Lecture notes in Heat Transfer pdf

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Heat Transfer 2013 Thomas Rodgers1.1. INTRODUCTION 3 1.1 Introduction Heat transfer is one of the most useful things you will ever learn as a process engineer, for the following reasons:  Many chemical reactions are carried out at high temperatures; attaining and main- taining these temperatures for optimum operation requires a knowledge of heat transfer.  Energy comprises a major economic cost in the processing industries and in do- mestic situations, with energy often lost as heat (the lowest form of energy); energy conservation and recovery requires an understanding of heat transfer principles.  Heat transfer is a good example of transport phenomena (of which the other two are mass transfer and momentum transfer), the basis of chemical engineering; a good understanding of heat transfer eases the understanding of these other transfer processes and of rate processes generally. Heat transfer is also a relatively easy subject to understand, conceptually, and one that is very familiar – in fact, the one subject in chemical engineering that we probably allude to every day. However, the downside of a subject that is conceptually easy to understand is that the theory for it is therefore well developed mathematically. So to become expert at it, you need to become skillful at difficult maths. Heat transfer is about transfer of energy, and you probably already know the following facts:  Unit of energy is the Joule.  Energy is conserved (First law of Thermodynamics).  Heat can only flow from a hotter material to a colder material (Second law of Ther- modynamics). This knowledge will actually form the basis for this course. Firstly we will consider Energy balances briefly, as energy balances are the foundation of heat transfer. Then we will consider Heat Transfer, i.e. the mechanisms by which heat is transferred from a hotter to a colder body, and how to calculate the rate at which this happens. Then we will finish the course by considering the Applications of Heat Transfer theory to some specific examples of industrial relevance, including heat losses from pipes, insulation, and heating up batch vessels. 1.2 Forms of energy Forms of energy include: Kinetic energy – energy arising from motion. This is important if the system is rapidly moving, such as a bullet. Most processes are fairly stationary, so the kinetic energies involved are negligible and can be ignored in the energy balance. But this might not be true if, for example, a stream enters or leaves the system with high velocity, such as a jet from a nozzle.4 CHAPTER1. INTRODUCTIONTOHEATTRANSFER Potential energy – energy arising from being moved against gravity. Most processes ˘ ´ occur at or near the earthâAZs surface, so potential energy is not a major considera- tion in an energy balance. However, when liquids are pumped to reasonable heights above the ground, the energy requirement to pump them may be substantial, and certainly is the energy that you would need to consider in sizing the pump motor. Internal energy – the energy of molecular motion (translation, vibration and rotation) and of intermolecular attraction and repulsion This is related to enthalpy, which we will talk about later, and is usually obtained from tables. Heat and Work – In many ways these are the forms of energy most familiar to us. In an important sense, heat and work are different from the other forms of energy described above, in that they are energy in transit, i.e. energy being transferred from one body to another. Possibly this is why they seem familiar to us. Looking at a brick, it is not evident that it contains internal energy, but if you drop it on your foot, the work it does on your foot is felt quite evidently. 1.2.1 Heat Heat is the most familiar form of energy. We know that a stove feels hot and ice feels cold. To describe this familiar phenomenon more carefully, what happens is that when we touch a stove, heat flows from the stove to our hand, and it therefore feels hot (relative to our hand). When we touch an ice cube, heat flows from our hand to the ice cube, and it feels cold (relative to our hand). From these familiar notions we can formulate two important ideas:  Heat is a form of energy which flows, or as we often say, is transferred from one object to another or between a system and its surroundings. Because of this heat flow, one object loses some energy, and the other object gains this energy. When we hold an ice cube, heat flows from our hand to the ice cube. So our hand loses some of its energy content, as shown by the decrease in its temperature. Conversely, the energy content of the ice cube increases, as shown by the fact that the ice cube melts. So heat is a form of energy in transit, a form which flows or is transferred as a result of a temperature difference.  