FUNDAMENTALS OF THERMODYNAMICS AND HEAT TRANSFER

FUNDAMENTALS OF THERMODYNAMICS AND HEAT TRANSFER Lecture 8: Heat transfer modes Pierwsza strona

What is heat transfer? Science about energy, its conversion and transfer Considered will be energy conversion to its useful forms. Power engineering Chemical engineering Electronics Wymiana Ciepła Space technology

Heat transfer processes Uses the elements of several disciplines: Classical thermodynamics Thermodynamics of irreversible processes Statistical thermodynamics Fluid mechanics Mathematics Physics Technology development design for operation at extreme parameters (space technology, microelectronics) Energy conversion degradation of natural environment Wymiana Ciepła

First Law of Thermodynamics Conservation of energy energy cannot be formed in the isolated system, can only change its form. Second Law of Thermodynamics Energy has quality and quantitiy. The quality can only be reduced in closed systems. Wymiana Ciepła

Wymiana Ciepła Flow of thermal energy is always from a body having a higher temperature to a body with lower temperature.

Fundamentals Ciepło sposób przepływu energii między ciałami, wynika z I ZT Nie może samorzutnie przechodzić od ciała o temperaturze niższej do ciała o wyższej temp. Wymiana ciepła związki między ilością przepływającego ciepła, a różnicą temperatur i czasem trwania zjawiska (złożony problem) Wymiana Ciepła Wymienniki ciepła urządzenia, w których zachodzi wymiana ciepła (rekuperatory, regeneratory, wymienniki mieszankowe)

Fundamentals Rate of heat Q & = dq dτ Heat flux density q = d 2 Q da dτ Heat transfer between the wall and a fluid convection Wymiana Ciepła Heat transfer between two fluids separated by the wall complex heat transfer

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Heat energy transfer mechanisms o Conduction o Convection o Radiation

Conduction This is the mechanism by which heat is transferred from one part of an object to another part through molecular collisions. If one part of an object is hotter than its neighboring part, the molecules in the hotter part have more energy and vibrate more vigorously than their neighbors. When they collide with their neighboring molecules which vibrate less vigorously, energy is transferred to the latter and the temperature of the colder part increases. Free electrons (electrons that has become detached from their parent molecules) also play a important part in the thermal conduction as they provide an effective mechanism for carrying heat energy from one part of an object to another part. Metals are good electrical conductors because they have a lot of free electrons. Hence they are also good conductors of heat.

Thermal conductivity The rate of heat transfer through thermal conduction is proportional to the cross-sectional area and the temperature difference, and inversely proportional to the thickness: Q T A t x For a slab of infinitestimal thickness dx and temperature difference dt, the thermal conduction equation can be written as: Q dt = ka t dx The proportional constant k is called the thermal conductivity. The negative sign is introduced to adopt the convention that the direction of heat flow is opposite to the direction of increasing temperature.

Conduction of heat Conduction: collissions and difussion of particles Wymiana Ciepła

Thermal conductivities of some substances Substance k (W/mK) Metals (at 25 o C) Silver 427 Copper 390 Aluminum 238 Iron 79.5 Nonmetals (approximate values) Concrete 0.8 Glass 0.8 Water 0.6 Rubber 0.2 Wood 0.08 Asbestos 0.08 Air (at 20 o C) 0.0234

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Steady state thermal conduction In steady state, the rate of heat flow through an insulated uniform rod can be determined using the equation: Q t = ka T T 1 2 where T 2 is the temperature at the hotter end of the rod, T 1 the temperature at the colder end of the rod, A the crosssectional area of the rod, and L its length. L

Heat transfer through compound slabs Often compound slabs consisting several layers of different materials are used for insulation. The rate of heat transfer through such a slab can be calculated from the thermal conductivities of the materials that make up of the slab. Consider a compound slab consisting of two materials of thicknesses L 1, L 2 and thermal conductivities k 1, k 2 respectively. The temperatures of the outer surfaces are T 1 and T 2, where T 2 > T 1. Let the temperature at the interface be T. Since the heat transfer rates through the two layers must be the same, we have k 1 A(T T 1 ) k 2 A(T 2 T) = L 1 L 2

Solving the equation for T, we have T = k 1 L 2 T 1 + k 2 L 1 T 2 k 1 L 2 + k 2 L 1 Substituting this expression for T in the heat transfer rate equation for either layer, we obtain Q t = A ( T 2 T 1 ) L 1 / k 1 + L 2 / k 2

