ME3122_Conduction Notes 2014

ME3122 HEAT TRANSFERNational University of SingaporeDepartment of Mechanical EngineeringAY 2014/15 Semester 1

Lecturers & Tutors*ME3122 Heat TransferLecturers

Adj Assoc Prof SG LeeOffice: E2-02-01 Email: [email protected]

Prof Andrew Tay AOOffice: EA-07-19Email: [email protected]

Tutors

Gong Feng

Hassanali Ghaedamini Harouni

ME3122 Heat Transfer

ObjectivesTo develop a basic understanding of the different modes of heat transfer (Conduction, Convection & Radiation) and their applications to Engineering problems.ME3122 Heat Transfer*


Major TopicsConduction (SG Lee)Steady, one-dimensional heat conduction with and without energy generationUnsteady heat conduction, lumped system analysis Extended surfacesRadiation (SG Lee) Black and gray body radiation Radiation between diffuse surfaces Convection (AO Tay) Hydrodynamic and thermal boundary layers Laminar and turbulent forced convection Reynolds analogy Free convectionHeat Exchangers (AO Tay)UA-LMTDEffective-NTUME3122 Heat Transfer*


GradingFinal Examination: 50%Continual Assessment: 50%Lab (2): 20%Quizzes (2): 30%*ME3122 Heat Transfer


Textbook & ReferencesTextbookHeat Transfer by J.P. HolmanMcGraw-Hill Book Company

ReferencesFundamentals of Heat and Mass Transfer by Incropera and DewittJohn Wiley & SonsHeat and Mass Transfer by Y. A. CengelMcGraw-Hill Book Company

ME3122 Heat Transfer*


INTRODUCTION TO HEAT TRANSFER*ME3122 Heat Transfer


Difference between Thermodynamics and Heat TransferThermodynamics tells usHow much heat is transferred?How much work is done?Final/equilibrium state of the system.

Heat transfer tells usHow (with what modes) heat is transferred?At what rate is heat transferred?Temperature distribution inside the body.ME3122 Heat Transfer*


What is Heat Transfer?*ME3122 Heat TransferHeat transfer is thermal energy in transit due to a temperature difference.Thermal energy is associated with the translation, rotation, vibration and electronic states of the atoms and molecules that comprise matter. It represents the cumulative effect of microscopic activities and is directly linked to the temperature of matter.


Terminology*ME3122 Heat Transfer

QuantityMeaningSymbolUnitsThermal EnergyEnergy associated with microscopic behavior of matterU or uJ or J/kgTemperatureA means of indirectly assessing the amount of thermal energy stored in matterTK or CHeat TransferThermal energy transport due to temperature gradientsHeatAmount of thermal energy transferred over a time interval t 0QJHeat RateThermal energy transfer per unit timeqW or J/sHeat FluxThermal energy transfer per unit time and per unit surface areaq"W/m2


Application Areas of Heat TransferME3122 Heat Transfer*Heat transfer is commonly encountered in engineering systems and other aspects of life.A knowledge of heat transfer is necessary in order to evaluate cost, the feasibility, and the size of the equipment to transfer a specified amount of heat in a given time.


Modes of Heat Transfer*ME3122 Heat TransferConduction: Heat transfer in a solid or a stationary fluid (gas or liquid) due to the random motion of its constituent atoms, molecules and /or electrons.Convection: Heat transfer due to the combined influence of bulk and random motion for fluid flow over a surface.Radiation: Energy that is emitted by matter due to changes in the electron configurations of its atoms or molecules and is transported as electromagnetic waves (or photons).


ConductionConduction is the transfer of energy from the more energetic to less energetic particles of a substance due to interactions between the particles. Conduction can take place in solids, liquids, or gases.In gases and liquids, conduction is due to the collisions and diffusion of the molecules during their random motion.In solids, it is due to the combinations of vibrations of the molecules in a lattice and the energy transport by free electrons.ME3122 Heat Transfer*


Consider steady conduction through a plane wall of thickness x = L and area A due to a temperature difference across the wall T = T2 T1.Experiments have shown that the rate of heat conduction through the plane wall isConduction (cont.)ME3122 Heat Transfer*


Conduction (cont.)ME3122 Heat Transfer*


ConvectionEnergy transfer by random molecular motion (as in conduction) and bulk (macroscopic) motion of the fluid.Advection: transport due solely to bulk fluid motion.Types of convectionForced convection: flow is caused by external means, e.g fan, pump,windNatural (free) convection: flow induced by buoyancy forces due to density differences arising from temperature variations in the fluid.Latent heat exchange associated with phase change boiling and condensation.ME3122 Heat Transfer*


Convection (cont.)Relation of convection to flow over a surface and development of velocity and thermal boundary layers:

Convective heat transfer between a surface and a fluid can be calculated by Newtons law of cooling:h: Convection heat transfer coefficient (W/m2K)

ME3122 Heat Transfer*Edge of BL


Convective heat transfer coefficientsThe heat transfer coefficient depends on surface geometry, nature of the fluid motion, as well as fluid properties.