Secondly, in order for there to be a flow of heat, there must be a temperature differ- ence or gradient (heat, like water, will only flow “downhill”). From these two ideas, we can define heat as “the form of energy which flows from one object or system to another as the result of a temperature difference”. And it is one of the laws of thermodynamics, and something that we know from our everyday experience, that the direction of the flow is from hotter bodies to colder bodies. 1.2.2 Enthalpy We noted earlier the concept of Internal Energy, that is, the energy that a material pos- sesses as a result of the motion and attractions and repulsions of its molecules. This energy depends on the composition of the material and its state, which is determined by the tem- perature and pressure. Related to the internal energy is the Enthalpy,h. The properties of enthalpy are as follows:1.3. SPECIFICHEATCAPACITY 5  For a given material, at constant pressure, the enthalpy depends only on the ma- terial’s temperature and physical state (i.e. liquid, solid, vapour) So, for example,   water at 100 C has less energy and less enthalpy than steam at 100 C.  At constant temperature and physical state, the change of enthalpy with pressure is zero for ideal gases and small for liquids and solids. This means that, for liquids, if you know the enthalpy at a given temperature and the corresponding vapour pressure then this is close enough for other pressures. So tables often give enthalpy data at a particular temperature and the corresponding vapour pres- sure. There are no absolute values for enthalpy. Instead, the enthalpy of a substance is given a value of zero at some arbitrary datum point, and all other enthalpies are quoted relative to  this reference point. For many substances the reference datum is set at 25 C and 1 atm pressure, with the substance in its physical state normal to those conditions. But other datum points can be taken – for example, for water, the datum point at which the enthalpy of liquid water is zero is often taken to be the triple point (the point at which solid, liquid and gaseous water can coexist) which occurs at a conveniently close to zero temperature  of 0.01 C, and its equilibrium vapour pressure of 611.2 Pa. 1.3 Specific heat capacity Which weighs more, a kg of water or a kg of air? Okay, then which will require more energy to heat it up? When you heat a material up, its enthalpy increases as the temperature increases. How  much energy (or enthalpy) does it take to raise the temperature of a material by, say, 1 C? This depends on the material. For water, for example, it takes about 4180 J to raise the  temperature of 1 kg by 1 C, while for air, it takes only about 1005 J (less than a quarter) to achieve the same temperature rise.  The amount of energy required to heat up a kilogram of a material by 1 C is called its specific heat capacity,C (“specific” refers to the fact that we are dealing with a kilogram p – if we were talking about the energy to heat up a mole of material, we would use the 1 1 molar heat capacity). The units of specific heat capacity are J kg K i.e. the amount of energy in Joules to raise the temperature of 1 kg by 1 K. Specific heat capacity is clearly just the slope of the graph of enthalpy against temperature (provided the material does not change its phase). dh =C p dT Specific heat capacity depends slightly on the temperature. When calculating the energy required to achieve a particular temperature change, use the specific heat capacity at the midpoint temperature (but not if there is a phase change involved – in this case, the en- ergy requirements above and below the phase change must be calculated separately). Or, for a more accurate calculation, look up the enthalpy of the material at the two tempera- tures, and subtract one from the other.6 CHAPTER1. INTRODUCTIONTOHEATTRANSFER Slope =C p h T Figure 1.1: Specific heat capacity. 