The rate of heat transfer through a compound slab consisting of n materials can be generalized from the 2- layer equation: Q t = A ( T 2 T 1 ) n Σ L i / k i i=1 L/k for a particular substance is often referred to as the R value of the material, and the above equation can be written in terms of the R values: Q t = A ( T 2 T 1 ) n Σ R i i=1

R values of some common building materials Material R value (mk/w) Hardwood siding (1 ) 0.25 Brick (4 ) 1.00 Fiberglass board (1 ) 1.10 Flat glass (0.125 ) 0.30 Air space (3.5 ) 0.25 Drywall (0.5 ) 0.10 Stagnant air layer* 0.04 *At any vertical surface open to the air, a very thin stagnant layer of air adheres to the surface and the R value of this stagnant air layer on an outside wall depends on wind speed. When determining the R value of a wall, one must consider the stagnant air layers on both sides of the wall.

Example: What is the R value of a wall consisting of a 4 brick layer, a 2 fiberglass board and a 0.5 dry wall? R 1 (outside stagnant air layer) 0.04 mk/w R 2 (brick) 1.00 R 3 (fiberglass board) 2.20 R 4 (dry wall) 0.12 R 5 (inside stagnant air layer) 0.04 ΣR i 3.38 mk/w

Radial heat flow Consider a steam pipe of radius a, surrounded by a layer of insulating material of outer radius b, length L and thermal conductivity k. If the temperature of the steam pipe is T 2 and that of the air outside the insulating material is T 1 (T 1 < T 2 ), What is the rate of energy transfer through the insulating material and what is the temperature at r (a < r < b) when a steady state has been reached? Law of thermal conduction: Q dt = ka t dr Insulating material Q is constant at steady state t Q T 1 Representing by H, the law can be written as: t H = k( 2πrL) dt dr Steam pipe 2a k 2b T 2 L

From the above equation: 1 dr r = 2πkL dt H πkl ln r = 2 T + H where c is an integration constant. c (equ. 1) At r = a, T = T 2 2 kl ln a = π T 2 + c (equ. 2) H At r = b, T = T 1 2 kl ln b = π T1 + c (equ. 3) H

Solving equ. 2 and equ 3 for H and c: H c = 2 1 πkl( T 2 T ln( b / a) ) = (This is the rate of energy transfer) T 2 ln b T1ln T 2 T 1 a Substituting the expressions for H and c in equ 1, we have T = T 2 ln( b / r) + T1ln( r ln( b / a) / a)

Convection This is the mechanism by which heat is transferred by actual motion of material. Natural convection: The material flows due to differences in density (caused by thermal expansion). Examples: Air flow at a beach. Water mixing in a lake when its surface is cooled. Forced convection: The material if forced to move by a blower of pump. Examples: Hot-water heating system. Flow of blood in the body.

Heat convection Energy transfer by means opf convection takes place by movement of hot particles upwards and falling of colder ones. Wymiana Ciepła

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Heat transfer by convection There is no simple equation for calculating the amount of heat transferred by convection. The heat lost or gained by a surface at one temperature in contact with a fluid at another temperature depends on many factors, such as the shape and orientation of the surface, the mechanical and thermal properties of the fluid and the nature of fluid flow (laminar or turbulent).

For practical calculations, the following equation is often used: Q = ha T t Q where, = rate of energy transfer or heat current t A = surface area T = temperature difference between the surface and the main body of the fluid h = convection coefficient

Convection coefficient The convection coefficient defined in the above equation is temperature dependent and needs to be determined experimentally. A situation of practical important is that of natural convection from a wall or a pipe that is at constant temperature and is surrounded by air at atmospheric pressure whose temperature is less than that of the wall or pipe by T. The convection coefficients applicable in this situations are shown below: h, (W m -2 K -1 ) Horizontal plate, facing upward 0.0595x10-4 ( T) 1/4 Horizontal plate, facing downward 0.0314x10-4 ( T) 1/4 Vertical plate 0.0424x10-4 ( T) 1/4 Horizontal or vertical pipe (dia. D) 1.00x10-4 ( T/D) 1/4

Złożona wymiana ciepła Przenikanie jako przykład złożonej wymiany ciepła w przypadku płaskiej ścianki można przedstawić następująco: T f1 δ konwekcja, α 1 T f2 przewodzenie, λ konwekcja, α 2 Wymiana Ciepła Q& = T T f 1 f 2 1 R R 1 = α A 1 δ + + λa α A 2

Radiation This is the mechanism by which heat is transferred by continual emission of energy from the surface of a body. This energy is called radiant energy and is in the form of electromagnetic waves. These waves travel with the speed of light and are transmitted through vacuum as well as through. When they fall on a body that is not transparent to them, such as the surface of one s hand or the walls of a room, they are absorbed, resulting in a transfer of heat to the absorbing material. The radiant energy emitted by a surface depends on the nature of the surface and on its temperature. At any temperature, the radiant energy emitted is a mixture of waves of different wavelengths, and the mixture is a function of temperature.