Flowh (W/m2K)Air, natural/free convection6 30Superheated steam or air, forced convection 30 300Oil, forced convection 60 1,800Water, forced convection 300 6,000Water, boiling 3,000 60,000Steam, condensing 6,000 120,000



RadiationThermal radiation is energy emitted by matter.Energy is transported by electromagnetic waves (or photons).Can occur from solid surfaces, liquids and gases.Does not require presence of a medium, e.g. solar energy travels through vaccumRadiation heat transfer at a gas/surface interface involves radiation emission, E, from the surface and the absorption of radiation incident from the surroundings (irradiation, G).ME3122 Heat Transfer*


Radiation (cont.)For an ideal emitter, or blackbodyEb = T4sStefan-Boltzmann lawwhere Ts: absolute temperature of surface (K) : Stefan-Boltzmann constant (5.67x10-8 W/m2K4)For a real (non-ideal) surfaceE = Eb = T4s where : surface emissivity (0 1)Energy absorption due to irradiation G:Gabs = G = T4sur where : surface absorptivity (0 1)For a gray surface, = ME3122 Heat Transfer*


Irradiation: Special case of surface exposed to large surroundings of uniform temperature, TsurRadiation (cont.)*ME3122 Heat Transfer


ME3122 Heat Transfer*In a manufacturing plant, the walls and ceiling of an oven are made of 200 mm thick fire-clay brick having a thermal conductivity of 1.5 W/mK. During steady-state operation, measurements reveal an inner surface temperature of 1200C and an outer surface temperature of 200C. The internal dimensions of the oven are as follows: Length = 4m, Width = 3m and the Height = 3m. What is the rate of heat input required to maintain steady-state temperature inside the oven?Example 1


Solution for Example 1ME3122 Heat Transfer*LL=Ti=To=


An insulated pipe supplying steam from a boiler runs through a room where the air and walls are at 30C. The outer diameter of the pipe is 100 mm and its surface temperature is 250C. The natural convection heat transfer coefficient from the surface to the air is 20 W/m2K. Find the rate of heat transfer from the surface due to convection and radiation per unit length of pipe. For radiation heat loss, the outer surface of the pipe may be treated as blackbody surface.ME3122 Heat Transfer* Example 2


Solution for Example 2ME3122 Heat Transfer* Ts=Steam


Example 3The temperature in a house located at latitude 40N is maintained at 23C with a temperature controller. The temperature of the inner surfaces of walls, floors and the ceiling of the house are found to be at an average temperature of 12C in winter and 27C in summer. A person with an external body surface area of 1.2 m2 and temperature of 32C remains in standing position for fifteen minutes inside the room, where the temperature is 23C. Find the rate of radiation exchange between the person and surrounding surfaces. ME3122 Heat Transfer*


Solution for Example 3

ME3122 Heat Transfer*Ts =32oC = 305 K, Tsur=27oC = 300 KTsur=12oC = 285 K


Summary of Heat Transfer ProcessesME3122 Heat Transfer*


Exercise 1In a cold climate, a house is heated either using electricity or gas or coal to maintain the desired temperature. The roof of such a house is 5 m long and 7 m wide, and 0.20 m thick, and is made of concrete having a thermal conductivity of 0.8 W/mK. On a winter night , the temperatures of the inner and outer surfaces of the roof are measured as 16C and 2C, respectively, for a period of 8 hours. Determine (i) the rate of heat loss through the roof and (ii) the cost of heat loss to the home owner if the cost of electricity is $0.17 per kWh.

An electrical heater, which consists of a rod 300 mm long and 10 mm in diameter, is placed in room at 12C in steady state operation. Heat is generated in the rod as a result of resistance heating and the surface temperature is 140C under steady state operation. The voltage drop and the current through the rod are measured and found to be 50 V and 2 A, respectively. Considering negligible heat losses by radiation, estimate the convective heat transfer coefficient between the outer surface of the rod and the air in the room.

