33CHAPTER CHAPTER 33St dSt d St t St t SteadySteady--State, State, OneOne--DimensionalDimensionalConductionConduction

Heat Transfer 1

ObjectivesObjectives

Determine expressions for the temperatureDetermine expressions for the temperature distribution and heat transfer rate in Plane Wall. Plane Wall. Introduce the concept of thermal resistancethermal resistance and pto show how thermal circuitsthermal circuits may be used to model heat flow.

Heat Transfer 2

I t d tiI t d tiIntroductionIntroduction

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Heat transfer problems are often classified as being steadysteady or

ii ( l ll d d )transienttransient (also called unsteady).

The term steadysteady implies no changeno changewith time at any point within the medium,

while transient transient implies variation variation with timewith time or time dependence.

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Heat transfer problems are also classified as being

• one-dimensional, • two-dimensional, ortwo dimensional, or • three-dimensional,

d di th l ti it d f h t t f t idepending on the relative magnitudes of heat transfer rates in different directions and the level of accuracy desired.

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The temperature in a medium, in some cases, varies mainly in two primarytwo primarydi idi i d th i ti fdirectionsdirections, and the variation of temperature in the third direction (and thus heat transfer in that direction) is negligible.

A heat transfer problem in that case is said to be Two-Dimensional.

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A heat transfer problem is said to be OneOne--DimensionalDimensional if theA heat transfer problem is said to be OneOne Dimensional Dimensional if the temperature in the medium varies in one direction only and thus heat is transferred in one direction, and the variation of temperature

d th h t t f i th di ti li ibland thus heat transfer in other directions are negligible or zero.

For example, (Figure).

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Methodology of a Conduction AnalysisSpecify appropriate form of the heat equation.

Solve for the temperature distribution.

Apply Fourier’s law to determine the heat flux.

C G t iCommon Geometries:

The Plane Wall: Described in rectangular (x) coordinate. Area

perpendicular to direction of heat transfer is constant (independent of x).

The Tube Wall: Radial conduction through tube wall.

The Spherical Shell: Radial conduction through shell wall.

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3.13.1The Plane WallThe Plane Wall__________________________________

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Steady Heat Conduction In Plane WallsSteady Heat Conduction In Plane Walls

IFthe temperature of the wall varies in one direction only.the surrounding temperatures remain constant. then heat transfer through the wall can be modeled as steady and one-dimensional

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In Fig. 3.1a, a plane wall In Fig. 3.1a, a plane wall separates two fluids of separates two fluids of different temperatures. different temperatures. Heat transfer occurs;Heat transfer occurs;Heat transfer occurs;Heat transfer occurs;

by convection from the by convection from the hot fluid at Thot fluid at T∞, 1∞, 1 to one to one ,,surface of the wall at surface of the wall at TTs, 1s, 1, , by conduction throughby conduction throughby conduction through by conduction through the wall, and the wall, and by convection from the by convection from the other surface of the other surface of the wall at Twall at Ts, 2s, 2 to the cold to the cold fluid at Tfluid at T 22..

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fluid at Tfluid at T∞, 2∞, 2. .

3.1.1 3.1.1 Temperature DistributionTemperature Distribution

The Standard ApproachThe Standard Approach

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The appropriate form of heat equation is Eq. 2.21

)1.3(0=⎟⎠⎞

⎜⎝⎛

dxdTk

dxd

The equation may be integrated twice to obtain the general solutiontwice to obtain the general solution

T(x) = C1x + C2 (3.2)

Apply BCs of the first kind at x = 0 and x = L, to obtain C1 and C2, inand x L, to obtain C1 and C2, in which case

