side of the cylinder, resulting in an increase in free-stream velocity on the front side of the cylinder and a decrease on the back side. The transverse velocity (that velocity parallel to the surface) would decrease from a value of at the outer edge of the boundary layer to zero at the surface. As the flow proceeds푢∞ to the back side of the cylinder, the pressure increase causes a reduction in velocity in the free stream and throughout the boundary layer.

Figure 5.10 Cylinder in cross flow. Figure 5.11 Velocity distributions indicating flow separation on a cylinder in cross flow.

The pressure increase and reduction in velocity are related through the Bernoulli equation written along a streamline:

= 2 푑푝 푢 휌 −푑 �2푔푐� When the velocity gradient at the surface becomes zero, the flow is said to have reached a separation point: seperation point at = 0 휕푢

휕푦� 푦=0 The drag coefficient for bluff bodies is defined by

= Drage force = 2 5.79 휌푢∞

퐷 퐷 푐 where is the퐹 drag퐶 coefficient퐴 2푔 and

A is the퐶퐷 frontal area of the body exposed to the flow, which, for a cylinder, is the product of diameter and length. =

퐴 The 퐿퐷values of the drag coefficient for cylinders and spheres are given as a function of the Reynolds number in Figures 5.12 and 5.13.

Figure 5.12: Drag coefficient for circular cylinders as a function of the Reynolds number.

Figure 5.13: Drag coefficient for spheres as a function of the Reynolds number.

5.9.1 Cylinder The resulting correlation for average heat-transfer coefficients in cross flow over circular cylinders is

= = 1 ( flow of gas) 5.80 푛 ℎ퐷 푢∞퐷 3

푁푢퐷푓 푘푓 퐶 � 푣푓 � 푃푟푓 where the constants and are tabulated in Table 5.1. Properties for use with

Equation (5.80) are evaluated퐶 푛at the film temperature as indicated by the subscript .

푓 Table 5.1 Constants for use with Equation (5.80),

푅푒0.4퐷푓–4 0.989퐶 0.330푛 4–40 0.911 0.385 40–4000 0.683 0.466 4000–40,000 0.193 0.618 40,000–400,000 0.0266 0.805 For non-circular cylinder table 5.2 used to evaluate the constants in equation 5.80. Table 5.2 Constants for heat transfer from noncircular cylinders for use with Equation (5.80).

Fand has shown that the heat-transfer coefficients from liquids to cylinders in cross flow may be better represented by the relation = (0.35 + 0.56 . ) . 10 > < 10 flow of liquid 5.81 0 52 0 3 −1 5 푁푢For퐷푓 equation 5.81 the properties푅푒퐷푓 푃푟 푓at 푅푒퐷푓

Still a more comprehensive relation푇푓 is given by Churchill and Bernstein that is applicable over the complete range of available data: For the flow of air, water, and liquid sodium

. 4 1 1 = 0.3 + 1 + 5 5 5.82 2 3 퐷 퐷 8 0 62푅푒. 푃푟 푅푒 퐷 1 푁푢 2 4 � �282000� � 0 4 3 �1+�푃푟� � For 10 < < 10 , > 0.2 2 7 Where: 푅푒퐷 푃푒퐷 =

푃푒퐷 푅푒퐷푃푟 = 0.8237 ln 1 −1 < 0.2 5.83 2 퐷 퐷 퐷 5.9.2푁푢 Spheres� − �푃푒 �� 푃푒

McAdams recommends the following relation for heat transfer from spheres to a flowing gas:

• Flow of gas . = = 0.37 17 < < 7 10 5.84 0 6 ℎ퐷 푢∞퐷 4 푘 푁푢퐷 � 푣 � 푅푒퐷 ∗ All properties of fluid at film temperature .

푇푓 Achenbach has obtained relations applicable over a still wider range of Reynolds numbers for air with Pr =0.71:

100 < < 3 10 = 2 + (0.25 + 3 10 . ) 5.84 1 = 0.71 5 −4 1 6 2 퐷 푁푢퐷 푅푒퐷 ∗ 푅푒퐷 � 푅푒 ∗ 3 10 <푃푟 < 5 10 = 430 + + + 5.85 5 = 0.71 6 2 3 퐷 푁푢퐷 푎 푅푒퐷 푏 푅푒퐷 푐 푅푒퐷 � ∗ 푅푒 ∗ 푃푟

All properties of fluid at film temperature .

