One-Dimensional, Steady-StateConduction without

Thermal Energy Generation

Lecture 5 - Chapter Three

Sections 3.1 through 3.4

Methodology

• Specify appropriate form of the heat equation.

• Solve for the temperature distribution.

• Apply Fourier’s law to determine the heat flux.

Simplest Case: One-Dimensional, Steady-State Conduction with No Thermal Energy

Generation.

• Common Geometries:

– The Plane Wall: Described in rectangular (x) coordinate. Area

perpendicular to direction of heat transfer is constant (independent of x).

– The Tube Wall: Radial conduction through tube wall.

– The Spherical Shell: Radial conduction through shell wall.

Methodology of a Conduction Analysis

• Consider a plane wall between two fluids of different temperature:

The Plane Wall

• The implications are:

0d dT

k Adx dx⎛ ⎞⋅ =⎜ ⎟⎝ ⎠

( )Heat flux is independent of .xq x′′

( )Assuming steady-state conditions and

no internal heat generation . ., 0 ,

then the 1-D heat conduction equation

reduces to:

i e q =

• Boundary Conditions: ( ) ( ),1 ,20 , s sT T T L T= =( )Heat rate is independent of .xq x

• Temperature Distribution

• shown for constant k

• For constant k and A

2

20

d T

dx=

The Plane Wall

( ) 1-D heat conduction equation for steady-state

conditions and no internal heat generation . ., 0 ,

is:

i e q

∴

=

for Boundary Conditions: ( ) ( ),1 ,20 , s sT T T L T= =

, for constant k and A2

20

d T

dx=

2

12

1 2

Integrate twice to get ( ):

( ) ( )

( )

T x

d T dTT x dx dx C dx

dxdxT x C x C

= = +

∴ = ⋅ +

∫ ∫ ∫

,1 2 ,1

,2 ,2 1 2 1 ,1

,2 ,11 1 2

,2 ,1 ,1

at 0, ( ) , and

at , ( ) , and

which gives , Substituting the values for and in to exp. for ( ):

( ) ( ) and apply Fouri

s s

s s s

s s

s s s

x T x T C T

x L T x T T C L C C L T

T TC C C T x

Lx

T x T T TL

∴ = = =

= = ∴ = ⋅ + = ⋅ +

−∴ =

∴ = − + er's Law to get heat transfer, xq

Plane Wall (cont.)

• Heat Flux and Heat Rate:

( ),1 ,2x s sdT k

q k T Tdx L

′′ = − = −

( ),1 ,2x s sdT kA

q kA T Tdx L

= − = −

• Thermal Resistances and Thermal Circuits⎛ ⎞Δ

=⎜ ⎟⎝ ⎠

tT

Rq

Conduction in a plane wall: ,t condL

RkA

=

Convection: ,1

t convRhA

=

Thermal circuit for plane wall with adjoining fluids:

1 2

1 1tot

LR

h A kA h A= + + ,1 ,2

xtot

T Tq

R∞ ∞−

=

Plane Wall (cont.)

• Thermal Resistance for Unit Surface Area:

,t condL

Rk

′′ = ,1

t convRh

′′ =

Units: K/WtR ↔ 2m K/WtR′′↔ ⋅

• Radiation Resistance:,

1t rad

r

Rh A

= ,1

t radr

Rh

′′ =

( )( )2 2r s sur s surh T T T Tεσ= + +

• Composite Wall with Negligible Contact Resistance:

Plane Wall (cont.) • Composite Wall with Negligible Contact Resistance:

,1 ,4x

t

T Tq

R∞ ∞−

=∑

(3.14)

1 4

1 1 1C totA Bt tot

A B C

L RL LR R

A h k k k h A

⎡ ⎤ ′′∑ = = + + + + =⎢ ⎥

⎣ ⎦

• Overall Heat Transfer Coefficient (U) :

A modified form of Newton’s Law of Cooling to encompass multiple resistances to heat transfer.

x overallq UA T= Δ (3.17)

1totR

UA= (3.19)

Plane Wall (cont.)

