CHAPTER Heat Transfer 1 · that this energy transfer is deﬁned as heat. The science of heat transfer seeks not merely to explain how heat energy may be transferred, but also to

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C H A P T E R

1 IntroductionHeat transfer is the science that seeks to predict the energy transfer that may take placebetween material bodies as a result of a temperature difference. Thermodynamics teachesthat this energy transfer is defined as heat. The science of heat transfer seeks not merelyto explain how heat energy may be transferred, but also to predict the rate at which theexchange will take place under certain specified conditions. The fact that a heat-transferrate is the desired objective of an analysis points out the difference between heat transferand thermodynamics. Thermodynamics deals with systems in equilibrium; it may be usedto predict the amount of energy required to change a system from one equilibrium state toanother; it may not be used to predict how fast a change will take place since the systemis not in equilibrium during the process. Heat transfer supplements the first and secondprinciples of thermodynamics by providing additional experimental rules that may be usedto establish energy-transfer rates. As in the science of thermodynamics, the experimentalrules used as a basis of the subject of heat transfer are rather simple and easily expanded toencompass a variety of practical situations.

As an example of the different kinds of problems that are treated by thermodynamicsand heat transfer, consider the cooling of a hot steel bar that is placed in a pail of water.Thermodynamics may be used to predict the final equilibrium temperature of the steelbar–water combination. Thermodynamics will not tell us how long it takes to reach thisequilibrium condition or what the temperature of the bar will be after a certain length oftime before the equilibrium condition is attained. Heat transfer may be used to predict thetemperature of both the bar and the water as a function of time.

Most readers will be familiar with the terms used to denote the three modes of heattransfer: conduction, convection, and radiation. In this chapter we seek to explain the mech-anism of these modes qualitatively so that each may be considered in its proper perspective.Subsequent chapters treat the three types of heat transfer in detail.

1-1 CONDUCTION HEAT TRANSFERWhen a temperature gradient exists in a body, experience has shown that there is an energytransfer from the high-temperature region to the low-temperature region. We say that theenergy is transferred by conduction and that the heat-transfer rate per unit area is proportionalto the normal temperature gradient:

qx

A∼ ∂T

∂x

1

Heat TransferChapter One

Introduction

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When the proportionality constant is inserted,

qx = −kA ∂T∂x

[1-1]

where qx is the heat-transfer rate and ∂T/∂x is the temperature gradient in the direction ofthe heat flow. The positive constant k is called the thermal conductivity of the material, andthe minus sign is inserted so that the second principle of thermodynamics will be satisfied;i.e., heat must flow downhill on the temperature scale, as indicated in the coordinate systemof Figure 1-1. Equation (1-1) is called Fourier’s law of heat conduction after the Frenchmathematical physicist Joseph Fourier, who made very significant contributions to theanalytical treatment of conduction heat transfer. It is important to note that Equation (1-1)is the defining equation for the thermal conductivity and that k has the units of watts permeter per Celsius degree in a typical system of units in which the heat flow is expressedin watts.

Figure 1-1 Sketch showingdirection of heat flow.

Temperatureprofile

T

qx

x

We now set ourselves the problem of determining the basic equation that governs thetransfer of heat in a solid, using Equation (1-1) as a starting point.

Consider the one-dimensional system shown in Figure 1-2. If the system is in a steadystate, i.e., if the temperature does not change with time, then the problem is a simple one,and we need only integrate Equation (1-1) and substitute the appropriate values to solve forthe desired quantity. However, if the temperature of the solid is changing with time, or ifthere are heat sources or sinks within the solid, the situation is more complex. We considerthe general case where the temperature may be changing with time and heat sources maybe present within the body. For the element of thickness dx, the following energy balancemay be made:

Energy conducted in left face + heat generated within element

= change in internal energy + energy conducted out right face

These energy quantities are given as follows:

Energy in left face = qx = −kA ∂T∂x

Energy generated within element = qA dx

Figure 1-2 Elemental volume forone-dimensional heat-conduction analysis.

A

qx

qx + dx

qgen = q•A dx

x dx

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Change in internal energy = ρcA ∂T∂τdx

Energy out right face = qx+dx = −kA ∂T

∂x

]x+dx

= −A[k∂T

∂x+ ∂

∂x

(k∂T

∂x

)dx

]

where

q= energy generated per unit volume, W/m3

c= specific heat of material, J/kg · ◦C

ρ= density, kg/m3

Combining the relations above gives

−kA ∂T∂x

+ qA dx= ρcA ∂T∂τdx−A

[k∂T

∂x+ ∂

∂x

(k∂T

∂x

)dx

]

or∂

∂x

(k∂T

∂x

)+ q = ρc ∂T

∂τ[1-2]

This is the one-dimensional heat-conduction equation. To treat more than one-dimensionalheat flow, we need consider only the heat conducted in and out of a unit volume in all threecoordinate directions, as shown in Figure 1-3a. The energy balance yields

qx + qy + qz+ qgen = qx+dx + qy+dy + qz+dz+ dE

dτ

and the energy quantities are given by

qx = −k dy dz ∂T∂x

qx+dx = −[k∂T

∂x+ ∂

∂x

(k∂T

∂x

)dx

]dy dz

qy = −k dx dz ∂T∂y

qy+dy = −[k∂T

∂y+ ∂

∂y

(k∂T

∂y

)dy

]dx dz

qz = −k dx dy ∂T∂z

qz+dz = −[k∂T

∂z+ ∂

∂z

(k∂T

∂z

)dz

]dx dy

qgen = q dx dy dzdE

dτ= ρc dx dy dz ∂T

∂τ

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Figure 1-3 Elemental volume for three-dimensional heat-conduction analysis:(a) cartesian coordinates; (b) cylindrical coordinates; (c) spherical coordinates.

y

xZ

Z

Z

qy + dy

qZ+dZ

qx + dx

qy

qZ

qx

dx

dydZ

dr

dr

dZ

qgen = q• dx dy dZ

y

x

y

x

φ

φ

φd

φd

d

θ

θ

r

(a)

(c)

(b)

r

so that the general three-dimensional heat-conduction equation is

∂

∂x

(k∂T

∂x

)+ ∂

∂y

(k∂T

∂y

)+ ∂

∂z

(k∂T

∂z

)+ q= ρc ∂T

∂τ[1-3]

For constant thermal conductivity, Equation (1-3) is written

∂2T

∂x2+ ∂2T

∂y2+ ∂2T

∂z2+ q

k= 1

α

∂T

∂τ[1-3a]

where the quantity α= k/ρc is called the thermal diffusivity of the material. The larger thevalue of α, the faster heat will diffuse through the material. This may be seen by examiningthe quantities that make up α. A high value of α could result either from a high value ofthermal conductivity, which would indicate a rapid energy-transfer rate, or from a low valueof the thermal heat capacity ρc. A low value of the heat capacity would mean that less of theenergy moving through the material would be absorbed and used to raise the temperature of

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the material; thus more energy would be available for further transfer. Thermal diffusivityα has units of square meters per second.

In the derivations above, the expression for the derivative at x+ dx has been written inthe form of a Taylor-series expansion with only the first two terms of the series employedfor the development.

Equation (1-3a) may be transformed into either cylindrical or spherical coordinates bystandard calculus techniques. The results are as follows:

Cylindrical coordinates:

∂2T

∂r2+ 1

r

∂T

∂r+ 1

r2

∂2T

∂φ2+ ∂2T

∂z2+ q

k= 1

α

∂T

∂τ[1-3b]

Spherical coordinates:

1

r

∂2

∂r2(rT)+ 1

r2 sin θ

∂

∂θ

(sin θ

∂T

∂θ

)+ 1

r2 sin2 θ

∂2T

∂φ2+ q

k= 1

α

∂T

∂τ[1-3c]

The coordinate systems for use with Equations (1-3b) and (1-3c) are indicated inFigure 1-3b and c, respectively.

Many practical problems involve only special cases of the general equations listedabove. As a guide to the developments in future chapters, it is worthwhile to show thereduced form of the general equations for several cases of practical interest.

Steady-state one-dimensional heat flow (no heat generation):

d2T

dx2= 0 [1-4]

Note that this equation is the same as Equation (1-1) when q= constant.

Steady-state one-dimensional heat flow in cylindrical coordinates (no heatgeneration):

d2T

dr2+ 1

r

dT

dr= 0 [1-5]

Steady-state one-dimensional heat flow with heat sources:

d2T

dx2+ q

k= 0 [1-6]

Two-dimensional steady-state conduction without heat sources:

∂2T

∂x2+ ∂2T

∂y2= 0 [1-7]

1-2 THERMAL CONDUCTIVITYEquation (1-1) is the defining equation for thermal conductivity. On the basis of this def-inition, experimental measurements may be made to determine the thermal conductivityof different materials. For gases at moderately low temperatures, analytical treatments inthe kinetic theory of gases may be used to predict accurately the experimentally observedvalues. In some cases, theories are available for the prediction of thermal conductivities in

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liquids and solids, but in general, many open questions and concepts still need clarificationwhere liquids and solids are concerned.

The mechanism of thermal conduction in a gas is a simple one. We identify the kineticenergy of a molecule with its temperature; thus, in a high-temperature region, the moleculeshave higher velocities than in some lower-temperature region. The molecules are in contin-uous random motion, colliding with one another and exchanging energy and momentum.The molecules have this random motion whether or not a temperature gradient exists in thegas. If a molecule moves from a high-temperature region to a region of lower temperature,it transports kinetic energy to the lower-temperature part of the system and gives up thisenergy through collisions with lower-energy molecules.

Table 1-1 lists typical values of the thermal conductivities for several materials toindicate the relative orders of magnitude to be expected in practice. More complete tabularinformation is given in Appendix A. In general, the thermal conductivity is stronglytemperature-dependent.

Table 1-1 Thermal conductivity of various materials at 0◦C.

Thermal conductivityk

Material W/m · ◦C Btu/h · ft · ◦F

Metals:Silver (pure) 410 237Copper (pure) 385 223Aluminum (pure) 202 117Nickel (pure) 93 54Iron (pure) 73 42Carbon steel, 1% C 43 25Lead (pure) 35 20.3Chrome-nickel steel (18% Cr, 8% Ni) 16.3 9.4

Nonmetallic solids:Diamond 2300 1329Quartz, parallel to axis 41.6 24Magnesite 4.15 2.4Marble 2.08−2.94 1.2−1.7Sandstone 1.83 1.06Glass, window 0.78 0.45Maple or oak 0.17 0.096Hard rubber 0.15 0.087Polyvinyl chloride 0.09 0.052Styrofoam 0.033 0.019Sawdust 0.059 0.034Glass wool 0.038 0.022Ice 2.22 1.28

Liquids:Mercury 8.21 4.74Water 0.556 0.327Ammonia 0.540 0.312Lubricating oil, SAE 50 0.147 0.085Freon 12, CCl2F2 0.073 0.042

Gases:Hydrogen 0.175 0.101Helium 0.141 0.081Air 0.024 0.0139Water vapor (saturated) 0.0206 0.0119Carbon dioxide 0.0146 0.00844

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1-3 CONVECTION HEAT TRANSFERIt is well known that a hot plate of metal will cool faster when placed in front of a fan thanwhen exposed to still air. We say that the heat is convected away, and we call the processconvection heat transfer. The term convection provides the reader with an intuitive notionconcerning the heat-transfer process; however, this intuitive notion must be expanded toenable one to arrive at anything like an adequate analytical treatment of the problem. Forexample, we know that the velocity at which the air blows over the hot plate obviouslyinfluences the heat-transfer rate. But does it influence the cooling in a linear way; i.e., ifthe velocity is doubled, will the heat-transfer rate double? We should suspect that the heat-transfer rate might be different if we cooled the plate with water instead of air, but, again,how much difference would there be? These questions may be answered with the aid ofsome rather basic analyses presented in later chapters. For now, we sketch the physicalmechanism of convection heat transfer and show its relation to the conduction process.

Consider the heated plate shown in Figure 1-7. The temperature of the plate is Tw, andthe temperature of the fluid is T∞. The velocity of the flow will appear as shown, beingreduced to zero at the plate as a result of viscous action. Since the velocity of the fluid layerat the wall will be zero, the heat must be transferred only by conduction at that point. Thuswe might compute the heat transfer, using Equation (1-1), with the thermal conductivityof the fluid and the fluid temperature gradient at the wall. Why, then, if the heat flows byconduction in this layer, do we speak of convection heat transfer and need to consider thevelocity of the fluid? The answer is that the temperature gradient is dependent on the rate atwhich the fluid carries the heat away; a high velocity produces a large temperature gradient,and so on. Thus the temperature gradient at the wall depends on the flow field, and we mustdevelop in our later analysis an expression relating the two quantities. Nevertheless, it mustbe remembered that the physical mechanism of heat transfer at the wall is a conductionprocess.

To express the overall effect of convection, we use Newton’s law of cooling:

q=hA (Tw− T∞) [1-8]

Here the heat-transfer rate is related to the overall temperature difference between thewall and fluid and the surface area A. The quantity h is called the convection heat-transfercoefficient, and Equation (1-8) is the defining equation.An analytical calculation of h may bemade for some systems. For complex situations it must be determined experimentally. Theheat-transfer coefficient is sometimes called the film conductance because of its relationto the conduction process in the thin stationary layer of fluid at the wall surface. FromEquation (1-8) we note that the units of h are in watts per square meter per Celsius degreewhen the heat flow is in watts.

In view of the foregoing discussion, one may anticipate that convection heat transferwill have a dependence on the viscosity of the fluid in addition to its dependence on the

Figure 1-7 Convection heat transfer from a plate.

Flow Free stream

Wall

qu

Tw

u T

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thermal properties of the fluid (thermal conductivity, specific heat, density). This is expectedbecause viscosity influences the velocity profile and, correspondingly, the energy-transferrate in the region near the wall.

If a heated plate were exposed to ambient room air without an external source of motion,a movement of the air would be experienced as a result of the density gradients near theplate. We call this natural, or free, convection as opposed to forced convection, whichis experienced in the case of the fan blowing air over a plate. Boiling and condensationphenomena are also grouped under the general subject of convection heat transfer. Theapproximate ranges of convection heat-transfer coefficients are indicated in Table 1-3.

Convection Energy Balance on a Flow Channel

The energy transfer expressed by Equation (1-8) is used for evaluating the convection lossfor flow over an external surface. Of equal importance is the convection gain or loss resultingfrom a fluid flowing inside a channel or tube as shown in Figure 1-8. In this case, the heatedwall at Tw loses heat to the cooler fluid, which consequently rises in temperature as it flows

Table 1-3 Approximate values of convection heat-transfer coefficients.

h

Mode W/m2 · ◦C Btu/h · ft2 · ◦F

Across 2.5-cm air gap evacuated to a pressureof 10−6 atm and subjectedto �T = 100◦C − 30◦C 0.087 0.015

Free convection, �T = 30◦CVertical plate 0.3 m [1 ft] high in air 4.5 0.79Horizontal cylinder, 5-cm diameter, in air 6.5 1.14Horizontal cylinder, 2-cm diameter,

in water 890 157Heat transfer across 1.5-cm vertical air

gap with �T = 60◦C 2.64 0.46Fine wire in air, d= 0.02 mm,�T = 55◦C 490 86

Forced convectionAirflow at 2 m/s over 0.2-m square plate 12 2.1Airflow at 35 m/s over 0.75-m square plate 75 13.2Airflow at Mach number = 3, p= 1/20 atm,T∞ = −40◦C, across 0.2-m square plate 56 9.9

Air at 2 atm flowing in 2.5-cm-diametertube at 10 m/s 65 11.4

Water at 0.5 kg/s flowing in 2.5-cm-diametertube 3500 616

Airflow across 5-cm-diameter cylinderwith velocity of 50 m/s 180 32

Liquid bismuth at 4.5 kg/s and 420◦Cin 5.0-cm-diameter tube 3410 600

Airflow at 50 m/s across fine wire,d= 0.04 mm 3850 678

Boiling waterIn a pool or container 2500–35,000 440–6200Flowing in a tube 5000–100,000 880–17,600

Condensation of water vapor, 1 atmVertical surfaces 4000–11,300 700–2000Outside horizontal tubes 9500–25,000 1700–4400

Dropwise condensation 170,000–290,000 30,000–50,000


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Figure 1-8 Convection in a channel.

Ti

m

qTe

from inlet conditions at Ti to exit conditions at Te. Using the symbol i to designate enthalpy(to avoid confusion with h, the convection coefficient), the energy balance on the fluid is

q= m(ie− ii)where m is the fluid mass flow rate. For many single-phase liquids and gases operating overreasonable temperature ranges �i= cp�T and we have

q= mcp(Te− Ti)which may be equated to a convection relation like Equation (1-8)

q= mcp(Te− Ti)=hA(Tw, avg − Tfluid, avg) [1-8a]

In this case, the fluid temperatures Te, Ti, and Tfluid are called bulk or energy averagetemperatures. A is the surface area of the flow channel in contact with the fluid. We shallhave more to say about the notions of computing convection heat transfer for external andinternal flows in Chapters 5 and 6. For now, we simply want to alert the reader to thedistinction between the two types of flows.

We must be careful to distinguish between the surface area for convection that isemployed in convection Equation (1-8) and the cross-sectional area that is used to calculatethe flow rate from

m= ρumeanAc

where Ac =πd2/4 for flow in a circular tube. The surface area for convection in this casewould be πdL, where L is the tube length. The surface area for convection is always thearea of the heated surface in contact with the fluid.

1-4 RADIATION HEAT TRANSFERIn contrast to the mechanisms of conduction and convection, where energy transfer through amaterial medium is involved, heat may also be transferred through regions where a perfectvacuum exists. The mechanism in this case is electromagnetic radiation. We shall limitour discussion to electromagnetic radiation that is propagated as a result of a temperaturedifference; this is called thermal radiation.

Thermodynamic considerations show∗ that an ideal thermal radiator, or blackbody, willemit energy at a rate proportional to the fourth power of the absolute temperature of thebody and directly proportional to its surface area. Thus

qemitted = σAT 4 [1-9]

*See, for example, J. P. Holman, Thermodynamics. 4th ed. New York: McGraw-Hill, 1988, p. 705.


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where σ is the proportionality constant and is called the Stefan-Boltzmann constant withthe value of 5.669 × 10−8 W/m2 · K4. Equation (1-9) is called the Stefan-Boltzmann lawof thermal radiation, and it applies only to blackbodies. It is important to note that thisequation is valid only for thermal radiation; other types of electromagnetic radiation maynot be treated so simply.

Equation (1-9) governs only radiation emitted by a blackbody. The net radiant exchangebetween two surfaces will be proportional to the difference in absolute temperatures to thefourth power; i.e.,

qnet exchange

A∝ σ(T 4

1 − T 42 ) [1-10]

We have mentioned that a blackbody is a body that radiates energy according to theT 4 law. We call such a body black because black surfaces, such as a piece of metal coveredwith carbon black, approximate this type of behavior. Other types of surfaces, such as aglossy painted surface or a polished metal plate, do not radiate as much energy as theblackbody; however, the total radiation emitted by these bodies still generally follows theT 4 proportionality. To take account of the “gray” nature of such surfaces we introduceanother factor into Equation (1-9), called the emissivity ε, which relates the radiation ofthe “gray” surface to that of an ideal black surface. In addition, we must take into accountthe fact that not all the radiation leaving one surface will reach the other surface sinceelectromagnetic radiation travels in straight lines and some will be lost to the surroundings.We therefore introduce two new factors in Equation (1-9) to take into account both situations,so that

q=FεFGσA (T 41 − T 4

2 ) [1-11]

where Fε is the emissivity function, and FG is the geometric “view factor” function. Thedetermination of the form of these functions for specific configurations is the subject of asubsequent chapter. It is important to alert the reader at this time, however, to the fact thatthese functions usually are not independent of one another as indicated in Equation (1-11).

