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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:
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.