A blackbody at 25C is exposed to solar radiation and the temperature increased to 95C. Estimate the increase in radiation heat transfer. ME3122 Heat Transfer*


Exercise 1.1In a cold climate, a house is heated either using electricity or gas or coal to maintain the desired temperature. The roof of such a house is 5 m long and 7 m wide, and 0.20 m thick, and is made of concrete having a thermal conductivity of 0.8 W/mK. On a winter night , the temperatures of the inner and outer surfaces of the roof are measured as 16C and 2C, respectively, for a period of 8 hours. Determine (i) the rate of heat loss through the roof and (ii) the cost of heat loss to the home owner if the cost of electricity is $0.17 per kWh.ME3122 Heat Transfer*


Solution Outline for Exercise 1.1ME3122 Heat Transfer*LL


Exercise 1.2An electrical heater, which consists of a rod 300 mm long and 10 mm in diameter, is placed in room at 12C in steady state operation. Heat is generated in the rod as a result of resistance heating and the surface temperature is 140C under steady state operation. The voltage drop and the current through the rod are measured and found to be 50 V and 2 A, respectively. Considering negligible heat losses by radiation, estimate the convective heat transfer coefficient between the outer surface of the rod and the air in the room.ME3122 Heat Transfer*


Solution Outline for Exercise 1.2Neglect radiation heat loss.Under steady state operation, heat loss from the surface by convection equals energy generated within the rod due to resistance heating.q = energy generated = V I = (voltage drop, V) (current, A), W

q = heat lost by convection = hAs(Ts Ta), W

h = q / [As(Ts Ta)] = ,W/m2K ME3122 Heat Transfer*


Exercise 1.3A blackbody at 25C is exposed to solar radiation and the temperature increased to 95C. Estimate the increase in radiation heat transfer.

Solution OutlineCalculate emissive power at both temperatures. Increase in radiation heat transfer is equal to the difference in emissive power.E1 = T14 , W/m2 E2 = T24 , W/m2

Increase in radiation heat transfer = E2 E1



CONSERVATION OF ENERGYME3122 Heat Transfer*


Conservation of Energy(First Law of Thermodynamics)An important tool in heat transfer analysis, often providing the basis for determining the temperature of a system.Alternative FormulationsTime Basis: At an instantOver a time intervalType of System:Control volumeControl surfaceME3122 Heat Transfer*


Application to a Control VolumeAt an Instant of Time:

ME3122 Heat Transfer*Note representation of system by a control surface (dashed line) at the boundaries.


Application to a Control Volume (cont.)Over a Time Interval:



Transient Process for a Closed System of Mass (M)Assuming Heat Transfer to the System (Inflow) and Work Done by the System (Outflow),

Special Case Closed System*ME3122 Heat Transfer


Special Case Open System*ME3122 Heat Transfer


For steady state conditions, no changes in kinetic or potential energy, no thermal energy generation, negligible pressure drop:

Special Case Open System (cont.)*ME3122 Heat Transfer


Surface Energy BalanceA special case for which no volume or mass is encompassed by the control surface.Applies for steady-state and transient conditions.With no mass and volume, energy storage and generation are not pertinent to the surface energy balance.Consider surface of wall with heat transfer by conduction, convection and radiation.*ME3122 Heat TransferL


FUNDAMENTAL CONCEPTS OF CONDUCTIONME3122 Heat Transfer*


IntroductionConduction refers to the transport of energy in a medium (solid, liquid or gas) due to a temperature gradient.The physical mechanism is random atomic or molecular activity.Governed by Fouriers law.

In this section we will learnThe definition of important transport properties and what governs thermal conductivity in solids, liquids and gasesThe general formulation of Fouriers law, applicable to any geometry and multiple dimensionsHow to obtain temperature distributions by using the heat diffusion equationHow to apply boundary and initial conditionsME3122 Heat Transfer*


Thermal ConductivityRecall from equation for heat conduction:

The proportionality constant is a transport property, known as thermal conductivity k (units W/mK).The thermal conductivity of a material is a measure of the ability of the material to conduct heat.It depends on the physical structure of matter, atomic and molecular, which is related to the state of the matter. It is also a function of the temperature of the material.Usually assumed to be isotropic (independent of the direction of transfer): kx = ky = kz = k.ME3122 Heat Transfer*