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Radiant energy from the Sun The amount of radiant energy reaching the top of the Earth s atmosphere from the Sun is approximately 1340 J per second. Some of this energy is reflected into the space and some absorbed by the atmosphere. The amount of the Sun s radiant energy that reaches the surface of the Earth is hundreds of times of all the energy needed on this planet. The Sun s radiant energy is primary visible and infrared light accompanied by a significant amount of ultraviolet radiation. The Sun influences the Earth s average temperature, ocean currents, agriculture, rain patterns, etc.

Stefan s law The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is know as Stefan s law: Q t = σ A e T 4 where σ = 5.6696x10-8 W m -2 K -4 (Stefan-Boltzmann constant) A = surface area of the object in m 2 e = emissivity of the object T = the surface temperature of the object in kelvins. This relation was deduced by Josef Stefan (1835-1893) on the basis of experimental measurements made by John Tyndall (1820-1893) and was later derived from theoretical considerations by Ludwig Boltzmann (1844-1906). Hence, it is also called Stefen-Boltzmann law.

Emissivity It is the fraction of the incoming radiation the surface absorbs. It has a value if the range 0 1. It depends on the properties of the surface of the object, generally larger for dark and rough surfaces than for light and smooth ones. A good radiant energy emitter has a large emissivity, hence it is also a good absorber and a poor reflector. A poor radiant energy emitter is also a poor absorber and a good reflector. An ideal absorber (e = 1) reflects no radiant energy, and hence would appear black in color. Such an ideal absorber is called blackbody.

As an object radiates electromagnetic energy, it also absorbs electromagnetic radiation, otherwise its temperature would reduce to absolute zero. The amount of electromagnetic energy absorbed by an object depends on the temperature of its surroundings. If an object at a temperature T and its surroundings are at temperature T o, then the rate of net energy gained (or lost) by the object due to emission and absorption of electromagnetic radiation is: Q t = σ A e (T 4 - T o4 ) Hence, if T > T o, The object radiates more energy than it absorbs, and its temperature decreases until equilibrium with its surroundings is reached.

Example: A metal block of dimensions 10 cm x 10 cm x 0.2 cm is heated by an electric heater to 800 o C in vacuum. If the emissivity of the metal block is 0.9, how much electric power input to the heater would be required to maintain the metal block at this temperature? To maintain the metal block at a fixed temperature, the electric power input to the heater must be equal to the rate of energy loss by radiation. Rate of radiant energy loss: Q t = σ A e T 4, where σ = 5.6696x10-8 W m-2 K-4 A = 2 x (0.1x0.1 + 0.1x0.002 + 0.1x0.002) = 0.0208 e = 0.9 T = 800 +273.15 = 1073 K Q t = 5.6696 x 0.0208 x 0.9 x (1073) 4 x 10-8 = 1407 W

The Dewar flask The Dewar flask (invented by James Dewar) is a container designed for storage of cold or hot liquids for long period of time. It is a Pyrex glass vessel and has a design as shown in the diagram below. It has two walls with evacuated space in between to minimize energy transfer by conduction and convection. The surfaces of the wall are coated with silver (a very good reflector with low emissivity) to minimize energy transfer by radiation.

Przewodzenie w ciałach o małym oporze Przewodzenie w ciałach o małym oporze cieplnym (Lumped Capacity Method) jest potężnym narzędziem w obliczeniach niestacjonarnej wymiany ciepła. Przyjmijmy, że ciało ma objętość V, powierzchnię A, gęstość właściwą ρ, oraz ciepło właściwe c. Jego temperatura T, jest jednakowa w całej objętości, i zmienia się na skutek wymiany ciepła z otaczającym je płynem o stałej w czasie temperaturze T. powierzchnia A objętość, V ρ, c, T(t) du dt Wymiana Ciepła olej w temp. = qa U = V c( T T ) q = α( T T ) ρ 0 T

Przewodzenie w ciałach o małym oporze Podstawiając powyższe wyrażenia do równania bilansu energii: Warunki brzegowe: Vρc dt dt ( T ) = Aα T dla t = 0 T = T 0 Po przekształceniach: dt T T αa = ρcv dt T T 0 dt T T αa = ρcv τ 0 dt Rozwiązanie równania ma postać: T ln T 0 T T αa = τ ρcv T T T 0 T = e αa τ ρcv = e τ τ 0 Cieplna stała czasowa: τ = 0 ρcv αa Wymiana Ciepła

