1/22/05me 2591 me 259 heat transfer lecture slides ii dr. gregory a. kallio dept. of mechanical...

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ME 259Heat Transfer

Lecture Slides II

Dr. Gregory A. Kallio

Dept. of Mechanical Engineering, Mechatronic Engineering & Manufacturing Technology

California State University, Chico

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Steady-State Conduction Heat Transfer

Incropera & DeWitt coverage:

– Chapter 2: General Concepts of Heat Conduction

– Chapter 3: One-Dimensional, Steady-State Conduction

– Chapter 4: Two-Dimensional, Steady-State Conduction

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General Concepts of Heat Conduction

Reading: Incropera & DeWitt

Chapter 2

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Generalized Heat Conduction

Fourier’s law, 1-D form:

Fourier’s law, general form:

- q” is the heat flux vector, which has three components; in Cartesian coordinates:

(magnitude)

dxdTkxq

Tkq

kqjqiqq zyxˆˆˆ

222zyx qqqq

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The Temperature Gradient

T is the temperature gradient, which is:– a vector quantity that points in direction of

maximum temperature increase – always perpendicular to constant

temperature surfaces, or isotherms

(Cartesian)

(Cylindrical)

(Spherical)

kz

Tj

y

Ti

x

TT ˆˆˆ

kz

Tj

T

ri

r

TT ˆˆ1ˆ

kT

rj

T

ri

r

TT ˆ

sin

1ˆ1ˆ

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Thermal Conductivity

k is the thermal conductivity of the material undergoing conduction, which is a tensor quantity in the most general case:

– most materials are homogeneous, isotropic, and their structure is time-independent; hence:

which is a scalar and usually assumed to be a constant if evaluated at the average temperature of the material

),,,,( Ttzyxkk�

),(Tkk

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Total Heat Rate

Total heat rate (q) is found by integrating the heat flux over the appropriate area:

k and T must be known in order to calculate q” from Fourier’s law– k is usually obtained from material property

tables– to find T, another equation is required;

this additional equation is derived by applying the conservation of energy principle to a differential control volume undergoing conduction heat transfer; this yields the general Heat Diffusion (Conduction) Equation

A

Adqq

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Heat Diffusion (Conduction) Equation

For a homogeneous, isotropic solid material undergoing heat conduction:

Cylindrical and spherical coordinate system forms given in text (p. 64-65)

This is a second-order, partial differential equation (PDE); its solution yields the temperature field, T(x,y,z,t), within a given solid material

t

Tcq

z

Tk

zy

Tk

yx

Tk

x

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Heat Diffusion (Conduction) Equation

For constant thermal conductivity (k):

For k = constant, steady-state conditions, and no internal heat generation

– this is known as Laplace’s equation, which appears in other branches of engineering science (e.g., fluids, electrostatics, and solid mechanics)

y)diffusivit (thermal

:where

, 1

2

2

2

2

2

2

c

k

t

T

k

q

z

T

y

T

x

T

00 22

2

2

2

2

2

Tz

T

y

T

x

T or ,

:)0( q

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Boundary Conditions and Initial Condition

Boundary Conditions: known conditions at solution domain boundaries

Initial Condition: known condition at t = 0 Number of boundary conditions required to

solve the heat diffusion equation is equal to the number of spatial dimensions multiplied by two

There is only one initial condition, which takes the form

– where Ti may be a constant or a function of x,y, and z

iTzyxT )0,,,(

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Types of Boundary Conditions for Conduction Problems

Specified surface temperature, e.g.,

Specified surface heat flux, e.g.,

Specified convection (h, T given), e.g.,

Specified radiation (, Tsur given), e.g.,

0),,,0( TtzyT

00

qx

Tk

x

),,,0(0

tzyTThx

Tk

x

),,,0(44

0

tzyTTx

Tk sur

x

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Solving the Heat Diffusion Equation

Choose a coordinate system that best fits the problem geometry.

Identify the independent variables (x,y,z,t), e,g, is it a S-S problem? Is conduction 1-D, 2-D, or 3-D? Justify assumptions.

Determine if k can be treated as constant and if

Write the general heat conduction equation using the chosen coordinates.

Reduce equation to simplest form based upon assumptions.

Write boundary conditions and initial condition (if applicable).

Obtain a general solution for T(x,y,z,t) by some method; if impossible, resort to numerical methods.

.0q

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Solving the Heat Diffusion Equation, cont.