Radiation in an Enclosure

A simple radiation problem is encountered when we have a heat-transfer surface at temper-ature T1 completely enclosed by a much larger surface maintained at T2. We will show inChapter 8 that the net radiant exchange in this case can be calculated with

q= ε1σA1 (T41 − T 4

2 ) [1-12]

Values of ε are given in Appendix A.Radiation heat-transfer phenomena can be exceedingly complex, and the calculations

are seldom as simple as implied by Equation (1-11). For now, we wish to emphasize the dif-ference in physical mechanism between radiation heat-transfer and conduction-convectionsystems. In Chapter 8 we examine radiation in detail.


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SolutionFrom Appendix A, the thermal conductivity for copper is 370 W/m · ◦C at 250◦C. From Fourier’slaw

q

A= −k dT

dx

Integrating gives

q

A= −k �T

�x= −(370)(100 − 400)

3 × 10−2= 3.7 MW/m2 [1.173 × 106 Btu/h · ft2]

Convection Calculation EXAMPLE 1-2

Air at 20◦C blows over a hot plate 50 by 75 cm maintained at 250◦C. The convection heat-transfercoefficient is 25 W/m2 · ◦C. Calculate the heat transfer.

SolutionFrom Newton’s law of cooling

q= hA (Tw− T∞)= (25)(0.50)(0.75)(250 − 20)

= 2.156 kW [7356 Btu/h]

Multimode Heat Transfer EXAMPLE 1-3

Assuming that the plate in Example 1-2 is made of carbon steel (1%) 2 cm thick and that 300 Wis lost from the plate surface by radiation, calculate the inside plate temperature.

SolutionThe heat conducted through the plate must be equal to the sum of convection and radiation heatlosses:

qcond = qconv + qrad

−kA �T�x

= 2.156 + 0.3 = 2.456 kW

�T = (−2456)(0.02)

(0.5)(0.75)(43)= −3.05◦C [−5.49◦F]

where the value of k is taken from Table 1-1. The inside plate temperature is therefore

Ti= 250 + 3.05 = 253.05◦C

Heat Source and Convection EXAMPLE 1-4

An electric current is passed through a wire 1 mm in diameter and 10 cm long. The wire issubmerged in liquid water at atmospheric pressure, and the current is increased until the water

EXAMPLE 1-1 Conduction Through Copper Plate

One face of a copper plate 3 cm thick is maintained at 400◦C, and the other face is maintained at100◦C. How much heat is transferred through the plate?


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boils. For this situation h= 5000 W/m2 · ◦C, and the water temperature will be 100◦C. How muchelectric power must be supplied to the wire to maintain the wire surface at 114◦C?

SolutionThe total convection loss is given by Equation (1-8):

q=hA (Tw− T∞)

For this problem the surface area of the wire is

A=πdL=π(1 × 10−3)(10 × 10−2)= 3.142 × 10−4 m2

The heat transfer is therefore

q= (5000 W/m2 · ◦C)(3.142 × 10−4 m2)(114 − 100)= 21.99 W [75.03 Btu/h]

and this is equal to the electric power that must be applied.

EXAMPLE 1-5 Radiation Heat Transfer

Two infinite black plates at 800◦C and 300◦C exchange heat by radiation. Calculate the heattransfer per unit area.

SolutionEquation (1-10) may be employed for this problem, so we find immediately

q/A= σ(T 41 − T 4

2 )

= (5.669 × 10−8)(10734 − 5734)

= 69.03 kW/m2 [21,884 Btu/h · ft2]

EXAMPLE 1-6 Total Heat Loss by Convection and Radiation

A horizontal steel pipe having a diameter of 5 cm is maintained at a temperature of 50◦C in a largeroom where the air and wall temperature are at 20◦C. The surface emissivity of the steel may betaken as 0.8. Using the data of Table 1-3, calculate the total heat lost by the pipe per unit length.

SolutionThe total heat loss is the sum of convection and radiation. From Table 1-3 we see that an estimatefor the heat-transfer coefficient for free convection with this geometry and air ish= 6.5 W/m2 · ◦C.The surface area is πdL, so the convection loss per unit length is

q/L]conv = h(πd)(Tw− T∞)= (6.5)(π)(0.05)(50 − 20)= 30.63 W/m

The pipe is a body surrounded by a large enclosure so the radiation heat transfer can becalculated from Equation (1-12). With T1 = 50◦C = 323◦K and T2 = 20◦C = 293◦K, we have

q/L]rad = ε1(πd1)σ(T41 − T 4

2 )

= (0.8)(π)(0.05)(5.669 × 10−8)(3234 − 2934)

= 25.04 W/m


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The total heat loss is therefore

q/L]tot = q/L]conv + q/L]rad

= 30.63 + 25.04 = 55.67 W/m

In this example we see that the convection and radiation are about the same. To neglect eitherwould be a serious mistake.

1-6 SUMMARYWe may summarize our introductory remarks very simply. Heat transfer may take place byone or more of three modes: conduction, convection, and radiation. It has been noted that thephysical mechanism of convection is related to the heat conduction through the thin layerof fluid adjacent to the heat-transfer surface. In both conduction and convection Fourier’slaw is applicable, although fluid mechanics must be brought into play in the convectionproblem in order to establish the temperature gradient.

Radiation heat transfer involves a different physical mechanism—that of propagationof electromagnetic energy. To study this type of energy transfer we introduce the concept ofan ideal radiator, or blackbody, which radiates energy at a rate proportional to its absolutetemperature to the fourth power.

It is easy to envision cases in which all three modes of heat transfer are present, asin Figure 1-9. In this case the heat conducted through the plate is removed from the platesurface by a combination of convection and radiation. An energy balance would give

−kA dT

dy

]wall

=hA (Tw− T∞)+FεFGσA (T 4w− T 4

s )

where

Ts = temperature of surroundings

Tw= surface temperature

T∞ = fluid temperature

To apply the science of heat transfer to practical situations, a thorough knowledge ofall three modes of heat transfer must be obtained.

Figure 1-9 Combination of conduction, convection, andradiation heat transfer.

Radiant energySurrounding at TS

Heat conductedthrough wall

Flow, T∞

TW

qconv = hA(TW − T∞)


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One Dimension

2-1 INTRODUCTIONWe now wish to examine the applications of Fourier’s law of heat conduction to calculationof heat flow in some simple one-dimensional systems. Several different physical shapesmay fall in the category of one-dimensional systems: cylindrical and spherical systems areone-dimensional when the temperature in the body is a function only of radial distanceand is independent of azimuth angle or axial distance. In some two-dimensional problemsthe effect of a second-space coordinate may be so small as to justify its neglect, and themultidimensional heat-flow problem may be approximated with a one-dimensional analysis.In these cases the differential equations are simplified, and we are led to a much easiersolution as a result of this simplification.

2-2 THE PLANE WALLFirst consider the plane wall where a direct application of Fourier’s law [Equation (1-1)]may be made. Integration yields

q= − kA�x

(T2 − T1) [2-1]

when the thermal conductivity is considered constant. The wall thickness is �x, and T1and T2 are the wall-face temperatures. If the thermal conductivity varies with temperatureaccording to some linear relation k= k0(1 +βT ), the resultant equation for the heat flowis

q= −k0A

�x

[(T2 − T1)+ β

2(T 2

2 − T 21 )

][2-2]

If more than one material is present, as in the multilayer wall shown in Figure 2-1, theanalysis would proceed as follows: The temperature gradients in the three materials areshown, and the heat flow may be written

q= −kAA T2 − T1

�xA= −kBA T3 − T2

�xB= −kCA T4 − T3

�xC

Note that the heat flow must be the same through all sections.

14

Steady- State Conduction One Dimention


Chapter Two


q = −kA ∂T∂x

((H.W))

composit


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15 2-3 Insulation and R Values

Figure 2-1 One-dimensional heat transfer through a composite wall and electrical analog.

q q

1 2 3 4

ATemperatureprofile

RA

ΔxA

kAA

T1 T2 T3 T4

qRB RC

ΔxB

kBAΔxC

kCAA B C

Solving these three equations simultaneously, the heat flow is written

q= T1 − T4

�xA/kAA+�xB/kBA+�xC/kCA [2-3]

At this point we retrace our development slightly to introduce a different conceptual view-point for Fourier’s law. The heat-transfer rate may be considered as a flow, and the combina-tion of thermal conductivity, thickness of material, and area as a resistance to this flow. Thetemperature is the potential, or driving, function for the heat flow, and the Fourier equationmay be written

Heat flow = thermal potential difference

thermal resistance[2-4]

a relation quite like Ohm’s law in electric-circuit theory. In Equation (2-1) the thermalresistance is�x/kA, and in Equation (2-3) it is the sum of the three terms in the denominator.We should expect this situation in Equation (2-3) because the three walls side by side act asthree thermal resistances in series. The equivalent electric circuit is shown in Figure 2-1b.

The electrical analogy may be used to solve more complex problems involving bothseries and parallel thermal resistances. A typical problem and its analogous electric circuitare shown in Figure 2-2. The one-dimensional heat-flow equation for this type of problemmay be written

q= �Toverall∑Rth

[2-5]

where the Rth are the thermal resistances of the various materials. The units for the thermalresistance are ◦C/W or ◦F · h/Btu.

It is well to mention that in some systems, like that in Figure 2-2, two-dimensionalheat flow may result if the thermal conductivities of materials B, C, and D differ by anappreciable amount. In these cases other techniques must be employed to effect a solution.

2-3 INSULATION AND R VALUESIn Chapter 1 we noted that the thermal conductivities for a number of insulating materials aregiven in Appendix A. In classifying the performance of insulation, it is a common practice


T1

T2T3

T4


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C H A P T E R 2 16

Figure 2-2 Series and parallel one-dimensional heat transfer through acomposite wall and electrical analog.

A

1 2 3 4 5

E

F

G

B

C

D

RB

RC

RDRERA

T1

q

T2 T3 T4 T5

RF

RG

in the building industry to use a term called the R value, which is defined as

R= �T

q/A[2-6]

The units for R are ◦C · m2/W or ◦F · ft2 · h/Btu. Note that this differs from the thermal-resistance concept discussed above in that a heat flow per unit area is used.

At this point it is worthwhile to classify insulation materials in terms of their applicationand allowable temperature ranges. Table 2-1 furnishes such information and may be usedas a guide for the selection of insulating materials.

2-4 RADIAL SYSTEMSCylinders

Consider a long cylinder of inside radius ri, outside radius ro, and length L, such as theone shown in Figure 2-3. We expose this cylinder to a temperature differential Ti− To andask what the heat flow will be. For a cylinder with length very large compared to diameter,it may be assumed that the heat flows only in a radial direction, so that the only spacecoordinate needed to specify the system is r. Again, Fourier’s law is used by inserting theproper area relation. The area for heat flow in the cylindrical system is

Ar = 2πrL

so that Fourier’s law is written

qr = −kAr dTdr

[2-7]

or

qr = −2πkrLdT

dr


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17 2-4 Radial Systems

Table 2-1 Insulation types and applications.

ThermalTemperature conductivity, Density,

Type range, ◦C mW/m · ◦C kg/m3 Application

1 Linde evacuated superinsulation −240–1100 0.0015–0.72 Variable Many2 Urethane foam −180–150 16–20 25–48 Hot and cold pipes3 Urethane foam −170–110 16–20 32 Tanks4 Cellular glass blocks −200–200 29–108 110–150 Tanks and pipes5 Fiberglass blanket for wrapping −80–290 22–78 10–50 Pipe and pipe fittings6 Fiberglass blankets −170–230 25–86 10–50 Tanks and equipment7 Fiberglass preformed shapes −50–230 32–55 10–50 Piping8 Elastomeric sheets −40–100 36–39 70–100 Tanks9 Fiberglass mats 60–370 30–55 10–50 Pipe and pipe fittings

10 Elastomeric preformed shapes −40–100 36–39 70–100 Pipe and fittings11 Fiberglass with vapor −5–70 29–45 10–32 Refrigeration lines

barrier blanket12 Fiberglass without vapor to 250 29–45 24–48 Hot piping

barrier jacket13 Fiberglass boards 20–450 33–52 25–100 Boilers, tanks,

heat exchangers14 Cellular glass blocks and boards 20–500 29–108 110–150 Hot piping15 Urethane foam blocks and 100–150 16–20 25–65 Piping

boards16 Mineral fiber preformed shapes to 650 35–91 125–160 Hot piping17 Mineral fiber blankets to 750 37–81 125 Hot piping18 Mineral wool blocks 450–1000 52–130 175–290 Hot piping19 Calcium silicate blocks, boards 230–1000 32–85 100–160 Hot piping, boilers,

chimney linings20 Mineral fiber blocks to 1100 52–130 210 Boilers and tanks

Figure 2-3 One-dimensional heat flowthrough a hollow cylinderand electrical analog.

L

q

rri

ro dr

Ti

R th =ln(ro/ri)2 kL π

q

To


To

Ti


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C H A P T E R 2 18

Figure 2-4 One-dimensional heat flow through multiple cylindrical sectionsand electrical analog.

T1

r1 r2

T2 r3

T3

r4

q

q

T4 T1 RA T2 RB T3 T4RCA

B

Cln(r2�r1)2 kAL π

ln(r3�r2)2 kBL π

ln(r4�r3)2 kCL π

with the boundary conditions

T = Tiat r= ri

T = Toat r= ro

The solution to Equation (2-7) is

q= 2πkL (Ti− To)ln (ro/ri)

[2-8]

and the thermal resistance in this case is

Rth = ln (ro/ri)

2πkL

The thermal-resistance concept may be used for multiple-layer cylindrical walls just as itwas used for plane walls. For the three-layer system shown in Figure 2-4 the solution is

q= 2πL (T1 − T4)

ln (r2/r1)/kA+ ln (r3/r2)/kB + ln (r4/r3)/kC[2-9]

The thermal circuit is also shown in Figure 2-4.

Spheres

Spherical systems may also be treated as one-dimensional when the temperature is a functionof radius only. The heat flow is then

q= 4πk (Ti− To)1/ri− 1/ro

[2-10]

The derivation of Equation (2-10) is left as an exercise.

Multilayer Conduction EXAMPLE 2-1

An exterior wall of a house may be approximated by a 4-in layer of common brick [k=0.7 W/m · ◦C] followed by a 1.5-in layer of gypsum plaster [k= 0.48 W/m · ◦C]. What thick-ness of loosely packed rock-wool insulation [k= 0.065 W/m · ◦C] should be added to reduce theheat loss (or gain) through the wall by 80 percent?


(B.Cs)

((H.W))


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196 2-4 Radial Systems

SolutionThe overall heat loss will be given by

q= �T∑Rth

Because the heat loss with the rock-wool insulation will be only 20 percent (80 percent reduction)of that before insulation

q with insulation

q without insulation= 0.2 =

∑Rth without insulation∑Rth with insulation

We have for the brick and plaster, for unit area,

Rb = �x

k= (4)(0.0254)

0.7= 0.145 m2 · ◦C/W

Rp = �x

k= (1.5)(0.0254)

0.48= 0.079 m2 · ◦C/W

so that the thermal resistance without insulation is

R= 0.145 + 0.079 = 0.224 m2 · ◦C/W

Then

R with insulation = 0.224

0.2= 1.122 m2 · ◦C/W

and this represents the sum of our previous value and the resistance for the rock wool

1.122 = 0.224 +RrwRrw= 0.898 = �x

k= �x

0.065

so that�xrw= 0.0584 m = 2.30 in

Figure Example 2-2

T1 = 600˚C

T2 = 100˚C

Stainless steel

Asbestos

r1 r2

r3

T1 T2

ln (r2�r1)ksL2π

ln (r3�r2)kaL2π

EXAMPLE 2-2 Multilayer Cylindrical System

A thick-walled tube of stainless steel [18% Cr, 8% Ni, k= 19 W/m · ◦C] with 2-cm inner diam-eter (ID) and 4-cm outer diameter (OD) is covered with a 3-cm layer of asbestos insulation[k= 0.2 W/m · ◦C]. If the inside wall temperature of the pipe is maintained at 600◦C, calculatethe heat loss per meter of length. Also calculate the tube–insulation interface temperature.

SolutionFigure Example 2-2 shows the thermal network for this problem. The heat flow is given by

q

L= 2π (T1 − )

ln (r2/r1)/ks+ ln(r3/r2)/ka= 2π (600 − 100)

(ln 2)/19 + (ln 52 )/0.2

= 680 W/m

This heat flow may be used to calculate the interface temperature between the outside tube walland the insulation. We have

q

L= −

ln (r3/r2)/2πka= 680 W/m


T3

T3T2

T3

T3

T3


the outside temperature is 100 C


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C H A P T E R 2 20

where is the interface temperature, which may be obtained as

= 595.8◦C

The largest thermal resistance clearly results from the insulation, and thus the major portion of thetemperature drop is through that material.

Convection Boundary Conditions

We have already seen in Chapter 1 that convection heat transfer can be calculated from

qconv =hA (Tw− T∞)

An electric-resistance analogy can also be drawn for the convection process by rewritingthe equation as

qconv = Tw− T∞1/hA

[2-11]

where the 1/hA term now becomes the convection resistance.

2-5 THE OVERALL HEAT-TRANSFERCOEFFICIENT

Consider the plane wall shown in Figure 2-5 exposed to a hot fluid A on one side and acooler fluid B on the other side. The heat transfer is expressed by

q=h1A (TA− T1)= kA

�x(T1 − T2)=h2A (T2 − TB)

The heat-transfer process may be represented by the resistance network in Figure 2-5, andthe overall heat transfer is calculated as the ratio of the overall temperature difference tothe sum of the thermal resistances:

q= TA− TB1/h1A+�x/kA+ 1/h2A

[2-12]

Figure 2-5 Overall heat transfer through a plane wall.

TA

TA

T1T1

T2

T2 TB

TB

q

1h1A

h1

h2

1h2A

ΔxkA

Flui

d A

Flui

d Bq


T2

T2

hot cold


= �T

R


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21 2-5 The Overall Heat-Transfer Coefficient

Figure 2-6 Resistance analogy for hollow cylinder with convection boundaries.

Flui

d B

Fluid A

1 2

q

TA Ti To TB

1 1hiAi hoAo

Observe that the value 1/hA is used to represent the convection resistance. The overallheat transfer by combined conduction and convection is frequently expressed in terms ofan overall heat-transfer coefficient U, defined by the relation

q=UA�Toverall [2-13]

where A is some suitable area for the heat flow. In accordance with Equation (2-12), theoverall heat-transfer coefficient would be

U = 1

1/h1 +�x/k+ 1/h2

The overall heat-transfer coefficient is also related to the R value of Equation (2-6) through

U = 1

R value

For a hollow cylinder exposed to a convection environment on its inner and outer surfaces,the electric-resistance analogy would appear as in Figure 2-6 where, again, TA and TB arethe two fluid temperatures. Note that the area for convection is not the same for both fluidsin this case, these areas depending on the inside tube diameter and wall thickness. Theoverall heat transfer would be expressed by

q= TA− TB1

hiAi+ ln (ro/ri)

2πkL+ 1

hoAo

[2-14]

in accordance with the thermal network shown in Figure 2-6. The termsAi andAo representthe inside and outside surface areas of the inner tube. The overall heat-transfer coefficientmay be based on either the inside or the outside area of the tube. Accordingly,

Ui= 11

hi+ Ai ln (ro/ri)

2πkL+ Ai

Ao

1

ho

[2-15]

Uo= 1Ao

Ai

1

hi+ Ao ln (ro/ri)

2πkL+ 1

ho

[2-16]

The general notion, for either the plane wall or cylindrical coordinate system, is that

UA= 1/�Rth = 1/Rth,overall


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C H A P T E R 2 22

Calculations of the convection heat-transfer coefficients for use in the overall heat-transfercoefficient are made in accordance with the methods described in later chapters. Some typi-cal values of the overall heat-transfer coefficient for heat exchangers are given in Table 10-1.Some values ofU for common types of building construction system are given in Table 2-2and may be employed for calculations involving the heating and cooling of buildings.

