A heat transfer textbook

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Heat transfer

Heat naturally flows from hot objects to cold ones, a concept historically understood as the movement of an invisible fluid called caloric In the 18th and early 19th centuries, scientists attributed various properties to caloric, including weight and the inability to be created or destroyed, though some of these ideas were inconsistent with nature The essential characteristic of caloric was its flow from hot to cold bodies, providing a useful framework for understanding heat While modern explanations of heat transfer are more refined, the notion of caloric flowing from hotter to cooler objects remains a helpful visualization.

Heat flow is a fundamental and constant phenomenon, affecting everything around us It transfers from our bodies to the air, warming our surroundings Even after leaving a room, air motion persists due to the walls not being perfectly isothermal This heat transfer is vital for all forms of life and occurs throughout the Earth, which has a hot core and cooler surface A truly isolated and isothermal environment would be devoid of any processes, representing a complete absence of life and activity.

The primary driver of heat flow processes in the universe is the cooling of thermal gradients The natural heat flows we experience are mainly due to the cooling of the sun, while the conductive cooling of Earth's core and the radiative cooling of other stars play a secondary role in our daily lives.

Life on Earth has evolved to adapt to natural energy flows, but while "natural man" exists in harmony with these systems, "technological man" has leveraged intellect and effort to manipulate energy flows that far exceed natural levels To illustrate this concept, we encourage readers to conduct a simple experiment.

To measure your power output, engage in activities like lifting weights or running your body weight up a stairwell while timing yourself with a stopwatch Express your results in watts (W), which typically should be less than 1 kW or 746 W (1 horsepower) You may find the actual values surprisingly lower than expected when collected in a group setting.

Turning on a 150 W light bulb utilizes an amount of energy far beyond what a human can produce through sustained effort Appliances like ovens, toasters, and hot water heaters consume energy levels that are significantly higher than our capabilities Furthermore, the energy required by automobiles can be three times greater If every person in the United States worked non-stop, their collective power output would still fall short of that produced by a single city power plant.

Our increasing demand for energy has significantly heightened the intensity of heat transfer processes, surpassing levels typically associated with life on Earth Until the mid-thirteenth century, the energy harnessed was primarily derived from the sun.

Some anthropologists propose the term Homo technologicus, or technological man, as a more accurate descriptor of humans compared to the traditional Homo sapiens This perspective highlights that while we may not be significantly wiser than other animals, our unique ability for sustained tool-making sets us apart Initially, heat transfer relied on gentle processes such as animal, wind, and water power, alongside wood combustion However, population growth and deforestation led to a shift towards coal usage in England by the late seventeenth century By the early eighteenth century, the advent of commercial steam engines dramatically increased coal consumption, a trend that subsequently spread to Europe and America.

The development of fossil energy sources has been a bit like Jules

In Jules Verne's "Around the World in Eighty Days," a crew resorts to burning the interior of a ship to fuel its steam engine in a race against time This scenario highlights the reliance on nonrenewable fossil fuels and, more recently, uranium fission, which have produced significant energy outputs in power generation Notably, the heat energy generated in a nuclear reactor can reach approximately one million watts per square meter, illustrating the immense power derived from these sources.

To effectively manage energy concentrations, a complex system of heat and work transfer processes is essential Understanding and controlling these processes is crucial for reducing intense heat flows to a manageable level For instance, in a town where coal is converted into fuel gas and coke, several heat transfer challenges must be addressed before we can enjoy a glass of iced tea As natural gas supplies potentially diminish, revisiting these power sources becomes increasingly relevant.

• A variety of high-intensity heat transfer processes are involved with combustion and chemical reaction in the gasiﬁer unit itself.

• The gas goes through various cleanup and pipe-delivery processes to get to our stoves The heat transfer processes involved in these stages are generally less intense.

The gas stove burns fuel to generate a flame, which effectively transfers heat to the bottom of the teakettle This process, though small in scale, is highly efficient, as boiling is an effective method for heat removal.

In a steam power plant, coke is combusted, generating intense heat transfer rates from the combustion chamber to the boiler, and subsequently from the boiler walls to the water inside.

Steam flows through a turbine, engaging in multiple heat transfer processes, including condensation during the final stages The exhausted steam is subsequently condensed using various heat transfer devices.

Effective cooling is essential at every stage of the electrical supply system, including the generator's windings and bearings, transformers, switches, power lines, and residential wiring.

Ice cubes for tea are produced in an electrical refrigerator, which operates through three primary heat exchange processes These include the condensation of refrigerant at room temperature to expel heat, the absorption of heat from inside the refrigerator via the evaporation of refrigerant, and the management of heat leakage from the surrounding room into the appliance.

Relation of heat transfer to thermodynamics

The First Law with work equal to zero

Thermodynamics, a fundamental topic in engineering education, frequently addresses the principles of heat transfer between systems The First Law of Thermodynamics, applicable to closed systems, emphasizes the relationship between heat transfer and energy conservation.

Figure 1.1 The First Law of Thermodynamics for a closed system. rate basis:

= Wk positive away from the system

+ dU dt positive when the system’s energy increases

The heat transfer rate (Q) and work transfer rate (Wk) are measured in joules per second (J/s) or watts (W) The rate of change of internal thermal energy (U) with respect to time (t) is represented by the derivative dU/dt.

This interaction is sketched schematically in Fig.1.1a.

Heat transfer processes can typically be analyzed independently of any work processes, although they may later be integrated with work in the examination of actual systems When p dV work is the sole type of work involved, the relevant equation is represented as eqn (1.1).

This equation has two well-known special cases:

Constant volume process: Q= dU dt =mc v dT dt (1.2b)

Constant pressure process: Q= dH dt =mc p dT dt (1.2c) where H ≡ U+pV is the enthalpy, andc v andc p are the speciﬁc heat capacities at constant volume and constant pressure, respectively.

When the substance undergoing the process is incompressible (so that

V is constant for any pressure variation), the two speciﬁc heats are equal:

8 Introduction §1.2 c v =c p ≡c The proper form of eqn (1.2a) is then

Since solids and liquids can frequently be approximated as being incompressible, we shall often make use of eqn (1.3).

If the heat transfer were reversible, then eqn (1.2a) would become 2

That might seem to suggest thatQcan be evaluated independently for inclusion in either eqn (1.1) or (1.3) However, it cannot be evaluated using

In engineering thermodynamics, real heat transfer processes are inherently irreversible, and the entropy (S) is not defined as a function of temperature (T) in such processes This field may be more accurately referred to as thermostatics, as it primarily focuses on describing the equilibrium states surrounding irreversible processes.

Since the rate of heat transfer cannot be predicted using T dS, how can it be determined? IfU (t)were known, then (whenWk= 0) eqn (1.3) would giveQ, butU (t)is seldom knowna priori.

To accurately predict heat transfer, it is essential to incorporate a new set of physical principles known as transport laws, which extend beyond traditional thermodynamics Key laws such as Fourier’s law, Newton’s law of cooling, and the Stefan-Boltzmann law play a crucial role in this context It is important to note that a comprehensive understanding of heat transfer necessitates the integration of these principles with the First Law of Thermodynamics.

Reversible heat transfer as the temperature gradient vanishes

Consider a wall connecting two thermal reservoirs as shown in Fig.1.2.

Heat will naturally flow from a higher temperature (T1) to a lower temperature (T2) in an irreversible manner, in line with the Second Law of Thermodynamics This process results in an increase in the universe's entropy As the temperatures converge (T2 approaches T1), the heat transfer process becomes more quasistatic and reversible, yet the rate of heat transfer will also decrease.

2 T = absolute temperature, S = entropy, V = volume, p = pressure, and “rev” denotes a reversible process. §1.2 Relation of heat transfer to thermodynamics 9

Irreversible heat flow occurs between two thermal reservoirs separated by a wall, with entropy generated in all real heat transfer processes due to the presence of a temperature difference If there is no temperature difference, heat transfer ceases, resulting in zero entropy production.

In a steady-state irreversible process, the properties of the wall remain constant over time, leading to a constant entropy for the wall This raises the question of how the entropy of the universe can still increase despite this constancy We will explore this intriguing dilemma further.

The entropy increase of the universe as the result of a process is the sum of the entropy changes ofall elements that are involved in that process.

The rate of entropy production of the universe, ˙S Un, resulting from the preceding heat transfer process through a wall is

= 0, since S wall must be constant

+S˙ res 2 (1.5) where the dots denote time derivatives (i.e., ˙x≡dx/dt) Since the reservoir temperatures are constant,

Now Q res 1 is negative and equal in magnitude to Q res 2, so eqn (1.5) becomes

The term in parentheses is positive, so ˙S Un >0 This agrees with Clau- sius’s statement of the Second Law of Thermodynamics.

The rate of heat transfer, Q, is influenced by the wall's resistance to heat flow, which plays a crucial role in the increase of the universe's entropy Interestingly, while the wall contributes to this entropy increase, its own entropy remains unchanged; only the entropies of the surrounding reservoirs are affected.

Modes of heat transfer

Figure1.3shows an analogy that might be useful in ﬁxing the concepts of heat conduction, convection, and radiation as we proceed to look at each in some detail.

Fourier’s law Joseph Fourier (see Fig 1.4) published his remarkable bookThéorie Analytique de la Chaleurin 1822 In it he formulated a very complete exposition of the theory of heat conduction.

The author introduces his treatise by presenting the empirical law named after him, which states that the heat flux (q in W/m²) due to thermal conduction is directly proportional to the temperature gradient and inversely directed This relationship can be expressed mathematically as q = -k(dT/dx), where k represents the constant of proportionality.

The constant,k, is called thethermal conductivity It obviously must have the dimensions W/mãK, or J/mãsãK, or Btu/hãftã ◦ F if eqn (1.8) is to be dimensionally correct.

The heat ﬂux is a vector quantity Equation (1.8) tells us that if temperature decreases withx,qwill be positive—it will ﬂow in thex-direction.

When temperature T increases with respect to position x, the heat flow q will be negative, indicating it moves in the opposite direction of x Regardless of the situation, heat will always transfer from areas of higher temperature to those of lower temperature The equation representing this principle in one dimension is known as Fourier's law, and its three-dimensional form is expressed as q = -k∇T.

3 The heat ﬂux, q, is a heat rate per unit area and can be expressed as Q/A, where A is an appropriate area.

Figure 1.3 An analogy for the three modes of heat transfer.

Baron Jean Baptiste Joseph Fourier (1768–1830) led an extraordinary dual life as a prominent government official in Napoleonic France while also making significant contributions as an applied mathematician His notable experiences included accompanying Napoleon during the Egyptian campaign.

From 1798 to 1801, Fourier served as the prefect of the Isère department in France, continuing his work on heat flow theory until Napoleon's first fall in 1814 In 1807, he submitted a comprehensive 234-page monograph on heat flow, which was reviewed by prominent figures such as Lagrange and Laplace They criticized his use of a series expansion proposed by Daniel Bernoulli in the 18th century The resolution of this controversy, which included Fourier's governing differential equation and the renowned "Fourier series," was not published until 1822.

To determine the heat flux (q) and heat transfer rate (Q) through a lead slab with a thermal conductivity of 35 W/m·K, we consider a slab with a front temperature of 110 °C and a back temperature of 50 °C Given the slab's area of 0.4 m² and thickness of 0.03 m, we can calculate the heat flux using Fourier's law The temperature difference across the slab is 60 °C, leading to a heat flux of approximately 700 W/m² Consequently, the total heat transfer rate (Q) through the slab is calculated to be 280 W.

Solution For the moment, we presume thatdT /dx is a constant equal to(T back −T front )/(x back −x front ); we verify this in Chapter 2.

In one-dimensional heat conduction, determining the direction of heat flow is straightforward, allowing for a simplified expression of Fourier’s law as q = k∆T, where q represents the heat transfer rate, k is the thermal conductivity, and ∆T is the temperature difference.

In equation (1.9), L represents the thickness in the direction of heat flow, while both q and ∆T are expressed as positive values It is essential to note that heat, denoted by q, always transfers from areas of high temperature to areas of low temperature.

Thermal conductivity values play a crucial role in understanding heat conduction, particularly in gases, where molecular velocity is temperature-dependent When examining the conduction process from a hot wall to a cold wall, it is essential to consider scenarios where gravity can be disregarded, as this simplifies the analysis of heat transfer.

When molecules near a hot wall collide with it, they gain energy and speed, subsequently transferring this kinetic energy to neighboring molecules This energy transfer continues until it reaches the cooler wall In solids, similar energy transfer occurs as molecules vibrate within their lattice structure, contributing to the overall vibration of the lattice Additionally, this process is mirrored in the movement of the electron "gas" within the solid.

Figure 1.5 Heat conduction through gas separating two solid walls. processes are more eﬃcient in solids than they are in gases Notice that

−dT dx = q k ∝ 1 k since, in steady conduction, q is constant

Solids typically exhibit higher thermal conductivities than gases, resulting in smaller temperature gradients for a given heat flux In gases, thermal conductivity is directly related to molecular speed and molar specific heat, while being inversely related to the cross-sectional area of the molecules This article primarily focuses on S.I units (Système International d’Unités), but it is essential to have conversion factors for thermal conductivity since reference materials will still be available in English units.

K Thus the conversion factor from W/mãK to its English equivalent, Btu/hã ftã ◦ F, is

Consider, for example, copper—the common substance with the highest conductivity at ordinary temperature: k Cu at room temp =(383 W/mãK)

1.731 W/mãKBtu/hãftã ◦ F =221 Btu/hãftã ◦ F

Thermal conductivities vary significantly, ranging from gases to diamond at room temperature, with a factor of about 10^5 This variation can reach up to 10^7 when including the effective conductivity of cryogenic superinsulations, which consist of evacuated powders, fibers, or multilayered materials Understanding the order of magnitude of thermal conductivities across different materials is crucial for avoiding errors in future calculations and making informed assumptions during problem-solving For actual numerical values of thermal conductivity, refer to Appendix A and Figs 2.2 and 2.3, which provide a comprehensive listing of relevant physical properties.

In a composite wall consisting of a 3 mm thick copper slab (thermal conductivity k = 372 W/m·K) sandwiched between two 2 mm thick layers of stainless steel (thermal conductivity k = 17 W/m·K), the temperature distribution can be analyzed given the boundary conditions of 400 °C on one side and 100 °C on the opposite side To determine the temperature gradient within the copper slab and calculate the heat conduction through the entire wall, one must apply Fourier's law of heat conduction and consider the thermal resistances of each material.

The temperature drop primarily occurs in the stainless steel, as its thermal conductivity is less than 1/20 that of copper Consequently, the copper remains nearly isothermal at an average temperature of approximately 400 degrees.

The heat conduction in a 4 mm slab of stainless steel can be estimated independently of the copper, applying Fourier’s law The calculated heat transfer rate is q = -k(dT/dx) = 17 W/m·K.

To enhance the precision of the initial calculation, it is essential to incorporate the effects of copper This involves determining the temperature differences, ∆T s.s and ∆T Cu, as illustrated in Fig 1.7 According to the principle of energy conservation, the steady heat flux must remain consistent across all three slabs, leading to the equation q = k∆T/L s.s = k∆T/L.

Cu §1.3 Modes of heat transfer 17

Figure 1.7 Temperature drop through a copper wall protected by stainless steel (Example1.2). but

Solving this, we obtain∆T Cu = 9.94 K So∆T s.s =(300−9.94)/2 =

145 K It follows thatT Cu, left =255 ◦ C andT Cu, right =245 ◦ C.

The heat ﬂux can be obtained by applying Fourier’s law to any of the three layers We consider either stainless steel layer and get q=17 W mãK

145 K 0.002 m =1233 kW/m 2 Thus our initial approximation was accurate within a few percent.

A look ahead

To demonstrate heat protection, find a small open flame that produces soot, such as a candle or kerosene lamp, avoiding clean blue flames that emit less heat Position your finger 1 to 2 cm away from the flame until it feels uncomfortably hot Next, dip a fine mesh screen in soapy water to fill its holes and place it between your finger and the flame You will observe that your finger remains protected from the heat until the water evaporates.

Water is relatively transparent to light What does this experiment show you about the transmittance of water to infrared wavelengths?

So far, we have only scratched the surface of heat transfer concepts, outlining fundamental mechanisms and providing some quantitative relationships However, this foundational knowledge is just the beginning when tackling real-world heat transfer challenges To effectively solve practical problems, three essential tasks need to be accomplished.

• The heat diﬀusion equation must be solved subject to appropriate boundary conditions if the problem involves heat conduction of any complexity.

• The convective heat transfer coeﬃcient,h, must be determined if convection is important in a problem.

• The factor F 1–2 orF 1–2 must be determined to calculate radiative heat transfer.

Any of these determinations can involve a great deal of complication, and most of the chapters that lie ahead are devoted to these three basic problems.

Before delving into the three key questions, we will first examine the fundamental applied problem of heat transfer: the design of a heat exchanger Chapter 2 establishes the essential analytical tools required for this process, while Chapter 3 demonstrates the practical application of these tools in heat exchanger design.

36 Introduction §1.5 design ifh is already known This will make it easier to see the importance of undertaking the three basic problems in subsequent parts of the book.

Problems

This book primarily utilizes S.I units, making it easier for students familiar with the metric system For those struggling with dimensional conversions, Appendix B offers valuable assistance While English units are rarely used, they do appear in select problems at the end of each chapter to help students practice converting back to these units, acknowledging their continued relevance in the U.S for the foreseeable future.

A common topic of debate in heat transfer courses is the distinction between "theoretical" and "practical" problems, often sparking discussions between students and teachers Students tend to perceive problems as "theoretical" when they don't involve numerical calculations or require the derivation of algebraic equations, highlighting a need for clarity in understanding the nuances of theoretical and practical applications in heat transfer.

The problems assigned in this book are all intended to be useful in that they do one or more of ﬁve things:

1 They involve a calculation of a type that actually arises in practice (e.g., Problems1.1,1.3,1.8to1.18, and1.21through1.25).

2 They illustrate a physical principle (e.g., Problems 1.2, 1.4 to 1.7, 1.9, 1.20, 1.32, and 1.39) These are probably closest to having a

In the text, you are encouraged to apply the methods presented to derive additional results essential for specific applied problems, such as Problems 1.10, 1.16, 1.17, and 1.21 These types of problems are often the most challenging yet hold significant value for your understanding and application of the concepts.

4 They anticipate development that will appear in subsequent chapters (e.g., Problems1.16,1.20,1.40, and1.41).

To excel in the tasks presented, it is essential to enhance your skills in numerical and algebraic computations This requirement is particularly emphasized in the problems outlined in Chapter 1, specifically Problems 1.6 to 1.9, as well as Problems 1.15 and 1.17.

Partial numerical answers to some of the problems follow them in brackets Tables of physical property data useful in solving the problems are given in AppendixA.

Actually, we wish to look at thetheory,analysis, andpracticeof heat transfer—all three—according to Webster’s deﬁnitions:

Theory: “a systematic statement of principles; a formulation of apparent relationships or underlying principles of certain observed phenom- ena.”

Analysis involves solving problems through equations and breaking down a whole into its components to understand their nature, function, and relationships.

Practice: “thedoing of something as an application of knowledge.”

A composite wall is constructed with alternating layers of fir (5 cm), aluminum (1 cm), lead (1 cm), and corkboard (6 cm), with an external temperature of 60 °C on the fir side and 10 °C on the corkboard side To analyze the thermal behavior of the wall, it is essential to plot the temperature gradient across its layers The resulting temperature profile may indicate potential simplifying assumptions for further analysis, such as neglecting heat transfer through certain materials or assuming steady-state conditions.

1.3 q=5000 W/m 2 in a 1 cm slab andT =140 ◦ C on the cold side.

Tabulate the temperature drop through the slab if it is made of

Indicate which situations would be unreasonable and why.

The heat diffusion equation illustrates that in transient conduction, temperature variations are influenced by thermal diffusivity, α, which accounts for the rate at which heat spreads through a material In contrast, steady conduction problems can be effectively solved using only the thermal conductivity, k, as demonstrated in Example 1.1 This distinction highlights the different approaches required for analyzing transient versus steady-state heat transfer scenarios.

In a steady-state system, a 1 m copper rod with a cross-section of 1 cm² connects a thermal reservoir at 200 °C to another at 0 °C The rates of change of entropy can be calculated for each component: (a) the first reservoir experiences a decrease in entropy, (b) the second reservoir sees an increase in entropy, (c) the rod itself undergoes a change in entropy, and (d) the overall entropy change for the universe is +0.0120 W/K This positive change in entropy indicates compliance with the Second Law of Thermodynamics, which states that the total entropy of an isolated system can never decrease over time.

1.6 Two thermal energy reservoirs at temperatures of 27 ◦ C and

In a system with a temperature difference of -43 °C and a slab of material measuring 10 cm in thickness and 930 cm² in cross-sectional area, the thermal conductivity of the slab is 0.14 W/m·K Operating under steady-state conditions, the rates of change of entropy for the higher temperature reservoir, the lower temperature reservoir, the slab, and the entire universe must be calculated Furthermore, it is essential to determine whether these results comply with the Second Law of Thermodynamics.

In this problem, we analyze the effects of replacing thermal energy reservoirs with adiabatic walls on a slab's final equilibrium temperature Given the slab's density of 26 lb/ft³ and a specific heat of 0.65 Btu/lb·°F, we can calculate the final temperature and the entropy change for the slab, which is found to be 30.81 J/K Additionally, we assess whether this process aligns with the Second Law of Thermodynamics, ensuring that the principles of energy conservation and entropy increase are upheld.

A copper sphere with a diameter of 2.5 cm and a uniform temperature of 40 °C is placed in a slow-moving air stream at 0 °C, which generates a convection heat transfer coefficient of 15 W/m²K Due to copper's high thermal conductivity, the temperature within the sphere remains uniform throughout the cooling process The instantaneous energy balance between the sphere and the surrounding air can be established, leading to a solution that plots the sphere's temperature as it cools from 40 °C to 0 °C over time.

1.9 Determine the total heat transfer in Problem1.8as the sphere cools from 40 ◦ C to 0 ◦ C Plot the net entropy increase resulting from the cooling process above, ∆S vs T (K) [Total heat transfer = 1123 J.]

1.10 A truncated cone 30 cm high is constructed of Portland ce- ment The diameter at the top is 15 cm and at the bottom is

7.5 cm The lower surface is maintained at 6 ◦ C and the top at

To calculate the rate of heat transfer from top to bottom in a one-dimensional system with one surface at 40 °C and the other insulated, we apply Fourier’s law of heat conduction Since the heat transfer, Q, remains constant across every cross-section, we can express this relationship mathematically By integrating Fourier's law from the top to the bottom of the system, we can derive a formula that connects the unknown heat transfer rate Q with the known temperatures at each end This approach allows for an accurate determination of the heat transfer rate in watts.

To maintain the temperature of 100 kg of water at 75 °C in a 20 °C room, while limiting heat loss to no more than 3 °C per hour, it is essential to select an appropriate insulating material and determine its necessary thickness Assuming negligible temperature drops in the steel casing and convective boundary layers simplifies the calculations By applying the principles of thermal insulation, one can identify suitable materials and their respective thicknesses to achieve the desired thermal performance, ensuring efficient heat retention in the hot water heater.

1.12 What is the temperature at the left-hand wall shown in Fig.1.17.

Both walls are thin, very large in extent, highly conducting, and thermally black [T right =42.5 ◦ C.]

1.13 Develop S.I to English conversion factors for:

• The speciﬁc heat per unit mass,c

In each case, begin with basic dimension J, m, kg, s, ◦ C, and check your answers against Appendix B if possible.

1.14 Three inﬁnite, parallel, black, opaque plates transfer heat by radiation, as shown in Fig.1.18 FindT 2.

1.15 Four inﬁnite, parallel, black, opaque plates transfer heat by radiation, as shown in Fig.1.19 FindT 2andT 3 [T 2 =75.53 ◦ C.]

In a system with two large, black, horizontal plates separated by a distance L, the top plate is maintained at a controllable warm temperature (T_h), while the bottom plate is kept at a specified cooler temperature (T_c) The gas between the plates remains stationary due to the temperature gradient, with the top being warm and the bottom cool The relationship between radiative heat transfer (q_rad) and conductive heat transfer (q_cond) can be expressed as q_rad/q_cond = f(N, Θ), where N is a dimensionless group that includes parameters such as σ (Stefan-Boltzmann constant), k (thermal conductivity), L, and T_c This equation can be analyzed by plotting N as a function of Θ (the ratio of T_h to T_c) for various values of q_rad/q_cond, specifically at 1, 0.8, and 1.2, along with other values for a comprehensive understanding.

Now suppose that you have a system in which L = 10 cm,

At a temperature of 100 K, the hydrogen gas has an average thermal conductivity (k) of 0.1 W/m·K To achieve a balance where the conduction and radiation heat fluxes are equal, it is essential to determine the specific operating point on the heat transfer curve Consequently, it is necessary to report the corresponding temperature (T h) that must be maintained for optimal performance.

1.17 A blackened copper sphere 2 cm in diameter and uniformly at

200 ◦ C is introduced into an evacuated black chamber that is maintained at 20 ◦ C.

• Write a diﬀerential equation that expresses T (t) for the sphere, assuming lumped thermal capacity.

• Identify a dimensionless group, analogous to the Biot number, than can be used to tell whether or not the lumped- capacity solution is valid.

• Show that the lumped-capacity solution is valid.

• Integrate your diﬀerential equation and plot the temperature response for the sphere.

The heat diﬀusion equation

When designing heat exchangers, a crucial concept to understand is the overall heat transfer coefficient, which indicates the heat exchanger's resistance to heat flow This coefficient is typically derived from analyzing the resistances of individual components It is essential to accurately predict and evaluate the conductive resistance of complex bodies beyond simple flat walls Detailed discussions on evaluating heat transfer coefficients will be addressed in Chapters 6 and 7; however, for now, h values should be treated as provided data in any related problems.

The heat conduction aspect of heat exchanger issues is significantly more intricate than the basic planar analyses discussed in Chapter 1 To address these complexities, it is essential to derive the heat conduction equation and understand the methods for solving it.

The temperature distribution within a three-dimensional body, influenced by heating from one side, results in a space- and time-dependent temperature field, represented as T = T(x, y, z, t) or T(r, t) This temperature field defines the instantaneous temperature at any point in the body.

50 Heat conduction, thermal resistance, and the overall heat transfer coeﬃcient §2.1

Figure 2.1 A three-dimensional, transient temperature ﬁeld. isothermal surfaces,T 1 ,T 2 , and so on.

We next consider a very important vector associated with the scalar,

T The vector that has both the magnitude and direction of the maximum increase of temperature at each point is called thetemperature gradient,

“Experience”—that is, physical observation—suggests two things about the heat ﬂow that results from temperature nonuniformities in a body. §2.1 The heat diﬀusion equation 51

This says that qand ∇T are exactly opposite one another in direction and

This says that the magnitude of the heat ﬂux is directly proportional to the temperature gradient

The heat flux is defined by both its direction and magnitude, as expressed in Fourier's law: q = -k∇T This fundamental principle breaks down into three components, with the x-component represented as q_x = -k∂T.

The coeﬃcient k—the thermal conductivity—also depends on position and temperature in the most general case: k=k[r , T (r , t)] (2.3)

Most materials are nearly homogeneous, allowing us to express thermal conductivity as a function of temperature, k = k(T) The key assumption is that thermal conductivity remains constant, but its validity must be assessed for each specific case.

The behavior of thermal conductivity (k) is temperature-dependent, consistently increasing with temperature (T) in gases at low pressures, while in metals and liquids, it can either rise or fall It is crucial to determine if k can be considered approximately constant within the specific temperature range of interest For instance, in the case of iron, k can be treated as a constant between 0 °C and 40 °C, although inaccuracies may arise outside this range.

If the thermal conductivity (k) varies linearly with temperature (T) and the heat transfer is steady and planar, the heat transfer rate (q) can be expressed as q = k∆T / L, where k is determined at the average temperature However, in cases of non-planar heat transfer or when k does not follow a simple linear relationship, defining a precise effective value of k becomes challenging Nonetheless, for small temperature differences (∆T), a reasonably accurate approximation can still be achieved by using a constant average value of k.

Figure 2.2 Variation of thermal conductivity of metallic solids with temperature

Figure 2.3 The temperature dependence of the thermal conductivity of liquids and gases that are either saturated or at 1 atm pressure.

Figure 2.4 Control volume in a heat-ﬂow ﬁeld.

Now that we have revisited Fourier’s law in three dimensions, we see that heat conduction is more complex than it appeared to be in Chapter1.

We must now write the heat conduction equation in three dimensions.

We begin, as we did in Chapter1, with the First Law statement, eqn (1.3):

In this section, we apply equation (1.3) to analyze a three-dimensional control volume, illustrated in Fig 2.4 This control volume represents a finite region within a conducting body, designated for our analysis The surface is labeled as S, while the volume and region are referred to as R; both are in a state of rest Within this context, we identify an element of the surface, dS, which features two vectors: the unit normal vector, n (where |n| = 1), and the heat flux vector, q = -k∇T, at that specific point on the surface.

We consider the potential for a volumetric heat release of ˙ q(r) W/m³ to be uniformly distributed throughout the area This heat release may arise from various sources, including chemical or nuclear reactions, electrical resistance heating, external radiation, or other factors.