• Contact Resistance:

,A B

tcx

T TR

q

−′′ =′′′′

= ,,

t ct c

c

RR

A

Values depend on: Materials A and B, surface finishes, interstitial conditions, and contact pressure (Tables 3.1 and 3.2)

tbl_03_01

tbl_03_02

Electrical Analogy of Thermal Circuits

To solve a parallel resistance network like that shown opposite, we can reduce the network to and equivalent resistance

For electrical circuits:

1R 4R3R

2R

4V1V

1 4Total parR R R R= + +

andTotal

VI

R

Δ=

ITotalR

VΔ

1V 4V

2 3

2 3par

R RR

R R

×=

+1R parR

4R

xqtR∑

TΔ,1 ,4x

t t

T TTq

R R∞ ∞−Δ

= =∑ ∑

,1T∞ ,4T∞For thermal circuits:

• Series – Parallel Composite Wall:

Note: error increases as there is a departure from the one-dimensional conditions if .F Gk k≠

Circuits based on assumption of isothermal surfaces normal to x direction or adiabatic surfaces parallel to x direction provide approximations for .xq

Example: Parallel resistances

These IR photos show that the heat flow though the built-up walls is more complex than indicated than predicted by a simple parallel-resistance, 1-D thermal model. However, it will still provide a rough estimate of the average, wall heat transfer rate. The images clearly show “thermal bridging” by the wall studs.

Approximate Solution

( ) ( )

( ) ( )

( )

" ( )

here is the average

insul studsaverage insul stud

wall wall

average insul area insul stud area studs

wall average wall indoor outdoor

wallwall wall wall indoor outdoor

wall

A AU U U

A A

U U F U F

q U A T T

q A q UA T T

UA

= × + ×

= × + ×

= ⋅ −

= = −

area weighted "U-factor"

or "U-value" for the wall, (i.e., )

( )

wall average

wallwall wall indoor outdoor

UA U

q UA A T T

=

= ⋅ −

It is common practice to estimate the heat loss rate through an insulated stud-wall by area-weighting the projected area+ of the studs and insulation cavity to arrive at an average thermal resistance or “U-factor”Note: + (normal to the direction of average heat conduction)

wallq

Alternative Wall Design

Example of resistance network with both a radiative and convective boundary

Other 1-D Geometries

• Heat Equation:

The Tube Wall

10

d dTkr

r dr dr⎛ ⎞ =⎜ ⎟⎝ ⎠

Is the foregoing conclusion consistent with the energy conservation requirement?

How does vary with ?rq′′ r

What does the form of the heat equation tell us about the variation of with

in the wall? rq

r

• Temperature Distribution for Constant :k ( ) ( ),1 ,2

,21 2 2

lnln /

s ss

T T rT r T

r r r

⎛ ⎞−= +⎜ ⎟⎜ ⎟

⎝ ⎠

Tube Wall (Cont.)

• Heat Flux and Heat Rate:

( ) ( )

( ) ( )

( ) ( )

,1 ,22 1

,1 ,22 1

,1 ,22 1

ln /

22

ln /

22

ln /

r s s

r r s s

r r s s

dT kq k T T

dr r r r

kq rq T T

r r

Lkq rLq T T

r r

ππ

ππ=

′′ = − = −

′ ′′= = −

′′ = − (3.27)

• Conduction Resistance:( )

( )

2 1,

2 1,

ln /Units K/W

2ln /

Units m K/W2

t cond

t cond

r rR

Lkr r

Rk

π

π

= ↔

′ = ↔ ⋅

(3.28)

Why is it inappropriate to base the thermal resistance on a unit surface area?Why is it inappropriate to base the thermal resistance on a unit surface area?

• Composite Wall with Negligible Contact Resistance

• Composite Wall with Negligible Contact Resistance

( ),1 ,4,1 ,4r

tot

T Tq UA T T

R∞ ∞

∞ ∞−

= = −

1

Note that

is a constant independent of radius.totUA R −=

But, U itself is tied to specification of an interface.

( ) 1i i totU A R

−= (3.32)

Spherical Shell

• Heat Equation

Spherical Shell

22

10

d dTr

dr drr

⎛ ⎞ =⎜ ⎟⎝ ⎠

What does the form of the heat equation tell us about the variation ofwith ? Is this result consistent with conservation of energy?rq r

How does vary with ? rq′′ r

• Temperature Distribution for Constant :k

( ) ( ) ( )( )

1/,1 ,1 ,2

1 2

1

1 /s s s

r rT r T T T

r r

−= − −

−

• Heat flux, Heat Rate and Thermal Resistance:

( ) ( )( ),1 ,22

1 21/ 1 /r s s

dT kq k T T

dr r r r′′ = − = −

⎡ ⎤−⎣ ⎦

( ) ( ) ( )2,1 ,2

1 2

44

1/ 1/r r s sk

q r q T Tr r

ππ ′′= = −−

• Composite Shell:

overallr overall

tot

Tq UA T

R

Δ= = Δ

1 ConstanttotUA R −= ↔

( ) 1Depends on i i tot iU A R A

−= ↔

( ) ( )1 2,

1 / 1 /

4t condr r

Rkπ

−=

Critical radius for insulation: see example 3.5 in Text

crk

rh

=

p_03_97

Summary

Summary

Problem 3.23: Assessment of thermal barrier coating (TBC) for protectionof turbine blades. Determine maximum blade temperaturewith and without TBC.

