part 2b steady state 1-d conduction

8/19/2019 Part 2b Steady State 1-D Conduction

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University of Ottawa, CHG 2314, B. Kruczek 2

Conduction Analysis

• Selection of appropriate form of the heat equation

• Solve heat equation subject to appropriate initial and boundary conditions for thetemperature distribution

• Application ofFourier’s law

to determine the heat flux

Three main steps of conduction analysis

Simplest conduction case: (1) One-Dimensional , (2) Steady-State , (3) with or without Thermal Energy Generation

Common geometries

• The Plane Wall : Described in rectangular (x) coordinate. Area perpendicular todirection of heat transfer is constant (independent of x).

• The Tube Wall : Radial conduction through tube wall

• The Spherical Shell : Radial conduction through shell wall




Steady-state 1-D conduction – Plane wall

Temperature distribution• Heat Equation for 1-D plane wall having constant k at steady state

Solving governing differential equation byintegrating it twice:

2 2 2

2 2 2 pT T T T

k q ct x y z

where : C 1 and C 2 are the integration constants, determinatio

nof which require specifying two boundary conditions

Boundary conditions: ,1 ,20 , s sT T T L T

1 2T x C x C

Substituting expressions for constants: ,2 ,1

,1 s s

sT T

T x T L

2

12 0

d T dT dx C

dxdx 1

dT dx C dx

dx

,2 1 ,1 1 ,2 ,1 s s s sT C L T C T T L

,1 1 2 2 ,10 s sT C C C T

2

20

d T

dx





Temperature distribution• Heat Equation for 1-D plane wall at steady state variable k

Integrating once:

0d dT

k dx dx

Let : ok k aT

Qualitative determination of temperature profile

pT T T T

k k k q c x x y y z z t

"1 x

dT k C qdx or simply Fourier’s law is recovered:

where k o and a are constants. 1 2 1 20 , andT T T L T T T BCs :

Unlike the case with constant k , double integration of the governing differential equationdoes not lead to explicit relation T = f ( x ), but intuitively, T does not change linearly with x

For a > 0, k increases with T. Since q x ” is constant,dT / dx must decrease with

"

x

dT q k dx

For a < 0, k decreases with T. Since q x ” is constant,

dT / dx must increase with T For a = 0, k is independent of T. Since q x

” is constant,dT / dx must be independent of T

Q: How would the profiles in this problem change if T 1 < T 2?

0 L





Thermal circuit• Thermal resistance

Recall from Part 1 that rate equations for modes ofheat transfer can be written as:

We could get above equation, by realizing that all resistances are in series and knowingthe expressions for individual resistances

,1 ,2 ,1 ,2 ,1 ,2"" " " "

1 2 , 1 , , 21 1 xt conv t cond t conv t

T T T T T T q

h L k h R R R R

xt

T q

Rwhere : R t is the thermal resistance

", ,ort cond t cond

L LR RkA k

"

, ,

1 1 ort conv t conv

R R Ah h

Going back to equation for heat flux:

NB: Problem analysis may start from the equivalent thermal circuit , which in this case is:


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Overall heat transfer coefficient U • Application of thermal circuit to a

multilayer plane wallRate equation for heat transfer fromhot to cold fluid:

,1 ,4x

tT T q

R

tottot

1 4

1 1 1

C A B

t A B C

L R L L R R

A h k k k h A

• Alternative expression for rate equation

,1 ,4

xt

T T q

R

,1 ,4x U q A T T

1

tot

U AR

A modified form of Newton’s law of cooling toencompass multiple resistances to heat transfer

Q: Assign k A, k B, k C from thelowest to the highest value?