Table 2-2 Overall heat transfer coefficients for common construction systems according to Jamesand Goss [12].

Description of construction system U, Btu/hr · ft2 · ◦F U, W/m2 · ◦C

1 2 × 3 in double-wood stud wall, 406 mm OC, polyisocyanurate 0.027 0.153(0.08-mm vapor retarder, 19-mm insulation), fiberglass battsin cavity, 12.7-mm plywood

2 2 × 4 in wood stud wall, 406 mm OC, polyisocyanurate 0.060 0.359foil-faced, fiberglass batts in cavity, 15-mm plywood

3 2 × 4 in wood stud wall, 406 mm OC, 38-mm polyisocyanurate, 0.039 0.221foil-faced, cellular polyurethane in cavity, 19-mm plywood

4 2 × 4 in wood stud wall, 406 mm OC, 15-mm exterior sheathing, 0.326 1.850.05-mm polyethylene vapor barrier, no fill in cavity

5 Nominal 4-in concrete-block wall with brick facade and 0.080 0.456extruded polystyrene insulation

6 2 × 4 in wood stud wall, 406 mm OC, fiberglass batt insulation 0.084 0.477in cavity, 16-mm plywood

7 2 × 4 in wood stud wall, 406 mm OC, fiberglass batt insulation 0.060 0.341in cavity, 16-mm plywood, clay brick veneer

8 2 × 4 in wood stud wall, 406 mm OC, fiberglass batt in cavity, 0.074 0.41713-mm plywood, aluminum or vinyl siding

9 2 × 4 in wood stud wall, 406 mm OC, polyurethane foam 0.040 0.228in cavity, extruded polystyrene sheathing, aluminum siding

10 2 × 4 in steel stud wall, 406 mm OC, fiberglass batts 0.122 0.691in cavity, 41-mm air space, 13-mm plaster board

11 Aluminum motor home roof with fiberglass insulation 0.072 0.41in cavity (32 mm)

12 2 × 6 in wood stud ceiling, 406 mm OC, fiberglass 0.065 0.369foil-faced insulation in cavity, reflective airspace (ε≈ 0.05)

13 8-in (203-mm) normal-weight structural concrete (ρ= 2270 kg/m3) 0.144 0.817wall, 18-mm board insulation, painted off-white

14 10-in (254-mm) concrete-block-brick cavity wall, 0.322 1.83no insulation in cavities

15 8-in (203-mm) medium-weight concrete block wall, 0.229 1.3perlite insulation in cores

16 8-in (203-mm) normal-weight structural concrete, 0.764 4.34(ρ= 2270 kg/m3) including steel reinforcement bars(Note: actual thickness of concrete is 211 mm.)

17 8-in (203-mm) lightweight structural concrete (ρ= 1570 kg/m3) 0.483 2.75including steel reinforcement bars(Note: Actual thickness of concrete is 210 mm.)

18 8-in (203-mm) low-density concrete wall (ρ= 670 kg/m3) 0.216 1.23including steel reinforcement bars(Note: Actual thickness of concrete is 216 mm.)

19 Corrugated sheet steel wall with 10.2-in (260-mm.) 0.030 0.17fiberglass batt in cavity

20 Corrugated sheet steel wall with (159-mm) fiberglass batt in cavity 0.054 0.3121 Metal building roof deck, 25 mm polyisocyanurate, foil-faced 0.094 0.535

(ε≈ 0.03), 203-mm reflective air space22 Metal building roof deck, 25-mm foil-faced polyisocyanurate, 0.065 0.366

38-mm fiberglass batts in cavity


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EXAMPLE 2-3 Heat Transfer Through a Composite Wall

“Two-by-four” wood studs have actual dimensions of 4.13 × 9.21 cm and a thermal conductivityof 0.1 W/m · ◦C. A typical wall for a house is constructed as shown Figure Example 2-3. Calculatethe overall heat-transfer coefficient and R value of the wall.

Figure Example 2-3 (a) Construction of a dwelling wall; (b) thermal resistancemodel.

Outside air convection, h = 15 W/m2 • ˚C

Common brick, k = 0.69

40.6 cm

8 cm

1.9 cm, k = 0.96

1.9 cm, k = 0.48

Insulation, k = 0.04

Gypsumsheath

Inside air convection, h = 7.5 W/m2 • ˚C

2 x 4 studs

(a)

Tairinside

R brick R convectioninside

R sheathoutside

R insul R sheathinside

R sheathoutside

R studR sheath

inside

Tairoutside

R convectionoutside

(b)

SolutionThe wall section may be considered as having two parallel heat-flow paths: (1) through the studs,and (2) through the insulation. We will compute the thermal resistance for each, and then combinethe values to obtain the overall heat-transfer coefficient.

1. Heat transfer through studs (A= 0.0413 m2 for unit depth). This heat flow occurs through sixthermal resistances:

a. Convection resistance outside of brick

R= 1

hA= 1

(15)(0.0413)= 1.614 ◦C/W

b. Conduction resistance in brick

R=�x/kA= 0.08

(0.69)(0.0413)= 2.807 ◦C/W

c. Conduction resistance through outer sheet

R= �x

kA= 0.019

(0.96)(0.0413)= 0.48 ◦C/W


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C H A P T E R 2 24

d. Conduction resistance through wood stud

R= �x

kA= 0.0921

(0.1)(0.0413)= 22.3 ◦C/W

e. Conduction resistance through inner sheet

R= �x

kA= 0.019

(0.48)(0.0413)= 0.96 ◦C/W

f. Convection resistance on inside

R= 1

hA= 1

(7.5)(0.0413)= 3.23 ◦C/W

The total thermal resistance through the wood stud section is

Rtotal = 1.614 + 2.807 + 0.48 + 22.3 + 0.96 + 3.23 = 31.39 ◦C/W [a]

2. Insulation section (A= 0.406 − 0.0413 m2 for unit depth). Through the insulation sec-tion, five of the materials are the same, but the resistances involve different area terms,i.e., 40.6 − 4.13 cm instead of 4.13 cm, so that each of the previous resistances must be mul-tiplied by a factor of 4.13/(40.6 − 4.13)= 0.113. The resistance through the insulation is

R= �x

kA= 0.0921

(0.04)(0.406 − 0.0413)= 6.31

and the total resistance through the insulation section is

Rtotal = (1.614 + 2.807 + 0.48 + 0.96 + 3.23)(0.113)+ 6.31 = 7.337 ◦C/W [b]

The overall resistance for the section is now obtained by combining the parallel resistances inEquations (a) and (b) to give

Roverall = 1

(1/31.39)+ (1/7.337)= 5.947 ◦C/W [c]

This value is related to the overall heat-transfer coefficient by

q=UA�T = �T

Roverall[d]

where A is the area of the total section = 0.406 m2. Thus,

U = 1

RA= 1

(5.947)(0.406)= 0.414 W/m2 · ◦C

As we have seen, the R value is somewhat different from thermal resistance and is given by

R value = 1

U= 1

0.414= 2.414◦C · m2/W

CommentThis example illustrates the relationships between the concepts of thermal resistance, the overallheat-transfer coefficient, and the R value. Note that the R value involves a unit area concept, whilethe thermal resistance does not.


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EXAMPLE 2-4 Cooling Cost Savings with Extra Insulation

A small metal building is to be constructed of corrugated steel sheet walls with a total wall surfacearea of about 300 m2. The air conditioner consumes about 1 kW of electricity for every 4 kWof cooling supplied.1 Two wall constructions are to be compared on the basis of cooling costs.Assume that electricity costs $0.15/kWh. Determine the electrical energy savings of using 260 mmof fiberglass batt insulation instead of 159 mm of fiberglass insulation in the wall.Assume an overalltemperature difference across the wall of 20◦C on a hot summer day in Texas.

SolutionConsulting Table 2-2 (Numbers 19 and 20) we find that overall heat transfer coefficients for thetwo selected wall constructions are

U(260-mm fiberglass) = 0.17 W/m2 · ◦C

U (159-mm fiberglass) = 0.31 W/m2 · ◦C

The heat gain is calculated from q=UA�T , so for the two constructions

q (260-mm fiberglass) = (0.17)(300)(20)= 1020 W

q (159-mm fiberglass) = (0.31)(300)(20)= 1860 W

Savings due to extra insulation = 840 W

The energy consumed to supply this extra cooling is therefore

Extra electric power required = (840)(1/4)= 210 W

and the cost isCost = (0.210kW)(0.15$/kWh)= 0.0315 $/hr

Assuming 10-h/day operation for 23 days/month this cost becomes

(0.0315)(10)(23)= $7.25/month

Both of these cases are rather well insulated. If one makes a comparison to a 2 × 4 wood stud wallwith no insulation (Number 4 in Table 2-2) fill in the cavity (U = 1.85 W/m2 · ◦C), the heatingload would be

q= (1.85)(300)(20)= 11,100 W

and the savings compared with the 260-mm fiberglass insulation would be

11,100 − 1020 = 10,080 W

producing a corresponding electric power saving of $0.378/h or $86.94/month. Clearly the insu-lated wall will pay for itself. It is a matter of conjecture whether the 260-mm of insulation willpay for itself in comparison to the 159-mm insulation.

1This is not getting something for nothing. Consult any standard thermodynamics text for the reason for thisbehavior.


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C H A P T E R 2 26

Overall Heat-Transfer Coefficient for a Tube EXAMPLE 2-5

Water flows at 50◦C inside a 2.5-cm-inside-diameter tube such that hi= 3500 W/m2 · ◦C. Thetube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m · ◦C. The outside ofthe tube loses heat by free convection with ho= 7.6 W/m2 · ◦C. Calculate the overall heat-transfercoefficient and heat loss per unit length to surrounding air at 20◦C.

SolutionThere are three resistances in series for this problem, as illustrated in Equation (2-14). WithL= 1.0 m, di= 0.025 m, and do= 0.025 + (2)(0.0008)= 0.0266 m, the resistances may be cal-culated as

Ri = 1

hiAi= 1

(3500)π(0.025)(1.0)= 0.00364 ◦C/W

Rt = ln (do/di)

2πkL

= ln(0.0266/0.025)

2π(16)(1.0)= 0.00062 ◦C/W

Ro= 1

hoAo= 1

(7.6)π(0.0266)(1.0)= 1.575 ◦C/W

Clearly, the outside convection resistance is the largest, and overwhelmingly so. This means that itis the controlling resistance for the total heat transfer because the other resistances (in series) arenegligible in comparison. We shall base the overall heat-transfer coefficient on the outside tubearea and write

q= �T∑R

=UAo�T [a]

Uo = 1

Ao∑R

= 1

[π(0.0266)(1.0)](0.00364 + 0.00062 + 1.575)

= 7.577 W/m2 · ◦C

or a value very close to the value of ho= 7.6 for the outside convection coefficient. The heattransfer is obtained from Equation (a), with

q=UAo �T = (7.577)π(0.0266)(1.0)(50 − 20)= 19 W (for 1.0 m length)

CommentThis example illustrates the important point that many practical heat-transfer problems involvemultiple modes of heat transfer acting in combination; in this case, as a series of thermal resis-tances. It is not unusual for one mode of heat transfer to dominate the overall problem. In thisexample, the total heat transfer could have been computed very nearly by just calculating the freeconvection heat loss from the outside of the tube maintained at a temperature of 50◦C. Becausethe inside convection and tube wall resistances are so small, there are correspondingly small tem-perature drops, and the outside temperature of the tube will be very nearly that of the liquid inside,or 50◦C.

2-6 CRITICAL THICKNESS OF INSULATIONLet us consider a layer of insulation which might be installed around a circular pipe, asshown in Figure 2-7. The inner temperature of the insulation is fixed at Ti, and the outer


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27 2-6 Critical Thickness of Insulation

Figure 2-7 Critical insulation thickness.

Ti T

h, T

ro

ln (ro�ri)2 kLπ

12 roLhπ

Ti

ri+

surface is exposed to a convection environment at T∞. From the thermal network the heattransfer is

q= 2πL (Ti− T∞)ln (ro/ri)

k+ 1

roh

[2-17]

Now let us manipulate this expression to determine the outer radius of insulation ro, whichwill maximize the heat transfer. The maximization condition is

dq

dro= 0 =

−2πL (Ti− T∞)(

1

kro− 1

hr2o

)[

ln (ro/ri)

k+ 1

roh

]2

which gives the result ro= k

h[2-18]

Equation (2-18) expresses the critical-radius-of-insulation concept. If the outer radius is lessthan the value given by this equation, then the heat transfer will be increased by adding moreinsulation. For outer radii greater than the critical value an increase in insulation thicknesswill cause a decrease in heat transfer. The central concept is that for sufficiently small valuesof h the convection heat loss may actually increase with the addition of insulation becauseof increased surface area.

EXAMPLE 2-6 Critical Insulation Thickness

Calculate the critical radius of insulation for asbestos [k= 0.17 W/m · ◦C] surrounding a pipeand exposed to room air at 20◦C with h= 3.0 W/m2 · ◦C. Calculate the heat loss from a 200◦C,5.0-cm-diameter pipe when covered with the critical radius of insulation and without insulation.

SolutionFrom Equation (2-18) we calculate ro as

ro= k

h= 0.17

3.0= 0.0567 m = 5.67 cm

The inside radius of the insulation is 5.0/2 = 2.5 cm, so the heat transfer is calculated from Equation(2-17) as

q

L= 2π (200 − 20)

ln (5.67/2.5)

0.17+ 1

(0.0567)(3.0)

= 105.7 W/m

Without insulation the convection from the outer surface of the pipe is

q

L=h(2πr)(Ti− To)= (3.0)(2π)(0.025)(200 − 20)= 84.8 W/m


for sphere ro= k

h((H.W))


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C H A P T E R 2 28

So, the addition of 3.17 cm (5.67 − 2.5) of insulation actually increases the heat transfer by25 percent.

As an alternative, fiberglass having a thermal conductivity of 0.04 W/m · ◦C might beemployed as the insulation material. Then, the critical radius would be

ro= k

h= 0.04

3.0= 0.0133 m = 1.33 cm

Now, the value of the critical radius is less than the outside radius of the pipe (2.5 cm), so additionof any fiberglass insulation would cause a decrease in the heat transfer. In a practical pipe insulationproblem, the total heat loss will also be influenced by radiation as well as convection from theouter surface of the insulation.

2-7 HEAT-SOURCE SYSTEMSA number of interesting applications of the principles of heat transfer are concerned withsystems in which heat may be generated internally. Nuclear reactors are one example;electrical conductors and chemically reacting systems are others. At this point we shallconfine our discussion to one-dimensional systems, or, more specifically, systems wherethe temperature is a function of only one space coordinate.

Plane Wall with Heat Sources

Consider the plane wall with uniformly distributed heat sources shown in Figure 2-8. Thethickness of the wall in the x direction is 2L, and it is assumed that the dimensions inthe other directions are sufficiently large that the heat flow may be considered as one-dimensional. The heat generated per unit volume is q, and we assume that the thermalconductivity does not vary with temperature. This situation might be produced in a practicalsituation by passing a current through an electrically conducting material. From Chapter 1,

Figure 2-8 Sketch illustratingone-dimensionalconduction problem withheat generation.

q = heat generated per unit volume

•

TW

TW

TO

L

L

x=

0

x


(105.7 - 84.8) / 84.8 = 25% percent increase


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29 2-7 Heat-Source Systems

the differential equation that governs the heat flow is

d2T

dx2+ q

k= 0 [2-19]

For the boundary conditions we specify the temperatures on either side of the wall, i.e.,

T = Twat x= ±L [2-20]

The general solution to Equation (2-19) is

T = − q

2kx2 +C1x+C2 [2-21]

Because the temperature must be the same on each side of the wall, C1 must be zero. Thetemperature at the midplane (x= 0) is denoted by T0 and from Equation (2-21)

T0 =C2

The temperature distribution is therefore

T − T0 = − q

2kx2 [2-22a]

orT − T0

Tw− T0=( xL

)2[2-22b]

a parabolic distribution. An expression for the midplane temperature T0 may be obtainedthrough an energy balance. At steady-state conditions the total heat generated must equalthe heat lost at the faces. Thus

2

(−kA dT

dx

]x=L

)= qA 2L

where A is the cross-sectional area of the plate. The temperature gradient at the wall isobtained by differentiating Equation (2-22b):

dT

dx

]x=L

= (Tw− T0)

(2x

L2

)]x=L

= (Tw− T0)2

L

Then

−k(Tw− T0)2

L= qL

and

T0 = qL2

2k+ Tw [2-23]

This same result could be obtained by substituting T = Tw at x=L into Equation(2-22a).

The equation for the temperature distribution could also be written in the alternativeform

T − TwT0 − Tw = 1 − x2

L2[2-22c]


∂2T

∂x2+ ∂2T

∂y2+ ∂2T

∂z2+ q

k= 1

α

∂T

∂τAssumptions: 1. one dimention 2. steady state 3. k = const


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C H A P T E R 2 30

2-8 CYLINDER WITH HEAT SOURCESConsider a cylinder of radius R with uniformly distributed heat sources and constant thermalconductivity. If the cylinder is sufficiently long that the temperature may be considered afunction of radius only, the appropriate differential equation may be obtained by neglectingthe axial, azimuth, and time-dependent terms in Equation (1-3b),

d2T

dr2+ 1

r

dT

dr+ q

k= 0 [2-24]

The boundary conditions areT = Twat r=R

and heat generated equals heat lost at the surface:

qπR2L= −k2πRLdT

dr

]r=R

Since the temperature function must be continuous at the center of the cylinder, we couldspecify that

dT

dr= 0 at r= 0

However, it will not be necessary to use this condition since it will be satisfied automaticallywhen the two boundary conditions are satisfied.

We rewrite Equation (2-24)

rd2T

dr2+ dT

dr= −qr

k

and note that

rd2T

dr2+ dT

dr= d

dr

(rdT

dr

)

Then integration yields

rdT

dr= −qr2

2k+C1

and

T = −qr2

4k+C1 ln r+C2

From the second boundary condition above,

dT

dr

]r=R

= −qR2k

= −qR2k

+ C1

R

ThusC1 = 0

We could also note thatC1 must be zero because at r= 0 the logarithm function becomesinfinite.

From the first boundary condition,

T = Tw= −qR2

4k+C2 at r=R


∂2T

∂r2+ 1

r

∂T

∂r+ 1

r2

∂2T

∂φ2+ ∂2T

∂z2+ q

k= 1

α

∂T

∂τAssumptions: 1. one dimention 2. steady state 3. k = const


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31 2-8 Cylinder with Heat Sources

so that

C2 = Tw+ qR2

4k

The final solution for the temperature distribution is then

T − Tw= q

4k(R2 − r2) [2-25a]

or, in dimensionless form,T − TwT0 − Tw = 1 −

( rR

)2[2-25b]

where T0 is the temperature at r= 0 and is given by

T0 = qR2

4k+ Tw [2-26]

It is left as an exercise to show that the temperature gradient at r= 0 is zero.For a hollow cylinder with uniformly distributed heat sources the appropriate boundary

conditions would be

T = Ti at r = ri (inside surface)

T = To at r= ro (outside surface)

The general solution is still

T = − qr2

4k+C1 ln r+C2

Application of the new boundary conditions yields

T − To= q

4k(r2o − r2)+C1 ln

r

ro[2-27]

where the constant C1 is given by

C1 = Ti− To+ q (r2i − r2

o)/4k

ln (ri/ro)[2-28]

EXAMPLE 2-7 Heat Source with Convection

A current of 200 A is passed through a stainless-steel wire [k= 19 W/m · ◦C] 3 mm in diameter.The resistivity of the steel may be taken as 70 μ� · cm, and the length of the wire is 1 m. Thewire is submerged in a liquid at 110 ◦C and experiences a convection heat-transfer coefficient of4 k W/m2 · ◦C. Calculate the center temperature of the wire.