푇푓 • Flow of liquid For flow of liquids past spheres, the data of Kramers may be used to obtain the Correlation . . = 0.97 + 0.68 1 < < 2000 5.86 0 5 −0 3 푢∞퐷 푁푢All 퐷properties푃푟 at film temperature� 푣 � . 푅푒퐷

heat transfer from spheres to oil and푇푓 water over a more extended range of Reynolds numbers from 1 to 200,000: . . = 1.2 + 0.53 . 1 < < 2 10 5.87 0 25 −0 3 휇푤 0 54 5 푁푢 퐷푃푟 �휇∞� 푅푒퐷 푅푒퐷 ∗ where all properties are evaluated at free-stream conditions , except , which is

evaluated at the surface temperature of the sphere . 푇∞ 휇푤 • Liquid and Gases 푇푤 All the above data have been brought together by Whitaker to develop a single equation for gases and liquids flowing past spheres:

. 3.5 < < 8 10 = 2 + 0.4 1 + 0.06 2 1 5.88 ∞ 4 0.7 < < 380 4 2 3 0 4 휇 퐷 퐷 푅푒 ∗ 푁푢 � 푅푒퐷 푅푒퐷� 푃푟 �휇푤� � 푃푟 Properties in Equation (5.88) are evaluated at the free-stream temperature .

푇∞ Example 5.11: Airflow Across Isothermal Cylinder Air at 1 atm and 35◦C flows across a 5.0-cm-diameter cylinder at a velocity of 50 m/s. The cylinder surface is maintained at a temperature of 150◦C. Calculate the heat loss per unit length of the cylinder. Solution: We first determine the Reynolds number and then find the applicable constants from Table 5.1 for use with Equation (5.80). The properties of air are evaluated at the film temperature: = = = 92.5 + 273 = 365.5 K 푇푤+푇∞ 150+35 푓 ( . ) 푇 = =2 2 = 0.966 kg/m3 ( )( . 5) 푝 1 0132∗10 휌 =푅푇0.695287 , 365= 25.14 10 kg/m.s, = 0.0312 W/m.˚C. ( . )( )( . −)5 = = = 1.129 10 푃푟 휇 . ∗ 푘 휌푢∞퐷 0 966 50 0 05 5 −5 From푅푒퐷 table휇 5.1 2=140∗.100266 , = 0.805∗

퐶 푛 = = 1 푛 ℎ퐷 푢∞퐷 3 푁푢퐷푓 푘푓 퐶 � 푣푓 � 푃푟푓 = (0.0266)(1.129 10 ) . (0.695) = 275.1 1 ℎ퐷 5 0 805 3 푓 ( . )( . ) 푘 = = 171∗ .7 2 . W/m .˚C. 275 1 0 0312 ℎ = (0 05 ) 푞 퐿 = ℎ(171휋퐷 .7푇푤)(−)푇(∞0.05)(150 35) = 3100 W/m 푞 퐿 휋 −

H.W

A fine wire having a diameter of 3.94*10−5 m is placed in a 1-atm airstream at 25◦C 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 50◦C. Calculate the heat loss per unit length. 1. Use equation 5.80 2. Use equation 5.82

Example 5.12: Heat Transfer from Sphere Air at 1 atm and 27◦C blows across a 12-mm-diameter sphere at a free-stream velocity of 4 m/s. Asmall heater inside the sphere maintains the surface temperature at 77◦C. Calculate the heat lost by the sphere. Solution Consulting Equation (5.88) we find that the Reynolds number is evaluated at the free- stream temperature.We therefore need the following properties: at =27◦C=300 K,

= 0.708 , 푇∞ = 2.14 10 kg/m.s, = 0.02624 W/m.˚C, 2 −5 푃푟= 15.96 10휇∞ m /s ∗ 푘 −6 At푣 = 77˚C∗ = 350 K, = 2.075 10 kg/m.s ( )( . ) −5 푤= = = 3059푤 푇 . 휇 ∗ 푢∞퐷 4 0 012 −6 퐷 푣 15 96∗10 푅푒 . = 2 + 0.4 1 + 0.06 2 1 4 2 3 0 4 휇∞ 푁푢퐷 � 푅푒퐷 푅푒퐷� 푃푟 �휇푤� . . = 2 + (0.4)(3059) + (0.06)(3059) (0.708) 1 = 31.4 1 2 . 4 2 3 0 4 1 8462 ( )( ) ( . )( . ) 푁푢=퐷 �= = 68.66 2 � � 2 075 � . W/m .˚C. 푁푢퐷 푘 31 4 0 02624 ℎ = 퐷 ( 0 012) ( 2)( )( ) ( ) 푞 = ℎ468휋.퐷66 푇4푤 − 푇0∞.006 77 27 = 1.553 W 2 푞 휋 − 5.10 Flow Across Tube Banks Cross-flow over tube banks is commonly encountered in practice in heat transfer equipment such as the condensers and evaporators of power plants, refrigerators, and air conditioners. In such equipment, one fluid moves through the tubes while the other moves over the tubes in a perpendicular direction. In a heat exchanger that involves a tube bank, the tubes are usually placed in a shell (and thus the name shell-and-tube heat exchanger), especially when the fluid is a liquid, and the fluid flows through the space between the tubes and the shell. There are numerous types of shell-and-tube heat exchangers. Because many heat-exchanger arrangements involve multiple rows of tubes, the heattransfer characteristics for tube banks are of important practical interest. The heat- transfer characteristics of staggered and in-line tube banks were studied by Grimson. Determination of Maximum Flow Velocity For flows normal to in-line tube banks the maximum flow velocity will occur through the minimum frontal area ( ) presented to the incoming free stream velocity