Series and parallel resistances• Total resistance ( Rtot )

Resistances in series:

• Examples of parallel resistances

Simultaneous (parallel) convection and radiation from a surface at T s

Resistances in parallel:

tot A B

R R R

1

1 1 1 1 1tot

tot A B A BRR R R R R

If T sur = T : R”

rad

R”

conv

T T s

1"

1 1

1 1

sc r s

c r

T T q h h T T

h h




Series and parallel resistances• Examples of parallel resistances

Simultaneous (parallel) convection and radiation from a surface at T s

If T sur ≠ T :

4 4"

c s r s sur c s s sur q h T T h T T h T T T T

R”

rad

R”

conv T

T s

T sur

Composite wall made of 4 different materials E,F,G, H

NB: If k f k G the system is no longer one-dimensional,but it can be treated as a 1-D system by assuming:

a) Surfaces normal to the x -directionare isothermal:

b) Surfaces parallel to the x -direction

are adiabatic:




Contact resistance• Ideal and real interfaces

An ideal interface is formed between twoperfectly smooth solid surfaces A and B, sothat at the interface T A = T BIn reality, there is no perfectly smoothsurface, and at the interface T

A ≠ T

Bi.e.,

there is discontinuity in temperature profiledue to contact resistance

• Quantification of contact resistanceAll three modes of heat transfer may be involved in contact resistance

There is no thickness and hence no volume associated with contact resistanceIf temperatures of the surfaces in contact, T A and T B were know, the contactresistance could be evaluated from steady state heat flux :

,

A Btc

x

T T R

q t c

t cc

R R

A,

,

where : is the contact resistance per unitsurface area and is the contact resistance

2, m K/Wt c R , K/Wt c R

Numerical values of R” t,c are in the order of 10 -6 -10 -4 [m 2 K/W] - Table 3.2

" " " ", , ,tot cond A t c cond B R R R R

Q: In which cases the contact resistance will likely be a factor in heat transfer problems?




Example 1 – Problem 3.30

The performance of a gas turbine engines may be improved by increasing the tolerance of the turbineblades to hot gases emerging from the combustor. One approach to achieving high operation temperatureinvolves application of a thermal barrier coating (TBC) to the exterior surface of the blade, while passingcooling air through the blade. Typically, the blade is made from a high-temperature superalloy, such asInconel ( k = 25 W/m K), while a ceramic, such as zirconia ( k = 1.3 W/m K), is used as a TBC.

Consider conditions for which hot gases at T ,o = 1700 K and cooling air at T ,i = 400 K provide outer andinner surface coefficients of ho = 1000 W/m 2 K and h i = 500 W/m 2 K, respectively. If a 0.5 mm-thick zirconiaTBC is attached to a 5-mm-thick Inconel blade by means of a metallic bonding agent, which provides aninterfacial thermal resistance of R” t,c = 10 -4 m 2 K/W, can the Inconel be maintained at a temperature that isbelow the maximum allowable value of 1250 K? Radiation effect may be neglected, and the turbine blademay be approximated as a plane wall.




Steady-state 1-D conduction – Radialsystems

• Appropriate for of Heat Equation

Cylindrical wall Spherical wall

2

2

10

d dT kr

dr dr r

10

d dT kr

r dr dr

Boundary conditions(in both cases):

1 ,1 2 ,2, s sT r T T r T

• Temperature profiles (assuming constant k )Solving governing differential equations subject to boundary conditions:

,1 ,2,2

1 2 2ln

ln / s s

sT T r

T r T r r r

1/,1 ,1 ,2

1 2

1

1 / s s s

r r T r T T T

r r





• Heat flux and heat rate – application of Fourier’s law Cylindrical wall Spherical wall

• Conduction resistance, recognizing that : t q T R

,1 ,22 1ln /r s s

dT k q k T T

dr r r r

,1 ,22 1

22 ln /r r s s

Lk q rLq T T r r

,1 ,22 1

22

ln /r r s sk

q rq T T r r

A ( r )

[W/m 2]

[W/m]

[W]

[W/m 2]

[W]

,1 ,22

1 21/ 1/

r s s

dT k q k T T

dr r r r

2 ,1 ,21 2

441 / 1 /r r s sk q r q T T

r r

Q1: Why there is no q’ r for spherical wall?

A ( r )

2 1,cond

ln /

2t

r r R

Lk

2 1,cond

ln /

2t

r r R

k

[K/W]

[K m/W]

1 2,cond

1/ 1 /

4t

r r R

k

Q2: Why A doesn’t appear in the expressions for the thermal resistance?

Q3: What would be the expressions for the resistances if r 1 → r 2?