SolutionAll the power generated in the wire must be dissipated by convection to the liquid:

P = I2R= q=hA (Tw− T∞) [a]

The resistance of the wire is calculated from

R= ρ LA

= (70 × 10−6)(100)

π(0.15)2= 0.099 �

where ρ is the resistivity of the wire. The surface area of the wire is πdL, so from Equation (a),

(200)2(0.099)= 4000π(3 × 10−3)(1)(Tw− 110)= 3960 W


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C H A P T E R 2 32

andTw= 215◦C [419◦F]

The heat generated per unit volume q is calculated from

P = qV = qπr2L

so that

q= 3960

π (1.5 × 10−3)2(1)= 560.2 MW/m3 [5.41 × 107 Btu/h · ft3]

Finally, the center temperature of the wire is calculated from Equation (2-26):

T0 = qr2o4k

+ Tw= (5.602 × 108)(1.5 × 10−3)2

(4)(19)+ 215 = 231.6◦C [449◦F]

2-9 CONDUCTION-CONVECTION SYSTEMSThe heat that is conducted through a body must frequently be removed (or delivered) bysome convection process. For example, the heat lost by conduction through a furnace wallmust be dissipated to the surroundings through convection. In heat-exchanger applicationsa finned-tube arrangement might be used to remove heat from a hot liquid. The heat transferfrom the liquid to the finned tube is by convection. The heat is conducted through thematerial and finally dissipated to the surroundings by convection. Obviously, an analysisof combined conduction-convection systems is very important from a practical standpoint.

We shall defer part of our analysis of conduction-convection systems to Chapter 10on heat exchangers. For the present we wish to examine some simple extended-surfaceproblems. Consider the one-dimensional fin exposed to a surrounding fluid at a temperatureT∞ as shown in Figure 2-9. The temperature of the base of the fin is T0. We approach theproblem by making an energy balance on an element of the fin of thickness dx as shown inthe figure. Thus

Energy in left face = energy out right face + energy lost by convection

The defining equation for the convection heat-transfer coefficient is recalled as

q=hA (Tw− T∞) [2-29]

where the area in this equation is the surface area for convection. Let the cross-sectionalarea of the fin be A and the perimeter be P. Then the energy quantities are

Energy in left face = qx = −kAdTdx

Energy out right face = qx+dx = −kA dT

dx

]x+dx

= −kA(dT

dx+ d2T

dx2dx

)

Energy lost by convection = hP dx (T − T∞)


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33 2-9 Conduction-Convection Systems

Figure 2-9 Sketch illustrating one-dimensionalconduction and convection through arectangular fin.

Bas

e

dqconv = h Pdx (T − T∞)

qx

dx

qx+dx

L

Z

A

x

t

Here it is noted that the differential surface area for convection is the product of the perimeterof the fin and the differential length dx. When we combine the quantities, the energy balanceyields

d2T

dx2− hP

kA(T − T∞)= 0 [2-30a]

Let θ= T − T∞. Then Equation (2-30a) becomes

d2θ

dx2− hP

kAθ= 0 [2-30b]

One boundary condition is

θ= θ0 = T0 − T∞at x= 0

The other boundary condition depends on the physical situation. Several cases may beconsidered:

CASE 1 The fin is very long, and the temperature at the end of the fin is essentiallythat of the surrounding fluid.

CASE 2 The fin is of finite length and loses heat by convection from its end.CASE 3 The end of the fin is insulated so that dT/dx= 0 at x=L.

If we let m2 =hP/kA, the general solution for Equation (2-30b) may be written

θ=C1e−mx +C2e

mx [2-31]

For case 1 the boundary conditions are

θ = θ0at x= 0

θ = 0 atx= ∞


A = z.t (cross sectional area)As =Af= 2(t+z).L (total surface area)

As=Af

Am = L.t (profile area)

To


Am=Lt


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C H A P T E R 2 34

and the solution becomesθ

θ0= T − T∞T0 − T∞

= e−mx [2-32]

For case 3 the boundary conditions are

θ = θ0at x= 0dθ

dx= 0 atx=L

Thus

θ0 =C1 +C2

0 =m(−C1e−mL+C2e

mL)

Solving for the constants C1 and C2, we obtain

θ

θ0= e−mx

1 + e−2mL+ emx

1 + e2mL[2-33a]

= cosh [m(L− x)]coshmL

[2-33b]

The hyperbolic functions are defined as

sinh x= ex − e−x2

cosh x= ex + e−x2

tanh x= sinh x

cosh x= ex − e−xex + e−x

The solution for case 2 is more involved algebraically, and the result is

T − T∞To− T∞

= coshm(L− x)+ (h/mk) sinhm(L− x)coshmL+ (h/mk) sinhmL

[2-34]

All of the heat lost by the fin must be conducted into the base at x= 0. Using theequations for the temperature distribution, we can compute the heat loss from

q= −kA dT

dx

]x=0

An alternative method of integrating the convection heat loss could be used:

q=∫ L

0hP(T − T∞) dx=

∫ L

0hP θ dx

In most cases, however, the first equation is easier to apply. For case 1,

q= −kA (−mθ0e−m(0))= √

hPkA θ0 [2-35]

For case 3,

q= −kAθ0m

(1

1 + e−2mL− 1

1 + e+2mL

)[2-36]

= √hPkA θ0 tanhmL


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35 2-10 Fins

The heat flow for case 2 is

q= √hPkA (T0 − T∞)

sinhmL+ (h/mk) coshmL

coshmL+ (h/mk) sinhmL[2-37]

In this development it has been assumed that the substantial temperature gradients occuronly in the x direction. This assumption will be satisfied if the fin is sufficiently thin. Formost fins of practical interest the error introduced by this assumption is less than 1 percent.The overall accuracy of practical fin calculations will usually be limited by uncertainties invalues of the convection coefficient h. It is worthwhile to note that the convection coefficientis seldom uniform over the entire surface, as has been assumed above. If severe nonuniformbehavior is encountered, numerical finite-difference techniques must be employed to solvethe problem. Such techniques are discussed in Chapter 3.

2-10 FINSIn the foregoing development we derived relations for the heat transfer from a rod or finof uniform cross-sectional area protruding from a flat wall. In practical applications, finsmay have varying cross-sectional areas and may be attached to circular surfaces. In eithercase the area must be considered as a variable in the derivation, and solution of the basicdifferential equation and the mathematical techniques become more tedious. We presentonly the results for these more complex situations. The reader is referred to References 1and 8 for details on the mathematical methods used to obtain the solutions.

To indicate the effectiveness of a fin in transferring a given quantity of heat, a newparameter called fin efficiency is defined by

Fin efficiency = actual heat transferredheat that would be transferred

if entire fin area wereat base temperature

= ηf

For case 3, the fin efficiency becomes

ηf =√hP kA θ0 tanhmL

hPLθ0= tanhmL

mL[2-38]

The fins discussed were assumed to be sufficiently deep that the heat flow could beconsidered one-dimensional. The expression for mL may be written

mL=√hP

kAL=

√h(2z+ 2t)

kztL

where z is the depth of the fin, and t is the thickness. Now, if the fin is sufficiently deep, theterm 2z will be large compared with 2t, and

mL=√

2hz

ktzL=

√2h

ktL

Multiplying numerator and denominator by L1/2 gives

mL=√

2h

kLtL3/2

Lt is called the profile area of the fin, which we define as

Am=Lt


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C H A P T E R 2 36

so that

mL=√

2h

kAmL3/2 [2-39]

We may therefore use the expression in Equation (2-39) to compute the efficiency of a finwith insulated tip as given by Equation (2-38).

Harper and Brown [2] have shown that the solution in case 2 may be expressed inthe same form as Equation (2-38) when the length of the fin is extended by one-half thethickness of the fin. In effect, lengthening of the fin by t/2 is assumed to represent thesame convection heat transfer as half the fin tip area placed on top and bottom of the fin. Acorrected length Lc is then used in all the equations that apply for the case of the fin withan insulated tip. Thus

Lc =L+ t

2[2-40]

The error that results from this approximation will be less than 8 percent when(ht

2k

)1/2

≤ 1

2[2-41]

If a straight cylindrical rod extends from a wall, the corrected fin length is calculatedfrom

Lc =L+ πd2/4

πd=L+ d/4 [2-42]

Again, the real fin is extended a sufficient length to produce a circumferential area equal tothat of the tip area.

Examples of other types of fins are shown in Figure 2-10. Figure 2-11 presents acomparison of the efficiencies of a triangular fin and a straight rectangular fin correspondingto case 2. Figure 2-12 shows the efficiencies of circumferential fins of rectangular cross-sectional area. Notice that the corrected fin lengths Lc and profile area Am have beenused in Figures 2-11 and 2-12. We may note that as r2c/r1 → 1.0, the efficiency of thecircumferential fin becomes identical to that of the straight fin of rectangular profile.

It is interesting to note that the fin efficiency reaches its maximum value for the trivialcase of L= 0, or no fin at all. Therefore, we should not expect to be able to maximize finperformance with respect to fin length. It is possible, however, to maximize the efficiencywith respect to the quantity of fin material (mass, volume, or cost), and such a maximizationprocess has rather obvious economic significance. We have not discussed the subject ofradiation heat transfer from fins. The radiant transfer is an important consideration in a

Figure 2-10 Different types of finned surfaces. (a) Straight fin ofrectangular profile on plane wall, (b) straight fin ofrectangular profile on circular tube, (c) cylindrical tubewith radial fin of rectangular profile, (d) cylindrical-spineor circular-rod fin.

(a) (b) (c) (d )


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37 2-10 Fins

Figure 2-11 Efficiencies of straight rectangular and triangular fins.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5

L

tLc �

L � 2 rectangular L triangular

Am � tLc rectangular

2 Lc triangulart

t

L

t

Lc3/2(h/kAm)1/2

Fin

effi

cien

cy,

ƒη

Figure 2-12 Efficiencies of circumferential fins of rectangularprofile, according to Reference 3.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3

t

L r1r2

Lc3�2(h�kAm)1�2

Lc = L +

r2c = r1 + Lc

L = r2 − r1

Am = tLc

Fin

effi

cien

cy,

fη

t2

1

2

34 5

r2c�r1

number of applications, and the interested reader should consult Siegel and Howell [9] forinformation on this subject.

In some cases a valid method of evaluating fin performance is to compare the heattransfer with the fin to that which would be obtained without the fin. The ratio of thesequantities is

q with fin

q without fin= ηfAfhθ0

hAbθ0


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C H A P T E R 2 38

where Af is the total surface area of the fin and Ab is the base area. For the insulated-tipfin described by Equation (2-36),

Af = PL

Ab =A

and the heat ratio would become

q with fin

q without fin= tanhmL√

hA/kP

This term is sometimes called the fin effectiveness.

Thermal Resistance for Fin-Wall Combinations

Consider a fin attached to a wall as illustrated in either Figure 2-11 or Figure 2-12. Wemay calculate a thermal resistance for the wall using either Rw=�x/kA for a plane wall,or Rw= ln (ro/ri)/2πkL for a cylindrical wall. In the absence of the fin the convectionresistance at the surface would be 1/hA. The combined conduction and convection resistanceRf for the fin is related to the heat lost by the fin through

qf = ηfAfhθo= θo

Rf[2-43]

or, the fin resistance may be expressed as

Rf = 1

ηfAfh[2-44]

The overall heat transfer through the fin-wall combination is then

qf = Ti− T∞Rwf +Rf [2-45]

Figure 2-13 Heatloss from fin-wallcombination.

Wall, Rw

Fin, h, T

Ao

Ab Af

where Ti is the inside wall temperature and Rwf is the wall resistance at the fin position.This heat transfer is only for the fin portion of the wall. Now consider the wall sectionshown in Figure 2-13, having a wall area Ab for the fin and area Ao for the open section ofthe wall exposed directly to the convection environment. The open wall heat transfer is

qo= Ti− T∞Rwo+Ro [2-46]

where now

Ro= 1

hAo[2-47]

and Rwo is the wall resistance for the open wall section. This value is Rwo=�x/kwAo fora plane wall, where �x is the wall thickness. A logarithmic form would be employed for acylindrical wall, as noted above. The total heat lost by the wall is therefore

qtotal = qf + qo [2-48]


( fin performance )


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39 2-10 Fins

which may be expressed in terms of the thermal resistances by

qtotal = (Ti− T∞)[

1

Rwf +Rf + 1

Rwo+Ro]

= (Ti− T∞)Rwo+Ro+Rwf +Rf(Rwf +Rf )(Rwo+Ro) [2-49]

Conditions When Fins Do Not Help

At this point we should remark that the installation of fins on a heat-transfer surface will notnecessarily increase the heat-transfer rate. If the value of h, the convection coefficient, islarge, as it is with high-velocity fluids or boiling liquids, the fin may produce a reduction inheat transfer because the conduction resistance then represents a larger impediment to theheat flow than the convection resistance. To illustrate the point, consider a stainless-steelpin fin that has k= 16 W/m · ◦C,L= 10 cm, d= 1 cm and that is exposed to a boiling-waterconvection situation with h= 5000 W/m2 · ◦C. From Equation (2-36) we can compute

q with fin

q without fin= tanhmL√

hA/kp

=tanh

{[5000π(1 × 10−2)(4)

16π(1 × 10−2)2

]1/2

(10 × 10−2)

}

[5000π(1 × 10−2)2

(4)(16)π(1 × 10−2)

]1/2

= 1.13

Thus, this rather large pin produces an increase of only 13 percent in the heat transfer.Still another method of evaluating fin performance is discussed in Problem 2-68. Kern

and Kraus [8] give a very complete discussion of extended-surface heat transfer. Somephotographs of different fin shapes used in electronic cooling applications are shown inFigure 2-14. These fins are obviously not one-dimensional, i.e., they cannot be characterizedwith a single space coordinate.

Cautionary Remarks Concerning Convection Coefficients for Fins

We have already noted that the convection coefficient may vary with type of fluid, flowvelocity, geometry, etc. As we shall see in Chapters 5, 6, and 7, empirical correlations forh frequently have uncertainties of the order of ±25 percent. Moreover, the correlations arebased on controlled laboratory experiments that are infrequently matched in practice. Whatthis means is that the assumption of constant h used in the derivation of fin performancemay be in considerable error and the value of h may vary over the fin surface. For theheat-transfer practitioner, complex geometries like those shown in Figure 2-14 must betreated with particular care. These configurations usually must be tested under near or actualoperating conditions in order to determine their performance with acceptable reliability.These remarks are not meant to discourage the reader, but rather to urge prudence whenestimating the performance of complex finned surfaces for critical applications.


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C H A P T E R 2 40

Figure 2-14 Some fin arrangements used in electronic cooling applications.

Source: Courtesy Wakefield Engineering Inc., Wakefield, Mass.

Influence of Thermal Conductivity on FinTemperature Profiles EXAMPLE 2-8

Compare the temperature distributions in a straight cylindrical rod having a diameter of 2 cmand a length of 10 cm and exposed to a convection environment with h= 25 W/m2 · ◦C, forthree fin materials: copper [k= 385 W/m · ◦C], stainless steel [k= 17 W/m · ◦C], and glass[k= 0.8 W/m · ◦C]. Also compare the relative heat flows and fin efficiencies.

SolutionWe have

hP

kA= (25)π(0.02)

kπ(0.01)2= 5000

k


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41 2-10 Fins

The terms of interest are therefore

Material hPkA

m mL

Copper 12.99 3.604 0.3604Stainless steel 294.1 17.15 1.715Glass 6250 79.06 7.906

These values may be inserted into Equation (2-33a) to calculate the temperatures at differentx locations along the rod, and the results are shown in Figure Example 2-8. We notice that theglass behaves as a “very long” fin, and its behavior could be calculated from Equation (2-32). Thefin efficiencies are calculated from Equation (2-38) by using the corrected length approximationof Equation (2-42). We have

Lc =L+ d

4= 10 + 2

4= 10.5 cm

Figure Example 2-8

1.0

0.8

0.6

0.4

0.2

2 4 6 8 10x, cm

θθo

L = 10 cm

d = 2 cmh = 25 W�m2 • ˚C

Copper, k = 385 W�m • ˚C

Stainless steel, k = 17 W�m • ˚C

Glass, k = 0.8 W� m • ˚C

The parameters of interest for the heat-flow and efficiency comparisons are now tabulated as

Material hPkA mLc

Copper 0.190 0.3784Stainless steel 0.0084 1.8008Glass 3.9 × 10−4 8.302

To compare the heat flows we could either calculate the values from Equation (2-36) for a unitvalue of θ0 or observe that the fin efficiency gives a relative heat-flow comparison because themaximum heat transfer is the same for all three cases; i.e., we are dealing with the same fin size,shape, and value of h. We thus calculate the values of ηf from Equation (2-38) and the abovevalues of mLc.


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C H A P T E R 2 42

q relative toMaterial ηf copper, %

Copper 0.955 100Stainless steel 0.526 53.1Glass 0.124 12.6

The temperature profiles in the accompanying figure can be somewhat misleading. The glass hasthe steepest temperature gradient at the base, but its much lower value of k produces a lowerheat-transfer rate.

Straight Aluminum Fin EXAMPLE 2-9

An aluminum fin [k= 200 W/m · ◦C] 3.0 mm thick and 7.5 cm long protrudes from a wall,as in Figure 2-9. The base is maintained at 300◦C, and the ambient temperature is 50◦C withh= 10 W/m2 · ◦C. Calculate the heat loss from the fin per unit depth of material.

SolutionWe may use the approximate method of solution by extending the fin a fictitious length t/2 andthen computing the heat transfer from a fin with insulated tip as given by Equation (2-36). Wehave

Lc = L+ t/2 = 7.5 + 0.15 = 7.65 cm [3.01 in]

m=√hP

kA=[h(2z+ 2t)

ktz

]1/2≈√

2h

kt

when the fin depth z� t. So,

m=[

(2)(10)

(200)(3 × 10−3)

]1/2= 5.774

From Equation (2-36), for an insulated-tip fin

q= (tanhmLc)√hPkA θ0

For a 1 m depthA= (1)(3 × 10−3)= 3 × 10−3 m2 [4.65 in2]

and

q= (5.774)(200)(3 × 10−3)(300 − 50) tanh [(5.774)(0.0765)]= 359 W/m [373.5 Btu/h · ft]

Circumferential Aluminum Fin EXAMPLE 2-10

Aluminum fins 1.5 cm wide and 1.0 mm thick are placed on a 2.5-cm-diameter tube to dissipate theheat. The tube surface temperature is 170◦, and the ambient-fluid temperature is 25◦C. Calculatethe heat loss per fin for h= 130 W/m2 · ◦C. Assume k= 200 W/m · ◦C for aluminum.

SolutionFor this example we can compute the heat transfer by using the fin-efficiency curves inFigure 2-12. The parameters needed are

Lc =L+ t/2 = 1.5 + 0.05 = 1.55 cm

r1 = 2.5/2 = 1.25 cm

r2c = r1 +Lc = 1.25 + 1.55 = 2.80 cm


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43 2-10 Fins

r2c/r1 = 2.80/1.25 = 2.24

Am= t(r2c − r1)= (0.001)(2.8 − 1.25)(10−2)= 1.55 × 10−5 m2

L3/2c

(h

kAm

)1/2= (0.0155)3/2

[130

(200)(1.55 × 10−5)

]1/2= 0.396

From Figure 2-12, ηf = 82 percent. The heat that would be transferred if the entire fin were at thebase temperature is (both sides of fin exchanging heat)

qmax = 2π(r22c − r21)h(T0 − T∞)= 2π(2.82 − 1.252)(10−4)(130)(170 − 25)

= 74.35 W [253.7 Btu/h]

The actual heat transfer is then the product of the heat flow and the fin efficiency:

qact = (0.82)(74.35)= 60.97 W [208 Btu/h]

EXAMPLE 2-11 Rod with Heat Sources

Arod containing uniform heat sources per unit volume q is connected to two temperatures as shownin Figure Example 2-11. The rod is also exposed to an environment with convection coefficient hand temperature T∞. Obtain an expression for the temperature distribution in the rod.

