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MEC551
THERMAL ENGINEERING
Mohd Hafiz Mohd Noh
T1-A18-6C 1
2.0 Conduction
Conduction Analysis
Main Objective of Conduction Analysis
To determine the temperature distribution in a medium
2
Steady versus Transient Heat Transfer
Steady implies no change with time at any point within the medium
Transient implies variation with time or time dependence
3
multidimensional heat transfer,
Three prime coordinate systems:
rectangular (T(x, y, z, t)) ,
cylindrical (T(r, f, z, t)),
spherical (T(r, f, q, t)).
4
Heat Conduction Equation (Rectangular Coordinates)
Cartesian Coordinates:
5
dQy dQ(y+dy)
dQz
dQ(z+dz)
dQ(x+dx)
dQx
dx
dz
dy
X
Z
Y
Heat Conduction Equation (Rectangular Coordinates)
Differential volume:
Heat conduction rate in x-direction (into element):
dzdydxdV
dx
dTdzdyk
dx
dTAkdQ
A
x
6
dQx dy
dz
Heat Conduction Equation (Rectangular Coordinates)
Taylor Series Expansion:
Substitute our equation:
HOTxfhxfhxf )()()(
HOTdxx
Qd
QddxxQdx
x
)(
7 7
small
0
2
2
)(
)(
x
Tdzdyk
x
Qd
xf
dxh
x
TdzdykQdxf
x
X
Heat Conduction Equation (Rectangular Coordinates)
Therefore the net rate of flow in the x-direction is:
2
2
2
2
x
Tdzdydxk
x
Tdydzdxk
x
Tdydzk
x
TdydzkQdQd dxxx
dxx
QdQd xx
x
Tdxdzdyk
x
TdzdykdxxQd
2
2
)(
8
Heat Conduction Equation (Rectangular Coordinates)
Likewise:
2
2
2
2
z
TdzdydxkQdQd
y
TdzdydxkQdQd
dzzz
dyyy
9
Heat Conduction Equation (Rectangular Coordinates)
In addition to the heat flow into and out of the element, there is also the possibilities of:
- Heat being generated within the element (e.g. due to the flow of electricity).
- Heat being stored within the element, as in the case of an unsteady-state condition.
10
Heat Conduction Equation (Rectangular Coordinates)
The rate of heat generated is:
The rate of heat storage is:
dzdydxqdVq
volumepergenerated
heatofrate
11
changeetemperatur
ofrateheatspecific
p
density
pt
TdzdydxC
t
TdVC
Heat Conduction Equation (Rectangular Coordinates)
Assuming the element may expand or contract freely at constant pressure, the energy balance is given by:
12
Rate of
Heat Storage
Net rate of
Heat into
Element
Rate of
Heat
Generation
= +
Heat Conduction Equation (Rectangular Coordinates)
In equation form this is:
Set:
Generation
HeatofRate
elementoHeatofRateNetStorageHeatofRate
P qz
T
y
T
x
Tk
t
TC
int
2
2
2
2
2
2
13
PC
k
Thermal diffusivity (i.e. the ratio of heat conduction to heat
storage).
It represents how fast heat propagates through a material.
~
Heat Conduction Equation (Rectangular Coordinates)
Substituting in, this equation becomes the general differential conduction equation for rectangular coordinates:
Generation
HeatofRateelementoHeatofRateNet
StorageHeatofRate
k
q
z
T
y
T
x
T
t
T
int
2
2
2
2
2
21
14
Heat Conduction Equation (Rectangular Coordinates)
Special cases: 1) Fourier Equation (no heat generation, q =0):
2) Poisson Equation (steady state, T/t =0):
3) Laplace Equation (steady state with no heat generation):
2
2
2
2
2
21
z
T
y
T
x
T
t
T
02
2
2
2
2
2
z
T
y
T
x
T
15
02
2
2
2
2
2
k
q
z
T
y
T
x
T
Heat Conduction Equation (Spherical Coordinates)
Now calculate the general conduction equation in spherical coordinates
16
y x
z z
y x
r
q
T(r,,q)
Heat Conduction Equation (Spherical Coordinates)
Now calculate the general conduction equation in spherical coordinates
17
rq q
z
rdq
q(q)
q(q+dq)
y x
r
q
d
rsinqd
rdq dq
dr
q(r+dr)
q(r)
q(+d)
q()
Heat Conduction Equation (Spherical Coordinates)
Differential volume:
Heat conduction rate in r-direction (in to element):
qqf
fqq
sin
sin
2
drddr
drrddrdV
18
r
Tddkr
r
TrddrkQd
A
r
qqf
qfq
sin
sin
2
q(r)
rdq
Heat Conduction Equation (Spherical Coordinates)
Heat conduction rate in r-direction (out of element):
19
2
22
2
)(
sinsin2
sin
r
Tddkr
r
Tddkrdr
r
Tddkr
r
Qd
drQQdr
rdrr
qqfqqf
qqf
q(r+dr)
rdq
Heat Conduction Equation (Spherical Coordinates)
Total heat conduction rate in r-direction:
2
22
)(2sin
r
Tr
r
TrdrddkQdQd
drrrqqf
20
2
22
22
)(
sinsin2
sinsin
r
Tddkr
r
Tddkrdr
r
Tddkr
r
TddkrQdQd
drrr
qqfqqf
qqfqqf
Heat Conduction Equation (Spherical Coordinates)
Heat conduction rate in q-direction (in to element):
qfq
qfqq
Tddrk
dr
TdrdrkQd
A
sin
sin
21
q(q)
dr
rsinqd
Heat Conduction Equation (Spherical Coordinates)
Heat conduction rate in q-direction (out of element):
qfq
fq
q
qqq
dTT
ddrkT
ddrk
Qd
dQdQdd
2
2
sincossin
22
q(q+dq) dr
rsinqd
Heat Conduction Equation (Spherical Coordinates)
Total heat conduction rate in q-direction:
2
2
sincosq
qqfqqqTT
dddrkQdQdd
qf
qfq
qfqqqq
dTT
ddrk
Tddrk
TddrkQdQd
d
2
2
sincos
sinsin
23
Heat Conduction Equation (Spherical Coordinates)
Heat conduction rate in -direction (in to element):
fq
q
fqqf
Tdrdk
r
TdrdrkQd
sin
sin
24
rdq q()
Heat Conduction Equation (Spherical Coordinates)
Heat conduction rate in -direction (out of element):
2
2
sinsin fq
fq
fq
q
ff
f
fff
TddrdkTdrdk
Qd
dQdQdd
25
rdq
q(+d)
Heat Conduction Equation (Spherical Coordinates)