[K/W]





• Consider a composite cylindrical or spherical wallTotal resistance and overall heat transfer coefficient

Heat is transferred from hot fluid inside to cold fluidoutside through a 3-layer composite wall

If contact resistances are negligible, there are fiveresistances in series, thus:

,1 ,4 ,1 ,4tot

r T T

q UA T T R

tot

1where: UA

R

At steady state R tot and UA are constant , but U depends on A , which varies with r

1

1 1 2 2 3 3 4 4 t U A U A U A U A R UA

1

31 2 1 1 4 1

1 1 2 3 4 4

1

1 1ln ln ln

A B C

U r r r r r r r

h k r k r k r r h

For cylindrical wall:

For spherical wall:

1 2 22 21 2 3 1 3 41 1 2 1

2

1 44

1

1 1 1 11 11 1

A B C

U r r r r r r r r r r

h k k k hr





• Consider a thin-walled copper tube of radius r i

Critical thickness of insulationNB: Intuitively, as thickness of insulation (wall) increases, the resistance to heat transfer

increases. This is the case for the plane wall, but not necessarily for the radial walls

'

'

r tot

iT T q R

The tube is to be used to transport a low-temperature refrigerant at T i such that T i < T

,T h

To answer question, we need to investigate function R’ tot ( r ). Let’s see if the 1 st derivativecan be equal to zero (why?)

The tube is to be insulated with a material ofthermal conductivity k

What should be the thickness of insulation , r – r i ,to minimize heat gain from the environment?

• Rate equation

where: ' ' ' ln / 1

2 2

i

tot cond conv

r r R R R

k rh

Q: Does R ’ tot increase with r ?

'

2

ln / 1 1 10

2 2 2 2

itot r r dR d

dr dr k rh kr hr

k r

h There a min or max R’ tot at r = k / h





• Further investigation of R’ tot ( r )Critical thickness of insulation

Evaluation of the 2 nd derivative of R’ tot ( r ) at r = k/h:

,T h

2 '

2 2 2 3

1 1 1 1

2 2 2tot d R d

dr kr dr hr kr r h

Sub-in r = k/h:

2 '

2 2 3 2

1 1 1 10

2 2tot

d R k

h k k dr k hk h

NB: Since d 2 ( R’ tot )/dr 2 > 0: R’ tot ( k/h ) = R ’ tot, min and r =k/h ≡ r cr is the critical radius. Thus,t = r cr – r i = t cr is the critical thickness of insulation

If for a given k ( of insulation ) and h, r i > r cr insulating thetube will lead to increase of thermal resistance

If for a given k ( of insulation ) and h, r i < r cr insulating thetube may lead to a decrease in thermal resistance

Q: Derive expression for r cr in spherical coordinates

The analysis become more complex recognizing that h depends on r , and also when radiation effects are included





A composite spherical shell of inner radius r 1 = 0.25 m is constructed from lead ofouter radius r 2 = 0.30 m and AISI 302 stainless steel of outer radius r 1 = 0.31 m.The cavity is filled with radioactive wastes that generate heat at the rate of 5 x 10 5 W/m 3 . It is proposed to submerge the container in oceanic waters that are at atemperature of T = 10 oC and provide a uniform convection coefficient of h = 500W/m 2 K at the outer surface. However, environmental activists are concerned that atthese conditions heat generation by the radioactive wastes may result in melting ofthe lead shell and leaking of radioactive wastes to ocean. Are their concerns

justified?




Steady-state 1-D conduction with heatgeneration

• Thermal energy is generated from conversion of different forms of energy(e.g., electrical, chemical, nuclear)

Origins of internal heat generation

Resistance to the flow of electrical current liberates thermal energy – Ohmic heating

Example: consider flow of electrical current through a solid system:

2 g eE qV I R where I is the electrical current [ A] and R e is the electrical resistance [ Ω]

Ohmic heating, similarly to nuclear decay, exothermic and endothermic reactions,generally leads to uniform volumetric heat generation

• Energy generation by absorption of external radiationA solid (stationary fluid) is semi-transparent to external radiation (alpha, gammarays, thermal, visible radiation, etc).