Figure Example 2-11

dx

qx A qx+dx

h, T∞

T1 T2

L

q•

SolutionWe first must make an energy balance on the element of the rod shown, similar to that used toderive Equation (2-30). We have

Energy in left face + heat generated in element

= energy out right face + energy lost by convection

or

−kA dTdx

+ qA dx= −kA(dT

dx+ d2T

dx2dx

)+hP dx (T − T∞)

Simplifying, we haved2T

dx2− hP

kA(T − T∞)+ q

k= 0 [a]


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C H A P T E R 2 44

or, with θ= T − T∞ and m2 =hP/kAd2θ

dx−m2θ+ q

k= 0 [b]

We can make a further variable substitution as

θ′ = θ− q/km2

so that our differential equation becomes

d2θ′dx2

−m2θ′ = 0 [c]

which has the general solutionθ′ =C1e

−mx+C2emx [d]

The two end temperatures are used to establish the boundary conditions:

θ′ = θ′1 = T1 − T∞ − q/km2 =C1 +C2

θ′ = θ′2 = T2 − T∞ − q/km2 =C1e−mL+C2e

mL

Solving for the constants C1 and C2 gives

θ′ = (θ′1e2mL− θ′2emL)e−mx+ (θ′2emL− θ′1)emxe2mL− 1

[e]

For an infinitely long heat-generating fin with the left end maintained at T1, the temperaturedistribution becomes

θ′/θ′1 = e−mx [ f ]

a relation similar to Equation (2-32) for a non-heat-generating fin.

CommentNote that the above relationships assume one-dimensional behavior, i.e., temperature dependenceonly on the x-coordinate and temperature uniformity across the area A. For sufficiently large heatgeneration rates and/or cross-section areas, the assumption may no longer be valid. In these cases,the problem must be treated as multidimensional using the techniques described in Chapter 3.

2-11 THERMAL CONTACT RESISTANCEImagine two solid bars brought into contact as indicated in Figure 2-15, with the sides of thebars insulated so that heat flows only in the axial direction. The materials may have differentthermal conductivities, but if the sides are insulated, the heat flux must be the same throughboth materials under steady-state conditions. Experience shows that the actual temperatureprofile through the two materials varies approximately as shown in Figure 2-15b. The tem-perature drop at plane 2, the contact plane between the two materials, is said to be the result ofa thermal contact resistance. Performing an energy balance on the two materials, we obtain

q= kAA T1 − T2A

�xA= T2A− T2B

1/hcA= kBA T2B − T3

�xB

or

q= T1 − T3

�xA/kAA+ 1/hcA+�xB/kBA [2-50]


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45 2-11 Thermal Contact Resistance

where the quantity 1/hcA is called the thermal contact resistance and hc is called the contactcoefficient. This factor can be extremely important in a number of applications because ofthe many heat-transfer situations that involve mechanical joining of two materials.

The physical mechanism of contact resistance may be better understood by examininga joint in more detail, as shown in Figure 2-16. The actual surface roughness is exaggeratedto implement the discussion. No real surface is perfectly smooth, and the actual surfaceroughness is believed to play a central role in determining the contact resistance. There aretwo principal contributions to the heat transfer at the joint:

1. The solid-to-solid conduction at the spots of contact2. The conduction through entrapped gases in the void spaces created by the contact

The second factor is believed to represent the major resistance to heat flow, because thethermal conductivity of the gas is quite small in comparison to that of the solids.

Figure 2-15 Illustrations of thermal-contact-resistanceeffect: (a) physical situation;(b) temperature profile.

A Bq q

ΔxA ΔxB

T

T1

T3

T2A

T2B

1 2 3 x

(a)

(b)

Figure 2-16 Joint-roughness model for analysis ofthermal contact resistance.

B

A

T2B

T2A

Lg


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C H A P T E R 2 Steady-State Conduction—One Dimension 46

Table 2-3 Contact conductance of typical surfaces.

1/hc

Roughness Temperature, Pressure, h · ft2 · ◦F/

Surface type μ in μm ◦C atm Btu m2 · ◦C/W × 104

416 Stainless, ground, air 100 2.54 90–200 3–25 0.0015 2.64304 Stainless, ground, air 45 1.14 20 40–70 0.003 5.28416 Stainless, ground, with 100 2.54 30–200 7 0.002 3.52

0.001-in brass shim, airAluminum, ground, air 100 2.54 150 12–25 0.0005 0.88

10 0.25 150 12–25 0.0001 0.18Aluminum, ground, with 100 2.54 150 12–200 0.0007 1.23

0.001-in brass shim, airCopper, ground, air 50 1.27 20 12–200 0.00004 0.07Copper, milled, air 150 3.81 20 10–50 0.0001 0.18Copper, milled, vacuum 10 0.25 30 7–70 0.0005 0.88

Designating the contact area by Ac and the void area by Av, we may write for the heatflow across the joint

q= T2A− T2B

Lg/2kAAc +Lg/2kBAc + kfAv T2A− T2B

Lg= T2A− T2B

1/hcA

where Lg is the thickness of the void space and kf is the thermal conductivity of the fluidwhich fills the void space. The total cross-sectional area of the bars is A. Solving for hc, thecontact coefficient, we obtain

hc = 1

Lg

(Ac

A

2kAkBkA+ kB + Av

Akf

)[2-51]

In most instances, air is the fluid filling the void space and kf is small compared with kA andkB. If the contact area is small, the major thermal resistance results from the void space. Themain problem with this simple theory is that it is extremely difficult to determine effectivevalues of Ac, Av, and Lg for surfaces in contact.

From the physical model, we may tentatively conclude:

1. The contact resistance should increase with a decrease in the ambient gas pressure whenthe pressure is decreased below the value where the mean free path of the moleculesis large compared with a characteristic dimension of the void space, since the effectivethermal conductance of the entrapped gas will be decreased for this condition.

2. The contact resistance should be decreased for an increase in the joint pressure sincethis results in a deformation of the high spots of the contact surfaces, thereby creating agreater contact area between the solids.

A very complete survey of the contact-resistance problem is presented in References 4,6, 7, 10, 11. Unfortunately, there is no satisfactory theory that will predict thermal contactresistance for all types of engineering materials, nor have experimental studies yieldedcompletely reliable empirical correlations. This is understandable because of the manycomplex surface conditions that may be encountered in practice.

Radiation heat transfer across the joint can also be important when high temperatures areencountered. This energy transfer may be calculated by the methods discussed in Chapter 8.

For design purposes the contact conductance values given in Table 2-3 may be usedin the absence of more specific information. Thermal contact resistance can be reducedmarkedly, perhaps as much as 75 percent, by the use of a “thermal grease” like Dow 340.


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47 List of Worked Examples

EXAMPLE 2-12

Influence of Contact Conductanceon Heat Transfer

Two 3.0-cm-diameter 304 stainless-steel bars, 10 cm long, have ground surfaces and are exposedto air with a surface roughness of about 1 μm. If the surfaces are pressed together with a pressureof 50 atm and the two-bar combination is exposed to an overall temperature difference of 100◦C,calculate the axial heat flow and temperature drop across the contact surface.

SolutionThe overall heat flow is subject to three thermal resistances, one conduction resistance for eachbar, and the contact resistance. For the bars

Rth = �x

kA= (0.1)(4)

(16.3)π(3 × 10−2)2= 8.679◦C/W

From Table 2-2 the contact resistance is

Rc = 1

hcA= (5.28 × 10−4)(4)

π(3 × 10−2)2= 0.747◦C/W

The total thermal resistance is therefore∑Rth = (2)(8.679)+ 0.747 = 18.105

and the overall heat flow is

q= �T∑Rth

= 100

18.105= 5.52 W [18.83 Btu/h]

The temperature drop across the contact is found by taking the ratio of the contact resistance tothe total thermal resistance:

�Tc = Rc∑Rth

�T = (0.747)(100)

18.105= 4.13◦C [39.43◦F]

In this problem the contact resistance represents about 4 percent of the total resistance.

REVIEW QUESTIONS1. What is meant by the term one-dimensional when applied to conduction problems?2. What is meant by thermal resistance?3. Why is the one-dimensional heat-flow assumption important in the analysis of fins?4. Define fin efficiency.5. Why is the insulated-tip solution important for the fin problems?6. What is meant by thermal contact resistance? Upon what parameters does this resistance

depend?

LIST OF WORKED EXAMPLES2-1 Multilayer conduction2-2 Multilayer cylindrical system2-3 Heat transfer through a composite wall2-4 Cooling cost savings with extra insulation2-5 Overall heat-transfer coefficient for a tube


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C H A P T E R 2 Steady-State Conduction—One Dimension 48

2-6 Critical insulation thickness2-7 Heat source with convection2-8 Influence of thermal conductivity on fin temperature profiles2-9 Straight aluminum fin

2-10 Circumferential aluminum fin2-11 Rod with heat sources2-12 Influence of contact conductance on heat transfer

PROBLEMS2-1 A wall 2 cm thick is to be constructed from material that has an average thermal

conductivity of 1.3 W/m · ◦C. The wall is to be insulated with material having anaverage thermal conductivity of 0.35 W/m · ◦C, so that the heat loss per square meterwill not exceed 1830 W. Assuming that the inner and outer surface temperatures ofthe insulated wall are 1300 and 30◦C, calculate the thickness of insulation required.

2-2 A certain material 2.5 cm thick, with a cross-sectional area of 0.1 m2, has one sidemaintained at 35◦C and the other at 95◦C. The temperature at the center plane ofthe material is 62◦C, and the heat flow through the material is 1 kW. Obtain anexpression for the thermal conductivity of the material as a function of temperature.

2-3 Acomposite wall is formed of a 2.5-cm copper plate, a 3.2-mm layer of asbestos, anda 5-cm layer of fiberglass. The wall is subjected to an overall temperature differenceof 560◦C. Calculate the heat flow per unit area through the composite structure.

2-4 Find the heat transfer per unit area through the composite wall in Figure P2-4.Assume one-dimensional heat flow.

Figure P2-4

T = 370˚C

T = 66˚C

A

B

D

C

Ac = 0.1 m2

2.5 cm 7.5 cm 5.0 cm

kA = 150 W/m•˚CkB = 30kC = 50kD = 70AB = AD q

2-5 One side of a copper block 5 cm thick is maintained at 250◦C. The other side iscovered with a layer of fiberglass 2.5 cm thick. The outside of the fiberglass is main-tained at 35◦C, and the total heat flow through the copper-fiberglass combination is52 kW. What is the area of the slab?

2-6 An outside wall for a building consists of a 10-cm layer of common brick and a2.5-cm layer of fiberglass [k= 0.05 W/m · ◦C]. Calculate the heat flow through thewall for a 25◦C temperature differential.


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C H A P T E R

44-1 INTRODUCTIONIf a solid body is suddenly subjected to a change in environment, some time must elapsebefore an equilibrium temperature condition will prevail in the body. We refer to the equilib-rium condition as the steady state and calculate the temperature distribution and heat transferby methods described in Chapters 2 and 3. In the transient heating or cooling process thattakes place in the interim period before equilibrium is established, the analysis must bemodified to take into account the change in internal energy of the body with time, and theboundary conditions must be adjusted to match the physical situation that is apparent in theunsteady-state heat-transfer problem. Unsteady-state heat-transfer analysis is obviously ofsignificant practical interest because of the large number of heating and cooling processesthat must be calculated in industrial applications.

To analyze a transient heat-transfer problem, we could proceed by solving the generalheat-conduction equation by the separation-of-variables method, similar to the analyticaltreatment used for the two-dimensional steady-state problem discussed in Section 3-2. Wegive one illustration of this method of solution for a case of simple geometry and then referthe reader to the references for analysis of more complicated cases. Consider the infiniteplate of thickness 2L shown in Figure 4-1. Initially the plate is at a uniform temperature Ti,and at time zero the surfaces are suddenly lowered to T = T1. The differential equation is

∂2T

∂x2= 1

α

∂T

∂τ[4-1]

The equation may be arranged in a more convenient form by introduction of the variableθ= T − T1. Then

∂2θ

∂x2= 1

α

∂θ

∂τ[4-2]

with the initial and boundary conditions

θ= θi= Ti− T1 at τ= 0, 0 ≤ x≤ 2L [a]

θ= 0 at x= 0, τ> 0 [b]

θ= 0 at x= 2L, τ> 0 [c]

Unsteady-State ConductionChapter Four

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Figure 4-1 Infinite platesubjected to suddencooling of surfaces.

2L

x

T

Ti

T1

Assuming a product solution θ(x, τ)=X(x)H(τ) produces the two ordinary differentialequations

d2X

dx2+ λ2X= 0

dHdτ

+αλ2H = 0

whereλ2 is the separation constant. In order to satisfy the boundary conditions it is necessarythat λ2> 0 so that the form of the solution becomes

θ= (C1 cos λx+C2 sin λx)e−λ2ατ

From boundary condition (b), C1 = 0 for τ> 0. Because C2 cannot also be zero, we findfrom boundary condition (c) that sin 2Lλ= 0, or

λ= nπ

2Ln= 1, 2, 3, . . .

The final series form of the solution is therefore

θ=∞∑n=1

Cne−[nπ/2L]2ατ sin

nπx

2L

This equation may be recognized as a Fourier sine expansion with the constants Cn deter-mined from the initial condition (a) and the following equation:

Cn= 1

L

∫ 2L

0θi sin

nπx

2Ldx= 4

nπθi n= 1, 3, 5, . . .

The final series solution is therefore

θ

θi= T − T1

Ti− T1= 4

π

∞∑n=1

1

ne−[nπ/2L]2ατ sin

nπx

2Ln= 1, 3, 5 . . . [4-3]

We note, of course, that at time zero (τ= 0) the series on the right side of Equation (4-3)must converge to unity for all values of x.

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In Section 4-4, this solution will be presented in graphical form for calculation purposes.For now, our purpose has been to show how the unsteady-heat-conduction equation can besolved, for at least one case, with the separation-of-variables method. Further informationon analytical methods in unsteady-state problems is given in the references.

LUMPED-HEAT-CAPACITY SYSTEMWe continue our discussion of transient heat conduction by analyzing systems that may beconsidered uniform in temperature. This type of analysis is called the lumped-heat-capacitymethod. Such systems are obviously idealized because a temperature gradient must existin a material if heat is to be conducted into or out of the material. In general, the smallerthe physical size of the body, the more realistic the assumption of a uniform temperaturethroughout; in the limit a differential volume could be employed as in the derivation of thegeneral heat-conduction equation.

If a hot steel ball were immersed in a cool pan of water, the lumped-heat-capacitymethod of analysis might be used if we could justify an assumption of uniform ball tem-perature during the cooling process. Clearly, the temperature distribution in the ball woulddepend on the thermal conductivity of the ball material and the heat-transfer conditionsfrom the surface of the ball to the surrounding fluid (i.e., the surface-convection heat-transfer coefficient). We should obtain a reasonably uniform temperature distribution in theball if the resistance to heat transfer by conduction were small compared with the convectionresistance at the surface, so that the major temperature gradient would occur through thefluid layer at the surface. The lumped-heat-capacity analysis, then, is one that assumes thatthe internal resistance of the body is negligible in comparison with the external resistance.

The convection heat loss from the body is evidenced as a decrease in the internal energyof the body, as shown in Figure 4-2. Thus,

q=hA(T − T∞)= −cρV dTdτ

[4-4]

where A is the surface area for convection and V is the volume. The initial condition iswritten

T = T0 at τ= 0

so that the solution to Equation (4-4) is

T − T∞T0 − T∞

= e−[hA/ρcV ]τ [4-5]

Figure 4-2 Nomenclature for single-lump heat-capacityanalysis.

q = hA (T – T∞) = –c Vρ

ρ

dTdττ

T0

T∞

S

1hA

(b)(a)

Cth = cV

( )

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where T∞ is the temperature of the convection environment. The thermal network for thesingle-capacity system is shown in Figure 4-2b. In this network we notice that the thermalcapacity of the system is “charged” initially at the potential T0 by closing the switch S.Then, when the switch is opened, the energy stored in the thermal capacitance is dissipatedthrough the resistance 1/hA. The analogy between this thermal system and an electric systemis apparent, and we could easily construct an electric system that would behave exactly likethe thermal system as long as we made the ratio

hA

ρcV= 1

RthCthRth = 1

hACth = ρcV

equal to 1/ReCe, where Re and Ce are the electric resistance and capacitance, respectively.In the thermal system we store energy, while in the electric system we store electric charge.The flow of energy in the thermal system is called heat, and the flow of charge is calledelectric current. The quantity cρV/hA is called the time constant of the system because ithas the dimensions of time. When

τ= cρV

hA

it is noted that the temperature difference T − T∞ has a value of 36.8 percent of the initialdifference T0 − T∞.

The reader should note that the lumped-capacity formulation assumes essentially uni-form temperature throughout the solid at any instant of time so that the change in internalenergy can be represented by ρcVdT/dτ. It does not require that the convection boundarycondition have a constant value of h. In fact, variable values of h coupled with radiationboundary conditions are quite common. The specification of “time constant” in terms ofthe 36.8 percent value stated above implies a constant boundary condition.

For variable convection or radiation boundary conditions, numerical methods (seeSection 4-6) are used to advantage to predict lumped capacity behavior. A rather generalsetup of a lumped-capacity solution using numerical methods and Microsoft Excel is givenin Section D-6 of the Appendix. In some cases, multiple lumped-capacity formulations canbe useful. An example involving the combined convection-radiation cooling of a box ofelectronic components is also given in this same section of the Appendix.

Applicability of Lumped-Capacity Analysis

We have already noted that the lumped-capacity type of analysis assumes a uniform temper-ature distribution throughout the solid body and that the assumption is equivalent to sayingthat the surface-convection resistance is large compared with the internal-conduction resis-tance. Such an analysis may be expected to yield reasonable estimates within about 5 percentwhen the following condition is met:

h(V/A)

k< 0.1 [4-6]

where k is the thermal conductivity of the solid. In sections that follow, we examine thosesituations for which this condition does not apply. We shall see that the lumped-capacityanalysis has a direct relationship to the numerical methods discussed in Section 4-7. If oneconsiders the ratio V/A= s as a characteristic dimension of the solid, the dimensionlessgroup in Equation (4-6) is called the Biot number:

hs

k= Biot number = Bi

The reader should recognize that there are many practical cases where the lumped-capacitymethod may yield good results. In Table 4-1 we give some examples that illustrate therelative validity of such cases.

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Table 4-1 Examples of lumped-capacity systems.

Approximatevalue of h,

Physical situation k, W/m · ◦C W/m2 · ◦Ch(V/A)

k

1. 3.0-cm steel cube cooling in room air 40 7.0 8.75 × 10−4

2. 5.0-cm glass cylinder cooled by a 50-m/s airstream 0.8 180 2.813. Same as situation 2 but a copper cylinder 380 180 0.0064. 3.0-cm hot copper cube submerged in water 380 10,000 0.132

such that boiling occurs

We may point out that uncertainties in the knowledge of the convection coefficient of±25 percent are quite common, so that the condition Bi =h(V/A)/k< 0.1 should allow forsome leeway in application.

Do not dismiss lumped-capacity analysis because of its simplicity. Because of uncer-tainties in the convection coefficient, it may not be necessary to use more elaborate analysistechniques.

