steady state

Download STEADY STATE

Post on 18-Jan-2016

258 views

Category:

Documents

4 download

Embed Size (px)

DESCRIPTION

one dimension

TRANSCRIPT

  • CHAPTER 3

    STEADY-STATE ONE-DIMENSIONAL CONDUCTION

  • The term one-dimensional refers to the fact that only one coordinate is needed to describe the spatial variation of the dependent variables.

    In one dimensional system, temperature gradients exists along only a single coordinate direction.

    In this chapter we will

    - Learn how to obtain temperature profiles for common geometries with and without heat generation.

    - Introduce the concept of thermal resistance and

    thermal circuits

    Steady-State, One-Dimensional Conduction

  • The Plane Wall

    Consider a simple case of one-dimensional conduction in a plane

    wall, separating two fluids of different temperature, without energy

    generation

    Temperature is a function of x

    Heat is transferred in the x-direction

    Must consider

    Convection from hot fluid to wall

    Conduction through wall

    Convection from wall to cold fluid

    qx

    1,T

    1,sT

    2,sT

    2,T

    x

    x=0 x=L

    11, ,hT

    22, ,hT

    Hot fluid

    Cold fluid

  • Temperature Distribution Heat diffusion equation in the x-direction for steady-state

    conditions, with no energy generation:

    Integrate twice to obtain the general solution:

    T(x) = C1x + C2

    To obtain C1 and C2, boundary conditions must be introduced

    at x=0, T(0) = Ts,1

    at x=L, T(L) = Ts,2

    Apply the condition at x=0 to the general solution

    Ts,1 = C2

    Apply the condition at x=L to the general solution

    Ts,2 = C1L + C2 = C1L + Ts,1

    C1 = Ts,2 Ts,1

    L

    0

    dx

    dTk

    dx

    d qx is constant

  • Temperature Distribution

    Apply to the general solution, the temperature profile, assuming

    constant k:

    From Fouriers Law, we can determine the conduction heat transfer rate:

    qx = -kAdT = kA (Ts,1 Ts,2)

    dx L

    And heat flux:

    qx = qx = k (Ts,1 Ts,2)

    A L

    1,1,2, )()( sss TL

    xTTxT

  • Problem 3.1

    Consider the plane wall, separating hot and cold fluids at temp. T,1 and T,2 respectively. Using surface energy balances as boundary conditions at x=0 and x=L, obtain the temp. distribution within the wall and the heat flux in terms of T,1 , T,2 , h1 , h2 , k and L.

  • Thermal Resistance

    Based on the previous solution, the conduction hear transfer rate can be calculated:

    Similarly for heat convection, Newtons law of cooling applies:

    And for radiation heat transfer:

    Recall electric circuit theory - Ohms law for electrical resistance:

    kAL

    TTTT

    L

    kA

    dx

    dTkAq

    ssssx

    /

    2,1,2,1,

    hA

    TTTThAq SSx

    /1

    )()(

    Ah

    TTTTAhq

    r

    surssursrrad

    /1

    )()(

    Resistance

    Difference Potential=currentElectric

  • Thermal Resistance

    We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).

    The temperature difference is the potential or driving force for the heat flow

    The combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance

    R

    Tq overall

    Resistance

    Force DrivingOverall

    AhR

    hAR

    kA

    LR

    rradtconvtcondt

    1,

    1, ,,,

    CONDUCTION CONVECTION RADIATION

  • Thermal Resistance for Plane Wall

    In terms of overall temperature difference:

    qx

    1,T

    1,sT

    2,sT

    2,T

    x x=0

    x=L

    11, ,hT

    22, ,hT

    Hot fluid

    Cold fluid

    AhkA

    L

    AhR

    R

    TTq

    tot

    totx

    21

    2,1,

    11

    Ah

    TT

    kAL

    TT

    Ah

    TTq

    ssssx

    2

    2,2,2,1,

    1

    1,1,

    /1//1

  • Thermal Resistance for Composite Walls

  • Thermal Resistance for Composite Walls

    Heat transfer rate for composite wall:

    qx = T,1 - T ,4

    Rt

    = T,1 - T ,4

    [(1/h1A) + (LA/kAA) + (LB/kBA) + (LC/kCA) + (1/h4A)]

    -Alternatively:

    qx= T,1 Ts,1 = Ts,1 T2 = T2 T3 = U = overall heat transfer coefficient

    (1/h1A) (LA/kAA) (LB/kBA) T = temperature difference (overall)

    )]/1()/()/()/()/1[(

    11

    41 hkLkLkLhARU

    CCBBAAtot

    UAq

    TRR

    TUAq

    ttot

    x

    1

  • Composite Walls

    For resistances in series: Rtot=R1+R2++Rn

    For resistances in parallel:

    1/Rtot=1/R1+1/R2++1/Rn

  • Contact Resistance

    In composite systems, the temperature drop across the interface between materials may be appreciable, due to surface roughness effects.