The intensity of external radiation gradually decreases (decays) as it passes thoughthe system and as it decreases, heat is generated

Non-uniform heat generation, typically described by:

xq e where is a constant depending on the nature of radiation and x is adistance from the surface through which radiation enters the system




Steady-state 1-D conduction with heatgeneration in plane wall

• Heat eq. for 1-D plane wall having constant k and uniform heat generationTemperature distribution

2 2 2

2 2 2 pT T T T

k q ct x y z

2

20

d T k q

dx

Solving governing differential equation by integrating it twice:2

12

d T q dT qdx dx x C

k dx k dx 1

dT qdx x C dx

dx k 2

1 22

qT x x C x C

k

where : C 1 and C 2 are the integration constants, determination ofwhich require specifying two boundary conditions

NB1 : Regardless how boundary conditions are specified,

temperature profile has a parabolic shape

Consider asymmetric boundary conditions:

,1

,2

BC1: 0

BC2:

s

s

T x T

T x L T

,2 ,12

,12 2 s s

s

T T q qT x x L x T

k L k

0 L





• Effect of the location of origin in coordinate systemTemperature distribution

2

1 22

qT x x C x C

k

,1

,2

BC1: 0

BC2:

s

s

T x T

T x L T

,2 ,12

,12 2 s s

s

T T q qT x x L x T

k L k

0 L

Consider previously derived temperaturedistribution for asymmetric BCs:

,1

,2

BC1:

BC2:

s

s

T x L T

T x L T

Origin at one of

two surfacesThickness = L

Origin at mid-planeThickness = 2 L

2 2

,2 ,1 ,1 ,2

21

2 2 2 s s s sT T T T qL x x

T xk L L





• Application of Fourier’s law to previously derived temperature distributions Heat flux (flow)

,2 ,1" 2

,1

,2 ,1

2 2

2

s s x s

s s

T T d q qq k x L x T

dx k L k k q

qx T T L L

0 L

Origin at one oftwo surfacesThickness = L

Origin at mid-planeThickness = 2 L

2 2,2 ,1 ,1 ,2"

2

,2 ,1

1

2 2 2

2

s s s s x

s s

T T T T d qL x xq k

dx k L Lk

qx T T L

Although A does not change in the direction of heat flow, heat flux (flow) is not constant;it varies linearly with x

NB: Since heat flux (flow) varies with x a wall with internal heat generation cannot be a part of

equivalent thermal circuit





• Convective boundary conditionsTemperature distribution with unknown surface temperatures

Suppose that T s ,1 and T s ,2 are unknown , but T ,1 and T ,2 alongwith h 1 and h 2 are known, then:

Thus : Solving simultaneously Eqs. (1) and (2) allows determination of the unknown T s ,1 and T s ,2

Differentiating the derived temperature distribution forthe asymmetric BCs:

1 ,1 ,1

2 ,2 ,2

BC1: at 0 :

BC2: at :

s

s

dT x k h T T

dx

dT x L k h T T dx

0 L

,2 ,1 ,2 ,12

,12 2 2 s s s s

s

T T T T dT d q q q q x L x T x L

dx dx k L k k L k

Using BC1 (at x = 0): ,2 ,1 1,1 ,1 (1)

2 s s

s

T T hq L T T

L k k

Using BC2 (at x = L): ,2 ,1 2,2 ,2 (2)

2 s s

s

T T hq L T T

L k k





• Symmetrical boundary conditionsTemperature distribution and heat flux

T s ,1 = T s ,2 = T s as a result of T ,1 = T ,2 = T and h 1 = h 2 = h

Since T s ,1 = T s ,2 only one surface boundary condition canbe specified – additional BC is needed

Due to thermal symmetry, the position x = 0

corresponds to the mid-plane, where:

0dT

dx

Thermal symmetry plane is equivalent to an adiabaticsurface (perfectly insulated surface)

No heat transfer through aninsulated surface, thus:

0 0dT dT

k dx dx

BC1: at 0 : 0

BC2: at : s

dT x

dx x L T T

2 2

21

2 sqL xT x T k L

• Determination of T s Overall energy balance on the wall:

out 0 g E E 0 s s shA T T q A L

s

qLT T

h

2 2

"