With reference to Fig.2.4, we can write the heat conductedout ofdS, in watts, as

The heat generated (or consumed) within the regionRmust be added to the total heat ﬂowintoS to get the overall rate of heat addition toR:

1 Figure 2.4 is the three-dimensional version of the control volume shown in Fig 1.8. §2.1 The heat diﬀusion equation 55

The rate of energy increase of the regionRis dU dt =

∂t dR (2.6) where the derivative of T is in partial form because T is a function of bothrandt.

Finally, we combineQ, as given by eqn (2.5), anddU /dt, as given by eqn (2.6), into eqn (1.3) After rearranging the terms, we obtain

To simplify the left-hand side, we apply Gauss's theorem, which transforms a surface integral into a volume integral According to Gauss's theorem, for any continuous function of position, the relationship holds true.

Therefore, if we identifyAwith(k∇T ), eqn (2.7) reduces to

Next, since the regionRis arbitrary, the integrand must vanish identically 2 We therefore get theheat diﬀusion equationin three dimensions:

The limitations on this equation are:

• Incompressible medium (This was implied when no expansion work term was included.)

• No convection (The medium cannot undergo any relative motion.

However, itcanbe a liquid or gas as long as it sits still.)

To satisfy the equation $ \int f(x) \, dx = 0 $, the function $ f(x) $ must equal zero across the entire range of integration For instance, if $ f(x) = \sin x $, this condition holds true only over specific intervals, such as $ 2\pi $ and its multiples Therefore, for the equation to be valid for any chosen interval, the expression within the parentheses must equal zero universally.

If the variation ofkwithTis small,kcan be factored out of eqn (2.10) to get

The enhanced heat conduction equation includes the term α, representing thermal diffusivity, previously discussed after equation (1.14) In this context, the term ∇²T, also expressed as ∇ · ∇T, is known as the Laplacian, which is derived within a Cartesian coordinate system.

The Laplacian can also be expressed in cylindrical or spherical coordinates The results are:

(2.14b) where the coordinates are as described in Fig.2.5.

Figure 2.5 Cylindrical and spherical coordinate schemes.

Solutions of the heat diﬀusion equation

The heat diffusion equation allows us to calculate temperature distribution and heat flux in various bodies Initially, we determine the temperature T(r, t) To obtain the heat flux, we differentiate T according to Fourier’s law.

The heat diffusion equation is a partial differential equation (PDE) that can be effectively solved using basic mathematical tools In one-dimensional steady-state scenarios, it simplifies to an ordinary differential equation (ODE) Additionally, its linear nature makes it manageable to work with We can outline a systematic approach to solving it, demonstrated through a practical example.

A large, thin concrete slab with a thickness of Lis undergoes a setting process, which is exothermic and releases ˙q W/m³ During this process, the external surfaces of the slab are maintained at ambient temperature, where T_w equals T_∞ The maximum internal temperature of the slab can be determined based on these conditions.

Step 1 Pick the coordinate scheme that best ﬁts the problem and identify the independent variables that determine T In the example,

Twill is expected to vary primarily in the thin dimension, referred to as the x-direction It is essential to ensure that the edges are insulated and that the length (L) is significantly smaller than both the width and height.

If they are, this assumption should be quite good.) Since the interior temperature will reach its maximum value when the process becomes steady, we writeT =T (x only).

Step 2 Write the appropriate d.e., starting with one of the forms of eqn (2.11).

Therefore, since T = T (xonly), the equation reduces to the §2.2 Solutions of the heat diﬀusion equation 59 ordinary d.e. d 2 T dx 2 = −q˙ k

Step 3 Obtain the general solution of the d.e (This is usually the easiest step.) We simply integrate the d.e twice and get

Step 4 involves specifying the "side conditions" for the differential equation, including initial and boundary conditions This aspect often proves to be the most challenging for beginner students, as it significantly tests their practical understanding of the problems at hand.

To eliminate the constants of integration in the general solution, it is essential to specify two temperature conditions for each spatial coordinate and one for the time coordinate.

(These matters are discussed at greater length in Chapter4.)

In this case there are two boundary conditions:

It is crucial to avoid introducing inaccessible information in boundary or initial conditions Always pause to consider whether you can obtain a numerical value for the specified temperature or other data at a particular position or time If you cannot access this information, your results will be rendered meaningless.

In Step 5, substitute the general solution into the boundary and initial conditions to determine the constants, a process that can become complex in transient and multidimensional scenarios Typically, Fourier series methods are employed to tackle these challenges Conversely, steady one-dimensional problems are generally simpler to solve For instance, by evaluating the solution at x=0 and x=L, we can obtain the necessary results.

Figure 2.6 Temperature distribution in the setting concrete slab Example2.1.

Step 6 Put the calculated constants back in the general solution to get the particular solution to the problem In the example problem we obtain:

This should be put in neat dimensionless form:

In Step 7, examine the solution closely to uncover its insights Conduct various checks to ensure its accuracy For instance, plotting equation (2.15) in Fig 2.6 reveals a parabolic temperature distribution, which is symmetrical as anticipated.

The results satisfy the boundary conditions at the wall while maximizing at the center, allowing for a simplified representation of all scenarios through a single curve This approach is particularly beneficial in complex calculations, as it condenses the dependence of the solution on five different variables into a unified curve on a two-coordinate graph.

Finally, we check to see if the heat ﬂux at the wall is correct: q wall = −k∂T

Thus, half of the total energy generated in the slab comes out of the front side, as we would expect The solution appears to be correct.

In Step 8, once the temperature field is accurately established, you have the option to calculate the heat flux at any point within the body by substituting T(r, t) back into Fourier’s law This process was previously demonstrated in Step 7 to verify our solution.

We shall run through additional examples in this section and the following one In the process, we shall develop some important results for future use.

In a steady-state condition, a slab depicted in Fig 2.7 exhibits varying temperatures on its opposing sides, with no internal heat generation present The objective is to determine the temperature distribution across the slab and the corresponding heat flux.

Solution These can be found quickly by following the steps set down in Example2.1:

Figure 2.7 Heat conduction in a slab (Example2.2).

Step 1 T =T (x)for steadyx-direction heat ﬂow

Step 2 d 2 T dx 2 =0, the steady 1-D heat equation with no ˙q Step 3 T =C 1 x+C 2 is the general solution of that equation

Step 7 We note that the solution satisﬁes the boundary conditions and that the temperature proﬁle is linear.

This result, which is the simplest heat conduction solution, calls to mind Ohm’s law Thus, if we rearrange it:

In thermal analysis, R_t represents thermal resistance, measured in K/W, analogous to an electric circuit This concept illustrates heat flow through a slab, as depicted in Figure 2.8.

Thermal resistance and the electrical analogy

Fourier’s, Fick’s, and Ohm’s laws

Fourier’s law has significant analogies in various physical behaviors, with the electrical analogy being one of the most notable These analogous processes offer valuable insights for solving heat transfer problems, and heat conduction analyses can frequently be adapted to describe these processes effectively.

Figure 2.8 Ohm’s law analogy to conduction through a slab.

Let us ﬁrst considerOhm’s law in three dimensions:

Amperes represent the vectorial electrical current, while A denotes an area perpendicular to this current vector J signifies the current density or flux of current, γ indicates electrical conductivity measured in cm/ohm·cm², and V represents voltage.

To apply eqn (2.16) to a one-dimensional current ﬂow, as pictured in

Fig.2.9, we write eqn (2.16) as

L , (2.17) but ∆V is the applied voltage,E, and the resistance of the wire is R ≡

R (2.18) which is the familiar, but restrictive, one-dimensional statement of Ohm’s law.

Fick’s law is another analogous relation It states that during mass diffusion, the flux,j 1, of a dilute component, 1, into a second fluid, 2, is

Figure 2.9 The one-dimensional flow of current. proportional to the gradient of its mass concentration,m 1 Thus j 1 = −ρD 12 ∇m 1 (2.19) where the constantD 12 is the binary diffusion coefficient.

In a 1-meter-long thin tube filled with air, a small water leak at one end leads to a water vapor concentration reaching a mass fraction of 0.01 Meanwhile, a desiccator on the opposite end keeps the water vapor concentration at zero This setup creates a steady flux of water vapor from the side with the leak to the desiccator, highlighting the effects of concentration gradients on mass transfer in gaseous systems.

One place in which the usefulness of the electrical resistance analogy becomes immediately apparent is at the interface of two conducting media.

When two solid surfaces are pressed together, they can never achieve perfect thermal contact due to inherent roughness, resulting in tiny air gaps at the contact plane.

Figure 2.10 Heat transfer through the contact plane between two solid surfaces.

Heat transfer at an interface occurs through two primary mechanisms Effective conduction takes place at solid-to-solid contact points, while conduction through gas is less efficient.

ﬁlled interstices, which have low thermal conductivity, can be very poor.

Thermal radiation across the gaps is also ineﬃcient.

To enhance the thermal performance of contact surfaces, we introduce an interfacial conductance, denoted as h_c, which is positioned in series with the conductive materials on both sides This coefficient, h_c, functions similarly to a heat transfer coefficient and shares identical units of W/m²K.

∆T is the temperature diﬀerence across an interface of area A, thenQ=

Ah c ∆T It follows thatQ=∆T /R t for a contact resistanceR t =1/(h c A) in K/W.

The interfacial conductance,h c , depends on the following factors:

• The surface ﬁnish and cleanliness of the contacting solids.

• The materials that are in contact.

• The pressure with which the surfaces are forced together This may vary over the surface, for example, in the vicinity of a bolt.

• The substance (or lack of it) in the interstitial spaces Conductive shims or ﬁllers can raise the interfacial conductance.

• The temperature at the contact plane.

The inﬂuence of contact pressure is usually a modest one up to around

10 atm in most metals Beyond that, increasing plastic deformation of

Table 2.1 Some typical interfacial conductances for normal surface ﬁnishes and moderate contact pressures (about 1 to 10 atm) Air gaps not evacuated unless so indicated.

Stainless steel/stainless steel (evacuated interstices)

Aluminum/aluminum (low pressure and evacuated interstices)

At high pressure, the local contact points significantly increase contact resistance, as demonstrated in Table 2.1, which presents typical values supporting this observation These values are adapted from sources [2.1, Chpt 3] and [2.2] For a deeper understanding, theories related to contact resistance can be found in sources [2.3] and [2.4].

Heat ﬂows through two stainless steel slabs (k=18 W/mãK) that are pressed together The slab area is A = 1 m 2 How thick must the slabs be for contact resistance to be negligible?

Solution With reference to Fig.2.11, we can write

3000 =0.00033Thus,Lmust be large compared to 18(0.00033)/2 = 0.003 m if contact resistance is to be ignored IfL=3 cm, the error is about 10%. §2.3 Thermal resistance and the electrical analogy 67

Figure 2.11 Conduction through two unit-area slabs with a contact resistance.

Resistances for cylinders and for convection

In our ongoing development of methods for addressing one-dimensional heat conduction problems, we discover that various pathways of heat flow can be represented as thermal resistances, which can be integrated into our solutions Furthermore, once we solve the heat conduction equation, the resulting data can also serve as new thermal resistances.

Example 2.5 Radial Heat Conduction in a Tube

Find the temperature distribution and the heat ﬂux for the long hollow cylinder shown in Fig.2.12.

∂r =C 1 ; integrate again: T =C 1 lnr +C 2 Step 4 T (r =r i )=T i andT (r =r o )=T o

Figure 2.12 Heat transfer through a cylinder with a ﬁxed wall temperature (Example2.5).

Step 6 T =T i − ∆T ln(r o /r i )(lnr −lnr i )or

The solution illustrated in Fig 2.12 reveals a logarithmic temperature profile that meets both boundary conditions Notably, when the cylinder wall is extremely thin, or when the ratio of the inner radius to the outer radius (r_i/r_o) approaches 1, the relationship can be expressed as ln(r/r_i) ≈ (r - r_i)/r_i and ln(r_o/r_i) ≈ (r_o - r_i)/r_i.

T o −T i = r −r i r o −r i which is a simple linear proﬁle This is the same solution that we would get in a plane wall.

Step 8 At any station,r: q radial = −k∂T

The heat flux decreases inversely with the radius, which is logical because the same amount of heat must traverse a larger surface area as the radius expands This principle can be examined in the context of a cylinder with a specified length.

Finally, we again recognize Ohm’s law in this result and write the thermal resistance for a cylinder:

(2.22) This can be compared with the resistance of a plane wall:

Both resistances are inversely proportional to k, but each re-

In the preceding examples, the boundary conditions were all the same

In this scenario, we consider a temperature defined at the outer edge of a body, while also taking into account the ambient temperature surrounding it Additionally, there exists a heat transfer coefficient that facilitates the thermal interaction between the environment and the body.

Figure 2.13 Heat transfer through a cylinder with a convective boundary condition (Example2.6).

The convective heat transfer coefficient surrounding the cylinder, as illustrated in Example 2.5 and Figure 2.13, establishes thermal resistance between the cylinder and the surrounding environment at temperature T = T ∞ This analysis aims to determine the temperature distribution and heat flux in this scenario.

Step 1 through 3 These are the same as in Example2.5.

Step 4 The ﬁrst boundary condition isT (r = r i )= T i The second boundary condition must be expressed as an energy balance at the outer wall (recall Section1.3). q convection =q conduction at the wall or h(T−T ∞ ) r =r o = −k ∂T

Step 5 From the ﬁrst boundary condition we obtainT i = C 1 lnr i +

In Section 2.3, titled "Thermal Resistance and the Electrical Analogy," we will carefully examine the potential errors that can arise when substituting the general solution into the second boundary condition.

A common error is to substituteT = T o on the lefthand side instead of substituting the entire general solution That will do no good, becauseT o is not an accessible piece of information.

Equation (2.23) reduces to: h(T ∞ −C 1lnr o −C 2 )= kC 1 r o

When we combine this with the result of the ﬁrst boundary condition to eliminateC 2 :

This can be rearranged in fully dimensionless form:

In Step 7, we set the ratio of outer radius to inner radius (r_o/r_i) to 2 and plotted equation (2.24) for various Biot numbers The findings are illustrated in Fig 2.13, revealing several significant insights from the plot.

When the Biot number (Bi) approaches 1, the solution aligns with that presented in Example 2.5, indicating negligible convective resistance to heat flow This observation confirms our earlier expectations from Section 1.3 regarding scenarios with large Bi values Conversely, when Bi is significantly less than 1, the temperature difference (T - T_i) behaves in a markedly different manner.

Overall heat transfer coeﬃcient, U

We often want to transfer heat through composite resistances, as shown in Fig.2.18 It is very convenient to have a number, U, that works like this 4 :

The overall heat transfer coefficient is primarily determined by the system itself and often remains unaffected by the system's operating conditions For instance, in Example 2.6, we can utilize the value of Q provided by equation (2.25) to derive relevant results.

We have basedUon the outside area,A o =2π r o l, in this case We might instead have based it on inside area,A i =2π r i l, and obtained

4 This U must not be confused with internal energy The two terms should always be distinct in context. §2.4 Overall heat transfer coeﬃcient, U 79

Figure 2.18 A thermal circuit with many resistances.

It is therefore important to remember which area an overall heat transfer coeﬃcient is based on It is particularly important that A andU be consistent when we writeQ=U A∆T.

Estimate the overall heat transfer coeﬃcient for the tea kettle shown in Fig.2.19 Note that the ﬂame convects heat to the thin aluminum.

The heat is then conducted through the aluminum and ﬁnally convected by boiling into the water.

Solution We need not worry about deciding which area to baseA on because the area normal to the heat ﬂux vector does not change.

We simply write the heat ﬂow

1 hA+ L k Al A + 1 h b A and apply the deﬁnition ofU

Let us see what typical numbers would look like in this example: h might be around 200 W/m 2 K;L k Al might be 0.001 m/(160 W/mãK) or 1/160,000 W/m 2 K; and h b is quite large— perhaps about 5000

Figure 2.19 Heat transfer through the bottom of a tea kettle.

It is clear that the ﬁrst resistance is dominant, as is shown in Fig.2.19. Notice that in such cases

U A →1/R t dominant (2.35) whereAis any area (inside or outside) in the thermal circuit.

Boil water in a paper cup over an open ﬂame and explain why you can do so [Recall eqn (2.35) and see Problem2.12.]

A wall consists of alternating layers of pine and sawdust, as shown in Fig.2.20) The sheathes on the outside have negligible resistance andhis known on the sides ComputeQandUfor the wall.

As long as the thermal conductivity of the wood and sawdust remains similar, we can model the wall as a parallel resistance circuit, as illustrated in the figure.

5 For this approximation to be exact, the resistances must be equal If they diﬀer radically, the problem must be treated as two-dimensional. §2.4 Overall heat transfer coeﬃcient, U 81

Figure 2.20 Heat transfer through a composite wall. total thermal resistance of the circuit is

The method demonstrated in this example is commonly employed to calculate U-values for the walls and roofs of houses and buildings By combining the thermal resistances of various structural components—such as insulation, studs, siding, doors, and windows—one can determine the total U or R value This total is then utilized alongside weather data to estimate the heating and cooling loads required for efficient climate control.

Table 2.2 Typical ranges or magnitudes ofU

Walls and roofs dwellings with a 24 km/h outdoor wind:

Air to heavy tars and oils As low as 45 Air to low-viscosity liquids As high as 600

Steam or water to oil 60−340

Liquids in coils immersed in liquids 110−2,000

Steam-jacketed, agitated vessels 500−1,900 Shell-and-tube ammonia condensers 800−1,400 Steam condensers with 25 ◦ C water 1,500−5,000 Condensing steam to high-pressure boiling water

† Main heat loss is by inﬁltration.

A heat exchanger is defined as a device situated between two fluid masses with differing temperatures, serving the purpose of either facilitating or hindering heat transfer When designed to enhance heat exchange, the overall heat transfer coefficient (U) typically exceeds 40 W/m²K, while configurations aimed at reducing heat flow result in a U value below this threshold.

When considering insulation effectiveness, a U-value of 10 W/m²K signifies nearly perfect insulation It is essential to understand the relative values of U, which can be found in Table 2.2, for a better grasp of insulation performance.

Fluids with low thermal conductivities, including tars, oils, and various gases, typically result in low heat transfer coefficients (h) Consequently, when these fluids flow on one side of a heat exchanger, the overall heat transfer coefficient (U) is generally reduced.

• Condensing and boiling are very eﬀective heat transfer processes.

They greatly improve U but they cannot override one very small value ofhon the other side of the exchange (Recall Example2.10.)

• For a highU,all resistances in the exchanger must be low.

• The highly conducting liquids, such as water and liquid metals, give high values ofhandU.

Figure2.21shows one of the simplest forms of a heat exchanger—a pipe.

The left side of the pipe is new and clean, while the right side has accumulated a layer of scale, often consisting of magnesium sulfate (MgSO4) or calcium sulfate (CaSO4) in conventional freshwater preheaters This scale precipitates onto the pipe wall over time, necessitating the inclusion of an additional empirical resistance in the calculation of U to account for the impact of these buildups.

Figure 2.21 The fouling of a pipe.

Table 2.3 Some typical fouling resistances for a unit area.

Fluid and Situation Fouling Resistance

Treated boiler feedwater 0.0001−0.0002 Clean river or lake water 0.0002−0.0006 About the worst waters used in heat exchangers

Steam, oil-bearing (e.g., turbine exhaust)

Refrigerant vapors (oil-bearing) 0.0040 whereR f is a fouling resistance for a unit area of pipe (in m 2 K/W) And clearly

Table 2.3 presents typical values of R f, adapted from sources [2.6] and [2.7] It is important to note that fouling introduces an additional resistance of approximately 10^-4 m² K/W, similar to incorporating another heat transfer coefficient, h f, around 10,000 W/m² K, which acts in series with the other resistances in the heat exchanger.

The tabulated R f values are presented with only one significant figure due to their approximate nature, as exact values depend on specific heat exchanger configurations, fluid types, fluid velocities, operating temperatures, and equipment age Generally, resistance decreases with increased fluid velocity while it rises with temperature and age The values in the table are derived from typical maintenance practices and conventional shell-and-tube heat exchangers.

With misuse, a given heat exchanger can yield much higher values ofR f

In a water-to-water heat exchanger with a heat transfer coefficient (U) around 2000 W/m²K, fouling is insignificant as it contributes a minimal resistance in the system.

In a finned-tube heat exchanger, where hot gas flows through the tubes and cold gas moves across the fins, the overall heat transfer coefficient (U) can reach approximately 200 W/m²K Under these conditions, fouling is typically considered insignificant.

If your house features unpainted aluminum siding, the engineer's heat loss calculation is based on a U-value of 5 W/m²K However, due to air pollution, the thermal resistance (R_f) of the siding is measured at 0.0005 m²K/W.

Should the engineer redesign the siding?

Solution From eqn (2.36) we get

U uncorrected +R f =0.2000+0.0005 m 2 K/W Therefore, fouling is entirely irrelevant to domestic heat loads.

Since the engineer did not fail you in the preceding calculation, you entrust him with the installation of a heat exchanger at your plant.

He installs a water-cooled steam condenser withU = 4000 W/m 2 K.

The use of water-side fouling resistance for distilled water raises questions, especially when the water flowing through the tubes appears unclear This discrepancy prompts an investigation into the methods employed to achieve this outcome.

Solution Equation (2.36) and Table2.3give

=0.00085 to 0.00225 m 2 K/W Thus, U is reduced from 4,000 to between 444 and 1,176 W/m 2 K.

Fouling is crucial in this case, and the engineer was in serious error.

86 Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coeﬃcient

Summary

Four things have been done in this chapter:

The heat diffusion equation has been formulated, and a method for solving it in straightforward scenarios has been developed, yielding significant results Further discussions on solving the heat diffusion equation will be provided in Part II of this book.

In our exploration of the electric analogy to steady heat flow, we focused on the crucial concept of thermal resistance By leveraging this analogy, we effectively addressed heat transfer challenges in a manner similar to solving electrical circuit problems.

• The overall heat transfer coeﬃcient has been deﬁned, and we have seen how to build it up out of component resistances.

• Some practical problems encountered in the evaluation of overall heat transfer coeﬃcients have been discussed.

Three very important things havenot been considered in Chapter2:

In evaluating U, we have consistently treated the values of h as predetermined However, in practical scenarios, it is essential to establish accurate values of h tailored to the specific context Part III addresses the process of determining these values effectively.

Heat exchangers facilitate the transfer of energy as fluids flow through them, resulting in a varying driving temperature difference across the exchanger This complexity in energy exchange presents challenges in the design of heat exchangers, which we will address in Chapter 3.

• The heat transfer coeﬃcients themselves vary with position inside many types of heat exchangers, causingUto be position-dependent.

To demonstrate that the thermal conductivity $ k $ varies linearly with temperature $ T $ in a slab, it is essential to consider a one-dimensional and steady heat transfer scenario Under these conditions, the heat flux $ q $ can be accurately determined by evaluating the thermal conductivity $ k $ at the average temperature within the slab This relationship underscores the significance of mean temperature in calculating heat transfer in materials exhibiting linear thermal conductivity behavior.

To calculate the steady heat flux through a plane wall with a temperature-dependent thermal conductivity k(T), we can develop a numerical method that approximates the heat transfer For an iron slab with a thickness of 1 cm, where the temperature ranges from -100 °C on the left side to 400 °C on the right side, we can apply this method to predict the heat flux By discretizing the temperature profile and using appropriate boundary conditions, we can effectively estimate the heat flux through the slab, ensuring that the calculations adhere to the principles of thermal conductivity and heat transfer.

How far would you have erred if you had taken k average =

2.3 The steady heat ﬂux at one side of a slab is a known valueq o

The thermal conductivity varies with temperature in the slab, and the variation can be expressed with a power series as k= i=n $ i = 0

(a) Start with eqn (2.10) and derive an equation that relates

T to position in the slab, x (b) Calculate the heat ﬂux at any position in the wall from this expression using Fourier’s law.

Is the resultingqa function ofx?

2.4 Combine Fick’s law with the principle of conservation of mass

(of the dilute species) in such a way as to eliminate j 1, and obtain a second-order diﬀerential equation inm 1 Discuss the importance and the use of the result.

To determine the temperature distribution in a thick-walled pipe, it is essential to know the bulk interior temperature (T ∞ i) and the exterior air temperature (T ∞ o) Additionally, the interior and exterior heat transfer coefficients, denoted as h i and h o, respectively, play a crucial role in the analysis By employing the methodology outlined in Example 2.6, the results can be expressed in a dimensionless form, facilitating a clearer understanding of the thermal behavior within the pipe.

To convert the boundary conditions from Problem 2.5 into dimensionless form, incorporate the Biot numbers, allowing them to approach infinity This process will revert to the boundary conditions established in Example 2.5 Consequently, the solution derived in Problem 2.5 should align with the solution from Example 2.5.

Example 2.5when the Biot numbers approach inﬁnity Show that this is the case.

The critical radius of insulation is the point where adding more insulation actually starts to make things less efficient Imagine wrapping a warm drink in a cozy blanket; at first, it keeps the drink hot, but if you wrap it too tightly, it can trap heat and cause the drink to cool down faster So, there’s a perfect amount of insulation that keeps things warm without letting heat escape too quickly, and that’s what we call the critical radius.

The slab depicted in Fig 2.22 is insulated on five sides, with the sixth side exposed to ambient temperature and a defined heat transfer coefficient It generates heat at a rate of 1.0 kW/m³ and has a thermal conductivity of 0.2 W/m·K To determine the temperature distribution within the slab, it is essential to outline the assumptions made and clearly specify the boundary conditions Additionally, the temperatures at both the front and back faces of the slab need to be evaluated, and it is necessary to demonstrate that the solution aligns with the expected heat fluxes at these surfaces.

In the analysis of the composite wall depicted in Fig 2.23, both the concrete and brick sections possess equal thicknesses To determine the temperatures T1 and T2, as well as the heat flow rate q, we will approximate the heat transfer as one-dimensional It is essential to illustrate the thermal circuit for the wall and identify all four thermal resistances prior to calculations Additionally, we will calculate the percentage of heat flow q that passes through the brick section.

2.10 Compute QandU for Example 2.11if the wall is 0.3 m thick.

Five (each) pine and sawdust layers are 5 and 8 cm thick, re-

Problems 89 spectively; and the heat transfer coeﬃcients are 10 on the left and 18 on the right T ∞ 1 =30 ◦ C andT ∞ r =10 ◦ C.

2.11 ComputeU for the slab in Example1.2.

In Example 2.10, we analyze a tea kettle containing 1 kg of water, approximately 1 liter, with a flame impacting an area of 0.02 m² at the bottom We need to determine the rate at which the water temperature increases upon reaching its boiling point and calculate the temperature of the kettle's bottom, just below the water, assuming the gases from the flame are at a specified temperature.

500 ◦ C when they touch the bottom of the kettle Assume that the heat capacitance of the aluminum kettle is negligible (b)

An intriguing parlor trick demonstrates that placing a paper cup filled with water over an open flame can boil the water without igniting the paper This phenomenon can be explained through an electrical analogy, where the rate of temperature change (dT/dt) is measured at 0.37 °C/s, illustrating how the heat transfer from the flame to the water is efficient enough to raise the temperature without exceeding the paper's combustion point.

Copper plates with thicknesses of 2 mm and 3 mm are subjected to light processing On one side, non-oil-bearing steam condenses under pressure at a saturation temperature of 200 °C, exhibiting a heat transfer coefficient of 12,000 W/m²K Conversely, methanol boils under pressure at a temperature of 130 °C on the opposing side.

9000 W/m 2 K) EstimateU and q initially and after extended service List the relevant thermal resistances in order of decreasing importance and suggest whether or not any of them can be ignored.

2.14 0.5 kg/s of air at 20 ◦ C moves along a channel that is 1 m from wall to wall One wall of the channel is a heat exchange surface

In a thermal system with a heat transfer coefficient of U = 300 W/m²K and steam condensing at 120 °C on its back, the following parameters need to be determined: (a) the heat transfer rate at the entrance, (b) the rate of temperature increase of the fluid with respect to position x at the entrance, and (c) the temperature and heat flux 2 meters downstream, where the temperature is measured to be T 2m = 89.7 °C.

Function and conﬁguration of heat exchangers

Heat exchangers are designed to transfer energy between two fluid masses, typically separated by a simple or composite wall that creates thermal resistance An exception to this standard configuration is the direct-contact heat exchanger, which allows for direct interaction between the fluids.

3.2shows one such arrangement in which steam is bubbled into water.

Steam condensation occurs simultaneously with water heating, while in other configurations, immiscible fluids may interact, or noncondensable gases can be introduced into liquids.

This article focuses on heat exchangers featuring a dividing wall between two fluids While there are numerous configurations available, most commercial heat exchangers can be categorized into three fundamental types, as illustrated in Figure 3.3.

• The simple parallel or counterflow configuration These arrangements are versatile Figure3.4shows how the counterflow arrangement is bent around in a so-called Heliflow compact heat exchanger configuration.

• The shell-and-tube conﬁguration Figure3.5 shows the U-tubes of a two-tube-pass, one-shell-pass exchanger being installed in the

Figure 3.1 Heat exchange. supporting baﬄes The shell is yet to be added Most of the really large heat exchangers are of the shell-and-tube form.

The cross-flow configuration, illustrated in Figure 3.6, features typical cross-flow units where both flows remain unmixed In Figures 3.6a and 3.6c, each flow is confined to a designated path within the exchanger, preventing any lateral mixing Additionally, Figure 3.6b depicts a standard plate-fin cross-flow element, which similarly maintains unmixed flows.

Figure3.7, taken from the standards of the Tubular Exchanger Manu- facturer’s Association (TEMA) [3.1], shows four typical single-shell-pass heat exchangers and establishes nomenclature for such units.

