ME 259 Heat Transfer Lecture Slides II

1/22/05

Dr. Gregory A. Kallio Dept. of Mechanical Engineering, Mechatronic Engineering & Manufacturing Technology California State University, Chico

ME 259 1

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 1/22/05 ME 259 2

General Concepts of Heat Conduction

Reading: Incropera & DeWitt Chapter 2

1/22/05 ME 259 3

Generalized Heat Conduction



Fourier’s law, 1-D form:



q

  

k dT dx

Fourier’s law, general form:



q

  

k



T

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

q

 

q x



i

ˆ 

q y

 

q z



k

ˆ 

q

 

q x

 2 

q y

 2 

q z

 2 (magnitude) 1/22/05 ME 259 4

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 

T

 

T



x i

ˆ  

T



y j

ˆ  

T



z k

ˆ (Cartesian) 

T

 

T



r i

ˆ  1

r



T

 

j

ˆ  

T



z k

ˆ (Cylindrical) 

T

 

T



r i

ˆ  1

r



T

 

j

ˆ  1

r

sin  

T

 

k

ˆ (Spherical) 1/22/05 ME 259 5

Thermal Conductivity

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

k



k

 (

x

,

y

,

z

,

t

,

T

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

k



k

(

T

), which is a scalar and usually assumed to be a constant if evaluated at the average temperature of the material 1/22/05 ME 259 6

Total Heat Rate



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

q

 

A



q

  

d A



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

1/22/05 ME 259 7

Heat Diffusion (Conduction) Equation



For a homogeneous, isotropic solid material undergoing heat conduction:

 

x

  

k



T



x

     

y

  

k



T



y

     

z

  

k



T



z

   

q

  

c



T



t

 

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

1/22/05 ME 259 8

Heat Diffusion (Conduction) Equation



For constant thermal conductivity (k):

 2

T



x

2   2

T



y

2   2

T



z

2 

q k

  1  

T



t

, where :  

k



c

(thermal diffusivit y) 

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

(

q

  0 ) :  2

T



x

2   2

T



y

2   2

T



z

2  0 , or  2

T

 0 – this is known as Laplace’s equation, which appears in other branches of engineering science (e.g., fluids, electrostatics, and solid mechanics) 1/22/05 ME 259 9

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

T

(

x

,

y

,

z

, 0 ) 

T i

– where T

i

x,y, and z may be a constant or a function of 1/22/05 ME 259 10

Types of Boundary Conditions for Conduction Problems



Specified surface temperature, e.g.,

T

( 0 ,

y

,

z

,

t

) 

T

0 

Specified surface heat flux, e.g.,



k



T



x x

 0 

q

0  

Specified convection (h, T



given), e.g.,



k



T



x x

 0 

h



T

 

T

( 0 ,

y

,

z

,

t

)  

Specified radiation (



, T sur

given), e.g.,



k



T



x x

 0    4

T sur



T

4 ( 0 ,

y

,

z

,

t

)  ME 259 1/22/05 11

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

q

  0 .

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.

1/22/05 ME 259 12

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

1/22/05 ME 259 13

Example:

  GIVEN: Rectangular copper bar of dimensions L x W x H is insulated on the bottom and initially at T

T 1

and T

2

known h, T

i

throughout . Suddenly, the ends are subjected and maintained at temperatures , respectively, and the other three sides are exposed to forced convection with  . FIND: Governing heat equation, BCs, and initial condition 1/22/05 ME 259 14

One-Dimensional, Steady State Heat Conduction

Reading: Incropera & DeWitt, Chapter 3

1/22/05 ME 259 15

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



Plane Wall

L x

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

d

2

T dx

2  0 or

d dx dT dx

 0 – separating variables and integrating yields

dT dx



C

1 and then

T

(

x

) 

C

1

x



C

2 1/22/05 – where T(x) is the general solution; C

1 C 2

are integration constants that are determined from boundary conditions and ME 259 16

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



Plane Wall, cont.

– suppose the boundary conditions are

T

(

x

 0 ) 

T s

1 and

T

(

x



L

) 

T s

2 – integration constants are then found to be

C

1 

T s

2 

T s

1 and

C

2

L



T s

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

T

(

x

)  (

T s

2 

T s

1 )

x L



T s

1 1/22/05 ME 259 17

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:

q

 

kA dT dx

 

kAC

1 

kA



T s

1

L



T s

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

k

(

T

)

dT



q



x



C

» note that the temperature distribution would be nonlinear, in general 1/22/05 ME 259 18

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



Electric Circuit Analogy

– heat rate in plane wall can be written as

q

 (

T s

1 

T s

2 ) 

L

/

kA

temperatur e difference material constant – in electrical circuits we have Ohm’s law:

i

 

V R

– analogy:

q

(heat rate) 

i

(current)

T

(temperatu re) 

V

(voltage)

L kA

(thermal resistance ) 

R

(electric resistance ) 1/22/05 ME 259 19

Thermal Circuits for Plane Walls



Series Systems



Parallel Systems

1/22/05 ME 259 20

Thermal Circuits for Plane Walls, cont.