21

2 x sd qL xq k T qxdx k L




Steady-state 1-D conduction with heatgeneration in radial systems

• For 1-D radial systems at steady state with constant k and heat generation Heat Equation and boundary conditions

Cylindrical Coordinates Spherical Coordinates

0k d dT

r qr dr dr

2

2 0

k d dT r q

r dr dr

Cylindrical wall Solid cylinder Spherical wall Solid sphere

• Boundary conditionsSolid cylinder and sphere are analogous to a symmetrical wall;there require one surface boundary condition , while the otherone is a result of a max (min) temperature at centerline (point)

0 0dT

r dr

Radial walls require two surface boundary conditions; radial walls may have onesurface insulated

Q : Can solid cylinder (sphere) have “one” surface insulated?




Steady-state 1-D conduction with heatgeneration in radial systems

• Consider example of a solid cylinder Temperature distribution and heat flux

0k d dT

r qr dr dr

Solid cylinder

Heat Equation:

Integrating once: Boundary conditions:

BC1: at 0 : 0

BC2: at : o s

dT r dr

r r T T

2

12

d dT qr dT qr r dr dr r C dr dr k dr k

2

22 4

dT qr qr T r C

dr k k

by BC1: C 1 = 0

Rearranging and integrating again:

by BC2: 2

24

o s

r T C

qk

Therefore: 2 2

21

4o

s

o

qr r T r T

k r

• Application of Fourier law: 2 2

"

21

4 2o

r s

o

qr d r qr q k T dr k r

Heat rate:" 2 ' " ' 2 orr r r r r r q q A q Lr q q A q r

NB: A summary of temperature distributions, heat fluxes and heat rates for plane, cylindricaland spherical walls, as well as for solid cylinder and sphere are provided in Appendix C .





A high-temperature, gas-cooled nuclear reactor consists of a composite cylindrical wall forwhich a thorium fuel element ( k = 57 W/m K) is encased in graphite ( k = 3 W/m K) andgaseous helium flows through an annular coolant channel. Consider conditions for which thehelium temperature is T = 600 K and the convection coefficient at the outer surface of thegraphite is h o = 2000 W/m 2 K.

If thermal energy is uniformly generated in the fuel element at the rate of 10 8 W/m 3, whatare the temperatures T 1 and T 2 at the inner and outer surfaces, respectively, of the fuelelement?




Steady-state 1-D conduction – Extendedsurfaces

• The main purpose of extended surfaces ( fins ) is to: Rationale of Extended Surfaces

OR

sq hA T T

Q: If heat transfer from the original and extended surface is byconvection, why do we need conduction analysis?

Increase the rate of heat transfer for a given T Decrease T for a given rate of heat transfer

Original surface

Extended surface

Fins

• Rate of heat transfer Original surface by convection:

Extended surface also byconvection: e b b b f f f q h A T T h A T T

Conduction analysis are required to find how T changeswith the distance from the original surface

( T - T ) is the local driving force for convection from thefin

• Key features/assumptions in conduction analysis

Convection (radiation) from the extended surface is in adirection transverse to that of conduction

One dimensional conduction in the direction away (ortowards) the original surface





• Heat exchangers with finned

tubes

Configurations and applicationsa) Straight fin with

uniform cross sectionb) Straight fin with non

uniform cross sectionc) Annular fin d) Pin fin

Always non-uniformcross section

uniform ornon-uniform cross section

• Heat sinks used in microelectronics

a) c)

a)

d) With uniformcross section





Objective: Incorporate extended surface (fins) in thermal circuit• Approach to achieve the main objective:

Derive the governing differential equation for appropriate fin configuration

Identify boundary conditions and solve the differential equation

Rate of heat transfer from the fin based on temperature distributionApplication of Fourier’s law at the base

Application of Newton’s cooling

Fin efficiency

Efficiency of finned surface

Resistance to heat transfer from fin

Resistance to heat transfer from finned surface

Using tabulated expressions for fin efficiency of fins with variable cross-section

