steady heat conduction - lunds tekniska högskola

Steady Heat Conduction(Chapter 3)Zan Wu zan.wu@energy.lth.se Room: 5113

Objectives

Steady-state heat conduction• Without internal heat generation

- Derive temperature profile for a plane wall

- Derive temperature profile for a circular layer

- Interpret thermal resistance and apply this concept to calculate heat transfer rate

• With internal heat generation

- Derive temperature profile for a plane wall with Q’

Heat conduction equation (isotropic material)

If constant

Thermal diffusivity

+ + 1 18

1 19

Simple plane wall

BC: x = 0, t = t1; x = b, t = t2

General solution: t = c1x + c2

0

0

0

constant0 (1-19)

Simple plane wall

Heat flow, Fourier’s law

Alternate formulation

potential = resistance · current

)33(xb

tttt 121

(3-4)

·

Composite wall

“Serial circuit” 1 4 ( thermal resistence)t t Q

QA

bA

bA

btt3

3

2

2

1

141

)63(

Ab

Ab

Ab

ttQ

3

3

2

2

1

141

Convective thermal resistance

A

Newton’s law of cooling

1A ·

Convective thermal resistance:

Composite wall, convective BC

tf2

tf1

b1

1 2 3

b2 b3

1

2

)73(

A1

Ab

A1

ttQ

2

3

li i

i

1

2ff1

Hot liquid

Cold liquid

Circular tube or layer (Shell)

(3 10)

1 2 lni o

o i

t tQL r r

Composite circular walls

i of f

32

1 1 1 2 2 3 0

1 1 1 1ln ln2 2 2 2i

t tQ rr

r L L r L r r L

Contact resistanceTemperature drop due to thermal contact resistance

Fouling

The accumulation and formation of unwanted materials on the surfaces of processing equipmentOne of the major unsolved problems in heat transfer

Plane wall with internal heat generation

b

x

btf tf

Q'

Uniform heat generation per unit volume

Governing equation

)181(

Qzt

z

yt

yxt

xtc

2

2 0d t Qdx

21 2

'2Qt x c x c

General solution

Boundary conditions

0x 0dtdx

bx wall fdt t tdx

At the plane of symmetry

Adiabatic or insulated BC

Solution

2 2f( )

2Q Q bt b x t

bQbQtt 2

fmax 2

Recap: steady-state heat conduction

• Start with the heat conduction equation, simply it with proper assumptions

• Then get a general solution, combining with BCs to obtain a specific solution for temperature distribution

• Use the Fourier’s Law to obtain the heat transfer rate based on the temperature distribution

Heat Transfer from Fins, Extended Surfaces

Objectives

• Derive governing equations and formulate boundary conditions for rectangular and triangular fins

• Calculate fin efficiency and fin effectiveness

• Understand optimal fin criteria for rectangular and triangular fins

• Apply fin efficiency in heat transfer rate calculations

Example fins

(a) Individually finned tubes(b) flat (continuous) fins on an array of tubes

Example fins

Example microfins

Microfin copper tubeCarbon nanotube microfins

on a chip surface

Fins on Stegosaurus

Absorb radiation from the sun or cool the blood?

Rectangular fin

Rectangular fin

Boundary conditions:

long and thin fin, heat transfer at the fin tip is negligible

)313(0AC

dxd

2

2

)tt( fb

2bZ

Z2ACm 2

)tt(dxdt:Lx fLx

0dxdt

Lx

f111 tttt:0x

x d x

L

t 1

Q 1

.

t f

bZ

Rectangular fin

General solution:

(co sh2

sin h )2

m x m x

m x m x

e em x

e em x

1 2

3 4c o s h s in h

m x m xC e C e

C m x C m x

Hyperbolic functions

For x = L = 2

1 1

c o s h ( ) ( 3 3 8 )c o s h

f

f

t t m L xt t m L

2

1

1c o s h m L

heat transfer from the fin ?Q

1 10

s in h ( )co shx

d m LQ A A mdx m L

CmA

Rectangular fin

· tanh 2 · tanh 3 40

Rectangular fin

= 25 W/m2K, b = 2 cm, L = 10 cmEq. (3-38)

Fin performance evaluation

1: Fin effectiveness

2: Fin efficiency

1

1

from the finfrom the base area w ithout the fin

QQ

fromthefinfromasimilarfinbutwithλ ∞

Optimal fin: maximum heat transfer at a given fin weight

M = b L Z = Z A1

A1 = b L, Z, are given.

Find maximum for A1 = bL, constant.

(3-40)

C 2Z , A = bZ (3-35)

(3-52)

mLtanhACQ 11

b2

ACm 2

1Q

b

Ab

2tanhZb2Q 111

LZb

Optimal rectangular fin

Condition

1 0 gives optimum

after some algebra one obtains

21.419 (3 55)/ 2

dQdb

Lb b

Straight triangular fin

= t tf

Heat balance

Solution:

K0 as x 0 B = 0 because is finite for x = 0

x = L = 1

)623(0bL2

x1

dxd

x1

dxd

2

2

bL2

)x2(BKx2AI 00

L2AI 01

LxbZA

L

d x

x

b t 1

t f

1Q

Bessel differential equation

I0 and K0 are the modified Bessel functions of order zero

Triangular fin

)L2(IA

0

1

)653(

)L2(I)x2(I

0

0

1

LxdxdtAQ

1

Lx

0

011 dx

)x2(dI)L2(I

1bZQ

)663()L2(I)L2(Ib2ZQ

0

111

Table 3.2 for numerical values of I0 and I1

)673(b

2309.12/b

L

Optimal triangular fin: maximum heat transfer at given weight

Summary of formulae for rectangular and triangular fins

optimal fin optimal fin

)383(mLcosh

)xL(mcosh

ft1tftt

1

b22m

)403(mLtanh1Zb21Q

mLtanhb

2

mLmLtanh

)553(b

2419.12/b

L

)653()L2(0I)x2(0I

1

bL2

)L2(0I)L2(1I

1Zb21Q

)L2(0I)L2(1I

b2

L)L2(0I/)L2(1I

)673(b

2309.12/b

L

Formulas for fin performance

Some simple calculations give:

Rectangular fin

Triangular fin

mLtanhb

2

)L2(I)L2(I

b2

0

1

L)L2(I/)L2(I 01

mLmLtanh

Fin effectiveness

Fin efficiency

37

Circular or annular fins

Heat conducting area

A = 2r b

Convective perimeter

C = 2 2r = 4r

r 1 r 2

b

Fin efficiency for circular fins

How to use the fin efficiency in engineering calculations

s

flänsarareaoflänsad

QQQ

QQQ

finareaunfinned

( )

( )

fins b f

b b f fins

A t t

Q A t t Q

a

( ) (3 7 1)b f b f in sQ t t A A a