OneOne--Dimensional, SteadyDimensional, Steady--StateStateConduction withoutConduction without

Thermal Energy GenerationThermal Energy Generation

• 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

• Alternative conduction analysis

• 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

• Implications:

0d dTkdx dx

⎛ ⎞ =⎜ ⎟⎝ ⎠

(3.1)

• Heat Equation:

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

( )Heat rate is independent of .xq x

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

• Temperature Distribution for Constant :

( ) ( ),1 ,2 ,1s s sxT x T T TL

= + − (3.3)

k

• Heat Flux and Heat Rate:

( ),1 ,2x s sdT kq k T Tdx L

′′ = − = − (3.5)

( ),1 ,2x s sdT kAq kA T Tdx L

= − = − (3.4)

• Thermal Resistances and Thermal Circuits:tT

Rq

⎛ ⎞Δ=⎜ ⎟

⎝ ⎠

Conduction in a plane wall: ,t condLRkA

= (3.6)

Convection: ,1

t convRhA

= (3.9)

Thermal circuit for plane wall with adjoining fluids:

1 2

1 1tot

LRh A kA h A

= + + (3.12)

,1 ,2x

tot

T Tq

R∞ ∞−

= (3.11)

• Thermal Resistance for Unit Surface Area:

,t condLRk

′′ = ,1

t convRh

′′ =

Units: W/KtR ↔ 2m K/WtR′′↔ ⋅

• Radiation Resistance:

,1

t radr

Rh A

= ,1

t radr

Rh

′′ =

( )( )2 2r s sur s surh T T T Tεσ= + + (1.9)

• Contact Resistance:

,A B

tcx

T TRq−′′ =′′

′′= t c

t cc

RR

A,

,

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

• Composite Wall with Negligible Contact Resistance:

,1 ,4x

tot

T Tq

R∞ ∞−

= (3.14)

1 4

1 1 1C totA Btot

A B C

L RL LRA 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)

• Series – Parallel Composite Wall:

• Note departure from one-dimensional conditions for .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

ALTERNATIVE CONDUCTION ANALYSIS:

• STEADY STATE

• NO HEAT GENERATION

• NO HEAT LOSS FROM THE SIDES

• A(x) and k(T)

dxxx qq +=IS TEMPERATURE DISTRIBUTION ONE-DIMENSIONAL?

IS IT REASONABLE TO ASSUME ONE-DIMENSIONAL TEMPERATURE DISTRIBUTION IN x?

FROM THE FOURIER’S LAW:

dxdTTkxAqx )()(−=

∫∫ −=T

T

x

x dTTkxA

dxq00

)()(

Tube Wall

• Heat Equation:

The Tube Wall

1 0d dTkrr dr dr

⎛ ⎞ =⎜ ⎟⎝ ⎠

(3.23)

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 within the wall?

rqr

• Temperature Distribution for Constant :k

( ) ( ),1 ,2

,21 2 2

lnln /

s ss

T T rT r Tr r r

⎛ ⎞−= +⎜ ⎟⎜ ⎟

⎝ ⎠

(3.26)

• 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 Tdr r r r

kq rq T Tr r

Lkq rLq T Tr 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?

• Composite Wall with Negligible Contact Resistance

( ),1 ,4,1 ,4r

tot

T Tq UA T T

R∞ ∞

∞ ∞−

= = −

(3.30)

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)

• Heat Equation

Spherical Shell

22

1 0d dTrdr 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 sr r

T r T T Tr r

−= − −

−

• Heat flux, Heat Rate and Thermal Resistance:

( ) ( )( ),1 ,22

1 21/ 1/r s sdT kq k T Tdr r r r

′′ = − = −⎡ ⎤−⎣ ⎦

(3.35)( ) ( ) ( )2,1 ,2

1 2

441/ 1/r r s s

kq r q T Tr r

ππ ′′= = −−

• Composite Shell:overall

r overalltot

Tq UA TR

Δ= = Δ

1 ConstanttotUA R −= ↔

( ) 1 Depends on i i tot iU A R A−= ↔

( ) ( )1 2,

1/ 1/4t cond

r rR

kπ−

= (3.36)

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

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− −′′ ′′= + + + +

( )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− − − − − −′′ = + × + + × + × ⋅ = × ⋅

With a heat flux of

,o ,i 5 2w 3 2tot,w

T T 1300 Kq 3.52 10 W m

R 3.69 10 m K W

∞ ∞−

−′′ = = = ×

′′ × ⋅

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 1104 K= + × ⋅ =

( ) ( )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∞ ′′= + +⎡ ⎤⎣ ⎦

Without the TBC,

( )1 1 3 2tot, wo o iInR h L k h 3.20 10 m K W− − −′′ = + + = × ⋅

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

The inner and outer surface temperatures of the Inconel are then

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

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

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.

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

SCHEMATIC:

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 rR 3

∞− ⎡ ⎤= ⎢ ⎥⎣ ⎦& π

The thermal resistances are:

( )Pb1 1R 1/ 4 35.3 W/m K 0.00150 K/W

0.25m 0.30m⎡ ⎤⎡ ⎤= × ⋅ − =⎣ ⎦ ⎢ ⎥⎣ ⎦

π

( )St.St.1 1R 1/ 4 15.1 W/m K 0.000567 K/W

0.30m 0.31m⎡ ⎤⎡ ⎤= × ⋅ − =⎣ ⎦ ⎢ ⎥⎣ ⎦

π

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

π

totR 0.00372 K/W.=

The heat rate is then ( )( )35 3q=5 10 W/m 4 / 3 0.25m 32,725 W× =π

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.