Solve for the constants in the general solution by applying the boundary conditions and initial condition to obtain a particular solution.

Check solution for correctness (e.g., at boundaries or limits such as x = 0, t = 0, t , etc.)

Calculate heat flux or total heat rate using Fourier’s law, if required.

Optional: rearrange solution into a nondimensional form

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Example:

GIVEN: Rectangular copper bar of dimensions L x W x H is insulated on the bottom and initially at Ti throughout . Suddenly, the ends are subjected and maintained at temperatures T1 and T2 , respectively, and the other three sides are exposed to forced convection with known h, T.

FIND: Governing heat equation, BCs, and initial condition

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One-Dimensional, Steady-State Heat Conduction

Reading: Incropera & DeWitt,

Chapter 3

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1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Plane Wall

– if k = constant, general heat diffusion equation reduces to

– separating variables and integrating yields

– where T(x) is the general solution; C1 and C2 are integration constants that are determined from boundary conditions

002

2

dx

dT

dx

d

dx

Td or

211 )( CxCxTCdx

dT then and

x

L

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1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Plane Wall, cont.– suppose the boundary conditions are

– integration constants are then found to be

– the particular solution for the temperature distribution in the plane wall is now

21 )( and )0( ss TLxTTxT

1212

1 sss TC

L

TTC

and

112 )()( sss TL

xTTxT

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1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Plane wall, cont.– The conduction heat rate is found from

Fourier’s law:

– If k were not constant, e.g., k = k(T), the analysis would yield

» note that the temperature distribution would be nonlinear, in general

211 ss TTL

kAkAC

dx

dTkAq

CxqdTTk )(

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1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Electric Circuit Analogy

– heat rate in plane wall can be written as

– in electrical circuits we have Ohm’s law:

– analogy:

constant materialdifference etemperatur

kAL

TTq ss

/

)( 21

R

Vi

)resistance (electric )resistance (thermal

(voltage) re)(temperatu (current) rate)(heat

RkA

L

VT

iq

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Thermal Circuits for Plane Walls

Series Systems

Parallel Systems

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Thermal Circuits for Plane Walls, cont.

Complex Systems

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Thermal Resistances for Other Geometries Due to Conduction

Cylindrical Wall

Spherical Wall

krr

Rt 2)/ln( 12

k

rrRt 4

/1/1 21

lr1

r2

r2

r1

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Convective & Radiative Thermal Resistance

Convection

Radiation

)resistance thermal e(convectiv convt

ss

RhA

hA

TTTThAq

,

1/1

)(

)resistance thermal (radiative

where

radtr

ssr

r

ssr

RAh

TTTTh

Ah

TTTTAhq

,

22

1

))((

/1)(

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Critical Radius Concept

Since the surface areas of cylinders and spheres increase with r, there exist competing heat transfer effects with the addition of insulation under convective boundary conditions (see Example 3.4)

A critical radius (rcr) exists for radial systems, where:

– adding insulation up to this radius will increase heat transfer

– adding insulation beyond this radius will decrease heat transfer

For cylindrical systems, rcr = kins/h

For spherical systems, rcr = 2kins/h

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Thermal Contact Resistance

Thermal contact resistance exists at solid-solid interfaces due to surface roughness, creating gaps of air or other material:

K/W)(marea unit per resistance thermal

areacontact apparent where2

,

,,

ct

c

c

ct

c

BAct

R

A

A

R

qA

TTR

A

B q

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Thermal Contact Resistance

R”t,c is usually experimentally measured and depends upon– thermal conductivity of solids A and B– surface finish & cleanliness– contact pressure– gap material– temperature at contact plane

See Tables 3.1, 3.2 for typical values

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EXAMPLE

Given: two, 1cm thick plates of milled, cold-rolled steel, 3.18m roughness, clean, in air under 1 MPa contact pressure

Find: Thermal circuit and compare thermal resistances

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1-D, S-S Conduction in Simple Geometries with Heat Generation

Thermal energy can be generated within a material due to conversion from some other energy form:– Electrical

– Nuclear

– Chemical

Governing heat diffusion equation if k = constant:

systems Cartesian for

where

2

22

2 0/

dx

TdT

kqT

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S-S Heat Transfer from Extended Surfaces (i.e., fins)

Consider plane wall exposed to convection where Ts>T:

How could you enhance q ?– increase h– decrease T

– increase As (attach fins)