T(0)=Ts,1 and T(L)=Ts,2

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T(x) = C1x + C2 (3.2)

l i h di i h l l i i

T(0)=Ts,1 and T(L)=Ts,2

Applying the condition at x = 0 to the general solution, it follows that

T( ) T(0) T C *0 + CT(x) = T(0) = Ts,1= C1*0 + C2

Ts,1= C2

Similarly, at x = L,

T(x) = T(L) = Ts,2 = C1L + C2= C1L+Ts,1

in which case

( ) ( ) s,2 1 2 1 s,1

11,2, C

TT ss =−

Heat Transfer 14

1CL

T(x) = C1x + C2 (3 2)T(x) C1x + C2 (3.2)

LTT

C ss 1,2,1

−= C2=Ts,1

Substituting into the general solution, the temperature distribution is then

L

)3.3()()( 1,1,2, sss TTTLxxT +−=

In a plane wall, the temperature varies linearly with x.Assumptions:

1D,Steady-State conduction,No heat generation and

Heat Transfer 15

No heat generation and, Constant k

)33()()( TTTxxT +

Now that we have the temperature distribution, we may use Fourier’s law to determine the conduction heat transfer

)3.3()()( 1,1,2, sss TTTL

xT +−=

use Fourier s law to determine the conduction heat transfer rate. That is

)43()( TTkAdTkA

The heat flux is

)4.3()( 2,1, ssx TTLdx

kAq −=−=

)5.3()( 2,1,"

ssx

x TTLk

Aq

q −==

Eqs. 3.4 & 3.5 indicate that both heat transfer rate and heat flux are constants, independent of x.

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3.1.2 3.1.2 Thermal ResistanceThermal Resistance

Heat Transfer 17

)4.3()( 21 ssx TTLkA

ddTkAq −=−=

A very important concept is suggested by Eq. 3.4. In

)()( 2,1, ssx Ldxq

gg y qparticular, there exists an analogy between the diffusion of heat anddiffusion of heat and electrical charge.

Just as an electricalJust as an electrical resistance is associated with the conduction of l i i h lelectricity, a thermal

resistance may be associated with the

Heat Transfer 18

conduction of heat.

Defining resistance as the ratio of a driving potential to theDefining resistance as the ratio of a driving potential to the corresponding transfer rate, it follows from Eq. 3.4 that the thermal resistance for conduction in a plane wall is

)6.3(2,1,, kA

Lq

TTR

x

sscondt =

−=

qx

Similarly, for electrical conduction in the same system, Ohm’s law provides an electrical resistance of the formp

)7.3(2,1,

AL

IEE

R sse σ

=−

=AI σ

The analogy between Eqs. 3.6 and 3.7 is obvious.

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A thermal resistance may also be associated with heatA thermal resistance may also be associated with heat transfer by convection at a surface. From Newton’s law of cooling,

)8.3()( ∞−= TThAq s

The thermal resistance for convection is then

1TT − ∞ )9.3(1, hAq

TTR s

convt == ∞

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Circuit representations provide a useful t l f b th t li i dtool for both conceptualizing and quantifying heat transfer problems.

The equivalent thermal circuit for the plane wall with convection surface conditions is shown in Fig. 3.1b.

The heat transfer rate may be determined from separate consideration of each element in the network. Since qx is constant throughout the network, it follows that

)103()()()( 2,2,2,1,1,1, TTTTTT

q ssss ∞∞ −−−

Heat Transfer 21

)10.3(/1//1 2

,,,,

1

,,

AhkALAhqx ===

In terms of the overall temperature difference, T∞,1 – T∞,2 and the total thermal resistance R the heat transfer rate ma also betotal thermal resistance, Rtot, the heat transfer rate may also be expressed as

)113()( 21 TT ∞∞ −

Because the conduction and convection resistances are in series

)11.3()( 2,1,

totx R

q ∞∞=

and may be summed, it follows that

)12.3(11 LRtot ++=

A thermal resistance for radiation may be defined by reference to Eq 1 8:

)(21 AhkAAhtot

to Eq. 1.8:

)13.3(1, Ahq

TTR

rrad

sursradt =

−=

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)9.1(.))(( 22 EqseeTTTThwhere

q

surssursr

rrad

++= εσ

3.1.3 3.1.3 The Composite WallThe Composite Wall

Heat Transfer 23

Composite walls mayComposite walls may involve any number of series and parallel thermal

i t d t l fresistances due to layers of different materials.