Thermal Conductivity: SolidsSolid comprised of free electrons and atoms bound in lattice.Thermal energy transported throughMigration of free electrons, keLattice vibrational waves (phonons), kphk = ke + kphTo a first approximation, ke is inversely proportional to the electrical resistivity, e

Pure metals: e is low, ke >> kphAlloys: e is substantially larger, contribution of kph to k is no longer negligible.Non-metallic solids: k is determined primarily by kph.Crystalline solids: regularity of the lattice arrangement has an important effect on kph.ME3122 Heat Transfer*


Thermal Conductivity: FluidsThe fluid state includes both liquids and gases.Intermolecular spacing is much larger.Molecular motion is random.Thermal energy transport less effective than in solids thermal conductivity is lower.Kinetic theory of gases:

The thermal conductivity of a gasincreases with increasing temperature and decreasing molecular weight because of the associated increase in mean molecular speed. is generally independent of pressure because density and mean free path are directly and inversely proportional to the gas pressure, respectively.ME3122 Heat Transfer*


Thermal Conductivity: Fluids (cont.)Physical mechanisms controlling thermal conductivity in the liquid state is not well understood.The thermal conductivity of a liquidgenerally decreases with increasing temperature (water is an exception).is usually insensitive to pressure except near the critical point.generally decreases with increasing molecular weight.ME3122 Heat Transfer*


Thermal Conductivity: Insulation SystemsThermal insulations consist of low thermal conductivity materials combined to achieve an even lower system thermal conductivity.Types of insulation systemsFiber, powder and flake type insulations: solid material finely dispersed throughout an air space.Cellular insulations: when small voids or hollow spaces are sealed from each other. Examples: foamed systems made from plastic and glass materials. Several modes of heat transfer involved (conduction, convection, radiation).Effective thermal conductivity: depends on the thermal conductivity and radiative properties of solid material, volumetric fraction of the air space, structure/morphology (open vs. closed pores, pore volume, pore size etc.).Bulk density (solid mass/total volume) depends strongly on the manner in which the solid material is interconnected.ME3122 Heat Transfer*


Thermal Conductivity (cont.)ME3122 Heat Transfer*


Thermal DiffusivityIn heat transfer analysis, the ratio of the thermal conductivity to the heat capacity is an important property termed the thermal diffusivity .

Thermal diffusivity measures the ability of a material to conduct thermal energy relative to its ability to store thermal energy.Materials with larger will respond quickly to changes in their thermal environment, while materials of small will respond more sluggishly, taking longer to reach a new equilibrium condition.It is an important parameter in Transient heat transfer analysisME3122 Heat Transfer*


Fouriers LawA rate equation that allows determination of the conduction heat flux from knowledge of the temperature distribution in a medium.Its most general (vector) form for multi-dimensional conduction is:

Implications:Heat transfer is in the direction of decreasing temperature (basis for minus sign).Fouriers Law serves to define the thermal conductivity of the medium

Direction of heat transfer is perpendicular to lines of constant temperature (isotherms).Heat flux vector may be resolved into orthogonal components.



Fouriers Law Heat Flux Components*ME3122 Heat Transfer


Fouriers Law Heat Flux Components (cont.)*ME3122 Heat Transfer


Heat EquationA differential equation whose solution provides the temperature distribution in a stationary medium.Based on applying conservation of energy to a differential control volume through which energy transfer is exclusively by conduction.Cartesian Coordinates:ME3122 Heat Transfer*


Heat Equation (cont.)ME3122 Heat Transfer*


Heat Equation (cont.)ME3122 Heat Transfer*and simplifying:


Heat Equation (cont.)Cartesian Coordinates:

At any point in the medium the rate of energy transfer by conduction into a unit volume plus the volumetric rate of thermal energy generation must equal the rate of change of thermal energy stored within the volume.ME3122 Heat Transfer*Net conduction of heat into the CVRate ofenergygenerationper unitvolumeTime rate of change of thermal energy per unit volume


For Cartesian CoordinatesIf k = constant

For steady state conditions

For steady state conditions, one-dimensional conduction in x-direction, with constant properties and no energy generationHeat Equation Simplified Forms*ME3122 Heat Transfer


Heat Equation (cont.)*ME3122 Heat Transferx= r cosy= r sin


Heat Equation (cont.)*ME3122 Heat Transferx= r sin cosy= r sin sinz= r cos


Heat equation is a differential equation.For transient conduction, heat equation is first order in time, requiring specification of an initial temperature distribution:Since heat equation is second order in space, two boundary conditions must be specified. Some common cases: B.C. of First Kind (Dirichlet condition): Constant Surface TemperatureExample: When surface is in contact with a melting solid or a boiling liquid.