The images illustrate the challenges of translating basic concepts into hardware design Figure 3.7 depicts an exchanger with a single tube pass, where the shell flow is baffled to create a crisscross pattern over the tubes Despite this configuration, the flow still moves from the hot end to the cold end, resembling a simple parallel or counter-flow unit Additionally, the kettle reboiler shown in Figure 3.7d features a divided shell-pass flow configuration across two tube passes, facilitating flow from left to right and then returning.

The isothermal shell flow in a heat exchanger can move in any direction without affecting the tube flow, making it comparable to both simple parallel and counterflow configurations.

Figure 3.2 A direct-contact heat exchanger.

Notice that a salient feature of shell-and-tube exchangers is the presence of baffles Baffles serve to direct the flow normal to the tubes We

In Part III, it is demonstrated that heat transfer from a tube to a flowing fluid is more efficient when the fluid flows across the tube rather than along it This enhancement in heat transfer provides the complex shell-and-tube exchanger with a distinct advantage over simpler single-pass parallel and counterflow exchangers.

Baffles can introduce several challenges in fluid dynamics, as they create complex flow patterns that are difficult to analyze Fluid on the shell side may unpredictably leak through baffle holes in the axial direction or bypass the baffles near the wall Additionally, certain shell-flow configurations can inadvertently excite unexpected vibrational modes in the tubes, while many cross-flow setups also experience baffle-related complications.

fluid so as to move it across a tube bundle The plate-and-fin configuration (Fig.3.6b) is such a cross-flow heat exchanger.

The diverse arrangements of heat exchangers highlight the significant human ingenuity invested in enhancing heat transfer between fluid flows The endless variations in design and functionality become evident when exploring Experiment 3.1.

Throughout a day, take a notebook and document every heat exchanger you come across in your home, university, or vehicle Categorize each heat exchanger by type and highlight any unique enhancement features observed.

The analysis of heat exchangers ﬁrst becomes complicated when we account for the fact that two ﬂow streams change one another’s temper-

Figure 3.3 The three basic types of heat exchangers.

Evaluation of the mean temperature diﬀerence in a heat

Figure 3.4 Heliﬂow compact counterﬂow heat exchanger.

(Photograph coutesy of Graham Manufacturing Co., Inc.,

In Batavia, New York, we focus on the challenge of accurately predicting the mean temperature difference, which is discussed in Section 3.2 Additionally, Section 3.3 outlines a strategy for situations where the mean temperature cannot be initially established.

3.2 Evaluation of the mean temperature diﬀerence in a heat exchanger

Logarithmic mean temperature diﬀerence (LMTD)

In compact single-phase heat exchangers, the overall heat transfer coefficient (U) can be considered constant However, in larger exchangers, such as shell-and-tube configurations and large condensers, U may vary with position and local temperature When U remains relatively constant, we can effectively address the varying temperatures of fluid streams by expressing the overall heat transfer in terms of a mean temperature difference between the two fluid streams.

Figure 3.5 Typical commercial one-shell-pass, two-tube-pass heat exchangers.

The 1980 Chevette radiator features a cross-flow design that prevents flow mixing, showcasing flat vertical tubes at its edges In contrast, an automotive air conditioning condenser allows for mixed flow through its horizontal wavy fins while maintaining unmixed two-pass flow in the U-tubes Additionally, a basic module for a waste heat recuperator, measuring 1 ft × 1 ft × 2 ft., employs a plate-fin, gas-to-air cross-flow heat exchanger that also keeps both flows unmixed.

Figure 3.6 Several commercial cross-ﬂow heat exchangers.

(Photographs courtesy of Harrison Radiator Division, General

Figure 3.7 Four typical heat exchanger conﬁgurations (continued on next page) (Drawings courtesy of the Tubular Exchan- ger Manufacturers’ Association.)

106 §3.2 Evaluation of the mean temperature diﬀerence in a heat exchanger 107

To solve our problem, we need to determine the correct mean temperature difference that satisfies the equation This analysis will focus on the basic parallel and counterflow configurations, as illustrated in Fig 3.8.

Figure 3.8 illustrates the temperature variations of both streams in single-pass arrangements—parallel and counterflow configurations—against the length of travel In the parallel-flow setup, temperatures fluctuate more rapidly with distance, requiring less length for effective heat exchange Conversely, the counterflow arrangement facilitates a more thorough heat exchange between the two flows.

Figure3.9shows another variation on the single-pass conﬁguration.

This is a condenser in which one stream ﬂows through with its tempera-

Figure 3.8 illustrates the temperature variation in single-pass heat exchangers, highlighting that one fluid experiences temperature changes while the other condenses at a consistent temperature This setup possesses unique characteristics that will be discussed shortly.

The determination of the mean temperature difference (∆T mean) in heat exchangers is based on the differential heat transfer, represented by the equation dQ=U∆T dA= −(mc˙ p ) h dT h = ±(mc˙ p ) c dT c In this equation, the subscripts 'h' and 'c' refer to the hot and cold streams, respectively, while the signs indicate whether the arrangement is parallel or counterflow The variable dT signifies the temperature change across the exchanger Additionally, we denote the total heat capacities of the hot and cold streams for further analysis.

In a heat exchanger, the heat transfer rates are defined as C_h = (mċ_p)h W/K and C_c = (mċ_p)c W/K The relationship between the hot and cold sides can be expressed by the equation ∓C_h dT_h = C_c dT_c This equation can be integrated, considering the initial temperatures T_h in and T_c in, to evaluate the mean temperature difference in the heat exchanger.

Figure 3.9 The temperature distribution through a condenser. parallel flow orT h =T h in andT c =T c out for counterflow, to some arbitrary point inside the exchanger The temperatures inside are thus: parallel flow: T h =T h in − C c

(3.4) whereQis the total heat transfer from the entrance to the point of interest Equations (3.4) can be solved for the local temperature diﬀerences:

Substitution of these indQ=C c dT c =U∆T dAyields

Equations (3.6) can be integrated across the exchanger:

IfU andC c can be treated as constant, this integration gives parallel: ln

The variability of U must be considered in the integration from equation (3.7) to equations (3.8), as it can significantly complicate the latter Assuming the validity of equations (3.8), we can simplify them using the definitions of ∆T a.

∆T b , given in Fig.3.8: parallel: ln (1+C c /C h )(T c in −T c out )+∆T b

(3.9) Conservation of energy (Q c =Q h ) requires that

T c out −T c in (3.10) §3.2 Evaluation of the mean temperature diﬀerence in a heat exchanger 111

Then eqn (3.9) and eqn (3.10) give parallel: ln

Finally, we write 1/C c =(T c out −T c in )/Qand 1/C h =(T h in −T h out )/Qon the right-hand side of either of eqns (3.11) and get for either parallel or counterﬂow,

The appropriate∆T mean for use in eqn (3.11) is thus thelogarithmic mean temperature diﬀerence(LMTD):

The idea of a logarithmic mean diﬀerence is not new to us We have already encountered it in Chapter 2 Suppose that we had asked,

To determine the mean radius of a pipe that enables us to calculate the conduction through its wall as if it were a slab with a thickness of L = r_o - r_i, we need to analyze the relationship between the inner and outer radii of the pipe This comparison will help us understand the effective thermal conduction characteristics of the pipe wall.

Figure 3.10 Calculation of the mean radius for heat conduction through a pipe.

It follows that r mean = r o −r i ln(r o /r i ) =logarithmic mean radius

Suppose that the temperature diﬀerence on either end of a heat exchanger,∆T a , and∆T b , are equal Clearly, the eﬀective∆Tmust equal

∆T a and∆T b in this case Does the LMTD reduce to this value?

Solution If we substitute∆T a =∆T b in eqn (3.13), we get

0 =indeterminate Therefore it is necessary to use L’Hospital’s rule: limit

=∆T a =∆T b §3.2 Evaluation of the mean temperature diﬀerence in a heat exchanger 113

It follows that the LMTD reduces to the intuitively obvious result in the limit.

Water enters the tubes of a small single-pass heat exchanger at 20 ◦ C and leaves at 40 ◦ C On the shell side, 25 kg/min of steam condenses at

60 ◦ C Calculate the overall heat transfer coeﬃcient and the required

ﬂow rate of water if the area of the exchanger is 12 m 2 (The latent heat,h fg , is 2358.7 kJ/kg at 60 ◦ C.)

60 =983 kJ/s and with reference to Fig 3.9, we can calculate the LMTD without naming the exchanger “parallel” or “counterﬂow”, since the conden- sate temperature is constant.

Extended use of the LMTD

Limitations There are two basic limitations on the use of an LMTD.

The ﬁrst is that it is restricted to the single-pass parallel and counter-

ﬂow conﬁgurations This restriction can be overcome by adjusting the

LMTD for other conﬁgurations—a matter that we take up in the following subsection.

Figure 3.11 A typical case of a heat exchanger in which U varies dramatically.

The second limitation is the assumption of a constant value of U, which is significant because U must be minimally affected by temperature (T) for the integration of equation (3.7) to hold true Even if U is not a function of T, variations in flow configuration and temperature can lead to substantial fluctuations in U within a heat exchanger For instance, as illustrated in Figure 3.11, the variation of U can be considerable, particularly when the liquid is completely vaporized, altering the heat exchange mechanism on the water side If U were consistent throughout the heat exchanger, it could be analyzed as two distinct exchangers operating in series.

Designing heat exchangers presents challenges, particularly when the overall heat transfer coefficient (U) varies continuously with position This issue is most pronounced in large industrial shell-and-tube configurations, while compact heat exchangers, with their reduced surface area, experience less variability When U is location-dependent, it is essential to conduct analyses using an average U, as outlined in previous equations (3.1 to 3.13).

1 Actual heat exchangers can have areas well in excess of 10,000 m 2 Large power plant condensers and other large exchangers are often remarkably big pieces of equipment.

The PFT-type integral-furnace boiler features a heat exchange surface area of 4560 m², with approximately 88% of this area dedicated to furnace tubing and the remaining 12% allocated to the boiler itself Despite its specifications, this boiler is considered relatively compact in size (Photograph courtesy of Babcock and Wilcox Co.)

The LMTD correction factor, F, is essential for analyzing heat exchangers with complex configurations, such as multiple passes or cross-flow systems, where the overall heat transfer coefficient, U, can be considered constant In these scenarios, it is crucial to rederive the mean temperature difference similarly to how the LMTD was initially derived Each specific configuration requires individual analysis, leading to results that are typically more complex than the standard equations used for simpler systems.

In the early twentieth century, calculations for heat exchanger configurations were performed on an ad hoc basis However, in 1940, Bowman, Mueller, and Nagle organized these calculations systematically for common heat exchanger designs.

Heat exchanger eﬀectiveness

Understanding the performance of an exchanger is possible once we have information about its configuration and the imposed differences However, it is often the case that we lack sufficient knowledge about the system prior to finalizing the design.

To calculate the heat transfer (Q) in a system, we typically start with initial data, as illustrated in Fig 3.15 This process involves estimating an exit temperature to ensure that the heat transfer from the hot side (Qh) equals the heat transfer to the cold side (Qc), represented by the equation Qh = Qc = Ch ∆Th.

C c ∆T c Then we could calculate Qfrom U A(LMTD) or UAF(LMTD) and check it againstQ h The answers would diﬀer, so we would have to guess new exit temperatures and try again.

Such problems can be greatly simplified with the help of the so-called effectiveness-NTU method This method was first developed in full detail

In a single shell-pass heat exchanger, the R and P lines do not intersect, as illustrated in Fig 3.14(a) Consequently, achieving the desired temperatures is not possible with a single-shell exchanger.

In their 1955 book "Compact Heat Exchangers," Kays and London highlight a design problem where the Log Mean Temperature Difference (LMTD) cannot be calculated a priori This issue is particularly relevant for compact heat exchangers, as the overall heat transfer coefficient tends to remain relatively uniform, making the current method more applicable.

The heat exchanger eﬀectiveness is deﬁned as ε≡ C h (T h in −T h out )

C min (T h in −T c in ) = C c (T c out −T c in )

The effectiveness of a heat exchanger can be expressed as ε = actual heat transferred / maximum heat that could possibly be transferred In this context, C min represents the smaller value between C c and C h, highlighting the significance of the minimum heat capacity in determining the system's performance.

A second deﬁnition that we will need was originally made by E.K.W.

Nusselt, whom we meet again in PartIII This is the number of transfer units (NTU):

This dimensionless group can be viewed as a comparison of the heat capacity of the heat exchanger, expressed in W/K, with the heat capacity of the ﬂow.

We can immediately reduce the parallel-ﬂow result from eqn (3.9) to the following equation, based on these deﬁnitions:

We solve this forεand, regardless of whetherC minis associated with the hot or cold ﬂow, obtain for the parallel single-pass heat exchanger: ε≡ 1−exp[−(1+C min /C max )NTU]

(3.20) The corresponding expression for the counterﬂow case is ε= 1−exp[−(1−C min /C max )NTU]

Equations (3.20) and (3.21) illustrate the effectiveness of heat exchangers, as depicted in Fig 3.16 Additional calculations yield effectiveness results for various heat exchanger configurations, which are presented in Fig 3.17 To demonstrate the practical application of effectiveness in design, we will explore two illustrative examples.

Consider the following parallel-flow heat exchanger specification: cold flow enters at 40 ◦ C: C c =20,000 W/K hot flow enters at 150 ◦ C: C h =10,000 W/K

Determine the heat transfer and the exit temperatures.

In situations where exit temperatures are unknown, calculating the Log Mean Temperature Difference (LMTD) becomes unfeasible Instead, one can refer to the parallel-flow effectiveness chart or utilize equation (3.20) for further analysis.

Figure 3.16 The effectiveness of parallel and counterflow heat exchangers (Data provided by A.D Krauss.) and we obtainε=0.596 Now from eqn (3.17), we find that

Finally, from energy balances such as are expressed in eqn (3.4), we get

In this scenario, we analyze a heat exchanger similar to the one described in Example 3.5, where the area is treated as an unspecified design variable Our objective is to determine the optimal area required to achieve the desired temperature for the hot fluid exiting the exchanger.

Solution Once the exit cold ﬂuid temperature is known, the problem can be solved with equal ease by either the LMTD or the eﬀective-

Figure 3.17 The eﬀectiveness of some other heat exchanger conﬁgurations (Data provided by A.D Krauss.)

124 §3.3 Heat exchanger eﬀectiveness 125 ness approach.

2(150−90)=70 ◦ C Then, using the eﬀectiveness method, ε= C h (T h in −T h out )

10,000(150−40) =0.5455 so from Fig.3.16we read NTU 1.15 =U A/C min Thus

We could also have calculated the LMTD:

LMTD= (150−40)−(90−70) ln(110/20) =52.79 K so fromQ=U A(LMTD), we obtain

500(52.79) =22.73 m 2 The answers diﬀer by 1%, which reﬂects graph reading inaccuracy.

When analyzing heat transfer in a heat exchanger, a uniform temperature in either fluid simplifies the process significantly In this scenario, no correction is required for the Logarithmic Mean Temperature Difference (LMTD) since only one fluid experiences a temperature change Consequently, the configuration of the heat exchanger becomes irrelevant, allowing it to be treated as a single fluid stream flowing through an isothermal pipe.

In cases where all heat exchangers are equivalent, the effectiveness equation for any configuration converges to a common expression as C max approaches infinity This scenario can occur if the volumetric heat capacity rate becomes infinite due to a high flow rate or specific heat, or if the flow is involved in absorbing or releasing latent heat The limiting effectiveness expression can also be derived from energy-balance considerations, but here, we achieve it by allowing C max to approach infinity in the relevant equations.

3 We make use of this notion in Section 7.4, when we analyze heat convection in pipes and tubes.

Eqn (3.22) deﬁnes the curve forC min /C max =0 in all six of the eﬀective- ness graphs in Fig.3.16 and Fig.3.17.

Heat exchanger design

The previous sections outlined effective methods for designing heat exchangers, particularly compact cross-flow models commonly used in transportation equipment However, larger shell-and-tube exchangers present challenges related to the overall heat transfer coefficient (U) These challenges include the variation of U throughout the exchanger and the difficulty in predicting convective heat transfer coefficients due to the complex flow patterns in a baffled shell.

In Part III, we will achieve significant advancements in analyzing predictions for various convective flows The determination of heat transfer coefficients in baffled shells remains analytically unsolvable, often relying on empirical correlations or extensive commercial software that incorporates experimental data Additionally, predicting heat transfer during boiling or condensing flows presents even greater challenges, prompting ongoing research focused on refining these empirical prediction methods.

In designing heat exchangers, it is essential to not only predict heat transfer but also to focus on minimizing pumping power and reducing fixed costs.

The pumping power calculation is rooted in fundamental principles of fluid mechanics, typically covered in introductory courses on the subject For each fluid stream passing through a heat exchanger, pumping power can be generally calculated using the formula: pumping power = ˙m * ΔP / (ρ * g * η), where ˙m represents mass flow rate, ΔP is pressure difference, ρ is fluid density, g is acceleration due to gravity, and η is pump efficiency.

In heat exchanger design, the mass flow rate (˙m), pressure drop (∆p), and fluid density (ρ) are crucial factors Calculating the pressure drop can be simple in a single-pass pipe-in-tube heat exchanger, but it becomes significantly more complex in configurations like shell-and-tube exchangers For instance, the pressure drop in a straight section of pipe can be determined using specific formulas tailored for these systems.

The equation 2 (3.24) represents the relationship between the length of the pipe (L), hydraulic diameter (D h), mean flow velocity (u av), and the Darcy-Weisbach friction factor (f), as illustrated in Fig 7.6.

Optimizing the design of an exchanger is not just a matter of making

To enhance heat exchange efficiency, it is beneficial to incorporate fins or roughening elements in an exchanger, as discussed in Chapter 4 (refer to Fig 4.6) While these modifications may lead to an increase in pressure drop, they can also lower the fixed costs of the exchanger by improving the overall heat transfer coefficient (U) and decreasing the necessary surface area Additionally, these enhancements can reduce the required coolant flow rate by increasing the system's effectiveness, thereby balancing the pressure drop increase as outlined in equation (3.23).

To navigate the complexities of the design process for a large shell-and-tube exchanger, it is essential to consider the various trade-offs between advantages and disadvantages, as outlined in Taborek's list of design considerations.

• Decide which ﬂuid should ﬂow on the shell side and which should

When determining the flow direction in tubes, the primary goal is to minimize pumping costs For instance, in applications where water is used to cool oil, the more viscous oil typically flows through the shell side Additionally, factors such as corrosion behavior, fouling, and the challenges associated with cleaning fouled tubes significantly influence this decision.

• Early in the process, the designer should assess the cost of the calculation in comparison with:

(a) The converging accuracy of computation.

(b) The investment in the exchanger.

To accurately size a heat exchanger, begin by making a rough estimate using U values from Table 2.2 and any relevant experience This initial estimation will guide subsequent trial-and-error calculations, assist in determining flow rates, and help anticipate temperature variations, ultimately minimizing the risk of errors later in the process.

When assessing heat exchangers for a specific application, it is essential to evaluate their heat transfer efficiency, pressure drop, and associated costs This evaluation is typically performed using advanced computer programs that are continually updated to incorporate the latest research findings, ensuring accurate and reliable results.

The computer runs suggested by this procedure are normally very complicated and might typically involve 200 successive redesigns, even when relatively eﬃcient procedures are used.

Most students studying heat transfer will primarily focus on designing smaller heat exchangers, typically ranging from 0.1 to 10 m² Effective heat transfer calculations can be performed using the methods outlined in this chapter For guidance on pressure drop calculations, valuable resources include the Heat Exchanger Design Handbook, Idelchik’s collection, the TEMA design book, and additional references provided at the end of this chapter.

When calculating heat transfer in a system, we begin with one fluid for heating and another for cooling, often knowing the flow heat capacity rates (C_c and C_h), specific temperatures, and the heat transfer amount The challenge lies in narrowing down the problem, typically starting with the specification of an exchanger configuration, which requires experience This chapter offers foundational insights, while references [3.5, 3.7, 3.9, 3.10, 3.11, 3.12] provide deeper knowledge Additionally, manufacturer catalogs serve as valuable resources for advanced information.

Once the exchanger configuration is established, the overall heat transfer coefficient (U) can be approximated, making the area the primary design variable The design process can then follow the guidelines outlined in Sections 3.2 or 3.3 Starting with a comprehensive specification of the inlet and outlet temperatures is advantageous for effective design.

ThenA can be calculated and the design completed Usually, a reevalu- ation ofU and some iteration of the calculation is needed.

When starting without complete knowledge of outlet temperatures, we typically rely on a trial-and-error approach to determine the optimal area This process may involve a more complex series of tests aimed at optimizing both pressure drop and costs through adjustments in configuration.

Problems 129 as well If the C’s are design variables, the U will change signiﬁcantly, because h’s are generally velocity-dependent and more iteration will be needed.

We conclude PartIof this book facing a variety of incomplete issues.

The well-posed problem

The heat diffusion equation, introduced in Section 2.1, serves as a foundation for understanding heat conduction problems To effectively tackle these issues, it is crucial to articulate them clearly, especially as we progress to more complex transient and multidimensional heat conduction scenarios that have not been addressed previously.

A well-posed heat conduction problem contains all necessary information to achieve a unique solution Such a problem is defined clearly and is solvable, ensuring that the parameters and conditions are adequately specified for accurate results.

∂t for 0< t < T (where T can → ∞), and for(x, y, z)belonging to

142 Analysis of heat conduction and some steady one-dimensional problems §4.1 some region,R, which might extend to inﬁnity 1

2 T =T i (x, y, z) at t=0 This is called aninitial condition, or i.c.

(a) Condition 1 above is not imposed att=0.

In steady-state heat transfer, only one internal condition (i.c.) is necessary, as represented by the equation ∇ ã(k∇T ) + q˙ = 0 Additionally, for periodic heat transfer scenarios where the heat flux or boundary conditions fluctuate periodically over time, the internal condition can be disregarded, provided that the initial transient behavior is not considered.

3 T must also satisfy twoboundary conditions, or b.c.’s, for each coordinate The b.c.’s are very often of three common types.

(a) Dirichlet conditions, or b.c.’s of theﬁrst kind:

T is speciﬁed on the boundary of R for t > 0 We saw such b.c.’s in Examples2.1,2.2, and2.5.

(b) Neumann conditions, or b.c.’s of thesecond kind:

The boundary condition for the temperature derivative normal to the boundary is defined for values of time greater than zero This condition occurs when the heat flux, represented as k(∂T/∂x), is specified at the boundary or when insulation is applied, resulting in ∂T/∂x being equal to zero.

The derivative of temperature in a direction normal to a boundary is proportional to the temperature at that boundary, a condition often encountered during convection processes.

When the body is positioned to the left of the boundary on the x-coordinate, the boundary condition is represented as ∂x bndry = h(T − T ∞ ) bndry This boundary condition was previously utilized in Step 4 of Example 2.6 and has been discussed in Section 1.3.

1 (x, y, z) might be any coordinates describing a position r : T (x, y, z, t) = T ( r , t).

2 Although we write ∂T /∂x here, we understand that this might be ∂T /∂z, ∂T /∂r, or any other derivative in a direction locally normal to the surface on which the b.c is speciﬁed.

The general solution

Figure 4.1 The transient cooling of a body as it might occur, subject to boundary conditions of the ﬁrst, second, and third kinds.

This list of b.c.’s is not complete, by any means, but it includes a great number of important cases.

Figure 4.1 illustrates the transient cooling of a body from a consistent initial temperature, influenced by the three boundary conditions discussed earlier It is important to note that the initial temperature distribution is independent of the boundary condition, as previously mentioned in section 2(a).

The eight-point procedure described in Section 2.2 for solving the heat diffusion equation was designed to ensure that the problem adheres to the necessary requirements and is well-posed.

Once the heat conduction problem has been posed properly, the first step in solving it is to find the general solution of the heat diffusion equation.

We have remarked that this is usually the easiest part of the problem.

We next consider some examples of general solutions.

144 Analysis of heat conduction and some steady one-dimensional problems §4.2

One-dimensional steady heat conduction

Problem 4.1 highlights the ease of obtaining general solutions for linear ordinary differential equations by requesting a comprehensive table of general solutions related to one-dimensional heat conduction problems To illustrate the process, we will derive some of these results, starting with the heat diffusion equation, incorporating constant parameters k and ˙q.

Cartesian coordinates: Steady conduction in the y-direction Equation (2.11) reduces as follows:

Therefore, d 2 T dy 2 = −q˙ k which we integrate twice to get

Cylindrical coordinates with a heat source: Tangential conduction.

In this discussion, we examine the heat flow that generates a ring when two points are maintained at varying temperatures We will express equation (2.11) in cylindrical coordinates, utilizing equation (2.13) for clarity.

The temperature distribution in the thin ring depicted in Fig 4.2 is influenced by boundary conditions, which include specified temperatures at two angular locations.

Figure 4.2 One-dimensional heat conduction in a ring.

IfT is spatially uniform, it can still vary with time In such cases

∂t and∂T /∂tbecomes an ordinary derivative Then, sinceα=k/ρc, dT dt = q˙ ρc (4.2)

This result is consistent with the lumped-capacity solution described in

When the Biot number is low and internal resistance is negligible, the convective heat removal from a body's boundary can be averaged over its volume, leading to the effective heat transfer equation: ˙ q effective = −h(T body −T ∞ )A volume W/m³ Consequently, the heat diffusion equation simplifies to dT/dt = −hA/(ρcV)(T−T ∞ ).

The general solution in this situation was given in eqn (1.21) [A particular solution was also written in eqn (1.22).]

Separation of variables: A general solution of multidimensional problems

In scenarios where only one spatial derivative is significant in a heat diffusion equation, we can focus on predicting the transient cooling within a slab based on its location In the absence of heat generation, the governing equation for heat diffusion simplifies, allowing for more straightforward analysis of temperature changes over time.

A common trick is to ask: “Can we ﬁnd a solution in the form of a product of functions of t and x: T = T(t)ã X(x)?” To ﬁnd the answer, we substitute this in eqn (4.5) and get

X T = 1 αT X (4.6) where each prime denotes one diﬀerentiation of a function with respect to its argument Thus T = dT/dt and X = d 2 X/dx 2 Rearranging eqn (4.6), we get

The intriguing outcome reveals that the left-hand side is solely dependent on x, while the right-hand side relies exclusively on t Consequently, we equate both sides to a common constant, denoted as -λ², rather than λ, for reasons that will become evident shortly.

It follows that the diﬀerential eqn (4.7a) can be resolved into two ordinary diﬀerential equations:

The general solution of both of these equations are well known and are among the ﬁrst ones dealt with in any study of diﬀerential equations. They are:

X(x)=Ax+B for λ=0 (4.9) §4.2 The general solution 147 and

The general solution of equation (4.5) can be expressed as a product, which can be verified by substituting back into equation (4.8) For λ=0, the relationship T(t)=C holds, where capital letters represent constants of integration.

The applicability of this result hinges on its compatibility with the boundary conditions (b.c.’s) and initial conditions (i.c.) In this context, we reformulated the function X(t) using sines and cosines, incorporating a negative sign before λ² This approach facilitates the fitting of the boundary conditions through Fourier series techniques While this book does not delve deeply into these general methods, a comprehensive Fourier series solution is provided for a specific problem in Section 5.3.

The method of separation of variables is an effective technique for finding general solutions to linear partial differential equations (PDEs), applicable to various types of linear PDEs For instance, it can be utilized in two-dimensional steady heat conduction scenarios where there are no heat sources present.

Y = −λ 2 whereλcan be an imaginary number Then

Figure 4.3 A two-dimensional slab maintained at a constant temperature on the sides and subjected to a sinusoidal variation of temperature on one face.

A long slab is uniformly cooled to 0 °C on both sides, while a blowtorch heats the top edge, creating a sinusoidal temperature distribution along the surface This setup allows for the analysis of the temperature distribution within the slab.

The general solution is defined by equation (4.13), requiring the identification of suitable boundary conditions (b.c.’s) to apply it effectively Specifically, the boundary condition on the top surface is expressed as T(x,0) = Asin(πx).

L on the sides : T (0 orL, y)=0 asy → ∞: T (x, y → ∞)=0 Substitute eqn (4.13) in the third b.c.:

The only way that this can be true for all x is if G = 0 Substitute eqn (4.13), withG=0, into the second b.c.:

(O+F )e − λy =0 §4.2 The general solution 149 soFalso equals 0 Substitute eqn (4.13) withG=F =0, into the ﬁrst b.c.:

It follows that A = E and λ = π /L Then eqn (4.13) becomes the particular solution that satisﬁes the b.c.’s:

The temperature at the top of the slab exhibits a sinusoidal variation that decreases exponentially with depth Specifically, at a depth of y = 2L below the surface, the temperature can be expressed as T = 0.0019A sin(πx/L) Although the temperature distribution along the x-direction remains sinusoidal, its amplitude at y = 0 is reduced to less than 1/500 of the original value.

Consider some important features of this and other solutions:

The boundary condition at y = 0 is particularly effective for this general solution, as fitting the equation to a general temperature distribution, T(x, y=0) = fn(x), presents challenges This issue is similar to one that Fourier addressed using his Fourier series method, which we will explore further in Chapter 5.

Dimensional analysis

Many universities offer heat transfer courses following an introduction to fluid mechanics, which typically incorporates dimensional analysis While the conventional method of indices is commonly used for this analysis, it can be cumbersome and occasionally misleading For a clearer understanding, refer to the comprehensive presentation in [4.1].

