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FUNDAMENTALS OF THERMODYNAMICS AND HEAT TRANSFER

Lecture 8: Heat transfer modes

Wymiana Ciepła

What is heat transfer?

Science about energy, itsconversion and transfer

Considered will be energyconversion to its usefulforms.

Power engineering

Chemical engineering

Electronics

Space technology

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Flow of thermal energy is always from a body having a higher temperature to a body with lowertemperature.

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Heat transfer processes

Uses the elements of several disciplines:Classical thermodynamicsThermodynamics of irreversible processesStatistical thermodynamicsFluid mechanicsMathematicsPhysics

Technology development – design for operation at extremeparameters (space technology, microelectronics)

Energy conversion – degradation of natural environment

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First Law of Thermodynamics

Conservation of energy – in the isolated system energy cannot be formed, can onlychange its form.

Second Law of ThermodynamicsEnergy has quality and quantitiy. The quality can only be reduced in closedsystems.

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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 ilościąprzepływającego ciepła, a różnicą temperatur i czasem trwania zjawiska (złożony problem)

Wymienniki ciepła – urządzenia, w których zachodzi wymiana ciepła (rekuperatory, regeneratory, wymienniki mieszankowe)

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Fundamentals

Rate of heat

Heat flux density

τddQQ =&

τddAQdq

2

=

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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 conductivityThe rate of heat transfer through thermal conduction is proportional to the cross-sectional area and the temperature difference, and inversely proportional to the thickness:

xTA

tQ

ΔΔ

∝Δ

For a slab of infinitestimal thickness dx and temperature difference dT , the thermal conduction equation can be written as:

dxdTkA

tQ

−=Δ

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.

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Conduction of heatConduction: collissions and difussion of particles

Thermal conductivities of some substancesSubstance k (W/mK)Metals (at 25oC)Silver 427Copper 390Aluminum 238Iron 79.5

Nonmetals (approximate values)Concrete 0.8Glass 0.8Water 0.6Rubber 0.2Wood 0.08Asbestos 0.08

Air (at 20oC) 0.0234

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Steady state thermal conductionIn steady state, the rate of heat flow through an insulated uniform rod can be determined using the equation:

LkA

tQ TT 12 −=Δ

where T2 is the temperature at the hotter end of the rod, T1the temperature at the colder end of the rod, A the cross-sectional area of the rod, and L its length.

Heat transfer through compound slabsOften 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 L1, L2and thermal conductivities k1, k2respectively. The temperatures of the outer surfaces are T1 and T2, where T2> T1. Let the temperature at the interface be T. Since the heat transfer rates through the two layers must be the same, we have

k1A(T – T1) k2A(T2 – T) L1 L2

=

Solving the equation for T, we have

k1L2T1 + k2L1T2

k1L2 + k2 L1

T =

Substituting this expression for T in the heat transfer rate equation for either layer, we obtain

A ( T2 – T1 )

L1 / k1 + L2 / k2=

QΔt

The rate of heat transfer through a compound slab consisting of n materials can be generalized from the 2-layer equation:

A ( T2 – T1 )

Σ Li / ki

=QΔt

i=1

n

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:

A ( T2 – T1 )=

QΔt Σ Rii=1

n

R values of some common building materials

Material R value (mK/W)

Hardwood siding (1”) 0.25Brick (4”) 1.00Fiberglass board (1”) 1.10Flat glass (0.125”) 0.30Air space (3.5”) 0.25Drywall (0.5”) 0.10Stagnant 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?

R1 (outside stagnant air layer) 0.04 mK/WR2 (brick) 1.00R3 (fiberglass board) 2.20R4 (dry wall) 0.12R5 (inside stagnant air layer) 0.04

ΣRi 3.38 mK/W

Radial heat flowConsider 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 T2 and that of the air outside the insulating material is T1 (T1 < T2), 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?

drdTkA

tQ

−=Δ

Law of thermal conduction:

QΔt

is constant at steady state

Representing by H, the law can be written as:QΔt

drdTrLkH )2( π−=

Steam pipe

Insulatingmaterial

T1

T2

2a

2b

Lk

where c is an integration constant.

cTHkLr +π

−=2ln

⇒

(equ. 1)

At r = a, T = T2 cTHkLa +π

−= 22ln (equ. 2)

From the above equation: dTHkLdr

r ∫∫π

−=21

At r = b, T = T1 cTHkLb +π

−= 12ln (equ. 3)

Solving equ. 2 and equ 3 for H and c:

)/ln()(2 12

abTTkLH −π

=

12

12 lnlnTT

aTbTc−−

=

(This is the rate of energy transfer)

Substituting the expressions for H and c in equ 1, we have

)/ln()/ln()/ln( 12

abarTrbTT +

=

ConvectionThis 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.

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Heat convection

Energy transfer by means opfconvection takes place by movement of hot particlesupwards and falling of colderones.

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Heat transfer by convectionThere 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:

QΔt = h A ΔT

where, = rate of energy transfer or heat currentQΔt

A = surface areaΔT = temperature difference between the surface

and the main body of the fluidh = convection coefficient

The convection coefficient defined in the above equation is temperature dependent and needs to be determined experimentally.

