Steady Heat Conduction (Chapter 3) - Värmeöverfö · PDF fileAgenda •...

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Steady Heat Conduction (Chapter 3) Zan Wu [email protected] Room: 5123

Transcript of Steady Heat Conduction (Chapter 3) - Värmeöverfö · PDF fileAgenda •...

Steady Heat Conduction (Chapter 3) Zan Wu [email protected] Room: 5123

Agenda

• Steady-state heat conduction - without internal heat generation

- with internal heat generation

• Fins, extended surfaces - Rectangular fin

- Triangular fin

- Circular fin

Heat conduction equation isotropic material

If λ constant ⇒ homogeneous and isotropic material

)181( −′+

∂∂

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

=∂∂

Qzt

z

yt

yxt

xtc

λ

λλτ

ρ

2 2 2

2 2 2 (1 19)t t t t Qax y z cτ ρ

′ ∂ ∂ ∂ ∂= + + + − ∂ ∂ ∂ ∂

a thermal diffusivitycλρ

=

Simple Plane Wall

Steady heat conduction

One dimensional case No internal heat generation, Q’ = 0; Constant λ (1-19) Solution: t = c1x + c2

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

0=τ∂∂

0zy≡

∂∂

=∂∂

0dx

td2

2=

Simple Plane Wall

Heat flow

Alternate formulation

dxdtAQ λ−=

)33(xb

tttt 121 −

−+=

)43()tt(b1AQ 21 −−λ=

1 2

currentpotential resistance

bt t QAλ

− = ⋅

Composite Wall

“Serial circuit”

Convection resistance

1 4 ( thermal resistence)t t Q− = ⋅∑

QA

bA

bA

btt3

3

2

2

1

141 ⋅

λ

=−

)63(

Ab

Ab

Ab

ttQ

3

3

2

2

1

141 −

λ+

λ+

λ

−=

1( ) Convective resistance: f wQ A t tA

αα

= − ⇒

Composite Wall, convective BC

tf2

tf1

b1

λ1 λ2 λ3

b2 b3

α1

α2

)73(

A1

Ab

A1

ttQ

2

3

li i

i

1

2ff1 −

α+

λ+

α

−=

∑=

Circular Tube or Layer (Shell)

( ) ( )(3 10)

1 2 lni o

o i

t tQL r rπλ−

= −

Composite Circular Wall (Shell)

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 Lπ α π λ π λ π α

−=

+ + +

Contact Resistance Temperature drop due to thermal contact resistance

Fouling

The accumulation and formation of unwanted materials on the surfaces of processing equipment One 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

λ

λλτ

ρ

2

2 0d t Qdx λ

′+ =

21 2

'2Qt x c x cλ

= − + +General solution

Boundary conditions

0=x 0dtdx

=

bx = ( )wall fdt t tdx

λ α− = −

At the plane of symmetry

Adiabatic or insulated BC

Solution

2 2f( )

2Q Q bt b x tλ α′ ′

= − + +

α′

+λ′

+=bQbQtt 2

fmax 2

Heat Transfer from Fins, Extended Surfaces

Fin

• A fin is a surface that extends from an object to increase the rate of heat transfer to or from the environment

• Typically, the fin material has a relatively high thermal conductivity

• Used in various applications to increase the heat transfer from surfaces.

• Increase surface area

• Microfin

Example fins

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

Example fins

Example microfins

Microfin copper tube Carbon 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, the heat transferred at the fin tip is negligible

)313(0AC

dxd

2

2−=ϑ

λα

−ϑ

)tt( f−=ϑb

2bZ

Z2ACm2

λα

=λ⋅α

≈λα

=

)tt(dxdt:Lx fLx −α′=λ−= =

0dxdt

Lx=

=

f111 tttt:0x −=ϑ=ϑ⇒==

x dx

L

t1

Q1

.

tf

b Z

Rectangular fin

Solution:

(cosh2

sinh )2

mx mx

mx mx

e emx

e emx

+=

−=

1 2

3 4cosh sinh

mx mxC e C e

C mx C mx

ϑ −= + =

= +

Hyperbolic functions

For x = L ϑ = ϑ2

1 1

cosh ( ) (3 38)cosh

f

f

t t m L xt t mL

ϑϑ

− −= = −

2

1

1cosh mL

ϑϑ

=

heat transfer from the fin ?Q

1 10

sinh ( )coshx

d mLQ A A mdx mL

CmA

ϑλ λ ϑ

αλ

=

= − = − ⋅ ⋅ −

= ⇒

1 1 1tanh 2 tanh (3 40)Q C A mL b Z mLα λ ϑ α λ ϑ= ⋅ = −

Rectangular fin

Rectangular fin

α = 25 W/m2K, b = 2 cm, L = 10 cm

Rectangular fin

If the condition below is used

one has

and

and

LxLxdx

d=

=ϑα′=

ϑ

λ−

)413(mLsinh

mmLcosh

)xL(msinhm

)xL(mcosh

1−

λα′

+

−λα′

+−=

ϑϑ

)423(mLsinh

mmLcosh

11

2 −

λα′

+=

ϑϑ

)433(mLtanh

m1

mLtanhmAmQ 11 −

λα′

+

+λα′

ϑλ=

Rectangular fin

Fig. 3-13. Arrangement of rectangular fins

1

preferable if

0dQdL

>

1 ( )Q function L=L

.

