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7 PARTIAL DIFFERENTIAL EQUATIONS
Introduction
Elliptic Equations
Parabolic Equations
Hyperbolic Equations
Finite Element Method — An Introduction
Boundary Element Method — An Introduction
Chapter 7 Partial Differential Equations / 2
7.1 Introduction
• Partial Differential Equations (PDE) are differential equations which have at least two independent variables, e.g.
34 2
2
2
2
=+∂∂
+∂∂ u
yuxy
xu
second order & linear
xyxu
xu
=∂∂
∂+⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
2
33
2
2
6 third order & nonlinear
• For a second order linear two-dimensional equation, a general equation is
0,,,,2
22
2
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
∂∂
+∂∂
+∂∂
∂+
∂∂
yu
xuuyxD
yuC
yxuB
xuA (7.1)
which can be divided into three types.
TABLE 7.1 Types of second order linear PDEs
ACB 42 − Type Examples
< 0 Elliptic 02
2
2
2
=∂∂
+∂∂
yT
xT
Laplace equation
= 0 Parabolic 2
2
xTk
tT
∂∂
=∂∂
Heat conduction equation
> 0 Hyperbolic 2
2
22
2 1x
yct
y∂∂
=∂∂
Wave equation
Chapter 7 Partial Differential Equations / 3
7.2 Elliptic Equations
• Elliptic PDEs are generally related to steady-state problems with diffusivity having boundary conditions, e.g. the Laplace equation.
• Consider a steady-state 2-D heat conduction problem:
Δz
Δx
Δyq(x)
q(y)
q(x + Δx)
q(y + Δy)
x
y
FIGURE 7.1 Thin plate having a thickness Δz with temperatures at the boundary
Taking q(x) as the heat flux [J/m2⋅s], the heat flow over the element can be written in a time interval of Δt as:
( ) ( ) ( ) ( ) tzxyyqtzyxxqtzxyqtzyxq ΔΔΔΔ++ΔΔΔΔ+=ΔΔΔ+ΔΔΔ
which can be summarised into
( ) ( )( ) ( ) ( )( )( ) ( ) ( ) ( ) 0
0
=Δ
Δ+−+
ΔΔ+−
=ΔΔ+−+ΔΔ+−
yyyqyq
xxxqxq
xyyqyqyxxqxq
For 0, →ΔΔ yx :
0=∂
∂−
∂∂
−y
qxq yx (7.2)
Chapter 7 Partial Differential Equations / 4
• In conduction heat transfer, the relationship between heat flux and temperature is given by the Fourier Law:
yTCkq
xTCkq yx ∂
∂−=
∂∂
−= ρρ , (7.3)
where k is heat diffusivity [m2/s], ρ is density [kg/m3] and C is heat capasity [J/kg°C]. Combining Eq. (7.2) with Eq. (7.3) gives the Laplace equation:
02
2
2
2
=∂∂
+∂∂
yT
xT
(7.4)
• If a heat source or heat generation of ),( yxQ is present in the domain, a Poisson equation can be formed:
( ) 0,2
2
2
2
=+∂∂
+∂∂ yxQ
yT
xT
(7.5)
• To solve Eq. (7.4), use the central differencing:
( )
( )21,,1,
2
2
2,1,,1
2
2
2
2
y
TTTyT
xTTT
xT
jijiji
jijiji
Δ
+−=
∂∂
Δ
+−=
∂∂
−+
−+
Hence, Eq. (7.4) can be written in an algebraic form:
( ) ( )0
222
1,,1,2
,1,,1 =Δ
+−+
Δ
+− −+−+
yTTT
xTTT jijijijijiji
Taking Δx = Δy gives
04 ,1,1,,1,1 =−+++ −+−+ jijijijiji TTTTT (7.6)
Chapter 7 Partial Differential Equations / 5
0,0 x
y
m+1,0
0,n+1
i,ji,j+1
i−1,j
i,j−1i+1,j
FIGURE 7.2 Finite difference grid for solving the Laplace equation
• Eq. (7.6) needs boundary conditions (BC), which may be in the form of:
1. Fixed value — Dirichlet (see Fig. 7.3), e.g. Fixed temperature 2. First derivative, or gradient —Neumann, e.g. Fix heat flux or insulated
75°C
T11 T21 T31
T12 T22 T32
T13 T23 T33
100°C
50°C