Convection coefficient

h, (W·m-2 ·K-1)Horizontal plate, facing upward 0.0595x10-4(ΔT)1/4

Horizontal plate, facing downward 0.0314x10-4(ΔT)1/4

Vertical plate 0.0424x10-4(ΔT)1/4

Horizontal or vertical pipe (dia. D) 1.00x10-4(ΔT/D)1/4

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 ΔT. The convection coefficients applicable in this situations are shown below:

Wymiana Ciepła

Przenikanie jako przykład złożonej wymiany ciepła w przypadku płaskiej ścianki można przedstawićnastępująco:

ZZłłoożżona wymiana ciepona wymiana ciepłłaa

konwekcja, α1

konwekcja, α2

przewodzenie, λ

Tf2

Tf1

AAAR

RTT

Q ff

21

21 11αλ

δα

++=−

=&

δ

RadiationThis 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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• 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.

Radiant energy from the Sun

Stefan’s lawThe 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

Δt= σ A e T4

where σ = 5.6696x10-8 W·m-2·K-4 (Stefan-Boltzmann constant)A = surface area of the object in m2

e = emissivity of the objectT = 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.

• 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.

Emissivity

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 To, then the rate of net energy gained (or lost) by the object due to emission and absorption of electromagnetic radiation is:

Q

Δt = σ A e (T 4 - To4)

Hence, if T > To , 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 oC 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.

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 = 0.9T = 800 +273.15 = 1073 K

Q

Δt = σ A e T 4,Rate of radiant energy loss:

Q

Δt= 5.6696 x 0.0208 x 0.9 x (1073)4 x 10-8

= 1407 W

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.

The Dewar flask

• 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.

Wymiana Ciepła

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∞.

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym oporze

powierzchnia A

objętość, V

ρ, c, T(t)

olej w temp. ∞TqA

dtdU

−=

( ) ( )∞−=−= TTqTTcVU αρ 0

Wymiana Ciepła

Podstawiając powyższe wyrażenia do równania bilansu energii:

Warunki brzegowe:

Po przekształceniach:

Rozwiązanie równania ma postać:

( )∞−−= TTAdtdTcV αρ

00 TTtdla ==

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym oporze

dtcVA

TTdT

ρα

−=− ∞

∫∫ −=− ∞

τ

ρα

00

dtcVA

TTdTT

T

τραcVA

TTTT

−=−−

∞

∞

0

ln

0

0

ττ

τρα −−

∞

∞ ==−− ee

TTTT cV

A

AcVαρτ =0Cieplna stała czasowa:

Wymiana Ciepła

Zdefiniujmy bezwymiarową temperaturę oraz czas:

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym oporze

cVAρ

τατττ ==

0

*

∞

∞

−−

=TTTTT

0

*

cVAρτα

τττ ==0

*

∞

∞

−−

=TTTT

T0

*Umożliwia to nam analizęprzypadków gdzie występuje gwałtowna zmiana temperatury

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

Zdefiniujmy bezwymiarową liczbę Biota:

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym oporze

λαlBi =

( )( ) λ

ααλ lAAL

=≈/1/

zewnąewnnakonwekcjiopórwewnąewniaprzewodzenopórCiała stałe

Mieszalnik ( )( ) αα UUA

A=≈

/1/1

zewnąewnnakonwekcjiopórwewnąewniaprzewodzenopór

Wymiar charakterystyczny, l, jest wyrażony stosunkiem V/ATeorię można stosować w przypadku, gdy Bi<0.1 dla płaskich płyt, walców, kul.

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

Wymiana Ciepła

Zdefiniujmy bezwymiarową liczbę Biota: i Fouriera

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym oporze

λαlBi =

Wymiar charakterystyczny, l, jest wyrażony stosunkiem V/A

Teorię można stosować w przypadku, gdy Bi<0.1 dla płaskich płyt, walców, kul

Równanie można przedstawić w postaci zależności pomiędzy liczbami podobieństwa. W tym celu wykładnik liczby e należy przekształcić do postaci:

LlFoBi

AVl

lal

VcA

p

))((2 ==τ

λα

ρτα

2laFo τ

=

LlFoBieTTTT /)()(

0

−

∞

∞ =−−

Wymiana Ciepła

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym oporzeWyznaczmy 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:

Całkowita ilość ciepła wymienioną przez ciało, w czasie do dowolnej chwili:

τρ

ddTVcQ p=&

τρα

ρα

τVc

A

p

peVc

ATTddT −

∞ −= )( 0

⎥⎦

⎤⎢⎣

⎡−−==

−

∞∫ LlFoBi

eTTAdQQ))((

0 1)(ατ&

Wymiana Ciepła

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym 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/m2K, ile czasu potrzeba aby schłodzić płytędo temperatury 100°C? Założyć własności fizyczne materiału λ, ρ, c jak dla stali, czyli 50 W/mK, 7800 kg/m3, oraz 450 J/kg K, odpowiednio.

Dane: Płyta stalowa hartowana w oleju.Szukane: Czas schłodzenia z 600°C do 100°C.Założenia: Ciało o małym oporze przewodzenia.

Wymiana Ciepła

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym 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

Znajdźmy stałą czasową zagadnienia

Podstawiając dane zadania, tj.: T0=600oC, Tfinal=100oC, T∞=30oC

Rozwiązujemy ze względu na czas:

Przewodzenie w ciaPrzewodzenie w ciałłach o maach o małłym oporzeym oporze