Total heat conduction rate in -direction:
2
2
sin
sinsin
fq
fq
fq
q
fq
qfff
Tddrdk
TdrdkTdrdkdQQd d
2
2
sin fq
fqfff
TddrdkQdQd
d
26
Heat Conduction Equation (Spherical Coordinates)
Rate of Heat Generation:
Rate of Heat Storage:
qqf sin2
dddrrqdVq
t
TdddrrC
t
TdVC PP
qqf sin2
27
Heat Conduction Equation (Spherical Coordinates)
The energy balance is given by:
28
Rate of
Heat Storage
Net rate of
Heat into
Element
Rate of
Heat
Generation
= +
Heat Conduction Equation (Spherical Coordinates)
qqf
fq
qf
qqf
sin
sin
1
sincos
sinsin2
sin
2
2
2
2
2
2
22
2
dddrrq
T
TT
r
Tr
r
Tr
dddrk
t
TdddrrCP
29
Heat Conduction Equation (Spherical Coordinates)
k
qT
r
T
r
T
rr
T
r
T
r
t
T
k
CP
2
2
222
2
222
2
sin
111
sin
cos2
1
fqqqq
q
PC
k
30
Now by r2, sin(q), and k:
Recall: ~ thermal diffusivity
Heat Conduction Equation (Spherical Coordinates)
k
qT
r
T
r
T
rr
T
r
T
rt
T
T
rr
Tr
rr
2
2
22
sinsin
1
2
2
22
1
2
2
sin
1
11
sin
cos21
2
2
2
fq
qqq
q
31
Now simplify:
Heat Conduction Equation (Spherical Coordinates)
k
qT
r
T
rr
Tr
rr
t
T
2
2
222
2
2 sin
1sin
sin
11
1
fqqq
32
Therefore the conduction equation in spherical coordinates is:
Heat Conduction Equation (Cylindrical Coordinates)
Homework #2 Derive the equations for cylindrical coordinates
33
Heat Conduction Equation (Cylindrical Coordinates)
Equations for cylindrical coordinates:
k
q
z
TT
rr
Tr
rrt
T
k
q
z
TT
rr
T
rr
T
t
T
r
Tr
rr
2
2
2
2
2
2
2
2
2
2
1
2
2
111
111
q
q
34
Boundary and Initial Conditions
Temperature distribution in a medium can be determined from the solution of appropriate heat conduction equation. But the solution depends on the boundaries of the medium.
For cases in which the medium is time dependent, conditions at an initial time are also essential.
35
Boundary and Initial Conditions
The 4 most common boundary conditions are:
1) Constant Surface Temperature:
T(0,t) = Ts
36
x
T
Ts
T(x,t)
Boundary and Initial Conditions
The 4 most common boundary conditions are:
2) Constant and finite heat flux (heat transfer rate per unit area, W/m2):
s
x
qx
Tk
0
37
x
T
T(0,t)
qs
qs
Boundary and Initial Conditions
The 4 most common boundary conditions are:
3) Adiabatic or insulated surface:
00
xx
T
38
x
T
T(0,t)
T(x,t)
Boundary and Initial Conditions
The 4 most common boundary conditions are:
4) Convection surface condition:
tTThx
Tk
x
,00
39
x
T
T(x,t)
q
T, h
1-D Steady State Conduction (Example 2.1)
Example 2.1 - One-dimensional steady-state heat conduction (no heat generation):
Develop the expressions: 1) Temperature distribution T(x) within the slab.
2) Heat flow (Q), through the area (A) of the slab. 40
X
T
X=0 X=L
T1
T2
A, k
A slab (of thickness L)
with no energy
generation (q=0) has the
following boundary
conditions:
X= 0 ; T(0)= T1
X= L ; T(L)= T2
1-D Steady State Conduction (Example 2.1)
(I): Integrate and apply b.cs and solve constants C1 & C2 to find the temperature distribution T(x).
21)( CxCxT
02
2
x
T
41
X
T
X=0 X=L
T1
T2
A, k ~ 1-D Laplace Equation
(rectangular coordinates
112)( Tx
L
TTxT
L
TTC
TLCTLx
CTx
121
112
210
1-D Steady State Conduction (Example 2.1)
(II): Solve for Q
Ak
LRwhere
R
TT
L
TTkA
dx
dTkAQ
:
1212
L
TT
TxL
TT
dx
d
dx
dT
12
112
42
X
T
X=0 X=L
T1
T2
A, k
1-D Steady State Conduction (Analysis Procedure)
Slab (Plane Wall)
Consider a slab of isotropic (invariable) thermal conductivity material (k) with an heat generation rate of q(x) [W/m3].
Isotropic means: having properties that are identical in all directions.
43
T
X
1
2
T1
T2
dx
dT k, A
q(x) q(x)
T2
1-D Steady State Conduction (Analysis Procedure)
The general equation for conduction with heat generation is (assuming constant energy generation):
k
q
x
T
2
2
0
44
k
q
z
T
y
T
x
T
t
T
2
2
2
2
2
21
0, steady state 0, 1D 0, 1D
For: 0 X L
1-D Steady State Conduction (Analysis Procedure)
Solving differential equations:
212
1
2
2
2CxCx
k
qxT
Cxk
q
x
T
k
q
x
T
45
1-D Steady State Conduction (Analysis Procedure)
Solve C1 and C2 by using the boundary conditions:
(i) x=0, T(x)= T1 (ii) x=L, T(x)= T2
L
k
q
L
TTC
TLCLk
qTii
TCi
2
2)
)
121
11
2
2
12
212
2CxCx
k
qxT
46
T
X
1
2
T1
T2
dx
dT k, A
T2
1-D Steady State Conduction (Analysis Procedure)
Substituting in the constants gives:
The temperature distribution T(x) in the slab can now be found for a known heat generation rate q(x), thickness (L), and thermal conductivity coefficient (k) of the material.
12
22TxL
k
q
L
Tx
k
qxT
47
1-D Steady State Conduction (Analysis Procedure)
Once the temperature distribution T(x) in the slab is established from the solution to this equation, the heat flux q(x) anywhere in the slab can be determined from the Fouriers equation.
dx
dTk
A
48
Thermal conductivity
In a gas, conduction is due to the collisions and diffusion of the molecules due to their random motion.
In solids, it is due to the combination of vibrations of the molecules in their lattice and the energy transport of free electrons.