1.4 Rate of Energy Uptake Specific heat capacity is the amount of energy required to raise the temperature of 1 kg of  1 1 material by 1 C; it has units of J kg K . Looking at the units, if we multiply specific heat capacity by a temperature change and a mass flow rate, we will therefore get units of 1 _ J s , i.e. Watts, or rate of heat transfer. The symbol for rate of heat transfer isQ (the dot indicates a flow rate): _ Q =mC _ (T T ) p out in J kg J W = =   K s s kg K So if we know the heat transfer rate, mass flow rate and specific heat capacity, we can calculate the temperature change of a material from this equation. The above equation is one of the most useful equations you’ll learn in this course. There is no need to memorise it, as you can work it out logically just by considering the units of specific heat capacity. But you do need to understand it – and by understanding it, naturally you’ll remember it. For a batch operation the equation is much the same but without the dots, indicating the total quantity of energy required to change the temperature of a given mass: Q =mC (T T ) p final initial J J = kg  K kg K Alternatively, the rate of change of temperature can be related to the heat transfer rate: dT _ Q =mC p dt J K W = kg  kg K s 1.5 Latent heat of vaporisation/condensation (h ) and fg latent heat of fusion/melting (h ) sf   Water at atmospheric pressure boils when it reaches 100 C, to form steam at 100 C. Both are at the same temperature, but steam evidently contains more energy than water.1.6. THERMOPHYSICALPROPERTIESOFWATER,STEAM,ANDAIR 7 To put it another way, it takes energy to convert water into steam, and the question is, how much? The amount of energy required to convert water into steam is called the latent heat of vaporisation. Similarly, when steam gives up this energy and condenses into water, this is called the latent heat of condensation. The magnitude of the energy change is the same  whether we are evaporating water or condensing steam. For water/steam at 100 C and 1 atmospheric pressure, the latent heat of vaporisation/condensation is 2257 kJ kg . Com-  1 paring this with the enthalpy difference between water at 0 and 100 C of 419.2 kJ kg , we can see that it takes a lot more energy to convert water into steam than it does to  heat water from 0 to 100 C. So, for example, if we were drying a product, we would be much better, in energy terms, to remove as much water as possible mechanically (e.g. by squeezing or filtering) before evaporating off the rest. (Water can evaporate even be-  low 100 C, and still requires energy to do so. During drying, this energy comes from  1 the air. At 25 C, the latent heat of vaporisation is 2442 kJ kg , slightly greater than at  100 C.) Similarly, it takes energy to melt ice, and water turning to ice gives up energy. This is 1 called the latent energy of melting or fusion, and has a magnitude of 333.5 kJ kg at  0 C. Figure 1.2 shows the change in enthalpy with temperature for water, as it changes from ice to liquid water to steam, showing that the largest change arises due to the latent heat of vaporisation. Figure 1.2: Variation with temperature of enthalpy and specific heat capacity of water and steam at atmospheric pressure. 1.6 Thermophysical Properties of Water, Steam, and Air The most commonly encountered materials in the process industries are water, steam and air, and if you know how to read enthalpy tables for these, you will be able to do it for other substances. Also, it is important as engineers that you remember some of the important thermophysical properties of water, steam, and air, so that you can do rough8 CHAPTER1. INTRODUCTIONTOHEATTRANSFER calculations in your head or on an envelope. Table 1.1 shows typical values of important thermophysical properties for water, steam and air. Note that values are given for different  temperatures, either 0, 25, or 100 C. The shaded values you ought to remember, at least roughly – the others are there for reference. The density of air is not usually tabulated as it is easily calculated from the ideal gas law. Thermal conductivities and diffusivities are given for reference – we will introduce these later. Dynamic viscosities are also given, just for interest. Table 1.1: Typical values of important thermophysical properties of water, ice, steam, and air. Property Water Ice Steam Air a c b a Specific heat capacity, C 4:180 2:101 2:034 1:005 p 1 1 (kJ kg K ) b b Latent heat of vaporisation/- 2257 2257 1 condensation,h (kJ kg ) lg c c Latent heat of fusion/melting, 333:5 333:5 1 h (kJ kg ) sl 3 a b c b a, Density, (kg m ) 997 – 958 917 0:6 1:186 a c b a Thermal conductivity,  or k 0:611 2:240 0:0248 0:025 1 1 (W m K ) 6a 6b 6a Viscosity, (Pa s) 891 10 12:06 10 18:3 10 6a 6c 6b 6a Thermal diffusivity, 0:147 10 1:17 10 20:3 10 21:0 10 2 1 (m s ) 2 1 6a 6b 6a Dynamic viscosity,v (m s ) 0:894 10 20:1 10 15:4 10 (= momentum diffusivity) momentum diffusivity a b a Ratio 6:1 0:99 0:73 thermal diffusivity a  b  c  25 C, 100 C, 0 C, From ideal gas law. Shaded values you ought to remember, at least approximately. Figure 1.3 presents an abridged steam table, which list the important thermophysical prop- erties of liquid water and steam at selected temperatures and the corresponding saturated vapour pressure. Example: What is the energy requirement to raise the temperature of 60 kg of water from   10 C to 80 C? Calculate the answer in two ways – by using the specific heat capacity at   the midpoint temperature (45 C), and by subtracting the enthalpy of water at 10 C from  the enthalpy of water at 80 C. Why are the two answers different? Which answer is more accurate?  