    This temperature change is attribute to thermal contact resistance:

    "

    ",

    x

    BAct

    q

    TTR

    See tables 3.1, 3.2 for typical values of Rt,c

  • Problem 3.4 In a manufacturing process, a transparent film is being bonded to a

    substrate. To cure the bond at a temperature To, a radiant source is used to provide a heat flux qo (W/m

    2), all of which is absorbed at the bonded surface. The back of the substrate is maintained at T1 while the free surface of the film is exposed to air at T and a convection heat transfer coefficient, h.

    a) Show a thermal circuit representing the steady

    state heat transfer situation.

    b) Assume the following conditions:

    T = 20C, h = 50 W/m2K, T1 = 30C

    Calculate the heat flux qo that is required to

    maintain the bonded surface at To = 60C

  • Problem 3.15

    Consider a composite wall that includes an 8-mm thick hardwood siding (A), 40-mm by 130-mm hardwood studs (B) on 0.65-m centers with glass fiber insulation (D) (paper faced, 28 kg/m3) and a 12-mm layer of gypsum (vermiculite) wall board (C).

    What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high?)

    (Note: Consider the direction of heat transfer to be downwards, along the x-direction)

    Given: PROPERTIES from Table A-3 (T 300K): Hardwood siding, kA = 0.094 W/mK Hardwood, kB = 0.16 W/mK Gypsum, kC = 0.17 W/mK Insulation (glass fiber paper faced, 28 kg/m3), kD = 0.038W/mK.

  • Problem 3.20

    A composite wall separates combustion gases at 2600C

    from a liq coolant at 100C with gas and liq side convection

    coefficients of 50 and 1000 W/m2K. The wall is composed of

    a 10mm thick layer of beryllium oxide on the gas side. The

    contact resistance between the oxide and the steel is 0.05

    m2K/W.

    a) What is the heat loss per unit surface area of the composite?

    b) Sketch the temp. distribution from the gas to the liquid.

    Assume temperature of beryllium oxide at 1500 K

    and stainless steel at 1000 K.

  • Problem 3.22

    Consider a plane composite wall that is composed of two

    materials of thermal conductivities kA = 0.1 w/m.k and kB =

    0.04 w/m.k and thicknesses LA = 10mm and LB = 20mm.

    The contact resistance at the interface between the two

    mterials is known to be 0.30 m2k/w. Material A adjoins a

    fluid at 200c for which h = 10w/m2k and material B

    adjoins a fluid at 40c for which h = 20w/m2k

    a) What is the rate of heat transfer through a wall that is 2m high by 2.5m wide

    b) Sketch the temp. distribution

  • Radial Systems-Cylindrical Coordinates

    Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures

    dr

    dT)rL2(k=

    dr

    dTkA=qr - -

  • Radial Systems-Cylindrical Coordinates

    Heat diffusion equation in the r-direction for steady-state conditions, with no energy generation (for cylinder):

    Integrate twice to obtain general soution

    T(r) = C1 ln r + C2

    To obtain C1 and C2 , boundary conditions:

    T(r1) = Ts,1 and T(r2) = Ts,2

    Apply to the general solution:

    Ts,1 = C1 ln r + C2

    Ts,2 = C1 ln r + C2

    01

    dr

    dTkr

    dr

    d

    r

  • Solve C1 and C2 and substitute into general solution: The conduction heat transfer rate can be calculated: - Fouriers Law : In terms of equivalent thermal circuit:

    2,221

    2,1,ln

    )/ln(

    )()( s

    ssT

    r

    r

    rr

    TTrT

    condt

    ssssssr

    R

    TT

    Lkrr

    TT

    rr

    TTLkq

    ,

    2,1,

    12

    2,1,

    12

    2,1,

    )2/()/ln()/ln(

    2

    )2(

    1

    2

    )/ln(

    )2(

    1

    22

    12

    11

    2,1,

    LrhkL

    rr

    LrhR

    R

    TTq

    tot

    totr

    constdr

    dTrLk

    dr

    dTkAqr )2(

  • Composite Walls

  • The heat transfer rate may be expressed as

    qr = T,1 - T ,4

    1 + ln (r2/r1) + ln (r3/r2) + ln (r4/r3) + 1

    2r1Lh1 2kAL 2kBL 2kcL 2r4Lh4

    The heat transfer rate may also be expressed in terms of an overall heat

    transfer coefficient:

    qr = T,1 - T ,4 = UA(T,1 - T ,4)

    Rtot where U is the overall heat transfer coefficient. If A = A1 = 2r1L

    44

    1

    3

    41

    2

    31

    1

    2