49 Gas Liquid Solid
Thermal conductivity
50
Thermal conductivity (k) of a material is the measure of the ability of the material to conduct heat.
Thermal conductivity
Thermal conductivity is temperature dependent
51
Thermal conductivity
52
1-D Steady State Conduction
One-dimensional (1D) heat conduction
Implies that the temperature gradient exists in only one direction.
Steady state systems (SS)
The temperature within the solid is assumed not to be time dependent.
53
Thermal Resistance Method
1D/SS analysis can be applied to problems to determine the temperature distribution and heat flow in a solid, slab, cylinder, or sphere.
The thermal resistance approach (similar to Ohms Law) is a technique that simplifies complicated problems which involve multi-layered mediums when there is no heat generation (q=0).
54
Thermal Resistance Method (Analysis Procedure)
If q(x)=0 (no heat generation) then the rate of flow of heat energy normal to the area (A) is given by:
L Thickness of the slab
A Area normal to the direction of heat flow
K Thermal conductivity coefficient
T Temperature difference (gradient)
R Thermal resistance against heat conduction or conduction resistance of the wall.
kA
LRwhere
R
T
L
TkA
x
TkAQ
:
55
Thermal Resistance Method (Analysis Procedure)
This is like Ohms Law:
Therefore, circuit representations can provide a useful tool for both conceptualizing and calculating heat transfer problems.
)(tanRe
)()(
Rcesis
VDifferencePotentialICurrent
56
Analogy to Electrical Current Flow
Eq. 3-5 is analogous to the relation for electric current flow I, expressed as
Heat Transfer Electrical current flow
Rate of heat transfer Electric current
Thermal resistance Electrical resistance
Temperature difference Voltage difference
57
(3-6) 1 2
eR
V VI
Thermal Resistance Method (Analysis Procedure)
Ak
LR
58
T
X
1
2
T1
T2
dx
dT k, A
q(x) q(x)
T2
Q Q T1 T2
Thermal Resistance Method (Example 2.2)
AAAA CBA
59
Example 2.2 - Multi-Layer Wall: Determine Q.
Q(x)
T
X
A
T1
T3
kC, AC
T2
T4
kB, AB kA, AA
C
B
Q Q T2 T3 T1 T4 RA RB RC
Thermal Resistance Method (Example 2.2)
C
C
B
B
A
Ax
TTAk
x
TTAk
x
TTAkQ
342312
Ak
xR
Ak
xR
Ak
xR
C
CC
B
BB
A
AA
;;
60
Q Q T2 T3 T1 T4 RA RB RC
Thermal Resistance Method (Example 2.2)
ceresisthermal
differencepotentialthermal
R
T
Ak
x
Ak
x
Ak
x
TTQ
overall
C
C
B
B
A
A
tan
41
61
Q Q T2 T3 T1 T4 RA RB RC
Generalized Thermal Resistance Network
The thermal resistance concept can be used to solve steady heat transfer problems that involve parallel layers or combined series-parallel arrangements.
The total heat transfer of two parallel layers
1 2 1 21 2 1 21 2 1 2
1 1T T T TQ Q Q T T
R R R R
62
(3-29) 1
totalR
1 2
1 2 1 2
1 1 1 = total
total
R RR
R R R R R
(3-31)
Combined Series-Parallel Arrangement
The total rate of heat transfer through
the composite system
where
31 21 2 3
1 1 2 2 3 3 3
1 ; ; ; conv
LL LR R R R
k A k A k A hA
63
(3-32) 1
total
T TQ
R
1 212 3 3
1 2
total conv convR R
R R R R R RR R
(3-33)
(3-34)
Thermal Resistance Concept- Conduction Resistance
Equation 33 for heat conduction through a plane wall can be rearranged as
Where Rwall is the conduction resistance expressed as
64
(3-4) 1 2
, (W)cond wallwall
T TQ
R
(3-5) ( C/W)wallL
RkA
Thermal Resistance Concept- Convection Resistance
Thermal resistance can also be applied to convection processes.
Newtons law of cooling for convection heat transfer rate ( ) can be rearranged as
Rconv is the convection resistance-
It is the thermal resistance of the
surface Against heat convection
conv s sQ hA T T
65
(W)sconvconv
T TQ
R
1 ( C/W)conv
s
RhA
Thermal Resistance Concept- Radiation Resistance
The rate of radiation heat transfer between a surface and the surrounding
2 2 2 (W/m K)( )
radrad s surr s surr
s s surr
Qh T T T T
A T T
66
4 4 ( ) (W)s surrrad s s surr rad s s surrrad
T TQ A T T h A T T
R
1 ( /W)rad
rad s
R Kh A
Radiation Resistance- It is the thermal resistance Of a surface against radiation
Thermal Resistance Concept- Radiation and Convection Resistance
A surface exposed to the surrounding might involves convection and radiation simultaneously.
The convection and radiation resistances are parallel to each other.
When TsurrT, the radiation
effect can properly be
accounted for by replacing h
in the convection resistance
relation by
hcombined = hconv+hrad (W/m2K)
67
Thermal Resistance Network
consider steady one-dimensional heat transfer through a plane wall that is exposed to convection on both sides.
Under steady conditions we have
or
68
Rate of
heat convection
into the wall
Rate of
heat conduction
through the wall
Rate of
heat convection
from the wall = =
1 ,1 1
1 22 2 ,2
Q h A T T
T TkA h A T T
L
Thermal Contact Resistance
In reality surfaces have some roughness.
When two surfaces are pressed against each other, the peaks form good material contact but the valleys form voids filled with air.
As a result, an interface contains
numerous air gaps of varying sizes
that act as insulation because of the
low thermal conductivity of air.
Thus, an interface offers some
resistance to heat transfer, which
is termed the thermal contact
resistance, Rc.
69
Multilayer Plane Walls
In practice we often encounter plane walls that consist of several layers of different materials.
The rate of steady heat transfer through this two-layer composite wall can be expressed through where the total thermal resistance is
,1 ,1 ,2 ,2
1 2
1 1 2 2
1 1
total conv wall wall convR R R R R
L L
h A k A k A h A
70
Thermal Resistance Method (Example 2.3)
AAAA CBA
71
T
X
A
T1
T3
k3
T2
T4
k2
C B
k1
L1 L2 L3
Tf1
Tf4
Hot
Fluid
Tf1, h1
Cold
Fluid
Tf4, h4
T2 T3 T1 RA RB RC Tf1 Tf4 T4 Rf1 Rf4
Example 2.3 - Composite Wall with convection
surface conditions. Determine Q.