Finned surface with integral and attached fins





Governing differential equation

• Straight fin with variable cross-section

OR in out g st E E E E 0

in out E E

in cond ,x cdT

E q kA

dx

cond ,xout cond ,x dx conv cond ,x conv

dq E q q q dx q

dx

cond,x dxq

2

2c

out c C s c c s

dAdT d dT dT dT d T E kA k A dx hdA T T kA k A dx hdA T T

dx dx dx dx dx dx dx

Substituting rate equationsfor the energy terms:

and

NB: Up till now we started from appropriate for of Heat Equation derived in Chapter 2,but this approach is not applicable for conduction analysis in extended surfaces

convq

1 st law for differential element oflength dx (closed system):






OR

2

0cc s

dA dT d T k A dx hdA T T

dx dx dx

Substituting expressions for Ė in andĖ out into 1 st law:

• Straight fin with variable cross-section

Rearranging: 2

2

10c s

c c

dA dAd T dT hT T

dx A dx dx A k dx

• Annular fin

Starting from 1 st law for closed systems, identifying energy terms,and rearranging it can be show that

2

2

1 20

d T dT hT T

dr r dr kt where t = fin thickness






OR

• Simplification for a straight fin with uniform cross-section

2

2

10c s

c c

dA dAd T dT hT T

dx A dx dx A k dx

• Excess temperature – driving force convection:

Note the units: m 2 [1/m 2 ]

Recognizing that: s s

dAdA Pdx P

dx

Therefore: 2

20

c

d T hP T T

dx A k perimeter

x T x T

• Fin constant: 2 orc c

hP hP m m

A k A k

For pin fin:2

and4c

D A P D

thus:

2 4

hm

kD

For rectangular fin: and 2 2 2c A tw P w t w thus:

2 2

hm

kt

Therefore: 2

22 0

d m

dx





General solution of the governing differential equation

OR

• Fin with constant cross-section

22

2 0

d m

dx

• Annular fin

Looking for a functionwhose 2 nd derivative equalsto the original function

Because of 2 nd order thesolution must have twoconstants

1 2

mx mxx C e C e

Check:

1 2

mx mxdmC e mC e

dxand

2

2 2

1 22

mx mxdm C e m C e

dx

Therefore, the general solution satisfies the governing ODE

2

2

1 20

d T dT hT T

dr r dr kt

22

2

10

d d m

dr r dr

Modified Bessel

equation of zero order 1 0 2 0r C I mr C K mr

Where: I 0 and K 0 are the modified,zero-order Bessel functions of the firstand second kinds, respectively

NB: Bessel and modified Bessel aretabulated in Appendix B





Boundary conditions• The specific solutions for the temperature distribution in fins require determination

of constants C 1 and C 2 , which in turn require 2 boundary conditions

x L 0

• BC1 at x = 0

0 0orb b bx xT T T T

• BC2 at x = L

Can be specified in different waysdepending on problem specification

a) In the most general case (as in figure above) – convection from the tip:

orc c

x L x L

dT dhA T L T kA h L kdx dx

b) If the fin is sufficiently long T ( L) T - infinite fin:

or 0

L LT x L T





Boundary conditions• BC2 at x = L

c) Heat transfer from the tip is negligible – adiabatic tip:

c shA T L T hA T T

0 or 0x L x L

dT d

dx dx

Q: Is it possible to have the exact adiabatic tip conditions for a fin which is not infinitely long?

NB: An infinite fin has adiabatic tip, but a fin having an adiabatic tip is not necessarily infinite

x L 0

T , h T 1 T 2

If T 1 = T 2 ≠ T then:2

0L

x

dT

dx

The rod can be seen as two fins oflength L /2 with a common adiabatic tip

If T 1 > T 2 > T , or T 1 < T 2 < T ,there exists x’ such that0 < x’ < L at which: '

0x x

dT

dx

• Consider a rod connecting two walls





Boundary conditions• BC2 at x = L

d) Tip is in contact with a surface of knowntemperature – prescribed tip temperature:

x L 0

T , h T b T L

T b > T > T L

L L L x L x LT T or T T

This BC2 exists only if there is no x (0< x < L) for which dT / dx = 0 , which willhappen when:

T b < T < T L OR

If there is x’ (0< x’ < L) for which dT / dx = 0 , there are two fins (one havinglength x’ and the other one having length L- x’ ) with a common tip at x’