Steel Ball Cooling in Air EXAMPLE 4-1

A steel ball [c= 0.46 kJ/kg · ◦C, k= 35 W/m · ◦C] 5.0 cm in diameter and initially at a uniformtemperature of 450◦C is suddenly placed in a controlled environment in which the temperatureis maintained at 100◦C. The convection heat-transfer coefficient is 10 W/m2 · ◦C. Calculate thetime required for the ball to attain a temperature of 150◦C.

SolutionWe anticipate that the lumped-capacity method will apply because of the low value of h and highvalue of k. We can check by using Equation (4-6):

h(V/A)

k= (10)[(4/3)π(0.025)3]

4π(0.025)2(35)= 0.0023< 0.1

so we may use Equation (4-5). We have

T = 150◦C ρ= 7800 kg/m3 [486 lbm/ft3]

T∞ = 100◦C h= 10 W/m2 · ◦C [1.76Btu/h · ft2 · ◦F]T0 = 450◦C c= 460 J/kg · ◦C [0.11 Btu/lbm · ◦F]

hA

ρcV= (10)4π(0.025)2

(7800)(460)(4π/3)(0.025)3= 3.344 × 10−4 s−1

T − T∞T0 − T∞

= e−[hA/ρcV ]τ

150 − 100

450 − 100= e−3.344×10−4τ

τ = 5819 s = 1.62 h

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5-1 INTRODUCTIONThe preceding chapters have considered the mechanism and calculation of conduction heattransfer. Convection was considered only insofar as it related to the boundary conditionsimposed on a conduction problem. We now wish to examine the methods of calculatingconvection heat transfer and, in particular, the ways of predicting the value of the convectionheat-transfer coefficienth.The subject of convection heat transfer requires an energy balancealong with an analysis of the fluid dynamics of the problems concerned. Our discussion inthis chapter will first consider some of the simple relations of fluid dynamics and boundary-layer analysis that are important for a basic understanding of convection heat transfer. Next,we shall impose an energy balance on the flow system and determine the influence of theflow on the temperature gradients in the fluid. Finally, having obtained a knowledge of thetemperature distribution, the heat-transfer rate from a heated surface to a fluid that is forcedover it may be determined.

Our development in this chapter is primarily analytical in character and is concernedonly with forced-convection flow systems. Subsequent chapters will present empirical rela-tions for calculating forced-convection heat transfer and will also treat the subjects of naturalconvection and boiling and condensation heat transfer.

5-2 VISCOUS FLOWConsider the flow over a flat plate as shown in Figures 5-1 and 5-2. Beginning at theleading 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 τ between the fluid layers.If this stress is assumed to be proportional to the normal velocity gradient, we have thedefining equation for the viscosity,

τ=μdudy

[5-1]

The constant of proportionality μ is called the dynamic viscosity. A typical set of units isnewton-seconds per square meter; however, many sets of units are used for the viscosity,and care must be taken to select the proper group that will be consistent with the formulationat hand.

The region of flow that develops from the leading edge of the plate in which the effectsof viscosity are observed is called the boundary layer. Some arbitrary point is used to

CHAPTER FIVECONVECTION HEAT TRANSFER

Dr. Arif Al-Qassar

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Figure 5-1 Sketch showing different boundary-layer flow regimes on aflat plate.

Laminarsublayer

u∞

u

u∞

u

Laminar region Transition Turbulent

x

y

Figure 5-2 Laminar velocity profile

on a flat plate.

dudy

= μτ

y

x

u∞

u

designate the y position where the boundary layer ends; this point is usually chosen as they coordinate where the velocity becomes 99 percent of the free-stream value.

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

The transition from laminar to turbulent flow occurs whenu∞xν

= ρu∞xμ

> 5 × 105

where

u∞ = free-stream velocity, m/s

x= distance from leading edge, m

ν=μ/ρ= kinematic viscosity, m2/s

This particular grouping of terms is called the Reynolds number, and is dimensionless if aconsistent set of units is used for all the properties:

Rex = u∞xν

[5-2]

Although the critical Reynolds number for transition on a flat plate is usually taken as5 × 105 for most analytical purposes, the critical value in a practical situation is stronglydependent on the surface-roughness conditions and the “turbulence level” of the free stream.The normal range for the beginning of transition is between 5 × 105 and 106. With very large

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disturbances present in the flow, transition may begin with Reynolds numbers as low as105, and for flows that are very free from fluctuations, it may not start until Re = 2 × 106 ormore. In reality, the transition process is one that covers a range of Reynolds numbers, withtransition being complete and with developed turbulent flow usually observed at Reynoldsnumbers twice the value at which transition began.

The relative shapes for the velocity profiles in laminar and turbulent flow are indicatedin Figure 5-1. The laminar profile is approximately parabolic, while the turbulent profilehas a portion near the wall that is very nearly linear. This linear portion is said to be dueto a laminar sublayer that hugs the surface very closely. Outside this sublayer the velocityprofile is relatively flat in comparison with the laminar profile.

The physical mechanism of viscosity is one of momentum exchange. Consider thelaminar-flow situation. Molecules may move from one lamina to another, carrying with thema momentum corresponding to the velocity of the flow. There is a net momentum transportfrom regions of high velocity to regions of low velocity, thus creating a force in the directionof the flow. This force is the viscous-shear stress, which is calculated with Equation (5-1).

The rate at which the momentum transfer takes place is dependent on the rate at whichthe molecules move across the fluid layers. In a gas, the molecules would move aboutwith some average speed proportional to the square root of the absolute temperature since,in the kinetic theory of gases, we identify temperature with the mean kinetic energy of amolecule. The faster the molecules move, the more momentum they will transport. Hencewe should expect the viscosity of a gas to be approximately proportional to the square root oftemperature, and this expectation is corroborated fairly well by experiment. The viscositiesof some typical fluids are given in Appendix A.

In the turbulent-flow region, distinct fluid layers are no longer observed, and we areforced to seek a somewhat different concept for viscous action. A qualitative picture of theturbulent-flow process may be obtained by imagining macroscopic chunks of fluid trans-porting energy and momentum instead of microscopic transport on the basis of individualmolecules. Naturally, we should expect the larger mass of the macroscopic elements of fluidto transport more energy and momentum than the individual molecules, and we should alsoexpect a larger viscous-shear force in turbulent flow than in laminar flow (and a larger ther-mal conductivity as well). This expectation is verified by experiment, and it is this largerviscous action in turbulent flow which causes the flat velocity profile indicated in Figure 5-1.

Consider the flow in a tube as shown in Figure 5-3. A boundary layer develops at theentrance, as shown. Eventually the boundary layer fills the entire tube, and the flow is saidto be fully developed. If the flow is laminar, a parabolic velocity profile is experienced, asshown in Figure 5-3a. When the flow is turbulent, a somewhat blunter profile is observed,as in Figure 5-3b. In a tube, the Reynolds number is again used as a criterion for laminarand turbulent flow. For

Red = umd

ν> 2300 [5-3]

the flow is usually observed to be turbulent d is the tube diameter.Again, a range of Reynolds numbers for transition may be observed, depending on the

pipe roughness and smoothness of the flow. The generally accepted range for transition is

2000<Red < 4000

although laminar flow has been maintained up to Reynolds numbers of 25,000 in carefullycontrolled laboratory conditions.

The continuity relation for one-dimensional flow in a tube is

m= ρumA [5-4]

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Figure 5-3 Velocity profile for (a) laminar flow in a tube and (b) turbulent tube flow.

Starting length

(b)

Fully developedflow

Uniforminletflow

Boundary layer

Laminar sublayer

Turbulentcore

(a)

where

m= mass rate of flow

um= mean velocity

A= cross-sectional area

We define the mass velocity as

Mass velocity =G= m

A= ρum [5-5]

so that the Reynolds number may also be written

Red = Gd

μ[5-6]

Equation (5-6) is sometimes more convenient to use than Equation (5-3).

INVISCID FLOWAlthough no real fluid is inviscid, in some instances the fluid may be treated as such, and it isworthwhile to present some of the equations that apply in these circumstances. For example,in the flat-plate problem discussed above, the flow at a sufficiently large distance from theplate will behave as a nonviscous flow system. The reason for this behavior is that thevelocity gradients normal to the flow direction are very small, and hence the viscous-shearforces are small.

If a balance of forces is made on an element of incompressible fluid and these forcesare set equal to the change in momentum of the fluid element, the Bernoulli equation forflow along a streamline results:

p

ρ+ 1

2

V 2

gc= const [5-7a]

5-3

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or, in differential form,dp

ρ+ V dV

gc= 0 [5-7b]

where

ρ= fluid density, kg/m3

p= pressure at particular point in flow, Pa

V = velocity of flow at that point, m/s

The Bernoulli equation is sometimes considered an energy equation because the V 2/2gcterm represents kinetic energy and the pressure represents potential energy; however, itmust be remembered that these terms are derived on the basis of a dynamic analysis, so thatthe equation is fundamentally a dynamic equation. In fact, the concept of kinetic energy isbased on a dynamic analysis.

When the fluid is compressible, an energy equation must be written that will take intoaccount changes in internal thermal energy of the system and the corresponding changesin temperature. For a one-dimensional flow system this equation is the steady-flow energyequation for a control volume,

i1 + 1

2gcV 2

1 +Q= i2 + 1

2gcV 2

2 +Wk [5-8]

where i is the enthalpy defined by

i= e+pv [5-9]

and where

e= internal energy

Q= heat added to control volume

Wk= net external work done in the process

v= specific volume of fluid

(The symbol i is used to denote the enthalpy instead of the customary h to avoid confusionwith the heat-transfer coefficient.) The subscripts 1 and 2 refer to entrance and exit conditionsto the control volume. To calculate pressure drop in compressible flow, it is necessary tospecify the equation of state of the fluid, for example, for an ideal gas,

p= ρRT �e= cv�T �i= cp�T

The gas constant for a particular gas is given in terms of the universal gas constant � as

R= �M

where M is the molecular weight and � = 8314.5 J/kg · mol · K. For air, the appropriateideal-gas properties are

Rair = 287 J/kg · K cp,air = 1.005 kJ/kg · ◦C cv,air = 0.718 kJ/kg · ◦C

To solve a particular problem, we must also specify the process. For example, reversibleadiabatic flow through a nozzle yields the following familiar expressions relating the prop-erties at some point in the flow to the Mach number and the stagnation properties, i.e., the

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properties where the velocity is zero:

T0

T= 1 + γ − 1

2M2

p0

p=(

1 + γ − 1

2M2

)γ/(γ−1)

ρ0

ρ=(

1 + γ − 1

2M2

)1/(γ−1)

where

T0, p0, ρ0 = stagnation properties

γ = ratio of specific heats cp/cvM= Mach number

M= V

a

where a is the local velocity of sound, which may be calculated from

a=√γgcRT [5-10]

for an ideal gas.† For air behaving as an ideal gas this equation reduces to

a= 20.045√T m/s [5-11]

where T is in degrees Kelvin.

EXAMPLE 5-1 Water Flow in a Diffuser

Water at 20◦C flows at 8 kg/s through the diffuser arrangement shown in Figure Example 5-1. Thediameter at section 1 is 3.0 cm, and the diameter at section 2 is 7.0 cm. Determine the increase instatic pressure between sections 1 and 2. Assume frictionless flow.

Figure Example 5-1

Flow

1 2

SolutionThe flow cross-sectional areas are

A1 = πd21

4= π(0.03)2

4= 7.069 × 10−4 m2

A2 = πd22

4= π(0.07)2

4= 3.848 × 10−3 m2

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C H A P T E R 5 Principles of Convection 221

The density of water at 20◦C is 1000 kg/m3, and so we may calculate the velocities from themass-continuity relation

u= m

ρA

u1 = 8.0

(1000)(7.069 × 10−4)= 11.32 m/s [37.1 ft/s]

u2 = 8.0

(1000)(3.848 × 10−3)= 2.079 m/s [6.82 ft/s]

The pressure difference is obtained from the Bernoulli equation (5-7a):

p2 −p1

ρ= 1

2gc(u2

1 − u22)

p2 −p1 = 1000

2[(11.32)2 − (2.079)2]

= 61.91 kPa [8.98 lb/in2 abs]

Isentropic Expansion of Air EXAMPLE 5-2

Air at 300◦C and 0.7 MPa pressure is expanded isentropically from a tank until the velocity is300 m/s. Determine the static temperature, pressure, and Mach number of the air at the high-velocity condition. γ = 1.4 for air.

SolutionWe may write the steady-flow energy equation as

i1 = i2 + u22

2gc

because the initial velocity is small and the process is adiabatic. In terms of temperature,

cp(T1 − T2)= u22

2gc

(1005)(300 − T2)= (300)2

(2)(1.0)

T2 = 255.2◦C = 528.2 K [491.4◦F]

We may calculate the pressure from the isentropic relation

p2

p1=(T2

T1

)γ/(γ−1)

p2 = (0.7)

(528.2

573

)3.5= 0.526 MPa [76.3 lb/in2 abs]

The velocity of sound at condition 2 is

a2 = (20.045)(528.2)1/2 = 460.7 m/s [1511 ft/s]

so that the Mach number is

M2 = u2

a2= 300

460.7= 0.651

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# 101675 Cust: McGraw-Hill Au: Holman

Title: Heat Transfer 10/e Server: Short / Normal / Long Publishing Services

LAMINAR BOUNDARY LAYER ON A FLAT PLATEConsider the elemental control volume shown in Figure 5-4. We derive the equation ofmotion 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.

We apply Newton’s second law of motion,∑

Fx = d(mV )x

dτ

Figure 5-4 Elemental control volume for force balance on laminar boundary layer.

dyu∞ dx

dx

y

x

v

p dy

u

v+ dvdy

dy

u+dx

dx

(p+ dx) dy

dx

dx[ dy]dudy dy

dudy

du

dydu

dx

dp

+∂ ∂∂∂ ∂

∂

∂∂

∂∂

∂∂

∂∂

μ

μ

dy

At x= 0, δ= 0, so that

δ= 4.64

√νx

u∞ [5-20]

This may be written in terms of the Reynolds number as

δ

x= 4.64

Re1/2x

where Rex = u∞xν

[5-21]

The exact solution of the boundary-layer equations as given in Appendix B yields

δ

x= 5.0

Re1/2x

[5-21a]

The boundary-layer thickness is calculated from Equation (5-21):

At x= 20 cm: δ= (4.64)(0.2)

(25,448)1/2= 0.00582 m [0.24 in]

At x= 40 cm: δ= (4.64)(0.4)

(50,897)1/2= 0.00823 m [0.4 in]

Mass Flow and Boundary-Layer Thickness EXAMPLE 5-3

Air at 27◦C and 1 atm flows over a flat plate at a speed of 2 m/s. Calculate the boundary-layerthickness at distances of 20 cm and 40 cm from the leading edge of the plate. Calculate the massflow that enters the boundary layer between x= 20 cm and x= 40 cm. The viscosity of air at 27◦Cis 1.85 × 10−5 kg/m · s. Assume unit depth in the z direction.

SolutionThe density of air is calculated from

ρ= p

RT= 1.0132 × 105

(287)(300)= 1.177 kg/m3 [0.073 lbm/ft

3]

The Reynolds number is calculated as

At x= 20 cm: Re = (1.177)(2.0)(0.2)

1.85 × 10−5 = 25,448

At x= 40 cm: Re = (1.177)(2.0)(0.4)

1.85 × 10−5 = 50,897

The boundary-layer thickness is calculated from Equation (5-21):

At x= 20 cm: δ= (4.64)(0.2)

(25,448)1/2= 0.00582 m [0.24 in]

At x= 40 cm: δ= (4.64)(0.4)

(50,897)1/2= 0.00823 m [0.4 in]

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# 101675

Title: Heat Transfer 10/e

THE THERMAL BOUNDARY LAYER

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

Consider the system shown in Figure 5-7. The temperature of the wall is Tw, thetemperature of the fluid outside the thermal boundary layer is T∞, and the thickness of thethermal boundary layer is designated as δt . At the wall, the velocity is zero, and the heattransfer into the fluid takes place by conduction. Thus the local heat flux per unit area, q′′, is

q

A= q′′ = −k ∂T

∂y

]wall

[5-27]

From Newton’s law of cooling [Equation (1-8)],

q′′ =h(Tw− T∞) [5-28]

Figure 5-7 Temperature profile in thethermal boundary layer.

T∞

y

x

δ t

Twqw�A=–k

w y∂ T∂

Figure 5-9 Hydrodynamic and thermal boundary layers on aflat plate. Heating starts at x= x0.

δtδδ

x0

u∞

T∞

In the foregoing analysis the assumption was made that ζ< 1. This assumption issatisfactory for fluids having Prandtl numbers greater than about 0.7. Fortunately, mostgases and liquids fall within this category. Liquid metals are a notable exception, however,since they have Prandtl numbers of the order of 0.01.

The Prandtl number ν/α has been found to be the parameter that relates the relativethicknesses of the hydrodynamic and thermal boundary layers. The kinematic viscosity ofa fluid conveys information about the rate at which momentum may diffuse through thefluid because of molecular motion. The thermal diffusivity tells us the same thing in regardto the diffusion of heat in the fluid. Thus the ratio of these two quantities should expressthe relative magnitudes of diffusion of momentum and heat in the fluid. But these diffusionrates are precisely the quantities that determine how thick the boundary layers will be for agiven external flow field; large diffusivities mean that the viscous or temperature influenceis felt farther out in the flow field. The Prandtl number is thus the connecting link betweenthe velocity field and the temperature field.

The Prandtl number is dimensionless when a consistent set of units is used:

Pr = ν

α= μ/ρ

k/ρcp= cpμ

k[5-39]

In the SI system a typical set of units for the parameters would beμ in kilograms per secondper meter, cp in kilojoules per kilogram per Celsius degree, and k in kilowatts per meterper Celsius degree. In the English system one would typically employ μ in pound mass perhour per foot, cp in Btu per pound mass per Fahrenheit degree, and k in Btu per hour perfoot per Fahrenheit degree.

dimensionless group on the left side,

Nux = hxx

k[5-42]

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Substituting for the hydrodynamic-boundary-layer thickness from Equation (5-21) andusing Equation (5-36) gives

hx = 0.332k Pr1/3(u∞νx

)1/2[

1 −(x0

x

)3/4]−1/3

[5-41]

The equation may be nondimensionalized by multiplying both sides by x/k, producing thedimensionless group on the left side,

Nux = hxx

k[5-42]

called the Nusselt number after Wilhelm Nusselt, who made significant contributions to thetheory of convection heat transfer. Finally,

Nux = 0.332Pr1/3 Re1/2x

[1 −

(x0

x

)3/4]−1/3

[5-43]

or, for the plate heated over its entire length, x0 = 0 and

Nux = 0.332Pr1/3 Re1/2x [5-44]

Equations (5-41), (5-43), and (5-44) express the local values of the heat-transfer coefficientin terms of the distance from the leading edge of the plate and the fluid properties. For thecase where x0 = 0 the average heat-transfer coefficient and Nusselt number may be obtainedby integrating over the length of the plate:

h=∫ L

0 hx dx∫ L0 dx

= 2hx=L [5-45a]

For a plate where heating starts at x= x0, it can be shown that the average heat transfercoefficient can be expressed as

hx0−Lhx=L

= 2L1 − (x0/L)

3/4

L− x0[5-45b]

In this case, the total heat transfer for the plate would be

qtotal =hx0−L(L− x0)(Tw− T∞)

assuming the heated section is at the constant temperature Tw. For the plate heated over theentire length,

NuL= hL

k= 2 Nux=L [5-46a]

or

NuL= hL

k= 0.664 Re1/2

L Pr1/3 [5-46b]

where ReL= ρu∞Lμ

The reader should carry out the integrations to verify these results.The foregoing analysis was based on the assumption that the fluid properties were

constant throughout the flow. When there is an appreciable variation between wall andfree-stream conditions, it is recommended that the properties be evaluated at the so-calledfilm temperature Tf , defined as the arithmetic mean between the wall and free-streamtemperature,

Tf = Tw+ T∞2

[5-47]

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Constant Heat Flux

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

Nux = hx

k= 0.453 Re1/2

x Pr1/3 [5-48]

which may be expressed in terms of the wall heat flux and temperature difference as

Nux = qwx

k(Tw− T∞)[5-49]

The average temperature difference along the plate, for the constant-heat-flux condition,may be obtained by performing the integration

Tw− T∞ = 1

L

∫ L

0(Tw− T∞) dx= 1

L

∫ L

0

qwx

k Nuxdx

= qwL/k

0.6795 Re1/2L Pr1/3

[5-50]

or

qw= 32hx=L(Tw− T∞)

In these equations qw is the heat flux per unit area and will have the units of wattsper square meter (W/m2) in SI units or British thermal units per hour per square foot(Btu/h · ft2) in the English system. Note that the heat flux qw= q/A is assumed constantover the entire plate surface.