Example: What is the energy requirement to convert 1 kg of water at 10 C into steam  at 150 C? Calculate your answer by adding the energy required to raise the water from    10 C to 100 C (using the specific heat capacity at the midpoint temperature of 55 C),  adding the latent heat of vaporisation, the adding the energy to raise the steam from 100 C  to 150 C (again using the appropriate mid-point temperature to look up the specific heat capacity). Then calculate the energy requirement by subtracting the enthalpy of water at   10 C from the enthalpy of steam at 150 C. How do the two answers compare?1.7. CONSERVATIONOFENERGYANDENERGYBALANCES 9 Figure 1.3: Abridged steam tables. 1.7 Conservation of energy and energy balances Energy is expensive. An energy balance accounts for all the energy entering and leaving a system. If we overlook one of the forms of energy entering or leaving the system, we will not calculate the energy balance correctly. 1.7.1 Energy balances Conservation of energy is one of the fundamental laws of the universe, along with con- servation of mass, except where nuclear reactions are involved, in which case mass can be converted to energy via the equation 2 E =mc 2 Because c is such a large term, a very small amount of mass is converted into vast amounts of energy – this is why nuclear energy is such an attractive prospect.10 CHAPTER1. INTRODUCTIONTOHEATTRANSFER 1.7.2 Statement of the Law of Conservation But in most process industries, except the nuclear industry, we don’t get nuclear reactions occurring, so we can take it as fundamentally true that energy, and mass, are conserved. This means that all the energy that enters a system must leave, one way or another, or else must accumulate in the system. We write this as follows: IN OUT + GENERATION DISSIPATION = ACCUMULATION This applies for heat transfer, mass transfer and momentum transfer. This is a fundamen- tal statement for chemical engineers, and it applies to all situations, from overall plant balances to individual unit operations and to small elements within equipment. Applying this law to conservation of energy within process equipment results in sets of equations, either algebraic or differential, which describe the variation of temperature or heat flow within the equipment. If the process or operation involves no generation or dissipation of heat (e.g. no reactions producing or removing energy), then the above equation simplifies to IN OUT = ACCUMULATION If the system is also operating at steady state i.e. nothing is changing, then there cannot be any accumulation within the system, therefore IN OUT = 0 or TOTAL ENERGY IN = TOTAL ENERGY OUT We must remember to include all forms of energy involved, and recognise that a particular form of energy is not necessarily conserved, as energy can be transformed, e.g. from mechanical work into heat. It helps enormously when performing an energy balance (or a mass balance) to draw a dashed line around the system of interest, whether the system is an entire process, some section of it, a single unit operation, or a differential element within an item of process equipment. This helps to define clearly what energy flows are entering and leaving the system, and helps you to avoid overlooking any. Mass and energy balance examples can be deceptively easy, but more difficult examples require a systematic approach. Five helpful steps to performing mass and energy balances are as follows: 1. Draw a picture, with all streams (mass and energy) entering and leaving the system. Draw a dashed line to indicate the boundaries of the system. Label the streams (with numbers if necessary). 2. Decide a basis on which to perform the calculations. 3. Draw up two balance tables, one for the mass balance, the other for the energy balance. 4. Perform preliminary calculations – fill in the balance tables. 