Thermal Resistance Method (Example 2.3)
43211
4
44
3
3
43
2
2
32
1
1
21
1
11
11
ff R
f
RRRR
f
x
Ah
TT
Ak
L
TT
Ak
L
TT
Ak
L
TT
Ah
TTQ
R
TT
AhAk
L
Ak
L
Ak
L
Ah
TTQ
ffff
x
41
43
3
2
2
1
1
1
41
11
72
T2 T3 T1 RA RB RC Tf1 Tf4 T4 Rf1 Rf4
Thermal Resistance Method (Example 2.3)
However, with composite systems it is often convenient to express the rate of heat transfer in terms of overall heat transfer coefficient (U).
This is defined by an expression similar to Newtons Law of Cooling:
43
3
2
2
1
1
1
11
11
hk
L
k
L
k
L
h
ARU
73
TAUQ x
Thermal Resistance Method (Example 2.4)
74
Example 2.4 Combined Heat Transfer: conduction,
convection, and radiation
take place simultaneously
on boiler tubes.
The hot gases of combustion products create
a thin film of gas on the
outer wall of the boiler tube
and water film within it.
Determine the total resistance (R):
Thermal Resistance Method (Example 2.4)
75
There are parallel circuits in the gas film section due to both convection and radiation acting there.
T
X
T1
T3
T2
T4
L
Hot gas Water inside tube
Ra
dia
tio
n
Gas Film Water Film
Tube wall
T3 T4 T2 R2 R3, conv
R1,conv
T1
R1,rad
Thermal Resistance Method (Example 2.4)
3
43
2
32
1
21
R
TT
R
TT
R
TT
A
76
T3 T4 T2 R2 R3
R1,conv
T1
R1,rad
Thermal Resistance Method (Example 2.4)
21
4
2
4
121
,1
4
2
4
121
,1
21
1
TT
TTF
R
TTFR
TT
rad
FactorShaperad
77
T3 T4 T2 R2 R3
R1,conv
T1
R1,rad
The radiation thermal resistance in the gas R1,rad is given by:
Thermal Resistance Method (Example 2.4)
AhR conv
1
,1
1
kA
LR 2
78
The convection thermal resistance in the hot gas R1,conv is given by:
The conduction thermal resistance in the wall R2 is given by:
T3 T4 T2 R2 R3
R1,conv
T1
R1,rad
Thermal Resistance Method (Example 2.4)
AhR
4
3
1
79
The convection thermal resistance in the water (boiler) or ambient air (furnace) R3 is given by:
T3 T4 T2 R2 R3
R1,conv
T1
R1,rad
Thermal Resistance Method (Example 2.4)
AhkA
LAh
TT
TTF
RRRR
RRRR
convrad
4
1
1
21
4
2
4
121
32
1
,1,1
321
1
11
80
The total resistance is:
T3 T4 T2 R2 R3
R1,conv
T1
R1,rad
Radial Systems
Cylindrical and spherical systems often experience temperature gradients in the radial direction only and in the case can be treated as one-dimensional.
81
r
r
1-D, SS Heat Conduction (Cylindrical Coordinates)
Steady State Condition:
1-D Conduction:
Therefore:
0
t
T
0,02
2
2
2
z
TT
q
82
k
q
z
TT
rr
Tr
rrt
T
2
2
2
2
2
111
q
1-D, SS Heat Conduction (Cylindrical Coordinates)
The equation for 1-D, steady state heat conduction then becomes:
83
rk
q
dr
dTr
dr
d
k
q
r
Tr
rr
0
1
1-D, SS Heat Conduction (Cylindrical Coordinates)
For constant heat generation:
Solve for T(r), by integrating twice.
0gq
2120
10
ln4
2
CrCrk
gT
r
Cr
k
g
dr
dT
84
1-D, SS Heat Conduction (Cylindrical Coordinates)
Clearly, two boundary conditions are required to determine C1 and C2. Typically one of these will be the boundary condition at the outer surface.
The other boundary condition will be at the center (r=0) where the temperature is symmetric, such that:
22)( rratTrT
00)0(
ratdr
dT
85
1-D, SS Heat Conduction (Cylindrical Coordinates)
An alternative boundary condition at the center (r=0) can be obtained if the temperature is finite.
0)0( 0 ratTT
86
1-D, SS Heat Conduction (Example 2.5)
Example 2.5: Hollow Cylinder (Tube) (with convective surface conditions): Find all of the thermal resistances.
87
No heat generation
Cold Fluid, Tf2, h2
r2
r1
L
T1
T2
Hot fluid
Tf1, h1
Hot fluid
Tf1, h1
0q
1-D, SS Heat Conduction (Example 2.5)
Solve using the thermal resistance method:
The resistances Rf1 and
Rf2 can be found from
Newtons Law of
Cooling:
22
2
11
1
2
1
2
1
hLrR
hLrR
CylinderofAreaSurfaceOuter
f
CylinderofAreaSurfaceInterior
f
88
r2
r1
L
T1
T2
Hot fluid
Tf1, h1
Hot fluid
Tf1, h1
Cold Fluid, Tf2, h2
T2 Tf2 T1 RA Rf2 Tf1 Rf1
1-D, SS Heat Conduction (Example 2.5)
1
2
21
2
ln
2
1
2
1
2
2
1
2
1
2
1
rr
TTkLQ
dTkdrrL
Q
dTkdrA
Q
dr
dTkAQ
cond
T
T
r
r
cond
T
T
r
rrL
cond
cond
89
The resistance RA can be found from Fouriers Law:
1-D, SS Heat Conduction (Example 2.5)
A
condR
TTQ 21
1
1
2
2
ln
kL
rr
RA
90
If the heat transfer rate is constant, this can be further simplified:
Thermal
Resistance
1-D, SS Heat Conduction (Example 2.6)
Example 2.6: Composite Cylindrical Wall. Solve Q in terms of the overall thermal resistance (Rtot) and overall heat transfer coefficient (U).
