1

MECH 375, Heat TransferHandout #2: Steady State Conduction

Amir Maleki, Fall 2018

2

F O R T U N AT E N E W T O N , H A P P Y C H I L D H O O D O F S C I E N C E ! H E W H O H A S T I M E

A N D T R A N Q U I L L I T Y C A N B Y R E A D I N G T H I S B O O K [ O P T I C K S ] L I V E A G A I N

T H E W O N D E R F U L E V E N T S W H I C H T H E G R E AT N E W T O N E X P E R I E N C E D I N

H I S YO U N G D AY S . N AT U R E T O H I M W A S A N O P E N B O O K , W H O S E L E T T E R S

H E C O U L D R E A D W I T H O U T E F F O R T. T H E C O N C E P T I O N S W H I C H H E U S E D T O

R E D U C E T H E M AT E R I A L O F E X I S T E N C E T O O R D E R S E E M E D T O F L O W S P O N -

TA N E O U S LY F R O M E X P E R I E N C E I T S E L F, F R O M T H E B E A U T I F U L E X P E R I -

M E N T S W H I C H H E R A N G E D I N O R D E R L I K E P L AY T H I N G S A N D D E S C R I B E S

W I T H A N A F F E C T I O N AT E W E A LT H O F D E TA I L . O N O N E P E R S O N H E C O M -

B I N E D T H E E X P E R I M E N T E R , T H E T H E O R I S T, T H E M E C H A N I C A N D, N O T L E A S T,

T H E A R T I S T I N E X P O S I T I O N .

A L B E R T E I N S T E I N

Learning Objectives0 A large margin is intentionally lefthere for your notesBy the end of this handout, students will be

able to:

• Derive conduction equation in one dimensionusing the 1st law of thermodynamic

• Generalize 1D conduction equation to 3D

• Recognize the vectorial form of conductionequation

• Solve 1D conduction equation in Cartesian,cylindrical and spherical coordinates with var-ious boundary conditions.

• Simplify composite problems using the ther-mal resistance analogy.

• Explain what contact resistance is and how itcan be modeled.

• Illustrate why insulating layers over the cylin-drical or spherical objects have an optimumthickness.

1D Steady State Conduction(Cartesian Coordinate)

We assume the material is isotropic and has a constant conductivity.

Energy balance on a closed system here:

dEsys

dt= qnet − W

qnet = qx − qx+∆x + qgen =

k∂2Tdx2 + S = ρc

∂Tdt

5

Case 1: No source term, Steady state

kd2Tdx2 = 0⇒ T = C1x + C2

Find C1 and C2 using the boundary conditions:

Rcond =T1− T2

qx=

LkA

We can therefore think of a conductive material as a thermal resistance. The more conductive the material, the less thermally resistant it is.

The resistance analogy only applies when wehave 1D steady state conduction with constantconductivity and no heat generation.

6

Case 2: With source term, steady state, con-stant conductivity

kd2Tdx2 + S = 0⇒ T = −1

2Sk

x2 + C1x + C2

Find C1 and C2 using the boundary conditions:x = ±L, T = Tw.

T − Tw = 12

Sk L2

[1−

( xL

)2]

3D Steady State ConductionIn one dimension, we obtained

qx = −kA∂T∂x

kd2Tdx2 + S = ρc

dTdt

How do we extend this to three dimensions?

(qx, qy, qz) = −kA( )

+ S = ρc∂T∂t

or in vectorial form

Steady State ConductionCylindrical CoordinateUse the 3D vectorial form in cylindrical coordi-nate, impose axisymetry and assume the cylin-der is very long. i.e.

∂

∂θ=

∂

∂z= 0,

q = (qr, qθ, qz) = (qr, 0, 0)

qr = −k∂T∂r

A = −k∂T∂r

(2πrL)

So 1D conduction equation in cylindrical coor-dinate is simplified to

9

Case 1: Hollow cylinder without source term

1r

ddr

(rdTdr

)= 0⇒

find the coefficient C1 and C2 using the boundary conditions:

T − T2 =ln(r2/r)ln(r2/r1)

[T1− T2]

Rt,cond =T1− T2

qr=

ln(r2/r1)

2πLk

The resistance analogy only applies when we have 1D steady state conduc-tion with constant conductivity and no heat generation.

10

Case 2: cylinder with source term

k1r

ddr

(rdTdr

)+ S = 0⇒

find the coefficient C1 and C2 using the bound-ary conditions:

r = 0 →

r = R → T = Tw

T = Tw +S4k

R2(

1−( r

R

)2)

1D Steady State ConductionSpherical Coordinate

q = (qr, qθ, qφ) = (qr, 0, 0)

qr = −kdTdr

A = −kdTdr

(4πr2)

So the steady-state conduction equation is simplified to

1r2

ddr

(r2kdT

dr

)+ S = 0

12

Case 1: Hollow sphere without source termassume constant conductivity

1r2

ddr

(r2 dT

dr

)= 0⇒

r2 dTdr

= C1 ⇒dTdr

=C1

r2

T =C2

r+ C3

find the coefficient C1 and C2 using the bound-ary conditions:

r = r1 → T = T1

r = r2 → T = T2

Temperature profile is

T − T2 =1/r−1/r21/r1−1/r2

[T1− T2]

Rt,cond =T1− T2

qr=

14πk

(1r1− 1

r2

)

The resistance analogy only applies when we have 1D steady state conduction with constant conductivity and no heat generation.

13

Case 2: sphere with source term

k1r2

ddr

(r2 dT

dr

)+ S = 0 ⇒

find the coefficient C1 and C2 using the bound-ary conditions:

r = r0 → T = f inite

r = R → T = Tw

T = Tw +S6k

R2(

1−( r

R

)2)

Thermal ResistanceFor conduction we found these three correla-tions

Using the same definition of thermal resistance,

Rt,conv =Ts − T∞

q= , Rt,rad =

Ts − Tsur

q=

Composite walls

Rtot =

qx =

we define overall heat transfer coefficient U

U =1

RtotA=

qx = UA∆T ⇒

Rtot =

Rtot =

qx =T1,∞ − T4,∞

Rtot

Here we can define overall heat transfer coefficientU arbitrarily based on either surface areas:

qx = U1A1∆T = U2A2∆T = · · · ⇒

U1 =1

Rtot A1

Contact ResistanceIn composite walls, a perfect contact may not beachieved; i.e. contact spots are interspersed withgaps that are, in most instances, air filled This

may indeed lead to a temperature drop acrossthe interface. We model this with a contact resis-tance.

R′′t,c =TA − TB

q′′s

Example: Insulation optimal thicknessDesigning optimal insulation is a crucial step. In particular, the insulationthickness has two competing effects: i) increasing the thickness increasesthe conductive resistance, and therefore reduces heat transfer , but ii) itwill also increase surface area. The question is to find the optimum thick-ness that has the least heat transfer.

18

Practice: Find the critical radius for a spheri-cal object