Zdefiniujmy bezwymiarową temperaturę oraz czas: Przewodzenie w ciałach o małym oporze T * T T = T T 0 * τ αaτ τ = = τ ρcv 0 * T T T = T T 0 Umożliwia to nam analizę przypadków gdzie występuje gwałtowna zmiana temperatury * τ αa τ τ = = τ ρcv 0 Aby teoria przewodzenia ciepła miała zastosowanie musi być spełniony warunek, że opór przewodzenia w ciele stałym musi być dużo mniejszy od oporu przejmowania ciepła na zewnątrz Wymiana Ciepła

Przewodzenie w ciałach o małym oporze Ciała stałe opór przewodzenia wewnąewn L / opór konwekcji na zewnąewn 1 / ( λa) αl = ( αa) λ Mieszalnik Zdefiniujmy bezwymiarową liczbę Biota: opór przewodzenia wewnąewn 1/ opór konwekcji na zewnąewn 1 / αl Bi = λ ( αa ) U = ( UA) α Wymiar charakterystyczny, l, jest wyrażony stosunkiem V/A Teorię można stosować w przypadku, gdy Bi<0.1 dla płaskich płyt, walców, kul. Wymiana Ciepła Wymiary charakterystyczne dla różnych przypadków: 1. Płyta o grubości l: L=l/2 2. Walec o promieniu R: L=R/2 3. Kula o promieniu R L=R/3 4. Sześcian o krawędzi l L=l/6

Przewodzenie w ciałach o małym oporze Równanie można przedstawić w postaci zależności pomiędzy liczbami podobieństwa. W tym celu wykładnik liczby e należy przekształcić do postaci: αaτ c ρv p αl aτ l = = ( Bi)( Fo) 2 λ l V A l L Zdefiniujmy bezwymiarową liczbę Biota: i Fouriera αl Bi = λ Fo = aτ 2 l Wymiar charakterystyczny, l, jest wyrażony stosunkiem V/A Teorię można stosować w przypadku, gdy Bi<0.1 dla płaskich płyt, walców, kul T T ( Bi) ( Fo) l / L = e T 0 T Wymiana Ciepła

Przewodzenie w ciałach o małym oporze Wyznaczmy jeszcze strumień ciepła przepływający przez powierzchnię ciała. Jest on zmienny w czasie, gdyż pomimo stałej wartości współczynnika wnikania ciepła α, ulega zmianie różnica temperatur pomiędzy ciałem a otaczającym je płynem zna skutek zmiany temperatury ciała. Chwilowy strumień ciepła: Q& = c p ρv dt d τ dt αa = ( T T0 ) dτ c ρv p e αa τ c ρv p Całkowita ilość ciepła wymienioną przez ciało, w czasie do dowolnej chwili: Q = Qd & τ = αa( T T0 ) 1 e l ( Bi)( Fo) L Wymiana Ciepła

Przewodzenie w ciałach o małym oporze Hartowanie płyty stalowej Płyta stalowa o grubości 1 cm zostaje wyjęta z pieca o temperaturze 600 C i wrzucona do kąpieli olejowej o temperaturze 30 C. Jeżeli współczynnik przejmowania ciepła ma wartość 400 W/m 2 K, ile czasu potrzeba aby schłodzić płytę do temperatury 100 C? Założyć własności fizyczne materiału λ, ρ, c jak dla stali, czyli 50 W/mK, 7800 kg/m 3, oraz 450 J/kg K, odpowiednio. Dane: Szukane: Założenia: Płyta stalowa hartowana w oleju. Czas schłodzenia z 600 C do 100 C. Ciało o małym oporze przewodzenia. Wymiana Ciepła

Przewodzenie w ciałach o małym oporze Sprawdzamy wartość liczby Biota: V/A=WHL/2WH=L/2 Bi=α(L/2)/λ=400*0.005/50=0.04 czyli Bi = 0.04 < 0.1 Wynika stąd, że ciało ma mały opór cieplny przewodzenia i można skorzystać z omawianej teorii Wymiana Ciepła

Przewodzenie w ciałach o małym oporze Znajdźmy stałą czasową zagadnienia Podstawiając dane zadania, tj.: T 0 =600 o C, T final =100 o C, T =30 o C Rozwiązujemy ze względu na czas: Wymiana Ciepła