The method of functional replacement we present is significantly easier to use than traditional index methods, helping to safeguard against common mistakes.

Dimensional analysis plays a crucial role in understanding heat transfer, as illustrated in Example 2.6, which involves multiple variables Key components include the dependent variable, the temperature difference (T ∞ − T i), the primary independent variable, the radius (r), and five critical system parameters: r i, r o, h, k, and (T ∞ − T i).

By reorganizing the solution into dimensionless groups [eqn (2.24)], we reduced the total number of variables to only four:

 r r i , indep var. r o r i , Bi two system parameters

In our analysis, we refer to T i not as a variable but as a reference temperature that serves as the baseline for our calculations For instance, if T i were set at 0 °C, its subtraction from other temperatures would go unnoticed in our results.

This solution provided several benefits compared to the dimensional approach, notably allowing for the visualization of all possible solutions for a specific cylinder shape (r o /r i) within a single figure, as illustrated in Fig 2.13.

This study enabled us to analyze the simultaneous effects of parameters h, k, and r on the characteristics of the solution By integrating these parameters into a Biot number, we could determine the necessity of accounting for external convection even prior to solving the problem.

Nondimensionalization enables the simultaneous analysis of various heat conduction systems in cylinders, revealing that a large, highly conductive cylinder can exhibit behavior similar to that of a smaller cylinder with lower thermal conductivity.

Finally, we shall discover that, by nondimensionalizing a problembe- forewe solve it, we can often greatly simplify the process of solving it.

Our upcoming goal is to develop a strategy for nondimensionalizing problems prior to solving them or even formulating the necessary equations Central to this approach is the Buckingham Pi Theorem, which serves as a foundational principle in nondimensionalization.

The attention of scientiﬁc workers was apparently drawn very strongly toward the question of similarity at about the beginning of World War I.

In 1914, Buckingham introduced his renowned theorem in the Physical Review, further elaborating on the concept in the Transactions of the ASME the following year Around the same time, Lord Rayleigh also addressed the issue with notable clarity in 1915.

To grasp Buckingham’s theorem, it is essential to first address a key conceptual hurdle Once this hurdle is understood, the subsequent concepts will become significantly easier to comprehend.

Suppose thaty depends onr , x, zand so on: y=y(r , x, z, )

In a functional equation, any variable, such as x, can be multiplied or raised to a power with other variables without changing the equation's validity For example, y x can be expressed as y x x² r, x, xz, demonstrating that the relationship remains intact despite these modifications.

To see that this is true, consider an arbitrary equation: y =y(r , x, z)=r (sinx)e − z

This need only be rearranged to put it in terms of the desired modiﬁed variables andxitself(y/x, x 2 r , x, andxz): y x = x 2 r x 3 (sinx)exp −xz x

We can freely multiply or divide the powers of any variable without compromising the validity of any functional equation we formulate This fundamental principle underpins the significant example that follows.

In the heat exchanger problem illustrated in Fig 3.15, the primary unknown variable is one of the exit temperatures Even without a deep understanding of heat exchanger analysis, we can establish a functional equation grounded in our physical comprehension of the situation.

 (4.14) where the dimensions of each term are noted under the quotation.

To determine the number of dimensionless groups in equation (4.14), we can strategically choose one variable at a time to either divide or multiply into the other variables By selecting a variable with a specific dimension, we can manipulate the other variables with the same dimension to eliminate that dimension, thereby simplifying the equation and identifying the dimensionless groups.

We do this ﬁrst with the variable (T h in −T c in ), which has the dimension of K.

The equation's intriguing aspect lies in the remaining term, (T h in − T c in), which carries the units of K This term cannot exist independently in the equation, as achieving dimensional homogeneity necessitates an additional term in K for balance.

Therefore, we must remove it.

Now the equation has only two dimensions in it—W and m 2 Next, we multiplyU (T h in −T c in )byAto get rid of m 2 in the second-to-last term.

Accordingly, the termA(m 2 ) can no longer stay in the equation, and we have

Next, we divide the ﬁrst and third terms on the right by the second.

This leaves onlyC min (T h in −T c in ), with the dimensions of W That term must then be removed, and we are left with the completely dimensionless result:

Equation (4.15) has exactly the same functional form as eqn (3.21), which we obtained by direct analysis.

An illustration of dimensional analysis in a complex steady

pi-groups One wasξ=x/Land the other is a new one equal toΘ/Γ We call itΦ: Φ≡ T−T 1 qL˙ 2 /k =fn x L

And this is exactly the form of the analytical result, eqn (2.15).

Finally, we must deal with dimensions that convert into one another.

Newton’s Second Law of Motion establishes a relationship between kilograms (kg) and newtons (N), indicating they cannot be regarded as separate dimensions Similarly, joules (J) and newton-meters (N·m) both represent dimensions of energy; however, it is crucial to determine if a mechanism allows for their interchange If mechanical energy is distinct from thermal energy in a specific context, joules should not be equated with newton-meters.

Understanding dimensional analysis is crucial for various heat transfer problems, as demonstrated in the analyses of laminar convection in Section 6.4, natural convection in Section 8.3, film condensation in Section 8.5, and pool boiling burnout in Section 9.3 In these scenarios, heat transfer typically occurs without converting heat to work or vice versa, making it misleading to separate J into Nãm.

Additional examples of dimensional analysis appear throughout this book Dimensional analysis is, indeed, our court of ﬁrst resort in solving most of the new problems that we undertake.

4.4 An illustration of the use of dimensional analysis in a complex steady conduction problem

Heat conduction problems with convective boundary conditions can become complex quickly, making it essential to avoid errors Ensuring consistent dimensions throughout the solution process is crucial A highly effective strategy is to nondimensionalize the heat conduction equation prior to applying the boundary conditions, which also simplifies algebraic calculations This nondimensionalization should adhere to the principles of the pi-theorem We will demonstrate this approach using a detailed example.

Figure 4.5 Heat conduction through a heat-generating slab with asymmetric boundary conditions.

A slab shown in Fig.4.5has different temperatures and different heat transfer coefficients on either side and the heat is generated within it Calculate the temperature distribution in the slab.

Solution The diﬀerential equation is d 2 T dx 2 = −q˙ k and the general solution is

2k +C 1 x+C 2 (4.19) §4.4 An illustration of dimensional analysis in a complex steady conduction problem 161 with b.c.’s h 1 (T 1 −T ) x = 0 = −k dT dx x=0 , h 2 (T −T 2 ) x = L = −k dT dx x=L

There are eight variables involved in the problem: (T−T 2 ),(T 1 −T 2 ), x, L,k,h 1, h 2, and ˙q; and there are three dimensions: K, W, and m.

This results in 8−3 = 5 pi-groups For these we choose Π1 ≡Θ= T −T 2

2k(T 1 −T 2 ), whereΓ can be interpreted as a comparison of the heat generated in the slab to that which could ﬂow through it.

Under this nondimensionalization, eqn (4.19) becomes 5 Θ= −Γξ 2 +C 3 ξ+C 4 (4.21) and b.c.’s become

Bi1 (1−Θ ξ = 0 )= −Θ ξ = 0 , Bi2Θ ξ = 1 = −Θ ξ = 1 (4.22) where the primes denote diﬀerentiation with respect toξ Substitut- ing eqn (4.21) in eqn (4.22), we obtain

Substituting the ﬁrst of eqns (4.23) in the second we get

Rearranging dimensional equations into dimensionless form is a simple algebraic process If the outcomes presented are not clear, it may be helpful to sketch the calculations on paper for better understanding.

Achieving this complex result necessitated significant patience and precision, especially after simplifying the problem statement If the heat transfer coefficients are equal on both sides of the wall, then Bi 1 = Bi 2 ≡ Bi, leading to the simplification of equation (4.24) to Θ = 1 + Γ ξ - ξ² + 1/Bi.

1+2/Bi (4.25) which is a very great simpliﬁcation.

Equation (4.25) is plotted on the left-hand side of Fig.4.5for Bi equal to 0, 1, and∞ and for Γ equal to 0, 0.1, and 1 The following features should be noted:

• WhenΓ 0.1, the heat generation can be ignored.

• WhenΓ 1,Θ→Γ/Bi+Γ(ξ−ξ 2 ) This is a simple parabolic temperature distribution displaced upward an amount that depends on the relative external resistance, as reﬂected in the Biot number.

• If both Γ and 1/Bi become large,Θ →Γ/Bi This means that when internal resistance is low and the heat generation is great, the slab temperature is constant and quite high.

If T2 equals T1 in this scenario, Γ approaches infinity, necessitating a re-evaluation of the dimensional analysis The dimensional functional equation indicates that (T - T1) is a function of the variables x, L, k, h, and q̇ With six variables across three dimensions, three pi-groups emerge from this analysis.

T−T 1 ˙ qL/h =fn(ξ,Bi) where the dependent variable is like Φ [recall eqn (4.18)] multiplied by

Bi We can put eqn (4.25) in this form by multiplying both sides of it by h(T 1 −T 2 )/˙qδ The result is h(T −T 1 ) qL˙ = 1

The result is plotted on the right-hand side of Fig 4.5 The following features of the graph are of interest:

• Heat generation is the only “force” giving rise to temperature nonuni- formity Since it is symmetric, the graph is also symmetric.

Fin design

As the slab temperature of Bi 1 approaches a uniform value of T 1 + qL/2h, the complexity of the problem diminishes significantly In this scenario, employing a straightforward lumped-capacity heat balance would simplify the analysis, as it transitions away from a heat conduction issue.

When the Biot number (Bi) exceeds 100, the temperature distribution exhibits a pronounced parabolic shape, enhanced by an additional term In this scenario, the problem can effectively be addressed using first-kind boundary conditions, as the surface temperature remains closely aligned with the ambient temperature (T ∞).

Enhancing the convective heat removal from a surface can be significantly achieved by adding extensions that increase its area These extensions, known as fins, come in various designs and shapes For instance, commercial heat exchanger tubing can be effectively modified with different protrusions to optimize heat transfer efficiency.

Figure 4.7 illustrates a fascinating application of fins in heat exchanger design, as featured in a Scientific magazine article by Farlow, Thompson, and Rosner Their research presents compelling evidence regarding the unusual arrangements of these fins, highlighting their significance in enhancing heat transfer efficiency.

ﬁns on the back of the Stegosaurus were used to shed excess body heat after strenuous activity, which is consistent with recent suspicions that

The analysis of a straight fin extending from a wall reveals key characteristics of fin behavior, offering valuable insights applicable to various complex scenarios.

Analysis of a one-dimensional ﬁn

The equations presented in Figure 4.8 illustrate a one-dimensional fin extending from a wall, with the wall and the fin's roots maintained at a temperature $T_0$, which may be either higher or lower than the surrounding ambient temperature $T_\infty$ The fin's length is subjected to cooling or heating via a uniform heat transfer coefficient $h$ influenced by the ambient fluid It is important to note that while the assumption of a uniform heat transfer coefficient simplifies the analysis, it can lead to significant errors in certain scenarios, as discussed in Part III Additionally, the article provides eight examples of externally finned tubing to further illustrate these concepts.

The article discusses various types of commercial fins designed to enhance heat transfer efficiency It highlights typical circular fins of constant thickness, serrated circular fins, and dimpled spirally-wound circular fins, all aimed at improving convection Additionally, it mentions spirally-wound copper coils utilized both externally and internally, as well as bristle fins that are spirally wound and machined from base metal Finally, the article describes a spirally indented tube that further enhances convection and increases the surface area, showcasing an array of commercially available internally finned tubing.

Figure 4.6 Some of the many varieties of ﬁnned tubes.

The Stegosaurus, depicted in an etching by Daniel Rosner, may have possessed cooling fins that played a role in thermoregulation While the study of its physiology can provide insights, the accuracy of interpretations regarding its cooling mechanisms, whether through natural or forced convection, remains limited.

The tip may exchange heat with its surroundings via a heat transfer coefficient, hL, which typically differs from h The fin has a length of L, a uniform cross-sectional area of A, and a circumferential perimeter of P.

The characteristic dimension of the ﬁn in the transverse direction

(normal to thex-axis) is taken to beA/P Thus, for a circular cylindrical

ﬁn,A/P =π (radius) 2 /(2π radius)=(radius/2) We deﬁne a Biot number for conduction in the transverse direction, based on this dimension, and require that it be small:

Figure 4.8 The analysis of a one-dimensional ﬁn.

The condition indicates that the transverse variation of temperature (T) at any axial position (x) is significantly smaller than the difference between the surface temperature (T surface) and the ambient temperature (T ∞) Consequently, both the temperature T(x) and the heat flow can be effectively analyzed as one-dimensional.

An energy balance on the thin slice of the ﬁn shown in Fig.4.8gives

−kA dT dx x + δx +kA dT dx x +h(P δx)(T−T ∞ ) x =0 (4.28) but dT /dx| x + δx − dT /dx| x δx → d 2 T dx 2 = d 2 (T−T ∞ ) dx 2 (4.29) §4.5 Fin design 167 so d 2 (T−T ∞ ) dx 2 = hP kA(T−T ∞ ) (4.30)

The b.c.’s for this equation are

Alternatively, if the tip is insulated, or if we can guess that h L is small enough to be unimportant, the b.c.’s are

Before we solve this problem, it will pay to do a dimensional analysis of it The dimensional functional equation is

In our analysis, we represent kA, hP, and hLA as single variables to simplify the equation This approach is crucial as it eliminates any geometric complexities of the cross-section from consideration Consequently, the parameters P and A are only relevant in the context of their products with k, h, or hL.

In the absence of insulation, the analysis reveals that there are seven key variables related to W, K, and m, resulting in the formation of four distinct pi-groups.

 or if we rename the groups, Θ=fn(ξ, mL,Biaxial ) (4.33a) where we call 3 hP L 2 /kA≡mLbecause that terminology is common in the literature on ﬁns.

If the tip of the ﬁn is insulated,h L will not appear in eqn (4.32) There is one less variable but the same number of dimensions; hence, there will

The analysis of heat conduction reveals that there are only three significant pi-groups, with the Bi axial group, which relates to hL, being excluded Consequently, for the insulated fin, the temperature distribution can be expressed as Θ = fn(ξ, mL).

We put eqn (4.30) in these terms by multiplying it byL 2 /(T 0 −T ∞ ) The result is d 2 Θ dξ 2 =(mL) 2 Θ (4.34)

This equation is satisﬁed byΘ/i> ± (mL)ξ The sum of these two solutions forms the general solution of eqn (4.34): Θ=C 1 e mLξ +C 2 e − mLξ (4.35)

Temperature distribution in a one-dimensional ﬁn with the tip insulated The b.c.’s [eqn (4.31b)] can be written as Θ ξ = 0 =1 and dΘ dξ ξ = 1 =0 (4.36)

Substituting eqn (4.35) into both eqns (4.36), we get

To simplify the solution of equation (4.37) for C1 and C2, it's essential to revisit the key properties of hyperbolic functions, specifically the four fundamental functions defined as: sinh(x) ≡ e^x - e^(-x), which serves as the foundation for further calculations.

In fin design, the independent variable is denoted as x, and additional functions are established similarly to trigonometric functions The differential relations are formally articulated and bear resemblance to their trigonometric equivalents, such as the derivative of the hyperbolic sine function, where d/dx sinh(x) equals 1.

These are analogous to the familiar results, dsinx/dx = cosx and dcosx/dx= −sinx, but without the latter minus sign.

The solution of eqns (4.37) is then

2 coshmL (4.40) Therefore, eqn (4.35) becomes Θ= e −mL(1−ξ) +(2 coshmL)e −mLξ −e −mL(1+ξ)

2 coshmL which simpliﬁes to Θ= coshmL(1−ξ) coshmL (4.41) for a one-dimensional ﬁn with its tip insulated.

One of the most important design variables for a ﬁn is the rate at which it removes (or delivers) heat the wall To calculate this, we write

Fourier’s law for the heat ﬂow into the base of the ﬁn: 6

We multiply eqn (4.42) byL/kA(T 0 −T ∞ )and obtain, after substituting eqn (4.41) on the right-hand side,

QL kA(T 0 −T ∞ ) =mLsinhmL coshmL=mLtanhmL (4.43)

6 We could also integrate h(T − T ∞ ) over the outside area of the ﬁn to get Q The answer would be the same, but the calculation would be a little more complicated.

Figure 4.9 The temperature distribution, tip temperature, and heat ﬂux in a straight one-dimensional ﬁn with the tip insulated. which can be written

Introduction

James Watt did not invent the steam engine; instead, he significantly improved its efficiency by eliminating a wasteful heating and cooling process that consumed excessive energy By 1763, his innovations transformed the performance of steam engines, paving the way for advancements in industrial technology.

For over fifty years, Savery and Newcomen engines were utilized to pump water from Cornish mines and perform various tasks In that pivotal year, the young instrument maker James Watt was tasked with renovating the Newcomen engine model at the University of Glasgow, where it served as a demonstration for natural philosophy students Watt not only refurbished the machine but also identified and ultimately resolved its significant shortcomings.

Newcomen’s engine operated with a cold cylinder, causing steam to condense wastefully on the walls until they warmed up Once the cylinder was filled with steam, the steam valve closed, and jets of water were introduced to cool the cylinder, creating a vacuum that pulled the piston back Watt attempted to improve efficiency by insulating the cylinder to prevent initial steam condensation, but this approach inadvertently reduced the vacuum and diminished the engine's power during the working stroke.

194 Transient and multidimensional heat conduction §5.2

Then he realized that, if he led the steam outside to aseparate condenser, the cylinder could stay hot while the vacuum was created.

The separate condenser was the main issue in Watt’s ﬁrst patent

In 1769, James Watt's invention significantly improved steam engine thermal efficiency from 1.1% to 2.2% By his death in 1819, efficiencies had reached 5.7%, revolutionizing the world and powering the Industrial Revolution Today, the steam power cycles studied in thermodynamics courses are modeled as steady flow processes, reflecting the lasting impact of Watt's contributions.

Transient heat transfer is a critical factor in the design of food storage units, a lesson highlighted by the historical example of Newcomen's engine Modern designers recognize that while maintaining steady temperatures requires minimal energy, the significant costs arise from the energy needed to initially cool the food and the losses incurred when users frequently access the storage Understanding these transient processes is essential for optimizing energy efficiency in food storage systems.

We therefore turn our attention, ﬁrst, to an analysis of unsteady heat transfer, beginning with a more detailed consideration of the lumped- capacity system that we looked at in Section1.3.

Lumped-capacity solutions

We begin by looking brieﬂy at the dimensional analysis of transient conduction in general and of lumped-capacity systems in particular.

Dimensional analysis of transient heat conduction

We ﬁrst consider a fairly representative problem of one-dimensional transient heat conduction:

The solution of this problem must take the form of the following dimensional functional equation:

There are eight variables in four dimensions (K, s, m, W), so we look for

8−4=4 pi-groups We anticipate, from Section4.3, that they will include Θ≡ (T−T 1 ) (T i −T 1 ), ξ≡ x

L, and Bi≡ hL k , and we write Θ=fn(ξ,Bi,Π4 ) (5.1)

One possible candidate for Π4, which is independent of the other three, is Π4 ≡Fo=αt/L 2 (5.2) where Fo is theFourier number Another candidate that we use later is Π4 ≡ζ= x

If the problem involved only b.c.’s of the ﬁrst kind, the heat transfer coeﬃcient,h—and hence the Biot number—would go out of the problem.

Then the dimensionless function eqn (5.1) is Θ=fn(ξ,Fo) (5.4)

By the same token, if the b.c.’s had introduced diﬀerent values of h at x=0 andx=L,two Biot numbers would appear in the solution.

The lumped-capacity problem is intriguing in the context of dimensional analysis, as it simplifies the situation by excluding internal conduction effects, leading to the use of ρc instead of α This approach allows for the combination of ρ and c, since they only manifest as a product Additionally, the analysis incorporates the volume-to-external-area ratio, enhancing the understanding of the problem.

V /A, as a characteristic length Thus, for the transient lumped-capacity problem, the dimensional equation is

Figure 5.1 A simple resistance-capacitance circuit.

With six variables in the dimensions J, K, m, and s, only two pi-groups will appear in the dimensionless function equation. Θ=fn hAt ρcV

This is exactly the form of the simple lumped-capacity solution, eqn (1.22). Notice, too, that the groupt/T can be viewed as t

Electrical and mechanical analogies to the lumped-thermal-capacity problem

The concept of capacitance, borrowed from electrical circuit theory, is applied to heat transfer analysis In this context, a basic resistance-capacitance circuit is illustrated in Fig 5.1 Initially, the capacitor holds a voltage of E₀ Upon the sudden opening of the switch, the capacitor discharges through the resistor, causing the voltage to decrease according to the relationship dE/dt + E.

The solution of eqn (5.8) with the i.c.E(t=0)=E o is

E=E o e − t/RC (5.9) and the current can be computed from Ohm’s law, onceE(t)is known.

In heat conduction problems, thermal capacitance, represented as ρcV, is typically distributed throughout a material However, when the Biot number is low, the temperature T(t) becomes uniform within the body, allowing us to simplify the capacitance into a single circuit element The thermal resistance can be expressed as 1/hA, with the temperature difference (T − T∞) resembling E(t) Consequently, the thermal response can be described using an equation analogous to eqn (5.9) and referenced in eqn (1.22).

Notice that the electrical time constant, analogous toρcV /hA, isRC.

Now consider a slightly more complex system Figure 5.2 shows a spring-mass-damper system The well-known response equation (actually, a force balance) for this system is m

What is the mass analogous to? d 2 x dt 2 +c the damping coeﬃcient is analogous to R or to ρcV dx dt +k where k is analogous to 1/C or to hA x=F (t) (5.11)

A term analogous to mass would arise from electrical inductance, but we

A spring-mass-damper system, as illustrated in Figure 5.2, incorporates a forcing function that influences its behavior The mass in the system plays a crucial role by propelling it past its final equilibrium point Consequently, in an underdamped mechanical system, a specific response can be observed, as depicted in Figure 5.3, when the velocity at x = 0 is defined without any external forcing function applied.

Electrical inductance provides a similar eﬀect But the Second Law of

Thermodynamics does not permit temperatures to overshoot their equilibrium values spontaneously There are no physical elements analogous to mass or inductance in thermal systems.

Figure 5.3 Response of an unforced spring-mass-damper system with an initial velocity.

In exploring mechanical systems, we examine the thermal analogy of the forcing function, F, within a massless spring-damper setup This system typically features a time-dependent forcing function, prompting us to consider the characteristics and behavior of a corresponding thermal forcing function.

Lumped-capacity solution with a variable ambient temperature

To explore the thermal behavior of an object initially at temperature $ T_i $ and with a Biot number greater than 1, we place it into a cooling bath where the temperature increases over time, represented as $ T_{\infty}(t) = T_i + bt $ Consequently, the differential equation governing the temperature change simplifies to $ \frac{d(T - T_i)}{dt} = -(T - T_{\infty}) $.

T where we have arbitrarily subtractedT i under the diﬀerential Then d(T−T i ) dt +T−T i

To solve equation (5.12), it is essential to remember that the general solution of a linear ordinary differential equation with constant coefficients consists of the sum of a particular integral of the complete equation and the general solution of the homogeneous equation The general solution of the homogeneous equation is given by T - Ti = (constant)exp(-t/T) A particular integral can often be derived by making educated guesses for solutions and substituting them into the complete equation.

T−T i =bt−bT §5.2 Lumped-capacity solutions 199 satisﬁes eqn (5.12) Thus, the general solution of eqn (5.12) is

The solution for arbitrary variations ofT ∞ (t)is given in Problem5.52

The flow rates of hot and cold water are controlled within a mixing chamber, where we monitor the water temperature using a thermometer with a specific time constant, T Initially, the system contains cold water at temperature T_i As hot water is introduced, the outflow temperature increases linearly over time, following the equation T_exit_flow = T_i + bt This setup allows us to observe how the thermometer reflects the temperature changes in the mixing chamber.

Solution The initial condition in eqn (5.13), which describes this process, isT−T i =0 att =0 Substituting eqn (5.13) in the i.c., we get

0=C 1 −bT so C 1 =bT and the response equation is

Figure 5.4 illustrates the results, highlighting two distinct phases in the thermometer reading: a transient phase, represented by bTe − t/T, which diminishes over several time constants, and a steady phase, indicated by Ti + b(t−T), that continues thereafter Once the steady response is reached, the thermometer exhibits a temperature lag of bT in relation to the bath This constant error can be minimized by decreasing either the temperature T or the rate of temperature increase, b.

Second-order lumped-capacity systems

In this article, we explore the dynamics of two lumped-thermal-capacity systems connected in series, as illustrated in Fig 5.5 Heat transfer occurs through two slabs, which are separated by an interfacial resistance denoted as $ h - c_1 $ It is essential to ensure that the parameters $ h c L_1 / k_1 $, $ h c L_2 / k_2 $, and $ h L_2 / k_2 $ are all significantly less than one to maintain efficient thermal conduction between the slabs.

In analyzing the response of a thermometer to a linearly increasing ambient temperature, the thermal capacitance of each slab can be effectively lumped when the value is less than unity The temperature response of the slabs is governed by the following differential equations: for slab 1, the equation is $-(\rho c V)_1 \frac{dT_1}{dt} = h_c A (T_1 - T_2)$, and for slab 2, it is $-(\rho c V)_2 \frac{dT_2}{dt} = hA (T_2 - T_\infty) - h_c A (T_1 - T_2)$ The initial conditions for the temperatures $T_1$ and $T_2$ are also established to complete the analysis.

We next identify two time constants for this problem: 1

1 Notice that we could also have used (ρcV ) 2 /h c A for T 2 since both h c and h act on slab 2 The choice is arbitrary. §5.2 Lumped-capacity solutions 201

Figure 5.5 Two slabs conducting in series through an interfacial resistance. which we substitute in eqn (5.16) to get

+ h c hT 1 dT 1 dt =T 1 T 2 d 2 T 1 dt 2 −T 2 dT 1 dt or d 2 T 1 dt 2 +

=0 (5.19a) if we callT 1 −T ∞ ≡θ, then eqn (5.19a) can be written as d 2 θ dt 2 +bdθ dt +cθ=0 (5.19b)

Thus we have reduced the pair of ﬁrst-order equations, eqn (5.15) and eqn (5.16), to a single second-order equation, eqn (5.19b).

The general solution of eqn (5.19b) is obtained by guessing a solution of the formθ=C 1 e Dt Substitution of this guess into eqn (5.19b) gives

D 2 +bD+c =0 (5.20) from which we ﬁnd that D = −(b/2)±3

(b/2) 2 −c This gives us two values ofD, from which we can get two exponential solutions By adding

202 Transient and multidimensional heat conduction §5.2 them together, we form a general solution: θ=C 1 exp

To solve for the two constants we ﬁrst substitute eqn (5.21) in the ﬁrst of i.c.’s (5.17) and get

T i −T ∞ =θ i =C 1 +C 2 (5.22) The second i.c can be put into terms ofT 1 with the help of eqn (5.15):

We substitute eqn (5.21) in this and obtain

So we obtain at last:

The result is complex, particularly because it involves three algebraic terms, as noted in equation (5.19a) However, despite its complexity, the concept remains straightforward and easy to grasp.

A system involving three capacitances in series would similarly yield a third-order equation of correspondingly higher complexity, and so forth.

Transient conduction in a one-dimensional slab

Figure 5.6 The transient cooling of a slab;ξ =(x/L)+1.

5.3 Transient conduction in a one-dimensional slab

In scenarios where internal resistance significantly affects heat flow, the lumped capacitance assumption becomes inadequate To analyze temperature variations in a one-dimensional body over time and position, we need to solve the heat diffusion equation for T(x, t) We will begin by tackling this complex task in its simplest form and subsequently examine the outcomes of these calculations in various contexts.

A simple slab, depicted in Fig 5.6, begins at an initial temperature $ T_i $ When the surface temperature of the slab is abruptly altered to $ T_i $, we aim to determine the interior temperature profile over time.

The heat conduction equation is

In fully dimensionless form, eqn (5.24) and eqn (5.25) are

In the study of transient and multidimensional heat conduction, we nondimensionalize the problem using the equations Θ ≡ (T - T1) / (Ti - T1) and Fo ≡ αt / L², leading to boundary conditions Θ(0, Fo) = Θ(2, Fo) = 0 and Θ(ξ, 0) = 1 For ease of solving the equation, we redefine ξ as (x/L) + 1 instead of simply x/L.

The general solution of eqn (5.26) may be found using the separation of variables technique described in Sect.4.2, leading to the dimensionless form of eqn (4.11): Θ=e − ˆ λ 2 Fo !

Direct nondimensionalization of equation (4.11) reveals that ˆλ is equivalent to λL, as λ has units of (length) −1 Consequently, the solution introduces a fourth dimensionless group, ˆλ, which requires clarification The parameter λ, introduced during the separation-of-variables process, is referred to as an eigenvalue In this context, ˆλ=λL will ultimately represent a sequence of numbers that remain independent of the system parameters.

Substituting the general solution, eqn (5.28), in the ﬁrst b.c gives

0=e − ˆ λ 2 Fo (0+E) so E=0 and substituting it in the second yields

In the second case, we are presented with two choices The ﬁrst,

When G equals 0, it results in Θ being consistently 0 across all scenarios, leading to what mathematicians refer to as a trivial solution, which fails to satisfy the initial condition Conversely, selecting ˆλ n as nπ /2 produces a series of solutions, represented by the equation Θ = G n e − n 2 π 2 Fo/4 sin nπ.

