Islamic Azad University Karaj Branch

Dr. M. Khosravy

Chapter 3 One-Dimensional Steady-State Conduction

1

Dr. M. Khosravy 2

stoutgin EEEE !!!! +=+

Chapter 2: ! Need to obtain detailed temperature profiles: Energy conservation written for a differential volume

Conservation of Energy

Can be written for control volume or control surface ! Control volume and control surface: Convenient, but do not give any information on actual temperature distributions

When there are no contributions of kinetic, potential, internal energy and work

!! == outoutinin qEqE !! and

Course Road Map Chapter 1:

Dr. M. Khosravy 3

Chapter 2: Heat Diffusion Equation

stg EE !! ,

z

y

x

Energy Conservation Equation stoutgin EEEE !!!! +=+

qx+dx qx

qz

qz+dz

qy+dy

qy

tT

cqzT

kyy

Tk

yxT

kx p !

!=+"#$%

&'

!!

!!+""#

$%%&

'!!

!!+"

#$%

&'

!!

!! (!

! Involves conduction within the solid ! Convection is taken into account as a boundary condition

Dr. M. Khosravy 4

˙ E in + ˙ E g = ˙ E out + ˙ E st

tT

cqzT

kyy

Tk

yxT

kx p !

!"=+#$%&

'(

!!

!!+##$

%&&'

(!!

!!+#

$%&

'(

!!

!! !

Problems involving conduction: Chapters 2-3

Chapter 3: • We want to obtain temperature profiles for 1-D, SS conduction, with and without generation • Use temperature profiles to obtain expressions for heat transfer rate from Fourier’s law • For problems without generation, implement the thermal circuit approach to determine the heat transfer rate directly.

Energy Conservation

•  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

Dr. M. Khosravy 6

One-Dimensional Steady-State Conduction

•  Conduction problems may involve multiple directions and time-dependent conditions

•  Inherently complex – Difficult to determine temperature distributions

•  One-dimensional steady-state models can represent accurately numerous engineering systems

•  In this chapter we will !  Learn how to obtain temperature profiles for common geometries

with and without heat generation.

!  Introduce the concept of thermal resistance and thermal circuits

Dr. M. Khosravy 7

The Plane Wall Consider a simple case of one-dimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation

•  Temperature is a function of x •  Heat is transferred in the x-

direction Must consider

–  Convection from hot fluid to wall –  Conduction through wall –  Convection from wall to cold fluid

!  Begin by determining temperature distribution within the wall

qx

1,!T

1,sT

2,sT

2,!T

x

x=0 x=L

11, ,hT!

22, ,hT!

Hot fluid

Cold fluid

Dr. M. Khosravy 8

Temperature Distribution •  Heat diffusion equation (eq. 2.4) in the x-direction for

steady-state conditions, with no energy generation:

0=!"#$

%&dxdTk

dxd

•  Boundary Conditions: 2,1, )(,)0( ss TLTTT ==

•  Temperature profile, assuming constant k:

1,1,2, )()( sss TLx

TTxT +!=

"  Temperature varies linearly with x

"  qx is constant

(3.1)

Dr. M. Khosravy 9

Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated:

( ) ( )kALTT

TTLkA

dxdT

kAq ssssx /

2,1,2,1,

!=!=!=

!  Electric circuit theory - Ohm’s law for electrical resistance:

Similarly for heat convection, Newton’s law of cooling applies:

Resistancee DifferencPotential

current Electric =

hATT

TThAq SSx /1

)()( !!

"="=

And for radiation heat transfer:

AhTTTTAhqr

surssursrrad /1

)()( !=!=

(3.2a)

(3.2b)

(3.2c)

Dr. M. Khosravy 10

Thermal Resistance

! Compare with equations 3.2a-3.2c ! The temperature difference is the “potential” or driving

force for the heat flow and the combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance to this flow:

•  We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).

AhR

hAR

kAL

Rr

radtconvtcondt1,1, ,,, ===

!"==

RT

q overall

Resistance Force DrivingOverall

Dr. M. Khosravy 11

Thermal Resistance for Plane Wall

In terms of overall temperature difference: qx

1,!T

1,sT

2,sT

2,!T

x x=0

x=L

11, ,hT!

22, ,hT!

Hot fluid

Cold fluid

AhkAL

AhR

RTT

q

tot

totx

21

2,1,

11 ++=

!= ""

AhTT

kALTT

AhTT

q ssssx

2

2,2,2,1,

1

1,1,

/1//1!! "

="

="

=

Dr. M. Khosravy 12

True or False? •  The conduction resistance of a solid increases when its

conductivity increases.

•  The convection resistance of a fluid increases when the convection coefficient increases.

•  The radiation resistance increases when the surface emissivity decreases.

•  For one-dimensional, steady-state conduction in a plane wall with no heat generation, the heat flux is constant, independent of the direction of flow.

•  For the case above, the temperature distribution is parabolic.

False

False

True

True

False

Dr. M. Khosravy 13

Composite Walls ?  Express the following

geometry in terms of an equivalent thermal circuit.