Boundary and Initial ConditionsME3122 Heat Transfer*


B.C. of Second Kind (Neumann condition): Constant Surface Heat FluxExamples: When an electric heater is attached to a surface; if the surface is perfectly insulated.

B.C. of Third Kind: When convective heat transfer occurs at the surfaceApplied FluxInsulated SurfaceBoundary and Initial Conditions (cont.)ME3122 Heat Transfer*


Methodology of a Conduction AnalysisSpecify appropriate form of the heat equation.Solve for the temperature distribution (by applying the boundary and initial conditions).Apply Fouriers law to determine the heat flux.Simplest Case: One-Dimensional, Steady-State Conduction with No Thermal Energy Generation.Common Geometries:Plane Wall: Described in rectangular (x) coordinate. Area perpendicular to direction of heat transfer is constant (independent of x).Tube Wall: Radial conduction through tube wall.Spherical Shell: Radial conduction through shell wall.ME3122 Heat Transfer*


1-D, STEADY-STATE CONDUCTION WITHOUT THERMAL ENERGY GENERATIONME3122 Heat Transfer*


IntroductionConduction problems may involve multiple directions and time dependent conditions.Inherently complex difficult to determine temperature distributions.One-dimensional steady-state models can represent accurately numerous engineering systems.

In this section, we willLearn how to obtain temperature profiles for common geometries with and without heat generation.Introduce the concept of thermal resistance and thermal circuits.ME3122 Heat Transfer*


Plane WallME3122 Heat Transfer*Consider a plane wall between two fluids of different temperature.Assume 1-D, steady state conduction without heat generation.


Plane Wall (cont.)ME3122 Heat Transfer*


Thermal ResistanceME3122 Heat Transfer*


Thermal Resistance (cont.)ME3122 Heat Transfer*


Thermal Circuit for Plane WallME3122 Heat Transfer*


Composite Plane WallME3122 Heat Transfer*


Series Parallel Composite Wall:Note departure from one-dimensional conditions for .Circuits based on assumption of (a) isothermal surfaces normal to x direction, or (b) adiabatic surfaces parallel to x direction, Both provide approximations for and yield similar resultsComposite Plane Wall (cont.)*ME3122 Heat Transfer


Overall Heat Transfer Coefficient (U)A modified form of Newtons Law of Cooling to encompass multiple resistances to heat transfer.

ME3122 Heat Transfer*U : Overall heat transfer coefficient


Example 4A double glazed window (height = 1 m and width = 1.5 m consists of two 4 mm-thick layers of glass (k = 0.78 W/mK) separated by a 10mm thick stagnant air space (k = 0.026 W/mK). Determine the steady rate of heat transfer through this double-glazed window. The temperature inside the room is maintained at 22C while the ambient (outdoor) is 32C. The convective heat transfer coefficients of the inner and outer surfaces of the window are hi = 12 W/m2K and ho = 48 W/m2K, respectively, which include the effect of radiation.ME3122 Heat Transfer*


Solution for Example 4ME3122 Heat Transfer*


Exercise 2In an aluminum pan placed on a heater, heat is transferred steadily to the boiling water. Find the outer surface temperature of the bottom of the pan and the boiling heat transfer coefficient under the following conditions:The inner surface temperature of the bottom of the pan: 108CRate of heat transfer to the bottom of the pan: 600 WThermal conductivity of the aluminum pan material: 237 W/mKThe diameter of the pan: 250 mmTemperature of water inside the pan: 95C The thickness of the pan material: 5 mm (Ans: 940 W/m2K; 108.3C) ME3122 Heat Transfer*


Contact ResistanceThe temperature drop across the interface between materials may beappreciable, due to surface roughness effects.The associated thermal contact resistance for a unit area of the interface is defined as

Contact spots are interspersed with gaps (usually air filled). As the contact area is typically small, the major contribution to the resistance is made by the gaps (especially for rough surfaces).Thermal contact resistance values depend on: Materials A and B, surface finishes, interstitial conditions, and contact pressure.ME3122 Heat Transfer*