The term "eigenvalue" is a unique blend of the German word "eigenwert" and its English equivalent, "characteristic value." In the context of transient conduction in a one-dimensional slab, G_n represents the constant relevant to the nth solution of this phenomenon.

Despite the challenge that none of the equations in (5.29) satisfy the initial condition Θ(ξ,0)=1, we can address this issue by recalling that the sum of multiple solutions to a linear differential equation remains a valid solution.

In equation (5.30), we omit n=0 as it contributes nothing to the series Ultimately, our focus shifts to selecting the appropriate G_n values to ensure that equation (5.30) aligns with the initial condition, Θ(ξ,0)=.

The problem of picking the values ofG n that will make this equation true is called “making a Fourier series expansion” of the functionf (ξ)=

In this article, we focus on a specific approach to constructing Fourier series expansions rather than exploring general strategies We will demonstrate the process through a mathematical trick tailored to the particular problem we are addressing.

We multiply eqn (5.31) by sin(mπ /2), wherem may or may not equal n, and we integrate the result betweenξ=0 and 2.

The legitimacy of interchanging summation and integration is acknowledged, although this article does not provide proof for it Utilizing a table of integrals, we derive the necessary results.

Thus, when we complete the integration of eqn (5.32), we get

3 What is normally required is that the series in eqn (5.31) be uniformly convergent.

Substituting this result into eqn (5.30), we ﬁnally obtain the solution to the problem: Θ(ξ,Fo)= 4 π

Equation (5.33) admits a very nice simpliﬁcation for large time (or at large Fo) Suppose that we wish to evaluateΘat the outer center of the slab—atx=0 orξ=1 Then Θ(0,Fo)= 4 π ×

For values of Fo greater than 0.1, the solution can be simplified by using only the first term in the series, except near the boundaries This topic is further explored in Section 5.5 Before delving into this, we will examine the effects of applying boundary conditions of the third kind to the previous problem.

Suppose that the walls of the slab had been cooled by symmetrical convection such that the b.c.’s were h(T ∞ −T ) x =− L = −k∂T

∂x x = L or in dimensionless form, usingΘ≡(T−T ∞ )/(T i −T ∞ )andξ=(x/L)+1,

∂ξ ξ = 1 =0 §5.3 Transient conduction in a one-dimensional slab 207

Table 5.1 Terms of series solutions for slabs, cylinders, and spheres J 0 andJ 1 are Bessel functions of the ﬁrst kind.

Slab 2 sin ˆλ n ˆλ n +sin ˆλ n cos ˆλ n cos λˆ n x L cot ˆλ n = λˆ n

Sphere 2sin ˆλ n −ˆλ n cos ˆλ n λˆ n −sin ˆλ n cos ˆλ n r o λˆ n r sin ˆλ n r r o λˆ n cot ˆλ n =1−Bi r o

The solution is somewhat harder to ﬁnd than eqn (5.33) was, but the result is 4 Θ=

−ˆλ 2 n Fo 2 sin ˆλ n cos[ˆλ n (ξ−1)] λˆ n +sin ˆλ n cos ˆλ n

(5.34) where the values of ˆλ n are given as a function ofnand Bi =hL/kby the transcendental equation cot ˆλ n = ˆλ n

The positive roots of the equation, denoted as ˆλ n = λˆ1, ˆλ 2, ˆλ 3, etc., are influenced by the parameter Bi Consequently, the relationship can be expressed as Θ = fn(ξ, Fo, Bi) While this outcome is more complex than that for boundary conditions of the first kind, it simplifies to a single term when Fo is set to 0.2.

Similar series solutions can be constructed for cylinders and spheres that are convectively cooled at their outer surface,r =r o The solutions for slab, cylinders, and spheres all have the form Θ= T −T ∞

−ˆλ 2 n Fo f n (5.36) where the coeﬃcients A n , the functions f n , and the equations for the dimensionless eigenvalues ˆλ n are given in Table5.1.

4 See, for example, [5.1, §2.3.4] or [5.2, §3.4.3] for details of this calculation.

Temperature-response charts

Figure 5.7 illustrates the results of equation (5.34) for values of Fo ranging from 0 to 1.5 across six x-planes within the slab It is important to note that the x-coordinate extends from zero at the center to L at the boundary, while the variable ξ varies from 0 to 2, as indicated in the previous solution.

In the analysis of the graph, it is observed that, except for points where 1/Bi < 0.25 along the outer boundary, all curves appear as straight lines when Fo is set to 0.2 This linearity in the semilogarithmic coordinates highlights the significance of the lead term in equation (5.34), which remains the only relevant term By applying the logarithm to the simplified version of equation (5.34), we derive the expression lnΘ = ln.

2 sin ˆλ 1 cos[ˆλ 1 (ξ−1)] λˆ 1 +sin ˆλ 1 cos ˆλ 1 Θ -intercept at Fo = 0 of the straight portion of the curve

− ˆλ 2 1 Fo slope of the straight portion of the curve

If Fo is greater than 1.5, the following options are then available to us for solving the problem:

• Extrapolate the given curves using a straightedge.

• EvaluateΘusing the ﬁrst term of eqn (5.34), as discussed in Sect.5.5.

• If Bi is small, use a lumped-capacity result.

Figures 5.8 and 5.9 present graphs for cylinders and spheres that reflect the same principles discussed in relation to Figure 5.7 While these graphs are derived from different solutions, the numerical values differ slightly These charts, often referred to as Heisler charts, originate from [5.3, Chap 5] and are associated with a series of related charts published later by Heisler [5.4].

Another useful kind of chart derivable from eqn (5.34) is one that gives heat removal from a body up to a time of interest: t

Figure5.8Thetransienttemperaturedistributioninalongcylinderofradiusr o atsixpositions: r/r o =0isthecenterline;r/r o =1istheoutsideboundary.

Figure5.9Thetransienttemperaturedistributioninasphereofradiusr o atsixpositions:r/r o =0 isthecenter;r/r o =1istheoutsideboundary.

Dividing this by the total energy of the body aboveT ∞ , we get a quantity,Φ, which approaches unity ast→ ∞and the energy is all transferred to the surroundings: Φ≡ t

∂ξ surface dFo (5.37) where the volume, V = AL Substituting the appropriate temperature distribution [e.g., eqn (5.34) for a slab] in eqn (5.37), we obtainΦ(Fo,Bi) in the form of an inﬁnite series Φ(Fo, Bi)=1−

The coefficients $ D_n $ vary as functions of $ \hat{\lambda}_n $ and the Biot number (Bi) for different geometries, including slabs, cylinders, and spheres; for instance, for a slab, $ D_n = A_n \sin(\hat{\lambda}_n \hat{\lambda}) $ These functions enable the creation of a universal plot of $ \Phi(F_o, Bi) $, as illustrated in Figure 5.10.

The quantity Φ is closely linked to the mean temperature T(t) of a body at any given time Specifically, the energy lost as heat by time t influences the difference between the initial temperature and the mean temperature at that moment.

Thus, if we deﬁneΘas follows, we ﬁnd the relationship ofT (t)toΦ Θ≡ T (t)−T ∞

A dozen approximately spherical apples, 10 cm in diameter are taken from a 30 ◦ C environment and laid out on a rack in a refrigerator at

After one hour, the centers of the apples will reach a temperature of approximately 10 °C To achieve this temperature, it will take a specific duration influenced by the heat transfer properties, which is around 6 W/m²K due to natural convection The refrigerator will need to remove a calculated amount of heat to maintain the centers at 10 °C.

Figure 5.10 The heat removal from suddenly-cooled bodies as a function ofhand time 213

Furthermore, Bi − 1 = (hr o /k) − 1 = [6(0.05)/0.603] − 1 = 2.01 There- fore, we read from Fig.5.9in the upper left-hand corner: Θ=0.85 After 1 hr:

To ﬁnd the time required to bring the center to 10 ◦ C, we ﬁrst calculate Θ= 10−5

30−5 =0.2 and Bi − 1 is still 2.01 Then from Fig.5.9we read

Finally, we look upΦat Bi=1/2.01 and Fo=1.29 in Fig.5.10, for spheres: Φ=0.80= t

(25)(0.80)=43,668 J/apple Therefore, for the 12 apples, total energy removal=12(43.67)=524 kJ §5.4 Temperature-response charts 215

The temperature-response charts presented in Figures 5.7 to 5.10 are invaluable tools that can be adapted for various physical scenarios Numerous additional charts exist for different situations, many of which have been cataloged by Schneider For those dealing with complex heat conduction calculations, it is advisable to consult existing literature A highly recommended starting point is the comprehensive treatise on heat conduction by Carslaw and Jaeger.

A 1 mm diameter Nichrome wire, composed of 20% nickel and 80% chromium, serves dual purposes as both an electric resistance heater and a resistance thermometer in liquid flow Laboratory personnel are measuring the boiling heat transfer coefficient (h) by applying alternating current and comparing the average temperature of the heater (T av) with the liquid temperature (T ∞) They achieve a remarkable heat transfer coefficient of 30,000 W/m² K at a wire temperature of 100 °C However, a colleague raises the possibility that this high value may be due to rapid oscillations in surface temperature caused by the alternating current.

Heat generation is proportional to the product of voltage and current, represented as sin²(ωt), where ω denotes the current frequency in radians per second If the boiling action efficiently dissipates heat relative to the wire's heat capacity, the surface temperature can fluctuate significantly This transient conduction issue was initially addressed by Jeglic in 1962 and subsequently reformulated by Switzer and Lienhard in 1964, who provided response curves for analysis.

T av −T ∞ =fn(Bi, ψ) (5.41) where the left-hand side is the dimensionless range of the temperature oscillation, andψ = ωδ 2 /α, where δ is a characteristic length

[see Problem5.56] Because this problem is common and the solution is not widely available, we include the curves for ﬂat plates and cylinders in Fig.5.11and Fig.5.12respectively.

0.00000343 =27.5 and from the chart for cylinders, Fig.5.12, we ﬁnd that

A temperature ﬂuctuation of only 4% is probably not serious It therefore appears that the experiment was valid.

One-term solutions

As we have noted previously, when the Fourier number is greater than 0.2 or so, the series solutions from eqn (5.36) may be approximated using only their ﬁrst term: Θ≈A 1 ãf 1 ãexp

Likewise, the fractional heat loss, Φ, or the mean temperature Θ from eqn (5.40), can be approximated using just the ﬁrst term of eqn (5.38): Θ=1−Φ≈D 1exp

Table 5.2 presents the values of ˆλ 1, A 1, and D 1 for slabs, cylinders, and spheres based on the Biot number The error in the one-term solution for Θ is below 0.1% for spheres when the Fourier number (Fo) is greater than or equal to 0.28, as well as for slabs.

Fo≥0.43 These errors are largest for Biot numbers near unity If high accuracy is not required, these one-term approximations may generally be used whenever Fo≥0.2

Table 5.2 One-term coeﬃcients for convective cooling [5.1].

Transient heat conduction to a semi-inﬁnite region

In Bronowksi’s classic television series, The Ascent of Man, a captivating reenactment showcases the ancient Japanese technique of forging Samurai swords This intricate process involves repeatedly heating, folding, beating, and shaping the metal to achieve a blade known for its exceptional toughness and flexibility Once the blade reaches its final form, a tapered clay sheath is applied and baked onto its surface The red-hot blade is then rapidly quenched, resulting in a case-hardened structure that is hardest at the cutting edge and gradually less hard toward the blade's back.

Figure 5.13 The ceremonial case-hardening of a Samurai sword. §5.6 Transient heat conduction to a semi-inﬁnite region 221

Figure 5.14 The initial cooling of a thin sword blade Prior tot=t 4, the blade might as well be inﬁnitely thick insofar as cooling is concerned.

The blade features a tough and ductile core that resists breakage, complemented by a fine hard outer shell that can be honed to a sharp edge The clay sheath provides a thick cross section that effectively shields the blade from the rapid cooling effects of the quench The success of this process is contingent upon the cooling not penetrating the clay deeply and quickly.

Now we wish to ask: “How can we say whether or not the inﬂuence of a heating or cooling process is restricted to the surface of a body?”

To understand the conditions under which we can perceive the depth of a body as infinite in relation to the thickness of the region affected by heat transfer, we must consider specific parameters that influence this phenomenon.

In the absence of a clay retardant, the cooling process within the blade is significantly influenced by the sudden change in temperature at its boundary This phenomenon can be observed in any finite body experiencing a similar temperature shift, leading to a distinct temperature distribution throughout the material.

In Fig 5.14, the analysis is conducted over four sequential instances, revealing that only the fourth curve, corresponding to t = t4, shows a significant impact from the opposing wall Prior to this point, the wall's influence appears negligible, suggesting it could be considered to have infinite depth.

Since any body subjected to a sudden change of temperature is in-

ﬁnitely large in comparison with the initial region of temperature change, we must learn how to treat heat transfer in this period.

Solution aided by dimensional analysis

Calculating temperature distribution in a semi-infinite region presents challenges due to the ability to impose boundary conditions only at the exposed boundary However, dimensional analysis offers an effective solution to navigate this difficulty.

When the one boundary of a semi-inﬁnite region, initially atT = T i , is suddenly cooled (or heated) to a new temperature,T ∞ , as in Fig.5.14, the dimensional function equation is

T−T ∞ =fn[t, x, α, (T i −T ∞ )] where there isno characteristic length or time Since there are ﬁve variables in ◦ C, s, and m, we should look for two dimensional groups.

This exercise in dimensional analysis reveals that position and time can be combined into a single independent variable As a result, the heat conduction equation, along with its boundary conditions, can be simplified from a partial differential equation to an ordinary differential equation in the form of ζ = x√(αt) This transformation streamlines the analysis and solution of the equation.

∂t as follows, where we callT i −T ∞ ≡∆T:

Substituting the ﬁrst and last of these derivatives in the heat conduction equation, we get d 2 Θ dζ 2 = −ζ

Notice that we changed from partial to total derivative notation, since Θnow depends solely onζ The i.c for eqn (5.45) is

T (t=0)=T i or Θ(ζ→ ∞)=1 (5.46) §5.6 Transient heat conduction to a semi-inﬁnite region 223 and the one known b.c is

If we calldΘ/dζ ≡χ, then eqn (5.45) becomes the ﬁrst-order equation dχ dζ = −ζ

2 χ which can be integrated once to get χ≡ dΘ dζ =C 1 e −ζ 2 /4 (5.48) and we integrate this a second time to get Θ=C 1 ζ

The b.c is now satisﬁed, and we need only substitute eqn (5.49) in the i.c., eqn (5.46), to solve forC 1:

The deﬁnite integral is given by integral tables as√ π, so

Thus the solution to the problem of conduction in a semi-inﬁnite region, subject to a b.c of the ﬁrst kind is Θ= 1

The second integral in equation (5.50), derived through a change of variables, is known as the error function (erf) This term is linked to statistical issues associated with the Gaussian distribution, which models random errors Additionally, Table 5.3 provides values for both the error function and its complementary counterpart, defined as erfc(x) ≡ 1 − erf(x).

(5.50) is also plotted in Fig.5.15.

Table 5.3 Error function and complementary error function. ζ

In Figure 5.15, the early-time curves depicted in Figure 5.14 have converged into a single curve through a similarity transformation, expressed as ζ/2 = x/2√(αt) This transformation effectively illustrates the data consolidation, as confirmed by the accompanying table.

In other words, the local value of(T−T ∞ )is more than 99% of(T i −T ∞ ) for positions in the slab beyond farther from the surface than δ 99 =

For what maximum time can a samurai sword be analyzed as a semi- inﬁnite region after it is quenched, if it has no clay coating andh external

Solution First, we must guess the half-thickness of the sword (say,

3 mm) and its material (probably wrought iron with an average α

5 The transformation is based upon the “similarity” of spatial an temporal changes in this problem. §5.6 Transient heat conduction to a semi-inﬁnite region 225

Figure 5.15 Temperature distribution in a semi-infinite region. around 1.5×10 − 5 m 2 /s) The sword will be semi-infinite until δ 99 equals the half-thickness Inverting eqn (5.51), we find t δ 2 99

The quenching process affects the centerline of the sword in just 1/20 of a second Due to the thermal diffusivity of clay being approximately 30 times lower than that of steel, the coated steel's quench time must exceed 1 second before any temperature change in the steel occurs, assuming the clay and sword thicknesses are similar.

Equation (5.51) provides an interesting foretaste of the notion of a

The fluid boundary layer is a crucial concept in fluid dynamics, as illustrated in Figures 1.9 and 1.10, where the free stream flow around an object becomes disturbed in a thick layer due to fluid adherence This boundary layer's thickness increases downstream, with the thickness over a flat plate approximated by 4.92√νt, where t represents the time taken for the fluid to travel from the plate's leading edge to a specific point This relationship is analogous to equation (5.51), substituting thermal diffusivity α with kinematic viscosity ν, resulting in a slightly larger constant The resulting velocity profile is depicted in Figure 5.15.

If we repeated the problem with a boundary condition of the third kind, we would expect to getΘ=Θ(Bi, ζ), except that there is no length,

L, upon which to build a Biot number Therefore, we must replaceLwith

√αt, which has the dimension of length, so Θ=Θ ζ,h√ αt k

The term β ≡ h√ αt kis like the product: Bi√

Fo The solution of this problem (see, e.g., [5.6], §2.7) can be conveniently written in terms of the complementary error function, erfc(x)≡1−erf(x): Θ=erfζ

This result is plotted in Fig.5.16.

Most of us have passed our ﬁnger through an 800 ◦ C candle ﬂame and know that if we limit exposure to about 1/4 s we will not be burned. Why not?

Short exposure to flame causes only superficial heating, leading us to treat the finger as a semi-infinite region for calculations The burn threshold for human skin is approximately 65 °C, which is why water at 60 °C (140 °F) is deemed scalding To determine the time required for a finger's surface temperature to increase from the body temperature of 37 °C to the burn threshold, we will assume a heat transfer coefficient of 100 W/m²K Additionally, we will consider the thermal conductivity of human flesh to be equivalent to that of water, while using the thermal diffusivity value for beef.

37−800 =0.963 βζ= hx k =0 sincex=0 at the surface β 2 = h 2 αt k 2 = 100 2 (0.135×10 − 6 )t

The situation is quite far into the corner of Fig 5.16 We readβ 2

0.001, which corresponds with t 0.3 s For greater accuracy, we must go to eqn (5.53):

Figure 5.16 The cooling of a semi-inﬁnite region by an environment atT ∞ , through a heat transfer coeﬃcient,h.

By trial and error, we gett0.33 s In fact, it can be shown that Θ(ζ =0, β) 2

√π (1−β) forβ1 which can be solved directly for β = (1−0.963)√ π /2 = 0.03279, leading to the same answer.

Thus, it would require about 1/3 s to bring the skin to the burn point.

When you first immerse your hand in the subfreezing air of the freezer compartment, you experience a sharp, intense cold that is markedly different from the sensation of placing your finger in a still mixture of ice cubes and water, where the cold feels more uniform and less aggressive In contrast, swirling your finger in the ice-water mixture creates a dynamic sensation, as the movement enhances the cooling effect and increases contact with the cold surfaces The key variable that changes in these scenarios is the method of heat transfer: the still water provides a consistent temperature, while the swirling motion increases the rate of heat loss from your finger, intensifying the cold sensation.

Heat will be removed from the exposed surface of a semi-inﬁnite region, with a b.c of either the ﬁrst or the third kind, in accordance with Fourier’s law: q= −k ∂T

Diﬀerentiating Θ as given by eqn (5.50), we obtain, for the b.c of the ﬁrst kind, q= k(T ∞ −T i )

Thus,qdecreases with increasing time, ast − 1/2 When the temperature of the surface is ﬁrst changed, the heat removal rate is enormous Then it drops oﬀ rapidly.

When a specified input heat flux, $ q_w $, is suddenly applied at the boundary of a semi-infinite region, it leads to transient heat conduction By differentiating the heat diffusion equation with respect to $ x $, we can analyze the resulting temperature distribution over time.

When we substituteq= −k ∂T /∂x in this, we obtain α∂ 2 q

In our analysis, we established boundary conditions for the heat flux, specifically at x=0 for t > 0, where q(x=0, t > 0) equals either qw or qw - qq, and at t = 0, where q(x, 0) is either 0 or qw - qq This approach elegantly reformulates the challenge of predicting local heat flux into a problem akin to predicting local temperature in a semi-infinite medium subjected to a step change in wall temperature Consequently, the solution aligns with the established relationship: qw - qq = erf(x).

The temperature distribution is obtained by integrating Fourier’s law At the wall, for example:

Figure 5.17 A bubble growing in a superheated liquid.

Example 5.6 Predicting the Growth Rate of a Vapor Bubble in an Inﬁnite Superheated Liquid

Steady multidimensional heat conduction

or, if we takeα=0.139×10 − 6 m 2 /s (given in [5.14] for coarse, gravelly earth), x=2.356

As we dig deeper into the earth, temperatures decrease until reaching a minimum at approximately 2.8 meters, where it is at its coldest Beyond this depth, the temperature starts to rise slightly, although the increase is minimal, particularly during the midwinter season.

(Ω=π ), the reverse would be true.

The general equation for T (r ) during steady conduction in a region of constant thermal conductivity, without heat sources, is called Laplace’s equation:

Solving multidimensional heat conduction problems can be deceptively complex, as the Laplacian, ∇²T, consists of multiple second partial derivatives While we addressed a two-dimensional heat conduction example with tailored boundary conditions in Example 4.1, the analytical solutions for more complex scenarios can vary significantly in difficulty, ranging from straightforward calculations to substantial challenges, depending on one's mathematical background and the specifics of the problem.

The reader who wishes to study such analyses in depth should refer to

[5.6] or [5.15], where such calculations are discussed in detail.

Faced with a steady multidimensional problem, three routes are open to us:

• Find out whether or not the analytical solution is already available in a heat conduction text or in other published literature.

• Obtain the solution graphically if the problem is two-dimensional.

It is to the last of these options that we give our attention next.

Figure 5.20 The two-dimensional ﬂow of heat between two isothermal walls.

The flux plotting method effectively addresses steady planar problems where boundaries are maintained at two distinct temperatures or are insulated With some expertise, this approach can achieve an accuracy of a few percent, often surpassing the precision of specified boundary conditions and thermal conductivity Additionally, it offers a clear representation of the physical aspects of the problem.

Figure5.20shows heat ﬂowing from one isothermal wall to another in a regime that does not conform to any convenient coordinate scheme.

In our analysis, we identify multiple channels that each transport an identical heat flow, denoted as δQW/m Additionally, we incorporate a series of equally spaced isotherms, separated by a distance of δT, between the walls Given that the heat flux remains consistent across all channels, we can express this relationship with the equation δQ = kδT δnδs (5.64).

To ensure consistency in the flow field, we must arrange the parameters δQ, δT, and k to be identical for each rectangle Consequently, the ratio δs/δn must also remain uniform across all rectangles For simplicity, we define this ratio as equal to one, resulting in all elements appearing as distorted squares.

The objective then is to sketch the isothermal lines and the adiabatic, 7

In steady multidimensional heat conduction, heat flow occurs along specific lines, while lines perpendicular to them indicate the direction of heat transfer This relationship is essential for understanding the constraints involved in heat conduction analysis.

• Isothermal and adiabatic lines must intersect at right angles.

• They must subdivide the ﬂow ﬁeld into elements that are nearly square—“nearly” because they have slightly curved sides.

After sketching the grid, the temperature at any point within the field can be easily determined from the illustration Additionally, the heat flow per unit depth into the paper can be calculated.

I k∆T (5.65) where N is the number of heat ﬂow channels and I is the number of temperature increments,∆T /δT.

To create an effective flux plot, start by accurately outlining the region's boundaries using ink, which can be done with drafting software or a straight edge Next, gather a soft pencil, like a no 2 grade, along with a soft eraser An exemplary illustration from the renowned Heat Transfer Notes of the mid-twentieth century serves as a useful reference for this process, as depicted in Fig 5.21.

The example features an axis of symmetry, which we interpret as an adiabatic boundary that prevents heat transfer Consequently, the problem simplifies to sketching lines within just one half of the area This process is demonstrated in four clear steps.

Notice the following steps and features in this plot:

• Begin by dividing the region, by sketching in either a single isothermal or adiabatic line.

To create squares, draw lines perpendicular to the original line while allowing it to adjust for proper alignment This process will necessitate some erasing to achieve the desired outcome.

• Never make the original lines dark and ﬁrm.

To create an accurate final grid, it is essential to avoid excessive subdivision of the squares, as a very fine grid can obscure the visibility of inaccuracies in perpendicularity and squareness As shown in Step IV of Fig 5.21, the recommended grid should not be finer than what is depicted to maintain precision Additionally, it is crucial that there is no component of heat flow perpendicular to the grid; thus, the grid must remain adiabatic.

Figure 5.21 The evolution of a ﬂux plot.

To ensure that large, irregular regions are accurately represented, add an additional isotherm and adiabatic line This will help confirm that the areas are properly divided into suitable squares, as illustrated by the dashed lines in Fig 5.21.

• Fill in the ﬁnal grid, when you are sure of it, either in hard pencil or pen, and erase any lingering background sketch lines.

• Your ﬂow channels need not come out even Notice that there is an extra 1/7 of a channel in Fig.5.21 This is simply counted as 1/7 of a square in eqn (5.65).

• Never allow isotherms or adiabatic lines to intersect themselves.

When the sketch is complete, we can return to eqn (5.65) to compute the heat ﬂux In this case

When the authors of [5.3] did this problem, they obtainedN/I =3.00—a value only 2% below ours This kind of agreement is typical when ﬂux plotting is done with care.

Figure 5.22 A ﬂux plot with no axis of symmetry to guide construction.

When analyzing shapes, it's crucial to avoid misinterpreting symmetry As illustrated in Figure 5.22, a shape with unequal legs differs from previous examples, emphasizing that no lines should intersect or exit at the corners A and B.

Due to the lack of symmetry, there is no clear direction for the lines at the corners, which means that a line originating from point A will not reach point B.

The structure features metal walls spaced 8 cm apart, filled with insulating material that has a thermal conductivity of 0.12 W/m·K Protruding ribs, measuring 4 cm in length, extend from one wall every 14 cm and maintain the temperature of that wall To determine the heat flux through the wall, with the first wall at 40 °C and the ribbed wall at 0 °C, calculations are necessary Additionally, the temperature at the midpoint of the wall, located 2 cm from a rib, needs to be established.

Figure 5.23 Heat transfer through a wall with isothermal ribs. §5.7 Steady multidimensional heat conduction 241

Solution The ﬂux plot for this conﬁguration is shown in Fig.5.23.

For a typical section, there are approximately 5.6 isothermal increments and 6.15 heat ﬂow channels, so

5.6 (0.12)(40−0)=10.54 W/m where the factor of 2 accounts for the fact that there are two halves in the section We deduce the temperature for the point of interest,

A heat conduction shape factor S may be deﬁned for steady problems involving two isothermal surfaces as follows:

Transient multidimensional heat conduction

Figure 5.26 Resistance vanishes where two isothermal boundaries intersect.

The problem of locally vanishing resistance

Suppose that two diﬀerent temperatures are speciﬁed on adjacent sides of a square, as shown in Fig.5.26 The shape factor in this case is

Counting channels beyond N10 is impractical, as they infinitely multiply in the lower left corner This situation violates the principle that isotherms should not intersect, resulting in a 1/r singularity While the figure may appear accurate at a distance from the corner, the proximity of the isotherms will inevitably cause them to influence and distort each other, preventing any intersections.

S will never really be inﬁnite, as it appears to be in the ﬁgure.

Consider the cooling of a stubby cylinder, such as the one shown in

Fig 5.27a The cylinder is initially at T = T i , and it is suddenly subjected to a common b.c on all sides It has a length 2Land a radiusr o

Finding the temperature ﬁeld in this situation is inherently complicated.

It requires solving the heat conduction equation forT =fn(r , z, t)with b.c.’s of the ﬁrst, second, or third kind.

Figure 5.27a indicates that the scenario can be interpreted as a blend of an infinite cylinder and an infinite slab, allowing the problem to be analyzed from this perspective.

When a body experiences uniform boundary conditions of the first, second, or third kind, and starts with a uniform initial temperature, its temperature response can be expressed as the product of solutions from an infinite slab and an infinite cylinder, both sharing identical boundary and initial conditions In the scenario depicted in Fig 5.27a, if the cylinder initiates convective cooling into a surrounding medium at temperature T ∞ at time t = 0, the resulting dimensional temperature response is established.

In this analysis, the slab is defined by its characteristic length, L, which represents half of its thickness, while the cylinder's characteristic length, R, corresponds to its radius The dimensionless form of the equation can be expressed as Θ ≡ T(r, z, t) - T∞.

T i −T ∞ = Θinf slab (ξ,Fo s ,Bi s ) Θinf cyl (ρ,Fo c ,Bi c )

(5.70b) For the cylindrical component of the solution, ρ= r r o

, and Bi c = hr o k , while for the slab component of the solution ξ= z

The component solutions are none other than those discussed in Sec- tions5.3–5.5 The proof of the legitimacy of such product solutions is given by Carlsaw and Jaeger [5.6, §1.15].

In Figure 5.27b, we observe a point located within a one-eighth infinite region, positioned near the corner This scenario can be conceptualized as the interaction of three semi-infinite bodies To determine the temperature at this specific point, we define Θ as the difference between T(x₁, x₂, x₃, t) and T∞.