91
Cold Fluid, Tf2, h2
L
Hot fluid
Tf1, h1
Hot fluid
Tf1, h1
0qNo heat generation
1-D, SS Heat Conduction (Example 2.6)
Composite layers
92 rD
rC
rB
rA
TA
TB
TC
TD
C B A
1-D, SS Heat Conduction (Example 2.6)
Use the thermal resistance method:
21
21
2
1
2
ln
2
ln
2
ln
2
1
LhrLk
rr
Lk
rr
Lk
rr
Lhr
TTQ
DC
C
D
B
B
C
A
A
B
A
ff
r
93
rD
rC
rB
rA TA TB TC TD
TB TC TA RA RB RC Tf1 Tf4 TD Rf1 Rf4
T
r
TA TB
TC TD
Tf1
Tf4
1-D, SS Heat Conduction (Example 2.6)
Now express this in terms of U and Rtot
Where:
)( 21421121
ffDffA
tot
ff
r TTAUTTAUR
TTQ
11 2
1
2
ln
2
ln
2
ln
2
1
LhrLkLkLkLhrR
DC
r
r
B
r
r
A
r
r
A
totC
D
B
C
A
B
94
1-D, SS Heat Conduction (Example 2.6)
The overall heat transfer coefficient (based on the inner surface) where AA= 2rAL, is given by:
Or we could calculate U4 (based on the outer surface) where AD= 2rDL:
1
21
4
1lnlnln
1
hk
r
k
r
k
r
hr
rU
C
D
B
C
A
B
r
r
C
D
r
r
B
D
r
r
A
D
A
D
1
21
1
1lnlnln
1
hr
r
k
r
k
r
k
r
hU
D
A
r
r
C
A
r
r
B
A
r
r
A
A
C
D
B
C
A
B
95
1-D, SS Heat Conduction (Example 2.6)
The definition is arbitrary, the overall heat transfer coefficient may also be defined on any of the intermediate areas:
14321
totDCBA RAUAUAUAU
214
211
ffDr
ffAr
TTAUQ
TTAUQ
96
1-D, SS Heat Conduction (Example 2.6)
Check
21
111
1lnlnln
1
2
hr
r
k
r
k
r
k
r
h
LrAU
D
Ar
r
C
Ar
r
B
Ar
r
A
A
C
D
B
c
A
B
21
1lnlnln
1
2
1
1
hr
r
k
r
k
r
k
r
hLr D
Ar
r
C
Ar
r
B
Ar
r
A
A
AC
D
B
c
A
B
97
1-D, SS Heat Conduction (Example 2.6)
98
tot
DC
r
r
B
r
r
A
r
r
A
A
R
LhrLkLkLkhLr
AU
C
D
B
c
A
B
1
2
1
2
ln
2
ln
2
ln
2
1
1
21
1
Correct !
1-D, SS Heat Conduction (Example 2.7)
Example 2.7: Solid Cylinder Develop an expression for a 1-D, radial, steady state temperature
distribution T(r) and the flux q(r) for a solid cylinder of radius (r2) with an energy generation at a constant rate of q (W/m3) and temperature on the outer surface maintained at T2. Calculate the temperature at the center and the flux at the outer surface for r2= 1 cm, q= 2108 W/m3, k= 20 W/(mC), and T2= 100 C.
99
r2
T2= 100 C
1-D, SS Heat Conduction (Example 2.7)
Boundary Conditions
The general conduction equation for a cylinder is:
k
q
z
TT
rr
Tr
rrt
T
2
2
2
2
2
111
q
22;
0;0
TTrrat
dr
dTrat
100
r2
T2= 100 C
, steady state , 1-D , 1-D
1-D, SS Heat Conduction (Example 2.7)
21
2
1
)ln(4
)(
2
CrCk
rqrT
r
C
k
rq
r
T
drrk
q
r
Tr
01
k
q
r
Tr
rr
101
Integrating twice:
1-D, SS Heat Conduction (Example 2.7)
102
Applying boundary conditions at r= 0 gives:
Applying boundary conditions at r= r2 gives:
0
02
0,0
1
1
C
r
Cr
k
q
dr
dT
dr
dTrat
0
2
222
21
2
222
4
)ln(4
;
rk
qTC
CrCrk
qTTrrat
0
1-D, SS Heat Conduction (Example 2.7)
2
2
2
2
2
2
22
2
21
2
14
44
ln4
Tr
r
k
rq
k
rqTr
k
q
CrCrk
qT
103
The temperature distribution in the solid cylinder is:
1-D, SS Heat Conduction (Example 2.7)
22
rq
k
rqk
dr
dTkq
CC
m
Tr
r
k
rqT
Cm
W
m
W
350100204
01.0102
14
0
3
8
2
2
2
2
2
104
The heat flux in the cylinder is thus:
The temperature T(0) at r= 0 is:
1-D, SS Heat Conduction (Example 2.7)
2
3 6
8
22 10
2
01.0102
2)(
m
Wm
W mrqrq
105
The heat flux at the outer surface of the cylinder is:
1-D, SS Heat Conduction (Example 2.8)
Example 2.8: Hollow Cylinder A hollow cylinder is heated at the inner side at the rate of q0
(105W/m2) and dissipates heat from the outer surface into a fluid at Tf2. There is no energy generation and the conductivity (k) of the solid is assumed to be constant. Develop an expression for the temperature T1 and T2 at (the inner and outer surface) and calculate them for the following parameters.
106
Tf2= 100 C
h= 400 W/(m2C)
r2 =5 cm
r1 =3 cm
k= 15 W/(mC)
1-D, SS Heat Conduction (Example 2.8)
Since there is no heat generated in the cylinder, it is more convenient to solve the problem using the thermal resistance method.
Lk
rr
2
ln1
2
107
T2 Tf2 T1
Tf2= 100 C
h= 400 W/(m2C)
r2 =5 cm
r1 =3 cm
k= 15 W/(mC)
Lhr22
1
1-D, SS Heat Conduction (Example 2.8)
221
2
21
21
21
1000ln
2Lhrr
r
kL
fTTLrqAqQ
221
2
21
22
21
2110
ln2
hLr
f
r
r
LkA
TTTTLrq
108
T2 Tf2 T1
Lk
rr
2
ln1
2
Lhr22
1
Also:
1-D, SS Heat Conduction (Example 2.8)
Taking the first equality and solving for T1:
Taking the second equality and solving for T2:
201
1
2110
1
21
1
2
ln
ln
TqT
TTrq
r
r
k
r
r
r
k
2
22
102
22
22
10
f
f
Thr
rqT
hr
TTrq
109
1-D, SS Heat Conduction (Example 2.8)
Solving:
201 121 ln TqT rr
k
r
2
22
102 fT
hr
rqT
110
CCm
mm
Cm
Wm
W
2.35225003.0
05.0ln
15
03.010 2
5
CCm
m
Cm
Wm
W
25010040005.0
03.010
2
2
5
1-D, SS Heat Conduction (Spherical Coordinates)
Conduction in a spherical shell Consider heat conduction in a hollow sphere. In a steady state, one
dimensional system (without heat generation), the energy entering the differential control volume is equal to the energy leaving the differential control volume.