Other Relations

Equation (5-44) 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 [9] have correlated a large amount of data to give the following relationfor laminar flow on an isothermal flat plate:

Nux = 0.3387 Re1/2x Pr1/3

[1 +

(0.0468

Pr

)2/3]1/4

for Rex Pr> 100 [5-51]

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Isothermal Flat Plate Heated Over Entire Length EXAMPLE 5-4

For the flow system in Example 5-3 assume that the plate is heated over its entire length to atemperature of 60◦C. Calculate the heat transferred in (a) the first 20 cm of the plate and (b) thefirst 40 cm of the plate.

SolutionThe total heat transfer over a certain length of the plate is desired; so we wish to calculate averageheat-transfer coefficients. For this purpose we use Equations (5-44) and (5-45), evaluating theproperties at the film temperature:

Tf = 27 + 60

2= 43.5◦C = 316.5 K [110.3◦F]

From Appendix A the properties are

ν= 17.36 × 10−6 m2/s [1.87 × 10−4 ft2/s]k= 0.02749 W/m · ◦C [0.0159 Btu/h · ft · ◦F]

Pr = 0.7

cp= 1.006 kJ/kg · ◦C [0.24 Btu/lbm · ◦F]At x= 20 cm

Rex= u∞xν

= (2)(0.2)

17.36 × 10−6= 23,041

Nux= hxx

k= 0.332Re1/2

x Pr1/3

= (0.332)(23,041)1/2(0.7)1/3 = 44.74

hx= Nux

(k

x

)= (44.74)(0.02749)

0.2

= 6.15 W/m2 · ◦C [1.083 Btu/h · ft2 · ◦F]

The average value of the heat-transfer coefficient is twice this value, or

h= (2)(6.15)= 12.3 W/m2 · ◦C [2.17 Btu/h · ft2 · ◦F]

The heat flow isq=hA(Tw− T∞)

If we assume unit depth in the z direction,

q= (12.3)(0.2)(60 − 27)= 81.18 W [277 Btu/h]

At x= 40 cm

Rex= u∞xν

= (2)(0.4)

17.36 × 10−6= 46,082

Nux= (0.332)(46,082)1/2(0.7)1/3 = 63.28

hx= (63.28)(0.02749)

0.4= 4.349 W/m2 · ◦C

h= (2)(4.349)= 8.698 W/m2 · ◦C [1.53 Btu/h · ft2 · ◦F]q= (8.698)(0.4)(60 − 27)= 114.8 W [392 Btu/h]

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EXAMPLE 5-5 Flat Plate with Constant Heat Flux

A 1.0-kW heater is constructed of a glass plate with an electrically conducting film that producesa constant heat flux. The plate is 60 cm by 60 cm and placed in an airstream at 27◦C, 1 atm withu∞ = 5 m/s. Calculate the average temperature difference along the plate and the temperaturedifference at the trailing edge.

SolutionProperties should be evaluated at the film temperature, but we do not know the plate temperature.So for an initial calculation, we take the properties at the free-stream conditions of

T∞ = 27◦C = 300 K

ν= 15.69 × 10−6 m2/s Pr = 0.708 k= 0.02624 W/m · ◦C

ReL= (0.6)(5)

15.69 × 10−6= 1.91 × 105

From Equation (5-50) the average temperature difference is

Tw− T∞ = [1000/(0.6)2](0.6)/0.02624

0.6795(1.91 × 105)1/2(0.708)1/3= 240◦C

Now, we go back and evaluate properties at

Tf = 240 + 27 + 27

2= 147◦C = 420 K

and obtain

ν= 28.22 × 10−6 m2/s Pr = 0.687 k= 0.035 W/m · ◦C

ReL= (0.6)(5)

28.22 × 10−6= 1.06 × 105

Tw− T∞ = [1000/(0.6)2](0.6)/0.035

0.6795(1.06 × 105)1/2(0.687)1/3= 243◦C

At the end of the plate (x=L= 0.6 m) the temperature difference is obtained from Equations(5-48) and (5-50) with the constant 0.453 to give

(Tw− T∞)x=L= (243.6)(0.6795)

0.453= 365.4◦C

An alternate solution would be to base the Nusselt number on Equation (5-51).

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Plate with Unheated Starting Length EXAMPLE 5-6

Air at 1 atm and 300 K flows across a 20-cm-square plate at a free-stream velocity of 20 m/s. Thelast half of the plate is heated to a constant temperature of 350 K. Calculate the heat lost by theplate.

SolutionFirst we evaluate the air properties at the film temperature

Tf = (Tw+ T∞)/2 = 325 K

and obtainv= 18.23 × 10−6m2/s k= 0.02814 W/m · ◦C Pr = 0.7

At the trailing edge of the plate the Reynolds number is

ReL= u∞L/v= (20)(0.2)/18.23 × 10−6 = 2.194 × 105

or, laminar flow over the length of the plate.Heating does not start until the last half of the plate, or at a position x0 = 0.1 m. The local

heat-transfer coefficient for this condition is given by Equation (5-41):

hx= 0.332k Pr1/3(u∞/vx)1/2[1 − (x0/x)0.75]−1/3 [a]

Inserting the property values along with x0 = 0.1 gives

hx= 8.6883x−1/2(1 − 0.17783x−0.75)−1/3 [b]

The plate is 0.2 m wide so the heat transfer is obtained by integrating over the heated lengthx0<x<L

q= (0.2)(Tw− T∞)∫ L= 0.2

x0 = 0.1hxdx [c]

Inserting Equation (b) in Equation (c) and performing the numerical integration gives

q= (0.2)(8.6883)(0.4845)(350 − 300)= 421 W [d]

The average value of the heat-transfer coefficient over the heated length is given by

h= q/(Tw− T∞)(L− x0)W = 421/(350 − 300)(0.2 − 0.1)(0.2)= 421 W/m2 · ◦C

where W is the width of the plate.An easier calculation can be made by applying Equation (5-45b) to determine the average

heat transfer coefficient over the heated portion of the plate. The result is

h= 425.66 W/m2 · ◦C and q= 425.66 W

which indicates, of course, only a small error in the numerical integeration.

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EXAMPLE 5-7 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 heatedto a uniform temperature of 60◦C. Calculate the heat lost by the plate.

SolutionWe first evaluate the film temperature:

Tf = 20 + 60

2= 40◦C

The properties of engine oil are

ρ = 876 kg/m3 ν = 0.00024 m2/sk = 0.144 W/m · ◦C Pr = 2870

The Reynolds number is

Re = u∞Lν

= (1.2)(0.2)

0.00024= 1000

Because the Prandtl number is so large we will employ Equation (5-51) for the solution. We seethat hx varies with x in the same fashion as in Equation (5-44), that is, hx∝ x−1/2, so that weget the same solution as in Equation (5-45) for the average heat-transfer coefficient. EvaluatingEquation (5-51) at x= 0.2 gives

Nux= (0.3387)(1000)1/2(2870)1/3[1 +

(0.0468

2870

)2/3]1/4

= 152.2

and

hx= (152.2)(0.144)

0.2= 109.6 W/m2 · ◦C

The average value of the convection coefficient is

h= (2)(109.6)= 219.2 W/m2 · ◦C

so that the total heat transfer is

q=hA(Tw− T∞)= (219.2)(0.2)2(60 − 20)= 350.6 W

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THE RELATION BETWEEN FLUID FRICTION AND HEAT TRANSFER

We have already seen that the temperature and flow fields are related. Now we seek anexpression 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 Cf :

τw=Cf ρu2∞

2[5-52]

Equation (5-52) is the defining equation for the friction coefficient. The shear stress mayalso be calculated from the relation

τw=μ ∂u

∂y

]w

Using the velocity distribution given by Equation (5-19), we have

τw= 3

2

μu∞δ

and making use of the relation for the boundary-layer thickness gives

τw= 3

2

μu∞4.64

(u∞νx

)1/2[5-53]

Combining Equations (5-52) and (5-53) leads to

Cfx

2= 3

2

μu∞4.64

(u∞νx

)1/2 1

ρu2∞= 0.323 Re−1/2

x [5-54]

The exact solution of the boundary-layer equations yields

Cfx

2= 0.332 Re−1/2

x [5-54a]

Equation (5-44) may be rewritten in the following form:

NuxRex Pr

= hx

ρcpu∞= 0.332 Pr−2/3 Re−1/2

x

The group on the left is called the Stanton number,

Stx = hx

ρcpu∞

so that

Stx Pr2/3 = 0.332 Re1/2x [5-55]

and write

StxPr2/3 = Cfx

2[5-56]

and write

StxPr2/3 = Cfx

2[5-56]

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Equation (5-56), called the Reynolds-Colburn analogy, expresses the relation between fluidfriction and heat transfer for laminar flow on a flat plate. The heat-transfer coefficientthus could be determined by making measurements of the frictional drag on a plate underconditions in which no heat transfer is involved.

It turns out that Equation (5-56) can also be applied to turbulent flow over a flat plateand in a modified way to turbulent flow in a tube. It does not apply to laminar tube flow. Ingeneral, a more rigorous treatment of the governing equations is necessary when embarkingon new applications of the heat-transfer–fluid-friction analogy, and the results do not alwaystake the simple form of Equation (5-56). The interested reader may consult the referencesat the end of the chapter for more information on this important subject. At this point, thesimple analogy developed above has served to amplify our understanding of the physicalprocesses in convection and to reinforce the notion that heat-transfer and viscous-transportprocesses are related at both the microscopic and macroscopic levels.

EXAMPLE 5-8 Drag Force on a Flat Plate

For the flow system in Example 5-4 compute the drag force exerted on the first 40 cm of the plateusing the analogy between fluid friction and heat transfer.

SolutionWe use Equation (5-56) to compute the friction coefficient and then calculate the drag force. Anaverage friction coefficient is desired, so

St Pr2/3 = Cf

2[a]

The density at 316.5 K is

ρ= p

RT= 1.0132 × 105

(287)(316.5)= 1.115 kg/m3

For the 40-cm length

St = h

ρcpu∞= 8.698

(1.115)(1006)(2)= 3.88 × 10−3

Then from Equation (a)

Cf

2= (3.88 × 10−3)(0.7)2/3 = 3.06 × 10−3

The average shear stress at the wall is computed from Equation (5-52):

τw=Cf ρu2∞2

= (3.06 × 10−3)(1.115)(2)2

= 0.0136 N/m2

The drag force is the product of this shear stress and the area,

D= (0.0136)(0.4)= 5.44 mN [1.23 × 10−3 lbf ]

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TURBULENT-BOUNDARY-LAYER HEAT TRANSFER

Consider a portion of a turbulent boundary layer as shown in Figure 5-10. A very thin regionnear the plate surface has a laminar character, and the viscous action and heat transfer takeplace under circumstances like those in laminar flow. Farther out, at larger y distancesfrom the plate, some turbulent action is experienced, but the molecular viscous action andheat conduction are still important. This region is called the buffer layer. Still farther out,the flow is fully turbulent, and the main momentum- and heat-exchange mechanism is oneinvolving macroscopic lumps of fluid moving about in the flow. In this fully turbulent regionwe speak of eddy viscosity and eddy thermal conductivity. These eddy properties may be10 to 20 times as large as the molecular values.

The physical mechanism of heat transfer in turbulent flow is quite similar to that inlaminar flow; the primary difference is that one must deal with the eddy properties insteadof the ordinary thermal conductivity and viscosity. The main difficulty in an analyticaltreatment is that these eddy properties vary across the boundary layer, and the specificvariation can be determined only from experimental data. This is an important point. Allanalyses of turbulent flow must eventually rely on experimental data because there is nocompletely adequate theory to predict turbulent-flow behavior.

Figure 5-10 Velocity profile in turbulent boundary layer on a flat plate.

Buffer layerLaminarsublayer

Turbulenty

x

u∞

u

Stx Pr2/3 = 0.0296 Re−1/5x 5 × 105<Rex < 107 [5-81]

or

Stx Pr2/3 = 0.185(log Rex)−2.584 107<Rex < 109 [5-82]

The average heat transfer over the entire laminar-turbulent boundary layer is

St Pr2/3 = Cf

2[5-83]

For Recrit = 5 × 105 and ReL < 107, Equation (5-80) can be used to obtain

St Pr2/3 = 0.037 Re−1/5L − 871 Re−1

L [5-84]

Recalling that St = Nu/(ReL Pr), we can rewrite Equation (5-84) as

NuL= hL

k= Pr1/3(0.037 Re0.8

L − 871) [5-85]

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Turbulent Heat Transfer from Isothermal Flat Plate EXAMPLE 5-9

Air at 20◦C and 1 atm flows over a flat plate at 35 m/s. The plate is 75 cm long and is maintainedat 60◦C. Assuming unit depth in the z direction, calculate the heat transfer from the plate.

SolutionWe evaluate properties at the film temperature:

Tf = 20 + 60

2= 40◦C = 313 K

ρ= p

RT= 1.0132 × 105

(287)(313)= 1.128 kg/m3

μ= 1.906 × 10−5 kg/m · s

Pr = 0.7 k= 0.02723 W/m · ◦C cp= 1.007 kJ/kg · ◦C

The Reynolds number is

ReL= ρu∞Lμ

= (1.128)(35)(0.75)

1.906 × 10−5 = 1.553 × 106

and the boundary layer is turbulent because the Reynolds number is greater than 5 × 105. Therefore,we use Equation (5-85) to calculate the average heat transfer over the plate:

NuL= hL

k= Pr1/3(0.037 Re0.8

L − 871)

= (0.7)1/3[(0.037)(1.553 × 106)0.8 − 871] = 2180

h= NuLk

L= (2180)(0.02723)

0.75= 79.1 W/m2 · ◦C [13.9 Btu/h · ft2 · ◦F]

q=hA(Tw− T∞)= (79.1)(0.75)(60 − 20)= 2373 W [8150 Btu/h]

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Table 5-2 Summary of equations for flow over flat plates. Properties evaluated at Tf = (Tw+ T∞)/2 unless otherwise noted.

Flow regime Restrictions Equation Equation number

Heat transfer

Laminar, local Tw= const, Rex < 5 × 105, Nux= 0.332 Pr1/3Re1/2x (5-44)

0.6< Pr< 50

Laminar, local Tw= const, Rex < 5 × 105, Nux= 0.3387 Re1/2x Pr1/3[

1 +(

0.0468

Pr

)2/3]1/4 (5-51)

Rex Pr> 100

Laminar, local qw= const, Rex < 5 × 105, Nux= 0.453 Re1/2x Pr1/3 (5-48)

0.6< Pr< 50

Laminar, local qw= const, Rex < 5 × 105 Nux= 0.4637 Re1/2x Pr1/3[

1 +(

0.0207

Pr

)2/3]1/4 (5-51)

Laminar, average ReL < 5 × 105, Tw= const NuL= 2 Nux=L= 0.664 Re1/2L Pr1/3 (5-46)

Laminar, local Tw= const, Rex < 5 × 105, Nux= 0.564(Rex Pr)1/2

Pr � 1 (liquid metals)

Laminar, local Tw= const, starting at Nux= 0.332 Pr1/3 Re1/2x

[1 − ( x0

x

)3/4]−1/3(5-43)

x= x0, Rex < 5 × 105,0.6< Pr< 50

Turbulent, local Tw= const, 5 × 105<Rex < 107 Stx Pr2/3 = 0.0296 Re−0.2x (5-81)

Turbulent, local Tw= const, 107<Rex < 109 Stx Pr2/3 = 0.185(log Rex)−2.584 (5-82)Turbulent, local qw= const, 5 × 105<Rex < 107 Nux= 1.04 NuxTw=const (5-87)Laminar-turbulent, Tw= const, Rex < 107, St Pr2/3 = 0.037 Re−0.2

L − 871 Re−1L (5-84)

average Recrit = 5 × 105 NuL= Pr1/3(0.037 Re0.8L − 871) (5-85)

Laminar-turbulent, Tw= const, Rex < 107, NuL= 0.036 Pr0.43(Re0.8L − 9200)

(μ∞μw

)1/4(5-86)

average liquids, μ at T∞,μw at Tw

High-speed flow Tw= const, Same as for low-speed flow withq=hA(Tw− Taw) properties evaluated at

T ∗ = T∞ + 0.5(Tw− T∞)+ 0.22(Taw− T∞) (5-124)r= (Taw− T∞)/(To− T∞)= recovery factor= Pr1/2 (laminar)= Pr1/3 (turbulent)

Boundary-layer thickness

Laminar Rex < 5 × 105 δx = 5.0 Re−1/2

x (5-21a)

Turbulent Rex < 107, δx = 0.381 Re−1/5

x (5-91)δ= 0 at x= 0

Turbulent 5 × 105<Rex < 107, δx = 0.381 Re−1/5

x − 10,256 Re−1x (5-95)

Recrit = 5 × 105,δ= δlam at Recrit

Friction coefficients

Laminar, local Rex < 5 × 105 Cfx= 0.332 Re−1/2x (5-54)

Turbulent, local 5 × 105<Rex < 107 Cfx= 0.0592 Re−1/5x (5-77)

Turbulent, local 107<Rex < 109 Cfx= 0.37(log Rex)−2.584 (5-78)

Turbulent, average Recrit <Rex < 109 Cf = 0.455(log ReL)2.584 − A

ReL(5-79)

A from Table 5-1

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Table A-5 Properties of air at atmospheric pressure.†

The values of µ, k, cp , and Pr are not strongly pressure-dependentand may be used over a fairly wide range of pressures

ρ cp µ × 105 ν × 106 k α × 104

T,K kg/m3 kJ/kg · ◦C kg/m· s m2/s W/m · ◦C m2/s Pr

100 3.6010 1.0266 0.6924 1.923 0.009246 0.02501 0.770150 2.3675 1.0099 1.0283 4.343 0.013735 0.05745 0.753200 1.7684 1.0061 1.3289 7.490 0.01809 0.10165 0.739250 1.4128 1.0053 1.5990 11.31 0.02227 0.15675 0.722300 1.1774 1.0057 1.8462 15.69 0.02624 0.22160 0.708350 0.9980 1.0090 2.075 20.76 0.03003 0.2983 0.697400 0.8826 1.0140 2.286 25.90 0.03365 0.3760 0.689450 0.7833 1.0207 2.484 31.71 0.03707 0.4222 0.683500 0.7048 1.0295 2.671 37.90 0.04038 0.5564 0.680550 0.6423 1.0392 2.848 44.34 0.04360 0.6532 0.680600 0.5879 1.0551 3.018 51.34 0.04659 0.7512 0.680650 0.5430 1.0635 3.177 58.51 0.04953 0.8578 0.682700 0.5030 1.0752 3.332 66.25 0.05230 0.9672 0.684750 0.4709 1.0856 3.481 73.91 0.05509 1.0774 0.686800 0.4405 1.0978 3.625 82.29 0.05779 1.1951 0.689850 0.4149 1.1095 3.765 90.75 0.06028 1.3097 0.692900 0.3925 1.1212 3.899 99.3 0.06279 1.4271 0.696950 0.3716 1.1321 4.023 108.2 0.06525 1.5510 0.699

1000 0.3524 1.1417 4.152 117.8 0.06752 1.6779 0.7021100 0.3204 1.160 4.44 138.6 0.0732 1.969 0.7041200 0.2947 1.179 4.69 159.1 0.0782 2.251 0.7071300 0.2707 1.197 4.93 182.1 0.0837 2.583 0.7051400 0.2515 1.214 5.17 205.5 0.0891 2.920 0.7051500 0.2355 1.230 5.40 229.1 0.0946 3.262 0.7051600 0.2211 1.248 5.63 254.5 0.100 3.609 0.7051700 0.2082 1.267 5.85 280.5 0.105 3.977 0.7051800 0.1970 1.287 6.07 308.1 0.111 4.379 0.7041900 0.1858 1.309 6.29 338.5 0.117 4.811 0.7042000 0.1762 1.338 6.50 369.0 0.124 5.260 0.7022100 0.1682 1.372 6.72 399.6 0.131 5.715 0.7002200 0.1602 1.419 6.93 432.6 0.139 6.120 0.7072300 0.1538 1.482 7.14 464.0 0.149 6.540 0.7102400 0.1458 1.574 7.35 504.0 0.161 7.020 0.7182500 0.1394 1.688 7.57 543.5 0.175 7.441 0.730

†From Natl. Bur. Stand. (U.S.) Circ. 564, 1955.