Figure 5.27 Various solid bodies whose transient cooling can be treated as the product of one-dimensional solutions.

In the study of transient and multidimensional heat conduction, the semi-infinite body solution, represented by equation (5.53), accounts for convection at the boundary, while equation (5.50) applies when the boundary temperature changes instantaneously at time zero Additionally, various geometries can be modeled using product solutions, with the initial value of Θ at t=0 being one for each factor in the product.

A 4 cm square iron rod initially at 100 °C is immersed in a coolant at 20 °C, with a heat transfer coefficient of 800 W/m²K After 10 seconds, the temperature at a point 1 cm from one side and 2 cm from the adjacent side can be calculated.

Solution With reference to Fig 5.27c, see that the bar may be treated as the product of two slabs, each 4 cm thick We ﬁrst evaluate

(0.04 m/2) 2 = 0.565, and Bi 1 = Bi 2 = hL k = 800(0.04/2)/76 = 0.2105, and we then write Θ x L

= 0.93 from upper left-hand side of Fig 5.7 ×Θ2 x L

= 0.91 from interpolation between lower lefthand side and upper righthand side of Fig 5.7

Thus, at the axial line of interest, Θ=(0.93)(0.91)=0.846 so

Product solutions can also be used to determine the mean temperature, Θ, and the total heat removal, Φ, from a multidimensional object.

When combining multiple solutions, such as Θ1, Θ2, and potentially Θ3, the mean temperature of the multidimensional object, Θ, is determined by the product of the one-dimensional mean temperatures For two factors, this relationship is expressed as Θ = Θ1(Fo1, Bi1) × Θ2(Fo2, Bi2), and for three factors, it is represented as Θ = Θ1(Fo1, Bi1) × Θ2(Fo2, Bi2) × Θ3(Fo3, Bi3).

Since Φ =1−Θ, a simple calculation shows that Φcan found from Φ1, Φ2, andΦ3as follows: Φ=Φ1 +Φ2 (1−Φ1 ) for two factors (5.73a) Φ=Φ1 +Φ2 (1−Φ1 )+Φ3 (1−Φ2 ) (1−Φ1 ) for three factors (5.73b)

For the bar described in Example5.12, what is the mean temperature after 10 s and how much heat has been lost at that time?

Solution For the Biot and Fourier numbers given in Example5.12, we ﬁnd from Fig.5.10a Φ1 (Fo 1 =0.565,Bi 1 =0.2105)=0.10 Φ2 (Fo2 =0.565,Bi2 =0.2105)=0.10 and, with eqn (5.73a), Φ=Φ1 +Φ2 (1−Φ1 )=0.19 The mean temperature is Θ= T−20

252 Chapter 5: Transient and multidimensional heat conduction

5.1 Rework Example5.1, and replot the solution, with one change.

This time, insert the thermometer at zero time, at an initial temperature< (T i −bT).

When a body with a defined volume, surface area, and initial temperature $ T_i $ is suddenly placed in a bath where the temperature increases according to the equation $ T_{bath} = T_i + (T_0 - T_i)e^{t/\tau} $, we can analyze the temperature response of the body over time, where $ \tau = \frac{10\rho c V}{hA} $ and $ t $ is measured from the moment of immersion Given that the Biot number is small, the temperature of the body will approach the bath temperature gradually To visualize this, we can plot both the temperature response of the body and the bath temperature as functions of time, extending the analysis up to $ t = 2\tau $.

In a scenario where a body with a defined volume and surface area is submerged in a bath with a sinusoidally varying temperature at frequency ω around an average value, the heat transfer coefficient is established, and the Biot number is low After an extended period, the temperature variation of the body can be determined and plotted alongside the bath temperature, revealing intriguing dynamics in the thermal response of the body compared to the fluctuating bath conditions This analysis highlights the relationship between the body’s thermal inertia and the sinusoidal temperature changes of the surrounding medium.

A suggested program for solving this problem:

• Write the diﬀerential equation of response.

To determine the particular integral of the complete equation, assume that T - T mean equals C1 cos(ωt) + C2 sin(ωt) By substituting this expression into the differential equation, you can solve for the values of C1 and C2 that will satisfy the equation.

• Write the general solution of the complete equation It will have one unknown constant in it.

• Write any initial condition you wish—the simplest one you can think of—and use it to get rid of the constant.

• Let the time be large and note which terms vanish from the solution Throw them away.

• Combine two trigonometric terms in the solution into a term involving sin(ωt −β), where β = fn(ωT) is the phase lag of the body temperature.

5.4 A block of copper ﬂoats within a large region of well-stirred mercury The system is initially at a uniform temperature,T i

The heat transfer coefficient, denoted as h_m, is present on the interior surface of the thin metal container holding mercury, while another coefficient, h_c, exists between the copper block and the mercury This system experiences a sudden shift in ambient temperature from T_i.

To predict the temperature response of the copper block when the temperature of the surrounding environment is lower than that of the block, we can neglect the internal resistance of both the copper and the mercury It is essential to verify that the results align with the initial conditions and demonstrate the expected behavior as time approaches infinity.

5.5 Sketch the electrical circuit that is analogous to the second- order lumped capacity system treated in the context of Fig.5.5 and explain it fully.

A one-inch diameter copper sphere with a central thermocouple is immersed in water saturated at 211°F, as illustrated in Fig 5.28 The thermocouple readings during the quenching process reveal that if the Biot number is small, the center temperature reflects the uniform temperature of the sphere To analyze this, tangents to the curve are drawn and graphically differentiated, allowing for the calculation of dT/dt This data is then used to create a graph depicting the heat transfer coefficient as a function of the temperature difference between the sphere and the saturation temperature The results provide actual values of the heat transfer coefficient during boiling across various temperature differences, enabling an assessment of the maximum value obtained.

Biot number is too great to permit the use of lumped-capacity methods.

5.7 A butt-welded 36-gage thermocouple is placed in a gas ﬂow whose temperature rises at the rate 20 ◦ C/s The thermocouple steadily records a temperature 2.4 ◦ C below the known gas

ﬂow temperature Ifρc is 3800 kJ/m 3 K for the thermocouple material, what ishon the thermocouple? [h=1006 W/m 2 K.]

5.8 Check the point on Fig.5.7 at Fo = 0.2, Bi = 10, and x/L = 0 analytically.

5.9 Prove that when Bi is large, eqn (5.34) reduces to eqn (5.33).

5.10 Check the point at Bi= 0.1 and Fo =2.5 on the slab curve in

254 Chapter 5: Transient and multidimensional heat conduction

Figure 5.28 Conﬁguration and temperature response for Problem5.6

5.11 Sketch one of the curves in Fig.5.7,5.8, or5.9and identify:

• The region in which b.c.’s of the third kind can be replaced with b.c.’s of the ﬁrst kind.

• The region in which a lumped-capacity response can be assumed.

• The region in which the solid can be viewed as a semi- inﬁnite region.

Water flows over a flat slab of Nichrome, which is 0.05 mm thick and acts as a resistance heater powered by AC electricity The heat transfer coefficient (h) is measured at 2000 W/m²K This setup raises the question of the extent of surface temperature fluctuations that will occur.

Some introductory ideas

Joseph Black’s perception about forced convection (above) represents a very correct understanding of the way forced convective cooling works.

When cold air flows over a warm surface, it continuously removes the warm air that has become mixed with the surface and replaces it with colder air This chapter focuses on developing analytical descriptions of the processes involved in convective heating and cooling.

Our goal is to forecast heat transfer in fluids, starting with the analysis of fluid motion around heated or cooled bodies By accurately predicting this motion, we can determine the amount of heat exchanged when hot fluid is replaced with cold fluid and vice versa.

Fluids ﬂowing past solid bodies adhere to them, so a region of variable velocity must be built up between the body and the free ﬂuid stream, as

270 Laminar and turbulent boundary layers §6.1

The boundary layer (b.l.), depicted in Figure 6.1, is a thin region characterized by a thickness, δ It is defined as the distance from the wall where the flow velocity reaches within 1% of the free stream velocity (u ∞) Typically, the boundary layer is much thinner than the dimensions of the object immersed in the flow.

Before predicting h, the initial step involves mathematically describing the boundary layer, a concept first introduced by Prandtl and his students in 1904 Their work relied on simplifications that emerged from recognizing the thinness of the layer.

The dimensional functional equation for boundary layer thickness on a flat surface is expressed as δ = fn(u ∞, ρ, μ, x), where x represents the length along the surface In this equation, ρ denotes the fluid density in kg/m³, and μ signifies the dynamic viscosity in kg/m·s This equation incorporates five key variables that influence boundary layer behavior.

1 We qualify this remark when we treat the b.l quantitatively.

Ludwig Prandtl, who completed his doctorate at the Technical University of Munich in 1900, was appointed to a chair at the newly established fluid mechanics institute at Göttingen University in 1904 That same year, he published a groundbreaking paper on the boundary layer, which significantly influenced the development of modern fluid mechanics and aerodynamics His contributions during this period, prior to the rise of Hitler's regime, were crucial in establishing the foundations for the analysis of heat convection.

Ludwig Prandtl (1875–1953) significantly contributed to fluid mechanics, particularly through the introduction of the Reynolds number (Re x), which is defined as Re x = ρu ∞ x / ν, where ρ is the fluid density, u ∞ is the flow velocity, x is the characteristic length, and ν represents the kinematic viscosity The Reynolds number characterizes the relative influences of inertial and viscous forces in fluid dynamics, with the subscript indicating the specific length upon which it is based.

We discover shortly that the actual form of eqn (6.1) for a ﬂat surface, whereu ∞ remains constant, is δ x = 4.92

In scenarios where velocity is high or viscosity is low, the ratio δ/x is small, leading to enhanced heat transfer Conversely, when velocity decreases, the boundary layer becomes thicker, resulting in reduced heat transfer efficiency.

Reynolds was born in Ireland but he taught at the University of Manchester.

He was a signiﬁcant contributor to the subject of ﬂuid mechanics in the late

19th C His original laminar-to- turbulent ﬂow transition experiment, pictured below, was still being used as a student experiment at the University of Manchester in the 1970s.

Osborne Reynolds, known for his groundbreaking work on laminar-turbulent flow transition, conducted experiments that revealed how fluid behavior changes in a tube By injecting ink into a steady flow of water, he observed that when the average velocity exceeded a certain threshold, the once orderly ink streamline became unstable, leading to the formation of chaotic eddies and ultimately resulting in a complete mixing of the fluid This phenomenon is quantified by the Reynolds number, which plays a crucial role in understanding fluid dynamics.

Figure 6.4 Boundary layer on a long, ﬂat surface with a sharp leading edge. water, as is suggested in the sketch.

The transitional value of the average velocity, denoted as (u av) crit, is influenced by four key variables: pipe diameter (D), a, and ρ, measured in kilograms, meters, and seconds Consequently, there exists only one pi-group that encapsulates this relationship.

The critical Reynolds number for maintaining fully developed laminar flow in a pipe, irrespective of background noise, is 2100 While careful experimentation can extend laminar flow stability up to a Reynolds number of 10,000, with meticulous attention, it may reach even higher levels However, the definitive threshold below which flow remains laminar is consistently established at 2100.

In a boundary layer, fluid flow behaves similarly, as illustrated in Figure 6.4, which depicts fluid moving over a plate with a sharp leading edge The flow remains laminar until it reaches a transitional Reynolds number based on the distance x.

At larger values ofxthe b.l exhibits sporadic vortexlike instabilities over a fairly long range, and it ﬁnally settles into a fully turbulent b.l.

The critical Reynolds number (Re x critical) for the boundary layer is approximately 3.5×10^5, but it is significantly influenced by factors such as freestream turbulence, the leading edge shape, wall roughness, and vibrations On a flat plate, the boundary layer can remain laminar even under substantial disturbances if Re x is less than or equal to 6×10^4 In relatively undisturbed conditions, transition occurs within the Reynolds number range of 3×10^5 to 5×10^5, while in controlled laboratory settings, turbulent transition may be postponed until Re x reaches around 3×10^6 Typically, turbulent transition is nearly complete before Re x hits 4×10^6, often occurring much earlier.

The critical Reynolds number (Re) specifications apply only to flat surfaces; however, when the surface curves away from the flow, as illustrated in Fig 6.1, turbulence can occur at significantly lower Reynolds number values.

If the wall is at a temperatureT w , diﬀerent from that of the free stream,

T ∞ , there is athermal boundary layer thickness,δ t —diﬀerent from the

The thermal boundary layer thickness, denoted as δ, is illustrated in Figure 6.5 In this context, we equate the heat conducted away from the wall by the fluid to the heat transfer represented by a convective heat transfer coefficient.

Laminar incompressible boundary layer on a ﬂat surface 276

To predict the boundary layer flow field, we solve the equations governing the conservation of mass and momentum within the boundary layer The initial step involves formulating these essential equations.

Conservation of mass—The continuity equation

A velocity field can be represented in vector form as u = iu + jv + kw, where u, v, and w denote the x, y, and z components of velocity, respectively In a steady flow, individual particle paths become steady streamlines, which can be defined using a stream function, ψ(x, y) = constant Each constant value corresponds to a distinct streamline, as illustrated in the accompanying figure.

The velocity, denoted as u, aligns with the streamlines, preventing any flow from crossing them Consequently, adjacent streamlines exhibit characteristics similar to heat flow, particularly in the context of a laminar incompressible boundary layer on a flat surface.

Figure 6.7 A steady, incompressible, two-dimensional ﬂow

ﬁeld represented by streamlines, or lines of constantψ. channel in a ﬂux plot (Section5.7); such channels are adiabatic—no heat

In fluid dynamics, the conservation of mass is expressed by analyzing the inflow and outflow of mass across two faces of a triangular element with unit depth The equation representing this principle is given by ρv dx - ρu dy = 0, indicating that the mass flow rate remains constant across the element.

If the ﬂuid is incompressible, so thatρ=constant along each streamline, then

But we can also diﬀerentiate the stream function along any streamline, ψ(x, y)=constant, in Fig.6.7: dψ= ∂ψ

If we compare eqns (6.8) and (6.9), we immediately see that the coef-

ﬁcients ofdx anddy must be the same, so v= − ∂ψ

The two-dimensional continuity equation for incompressible flow mathematically represents the concept of continuous flow without breaks In three dimensions, this principle is further expressed through the continuity equation for incompressible fluids.

Fluid moves with a uniform velocity,u ∞ , in thex-direction Find the stream function and see if it gives plausible behavior (see Fig.6.8).

Solution u=u ∞ andv=0 Therefore, from eqns (6.10) u ∞ = ∂ψ

∂x y Integrating these equations, we get ψ=u ∞ y+fn(x) and ψ=0+fn(y)

Comparing these equations, we get fn(x) = constant and fn(y) = u ∞ y+constant, so ψ=u ∞ y+constant

The analysis reveals a series of uniformly spaced horizontal streamlines, consistent with expectations (refer to Fig 6.8) For clarity in the figure, we have set the arbitrary constant to zero.

Figure 6.8 Streamlines in a uniform horizontal ﬂow ﬁeld,ψ=u ∞ y.

The Navier-Stokes equation represents the momentum equation in viscous flow, characterized by its complex vectorial expression While a comprehensive derivation can be found in advanced fluid mechanics literature, this discussion will focus on a simplified derivation applicable solely to two-dimensional incompressible boundary layer flow, as illustrated in Figure 6.9.

Shear stresses continuously distort and rotate any element, as illustrated in the figure The enlarged section at the bottom highlights the horizontal shear stresses and pressure forces acting on the element, represented by bold arrows.

We also display, as lighter arrows, the momentum ﬂuxes entering and leaving the element.

Both x- and y-directed momentum interacts with an element, similar to a scenario where a man in a boxcar catches a baseball thrown by a child at a crossing When the man catches the ball, its momentum not only pushes him back but also causes a shift in his position toward the rear of the train due to relative motion Similarly, fluid particles entering an element will affect its motion, as their x-components of momentum are carried into the element by the flow.

To adhere to the principle of conservation of linear momentum, the velocities must be adjusted accordingly Therefore, it is essential that the sum of the external forces acting on control volume A in the x-direction is balanced by the rate at which control volume A exerts forces in the x-direction.

Stress, denoted by τ, is characterized by two subscripts: the first indicates the direction normal to the acting plane, while the second identifies the line of action When both subscripts match, the stress acts perpendicular to the surface, signifying either pressure or tension rather than shear stress.

Figure 6.9 Forces acting in a two-dimensional incompressible boundary layer. directed momentum out The external forces, shown in Fig.6.9, are τ yx +∂τ yx

∂y dy dx−τ yx dx+p dy− p+∂p

The rate at whichAlosesx-directed momentum to its surroundings is ρu 2 + ∂ρu 2

∂y dx dy §6.2 Laminar incompressible boundary layer on a ﬂat surface 281

We equate these results and obtain the basic statement of conservation ofx-directed momentum for the b.l.:

∂y dy dx−dp dxdx dy =

The shear stress in this result can be eliminated with the help of Newton’s law of viscous shear: τ yx =à∂u

∂y so the momentum equation becomes

Finally, we remember that the analysis is limited toρconstant, and we limit use of the equation to temperature ranges in whichàconstant.

The steady, two-dimensional, incompressible boundary layer momentum equation can be expressed in a more comprehensive form, applicable to compressible flow as well By multiplying the momentum equation by the velocity $ u $ and subtracting it from the left-hand side of the corresponding equation, we derive an alternative representation of the momentum equation: $ u\partial u $.

Equation (6.13) has a number of so-calledboundary layer approximations built into it:

The Bernoulli equation for the free stream ﬂow just above the boundary layer where there is no viscous shear, p ρ + u 2 ∞

2 = constant can be diﬀerentiated and used to eliminate the pressure gradient,

1 ρ dp dx = −u ∞ du ∞ dx so from eqn (6.12):

And if there is no pressure gradient in the flow—ifpandu ∞ are constant as they would be for flow past a flat plate—then eqns (6.12), (6.13), and (6.14) become

Predicting the velocity proﬁle in the laminar boundary layer without a pressure gradient

Two established strategies for solving the velocity profile in equation (6.15) have been widely utilized, with the first being developed by H Blasius, a student of Prandtl, prior to World War I This method is exact and involves the introduction of the stream function, ψ, into equation (6.15).

This reduces the number of dependent variables from two (uandv) to just one—namely,ψ We do this by substituting eqns (6.10) in eqn (6.15):

It turns out that eqn (6.16) can be converted into an ordinary d.e. with the following change of variables: ψ(x, y)≡√ u ∞ νx f (η) where η≡

Blasius gained significant recognition for his contributions to fluid mechanics but eventually distanced himself from the field, famously stating, “I decided that I had no gift for it; all of my ideas came from Prandtl.” In the context of laminar incompressible boundary layers on flat surfaces, a transformation is introduced involving an undetermined function f(η) This transformation parallels the approach used to derive an ordinary differential equation from the heat conduction equation After manipulating partial derivatives, the resulting equation is expressed as f d²f/dη² + 2d³f/dη³ = 0, where the relationship between the velocity u and the free stream velocity u∞ is given by u/u∞ = df/dη.

(6.19) The boundary conditions for this ﬂow are u(y =0)=0 or df dη η = 0 =0 u(y = ∞)=u ∞ or df dη η =∞ =1 v(y=0)=0 or f (η=0)=0

The solution of eqn (6.18) subject to these b.c.’s must be done numeri- cally (See Problem6.3.)

The Blasius problem solution is detailed in Table 6.1, with the dimensionless velocity components illustrated in Fig 6.10 The u component rises from zero at the wall (η=0) to 99% of u ∞ at η=4.92.

Thus, the b.l thickness is given by

3νx/u ∞ or, as we anticipated earlier [eqn (6.2)], δ x = 4.92

Concept of similarity The exact solution for u(x, y) reveals a most useful fact—namely, that u can be expressed as a function of a single variable,η: u u ∞ =f (η)=f y

Table 6.1 Exact velocity proﬁle in the boundary layer on a ﬂat surface with no pressure gradient y3 u ∞ /νx u u ∞ v3 x/νu ∞ η f (η) f (η) (ηf −f )

This is called asimilarity solution To see why, we solve eqn (6.2) for

9u ∞ νx = 4.92 δ(x) and substitute this inf (y3 u ∞ /νx) The result is f = u u ∞ =fn y δ(x)

The energy equation

To understand fluid dynamics in the boundary layer (b.l.), it is essential to adapt the heat conduction equation to account for fluid motion By solving this modified equation, we can determine the temperature distribution within the boundary layer This temperature field can then be utilized to calculate the heat transfer coefficient (h) using Fourier’s law, expressed as h = q.

To enhance our predictions, we build upon the analysis from Section 2.1 As illustrated in Figure 2.4, we consider a volume that encompasses a solid exposed to a temperature field In our revised model, this volume also includes fluid with a velocity field represented as u(x, y, z), as depicted in Figure 6.12 We proceed with several restrictive approximations to refine our analysis.

Pressure variations in fluid flow typically do not significantly impact thermodynamic properties According to thermodynamics, the specific internal energy (ˆu) is related to specific enthalpy (hˆ) by the equation hˆ = ˆu + p/ρ Additionally, the change in enthalpy can be expressed as dhˆ = c p dT + (∂h/∂p)ˆ T dp For most liquid flows and gas flows traveling at speeds below approximately one-third of the speed of sound, the effects of pressure changes on enthalpy, internal energy, and density can be reasonably neglected.

Under the specified conditions, density variations are primarily due to temperature changes, which are minimal Therefore, the flow can be considered nearly incompressible, as indicated by the fact that ∇·u = 0 for incompressible flow.

• Temperature variations in the flow are not large enough to changek significantly When we consider the flow field, we will also presume à to be unaffected by temperature change.

Changes in potential and kinetic energy are minimal compared to the fluctuations in thermal energy As the kinetic energy of a fluid can vary due to pressure gradients, this indicates that the variations in pressure are likely to be limited.

• The viscous stresses do not dissipate enough energy to warm theﬂuid signiﬁcantly. §6.3 The energy equation 293

Figure 6.12 Control volume in a heat-flow and fluid-flow field.

Just as we wrote eqn (2.7) in Section2.1, we now write conservation of energy in the form d dt

R ρu dRˆ rate of internal energy increase in R

S (ρh) ˆ uãn dS rate of internal energy and ﬂow work out of R

(−k∇T )ãn dS net heat conduction rate out of R

R ˙ q dR rate of heat generation in R

In the third integral, $ u \, dS $ denotes the volume flow rate across an element $ dS $ of the control surface Since the position of $ R $ remains constant over time, we can move the time derivative inside the first integral By applying Gauss's theorem to convert the surface integrals into volume integrals, equation (6.36) is derived.

Because the integrand must vanish identically (recall the footnote on pg.55in Chap.2) and becausekdepends weakly onT,

Since we are neglecting pressure eﬀects, we may introduce the following approximation: d(ρu)ˆ =d(ρh)ˆ −dp≈d(ρh)ˆ =ρdhˆ+h dρˆ Thus, collecting and rearranging terms ρ

The term involving density derivatives may be neglected on the basis that density changes are small and the ﬂow is nearly incompressible (but see Problem6.36for a more general result).

Upon substitutingdhˆ≈c p dT, we obtain our ﬁnal result: ρc p

The energy equation for a constant pressure flow field resembles the equation for a solid body, with the key difference being the inclusion of the enthalpy transport term, represented as ρc p uã ∇T.

Consider the term in parentheses in eqn (6.37):

The material derivative, denoted as DT/Dt, represents the rate of temperature change experienced by a fluid particle as it travels through a flow field This concept is extensively covered in fluid mechanics courses, highlighting its significance in understanding fluid behavior.

In a steady two-dimensional ﬂow ﬁeld without heat sources, eqn (6.37) takes the form u∂T

(6.39) Furthermore, in a b.l.,∂ 2 T /∂x 2 ∂ 2 T /∂y 2 , so the b.l form is u∂T

Heat and momentum transfer analogy

Consider a b.l in a fluid of bulk temperatureT ∞ , flowing over a flat surface at temperature T w The momentum equation and its b.c.’s can be written as u ∂

And the energy equation (6.40) can be written in terms of a dimensionless temperature,Θ=(T−T w )/(T ∞ −T w ), as u∂Θ

The issues of predicting u ∞ and Θ are fundamentally the same, with the only difference being that equation (6.41) includes ν, while equation (6.42) incorporates α If ν and α are equal, the temperature distribution in the boundary layer can be expressed as T - T w for ν = α.

T ∞ −T w =f (η) derivative of the Blasius function since the two problems must have the same solution.

In this case, we can immediately calculate the heat transfer coeﬃcient using eqn (6.5): h= k

∂y η = 0 but(∂ 2 f /∂η 2 ) η = 0 =0.33206 (see Fig.6.10) and∂η/∂y =3 u ∞ /νx, so hx k =Nu x =0.332063

Re x forν=α (6.43) Normally, in using eqn (6.43) or any other forced convection equation, properties should be evaluated at theﬁlm temperature,T f =(T w +T ∞ )/2.

Water ﬂows over a ﬂat heater, 0.06 m in length, at 15 atm pressure and 440 K The free stream velocity is 2 m/s and the heater is held at

460 K What is the average heat ﬂux?

Therefore,να, and we can use eqn (6.43) First, we must calculate the average heat ﬂux,q To do this, we set∆T ≡T w −T ∞ and write q= 1

Note that the average heat ﬂux is twice that at the trailing edge,x=L.

Usingk=0.674 W/mãK for water at the ﬁlm temperature, q=2(0.332)0.674

Equation (6.43) is clearly a very restrictive heat transfer solution We now want to ﬁnd how to evaluateqwhenνdoes not equalα.

The Prandtl number and the boundary layer thicknesses 296

The similarity between velocity and thermal boundary layers has significant implications for heat transfer analysis in laminar flow Dimensional analysis indicates that the heat transfer coefficient, denoted as h, is a function of several variables, including thermal conductivity (k), distance (x), density (ρ), specific heat capacity (c p), dynamic viscosity (μ), and free stream velocity (u ∞) Understanding these relationships is crucial for optimizing heat transfer in various applications.

Based on Newton's original hypothesis, we have excluded the term T w − T ∞, leading to the conclusion that the heat transfer coefficient (h) is not a function of the temperature difference (∆T) during forced convection This results in seven variables measured in J/K, m, kg, and s, which can be reduced to three dimensionless pi-groups It is important to note that there is no conversion between heat and work, so J should not be considered as Nãm but as a distinct unit The dimensionless groups identified are: Π1 = hx/k (Nusselt number), Π2 = ρu ∞ x à (Reynolds number), and a new group Π3 = àc p/k (Prandtl number), which can also be expressed as ν/α.

In forced convection flow scenarios, the Nusselt number, represented as Nu x, is defined by the equation Nu x = fn(Re x, Pr) While Equation (6.43) applies when the kinematic viscosity (ν) equals the thermal diffusivity (α) or when the Prandtl number (Pr) is equal to 1, it shares a similar structure to Equation (6.44) but does not account for the Prandtl number's influence on Nu x.

To better understand the physical meaning of the Prandtl number, let us brieﬂy consider how to predict its value in a gas.

In a gas, a small neighborhood around a point of interest reveals a velocity or temperature gradient, as illustrated in Figure 6.13 The mean free path of molecules between collisions is identified, with planes positioned at y±/2 that encapsulate the average travel of molecules at the plane y It is important to note that these planes should ideally be closer to y± for several nuanced reasons, which are elaborated on in detail in reference [6.4].

Shear stress (τ yx) is defined as the momentum change of all molecules crossing the y-plane of interest per unit area It is calculated by considering the mass flux of molecules moving from y−/2 to y+/2 and the corresponding change in fluid velocity.

The mass flux from top to bottom is proportional toρC, whereC, the mean molecular speed of the stationary fluid, is uorvin incompressible flow Thus, τ yx =C 1 ρC du dy

N m 2 and this also equalsàdu dy (6.45)

Figure 6.13 Momentum and energy transfer in a gas with a velocity or temperature gradient.

By the same token, q y =C 2 ρc v C dT dy and this also equals −kdT dy wherec v is the speciﬁc heat at constant volume The constants,C 1 and

C 2, are on the order of unity It follows immediately that à=C 1 ρC so ν=C 1

C γ whereγ ≡c p /c v is approximately a constant on the order of unity for a given gas Thus, for a gas,

Pr≡ ν α= a constant on the order of unity

A deeper exploration of the kinetic theory of gases provides valuable insights into the Prandtl number, and experimental evidence supports these findings, as detailed in Appendix A.

• For simple monatomic gases, Pr= 2 3 §6.4 The Prandtl number and the boundary layer thicknesses 299

• For diatomic gases in which vibration is unexcited (such as N 2 and

• As the complexity of gas molecules increases, Pr approaches an upper value of unity.

• Pr is most insensitive to temperature in gases made up of the simplest molecules because their structure is least responsive to temperature changes.

In a liquid, the physical mechanisms of molecular momentum and energy transport are much more complicated and Pr can be far from unity For example (cf TableA.3):

• For liquids composed of fairly simple molecules, excluding metals,

Pr is of the order of magnitude of 1 to 10.

• For liquid metals, Pr is of the order of magnitude of 10 − 2 or less.

• If the molecular structure of a liquid is very complex, Pr might reach values on the order of 10 5 This is true of oils made of long-chain hydrocarbons, for example.

The Prandtl number (Pr) can vary significantly, spanning nearly eight orders of magnitude in common fluids, yet it arises from similar mechanisms of heat and momentum transfer The numerical values of Pr, along with the underlying analogy, are rooted in fundamental processes of molecular transport.