Cylindrical WallME3122 Heat Transfer*


Cylindrical Wall (cont.)ME3122 Heat Transfer*


Cylindrical Wall (cont.)ME3122 Heat Transfer* Ts,1 Ts,2 Rt=


Cylindrical Wall (cont.)ME3122 Heat Transfer*for cylindrical wall


Composite Cylindrical WallME3122 Heat Transfer*


Example 5Steam at a temperature of 300C flows through a cast iron pipe (k = 75 W/mK) whose inner and outer diameters are 50mm and 55mm respectively. The pipe is covered with 25mm thick glass wool insulation (k = 0.05 W/mK). Heat is lost to the surroundings at 30C by convection and radiation, with a combined heat transfer coefficient of 25 W/m2K. The heat transfer coefficient at the inner wall of the pipe is 65 W/m2K. Find the rate of heat loss from the steam per unit length of the pipe. Find the temperature drop across the wall of the pipe and the insulation. Also find (UA) for the pipe.ME3122 Heat Transfer*


Solution Outline for Example 5Let Ti = Temperature of fluid inside the tubeT1 = Temperature of inner wall of the tubeT2 = Temperature of outer wall of the tube T3 = Temperature of the outer surface of the insulation To = Temperature of surrounding fluid*ME3122 Heat Transfer


Solution Outline for Example 5 (cont.)ME3122 Heat Transfer*


Critical Thickness of Insulation for Cylindrical pipeME3122 Heat Transfer*ToTiToT


Critical Thickness of Insulation (cont.)ME3122 Heat Transfer*


Critical Thickness of Insulation (cont.)r = rcrit: Resistance to heat transfer is minimum; heat transfer rate reaches a maximum.

r < rcrit: Addition of insulation increases heat transfer rate. Electric cables are designed for maximum heat dissipation, hence the insulation thickness should be around the critical value.

r > rcrit: Addition of insulation decreases heat transfer rate. To reduce heat losses, the insulation thickness should be much greater than the critical value.



Example 6An electric wire, diameter d = 3 mm and length L = 5 m, is tightly wrapped with a 2 mm-thick plastic cover (thermal conductivity, k = 0.15 W/mK). Measurements indicate that a current of 10 A passes through the wire causing a voltage drop of 8 V. The wire is exposed to an environment at 32C with a convective heat transfer coefficient, h = 12 W/m2K. Determine the temperature at the interface of the wire and the plastic cover in steady operation. Also, evaluate whether doubling the thickness of the plastic cover will increase or decrease this interface temperature.ME3122 Heat Transfer*


Solution for Example 6*ME3122 Heat Transfer


Solution for Example 6 (cont.)*ME3122 Heat TransferT1


Spherical WallME3122 Heat Transfer*


Spherical Wall (cont.)ME3122 Heat Transfer*


Spherical Wall (cont.)ME3122 Heat Transfer* Ts,1 Ts,2 Rt =for spherical wall


Summary of ResultsFor 1-D, steady state conduction with constant k and without heat generationME3122 Heat Transfer*

Plane WallCylindrical WallSpherical WallHeat equationTemperature distributionHeat flux (q)Heat rate (q)Thermal resistance (Rt)


1-D STEADY STATE CONDUCTION WITH ENERGY GENERATIONME3122 Heat Transfer*


Involves a local (volumetric) source of thermal energy due to conversion from another form of energy in a conducting medium.The source may be uniformly distributed, as in the conversion from electrical to thermal energy (Ohmic heating):

orIt may be non-uniformly distributed, as in the absorption of radiation passing through a semi-transparent medium. For a plane wall,

Generation causes the heat rate to vary with location. Accordingly, the concept of thermal resistance/circuit cannot be applied.Implications of Energy Generation*ME3122 Heat Transfer


Plane WallME3122 Heat Transfer*Consider one-dimensional, steady-state conduction in a plane wall of constant k, uniform generation, and asymmetric surface conditions:


Asymmetric Surface Conditions:Plane Wall (cont.)ME3122 Heat Transfer*


Plane Wall (cont.)ME3122 Heat Transfer*Symmetric Surface Conditions or One Surface Insulated:


Plane Wall (cont.)ME3122 Heat Transfer*


Radial Systems*ME3122 Heat Transfer


Solid CylinderME3122 Heat Transfer*


Solid Cylinder (cont.)ME3122 Heat Transfer*


Solid SphereME3122 Heat Transfer*


Solid Sphere (cont.)ME3122 Heat Transfer*


Application of Resistance ConceptWhen heat generation effects are present, the heat transfer rate is not a constant, nor independent of the spatial coordinate. Consequently, it would be incorrect to use the conduction resistance concepts.ME3122 Heat Transfer*