Schematic:

ASSUMPTIONS: (1) One-dimensional, steady-state conduction in a composite plane wall, (2) Constant properties, (3) Negligible radiation

Examples

Problem: Thermal Barrier (Cont.)

( )3 4 4 4 3 2 3 2tot,wR 10 3.85 10 10 2 10 2 10 m K W 3.69 10 m K W− − − − − −′′ = + × + + × + × ⋅ = × ⋅

( ) ( )3 4 2 5 2400 K 2 10 2 10 m K W 3.52 10 W m 1174 K− −= + × + × ⋅ × =( ) ( )s,o(w) ,i i wInT T 1 h L k q∞ ′′= + +⎡ ⎤⎣ ⎦

,o ,i 5 2w 3 2

tot,w

T T 1300 Kq 3.52 10 W m

R 3.69 10 m K W

∞ ∞−

−′′ = = = ×

′′ × ⋅

With a heat flux of

the inner and outer surface temperatures of the Inconel are

( )s,i(w) ,i w iT T q h∞ ′′= + ( )5 2 2400 K 3.52 10 W m 500 W m K / W 1104 K= + × ⋅ =

ANALYSIS: For a unit area, the total thermal resistance with the TBC is

( ) ( )1 1tot,w o t,c iZr InR h L k R L k h− −′′ ′′= + + + +

Problem: Thermal Barrier (Cont.) 3

Without the TBC,

( )1 1 3 23 20 10 m K Wtot ,wo o iInR h L k h .

− − −′′ = + + = × ⋅

The inner and outer surface temperatures of the Inconel are then

( ) 1212 Ks,i( wo ) ,i wo iT T q h∞ ′′= + =

( ) ( )[ ]1 1293 Ks ,o( wo ) ,i i woInT T h L k q∞′′= + + =

Use of the TBC facilitates operation of the Inconel below Tmax = 1250 K.

COMMENTS: Since the durability of the TBC decreases with increasing temperature, which increases with increasing thickness, limits to its thickness are associated with reliability considerations.

( )wo ,o ,i tot ,woq T T R∞ ∞′′ ′′= − = 4.06×105 W/m2

Problem: Radioactive Waste Decay

Problem 3.62: Suitability of a composite spherical shell for storingradioactive wastes in oceanic waters.

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties at 300K, (4) Negligible contact resistance.

PROPERTIES: Table A-1, Lead: k = 35.3 W/m⋅K, MP = 601K; St.St.: 15.1 W/m⋅K.

ANALYSIS: From the thermal circuit, it follows that

311

tot

T T 4q= q r

R 3π∞− ⎡ ⎤= ⎢ ⎥⎣ ⎦

Problem: Radioactive Waste Decay

The thermal resistances are:

( )Pb1 1

R 1/ 4 35.3 W/m K 0.00150 K/W0.25m 0.30m

π ⎡ ⎤⎡ ⎤= × ⋅ − =⎣ ⎦ ⎢ ⎥⎣ ⎦

( )St.St.1 1

R 1/ 4 15.1 W/m K 0.000567 K/W0.30m 0.31m

π ⎡ ⎤⎡ ⎤= × ⋅ − =⎣ ⎦ ⎢ ⎥⎣ ⎦

( )2 2 2convR 1/ 4 0.31 m 500 W/m K 0.00166 K/Wπ⎡ ⎤= × × ⋅ =⎢ ⎥⎣ ⎦

totR 0.00372 K/W.=

The heat rate is then

( )( )35 3q=5 10 W/m 4 / 3 0.25m 32,725 Wπ× =

and the inner surface temperature is ( )1 totT T R q=283K+0.00372K/W 32,725 W∞= + 405 K < MP = 601K.=

Hence, from the thermal standpoint, the proposal is adequate.

COMMENTS: In fabrication, attention should be given to maintaining a good thermal contact. A protective outer coating should be applied to prevent long term corrosion of the stainless steel.