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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:
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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(
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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.
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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.
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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.
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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?
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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! !"" ""= + + + +
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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#
$%
&
'(
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Problem: Radioactive Waste Decay
The thermal resistances are:
RPb = 1/ 4! ! 35.3 W/m "K( )#$ %& 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.