NB: In the case of fin with prescribed tip temperature, the heat transferred to the fin at the

base is lost by convection from the side surface and by conduction from the tip





Specific solution of the governing differential equation

• Consider a fin with constant cross-section and an adiabatic tip

General solution:

Where:

1 2

mx mxx C e C e 0BC1:

bx

BC2: 0x L

d

dxand

Using BC1: 1 2 1 2 (1)b b

C C C C

Using BC2: 1 2 1 20 0 (2)mx mx mL ml

x L

dC e C e mC e mC e

dx

Solving simultaneously Eqs. (1) and (2): 1

mL

b mL mL

eC

e e 2

mL

b mL mL

eC

e eand

Sub-in C 1 and C 2 into general solution and rearranging:

m L x m L xmL mx mL mx

b b bmL mL mL mL mL mLe e e e e ex

e e e e e e

sinh2

x xe exOther hyperbolic function:cosh2

x xe e

x

sinh

tanhcosh

x x

x x

x e ex

x e e

cosh

cothsinh

x x

x x

x e ex

x e e

1sech coshx x

1csch sinhx x

coshcoshb m L xx

mL





Heat transfer from fin

• Consider a fin with constant cross-section and an adiabatic tip

Where:

At steady state, all heat transferred to orfrom the fin must pass through the base:

, f cond bq q

,

0

f cond b cx

dT q q kAdx

Substituting expression for T ( x ):

, 0

0

coshsinh

cosh coshc b

cond b c b x

x

m L x kAdq kA T T T m m L xdx mL mL

c b c b c b

c

hP M kA m kA kA hPkA

, tanhcond bq M mL

Alternatively, all heat transferred to or from the fin is by convection: 0

L

f convq q hP T T dx

Substituting expression for T ( x ):

0

cosh

cosh

L

conv b

m L xq hP T T T T dx

mL

leads after integration and rearrangements to: ,tanhconv cond bq M mL q





Summary of expressions for T (x ) and q ffor fins with constant A

c

where:

• Corrected length for convective tip

The expressions for adiabatic tip can be used for fins with a convective tip byreplacing actual length ( L) with a corrected length ( Lc ), where:

For pin fin:

For rectangular fin:

4cL L D 2

cL L t





Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at T

= 1200 oC and maintains a convection coefficient of h = 250 W/m 2 K over the blade. The blades, which arefabricated from Inconel, k = 20 W/m K, have a length L = 50 mm. The blade profile has uniform cross-sectional area Ac = 6 x 10 -4 m 2 and a perimeter of P = 110 mm . A proposed blade-cooling scheme, whichinvolves routing air through the supporting disc, is able to maintain the base of each blade at a temperatureof T b = 300 oC.

a) If the maximum allowable blade temperature is 1050 oCand the blade tip may be assumed to be adiabatic, is theproposed cooling scheme satisfactory?

b) For the proposed cooling scheme, what is the rate atwhich heat is transferred from each blade to thecoolant?





Fin performance

• Fin effectiveness ( ε f )

The ratio of the fin heat transfer rate to the heattransfer rate that would exist without the fin:

,

f

b b f

c Aq

hIn the limiting case of infinitely long fin:

f c bq M hPkA

f c

kPhA NB: The use of fins is justified when ε f ≥ 2

• Fin efficiency, ( η f )

Ratio of the actual fin heat transfer rate tomaximum fin heat transfer rate:

max

f f

f

f

b

q qq h A

Fin efficiency in terms of temperature profile

q max is represented by the area of a rectangleunder the dashed red line, A max

q is represented by the area under theactual temperature profile, A

a L

max

f A

AThus:

f aL

Lor





Fin efficiency

• Consider a fin with constant cross-section

Regardless of type of the fin: f f f bq hA

Thus:

f c bq M hPkA

tanh tanhc b c

f b

hPkA mL mLkAhPL hP L

tanh f mLmL

For a fin with adiabatic tip :

For a fin with convective tip , the equation for the adiabatic tip isapplicable in which L is replaced by Lc

For infinite fin :

tanh tanh f c bq M mL hPkA mL

Thus:

1c b c f

b

hPkA kAhPL hP L

1

f mL

A f

Q: What is the efficiency oftruly infinitely long fin?