111
y x
z
dr
dTrk
dr
dTkAQ
A
r
drrr
24
1-D, SS Heat Conduction (Spherical Coordinates)
Separating variables:
Assuming constant k and integrating
2
1
2
1
)(
2)(
4
T
T
TfToffunction
r
r
r dTTkr
drQ
12211
4TTk
r
Q rr
r
112
1-D, SS Heat Conduction (Spherical Coordinates)
21
11
4
1
rrkR
2112
21
11
12 44
21
TTrr
rrk
TTkQ
rr
r
R
TTQr
12
113
where:
1-D, SS Heat Conduction (Spherical Coordinates)
The equation for the rate of heat transfer can also be done by simplifying the heat equation for spherical coordinates, which is recalling:
k
qT
r
T
rr
Tr
rrt
T
2
2
2222
2
2 sin
1sin
sin
111
fqqq
114
0, steady
state 0, 1D 0, 1D
0, no heat
generation
02
r
Tr
r
1-D, SS Heat Conduction (Spherical Coordinates)
Integrate twice:
21
2
1
Cr
CT
drr
CdT
115
1
2
2
0
0
Cdrr
Tr
r
Tr
r
1-D, SS Heat Conduction (Spherical Coordinates)
Apply boundary conditions:
At r = r1 T = T1 At r = r2 T = T2
1211
2
1
111 :
TCrC
Cr
CTrrat
1212
211
12
11222
2
221212
2
122
;
:
TTrr
rrC
rr
TrTrC
r
rCrCTC
r
CTrrat
116
1-D, SS Heat Conduction (Spherical Coordinates)
212
121
12
21
12
2122
12
1211
12
1122
12
22
12
21
12
112221
12
12
21
Trrr
rrrT
rrr
rrr
rrr
rrrrT
rrr
rrrrT
rr
TrTr
rrr
rrT
rrr
rrT
rr
TrTrTT
rrr
rr
Cr
CT
117
1-D, SS Heat Conduction (Spherical Coordinates)
1212
12
12
12
12
2
2
2
12
2
4
4
4
4
TTrr
rrk
TTrr
rr
r
kr
r
Ckr
dr
dTkrqAQ rr
118
1-D, SS Heat Conduction (Spherical Coordinates)
Spherical composites may be treated in much the same way as composite walls and cylinders, where appropriate forms of the total resistance (R) and overall heat transfer coefficient (U) may be determined.
119
y x
z
120
Critical Thickness of Insulation
Consider a tube, cable, or wire dissipating heat from the outer surface into the surrounding air by convection.
It is covered by a layer of insulation to minimize heat loss. In
many cases, the thermal resistance offered by a metal tube or wire is negligibly small in comparison to the insulation.
Critical Thickness of Insulation
Consider a tube, cable, or wire dissipating heat from the outer surface into the surrounding air by convection.
It is covered by a layer of insulation to minimize heat loss. In
many cases, the thermal resistance offered by a metal tube or wire is negligibly small in comparison to the insulation.
121
122 122
Critical Thickness of Insulation
The tube wall temperature (To) is nearly the same as the fluid.
insulation
insulation
ri
ro
ri
ro
(a) Rod or Wire (b) Pipe
To, ho
Ti Ti
123 123
Critical Thickness of Insulation
critoo
critoo
o
crito
rr
rrif
h
kr
,
,
,
Critical radius of insulation
Will decrease the rate of heat loss
expected. Good !
Will increase the heat loss
continuously. Maximum at the critical
thickness. Avoid !
Finite Differences (Numerical Methods)
m x increment
n y increment
124
x
y
b
x
y
Node
Finite Differences (Numerical Methods)
In the absence of a heat source or sink in the system, the rate of heat flow toward the nodal point must be equal to the rate of heat flow from it in steady state.
125
m, n
m,n+1
m,n-1
m+1,n m-1,n
Qm,n+1
Qm,n-1
Qm-1,n Qm+1,n
Finite Differences (Numerical Methods)
In the finite difference method the derivatives are replaced by differences.
Instead of taking the limit, the following approximation for the derivative can be used.
x
xfxxf
dx
xdf
x 0lim
x
xfxxf
dx
xdf
126
f(x+Dx) f(x)
x x+dx
Dx
Df
Finite Differences (Numerical Methods)
If the grid is subdivided into M sections of equal length.
directionxthein
M
Lx
127
Tm+1 Tm
Tm-1
m-1 m m+1
m- m+
Finite Differences (Numerical Methods)
x
TT
dx
dT
x
TT
dx
dT
mm
m
mm
m
1
2
1
1
2
1
128
Tm-1 Tm
m-1/2
Tm Tm+1
m+1/2
Finite Differences (Numerical Methods)
The 2nd derivative is simply:
2
11
2
2
2
11
21
21
x
TTT
x
xdx
Td
mmm
x
TT
x
TT
mdxdT
mdxdT
mmmm
129
Finite Differences (Numerical Methods)
Likewise:
2
11
2
2 2
y
TTT
dy
Td nnn
130
Finite Differences
Finite Differences of Plane Wall: The 1-D heat transfer through a plane wall is given by the following equation. Find the finite difference expression for:
This can be expressed in differential form as:
Where qm is the rate of heat generation per unit volume at node m.
02
2
k
q
dx
Td
131
3,2,102
2
11
mfork
q
x
TTT mmmm
Finite Differences (Example 2.12)
For 2-dimensions:
The finite difference formulation
is:
13,2,1
13,2,1
22 ,2
1,,1,
2
,1,,1
Nnfor
Mmfor
k
q
y
TTT
x
TTT nmnmnmnmnmnmnm
02
2
2
2
k
q
y
T
x
T
132
(m, n+1)
(m, n-1)
(m+1, n) (m, n) (m-1, n)
Dy Dx
Finite Differences (Example 2.12)
If x = y then:
Or since we are considering that k= constant, the heat flows may all be expressed in terms of temperature differentials and this same equation can be derived.
k
xqTTTTT
nm
nmnmnmnmnm
2
,
,1,1,,1,1 4
133
1
1
xAwheredy
dTkAQ
yAwheredx
dTkAQ
yyy
xxx
Finite Differences (Example 2.12)
Therefore the finite difference expressions for Q are:
y
TTxkQQ
y
TTxkQQ
x
TTykQQ
x
TTykQQ
nmnm
nmdowncond
nmnm
nmupcond
nmnm
nmrightcond
nmnm
nmleftcond
,1,
1,,
,1,
1,,
,,1
,1,
,,1
,1,
134
Finite Differences (Example 2.12)
Therefore the total heat transfer is:
Therefore if x = y:
0,1,1,,1,1
yx
nmnmnmnmnm AqQQQQ
135
yxq
y
TTx
y
TTx
x
TTy
x
TTy
k nmnmnmnmnm
nmnmnmnm
,
,1,,1,
,,1,,1
Finite Differences (Example 2.12)
Then:
k
xqTTTTT
nm
nmnmnmnmnm
2
,
,1,1,,1,1 4
136
Finite Differences
To use this numerical method, these equations must be written for each node within the material and the resultant system of equations solved for the temperature at the various nodes.
