01. introduction overview of mechanisms

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    1. INTRODUCTION

    Examples of Heat Transfer Problems

    (1) Slide Projector,

    (2) Ice Storage(3) Re-entry Aerodynamic Heating

    (4) Pot Handle

    (5) Under-window radiator of heating system

    (6) Refrigerator

    Focal Point in Heat Transfer

    Determination of the temperature distribution in a region

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    Modes of Heat Transfer

    (ii) Convection: by mass motion

    (iii)Radiation: by electromagnetic waves

    Conduction: Fourier's LawS= wall area

    L L = wall thickness

    siT T1= inside surface temperature

    soT T2= outside surface temperature

    Is proportional toS? How?Q&

    Is proportional toL ? How?x

    Q&

    Is proportional to (T1 - T2)? How?Q&

    (i) Conduction: by molecular or atomic activity

    Fig.1.1

    x

    xq

    L

    dx

    siTsoT

    A

    0

    Q&

    S [m2]

    T1 T2

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    is called thermal conductivity, a property of material Equation (1.1) is valid for: (i) steady state, (ii) one-

    dimensional conduction and (iii) constant .Reformulation of (1.1): Apply (1.1) to an element dx:

    dx

    T(x)dx)+T(xS=

    dx

    dx)+T(xT(x)S=Q

    x

    &

    dxT(x),T1 dx),T(xT2 +

    L

    TTSQ 21

    x

    =& 1.1.[W]

    Heat Transfer Rate (tepeln tok)

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    (1.2) becomes

    dxdTq

    x=& (1.4)

    For 3-D:

    Equation (1.5) is known as Fourier's law

    ,x

    Tq

    x

    =& ,

    y

    Tq

    y

    =&

    z

    Tq

    z

    =& (1.5)

    Definition: Heat flux (mrn tepeln tok)x

    q&

    (1.2)dx

    dTS=Q

    x& [W]

    S

    Qq x

    x

    &

    & = (1.3)[W/m2]

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    Convection: Newton's Law of Cooling

    Convection: Energy is transported by means of massmotion

    Classification:

    (1) Free convection

    (2) Forced convection

    Heat exchange between a surface and a fluid movingover it

    TTq

    ww&

    where

    wT = surface temperature

    T = fluid temperature far away from the surface

    wq& = surface (wall) flux [W/m2] (mrn tepeln tok)

    T

    sTw

    Twq&

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    Rewrite:

    (1.6)= TTq ww

    This isNewton's law of cooling.

    NOTE:(1) is called heat transfer coefficient

    (2) It is a defined quantity

    (3) It depends on geometry, fluid properties and

    motion

    (4) To determine , the temperature distribution in the

    fluid must be known

    (5) Major objective in convection: Determination of

    (souinitel pestupu tepla)

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    Typical Values of Table 1.1Typical values of

    5 - 30

    20 - 1000

    20 - 300

    50 - 20,000

    5,000 - 50,000

    2,000 - 100,000

    5,000 - 100,000

    Free Convection

    Gases

    Liquids

    Forced ConvectionGases

    Liquids

    Liquid metals

    Phase change

    Boiling liquids

    Condensation

    (W/m2 K)ProcessImportant:

    problems.solve

    tovaluestheseusenotDoonly.guide

    aastablethisUse

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    Radiation: Stefan-Boltzmann Law Transmission by electromagnetic waves

    No medium is needed. Best in a vacuum

    Maximum possible radiation: By an idealsurfacecalled blackbody.

    Stefan-Boltzmann law for blackbody radiation flux:

    T = surface temperature, measured in absolute degrees

    (Kelvin)

    =Stefan-Boltzmann constant

    (1.7)4

    0TE = [W]

    = blackbody radiation flux (zivost)0E

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    = 5.67 x 10-8

    W/m2

    -K4 (1.8)

    Real surface:

    E= radiation flux (zivost edho povrchu)

    Emissivity, , a surface property defined as(1.9)

    0E

    E=

    Combining (1.7) and (1.9)

    4TE= (1.10)

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    12Q& = Energy exchange between two surfaces

    Absorptivity (absorptance) a : Fraction of radiationincident on a surface which is absorbed

    Simplified model: Gray surface: = aSpecial case: A small gray surface enclosed by a much

    larger surface

    (1.11))T(TSQ 42

    4

    11112=&

    ( )1= small surface, ( )

    2= large surface

    Energy Exchange Between Two Bodies:

    A Simplified Model

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    Properties

    Heat transfer depends on material and surface

    properties as:

    Conductivity Density

    Viscosity

    Specific heat Emissivity

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    Example: Application of Problem SolvingMethodology

    Square transistor is mounted

    on a circuit board. Transistorsurface is cooled by convection.

    Size, dissipated power, heat

    transfer coefficient and ambienttemperature are known.

    (1) Observations

    (i) Schematic diagram

    Determine surface

    temperature.

    T sT

    board

    transistor

    1.3Fig.

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    (ii) Dissipated electric energy is removed by

    convection and radiation

    (iii) Surface temperature is higher than ambient

    temperature

    (iv) Increasing dissipated power increases surface

    temperature

    (2) Problem DefinitionFind the relationship between power and surface

    temperature

    (3) Solution Plan

    Apply Newton's law of cooling to the surface of the

    transistor.

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    (4) Plan Execution

    (i) Assumptions(1) Steady state

    (2) Dissipated electric energy leaves transistorsurface (no energy leaves from the back side)

    (3) Negligible heat loss by radiation

    (4) Uniform surface temperature

    (5) Uniform heat transfer coefficient

    (6) Constant ambient temperature

    (ii) Analysis

    Newton's law of cooling:

    T sT

    board

    transistor

    1.3Fig.

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    = heat transfer coefficient = 8 K][W/m2

    wq& = surface heat flux, 2W/m

    wT = ? surface temperature, Co

    T = ambient air temperature = 26 Co

    Conservation of energy and assumptions (2) and (3):

    surfacefromremovedHeatpowerDissipated =

    wqSP &= (b)

    S= surface area = 1 (cm) 1 (cm) = 1 = 0.0001 m2

    (a)

    = TTqww

    &

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    P= power dissipated in transistor = 0.04 W

    (b) into (a)

    )T(TS

    Pw = (c)

    Solving forw

    T

    S

    TT

    w

    +=

    (d)

    (iii) Computations

    (d) gives

    C768.0,0001

    0,0426

    wT

    o

    =+=

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    (iv) Checking

    Dimensional check: Equation (d) is dimensionallycorrect. Each term has units of temperature.

    Qualitative check: Equation (d) behaves correctly:

    wT Decreasing the surface area increases

    Decreasing increasesw

    T

    Limiting checks: Power off (P= 0): surface temperature = ambient

    temperature

    IncreasingPincreasesw

    T

    (d)S

    PTT

    w+=

    If = 0 ( no heat is removed) then wT

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    (5) Learning and Generalizing

    < 85

    o

    CwT(i)

    (ii) Examine assumptions (2) and (3):

    Heat removed from the back side lowerscalculated

    w

    T

    Radiation heat loss lowers calculatedw

    T

    (2) no energy leaves from the back side

    (3) negligible heat loss by radiation

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    )dt

    dum

    dt

    dU(

    dt

    dEEEEE akakoutgin ====+&&&& [W]

    Requirements for Energy Conservation

    a) Balance for Energy flux

    in every moment, [W].

    b) Balance for amount

    of heat, [J]. akoutginEEEE =+ [J]

    Energy Conservation for a Control Volume

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    Requirements for Energy Conservation

    Energy Conservation for a Control Surface

    0EEoutin

    = &&