. Thus, 푆푛 − 푑 ∞ = 푢 ( ) (in-line arrangement) 5.89 푠푛 For푢푚푎푥 staggered푢∞ � 푠 푛−푑 �

If + 1 2 < ( ) 2 2 푆푛 2 ��� 2 � 푆푝 � − 푑� ∗ 푆푛 − 푑 Then

= 1 (staggered arrangement) 5.90 2푠푛푢∞ 1 푚푎푥 2 푢 푆푛 2 2 �� 2 � +푆푝� −푑 If + 1 2 > ( ) 2 2 푆푛 2 ��� 2 � 푆푝 � − 푑� ∗ 푆푛 − 푑 = ( ) 푠푛 푢푚푎푥 푢∞ � 푠푛−푑 � The nomenclature for use with Table 5.3 is shown in Figure 5.14. The data of Table 5.3 pertain to tube banks having 10 or more rows of tubes in the direction of flow. For fewer rows the ratio of h for N rows deep to that for 10 rows is given in Table 5.4.

= = 1 (5.80) 푛 ℎ푑 푢∞푑 3

푑푓 푓 푓 푓 Table푁푢 5.3푘 Modified퐶 � 푣correlation� 푃푟 of Grimson for heat transfer in tube banks of 10 rows or more, , for use with Equation (5.80).

Figure 5.14 Nomenclature for use with Table 5.3: (a) in-line tube rows; (b) staggered tube rows.

Table 5.4 Ratio of h for N rows deep to that for 10 rows deep, for use with Equation (5.80).

Zukauskas has presented additional information for tube bundles that takes into account wide ranges of Reynolds numbers and property variations. The correlating equation takes the form . 10 < , < 10 = = , 1 5.91 ℎ�푑 푃푟 4 0.7 < < 500 6 푛 0 36 푅푒푑 푚푎푥 푁푢 푘 퐶푅푒푑 푚푎푥푃푟 �푃푟푤� � 푃푟 where all properties except are evaluated at and the values of the constants are given in Table 5.5 for greater푃푟푤 than 20 rows of 푇tubes.∞ For gases the Prandtl number ratio has little influence and is dropped. Once again, note that the Reynolds number is based on the maximum velocity in the tube bundle. For less than 20 rows in the direction of flow the correction factor in Table 5.6 should be applied. It is essentially the same as for the Grimson correlation.

Table 5.5 Constants for Zukauskas correlation [Equation (5.91)] for heat transfer in tube banks of 20 rows or more.

Table 5.6 Ratio of h for N rows deep to that for 20 rows deep for use with Equation (5.91).

Example 5.13: Heating of Air with In-Line Tube Bank

Air at 1 atm and 10◦C flows across a bank of tubes 15 rows high and 5 rows deep at a velocity of 7 m/s measured at a point in the flow before the air enters the tube bank. The surfaces of the tubes are maintained at 65◦C. The diameter of the tubes is 1 in [2.54 cm]; they are arranged in an in-line manner so that the spacing in both the normal and parallel directions to the flow is 1.5 in [3.81 cm]. Calculate the total heat transfer per unit length for the tube bank and the exit air temperature. Solution The constants for use with Equation (5.80) may be obtained from Table 5.3, using

The properties of air are evaluated at the film temperature, which at entrance to the tube bank is

= = 1 (5.80) 푛 ℎ퐷 푢∞퐷 3 푁푢퐷푓 푘푓 퐶 � 푣푓 � 푃푟푓

Because there are only 5 rows deep, this value must be multiplied by the factor 0.92, as determined from Table 5.4. The total surface area for heat transfer, considering unit length of tubes, is