!Rt = Rtot =

1A

1h1

+LAkA

+LBkB

+LCkC

+ 1h4

"

#$

%

&' =

((RtotA

qx =

T!,1 "T!,4

#Rt

Dr. M. Khosravy 14

Composite Walls

qx =T!,1 " T!,4

#Rt=$TRtot

qx =T!,1 " T!,4

[(1 / h1A) + (LA / kAA) + (LB / kBA) + (LC / kCA) + (1 / h4A)]

TUAqx !=

!  We can also write q in terms of an overall heat transfer coefficient, U

ARhkLkLkLhU

totCCBBAA

111

1

41

=++++

=)]/()/()/()/()/[(

where

Dr. M. Khosravy 15

Composite Walls

!  For resistances in series: Rtot=R1+R2+!+Rn

!  For resistances in parallel: 1/Rtot=1/R1+1/R2+!+1/Rn

q2

q3 T1 T2

q1

Dr. M. Khosravy 16

Example: Heat loss from the body

Determine the effect of a layer of fat on the heat loss from the body. Typical values of conductivity and thickness for the various tissues are given in the following table:

Tissue Conductivity (W/m.K)

Thickness (cm)

Skin 0.442 0.25

Fat 0.21 1.0

Muscle 0.42 2.0

Bone 0.50 0.75

Dr. M. Khosravy 17

Example (Problem 3.15 textbook) Consider a composite wall that includes an 8-mm thick hardwood siding (A), 40-mm by 130-mm hardwood studs (B) on 0.65-m centers with glass fiber insulation (D) (paper faced, 28 kg/m3) and a 12-mm layer of gypsum (vermiculite) wall board (C).

The width of each unit is 2.5 m. !  What is the thermal resistance associated with a wall that comprises 10 of

these studs stacked one next to each other? (Note: Consider the direction of heat transfer to be downwards, along the x-direction)

W=2.5 m

Dr. M. Khosravy 18

Contact Resistance The temperature drop across the interface between materials may be appreciable, due to surface roughness effects, leading to air pockets. We can define thermal contact resistance:

"",

x

BAct q

TTR

!=

See tables 3.1, 3.2 for typical values of Rt,c

Dr. M. Khosravy 19

Variations in Area

•  For steady-state conditions, no heat generation, one-dimensional heat transfer, qx is constant.

dxdT

xATkqx )()(!=

When area varies in the x direction and k is a function of temperature, the previous analysis for plane walls cannot be used

!  Fourier’s law can be written in its most general form:

!! "=#T

T

x

xx

oo

dTTkxAdx

q )()(

Dr. M. Khosravy 20

Example 3.3 Consider a conical section fabricated from pyroceram. It is of circular cross section, with the diameter D=ax, where a=0.25. The small end is at x1=50 mm and the large end at x2=250 mm. The end temperatures are T1=400 K and T2=600 K, while the lateral surface is well insulated.

1.  Derive an expression for the temperature distribution T(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution

2.  Calculate the heat rate, qx, through the cone.

T2 T1

x1 x2 x

Dr. M. Khosravy 21

Radial Systems-Cylindrical Coordinates Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures

Temperature distribution

Dr. M. Khosravy 22

Temperature Distribution •  Heat diffusion equation (eq. 2.5) in the r-direction for

steady-state conditions, with no energy generation:

01 =!"#$

%&

drdTkr

drd

r

•  Boundary Conditions: 2,21,1 )(,)( ss TrTTrT ==

•  Temperature profile, assuming constant k:

2,221

2,1, ln)/ln()(

)( sss T

rr

rrTT

rT +!!"

#$$%

&'= "  Logarithmic temperature distribution

(see previous slide)

•  Fourier’s law: constdrdT

rLkdrdT

kAqr =!"="= )2(

Dr. M. Khosravy 23

Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated:

( ) ( ) ( )condt

ssssssr R

TTLkrr

TTrrTTLk

q,

2,1,

12

2,1,

12

2,1,

)2/()/ln()/ln(2 !

="

!=

!"=

!  In terms of equivalent thermal circuit:

)2(1

2)/ln(

)2(1

22

12

11

2,1,

LrhkLrr

LrhR

RTT

q

tot

totr

!+

!+

!=

"= ##

•  Fourier’s law: constdrdT

rLkdrdT

kAqr =!"="= )2(

Dr. M. Khosravy 24

True or False? •  For one-dimensional, steady-state conduction in a cylidrical or

spherical shell without heat generation, the radial heat rate is independent of the radial coordinate, r.

•  For the same case as above, the radial heat flux is independent of radius.

•  The “thermal conduction resistance” as we derived it in class can be applied to a solid cylinder.

•  When adding more insulation in a pipe (i.e. a thicker layer of insulation), –  the conduction resistance increases –  the convection resistance decreases.