In natural convection, the fluid motion occurs by natural means such as buoyancy. Since the fluid velocity associated with natural convection is relatively low, the heat transfer coefficient encountered in natural convection is also low.

Mechanisms of Natural Convection

Consider a hot object exposed to cold air. The temperature of the outside of the object will drop (as a result of heat transfer with cold air), and the temperature of adjacent air to the object will rise. Consequently, the object is surrounded with a thin layer of warmer air and heat will be transferred from this layer to the outer layers of air.

Fig. 1: Natural convection heat transfer from a hot body.

The temperature of the air adjacent to the hot object is higher, thus its density is lower. As a result, the heated air rises. This movement is called the natural convection current. Note that in the absence of this movement, heat transfer would be by conduction only and its rate would be much lower.

In a gravitational field, there is a net force that pushes a light fluid placed in a heavier fluid upwards. This force is called the buoyancy force.

Fig. 2: Buoyancy force keeps the ship float in water.

The magnitude of the buoyancy force is the weight of the fluid displaced by the body.

Cool air

Warm air

Hot object

Water

Displaced volume

W

Buoyancy force

Ship

Natural (Free) Convection

Fbuoyancy = ρfluid g Vbody

where Vbody is the volume of the portion of the body immersed in the fluid. The net force is:

Fnet = W ‐ Fbuoyancy

Fnet = (ρbody ‐ ρfluid) g Vbody

Note that the net force is proportional to the difference in the densities of the fluid and the body. This is known as Archimedes’ principle.

We all encounter the feeling of “weight loss” in water which is caused by the buoyancy force. Other examples are hot balloon rising, and the chimney effect.

Note that the buoyancy force needs the gravity field, thus in space (where no gravity exists) the buoyancy effects does not exist.

Density is a function of temperature, the variation of density of a fluid with temperature at constant pressure can be expressed in terms of the volume expansion coefficient β, defined as:

Pconstant at 1

11

TT

KT P

It can be shown that for an ideal gas

T

1gas ideal

where T is the absolute temperature. Note that the parameter βΔT represents the fraction of volume change of a fluid that corresponds to a temperature change ΔT at constant pressure.

Since the buoyancy force is proportional to the density difference, the larger the temperature difference between the fluid and the body, the larger the buoyancy force will be.

Whenever two bodies in contact move relative to each other, a friction force develops at the contact surface in the direction opposite to that of the motion. Under steady conditions, the air flow rate driven by buoyancy is established by balancing the buoyancy force with the frictional force.

Grashof Number

Grashof number is a dimensionless group. It represents the ratio of the buoyancy force to the viscous force acting on the fluid:

TVg

Gr

2

Vg

forces viscous

forcesbuoyancy

It is also expressed as

2

3

TTg

Gr s

where g = gravitational acceleration, m/s

2 β = coefficient of volume expansion, 1/K δ = characteristic length of the geometry, m ν = kinematics viscosity of the fluid, m2

/s

The role played by Reynolds number in forced convection is played by the Grashof number in natural convection. The critical Grashof number is observed to be about 109 for vertical plates. Thus, the flow regime on a vertical plate becomes turbulent at Grashof number greater than 109. The heat transfer rate in natural convection is expressed by Newton’s law of cooling as: Q’conv = h A (Ts ‐ T∞)

Fig. 3: Velocity and temperature profile for natural convection flow over a hot vertical plate. Grcritical = 109

Natural Convection over Surfaces

Natural convection on a surface depends on the geometry of the surface as well as its orientation. It also depends on the variation of temperature on the surface and the thermophysical properties of the fluid.

The velocity and temperature distribution for natural convection over a hot vertical plate are shown in Fig. 3.

Note that the velocity at the edge of the boundary layer becomes zero. It is expected since the fluid beyond the boundary layer is stationary.

The shape of the velocity and temperature profiles, in the cold plate case, remains the same but their direction is reversed.

Natural Convection Correlations

The complexities of the fluid flow make it very difficult to obtain simple analytical relations for natural convection. Thus, most of the relationships in natural convection are based on experimental correlations.

The Rayleigh number is defined as the product of the Grashof and Prandtl numbers:

PrPr

2

3

TTg

GrRa s

The Nusselt number in natural convection is in the following form:

nCRa

k

hNu

where the constants C and n depend on the geometry of the surface and the flow. Table 14‐1 in Cengel book lists these constants for a variety of geometries.

These relationships are for isothermal surfaces, but could be used approximately for the case of non‐isothermal surfaces by assuming surface temperature to be constant at some average value.

Isothermal Vertical Plate

For a vertical plate, the characteristic length is L.

1393/1

944/1

10101.0

101059.0

RaRa

RaRaNu

Note that for ideal gases, β=1 / T∞

Isothermal Horizontal Plate

The characteristics length is A/p where the surface area is A, and perimeter is p.

a) Upper surface of a hot plate

1173/1

744/1

101015.0

101054.0

RaRa

RaRaNu

b) Lower surface of a hot plate 1154/1 101027.0 RaRaNu

Example 1: isothermal vertical plate

A large vertical plate 4 m high is maintained at 60°C and exposed to atmospheric air at 10°C. Calculate the heat transfer if the plate is 10 m wide.

Solution:

We first determine the film temperature as

Tf = (Ts + T∞) / 2 = 35°C = 308 K

The properties are:

β = 1 / 308 = 3.25x10‐3, k = 0.02685 (W/mK), ν = 16.5x10‐6

, Pr = 0.7

The Rayleigh number can be found:

11

26

3132

10743.37.0105.16

410601025.3/8.9Pr

mCKsm

GrRa

The Nusselt number can be found from:

7.72010743.31.01.03/1113/1 RaNu

The heat‐transfer coefficient is

KmWL

kNuh

2/84.44

02685.07.720

The heat transfer is

Q˚ = h A (Ts ‐ T∞) = 7.84 W/m°C2 (4 x 10 m2) (60 – 10 °C) = 9.675 kW

Natural Convection from Finned Surfaces

Finned surfaces of various shapes (heat sinks) are used in microelectronics cooling.

One of most crucial parameters in designing heat sinks is the fin spacing, S. Closely packed fins will have greater surface area for heat transfer, but a smaller heat transfer coefficient (due to extra resistance of additional fins). A heat sink with widely spaced fins will have a higher heat transfer coefficient but smaller surface area. Thus, an optimum spacing exists that maximizes the natural convection from the heat sink.

Fig. 4: A vertical heat sink.

Consider a heat sink with base dimension W (width) and L (length) in which the fins are assumed to be isothermal and the fin thickness t is small relative to fin spacing S. The optimum fin spacing for a vertical heat sink is given by Rohsenow and Bar‐Cohen as

4/1714.2

Ra

LSopt

where L is the characteristic length in Ra number. All the fluid property are determined at the film temperature. The heat transfer coefficient for the optimum spacing can be found from

optS

kh 31.1

Note: as a result of above‐mentioned “two opposing forces” (buoyancy and friction), heat sinks with closely spaced fins are not suitable for natural convection.

Example 2: Heat sink

A 12‐cm wide and 18‐cm‐high vertical hot surface in 25°C air is to be cooled by a heat sink with equally spaced fins of rectangular profile. The fins are 0.1 cm thick, 18 cm long in the vertical direction, and have a height of 2.4 cm from the base. Determine the optimum fin spacing and the rate of heat transfer by natural convection from the heat sink if the base temperature is 80°C.

Assumptions:

The fin thickness t is much smaller than the fin spacing S.

Solution:

L = 0.18

H = 2.4 cm

S

t = 1 mm

Ts =80°C

W = 0.12 m

T∞ = 25°C

The properties of air are evaluated at the film temperature:

Tf = (T∞ + Ts) / 2 = 52.5°C = 325.5 K

At this temperature, k = 0.0279 W /mK, ν = 1.82 x 10‐5 m2/s, Pr = 0.709, and assuming

ideal gas β = 1 / Tf = 1 / 325.5 K = 0.003072 1/K.

The characteristic length is L = 0.18 m.

7

2

3

10067.2Pr

LTTg

Ra s

The optimum fin spacing is determined

mmmRa

LSopt 2.70072.0714.2

4/1

The number of fins and the heat transfer coefficient for the optimum fin spacing case are

fins15

tS

Wn

mK

W

S

kh

opt

08.531.1

The rate of natural convection heat transfer becomes:

WTTnLHhQ s 2.362

Solved Problems - Natural (Free) Convection- Chapter Seven –Holman

-

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C H A P T E R

Radiation Heat Transfer

8-1 INTRODUCTIONPreceding chapters have shown how conduction and convection heat transfer may be cal-culated with the aid of both mathematical analysis and empirical data. We now wish toconsider the third mode of heat transfer—thermal radiation. Thermal radiation is that elec-tromagnetic radiation emitted by a body as a result of its temperature. In this chapter, weshall first describe the nature of thermal radiation, its characteristics, and the properties thatare used to describe materials insofar as the radiation is concerned. Next, the transfer ofradiation through space will be considered. Finally, the overall problem of heat transfer bythermal radiation will be analyzed, including the influence of the material properties andthe geometric arrangement of the bodies on the total energy that may be exchanged.

8-2 PHYSICAL MECHANISMThere are many types of electromagnetic radiation; thermal radiation is only one. Regardlessof the type of radiation, we say that it is propagated at the speed of light, 3 × 108 m/s. Thisspeed is equal to the product of the wavelength and frequency of the radiation,

c= λνwhere

c= speed of light

λ= wavelength

ν= frequency

The unit for λ may be centimeters, angstroms (1 Å = 10−8 cm), or micrometers(1 µm = 10−6 m). A portion of the electromagnetic spectrum is shown in Figure 8-1. Ther-mal radiation lies in the range from about 0.1 to 100 µm, while the visible-light portion ofthe spectrum is very narrow, extending from about 0.35 to 0.75 µm.

The propagation of thermal radiation takes place in the form of discrete quanta, eachquantum having an energy of

E=hν [8-1]

where h is Planck’s constant and has the value

h= 6.625 × 10−34 J · s

RADIATION HEAT TRANSFER

CHAPTER EIGHT

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Figure 8-1 Electromagnetic spectrum.

3 2 1 0 –1 –2 –3 –4 –5 –6 –7 –8 –9 –11 –12

Thermalradiation

Radiowaves

Infrared

Visible

Ultra-violet

X-rays

γ raysγ

1 A°1 µmµlog λ, mλ

–10

A very rough physical picture of the radiation propagation may be obtained by consideringeach quantum as a particle having energy, mass, and momentum, just as we considered themolecules of a gas. So, in a sense, the radiation might be thought of as a “photon gas” thatmay flow from one place to another. Using the relativistic relation between mass and energy,expressions for the mass and momentum of the “particles” could thus be derived; namely,

E=mc2 =hνm= hν

c2

Momentum = chν

c2= hν

c

By considering the radiation as such a gas, the principles of quantum-statistical thermody-namics can be applied to derive an expression for the energy density of radiation per unitvolume and per unit wavelength as†

uλ= 8πhcλ−5

ehc/λkT − 1[8-2]

where k is Boltzmann’s constant, 1.38066 × 10−23 J/molecule · K. When the energy densityis integrated over all wavelengths, the total energy emitted is proportional to absolutetemperature to the fourth power:

Eb = σT 4 [8-3]

Equation (8-3) is called the Stefan-Boltzmann law, Eb is the energy radiated per unit timeand per unit area by the ideal radiator, and σ is the Stefan-Boltzmann constant, which hasthe value

σ= 5.669 × 10−8 W/m2 · K4 [0.1714 × 10−8 Btu/h · ft2 · ◦R4]where Eb is in watts per square meter and T is in degrees Kelvin. In the thermodynamicanalysis the energy density is related to the energy radiated from a surface per unit timeand per unit area. Thus the heated interior surface of an enclosure produces a certain energydensity of thermal radiation in the enclosure. We are interested in radiant exchange withsurfaces—hence the reason for the expression of radiation from a surface in terms of itstemperature. The subscript b in Equation (8-3) denotes that this is the radiation from a black-body. We call this blackbody radiation because materials that obey this law appear blackto the eye; they appear black because they do not reflect any radiation. Thus a blackbody is

†See, for example, J. P. Holman, Thermodynamics, 4th ed. New York: McGraw-Hill, 1988, p. 350.

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also considered as one that absorbs all radiation incident upon it. Eb is called the emissivepower of a blackbody.

It is important to note at this point that the “blackness” of a surface to thermal radiationcan be quite deceiving insofar as visual observations are concerned. A surface coated withlampblack appears black to the eye and turns out to be black for the thermal-radiationspectrum. On the other hand, snow and ice appear quite bright to the eye but are essentially“black” for long-wavelength thermal radiation. Many white paints are also essentially blackfor long-wavelength radiation. This point will be discussed further in later sections.

8-3 RADIATION PROPERTIESWhen radiant energy strikes a material surface, part of the radiation is reflected, part isabsorbed, and part is transmitted, as shown in Figure 8-2. We define the reflectivity ρ asthe fraction reflected, the absorptivity α as the fraction absorbed, and the transmissivity τas the fraction transmitted. Thus

ρ+α+ τ= 1 [8-4]

Most solid bodies do not transmit thermal radiation, so that for many applied problems thetransmissivity may be taken as zero. Then

ρ+α= 1

Two types of reflection phenomena may be observed when radiation strikes a surface.If the angle of incidence is equal to the angle of reflection, the reflection is called spec-ular. On the other hand, when an incident beam is distributed uniformly in all directionsafter reflection, the reflection is called diffuse. These two types of reflection are depicted

Figure 8-2 Sketch showing effects ofincident radiation.

Incident radiation Reflection

Absorbed

Transmitted

Figure 8-3 (a) Specular (φ1 =φ2) and (b) diffuse reflection.

Reflectedrays

SourceSource

Mirror imageof source

2φ φ1

(a) (b)

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Figure 8-4 Sketch showing model used forderiving Kirchhoff’s law.

Sample

EA

qi Aαα

Blackenclosure

in Figure 8-3. Note that a specular reflection presents a mirror image of the source to theobserver. No real surface is either specular or diffuse. An ordinary mirror is quite specularfor visible light, but would not necessarily be specular over the entire wavelength range ofthermal radiation. Ordinarily, a rough surface exhibits diffuse behavior better than a highlypolished surface. Similarly, a polished surface is more specular than a rough surface. Theinfluence of surface roughness on thermal-radiation properties of materials is a matter ofserious concern and remains a subject for continuing research.

The emissive power of a bodyE is defined as the energy emitted by the body per unit areaand per unit time. One may perform a thought experiment to establish a relation between theemissive power of a body and the material properties defined above. Assume that a perfectlyblack enclosure is available, i.e., one that absorbs all the incident radiation falling upon it,as shown schematically in Figure 8-4. This enclosure will also emit radiation according tothe T 4 law. Let the radiant flux arriving at some area in the enclosure be qi W/m2. Nowsuppose that a body is placed inside the enclosure and allowed to come into temperatureequilibrium with it. At equilibrium the energy absorbed by the body must be equal to theenergy emitted; otherwise there would be an energy flow into or out of the body that wouldraise or lower its temperature. At equilibrium we may write

EA= qiAα [8-5]

If we now replace the body in the enclosure with a blackbody of the same size and shapeand allow it to come to equilibrium with the enclosure at the same temperature,

EbA= qiA(1) [8-6]

since the absorptivity of a blackbody is unity. If Equation (8-5) is divided by Equation (8-6),

E

Eb=α

and we find that the ratio of the emissive power of a body to the emissive power of ablackbody at the same temperature is equal to the absorptivity of the body. This ratio isdefined as the emissivity ε of the body,

ε= E

Eb[8-7]

so thatε=α [8-8]

Equation (8-8) is called Kirchhoff’s identity. At this point we note that the emissivities andabsorptivities that have been discussed are the total properties of the particular material;

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that is, they represent the integrated behavior of the material over all wavelengths. Realsubstances emit less radiation than ideal black surfaces as measured by the emissivity of thematerial. In reality, the emissivity of a material varies with temperature and the wavelengthof the radiation.

The Gray Body

Agray body is defined such that the monochromatic emissivity ελ of the body is independentof wavelength. The monochromatic emissivity is defined as the ratio of the monochromaticemissive power of the body to the monochromatic emissive power of a blackbody at thesame wavelength and temperature. Thus

ελ= Eλ

Ebλ

The total emissivity of the body may be related to the monochromatic emissivity by notingthat

E=∫ ∞

0ελ Ebλ dλ and Eb =

∫ ∞

0Ebλ dλ= σT 4

so that

ε= E

Eb=∫∞

0 ελ Ebλ dλ

σT 4[8-9]

where Ebλ is the emissive power of a blackbody per unit wavelength. If the gray-bodycondition is imposed, that is, ελ = constant, Equation (8-9) reduces to

ε= ελ [8-10]

The emissivities of various substances vary widely with wavelength, temperature, andsurface condition. Some typical values of the total emissivity of various surfaces are given inAppendix A. We may note that the tabulated values are subject to considerable experimentaluncertainty. A rather complete survey of radiation properties is given in Reference 14.

The functional relation for Ebλ was derived by Planck by introducing the quantumconcept for electromagnetic energy. The derivation is now usually performed by methodsof statistical thermodynamics, and Ebλ is shown to be related to the energy density ofEquation (8-2) by

Ebλ= uλc

4[8-11]

or

Ebλ= C1λ−5

eC2/λT − 1[8-12]

where

λ= wavelength, µm

T = temperature, K

C1 = 3.743 × 108 W ·µm4/m2 [1.187 × 108 Btu ·µm4/h · ft2]C2 = 1.4387 × 104 µm · K [2.5896 × 104 µm · ◦R]Aplot ofEbλ as a function of temperature and wavelength is given in Figure 8-5a. Notice

that the peak of the curve is shifted to the shorter wavelengths for the higher temperatures.These maximum points in the radiation curves are related by Wien’s displacement law,

λmaxT = 2897.6 µm · K [5215.6 µm · ◦R] [8-13]

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CHAPTER Heat Transfer 1 · that this energy transfer is deﬁned as heat. The science of heat transfer seeks not merely to explain how heat energy may be transferred, but also to