137
Finite Differences (Example 2.13)
Example 2.13: Finite Difference Modeling of a square plate. A small plate (1x1 m) and with a k= 10 W/(mC) has one face maintained at 500C and the rest at 100C.
Compute: (i) Temperature at various nodes.
(ii) Heat flow at the boundaries.
138
1 m
1 m
500C
100C
k
100C
100C
Finite Differences (Example 2.13)
my3
1
mx3
1
139
1
3
2
4
T=500C
T=100C
T=100C T=100C
Four
node
problem
Finite Differences (Example 2.13)
(i) The solution for finding the temperatures is (for
an interior node):
04100100:4
04100100:3
04500100:2
04500100:1
423
314
241
132
TCTTCNode
TCTCTNode
TTCTCNode
TTCCTNode
04 ,1,1,.1,1 nmnmnmnmnm TTTTT
140
1
3
2
4
Finite Differences (Example 2.13)
Rearranging equations:
04200
04200
04600
04600
432
431
421
321
TTT
TTT
TTT
TTT
141
Finite Differences (Example 2.13)
142
-4 1 1 0 T1 -600
1 -4 0 1 T2 -600
1 0 -4 1 T3 -200
0 1 1 -4 T4 -200
=
Finite Differences (Example 2.13)
Solve by Gaussian Elimination:
143
T1 T2 T3 T4 C
-4 1 1 0 -600
1 -4 0 1 -600
1 0 -4 1 -200
0 1 1 -4 -200
Finite Differences (Example 2.13)
144
T1 T2 T3 T4 C
-4 1 1 0 -600
4x1= -4x4= 4x0= 4x1= -600x4=
4 -16 0 4 -2400
1 0 -4 1 -200
0 1 1 -4 -200
X4
Finite Differences (Example 2.13)
145
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -16 0 4 -2,400
1 0 -4 1 -200
0 1 1 -4 -200
-4+(4)=0
Finite Differences (Example 2.13)
146
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 0 4 -2,400
1 0 -4 1 -200
0 1 1 -4 -200
1+(-16)=-15
Finite Differences (Example 2.13)
147
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -2,400
1 0 -4 1 -200
0 1 1 -4 -200
1+(0)=1
Finite Differences (Example 2.13)
148
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -2,400
1 0 -4 1 -200
0 1 1 -4 -200
0+(4)=4
Finite Differences (Example 2.13)
149
-600+(-2400)=-3000
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -3,000
1 0 -4 1 -200
0 1 1 -4 -200
Finite Differences (Example 2.13)
150
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -3,000
0 1 -15 4 -1,400
0 1 1 -4 -200
X4
Finite Differences (Example 2.13)
151
X15
-15+(1x15)=0 T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -3,000
0 0 -224 64 -24,000
0 1 1 -4 -200
Finite Differences (Example 2.13)
152
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -3,000
0 0 -224 64 -24,000
0 0 16 -56 -6,000 X15
Finite Differences (Example 2.13)
153
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -3,000
0 0 -224 64 -24,000
0 0 0 -720 -108,000 X =14 224
16
-224+(14x16)=0
Finite Differences (Example 2.13)
154
T1 T2 T3 T4 C
-4 1 1 0 -600
0 -15 1 4 -3,000
0 0 -224 64 -24,000
0 0 0 -720 -108,000
Finite Differences (Example 2.13)
000,108720
000,2464224
000,3415
6004
4
43
432
321
T
TT
TTT
TTT
155
CT 150720
000,1084
Solving for the unknowns
Finite Differences (Example 2.13)
CT
CT
CT
2504
150250600
25015
000,34150150
150224
64150000,24
1
2
3
156
Finite Differences (Example 2.13)
(ii) The heat rate is thus:
mW
x TTx
ykQ
000,4
1005002
110015010025010
1001002
1100500
2
1100100 310
157
1
3
2
4
-Qx=0
y
TxkQ
x
TykQ
y
x
Finite Differences (Example 2.13)
(ii) The heat rate is thus:
mW
x TTx
ykQ
000,4
1005002
110015010025010
1001002
1100500
2
1100100 421
158
1
3
2
4
-Qx=1
Finite Differences (Example 2.13)
(ii) The heat rate is thus:
mW
y TTy
xkQ
000,1
10015010015010
1001002
1100100
2
1100100 430
159
1
3
2
4
-Qy=0
Finite Differences (Example 2.13)
(ii) The heat rate is thus:
mW
y TTy
xkQ
000,9
20020050025050025010
5001002
1500100
2
1500500 211
160
1
3
2
4
+Qy=1
Finite Differences (Example 2.13)
Therefore:
Heat flowing into the plate = +9,000 W/m
Heat flow leaving the plate = -4000-4000-1000=-9,000 W/m 161
1
3
2
4
+9,000 W/m
-4,000 W/m -4,000 W/m
-1,000 W/m
Finite Differences (Example 2.14)
Example 2.14: Derive the heat equation for node 3 of the plate shown below.