5. Setx = the unknown quantity to be calculated. Solve forx.1.7. CONSERVATIONOFENERGYANDENERGYBALANCES 11 1.7.3 Temperature-Enthalpy Diagrams You will remember the enthalpy-temperature diagram for water, where we showed how enthalpy changed with temperature. We showed it this way, because it makes intuitive sense to consider, when we change the temperature of water, how its enthalpy would change. But it might actually be more sensible to look at it as, by adding energy to the water, we are changing its enthalpy, and are seeing how the temperature changes. In other words, it might make more sense to make temperature the dependent axis, and enthalpy the independent axis. This would make our temperature-enthalpy diagram for a pure component look like Figure 1.4. Figure 1.4: Temperature-enthalpy plot for a pure component at constant pressure. T is the temperature of evaporation, or the boiling point. For a multicomponent stream, E this would vary over the evaporation region from the bubble point to the dew point. T is the temperature of fusion, or the freezing point. For a multicomponent stream, this F would vary over the fusion region from the solidus point to the liquidus point. h is the enthalpy (or latent heat) of evaporation = fnT;P; compositionh is the en- lg sl thalpy (or latent heat) of fusion = fnT;P; composition h is the enthalpy (or latent sg heat) of sublimation =fnT;P; composition12Chapter 2 General Heat Transfer Equation Contents 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2 General Heat Transfer Equation . . . . . . . . . . . . . . . . . . . . 15 2.3 Mechanisms of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . 16 2.3.1 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.3.2 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.3.3 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.4 The Standard Engineering Equation, and Analogy with Electrical Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 13142.1. INTRODUCTION 15 2.1 Introduction We have talked a lot about energy balances, and that energy is conserved over a system, but may be transferred within the system. For example, in a heat exchanger the total energy entering and leaving the system is constant, but heat is transferred from the hot to the cold stream. And we established that the rate of change of enthalpy of either stream, 1 _ H & D , is equal in magnitude to the heat transfer rate,Q (in J s or Watts, W), through the heat exchanger walls. So energy balances tell us how much heat is transferred, but they do not tell us how (i.e. by what mechanisms) this heat is transferred, or how we can design our heat exchanger (or whatever) to achieve this rate of transfer. To decide these questions, we need to move from a study of energy balances to Heat Transfer. 2.2 General Heat Transfer Equation Let us start by considering what factors we might expect to affect heat transfer. If I have an external wall in my office, with a window, and I lose heat through the wall and the window, what factors might affect the rate at which I lose heat and therefore the size of the heater I need in my office?  size of the wall – Area,A  temperature difference between my office and the outside – T  thickness of the wall –x  what the wall is made of – its thermal conductivity – ork  rate of transfer of heat from air inside to the wall, and from the wall to the outside  Sunny, raining, windy  Open or closed window  etc. In a general sense, the thing that is causing heat transfer to occur is the temperature dif- ference, T If T is larger, the rate of heat transfer will be greater. The fact that there is a wall in the way means that there is a resistance to heat transferring. If we have a thicker wall, for example, the resistance to heat transfer will be greater, and heat will be transferred more slowly. So the rate of heat transfer could be described by an equation: t _ Q = R whereR incorporates all the factors that contribute to the resistance to heat transfer, such as the area of the wall, its thickness, what it’s made of, etc. The units of R are clearly 1 K W , in other words, the amount of temperature driving force required to cause heat to be transferred at a rate of 1 W. N.B. This is only true for steady state conditions. Alternatively, the rate of heat transfer could be described by: _ Q =UAT16 CHAPTER2. GENERALHEATTRANSFEREQUATION 1 1 where U represents some sort of overall heat transfer coefficient (units: W m K ), which would incorporate the thickness of the wall, what it’s made of, rate of heat transfer to it from the air inside and from it to the air outside. Clearly,R = 1=UA. _ _ This is the General Heat Transfer Equation, and ranks alongsideQ =MC (T T ) as p out in one of the most important heat transfer equations you will learn – memorise it and digest it The General Heat Transfer Equation can also be written as _ Q q_ = =UT A 2 whereq_ is called the heat flux or heat transfer per unit area (units: W m ). Often it is the heat flux that we are interested in. 2.3 Mechanisms of Heat Transfer Let’s start by identifying the major energy transfer mechanisms. Heat transfer Other energy transfer mechanisms Conduction Mechanical Convection Electrical Radiation Electromagnetic (e.g microwave) (Phase change) Chemical reaction Nuclear reaction The difference between the heat transfer list and the other mechanisms identified above for energy transfer is that in the first list, the driving force for energy transfer is a tem- perature difference. For heat transfer, energy will only flow if there is a temperature difference. The other mechanisms can generate thermal energy within a material without the requirement for a temperature gradient. To quote from Ozisik, page 1, “Since heat flow takes places whenever there is a tem- perature gradient in a system, a knowledge of the temperature distribution in a system is essential in heat transfer studies.” We will therefore aim, as much as possible, to focus on what the temperature distribution is in the systems we are considering, i.e. the temperature profile. 2.3.1 Conduction Conduction is the mechanism of heat transfer in which energy exchange takes place from a region of high temperature to one of lower temperature by the kinetic motion or direct impact of molecules and by the drift of electrons. The latter applies particularly to metals, which are both good electrical conductors and heat conductors. Essentially, conduction is heat transfer by molecular motion in solids or fluids at rest. The empirical law of heat conduction, based on experimental observation, was proposed by Joseph Fourier, who stated that the rate of heat flow by conduction in a given direction is proportional to the area normal to the direction of heat flow and to the gradient of the temperature in that direction:2.3. MECHANISMSOFHEATTRANSFER 17 dT _ Fourier’s law: Q =A x dx dT In terms of heat flux: q_ = x dx _ Q is the rate of heat flow through area A in the positive x-direction. Heat will only x flow in the positivex-direction if the temperature is decreasing in that direction – hence _ the negative sign, as dT=dx must be negative for Q to be positive – i.e. heat flows x “downhill”. The proportionality constant, (ork as is often used) is called the thermal conductivity of the material. Good conductors have high values of , good insulators 1 1 1 1 low values.  has SI units W m K , and varies from around 0.1 W m K for gases 1 1 and insulating materials to up to 1000 W m K for highly conducting metals such as copper or silver. If the temperature gradient through the material is uniform (as it would be in a slab of isotropic material) and at steady state, and the thermal conductivity does not change significantly with temperature, then Fourier’s equation can be written in its steady state form:  _ Q = AT x x =UAT T = R  x Therefore,U = andR = for pure conduction. x A Comparing with the General Heat Transfer Equation, we see that for pure conduction through a slab, U, the overall heat transfer coefficient, under steady state conditions is  x given by . If we think in terms of the resistance to heat transfer, then R = . This x A makes sense: as the thickness of the wall increases, so must its resistance to heat transfer. But if the thermal conductivity is very large, then heat is transferred easily, and the resis- tance to heat transfer is small. Similarly, if the area is very large, then a lot of heat will be lost through it. Figure 2.1 overleaf shows typical ranges of thermal conductivities for various materials. Note that it is a logarithmic scale, and that metals have thermal conductivities typically 1 000 10 000 times greater than insulators and gases. Figure 2.2 shows the effect of temperature on thermal conductivities of some representative materials. Table 2.1 below  gives thermal conductivities of various materials at 0 C.18 CHAPTER2. GENERALHEATTRANSFEREQUATION Question Example: Heat loss through a window. The window in my office is 2 metres by 3 metres. Assuming all other walls are well insulated and that heat loss only occurs through the window, calculate what size of heater I need in my office to maintain the temperature, if the inside surface   of the glass is at 11 C and the outside surface at 6 C. Sketch the temperature profile through the glass. 1 1 Data: Thickness of glass = 4 mm Thermal conductivity of glass = 0.78 W m K . Draw a picture:  _ Q = AT x x 0:78 6 (11 6) = 3 4 10 = 5850 W So I need nearly a 6 kW heater in my room to maintain it at a comfortable temper- ature (which is quite a lot). 