Complex Systems

1/22/05 ME 259 21

Thermal Resistances for Other Geometries Due to Conduction



Cylindrical Wall

r 1 r 2 R t

 ln(

r

2 2  /

k



r

1 ) 

Spherical Wall

r 2 r 1

l 1/22/05

R t

 1 /

r

1  1 4 

k

/

r

2 ME 259 22

Convective & Radiative Thermal Resistance



Convection

q



hA

(

T s



T

 ) 

T s



T

1 /

hA

 1

hA



R t

,

conv

(convectiv e thermal resistance ) 

Radiation

q



h r A

(

T s

where

h r



T

 ) 

T s



T

 1 /

h r A

  (

T s



T

 )(

T s

2 

T

2  ) 1

h r A



R t

,

rad

(radiative thermal resistance ) 1/22/05 ME 259 23

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 (r

cr

) 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, r

cr = k ins /h

 For spherical systems, r

cr = 2k ins /h

1/22/05 ME 259 24

Thermal Contact Resistance

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

A B q R t

,

c



T A



T B



R t

, 

c qA c A c

where

A c

 apparent contact area

R t

 ,

c

 thermal resistance per unit area (m 2 K/W) 1/22/05 ME 259 25

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 1/22/05 ME 259 26

EXAMPLE

  Given: two, 1cm thick plates of milled, cold rolled steel, 3.18

 m roughness, clean, in air under 1 MPa contact pressure Find: Thermal circuit and compare thermal resistances 1/22/05 ME 259 27

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:

 2

T



q

 /

k

 0 where  2

T



d

2

T dx

2 for Cartesian systems 1/22/05 ME 259 28

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

 Consider plane wall exposed to convection where T

s >T



:



How could you enhance q ?

– increase h – decrease T  – increase A

s

(attach fins) 1/22/05 ME 259 29

Fin Nomenclature

           x = longitudinal direction of fin L = fin length (base to tip)

L c

= fin length corrected for tip area W = fin width (parallel to base) t = fin thickness at base

A f A c

= fin surface area exposed to fluid = fin cross-sectional area, normal to heat flow Ap = fin (side) profile area P = fin perimeter that encompasses A

c

D = pin fin diameter

T b

= temperature at base of fin 1/22/05 ME 259 30

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 

q

 

q x i

ˆ (1 D conduction )  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 1/22/05 ME 259 31

Fin Performance



Fin Effectiveness



f

  HT from single fin HT from base area w/o fin

q f hA c

,

b

(

T b



T

 ) 

Fin Efficiency



f

 HT from single fin HT if entire fin were at

T b



q f q

max 

hA f

(

q f T b



T

 ) – for a straight fin of uniform cross-section: 

f

 tanh(

mL c

)

mL c

– where L

c

= L + t / 2 (corrected fin length) 1/22/05 ME 259 32

Calculating Single Fin Heat Rate from Fin Efficiency

  Calculate corrected fin length, L

c

Calculate profile area, A

p A p

,

rec



L c t

,

A p

,

tri

 1 2

Lt

,

A p

,

par

 1 3

Lt

,  Evaluate parameter

L

3

c

/ 2

h

/

kA p

 

mL c

/ 2 for rectangula r fins    Determine fin efficiency 

f

3.19, or Table 3.5

from Figure 3.18, Calculate maximum heat transfer rate from fin:

q f

, max 

hA f

(

T b



T

 )  Calculate actual heat rate:

q f

 

f q f

, max 1/22/05 ME 259 33

Maximum Heat Rate for Fins of Given Volume

 Analysis: Set

dq dL f

 0 with

A p

 constant  “Optimal” design results:

L

3

c

/ 2

h

/

kA p

 1.0035

for rectangula r profile  1.3094

for triangular profile  1.7536

for concave parabolic profile 

r

2 /

r

3  2 for annular, rectangula 1 r profile 1/22/05 ME 259 34

Fin Thermal Resistance

 Fin heat rate:

q f

 

f q f

, max  

f hA f

(

T b



T

 )  1

T b

/  

f T



hA f

 Define fin thermal resistance:

R t

,

f

 

f

1

hA f

 Single fin thermal circuit: 1/22/05 ME 259 35

Analysis of Fin Arrays

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

q t



Nq f



q b

 

N



f h



b



hA N



f



f b A f



hA b

 

A b



b

Thermal circuit:

– where

R t

,

c



R t

" ,

c NA c

,

b

,

R t

,

f



N



f

1

hA f

,

R b

,

conv

 1

hA b

1/22/05 ME 259 36

Analysis of Fin Arrays, cont.

 Overall thermal resistance:

R t

,

o

(

c

)  1 

o

(

c

)

hA t

where 

o

(

c

)

C

1   1  1  

f NA A t f hA f

  1

R t

, 

c

 / 

C

1

A f c

  ,

b A t



NA f



A b

(total surface area of array) then

q t



T b



T R t

,

o

(

c

)  1/22/05 ME 259 37

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/m 2 -K. Contact resistance of 2.75x10

-4 m 2 -K/W exists at base.

 Find: a) Total heat rate w/o and with fins b) Effect of R” t,c on heat rate 1/22/05 ME 259 38

Two-Dimensional, Steady State Heat Conduction

Reading: Incropera & DeWitt Chapter 4

1/22/05 ME 259 39

Governing Equation

 Heat Diffusion Equation reduces to:  2

T

 or 0 (Laplace' s equation)  2

T



x

2   2

T



y

2  0 (2 D, cartesian)  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.

1/22/05 ME 259 40

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) 1/22/05 ME 259 41

Conduction Shape Factor

 The heat rate in some 2-D geometries that contain two isothermal boundaries (T

1 , T 2

) with k = constant can be expressed as

q



Sk

(

T

1 

T

2 ) – where S = conduction shape factor (see Table 4.1)  Define 2-D thermal resistance:

R t

,

cond

( 2 

D

)  1

Sk

1/22/05 ME 259 42

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 1/22/05 ME 259 43