NB: For a fin with prescribed tip temperature, the concept of fin efficiency is applicableonly to the portion of heat transferred to/from the fin by convection





Efficiency of fins with variable cross-section

• Tabulated expressions for the efficiency of common fin shapes fromTable 3.5

Pin Fins





Efficiency of fins with variable cross-section• Tabulated expressions for the efficiency of common fin shapes from

Table 3.5 - continuedStraight Fins





Graphical representation of fin efficiency• Geometrical parameter of fin

Consider straight fins (rectangular, triangular, parabolic profiles) and annular finwith rectangular profile:

1/21/2

3/22c c c

c p

hP hmL L LkA kA

where A p is the fin profile area

Fin efficiency presented graphically as a function of: 1/2 3/2

p ch kA L

Q: Why Lc of triangular andparabolic fins is equal to L?





Graphical representation of fin efficiency• Efficiency of annular fin of rectangular profile





Incorporation of fins into thermal circuits

• Resistance to heat transfer from a fin ( R t , f )

Rate of heat transfer from a fin in terms of fin efficiency: f f f bq hA T T

Q: Is it possible to incorporate a fin with prescribed tip temperature to a thermal circuit?

Alternatively,

,

b f

t f

T T q

Rtherefore,

,

1 f

f t

f

RhA

• Finned surfaceIn practical applications individual fins are arranged in fin arrays

The concept of fin efficiency is extendedinto the concept of overall efficiency of

surface ( o ):

o

t

t b

qhA

Where: q t = the total rate of heat transferand A t = total surface area

NB: It is assumed that heat transfer coefficients from fins and prime surface are the same





Incorporation of fins into thermal circuits

• Relationship between fin and surface efficiency

Surface efficiency is a weighted average ofefficiency of fins and prime surface

t f f b b bq N hA hA

Where: N = number of fins, A b = prime surface area ( A b =A t - NA f )

Efficiency of prime surface is 100% (why?)

Sub-in for A b and rearranging:

1 1 f t t f b t o b

t

NAq hA hA

A

Thus:

1 1 f

t

o f NA

A

Since:

,

bt t o b

t o

q hAR ,

1o

tt

o

RhA

• Resistance of finned surface ( R t ,o )





Incorporation of fins into thermal circuits • Thermal circuit for finned surface

Fins attached to the base – contact resistance

,

1o

t

t

oR hA

Integral fins

1 , ,1 / f f t c c bC hA R A

,

1

o

o c

t c

t R

hA

where: 1

1 1

f f o c

t

NA

A C and




An isothermal silicon chip of width W = 20 mm on a side is soldered to an aluminum heat sink ( k = 180 W/mK) of equivalent width. The heat sink has a base thickness of Lb = 3 mm and an array of rectangular fins,each of length Lf = 15 mm. Airflow at T = 20 oC is maintained through channels formed by fins and a coverplate, and for a convection coefficient of h = 100 W/m 2 K, a minimum fin spacing of 1.8 mm is dictated bylimitations on the flow pressure drop. The solder joint has a thermal resistance of R” t,c = 2 x 10 -6 m 2 K/W.

a) Consider limitations for which the array has N = 11 fins, in whichcase values of the fin thickness t = 0.182 mm and pitch S = 1.982mm are obtained from the requirements that W = ( N – 1) S + t and S

– t = 1.8 mm. If the maximum allowable chip temperature is T c =85 oC, what is the corresponding value of the chip power? (Anadiabatic fin tip condition may be assumed, and airflow along theouter surfaces of the heat sink may be assumed to provide aconvection coefficient equivalent to that associated with airflow

through the channels.) b) With ( S -t ) and h fixed at 1.8 mm and 100 W/m 2 K, respectively,

explore the effect of increasing fin thickness by reducing the numberof fins. With N = 11 and S – t fixed at 1.8 mm, but relaxation of theconstraint on the pressure drop, explore the effect of increasing theairflow and hence the convection coefficient.

part 2b steady state 1-d conduction

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