2.3.2 Convection Convection is the mechanism of heat transfer between a flowing fluid and a solid body (or between a gas and a liquid at rest). Imagine a fluid flowing past a solid wall, Fig- ure 2.3. Clearly there will be a temperature profile between the bulk fluid and the wall – in the bulk the fluid will be fairly uniformly at its bulk average temperature, T , and near the f wall it will approach the wall temperature,T . w Imagine now a layer near the wall in which the temperature varies betweenT andT . If w f we were to consider heat conduction in this layer, we could describe it by  l _ Q = AT  where is the thermal conductivity of the liquid layer near the wall, and the thickness l of our imaginary layer. The trouble is, we don’t know the value of either or, and both l will change depending on the fluid and the flow conditions. So as we don’t know either,2.3. MECHANISMSOFHEATTRANSFER 19  Table 2.1: Thermal conductivities of various materials at 0 C. Material Thermal Conductivity 1 1 1 1  1 W m K Btu hr ft F Metals Silver (pure) 400 237 Copper (pure) 385 223 Aluminium (pure) 202 117 Nickel (pure) 93 54 Iron (pure) 73 42 Carbon steel, 1% 43 25 Lead (pure) 35 20.3 Stainless steel (15% Cr, 10% Ni) 19 11.3 Chrome-nickel steel (18% Cr, 8% Ni) 16.3 9.4 Non-metallic solids Quartz, parallel to axis 41.6 24 Magnesite 4.15 2.4 Marble 2:08 2:94 1:2 1:7 Ice 2.0 1.19 Sandstone 1.83 1.06 Mortar 1.16 0.69 Glass, window 0.78 0.45 Maple or oak 0.17 0.096 White pine 0.112 0.066 Corrugated cardboard 0.064 0.038 Sawdust 0.059 0.034 Glass wool 0.038 0.022 Liquids Mercury 8.21 4.74 Water 0.556 0.327 Ammonia 0.540 0.312 Lubricating oil, SAE50 0.147 0.085 Freon 12, CCl F 0.073 0.042 2 2 Gases Hydrogen 0.175 0.101 Helium 0.141 0.081 Air 0.024 0.0139 Water vapour (saturated) 0.0206 0.0119 Carbon dioxide 0.0146 0.00844 we may as well replace them both by a single term, which we will call the convective heat transfer coefficient,h: _ Q =hAT 2 1 To be dimensionally correct, h must have SI units of W m K . These are the same units asU; clearly, for pure convection,U is equal toh. The resistance to heat transfer by convection is given byR = 1=hA. The convective heat transfer coefficient varies with the type of flow (turbulent or laminar), the geometry of the system, the physical properties of the fluid, the average temperature, the position along the surface of the body, and time. So calculatingh is quite complicated.20 CHAPTER2. GENERALHEATTRANSFEREQUATION Figure 2.1: Typical range of thermal conductivity of various materials. Figure 2.2: Effect of temperature on thermal conductivity of materials. We will consider ways to calculate it later in the course.2.3. MECHANISMSOFHEATTRANSFER 21 Figure 2.3: Heat transfer by convection between a hot fluid and a wall. h also depends on whether the heat transfer is by forced convection or free convec- tion. Free (or natural) convection arises when the fluid motion is caused by buoyancy effects caused by density differences caused by temperature differences in the fluid. So a hot plate suspended in air causes the immediately adjacent air to heat up; this air rises, giving fluid motion relative to the solid plate. Forced convection, by contrast, arises when the fluid motion is artificially induced, e.g. by a pump or a fan which forces the fluid to flow over the surface of the solid. To summarise, the convective heat transfer coefficient,h, is affected by the fluid mechan- ics of the system, so is somewhat difficult to determine – once known, however, it is easy to use. Table 2.2 shows typical values ofh. Note that these values differ by 4 5 orders of magnitude. In practice, this means that if you have a wall with convective heat transfer on both sides, often the coefficient on one side will be the controlling one. Note too that condensing steam uses a heat transfer coefficient in the same way and with the same units, although the mechanism of heat transfer is not quite the same (involving phase change rather than conduction through an imaginary thin layer of fluid near the wall). The very high value of heat transfer coefficient for condensing steam is one reason that steam is a popular heat transfer fluid in the process industries. The two other major reasons are:  condensing steam has a very high latent heat of condensation, so delivers a lot of heat Table 2.2: Representative values of convective heat transfer coefficient. 2 1 Condition h (W m K ) Free convection, air 6 35 Forced convection, air 28 851 Free convection, water 170 1140 Forced convection, water 570 22 700 Boiling water 5 700 85 000 Condensing steam 57 000 170 000 Forced convection, sodium 113 000 227 000

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