Dr. M. Khosravy 25

Critical Radius of Insulation •  The rate of heat transfer from an

insulated pipe to the surrounding air is

convinscondr RR

TT

LrhLkrr

TTq

+!=

"+

"

!= ##

,

)()/ln(

1

2

12

1

21

2

q

qmax

qbare

hkr cylindercr =,

hkr spherecr2=,

convinscond RRTT

LrhLkrr

TTq

+!=

"+

"

#!= #

,

)()/ln(

1

2

12

1

21

2q

Dr. M. Khosravy 26

Composite Walls

?  Express the following geometry in terms of a an equivalent thermal circuit.

Dr. M. Khosravy 27

Composite Walls ?  What is the heat transfer rate?

where U is the overall heat transfer coefficient. If A=A1=2pr1L:

44

1

3

41

2

31

1

21

1

1 111

hrr

rr

kr

rr

kr

rr

kr

h

U

CBA

++++=

lnlnln

alternatively we can use A2=2pr2L, A3=2pr3L etc. In all cases:

!====

tRAUAUAUAU 144332211

)(

21

2)/ln(

2)/ln(

2)/ln(

21

4,1,4,1,

44

342312

11

4,1,

!!!!

!!

"="

=

#+

#+

#+

#+

#

"=

TTUARTT

q

LhrLkrr

Lkrr

Lkrr

Lhr

TTq

totr

CBA

r

Dr. M. Khosravy 28

Example (Problem 3.37 textbook) A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of 5°C. The tube wall has inner and outer radii of 25 and 75 mm respectively, and a thermal conductivity of 10 W/m.K. The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is R’t,c=0.01 m.K/W. The outer surface of the heater is exposed to a fluid of temperature –10°C and a convection coefficient of h=100 W/m2 .K.

!  Determine the heater power per unit length of tube required to maintain the heater at To=25°C.

Dr. M. Khosravy 29

Spherical Coordinates

•  Starting from Fourier’s law, acknowledging that qr is constant, independent of r, and assuming that k is constant, derive the equation describing the conduction heat transfer rate. What is the thermal resistance?

•  Fourier’s law:

drdTrk

drdTkAqr

)4( 2!"=

"=

)/1()/1()(4

21

2,1,

rrTTk

q ssr !

!"= !!"

#$$%

&'

(=

21,

1141

rrkR condt

Dr. M. Khosravy 30

Example (Problem 3.59 textbook) A spherical cryosurgical probe may be imbedded in diseased tissue, to freeze and destroy that tissue. Consider a probe of 3-mm diameter, whose surface is maintained at -30°C when imbedded in tissue that is at 37°C. A spherical layer of frozen tissue forms around the probe, with a temperature of 0°C existing at the interphase between the frozen and normal tissue. What is the thickness of the layer of frozen tissue?

Problem: Thermal Barrier Coating

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

Schematic:

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

Problem: Thermal Barrier (Cont.)

!!Rtot,w = 10"3+3.85#10"4+10"4+2#10"4+2#10"3( )m2 $K W = 3.69 #10"3m2 $K W

= 400K + 2!10"3+2!10"4( )m2 #K W 3.52!105W m2( ) = 1174KTs,o(w) = T!,i + 1 hi( )+ L k( )In"#

$% &&qw

!!qw =T",o#T",i

!!Rtot,w=

1300K

3.69$10#3m2%K W= 3.52 $105W m2

With a heat flux of

the inner and outer surface temperatures of the Inconel are

Ts,i(w) = T!,i + ""qw hi( ) = 400K + 3.52!105W m2 500W m2"K/W( ) = 1104K

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,

!!Rtot ,wo = ho"1 + L k( )In + hi

"1 = 3.20 # 10"3m2 $K W

The inner and outer surface temperatures of the Inconel are then

Ts ,i ( wo ) = T! ,i + ""qwo hi( ) = 1212KTs ,o( wo ) = T! ,i + 1 hi( )+ L k( )In"

#$% &&qwo = 1293K

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.

!!qwo = T" ,o #T" ,i( ) !!Rtot ,wo = 4.06!105 W/m2

Problem: Radioactive Waste Decay

Problem 3.62: Suitability of a composite spherical shell for storing radioactive 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

q=T1 !T"

Rtot= !q 4

3! r1

3#

$%

&

'(

Problem: Radioactive Waste Decay

The thermal resistances are:

RPb = 1/ 4! ! 35.3 W/m "K( )#$ %& 10.25m

' 10.30m

#

$(

%

&) = 0.00150 K/W

RSt.St. = 1/ 4! !15.1 W/m "K( )#$ %& 10.30m

' 10.31m

#

$(

%

&) = 0.000567 K/W

Rconv = 1/ 4! ! 0.312m2 !500 W/m2 "K( )#$%

&'(= 0.00166 K/W

Rtot = 0.00372 K/W.

The heat rate is then

q=5!105 W/m3 4! / 3( ) 0.25m( )3 = 32,725 W

and the inner surface temperature is T1 = T! + Rtot 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.

Dr. M. Khosravy 36

Summary

•  We obtained temperature distributions and thermal resistances for problems involving steady-state, one-dimensional conduction in orthogonal, cylindrical and spherical coordinates, without energy generation

•  Useful summary of one-dimensional, steady-state solutions to the heat equation with no generation is shown in Table 3.3.