Example 7A 2-kW resistance water heater is used to boil water in a kettle. The cylindrical heating element has a diameter of 5 mm and length 0.6 m, where thermal conductivity, k = 15 W/mK. The heater is submerged in water of 100C and experiences a convective heat transfer coefficient, h = 20,000 W/m2K. Calculate the temperature at the centre of the element.ME3122 Heat Transfer*


EXTENDED SURFACE/FINME3122 Heat Transfer*


Extended Surface/FinAn extended surface (also know as a combined conduction-convection system or a fin) is a solid within which heat transfer by conduction is assumed to be one dimensional, while heat is also transferred by convection (and/or radiation) from the surface in a direction transverse to that of conduction.

Extended surfaces may exist in many situations but are commonly used as fins to enhance heat transfer by increasing the surface area available for convection (and/or radiation). They are particularly beneficial when h is small, as for gas and natural convection.ME3122 Heat Transfer*


Some Typical Fin ConfigurationsME3122 Heat Transfer*Straight fins of (a) uniform and (b) non-uniform cross sections; (c) annular fin, and (d) pin fin of non-uniform cross section.


Some Innovative Fin Designs*ME3122 Heat Transfer


Fin Equation*ME3122 Heat Transfer


Fin Equation (cont.)*ME3122 Heat Transfer


Fin Equation (cont.)*ME3122 Heat Transfer(insulated tip)


Fin Equation Long Fin*ME3122 Heat Transfer


Fin Equation Adiabatic Fin Tip*ME3122 Heat Transfer


Fin Equation Convective Fin Tip*ME3122 Heat Transfer


Fin EfficiencyThe ratio of the actual heat transfer rate from the fin to the maximum rate at which a fin could dissipate energy.The maximum rate at which a fin could dissipate energy is the rate that would exist if the entire fin surface were at the base temperature.ME3122 Heat Transfer*< 1


Fin Efficiency (cont.)ME3122 Heat Transfer*


Fin EffectivenessThe ratio of the fin heat transfer rate to the heat transfer rate that would exist without the fin.

ME3122 Heat Transfer*> 1


Fin Effectiveness (cont.)ME3122 Heat Transfer*


Example 8A copper pin fin 2.5 mm in diameter protrudes from a wall at 100C into air at 28C. The heat transfer is mainly by natural convection with a heat transfer coefficient of 12 W/m2K. Calculate heat losses from the fin assuming the fin is infinitely long; the fin is 30 mm long and the tip of the fin is insulated; the fin 30 mm long having convective losses, with h = 12 W/m2K.ME3122 Heat Transfer*


Solution for Example 8 (cont.)*ME3122 Heat Transfer


Example 9An array of 10 aluminum alloy fins, each 3 mm wide, 0.4 mm thick, and 40 mm long, is used to cool a transistor. When the base is at 67C and the ambient is at 27C, how much power do they dissipate if the combined convection and radiation heat transfer coefficient is estimated to be 8 W/m2K? The alloy has a thermal conductivity of 175 W/mK. The heat transfer from the tip of the fin is negligible. Also, find the efficiency and effectiveness of the fin.ME3122 Heat Transfer*


Solution for Example 9 (cont.)ME3122 Heat Transfer*


TRANSIENT CONDUCTIONME3122 Heat Transfer*


IntroductionMany heat transfer problems are time dependent.Changes in operating conditions in a system cause temperature variation with time, as well as location within a solid, until a new steady state (thermal equilibrium) is achieved.In this section we will develop procedures for determining the time dependence of the temperature distribution.Solution techniques include Lumped Capacitance Method, Exact Solutions and Finite-Difference Method.We will focus on the Lumped Capacitance Method, which can be used for solids within which temperature gradients are negligible.ME3122 Heat Transfer*


Lumped Capacitance MethodBased on the assumption of a spatially uniform temperature distribution throughout the transient process. Hence, T(x, y, z, t) = T(t) Temperature is a function of time only.



*ME3122 Heat TransferLumped Capacitance Method (cont.)


*ME3122 Heat TransferLumped Capacitance Method (cont.)Rt=1/(hAs)Ct=Vc = time constant = Vc hAs


Lumped Capacitance Method (cont.)ME3122 Heat Transfer*


*ME3122 Heat TransferLumped Capacitance Method (cont.)