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Fin Nomenclature

x = longitudinal direction of fin L = fin length (base to tip) Lc = fin length corrected for tip area W = fin width (parallel to base) t = fin thickness at base Af = fin surface area exposed to fluid Ac = fin cross-sectional area, normal to heat flow Ap = fin (side) profile area P = fin perimeter that encompasses Ac

D = pin fin diameter Tb = temperature at base of fin

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1-D Conduction Model for Thin Fins

If L >> t and k/L >> h, then the temperature gradient in the longitudinal direction (x) is much greater than that in the transverse direction (y); therefore

Another way of viewing fin heat transfer is to imagine 1-D conduction with a negative heat generation rate along its length due to convection

)conduction D-(1 iqq xˆ

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Fin Performance

Fin Effectiveness

Fin Efficiency

– for a straight fin of uniform cross-section:

– where Lc = L + t / 2 (corrected fin length)

)(,

TThA

q

bbc

f

f

fin w/o area base from HTfin single from HT

)(max

TThA

q

q

q

T

bf

ff

bf

at were fin entire if HTfin single from HT

c

cf mL

mL )tanh(

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Calculating Single Fin Heat Rate from Fin Efficiency

Calculate corrected fin length, Lc

Calculate profile area, Ap

Evaluate parameter

Determine fin efficiency f from Figure 3.18, 3.19, or Table 3.5

Calculate maximum heat transfer rate from fin:

Calculate actual heat rate:

fins rrectangula for 2//2/3cpc mLkAhL

)(max, TThAq bff

max,fff qq

, , , 31

,21

,, LtALtAtLA parptripcrecp

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Maximum Heat Rate for Fins of Given Volume

Analysis:

“Optimal” design results:

constant withSet pf A

dL

dq0

profile rrectangula annular, for

profile parabolic concave for 1.7536

profile triangular for 1.3094

profile rrectangula for 1.0035

2/

3

/

12

2/3

rr

kAhL pc

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Fin Thermal Resistance

Fin heat rate:

Define fin thermal resistance:

Single fin thermal circuit:

ff

b

bfffff

hA

TT

TThAqq

/1

)(max,

ffft hA

R

1,

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Analysis of Fin Arrays

Total heat transfer = heat transfer from N fins + heat transfer from exposed base

Thermal circuit:

– where

bffb

bbbffbft

AANh

hAhANqNqq

bconvb

ffft

bc

ctct hA

RhAN

RNA

RR

11,,

,

",

, , ,

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Analysis of Fin Arrays, cont.

Overall thermal resistance:

)(,

,,1

1)(

)()(,

/1

11

1

cot

bt

bft

bcctff

f

t

fco

tcocot

R

TTq

ANAA

ARhAC

CA

NA

hAR

then

array) of area surface (total

where

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Example

Given: Annular array of 10 aluminum fins, spaced 4mm apart C-C, with inner and outer radii of 1.35 and 2.6 cm, and thickness of 1 mm. Temperature difference between base and ambient air is 180°C with a convection coefficient of 125 W/m2-K. Contact resistance of 2.75x10-4 m2-K/W exists at base.

Find: a) Total heat rate w/o and with fins b) Effect of R”t,c on heat rate

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Two-Dimensional, Steady-State Heat Conduction

Reading: Incropera & DeWitt

Chapter 4

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Governing Equation

Heat Diffusion Equation reduces to:

Solving the HDE for 2-D, S-S heat conduction by exact analysis is impossible for all but the most simple geometries with simple boundary conditions.

cartesian) D,-(2

or equation) s(Laplace'

0

0

2

2

2

2

2

y

T

x

T

T

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Solution Methods

Analytical Methods– Separation of variables (see section 4.2)– Laplace transform– Similarity technique – Conformal mapping

Graphical Methods– Plot isotherms & heat flux lines

Numerical Methods– Finite-difference method (FDM)– Finite-element method (FEM)

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Conduction Shape Factor

The heat rate in some 2-D geometries that contain two isothermal boundaries (T1, T2) with k = constant can be expressed as

– where S = conduction shape factor (see Table 4.1)

Define 2-D thermal resistance:

)( 21 TTSkq

SkR Dcondt

1)2(,

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Conduction Shape Factor, cont.

Practical applications:

– Heat loss from underground spherical tanks: Case 1

– Heat loss from underground pipes and cables: Case 2, Case 4

– Heat loss from an edge or corner of an object: Case 8, Case 9

– Heat loss from electronic components mounted on a thick substrate: Case 10