Boundary layer thicknesses, δ and δ t , and the Prandtl number

The exact solution of the boundary layer equations reveals that for Prandtl number Pr = 1, the relationship δ = δt holds true, resulting in dimensionless velocity and temperature profiles that are consistent on a flat surface Additionally, two other key observations can be easily identified.

• When Pr>1, δ > δ t , and when Pr δ t, which is ensured by the condition dT/dy = 0 at y = δ t Additionally, a fourth condition arises from applying equation (6.40) at the wall, where both u and v equal zero, leading to (∂²T/∂y²) at y = 0 = 0 These four conditions can be expressed in a dimensionless form.

Equations (6.48) provide enough information to approximate the temperature proﬁle with a cubic function.

(6.49) Substituting eqn (6.49) into eqns (6.48), we get a=1 −1=b+c+d 0=b+2c+3d 0=2c

302 Laminar and turbulent boundary layers §6.5 which gives a=1 b= − 3 2 c=0 d= 1 2 so the temperature proﬁle is

Predicting the heat ﬂux in the laminar boundary layer

To determine the thermal boundary layer thickness, δ t, we need to incorporate the temperature profile from equation (6.50) and the velocity profile from equation (6.29) into the integral form of the energy equation (6.47) This equation can be rewritten as u ∞ (T w − T ∞ ) d dx δ t, allowing us to calculate the unknown quantity δ t effectively.

If δt is less than δ, the integration can be completed without issues However, if δt exceeds δ, complications arise as the equation u/u∞ = 1 defines the velocity beyond y = δ, rather than following equation (6.29) For now, we will assume that the condition δt < δ is met By introducing φ ≡ δt / δ in equation (6.51) and defining y/δt ≡ η, we can proceed with our analysis.

Sinceφis a constant for any Pr [recall eqn (6.46)], we separate variables:

280φ 3 §6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface 303

Figure 6.14 The exact and approximate Prandtl number inﬂu- ence on the ratio of b.l thicknesses.

Integrating this result with respect tox and taking δ t =0 atx = 0, we get δ t =

Re x in the integral formulation [eqn (6.31b)] We divide by this value ofδto be consistent and obtain δ t δ ≡φ=0.9638

The unapproximated result above is shown in Fig.6.14, along with the results of Pohlhausen’s precise calculation (see Schlichting [6.3, Chap 14]).

It turns out that the exact ratio,δ/δ t , is represented with great accuracy

304 Laminar and turbulent boundary layers §6.5 by δ t δ =Pr − 1/3 0.6 Pr 50 (6.55)

So the integral method is accurate within 2.5% in the Prandtl number range indicated.

Figure 6.14 is limited to Prandtl numbers (Pr) less than 0.6 because the lowest Pr for pure gases is 0.67, while liquid metals have Pr values around 10^-2 At Pr = 0.67, the ratio of thermal boundary layer thickness to velocity boundary layer thickness (δt/δ) is 1.143, slightly breaching the assumption that δt is much less than δ In contrast, for mercury at 100 °C, Pr is 0.0162, resulting in δt/δ being 3.952, which significantly violates this condition Consequently, the theory is valid for gases and most liquids, excluding metallic liquids.

The ﬁnal step in predicting the heat ﬂux is to write Fourier’s law: q= −k ∂T

Using the dimensionless temperature distribution given by eqn (6.50), we get q= +kT w −T ∞ δ t

∆T = 3k 2δ t = 3 2 k δ δ δ t (6.57) and substituting eqns (6.54) and (6.31b) forδ/δ t andδ, we obtain

4.64 1.025 Pr 1/3 =0.3314 Re 1/2 x Pr 1/3 Considering the various approximations, this is very close to the result of the exact calculation, which turns out to be

This expression gives very accurate results under the assumptions on which it is based: a laminar two-dimensional b.l on a ﬂat surface, with

T w =constant and 0.6 Pr 50. §6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface 305

Figure 6.15 A laminar b.l in a low-Pr liquid The velocity b.l. is so thin thatuu ∞ in the thermal b.l.

Some other laminar boundary layer heat transfer equations

High Pr At high Pr, eqn (6.58) is still close to correct The exact solution is

In low-Prandtl number (low-Pr) conditions, liquid flow over a flat plate results in a thermal boundary layer where the velocity approaches that of the free stream, effectively negating the no-slip condition at the surface This simplification allows for the dimensional functional equation for heat transfer to be expressed as h = f(x, k, ρc_p, u_∞), highlighting the dependence on various physical properties and flow parameters.

There are ﬁve variables in J/K, m, and s, so there are only two pi-groups.

Nu x = hx k and Π2 ≡Re x Pr= u ∞ x α

The new group,Π2, is called aPéclét number, Pe x , where the subscript identiﬁes the length upon which it is based It can be interpreted as follows:

Pe x ≡ u ∞ x α = ρc p u ∞ ∆T k∆T = heat capacity rate of ﬂuid in the b.l. axial heat conductance of the b.l (6.61)

So long as Pe x is large, the b.l assumption that ∂ 2 T /∂x 2 ∂ 2 T /∂y 2 will be valid; but for small Pe x (i.e., Pe x 100), it will be violated and a boundary layer solution cannot be used.

The exact solution of the b.l equations gives, in this case:

General relationship Churchill and Ozoe [6.5] recommend the following empirical correlation for laminar ﬂow on a constant-temperature ﬂat surface for the entire range of Pr:

This relationship demonstrates high accuracy, closely aligning with equations (6.59) and (6.62) in both high and low Prandtl number limits The computation of an average Nusselt number for the general case is assigned as an exercise (Problem 6.10).

The boundary layer depicted in Figure 6.16 features a heated region that initiates at a distance $ x_0 $ from the leading edge The heat transfer in this scenario can be effectively calculated using integral methods, as outlined in Problem 6.41.

Average heat transfer coeﬃcient, h The heat transfer coeﬃcienth, is the ratio of two quantities,qand∆T, either of which might vary withx.

So far, we have only dealt with theuniform wall temperature problem.

Equations (6.58), (6.59), (6.62), and (6.63) can be utilized to calculate q(x) when the temperature difference (T_w - T_∞) is a constant, denoted as ∆T The following subsection will address the prediction of [T(x) - T_∞] when q is a constant, a scenario referred to as the uniform wall heat flux problem.

Figure 6.16 A b.l with an unheated region at the leading edge.

The termhis used to designate eitherq/∆T in the uniform wall temperature problem orq/∆T in the uniform wall heat ﬂux problem Thus, uniform wall temp.: h≡ q

The Nusselt number, denoted as Nu_L, is defined based on the heat transfer coefficient (h) and a characteristic length (L) It is important to note that this designation should not be interpreted as an average of Nu_x, as such an average would lack significance in these contexts.

Thus, for a ﬂat surface (withx 0 =0), we use eqn (6.58) in eqn (6.65) to get h= 1

(6.67) Thus,h=2h(x=L)in a laminar ﬂow, and

Nu L = hL k =0.664 Re 1/2 L Pr 1/3 (6.68) Likewise for liquid metal ﬂows:

In conclusion, the findings presented are limited to two-dimensional, incompressible, laminar boundary layers on flat, isothermal walls, applicable under conditions of moderate velocities.

• Re x or Re L is not above the turbulent transition value, which is typically a few hundred thousand.

• The Mach number of the ﬂow, Ma≡u ∞ /(sound speed), is less than about 0.3 (Even gaseous ﬂows behave incompressibly at velocities well below sonic.) A related condition is:

The Eckert number (Ec), defined as Ec ≡ u²∞ / cp (Tw - T∞), is significantly less than one, indicating that viscous dissipation heating is negligible in this analysis This assumption was inherently considered when J was treated as an independent unit during the dimensional analysis of the problem.

It is worthwhile to notice how h and Nu depend on their independent variables: horh∝ 1

As the height (h) approaches infinity, the Nusselt number (Nu x) diminishes at the leading edge, specifically at x = 0 However, an infinite value of h or shear stress is not realistically attainable at this point, as the boundary layer description fails in the immediate vicinity of x = 0.

Fluid properties such as thermal conductivity (k), density (ρ), specific heat capacity (c p), and dynamic viscosity (μ) can vary significantly with temperature within the boundary layer (b.l.) However, evaluating these properties at the average temperature, or film temperature (T f = (T w + T ∞ )/2), typically yields accurate results Additionally, while properties are listed at a single pressure in Appendix A, it is important to note that μ, k, and c p show minimal variation with pressure, particularly in liquids.

The Reynolds analogy

From eqn (6.72) we ﬁnd that

The analogy between heat and momentum transfer can be generalized to yield valuable insights Referring to equation (6.25), this analysis is applicable specifically to flat surfaces without a pressure gradient.

2 (6.25) and by rewriting eqns (6.47) and (6.51), we obtain for the constant wall temperature case: d dx φ δ

But the similarity of temperature and ﬂow boundary layers to one another

[see, e.g., eqns (6.29) and (6.50)], suggests the following approximation, which becomes exact only when Pr=1:

Substituting this result in eqn (6.74) and comparing it to eqn (6.25), we get

Finally, we substitute eqn (6.55) to eliminate φ from eqn (6.75) The result is one instance of theReynolds-Colburn analogy: 8 h ρc p u ∞ Pr 2/3 = C f

In 1874, Reynolds developed a significant analogy that was later utilized by Colburn in the 20th century This analogy is specifically applicable to flat plates with a Prandtl number range of 0.6 to 50, although the Prandtl number factor may vary for different flow conditions or other Prandtl number ranges.

In Reynolds' analogy, the coefficient of skin friction, C f, must be considered as a pure value The profile drag, caused by pressure variations around the body, does not influence heat transfer Therefore, the analogy is not applicable when profile drag is factored into C f.

The dimensionless group h/ρc p u ∞ is called the Stanton number It is deﬁned as follows:

Re x Pr The physical signiﬁcance of the Stanton number is

St = h∆T ρc p u ∞ ∆T = actual heat flux to the fluid heat flux capacity of the fluid flow (6.77)

The group St Pr 2/3 was dealt with by the chemical engineer Colburn, who gave it a special symbol: j≡ Colburnj-factor =St Pr 2/3 = Nu x

Does the equation for the Nusselt number on an isothermal ﬂat surface in laminar ﬂow satisfy the Reynolds analogy?

Solution If we rewrite eqn (6.58), we obtain

But comparison with eqn (6.33) reveals that the left-hand side of eqn (6.79) is preciselyC f /2, so the analogy is satisﬁed perfectly Like- wise, from eqns (6.68) and (6.34), we get

The Reynolds-Colburn analogy allows for the direct inference of heat transfer data from shear stress measurements and vice versa This analogy can also be applied to turbulent flow, which presents greater analytical challenges Further exploration of this topic will be addressed in Section 6.8.

Turbulent boundary layers

How much drag force does the air ﬂow in Example 6.5exert on the heat transfer surface?

Solution From eqn (6.80) in Example6.7, we obtain

Re L Pr 1/3 From Example6.5we obtain Nu L , Re L , and Pr 1/3 :

=0.2522 kg/mãs 2 and the force is τ yx A=0.2522(0.5) 2 =0.06305 kgãm/s 2 =0.06305 N

Big whirls have little whirls,

That feed on their velocity.

Little whirls have littler whirls,

And so on, to viscosity.

L F Richardson's insightful verse highlights the complex nature of turbulence in fluids, describing it as a spectrum of interrelated vortices In this dynamic system, kinetic energy from larger vortices is gradually transferred to smaller ones, continuing until the smallest whirls are ultimately subdued by viscous shear stresses.

The next time the weatherman shows a satellite photograph of North

America on the 10:00p.m.news, notice the cloud patterns There will be

The atmosphere is characterized by a hierarchy of vortices, starting with enormous continental-scale vortices that spawn smaller weather-making vortices hundreds of miles wide These larger systems can lead to cyclones and tornadoes, though they often exhibit less violence As these vortices interact with the ground, they break down into smaller whirls, which can be observed as dust devils on the Great Plains, where fewer obstacles exist This process of vortex dissipation continues down to millimeter and micrometer scales, where momentum exchange transitions from turbulence to a more viscous stretching of the fluid.

In turbulent pipe flow at high Reynolds numbers, as illustrated in Fig 6.17, turbulence is characterized by a range of coexisting vortices that vary significantly in size, from a substantial fraction of the pipe radius to micrometer dimensions The spectrum of these vortex sizes changes depending on their location within the pipe, with the size and intensity of vortices near the wall approaching zero due to the fluid velocity also reaching zero at that boundary.

Figure 6.17 illustrates the variation of velocity in a pipe, demonstrating how it changes with both location and time This variation is attributed to turbulent motions superimposed on the average local flow Similar fluctuations can occur in other flow variables, such as temperature (T) and density (ρ) For any given variable, we can express a local time-average value as u ≡ 1.

0 u dt (6.81) whereTis a time that is much longer than the period of typical ﬂuctua- tions 9 Equation (6.81) is most useful for so-calledstationary processes— ones for whichuis nearly time-independent.

If we substituteu=u+u in eqn (6.81), whereuis the actual local velocity and u is the instantaneous magnitude of the ﬂuctuation, we obtain u= 1

9 Take care not to interpret this T as the thermal time constant that we introduced in Chapter 1; we denote time constants are as T §6.7 Turbulent boundary layers 315

Figure 6.17 Fluctuation ofuand other quantities in a turbulent pipe ﬂow.

This is consistent with the fact that u or any other average fluctuation=0 (6.83) since the fluctuations are defined as deviations from the average.

To measure the size or length scale of turbulent vortices, one effective experimental approach involves positioning two velocity-measuring devices in close proximity within a turbulent flow field.

When velocity probes are positioned closely together, their measurements show a strong correlation As the probes are separated, the distance at which their measurements become unrelated indicates the average size of turbulent motions in the fluid.

Prandtl introduced the concept of mixing length, a measure of turbulence that represents the average distance a fluid parcel travels between interactions This concept shares physical significance with the molecular mean free path, although it presents challenges in obtaining a precise experimental measurement.

316 Laminar and turbulent boundary layers §6.7 correlation lengthscale of turbulence But we can still use the concept of to examine the notion of a turbulent shear stress.

Turbulence generates shear stresses through a momentum exchange process similar to that of molecular viscosity In laminar flow, the shear stress can be described by the equation τ yx = (constant) ρC ∂u, highlighting the relationship between kinetic calculations and fluid dynamics.

In turbulent flow, as illustrated in Fig 6.18, we can conceptualize Prandtl's fluid parcels as carriers of x-momentum, rather than focusing on individual molecules This perspective allows us to reformulate equation (6.45), considering the molecular mean free path and the velocity difference for a molecule that has traveled a distance within the mean velocity gradient.

• The shear stress τ yx becomes a ﬂuctuation in shear stress, τ yx , resulting from the turbulent movement of a parcel of ﬂuid

• changes from the mean free path to the mixing length

• C is replaced byv=v+v , the instantaneous vertical speed of the ﬂuid parcel

• The velocity ﬂuctuation, u , is for a ﬂuid parcel that moves a distance through the mean velocity gradient,∂u/∂y It is given by

The derivation of Equation (6.84) can be formally achieved using the Navier-Stokes equation, resulting in a constant value of -1 Consequently, the average fluctuating shear stress is represented as τ yx = -ρ.

Figure 6.18 The shear stress,τ yx , in a laminar or turbulent ﬂow.

Notice that, whileu =v =0, averages of cross products of ﬂuctuations

(such asu v oru 2 ) do not generally vanish Thus, the time average of the ﬂuctuating component of shear stress is τ yx = −ρv u (6.86)

In fluid dynamics, the mean shear stress is linked to the average velocity gradient, represented as à(∂u/∂y), alongside the varying shear stress This relationship aligns with Newton's law of viscous shear, highlighting the fundamental principles governing fluid flow behavior.

Calculating the value of $ v_u $ directly is not straightforward, so we will model it instead It is evident that $ v_u $ approaches zero when the velocity gradient $(\partial u/\partial y)$ is zero, and it increases with an increasing velocity gradient Therefore, we can assume that $ v_u $ is proportional to $(\partial u/\partial y)$ Consequently, the total time-average shear stress $\tau_{yx}$ can be expressed as the sum of mean flow and turbulent contributions, both of which are proportional to the mean velocity gradient, leading to the expression $\tau_{yx} = \alpha \partial u$.

∂y + some other factor, which reﬂects turbulent mixing

In the study of laminar and turbulent boundary layers, the concept of eddy diffusivity for momentum (ε_m) plays a crucial role in analyzing flow fields and heat transfer This eddy diffusivity can be related to the mixing length, particularly when the velocity (u) increases in the y-direction As a fluid parcel descends into a slower-moving region, it experiences a change in velocity proportional to the gradient (∂u/∂y) Conversely, when a parcel ascends into a faster-moving area, the velocity gradient's sign reverses The vertical velocity fluctuations (v) are positive for upward motion and negative for downward motion, indicating that, on average, the magnitudes of u and v for the eddies are comparable Thus, we can anticipate that the relationship between the density (ρ) and eddy diffusivity (ε_m) will reflect this balance.

In equation (6.88), the absolute value is essential to ensure the correct sign when the derivative ∂u/∂y is negative Both the derivatives ∂u/∂y and v u are measurable, allowing us to set the constant in equation (6.88) to one for a quantifiable definition of the mixing length Additionally, this leads to the derivation of the eddy diffusivity expression, represented as ε m = 2.

The primary concern in convective heat transfer is how flowing fluids cool solid surfaces, particularly focusing on turbulence near walls In turbulent boundary layers, steep gradients occur close to the wall, while farther away, larger eddies enhance turbulent mixing This differs from laminar boundary layers, where heat and momentum transfer relies on molecular diffusion rather than vortex motion The crucial processes in turbulent convection happen within mere millimeters of the wall, making the outer boundary layer less significant Analyzing turbulent flow near a wall reveals that time-averaged boundary layer momentum equations yield critical insights into this phenomenon.

In the innermost region of a turbulent boundary layer — y/δ 0.2, where δ is the b.l thickness — the mean velocities are small enough that the convective terms in eqn (6.90a) can be neglected As a result,

∂τ yx /∂y0 The total shear stress is thus essentially constant inyand must equal the wall shear stress: τ w τ yx =ρ (ν+ε m )∂u

Equation (6.91) shows that the near-wall velocity proﬁle does not depend directly uponx In functional form u=fn τ w , ρ, ν, y (6.92)

(Note thatε m does not appear because it is deﬁned by the velocity ﬁeld.)

Heat transfer in turbulent boundary layers

The turbulent thermal boundary layer consists of distinct inner and outer regions, with the inner part exhibiting weaker turbulent mixing where heat transport is primarily governed by conduction As one moves further from the wall, a logarithmic temperature profile emerges, and in the outermost sections of the boundary layer, turbulent mixing becomes the predominant mechanism for heat transport.

The boundary layer concludes where turbulence dissipates and uniform free-stream conditions are established, leading to identical thermal and momentum boundary layer thicknesses While this might imply that the Prandtl number has no influence on turbulent heat transfer, it actually affects the sublayers near the wall, where molecular viscosity and thermal conductivity govern the transport of heat and momentum.

The Reynolds-Colburn analogy for turbulent ﬂow

The eddy diffusivity for momentum was introduced by Boussinesq [6.11] in 1877 It was subsequently proposed that Fourier’s law might likewise be modified for turbulent flow as follows: q= −k∂T

∂y − another constant, which reﬂects turbulent mixing

∂y whereT is the local time-average value of the temperature Therefore, q= −ρc p (α+ε h ) ∂T

∂y (6.103) whereε h is called theeddy diﬀusivity of heat This immediately suggests yet another deﬁnition: turbulent Prandtl number, Pr t ≡ ε m ε h (6.104) §6.8 Heat transfer in turbulent boundary layers 323

Equation (6.103) can be written in terms of ν andε m by introducing Pr and Pr t into it Thus, q= −ρc p ν

Before trying to build a form of the Reynolds analogy for turbulent

ﬂow, we must note the behavior of Pr and Pr t :

The physical property of a fluid known as Prandtl number (Pr) is theoretically and practically close to unity for ideal gases However, for liquids, the Prandtl number can vary significantly, often differing from unity by several orders of magnitude.

The property of Pr t primarily relates to the flow field rather than the fluid itself, typically maintaining a numerical value close to unity, usually within a factor of 2 While it fluctuates throughout the boundary layer, nonmetallic fluids often exhibit a Pr t value around 0.85.

The time-average boundary-layer energy equation is similar to the time-average momentum equation [eqn (6.90a)] ρc p u∂T

In the near wall region, the convective terms are negligible, indicating that the heat flux remains constant in the y-direction and is equal to the wall heat flux, represented as $ q = q_w = -\rho c_p \nu $.

We may integrate this equation as we did eqn (6.91), with the result that

The constantAdepends upon the Prandtl number It reﬂects the thermal resistance of the sublayer near the wall As was done for the constant

Bin the velocity proﬁle, experimental data or numerical simulation may be used to determineA(Pr)[6.12,6.13] For Pr≥0.5,

To obtain the Reynolds analogy, we can subtract the dimensionless log-law, eqn (6.99), from its thermal counterpart, eqn (6.108):

In the outer part of the boundary layer,T (y)T ∞ andu(y)u ∞ , so

We can eliminate the friction velocity in favor of the skin friction coeﬃ- cient by using the deﬁnitions of each: u ∗ u ∞ =

C f =A(Pr)−B (6.110d) Rearrangment of the last equation gives q w

The lefthand side is simply the Stanton number, St=h

(ρc p u ∞ ) Upon substitutingB=5.5 and eqn (6.109) forA(Pr), we obtain the Reynolds-

Colburn analogy for turbulent ﬂow:

The results can be applied using equation (6.102) for the friction coefficient $C_f$ or with corresponding data to determine the local heat transfer coefficient in a turbulent boundary layer This equation is effective for both uniform wall temperature ($T_w$) and uniform heat flux ($q_w$) The thin, near-wall region of the boundary layer predominantly influences thermal resistance, and this layer is largely independent of the flow's upstream history.

Equation (6.111) applies to smooth walls with minimal or zero pressure gradients, where the term 12.8(Pr 0.68 −1) in the denominator represents the thermal resistance of the sublayer However, in the presence of rough walls, this term must be adjusted to reflect the impact of surface roughness, as detailed in Section 7.3.

Other equations for heat transfer in the turbulent b.l.

While equation (6.111) accurately predicts the local value of h in a turbulent boundary layer, several simplified approximations have been proposed in the literature Notably, when Prandtl numbers are close to unity and Reynolds numbers are slightly above transition, the laminar flow Reynolds-Colburn analogy can be effectively applied.

The best exponent for the Prandtl number in such an equation actually depends upon the Reynolds and Prandtl numbers For gases, an exponent of−0.4 gives somewhat better results.

A simplified expression for the drag coefficient $ C_f $ allows for a broader approximation in turbulent flow over a smooth flat plate, particularly in the low Reynolds number range, as demonstrated by Schlichting.

With this Reynolds number dependence, Žukauskas and coworkers [6.14,

Pr − 0.57 , 0.7≤Pr≤380 (6.113) so that when eqn (6.112) is used to eliminateC f

Somewhat better agreement with data, for 2×10 5 Re x 5×10 6 , is obtained by adjusting the constant [6.15]:

The average Nusselt number for uniform T w is obtained from eqn.

326 Laminar and turbulent boundary layers §6.8 where we ignore the fact that there is a laminar region at the front of the plate Thus,

This equation may be used for either uniformT w or uniformq w , and for

A ﬂat heater with a turbulent b.l on it actually has a laminar b.l be- tweenx =0 andx =x trans, as is indicated in Fig.6.4 The obvious way to calculatehin this case is to write h= 1

(6.117) wherex trans =(ν/u ∞ )Retrans Thus, we substitute eqns (6.58) and (6.114) in eqn (6.117) and obtain, for 0.6 Pr 50,

Re 0.8 trans −17.95 Pr 0.097 (Re trans ) 1/2 ;

If Re L Retrans, this result reduces to eqn (6.116).

Whitaker [6.16] suggested setting Pr 0.097 ≈ 1 and Retrans ≈ 200,000 in eqn (6.118):

The expression has been adjusted to reflect the variability of liquid viscosity using the factor (à ∞ /à w) 1/4, where à ∞ represents the viscosity at the free-stream temperature (T ∞) and à w denotes the viscosity at the wall temperature (T w) Other physical properties should be assessed at T ∞ When applying equation (6.119) for predicting heat transfer in a gaseous flow, the viscosity-ratio correction term should be omitted, and properties must be evaluated at the film temperature This is due to the fact that gas viscosity increases with temperature, making the correction inaccurate.

Finally, it is important to remember that eqns (6.118) and (6.119) should be used only when Re L is substantially above the transitional value. §6.8 Heat transfer in turbulent boundary layers 327

A correlation for laminar, transitional, and turbulent ﬂow

The previous equations fail to address heat transfer in the extended transition region, as both equations (6.118) and (6.119) rely on certain assumptions.

ﬂow abruptly passes from laminar to turbulent at a critical value of x, and we have noted in the context of Fig.6.4that this isnot what occurs.

The location of the transition depends upon such variables as surface roughness and the turbulence, or lack of it, in the stream approaching the heater.

Churchill [6.17] suggests correlating anyparticular set of data with

The value of φ u typically ranges between 10^5 and 10^7 and should be adjusted to fit specific data sets In systems characterized by minimal disturbances, φ u tends to be higher, while in more chaotic environments, it is generally lower An estimate for φ u can be derived from the Reynolds number at the conclusion of the turbulent transition region, represented as φ u ≈ φ(Re x = Re u).

The equation is for uniformT w , but it may be used for uniform q w if the constants 0.3387 and 0.0468 are replaced by 0.4637 and 0.02052, respectively.

Churchill also gave an expression for the average Nusselt number:

(6.120c) where φ is deﬁned as in eqn (6.120b), using Re L in place of Re x , and φ um ≈ 1.875φ(Re L = Re u ) This equation may be used for either uni- formT w or uniformq w

Equations (6.120a) and (6.120c) offer the benefit of accurately predicting heat transfer across various flow regimes, including laminar, transitional, and turbulent, once the parameters φ u or φ um are established.

As a ship departs from the river's mouth, where the water temperature is 24°C, it enters the ocean with a significantly colder temperature of 10°C The ship's hull features a sharp and relatively flat forward end When the vessel moves at a speed of 5 knots, the coefficient of friction (C_f) can be calculated at a distance of 1 meter from the forward edge of the hull.

Solution If we assume that the hull’s heat capacity holds it at the river temperature for a time, we can take the properties of water at

T f =(10+24)/2=17 ◦ C:ν=1.085ì10 − 6 m 2 /s,k=0.5927 W/mãK, ρ=998.8 kg/m 3 ,c p =4187 J/kgãK, and Pr=7.66.

One knot equals 0.5144 m/s, so u ∞ = 5(0.5144) = 2.572 m/s. Then, Re x =(2.572)(1)/(1.085×10 − 6 )=2.371×10 6 , indicating that the ﬂow is turbulent at this location.

We have given several diﬀerent equations for C f in a turbulent boundary layer, but the most accurate of these is eqn (6.102):

T b , n = 0.11 in eqn (7.44) Thus, with eqn (7.41) we have

0.12 =8,907 W/m 2 K The corrected friction factor, with eqn (7.46), is f =(0.0128) (7−1.74)/6=0.0122

Heat transfer surface viewed as a heat exchanger

and Lyon [7.23] recommends the following equation, shown in Fig.7.8:

In both these equations, properties should be evaluated at the average of the inlet and outlet bulk temperatures and the pipe ﬂow should have

L/D > 60 and Pe D > 100 For lower Pe D , axial heat conduction in the liquid metal may become signiﬁcant.

While equations (7.53) and (7.54) may be accurate for pure liquids, it is important to recognize that the liquid metals used in practice are rarely pure Lubarsky and Kaufman [7.20] highlighted this issue by drawing a line through the majority of the data presented in Fig 7.8.

Utilizing equation (7.55) for constant q w is generally less favorable compared to equation (7.54) It is advisable to employ this approach when a more conservative estimate is preferred.

7.4 Heat transfer surface viewed as a heat exchanger

In analyzing fluid flow through a pipe with a consistent wall temperature, we can effectively predict the heat transfer coefficient (h) across various conditions However, determining the net heat transfer for a pipe of a specified length becomes complex due to the variation of the bulk temperature (T_b) along the pipe's length.

A section of pipe acts as a heat exchanger with an overall heat transfer coefficient, U, between the wall and the bulk fluid To determine the required pipe surface area for increasing the bulk temperature from T_b in to T_b out, specific calculations can be performed.

T b out −T b in =hA(LMTD) or

By the same token, heat transfer in a duct can be analyzed with the effectiveness method (Sect.3.3) if the exiting ﬂuid temperature is unknown.

368 Forced convection in a variety of conﬁgurations §7.4

In scenarios where the boundary temperature $ T_b $ is unknown, we can establish an energy balance at any cross-section, as demonstrated in equation (7.8) This balance can be expressed as $ dQ = q_w P dx = h_P (T_w - T_b) dx = \dot{m}_P dT_b $.

Integration can be done fromT b (x=0)=T b in toT b (x=L)=T b out

We recognize in this the deﬁnition ofhfrom eqn (7.27) Hence, hP L mc˙ p = −ln

T w −T b in which can be rearranged as

This equation is applicable for both laminar and turbulent flow, allowing for the calculation of bulk temperature variation by substituting T_b out with T_b (x), replacing L with x, and appropriately adjusting h.