Surface energy balance:Validity of Lumped Capacitance MethodME3122 Heat Transfer*Ts,1Ts,2Tqcondqconv

Validity of Lumped Capacitance Method (cont.)ME3122 Heat Transfer*i.e. approx. uniform temperature in the solid


Validity of Lumped Capacitance Method (cont.)ME3122 Heat Transfer*


Biot and Fourier NumbersME3122 Heat Transfer*


Biot and Fourier Numbers (cont.)ME3122 Heat Transfer*


Example 10Steel ball bearings are required to be subjected to heat treatment to obtain the desired surface characteristics. The balls are heated to a temperature of 650C and then quenched in a pool of oil that has a temperature 55C. The ball bearings have a diameter of 40 mm. The convective heat transfer coefficient between the ball bearings and oil is 300 W/m2K. Determinethe length of time that the bearings must remain in the oil before their temperature drops to 200C,total amount of heat removed from each bearing during this time interval, and instantaneous heat transfer rate from the bearings when they are first placed in the oil and when they reach 200C.The properties of steel ball bearings are as follows:k = 50 W/mK; = k/cp = 1.310-5 m2/s *ME3122 Heat Transfer


Example 11The hot plate of a cooker has a surface area of 0.05 m2 and is made of steel (density 7820 kg/m3) having a total mass of 1.4 kg. The convection heat transfer coefficient is 17 W/m2K between the plate and its surroundings at 27C. How long after being switched on, would the plate take to attain a temperature of 117C? The plate heater is rated at 500 W and initially at the temperature of the surroundings. The specific heat of the plate is 461 J/kgK. ME3122 Heat Transfer*


Example 12The temperature of a stream of natural gas flowing through a pipe at 100C is to be measured by a thermocouple whose junction can be approximated as a 1-mm diameter sphere. The properties of the junction are as follows: k = 35 W/mK, = 8500 kg/m3 and c = 320 J/kgK. The convection heat transfer coefficient between the junction and the gas, h = 210 W/m2K. The thermocouple is initially at 28C. Determine the time constant of the thermocouple. Also, find the time taken to read 99% of the initial temperature difference.*ME3122 Heat Transfer


In the general case we may have convection, radiation, internal energy generation and an applied heat flux. The overall energy balance becomes:TsurT, hqsqradqconvAs,hAs(c,r)Numerical solutions are generally required, e.g. Finite Difference method, Finite Element methodSimplified solutions exist for no imposed heat flux or generation.*ME3122 Heat TransferGeneral Lumped Capacitance Analysis


Other transient problemsWhen the lumped capacitance analysis is not valid (e.g. Bi >>0.1), we must solve the partial differential equations analytically or numerically.Exact and approximate solutions may be used.Tabulated values of coefficients used in the solutions of these equations are available.Transient temperature distributions for commonly encountered problems involving semi-infinite solids can be found in the literature.*ME3122 Heat Transfer


*Examples include, power generation, heating and cooling, and many industrial and domestic applications.* = delta*******To simplify the form of this equation, transform the dependent variable by defining an excess temperature as (x) = T(x) - T where since T is a constant, d/dx = dT/dx. This equation is a linear, homogeneous, 2nd order differential equation with constant coefficients. Its general solution is of the form*The amount of heat transferred from the fin may be evaluated in 2 alternative ways. Conservation of energy dictates that the rate at which heat is transferred by convection from the fin must equal to the rate at which it is conducted through the base of the fin.*A simple, yet common, transient conduction problem is one for which a solid experiences a sudden change in its thermal environment. Consider a hot metal forging that is initially at a uniform temperature Ti and is quenched by immersing it in a liquid of lower temperature T. If the quenching is said to begin at time t = 0, the temperature of the solid will decrease for t > 0, until it eventually reaches T. *The difference between the solid and fluid temperatures must decay exponentially.**Any increase in Rt or Ct will cause a solid to response more slowly to changes in its thermal environment.*Q is related to the change in the internal energy of the solid.*There is a strong preference for using the lumped capacitance method as it is the simplest and most convenient method that can be used to solve transient heating and cooling problems. Hence it is important to determine under what condition it may be used with reasonable accuracy. To develop a suitable criterion consider S-S conduction through a plane wall of area A.**

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ME3122_Conduction Notes 2014