Consider the series composite wall of Figure 3 2 Th di i l3.2. The one-dimensional heat transfer rate for this system may expressed as:y y p

)14.3(4,1,

∑∞∞ −

=t

x R

TTq

where T∞ ,1 – T∞ ,4 is the overalltemperature difference and the s mmation incl des all the thermal

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∑ t summation includes all the thermal resistances.

)15.3()]/1()/()/()/()/1[(

4,1,

AhAkLAkLAkLAhTT

qx−

= ∞∞Hence )()]/1()/()/()/()/1[( 41 AhAkLAkLAkLAh

qCCBBAA

x ++++

Alternatively, the heat transfer rate can be relatedtransfer rate can be related to temperature difference and thermal resistance

i t d ith hassociated with each element.

−− TTTTTT

For example

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)16.3(.....)/()/()/1(

3221,

1

1,1, =−

=−

=−

= ∞

AkLTT

AkLTT

AhTT

qBBAA

ssx

With composite systems it is often convenient to work with an overall heat transfer coefficient, U, which is defined by an expression analogous to Newton’s law of cooling. Accordingly TTAccordingly

)17.3(TUAqx Δ=

The overall heat transfer coefficient is related to the total

)14.3(4,1,

∑∞∞ −

=t

x R

TTq

The overall heat transfer coefficient is related to the total thermal resistance, and from Eqs. 3.14 and 3.17 we see that

UA=1/Rtot

)18.3()]/1()/()/()/()/1[(

11hkLkLkLhAR

U++++

==

totHence for the composite wall of Fig.3.2

)]/1()/()/()/()/1[( 41 hkLkLkLhAR CCBBAAtot ++++

In general, we may write1TΔ

Heat Transfer 26

)19.3(1UAq

TRR ttot =Δ

==∑

S i P ll l C it W ll

Note departure from one-

Series – Parallel Composite Wall:

dimensional conditions for kF ≠ kG

Ci it b d ti fCircuits based on assumption of: (a) isothermal surfaces normal to x

direction, or (b) adiabatic surfaces parallel to x

direction provide approximations for q .for qx.

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3.1.4 Contact Resistance3.1.4 Contact Resistance

Heat Transfer 28

Thermal Contact Resistance

In the analysis of heat conduction through multilayer solids, we assumed “perfect contact” at the interface of two layers, and thus no temperature drop at the interface.

This would be the case when the surfaces are perfectly smooth and they produce a perfect contact at each pointperfect contact at each point.

Heat Transfer 29

Thermal Contact ResistanceIn reality, a surface is microscopically rough with numerous peaks and valleys, no matter how smooth it appears to besmooth it appears to be.

When two such surfaces are pressed against each other the peaks will form good material contactother, the peaks will form good material contact but the valleys will form voids filled with air.

As a result, an interface will contain numerous airAs a result, an interface will contain numerous air gaps of varying sizes that act as insulation because of the low thermal conductivity of air.

Thus, an interface offers some resistance to heat transfer, and this resistance per unit interface area is called thermal contact resistance, RC.

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How thermal contact resistance can be minimized?How thermal contact resistance can be minimized?

Thermal contact resistance can be minimized by l i h ll d i li id h fapplying a thermally conducting liquid on the surfaces

before they are pressed against each other, by replacing the air at the interface by a better conducting y p g y ggas such as helium or hydrogen, by increasing the interface pressure, and by inserting a soft metallic foil such as tin silver copperby inserting a soft metallic foil such as tin, silver, copper, nickel, or aluminum between the two surfaces.

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Example ProblemExample Problempp

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