The left-hand side of eqn (7.57) is the heat exchanger eﬀectiveness.

On the right side, we substitute U with h, noting that P L equals A, which represents the exchanger surface area We define C min as mc˙ p Given that T w is uniform, the associated stream must possess a significantly large capacity rate.

When the ratio of C min to C max is zero, we define the argument of the exponential as NTU = U A/C min This leads to the equation ε = 1 - exp(-NTU), which can also be derived from equations (3.20) or (3.21) by assuming C min/C max = 0 A heat exchanger characterized by one isothermal stream, resulting in C min/C max = 0, is referred to as a single-stream heat exchanger.

Equation7.57 applies to ducts of any cross-sectional shape We can cast it in terms of thehydraulic diameter, D h = 4A c /P, by substituting §7.4 Heat transfer surface viewed as a heat exchanger 369 ˙ m=ρu av A c :

For a circular tube, withA c =π D 2 /4 andP =π D,D h =4(π D 2 /4)

= D To use eqn (7.59) for a noncircular duct, of course, we will need the value ofhfor its more complex geometry We consider this issue in the next section.

Air at 20 ◦ C is fully thermally developed as it ﬂows in a 1 cm I.D pipe.

The average velocity is 0.7 m/s If the pipe wall is at 60 ◦ C , what is the temperature 0.25 m farther downstream?

1.70×10 − 5 =412 The ﬂow is therefore laminar, so

Heat transfer coeﬃcients for noncircular ducts

While our previous discussions have primarily centered on flows within circular tubes, it's important to recognize that various cross-sectional shapes can also be encountered For instance, air may flow through rectangular passages formed by heat exchanger fins, or through irregular openings created by fluid moving between objects In these scenarios, the fundamental concepts from earlier sections remain relevant; however, the Nusselt numbers specific to circular tubes are not applicable for calculating heat transfer rates in these alternative geometries.

The hydraulic diameter, which was introduced in connection with eqn (7.59), provides a basis for approximating heat transfer coeﬃcients in noncircular ducts Recall that the hydraulic diameter is deﬁned as

The hydraulic diameter, defined as the ratio of the cross-sectional area (A_c) to the wetted perimeter (P) of a passage, plays a crucial role in determining heat transfer coefficients in turbulent flow conditions This measurement effectively represents the fluid area per unit length of wall, and it significantly influences convection resistance, particularly within the sublayer adjacent to the wall Across various duct shapes, this ratio can determine the heat transfer coefficient with an accuracy of approximately ±20%.

In fully-developed laminar flow, the heat transfer coefficient is influenced by the duct's shape, making the hydraulic diameter insufficient on its own for defining it However, the hydraulic diameter serves as a suitable characteristic length for organizing laminar Nusselt numbers.

Figure 7.9 Flow in a noncircular duct. §7.5 Heat transfer coeﬃcients for noncircular ducts 371

The inclusion of the factor of four in the definition of hydraulic diameter (D h) guarantees it accurately reflects the diameter of a circular tube, as demonstrated by the relationship D h = D for a circular tube with diameter D Additionally, for a rectangular duct with width a and height b, the hydraulic diameter is calculated as D h = 4ab.

2a+2b = 2ab a+b (7.61a) an annular duct of inner diameterD i and outer diameterD o

=(D o −D i ) (7.61b) and, for very wide parallel plates, eqn (7.61a) witha b gives two parallel plates a distancebapart D h =2b (7.61c)

Turbulent ﬂow in noncircular ducts

When calculating turbulent heat transfer coefficients and bulk temperature changes, we can cautiously substitute the hydraulic diameter (D h) for the circular tube diameter (D) This substitution is particularly relevant for the Reynolds number, which is essential for determining the friction factor (f) and Nusselt number (Nu D h) using circular tube formulas It is important to note that the mass flow rate and bulk velocity should be derived from the actual cross-sectional area, which typically does not equal π D h² /4.

Problem7.46) The following example illustrates the procedure.

An air duct carries chilled air at an inlet bulk temperature ofT b in =

17 ◦ C and a speed of 1 m/s The duct is made of thin galvanized steel, has a square cross-section of 0.3 m by 0.3 m, and is not insulated.

A length of the duct 15 m long runs outdoors through warm air at

T ∞ =37 ◦ C The heat transfer coeﬃcient on the outside surface, due to natural convection and thermal radiation, is 5 W/m 2 K Find the bulk temperature change of the air over this length.

Solution The hydraulic diameter, from eqn (7.61a) witha=b, is simply

Using properties of air at the inlet temperature (290 K), the Reynolds number is

The Reynolds number for turbulent flow in a noncircular duct is approximately 19,011, indicating that the flow is turbulent This value aligns with the common threshold of around 2300 for turbulent transition in circular tubes The friction factor can be calculated using the equation $ f = 1.82 \log_{10}(19,011) - 1.64 $.

=0.02646 and the Nusselt number is found with Gnielinski’s equation, (7.43)

(0.713) 2/3 −1" =49.82 The heat transfer coeﬃcient is h=Nu D h k

To determine the bulk temperature change, we must consider the thermal resistances involved in heat transfer through a thin metal duct While the duct wall itself has minimal thermal resistance, the convection resistance from the surrounding air must be taken into account Heat transfers from the ambient air at temperature T ∞ to the duct wall via the outside heat transfer coefficient, and then from the duct wall to the flowing air through the inside heat transfer coefficient This process effectively involves two resistances in series, transitioning from the fixed temperature T ∞ to the increased temperature T b An overall heat transfer coefficient can be utilized to simplify the analysis of these series resistances.

U based on inside area to be

We then adapt eqn (7.59) by replacinghbyU andT w byT ∞ :

=0.3165 The outlet bulk temperature is therefore

T b out =[17+(37−17)(0.3165)] ◦ C=23.3 ◦ C §7.5 Heat transfer coeﬃcients for noncircular ducts 373

Substituting the hydraulic diameter (D h) for the diameter (D) in turbulent circular tube formulas yields results that are typically accurate within ±20%, frequently achieving ±10% accuracy However, less accurate results occur in duct cross-sections with sharp corners, like acute triangles To enhance precision, specialized equations for "effective" hydraulic diameters have been created for specific geometries, potentially improving accuracy to 5% or 10%.

When a section of a duct's cross-section is heated, such as one wall of a rectangular duct, the hydraulic diameter is calculated using the entire wetted perimeter rather than just the heated area This scenario often arises in annular ducts, where the inner tube serves as a heating element, leading to one-sided or uneven heating.

The hydraulic diameter method generally forecasts the heat transfer coefficient on the outer tube with an accuracy of ±10%, regardless of the heating configuration In contrast, the heat transfer coefficient on the inner surface is significantly influenced by both the diameter ratio and the specific heating arrangement.

For that surface, the hydraulic diameter approach is not very accurate, especially ifD i D o ; other methods have been developed to accurately predict heat transfer in annular ducts (see [7.3] or [7.8]).

Laminar ﬂow in noncircular ducts

Laminar velocity profiles in noncircular ducts evolve similarly to those in circular tubes, resulting in fully developed profiles that typically exhibit a paraboloidal shape.

ﬂow between parallel plates located aty=b/2 andy= −b/2, u u av = 3 2

The bulk velocity for the system is represented by equation (7.62), which parallels equation (7.15) for a circular tube, despite differences in constants and coordinates Additionally, examining the temperature profiles between parallel plates reveals that constant Nusselt numbers can be articulated in relation to the hydraulic diameter under different boundary conditions.

7.541 for fixed plate temperatures 8.235 for fixed flux at both plates 5.385 one plate fixed flux, one adiabatic

Some other cases are summarized in Table7.4 Many more have been considered in the literature (see, especially, [7.5]) The latter include

Table 7.4 Laminar, fully developed Nusselt numbers based on hydraulic diameters given in eqn (7.61)

The study explores parallel plates with dimensions ranging from 7.541 to 8.235, featuring various wall boundary conditions and an array of cross-sectional shapes, including triangles, circular sectors, trapezoids, rhomboids, hexagons, limaçons, and crescent moons It emphasizes the importance of boundary conditions in small ducts, where the conduction resistance of the tube wall may significantly influence convective resistance, leading to temperature or flux variations around the tube perimeter This variation can notably impact the laminar Nusselt number For instance, the rectangular duct values in Table 7.4, which assume uniform temperature around the tube perimeter, may not hold true for ducts with considerable conduction resistance, such as copper ducts heated at a constant rate Additionally, laminar entry length formulas for noncircular ducts are referenced from Shah and London.

Heat transfer during cross ﬂow over cylinders

Understanding heat transfer in cross flow is enhanced by examining fluid flow patterns around a cylinder with fluid flowing perpendicular to it As the Reynolds number (Re) increases from below 5 to nearly 10^7, the flow develops distinct characteristics Notably, the transition between different flow patterns occurs in a continuous manner, illustrating the complexity of heat transfer dynamics in this configuration.

Figure 7.10 Regimes of ﬂuid ﬂow across circular cylinders [7.24].

The relationship between the Strouhal and Reynolds numbers for circular cylinders reveals that the flow field experiences increasing disorder with each transition Interestingly, at the highest Reynolds number values, this disorder transforms back into a state of order.

The vortex-shedding frequency, denoted as f v, serves as a key indicator of the complexity within the flow field Dimensional analysis reveals that this frequency is related to a dimensionless parameter known as the Strouhal number, Str, which is influenced by the Reynolds number of the flow.

Figure7.11deﬁnes this relationship experimentally on the basis of about

The Strouhal numbers remain slightly above 0.2 across a wide range of Reynolds numbers (Re D), indicating that the vortex-shedding frequency increases almost linearly with velocity behind a given object.

On a breezy day, step outside and find a tall tree with a straight trunk or the vertical shaft of a water tower to observe the effects of heat transfer during cross flow over cylindrical structures.

Figure 7.12 Giedt’s local measurements of heat transfer around a cylinder in a normal cross ﬂow of air.

To estimate the vortex-shedding frequency, position a finger a couple of diameters downstream and approximately one radius off-center Utilize Strouhal number 0.21 to calculate the free stream velocity, u ∞ Assess whether the obtained value of u ∞ is reasonable.

Vortex shedding significantly complicates the heat removal process, as illustrated by Giedt’s data in Fig 7.12 This data demonstrates the variations in heat removal resulting from the continuously fluctuating motion of the fluid behind the cylinder.

Forced convection varies across different configurations, with the Nusselt number (Nu D) demonstrating a notable minimum at 110°C when the Reynolds number (Re D) is 71,000 Conversely, Nu D reaches its maximum at the same temperature under these conditions.

With a value of Re D = 140,000, direct predictions using the b.l methods discussed in Chapter 6 are not feasible Nonetheless, significant insights can still be gleaned from the data by employing relationships of this nature.

The comprehensive research conducted by Churchill and Bernstein [7.26] significantly advances the correlation of heat transfer data from cylinders, effectively encompassing the full spectrum of available data.

This expression underpredicts most of the data by about 20% in the range

20,000 < Re D 0 \) and $ N_{2,s} < 0 $ as depicted in Fig 11.16 These findings are a direct result of the steady-state conservation of species.

Recalling the general expression forN i , eqn (11.25), and introducing §11.7 Steady mass transfer with counterdiﬀusion 651

Fick’s law, eqn (11.34), we write

The termxN 1 represents vertical convective transport induced by mass transfer The total mole ﬂux, N, must also be constant at itss-surface value; by eqn (11.23), this is

Substituting this result into eqn (11.81), we obtain a diﬀerential equation forx 1: cD 12 dx 1 dy =N s x 1 −N 1,s (11.83)

In this equation,x 1is a function ofy, theN’s are constants, andcD 12 depends on temperature and pressure If the temperature and pressure are constant, so too iscD 12 Integration then yields

We need to ﬁx the constant and the two mole ﬂuxes,N 1,s andN s To do this, we apply the boundary conditions at either end of the layer The

ﬁrst boundary condition is the mole fraction of species 1 at the bottom of the layer x 1 =x 1,s at y=0 and it requires that constant = −ln(N s x 1,s −N 1,s ) (11.85) so

The second boundary condition is the mole fraction at the top of the layer x 1 =x 1,e at y=L

652 An introduction to mass transfer §11.7 which yields

The final boundary condition involves the ratio of N1,s to Ns, which may not equal one due to the potential for species 2 to penetrate the bottom layer This ratio is contingent upon the specific circumstances of each problem, as illustrated by the subsequent examples.

Find an equation for the evaporation rate of the liquid in the Stefan tube described at the beginning of this section.

In the evaporation process, species 1 represents the evaporating vapor, while species 2 is the stationary gas Only the vapor transfers through the surface, as the gas is not significantly absorbed into the already saturated liquid, resulting in N2,s = 0 and N s = N1,s = N vapor,s, which indicates the evaporation rate of the liquid The surface of evaporation is located just above the liquid, and the mole fraction of the evaporating liquid can be derived from solubility data; for instance, Raoult’s law can be applied when the gas is relatively insoluble in the liquid The e-surface is positioned at the tube's mouth, where the gas flow may contain a minimal concentration of vapor, typically close to zero The ratio N1,s / N s equals unity, confirming the rate of evaporation.

What is the evaporation rate in the Stefan tube if the gas is bubbled up to the liquid surface at some ﬁxed rate,N gas?

Solution Again, N 1,s = N vapor,s is the evaporation rate However, the total mole ﬂux is

N s =N gas +N 1,s §11.7 Steady mass transfer with counterdiﬀusion 653

(11.90) This equation ﬁxesN 1,s , but it must be solved iteratively.

Once we have found the mole ﬂuxes, we may compute the concentration distribution,x 1 (y), using eqn (11.86): x 1 (y)= N 1,s

Alternatively, we may eliminate N s between eqns (11.86) and (11.87) to obtain the concentration distribution in a form that depends only on the ratioN 1,s /N s : x 1 −N 1,s /N s x 1,s −N 1,s /N s = x 1,e −N 1,s /N s x 1,s −N 1,s /N s y/L

Find the concentration distribution of water vapor in a helium–water

Stefan tube at 325 K and 1 atm The tube is 20 cm in length Assume the helium stream at the top of the tube to have a mole fraction of water of 0.01.

In this solution, water is identified as species 1 and helium as species 2 The vapor pressure of liquid water closely approximates the saturation pressure corresponding to its temperature According to the steam tables, the vapor pressure (p_v) is calculated to be 1.341×10^4 Pa, leading to a mole fraction of water (x_1,s) of 1.341×10^4 Pa.

We use eqn (11.14) to evaluate the mole concentration in the tube: c = 101,325 8314.5(325) =0.03750 kmol/m 3

From eqn (11.42) we obtainD 12 (325 K,1 atm)=1.067×10 − 4 m 2 /s.

Then eqn (11.89) gives the molar evaporation rate:

654 An introduction to mass transfer §11.8

This corresponds to a mass evaporation rate: n 1,s =4.754ì10 − 5 kg/m 2 ãs The concentration distribution of water vapor [eqn (11.91)] is x 1 (y)=1−0.8677 exp(0.6593y) whereyis expressed in meters.

Stefan tubes are commonly utilized for measuring mass transfer coefficients by tracking liquid level changes over time and applying equation (11.89) for D12 However, these measurements can be affected by various experimental errors, such as the latent heat of vaporization cooling the gas mixture at the interface, leading to temperature gradients within the tube Additionally, vortices at the tube's top, where it interfaces with the gas stream, may induce further mixing, while density gradients can result in buoyant circulation For a comprehensive discussion on these errors and alternative measurement methods, refer to the work of Marrero and Mason [11.7].

This section addresses a problem that can also be approached using a mass-based solution, assuming a constant value of ρD 12, as illustrated in Problems 11.33 and 11.34 This mass-based approximation is crucial for analyzing high-rate convective mass transfer in the subsequent section.

Mass transfer coeﬃcients at high rates of mass transfer 654

In Section 11.6, we established an analogy between heat and mass transfer, enabling the calculation of mass transfer coefficients under low mass transfer rates, provided that the velocity field remains unaffected by mass transfer and the species involved are dilute However, when these conditions are not satisfied, the mass transfer coefficient may differ from the analogous value, potentially increasing or decreasing by a few percent to an order of magnitude or more, influenced by the concentrations of the diffusing species Additionally, convective transport can significantly impact the overall mass flux alongside the diffusive transport represented by the mass transfer coefficient.

Figure 11.17 The mass concentration boundary layer.

This section explores mass convection where the transferred species influences the velocity field and is not always in a dilute state We begin by defining the driving force for mass transfer, which determines the overall mass flux from the wall Furthermore, we establish a relationship between the mass transfer coefficient at high mass transfer rates and that at low mass transfer rates.

The mass transfer driving force

Figure 11.17 illustrates a boundary layer over a wall experiencing net mass transfer (n s ≡ m˙) of various species perpendicular to the surface Specifically, we will concentrate on species i, which has a concentration of m i,e in the free stream and m i,s at the wall.

The mass ﬂux ofileaving the wall is obtained from eqn (11.21): n i,s =m i,s m˙ +j i,s (11.93)

We seek to obtain ˙m in terms of the concentrationsm i,s andm i,e As before, we deﬁne the mass transfer coeﬃcient for speciesi,g m,i (kg/m 2 ãs), as g m,i =j i,s m i,s −m i,e (11.94) Thus, n i,s =m i,s m˙ +g m,i m i,s −m i,e (11.95)

The mass transfer coefficient is again based on thediffusivetransfer from the wall; however, it may now differ considerably from the value for low- rate transport.

11 In this context, we denote the total mass ﬂux through the wall as ˙ m , rather than n s , so as to be consistent with other literature on the subject.

Equation (11.95) may be rearranged as ˙ m =g m,i m i,e −m i,s m i,s −n i,s /m˙

The total mass flux of all species through the wall, denoted as ˙m, is expressed as the product of the mass transfer coefficient and the concentration ratio This concentration ratio is referred to as the mass transfer driving force for species i.

The ratio of mass ﬂuxes in the denominator is called themass fraction in the transferred state, denoted asm i,t : m i,t ≡n i,s /m˙ (11.98)

The mass fraction in the transferred state refers to the portion of the total mass flux, ˙m, that consists of species i Unlike traditional mass fractions, this value can vary widely, reflecting its unique role in the mass transfer process.

−∞to+∞, depending on the relative magnitudes of ˙m andn i,s If, for example, n 1,s −n 2,s in a binary mixture, then ˙m is very small and bothm 1,t andm 2,t are very large.

Equations (11.96), (11.97), and (11.98) provide a formulation of mass transfer problems in terms of the mass transfer coeﬃcient,g m,i , and the driving force for mass transfer,B m,i : m˙ =g m,i B m,i (11.99) where

Mass transfer rates in a mixture can be calculated using any arbitrary species, yielding consistent results across different species This principle is clearly demonstrated in binary mixtures, where the mass transfer coefficients for each species are equal, indicating that g m,1 equals g m,2 and B m,1 equals B m,2.

In many situations, only one species is transferred through the wall.

If speciesiis the only one passing through thes-surface, thenn i,s =m˙ , so thatm t,i =1 The mass transfer driving force is simply

In all the cases described in Section11.6, only one species is transferred.

A pan filled with water at a temperature of 75 °C is exposed to an air stream containing 5% water vapor The average mass transfer coefficient for water over the surface of the pan is denoted as gm,H2O This setup illustrates the interaction between the hot water and the surrounding air, highlighting the significance of mass transfer in thermal processes Understanding these dynamics is crucial for optimizing heat transfer efficiency in various applications.

0.0170 kg/m 2 ãs and the pan has a surface area of 0.04 m 2 , what is the evaporation rate?

Under normal conditions, only water vapor can pass through the liquid surface, as air is not significantly absorbed by water Therefore, we utilize equation (11.101) to determine the mass transfer driving force According to steam table data, the saturation pressure of water at 75 °C is 38.58 kPa, leading to a water vapor mole fraction of x H2O,s = 38.58/101.325 = 0.381 By substituting this value into equation (11.67), we find that the mass flow rate of water vapor, m H2O,s, is 0.277.

The eﬀect of mass transfer rates on the mass transfer coeﬃcient

Finding the mass transfer coefficient, g m,i, remains a challenge, and while we could apply methods similar to those used for determining heat transfer coefficients, such as solving momentum and species equations or correlating mass transfer data, these methods are often more complex due to the interaction between flow fields and mass transfer rates Simple solutions for mass transfer problems are not readily available Therefore, we utilize a widely accepted approximate method that enables us to calculate g m,i by adjusting the low-rate mass transfer coefficient with a correction for the effects of finite mass transfer rates.

To isolate the effect of ˙m on the mass transfer coefficient, we first define the mass transfer coefficient at zero net mass transfer,g ∗ m,i : g m,i ∗ ≡ lim ˙ m → 0 g m,i

The mass transfer coefficient, denoted as g ∗ m,i, represents the value for low mass transfer rates derived from the analogy between heat and mass transfer While g m,i is influenced by the rate of mass transfer, g ∗ m,i is solely determined by the flow configuration and the physical properties of the system.

In a boundary layer, fluid near the wall experiences a reduction in speed due to the no-slip condition, which can be modeled as a stagnant film—a stationary layer of fluid without horizontal gradients This approach effectively represents high-rate mass transfer effects on the local concentration boundary layer thickness, denoted as δ c.

The finite mass transfer rate across the film indicates the presence of vertical convection and counter-diffusion effects The stagnant film configuration, as illustrated in Fig 11.18, mirrors the setup discussed previously in Fig 11.16 Consequently, the solution derived earlier, represented by equation (11.88), accurately describes the mass transfer rate across the stagnant film while considering the influence of vertical convective transport.

In current mass-based analyses, it is practical to utilize the mass-based equivalent of the mole-based equation This equivalent can be expressed as m˙ = ρD im δ c ln.

1+ m i,e −m i,s m i,s −n i,s /m˙ which we may recast in the following, more suggestive form ˙ m = ρD im δ c ln(1+B m,i )

Comparing this equation with eqn (11.99), we see that g m,i = ρD im δ c ln(1+B m,i )

B m,i and when ˙m approaches zero, g m,i ∗ = lim ˙ m → 0 g m,i = lim

The optimal value of g m,i ∗ (or δ c) can be determined by solving the related low-rate mass transfer problem, utilizing the principles of heat and mass transfer analogy This value of g m,i ∗ is essential as it defines the effective concentration boundary layer thickness, δ c.

The blowing factor, defined as [ln(1+B m,i )]/B m,i, quantifies the impact of mass transfer on the velocity field When B m,i is greater than zero, it indicates mass flow away from the wall, known as blowing, which results in a blowing factor that is a positive value less than one, thereby reducing g m,i Conversely, when B m,i is less than zero, it signifies mass flow toward the wall, referred to as suction, leading to a blowing factor that is a positive value greater than one.

Simultaneous heat and mass transfer

11.9 Simultaneous heat and mass transfer

Numerous critical engineering processes involving mass transfer occur concurrently with heat transfer, exemplified by equipment such as cooling towers, dryers, and combustors that intricately link these two essential processes.

Coupling occurs when temperature-dependent mass transfer processes lead to heat exchange at a surface During evaporation, for instance, latent heat is absorbed at the liquid surface as vapor forms, cooling the surface and decreasing vapor pressure, which in turn reduces the evaporation rate Similarly, in carbon oxidation, heat is released during the oxidation process, and the oxidation rate is influenced by temperature The interplay between convective cooling and the reaction rate ultimately dictates the surface temperature of the burning carbon.

Simultaneous heat and mass transfer processes can be categorized into two distinct groups: low-rate and high-rate transfers In low-rate mass transfer scenarios, the impact on the velocity field is minimal, allowing heat transfer rates to be calculated independently of mass transfer Conversely, high-rate mass transfer necessitates a correction to the heat transfer coefficient to account for the effects of counterdiffusion, a crucial consideration in accurately predicting heat transfer rates in such systems.

Heat transfer at low rates of mass transfer

Low-rate heat and mass transfer is commonly observed in the evaporation of water into air at low or moderate temperatures, a typical example being the process utilized in an Asling psychrometer This device is specifically designed to measure air humidity, illustrating a fundamental application of heat and mass transfer principles in everyday measurement tools.

A sling psychrometer consists of a wet-bulb thermometer, which has a wet cloth wrapped around its bulb, and a dry-bulb thermometer mounted on a swivel handle These two thermometers are swung in a rotary motion until they stabilize, allowing for accurate measurement of humidity levels.

The wet-bulb thermometer cools as the latent heat from vaporized water is released, reaching a temperature where the cooling effect of evaporation equals the warming effect of the surrounding air This specific temperature is known as the wet-bulb temperature.

The wet bulb temperature measured by a sling psychrometer is closely linked to the moisture content in the air Typically, the highest ambient air temperatures we experience are relatively low, resulting in a minimal mass transfer rate To evaluate this, we can calculate an upper limit on the mass transfer of water vapor (B m,H2O) by using extreme conditions that enhance evaporation, specifically by assuming a maximum air temperature of 120°F and minimal humidity levels.

At 120°F, the vapor pressure on the wet bulb is lower than the saturation pressure due to the cooling effect of evaporation This relationship can be quantified as the ratio of the saturation pressure of water at this temperature (11,671 Pa) to the atmospheric pressure (101,325 Pa), resulting in a value of 0.115.

The wet-bulb temperature in air-water systems closely approximates the adiabatic saturation temperature, which is the temperature attained when an air-water mixture reaches saturation through the addition of water vapor without heat input This temperature is a key thermodynamic property of the air-water combination According to our established criterion for low-rate mass transfer, as indicated in equation (11.74), this relationship is satisfied, reinforcing the principles of simultaneous heat and mass transfer.

Alternatively, in terms of the blowing factor, eqn (11.104), ln(1+B m,H 2 O )

This means that under the worst normal circumstances, the low-rate theory should deviate by only 4 percent from the actual rate of evaporation.

To analyze the energy balance on the wick, we consider the u and s surfaces, as illustrated in Fig 11.20 At a steady temperature, there is no heat transfer through the u-surface into the wet bulb; however, liquid water travels through this surface to the wick's surface, where it evaporates By applying an energy balance between the u and s surfaces, we account for the enthalpy of water vapor leaving the system, represented as n H₂O,s hˆH₂O,s.

− q s heat convected to the wet bulb

= n H 2 O,u hˆH 2 O,u enthalpy of liquid water arriving

Since mass is conserved, n H 2 O,s = n H 2 O,u , and because the enthalpy change results from vaporization, ˆh H 2 O,s −hˆH 2 O,u =h fg Hence, n H 2 O,s h fg T wet-bulb =h(T e −T wet-bulb )

For low-rate mass transfer, n H 2 O,s j H 2 O,s , and this equation can be written in terms of the mass transfer coeﬃcient g m,H 2 O m H 2 O,s −m H 2 O,e h fg

The heat and mass transfer coeﬃcients depend on the geometry and

The wet-bulb temperature measured by a psychrometer is influenced by the specific device used, highlighting the interdependence of the two coefficients due to the relationship between heat and mass transfer In the context of forced convection in cross flow, the heat transfer coefficient can be expressed in the form hD k = CRe a Pr b, as discussed in Chapter 7.

666 An introduction to mass transfer §11.9 where C is a constant, and typical values of a and b are a 1/2 and b1/3 From the analogy, g m D ρD 12 =CRe a Sc b Dividing the second expression into the ﬁrst, we ﬁnd h g m c p

Bothα/D 12and Sc/Pr are equal to the Lewis number, Le Hence, h g m c p =Le 1 − b Le 2/3 (11.108)

The ratio of heat transfer coefficient (h) to mass transfer coefficient (g m) is primarily influenced by the physical properties of the gas mixture, specifically the Lewis number (Le) and specific heat capacity (c p), rather than by geometry or flow rate In air-water systems, the Lewis number is approximately 0.847, and since the concentration of water vapor is typically low, the specific heat capacity can often be approximated as that of air (c p air) This relationship between h and g m was initially established by researchers in the field.

In 1922, W K Lewis focused on air-water systems, leading to an approximation for the case when Le = 1 This work laid the groundwork for a more general equation, eqn (11.108), which is a Reynolds-Colburn type analogy, akin to eqn (6.76) This generalization was introduced by Chilton and Colburn in 1934.

Equation (11.107) may now be written as

The wet-bulb temperature can be determined iteratively using steam tables, which relate it to the dry-bulb temperature and ambient humidity This method allows for the generation of psychrometric charts commonly found in engineering and thermodynamics literature The wet-bulb temperature is crucial in various phase-change processes, as it represents the steady temperature a small body reaches when convective heating balances latent heat removal during evaporation or sublimation This temperature remains constant until the phase-change is complete, playing a significant role in processes such as water droplet evaporation, dry ice sublimation, and fuel spray combustion However, for larger bodies, achieving a steady state may take longer.

Stagnant ﬁlm model of heat transfer at high mass transfer rates

The multicomponent energy equation accounts for the enthalpy of each species in a mixture, denoted as ˆh i In flows involving mass transfer, species move at varying velocities, necessitating the inclusion of individual enthalpy transport in the energy equation alongside heat conduction within the fluid mixture For steady, low-speed flows devoid of internal heat generation or chemical reactions, the energy balance can be reformulated as indicated in equation (6.36).

Tiêu đề	A Heat Transfer Textbook
Tác giả	John H. Lienhard IV, John H. Lienhard V
Trường học	Cambridge
Chuyên ngành	Heat Transfer
Thể loại	textbook
Năm xuất bản	Third Edition
Thành phố	Cambridge

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Số trang	762
Dung lượng	8,59 MB
File đính kèm	A heat transfer textbook.rar (7 MB)