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 Fouriers 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 Fouriers 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 - Ohms law for electrical resistance:

Similarly for heat convection, Newtons 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 overallResistance

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 ssssx2

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

! Fouriers 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)

Fouriers 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

!+

!+

!=

"= ##

Fouriers 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 5C. 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 Rt,c=0.01 m.K/W. The outer surface of the heater is exposed to a fluid of temperature 10C 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=25C.

Dr. M. Khosravy 29

Spherical Coordinates

Starting from Fouriers 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?

Fouriers 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 -30C when imbedded in tissue that is at 37C. A spherical layer of frozen tissue forms around the probe, with a temperature of 0C 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( )#$ %& 1

0.25m' 1

0.30m#

$(

%

&) = 0.00150 K/W

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

0.30m ' 1

0.31m#

$(

%

&) = 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.