162
1
3
5
2
4
6
ins
ula
tio
n
Given:
k= constant
b= thickness
x= y
Steady state
Finite Differences (Example 2.14)
Also note the half areas:
y
163
1
3
5
4
2
y
2
x
Finite Differences (Example 2.14)
Since the heat transfer is steady state then Q=0 and the equation at node 3 is:
Note: because of the insulation
1,
21
1,
21
,1
353134
220
nm
x
nm
x
nm
y
Q
A
Q
AQ
Ay
TTbxk
y
TTbxk
x
TTbyk
164
0,1 nmQ
Finite Differences (Example 2.15)
Example 2.15: Steady 2-D Heat Conduction in an L-bar. Given: k = 15 W/(mC) h= 80 W/(m2C) T= 25C q= gn= 2x10
6 W/m3
165
ins
ula
tio
n
T= 90 C
Convection
h, T= 25C
qr= 5000 W/m2
1 2 3
4 5 6 7 8 9
10 11 12 13 14 15
x= y= L = 0.012m
Dx Dy
Finite Differences (Example 2.15)
Assumptions:
Heat transfer is steady and 2-D
Thermal conductivity (k) is constant
Heat generation q is constant
Radiation heat transfer is neglible
Form the volume elements by partitioning the region between nodes. Node 5 is the only completely interior node. Consider the volume element represented by Node 5 to be full size (e.g. x=y=1).
166
Finite Differences (Example 2.15)
Then the elements represented by a regular boundary node (i.e. Node 2) becomes half size (e.g. x=y/2=1) and a corner node (i.e. Node 1) is quarter size (e.g. x/2=y/2=1) .
167
ins
ula
tio
n
T= 90 C
Convection
h, T= 25C
qr= 5000 W/m2
1 2 3
4 5 6 7 8 9
10 11 12 13 14 15
x y
Finite Differences (Example 2.15)
Since the bottom surface is at a constant temperature of 90 C, then:
CTTTTTT 90151413121110 168
insu
lati
on
T= 90 C
Convection
h, T= 25C
qr= 5000 W/m2
1 2 3
4 5 6 7 8 9
10 11 12 13 14 15
x= y= L
x y
Finite Differences (Example 2.15)
Node 1 (Energy balance): Insulated on the left
Convection on top
Conduction on right and bottom
Cm
W
Cm
W
Cm
W
Cm
W
Cm
WmC
TTTm
152
012.0102
15
2580
15
012.0802
26
421
22
169
1 2
4 5
ins
ula
tio
n
Convection
h, T= 25C
Dy ______
2
Dx/2
2.11064.2 421 TTT
Lyx
222220 1
14121
yxg
y
TTxk
x
TTykTT
xh
Finite Differences (Example 2.15)
Node 2 (Energy balance): Convection on top
Conduction right, left, bottom
170
2 3
5 6
Dy
Dx
1
4
Convection
h, T= 25C
22
20
22125
232
yxg
x
TTyk
y
TTxk
x
TTykTTxh
225321
22
24 L
k
gT
k
hLTTT
k
hLT
4.222128.4 5321 TTTT
Finite Differences (Example 2.15)
Node 3 (Energy balance): Convection on top and right
Conduction at bottom and left
0
222
222
332
363
yxg
x
TTyk
y
TTxkTT
yxh
171
2 3
5 6
Dy
Dx
Convection
h, T= 25C
k
LgT
k
hLTT
k
hLT
2
222 3632
8.12128.2 632 TTT
Finite Differences (Example 2.15)
Node 4 (Energy balance): Insulated on left
Conduction at the top, right, bottom This node is on the insulated boundary and can
be treated as an interior node by replacing the insulation with a mirror. This puts a reflected image of node 5 to the left of node 4.
042
4410515
k
LgTTTTT
Interior
172
4 5
10 insu
lati
on
T= 90 C
11
1 2
Dy
Dx
2.1099024
2
4541
10
k
LgTTT
T
5
Finite Differences (Example 2.15)
Node 5 (Energy balance): Interior node
Conduction all sides
Can use the equation for an interior node
2.109904
04
2
55624
2
5511624
11
k
LgTTTT
k
LgTTTTT
T
173
ins
ula
tio
n
1 2 3
4 5 6
10 11 12
Dy
T= 90 C
Finite Differences (Example 2.15)
Node 6 (Energy balance): Convection upward right corner
Conduction everywhere else
174 T= 90 C
2 3
5 6 7
11 12 13
Dx
Dy
Convection
h, T= 25C
6 Qcond
Qcond
Qcond
Qconv
Qconv
Qcond
Finite Differences (Example 2.15)
04
3
2
222
663
65
612676
yxgy
TTxkTT
x
yk
y
TTxk
x
TTykTT
yxh
175 T= 90 C
2 3
5 6 7
11 12 13
Dx
Dy
Convection
h, T= 25C
of the internal
energy generation,
since only the
volume
0.212128.62 7653 TTTT
Finite Differences (Example 2.15)
Node 7 (Energy balance): Convection on top
Conduction right, left, and bottom
176
T= 90 C
6 7 8
12 13 14
x
y
Convection
h, T= 25C
0
2
2
776713
787
yxg
x
TTyk
y
TTxk
x
TTykTTxh
4.202128.4
2180
24
876
2
7876
TTT
k
LgT
k
hLTT
k
hLT
Finite Differences (Example 2.15)
Node 8 (Energy balance): Identical to Node 7
k
LgT
k
hLTT
k
hLT
2
8987
2180
24
177
T= 90 C
7 8 9
13 14 15
Dy
Convection
h, T= 25C
4.202128.4 987 TTT
Dx
Finite Differences (Example 2.15)
Node 9 (Energy balance): qr heat flow on right
Convection on top
Conduction on bottom and left
0222
222
998
9159
yxg
x
TTyk
y
TTxk
yqTT
xh R
178
qr= 5000
W/m2 8 9
14 15
Dx
Dy
Convection
h, T= 25C
T= 90 C
2.105064.2
2902
98
2
998
TT
k
LgT
k
hLL
k
qT
k
hLT R
Finite Differences (Example 2.15)
We now have 9 equations and 9 unknowns, so we can solve:
179
2.11064.2 421 TTT
4.222128.4 5321 TTTT
8.12128.2 632 TTT
2.10924 541 TTT
2.1094 5624 TTTT
0.212128.62 7653 TTTT
4.202128.4 876 TTT
4.202128.4 987 TTT
2.105064.2 98 TT
Node 1:
Node 2:
Node 3:
Node 4:
Node 5:
Node 6:
Node 7:
Node 8:
Node 9:
Finite Differences (Example 2.15)
Solving:
T1= 112.1 C
T2= 110.8 C
T3= 106.6 C
T4= 109.4 C
T5= 108.1 C
T6= 103.2 C
T7= 97.3 C
T8= 96.3 C
T9= 97.6 C
180
Finite Differences (Example 2.15)
181
insu
lati
on
T= 90 C
Convection
h, T= 25C
qr= 5000 W/m2
1 2 3
4 5 6 7 8 9
10 11 12 13 14 15
Temperature
(C)
Hi
Low
END OF CONDUCTION SECTION
182