Z

t1

b

1Q

Criterion for benefit:

0dLQd 1 >

mLtanhm

1

mLtanhmAmQ 11

λα′

+

+λα′

ϑλ=

2

2

2

21

1

N

mLtanhmmLcosh

mm

N

)mLtanhm

1(mLcosh

m

AmdLQd

+

λα′

λα′

−λα′

+⋅ϑλ=

⇒> 0dLQd 1

0mLtanhmm

mLtanhm

1 22

2>

λα′

−λ

α′−

λα′

+

0m

1 22

2>

λ

α′−

Rectangular fin

Assume α′ = α ⇒

or

rule of thumb:

b2

ACm2

λα

=λα

=

02

b1 2

2>

λ⋅α

λα−

λα

>2

b1

)463(1b

2−>

αλ

)473(5b

2−>

αλ

Fin effectiveness, fin efficiency

1:

2:

1

1

from the fin from the base area without the fin

QQ

η =

1

1

from the finfrom a similar fin but with λ

QQ

φ == ∞

Optimal fin

Criterion: maximum heat flow at a given weight M = ρ b L Z = ρ Z A1

A1 = b L, Z, ρ are given.

Find maximum for A1 = b⋅L, constant. (3-40): C ≈ 2Z , A = b⋅Z

mLtanhACQ 11 ϑλα=

b2

ACm2

λα

=λα

=

1Q

⋅λα

⋅ϑ⋅αλ=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

λα

=

= −

Fin arrangement

After some algebra one finds:

)523(b

Ab

2tanhZb2

mLtanhAmQ

11

11

λα

ϑαλ=

=ϑλ=

)583(b

Ab

2u 1 −⋅λα

=

)543(ucosh

u3utanh 2 −=

{ 1(3 54) from the condition / 0}dQ db− =

)a613(4

1utanh

uZ1QA 233

3

1

11 −

λα⋅

ϑ

=

Weight of the fin

M = ρ b L Z = ρ Z A1 =

ρ/λ is the material parameter

see Table 3-1.

Aluminum is applied instead of copper. ρ/λ Aluminum: 11.8; copper: 23.0

Why not Magnesium? ρ/λ Magnesium: 10.2

λρ⋅

α⋅

ϑ

= 232

3

1

1

41

utanhu

Z1Q

Straight triangular fin

ϑ = t − tf

Heat balance ⇒

Solution:

K0 → ∞ as x → 0 ⇒ B = 0 because ϑ is finite for x = 0

x = L ϑ = ϑ1 ⇒

)623(0bL2

x1

dxd

x1

dxd

2

2−=ϑ

λα

−ϑ

bL2

λα

( ) )x2(BKx2AI 00 β+β=ϑ

( )L2AI01 β=ϑ

LxbZA ⋅=

δ

L

dx

x

b t1

tf1Q

Bessel differential equation

I0 and K0 are the modified Bessel functions of order zero

Triangular fin

)L2(IA

0

ϑ=

)653()L2(I)x2(I

0

0

1−

ββ

=ϑϑ

Lx1 dx

dtAQ=

λ−=

Lx

0

011 dx

)x2(dI)L2(I

1bZQ=

β

βϑλ−=

Triangular fin

Introduce 2

/

xd d d dxdx d dx d

ξ βξ β

ξ ξ

=

⇒ = ⋅ =

=

ξξ

β=

β

== Lx

0

Lx

0d

)(dIL/dx

)x2(dI

)L2(Ib

2)L2(IL/ 11 βλα

=ββ=

)663()L2(I)L2(Ib2ZQ

0

111 −

ββ

αλϑ=⇒

Table 3.2 for numerical values of I0 and I1

Optimal triangular fin: maximum heat flux at given weight.

)673(b

2309.12/b

L−

αλ

=

Triangular fin

Summary of formulae for rectangular and triangular fins

optimal fin

optimal fin

)383(mLcosh

)xL(mcosh

ft1tftt

1−

−=

−−

=ϑϑ

b22mλα

=

)403(mLtanh1Zb21Q −ϑαλ=

mLtanhb

2αλ

mLmLtanh

)553(b

2419.12/b

L−

αλ

=

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

1−

ββ

=ϑϑ

bL2

λα

)L2(0I)L2(1I1Zb21Q

ββ

⋅ϑ⋅αλ=

)L2(0I)L2(1I

b2

ββ

⋅αλ

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

βββ

)673(b

2309.12/b

L−

αλ

=

Formulas for fin efficiencies

Some simple calculations give:

Rectangular fin

Triangular fin

mLtanhb

2αλ

)L2(I)L2(I

b2

0

1ββ

αλ

L)L2(I/)L2(I 01

βββ

mLmLtanh

Fin effectiveness

Fin efficiency

42

Circular or annular fins

Heat conducting area

A = 2πr b

Convective perimeter

C = 2 ⋅ 2πr = 4πr

r1 r2

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 Qλ

α

α φ=∞

= − + ⋅

{ }( ) (3 71)b f b finsQ t t A Aα φ= − + ⋅ −