0°C FIGURE 7.3 Grid/Node representatioan and BC for Fig. 7.1
For node (1,1) — T01 = 75°C and T10 = 0°C:
04 1110120121 =−+++ TTTTT
754 211211 =−− TTT
Chapter 7 Partial Differential Equations / 6
Hence, the complete system (can be assembled into a matrix equation) is:
15041004175450404
75450404
754
332332
33231322
231312
33322231
2332221221
13221211
323121
22312111
122111
=+−−=−+−−=−+−=−+−−=−−+−−=−−+−=−+−=−−+−=−−
TTTTTTT
TTTTTTT
TTTTTTTTT
TTTTTTT
TTT
• If the Gauss-Seidel method is used, at each node (i,j):
41,1,,1,1
,−+−+ +++
= jijijijiji
TTTTT (7.7)
which can be solve iteratively via relaxation approach:
( ) old,
new,
new, 1 jijiji TTT λλ −+= (7.8)
where λ is the relaxation parameter 21 ≤≤ λ , and is subjected to the termination criterion as followed:
%100new,
old,
new,
,×
−=
ji
jijia T
TTji
ε
Example 7.1
Solve the problem in Fig. 7.3 using the Gauss-Seidel method using the termination criterion %1≤aε dan the relaxation parameter λ = 1.5.
Solution
Let all the initial values be 0°C. For the first iteration:
75.184
007504
1012012111 =
+++=
+++=
TTTTT
( ) ( ) 125.2805.1175.185.1new11 =−+=T
03125.74
00125.2804
2022113121 =
+++=
+++=
TTTTT
Chapter 7 Partial Differential Equations / 7
( ) ( ) 54688.1005.1103125.75.1new21 =−+=T
13672.154
0054688.10504
3032214131 =
+++=
+++=
TTTTT
( ) ( ) 70508.205.1113672.155.1new31 =−+=T
and, for other nodes:
99554.9618579.34
46900.7445703.18
12696.8067188.38
33
23
23
22
13
12
==
==
==
TT
TT
TT
Error for node (1,1) — for the first iteration, all errors are 100%:
%100100125.28
0125.2811
=×−
=aε
For the second iteration:
68736.6786833.7160108.28
95872.8763333.6135718.22
21973.7595288.5751953.32
33
32
31
23
22
21
13
12
11
===
===
===
TTT
TTT
TTT
The process is repeated until the ninth iteration in which the termination criterion is fulfilled (εa < 1%):
71050.6933999.5288506.33
06402.7611238.5629755.33
58718.7821152.6300061.43
33
32
31
23
22
21
13
12
11
===
===
===
TTT
TTT
TTT
Chapter 7 Partial Differential Equations / 8
7.3 Parabolic Equations
• Parabolic PDEs are generally related to transient problems with diffusivity, e.g. the 1-D heat conduction equation.
• For a transient problem, there are three approaches:
1. Explicit method, 2. Implicit method, 3. Semi-implicit method — the Crank-Nicolson method.
• Consider a transient 1-D heat conduction problem:
input − output = storage
( ) ( ) TCzyxtzyxxqtzyxq ΔΔΔΔ=ΔΔΔΔ++ΔΔΔ ρ
x
Hot Cold
FIGURE 7.4 A rod with different temperature at its ends
Dividing with the volume zyx ΔΔΔ and the time interval Δt:
( ) ( )tTC
xxxqxq
ΔΔ
=Δ
Δ+− ρ
In the limits of 0, →ΔΔ tx :
tTC
xq
∂∂
=∂∂
− ρ (7.9)
By using the Fourier law, Eq. (7.3), Eq. (7.9) becomes
tT
xTk
∂∂
=∂∂
2
2
(7.10)
Chapter 7 Partial Differential Equations / 9
0,0 x
t
m+1,0
i,li,l+1
i−1,l
i,l−1i+1,l
FIGURE 7.5 Finite difference grid for the heat conduction equation
• For the explicit method, the right-hand side of Eq. (7.9) can be discretised via central difference as:
( )211
2
2 2x
TTTxT l
il
il
i
Δ+−
=∂∂ −+
and, the left-hand side can be discretised via forward difference as:
tTT
tT l
il
i
Δ−
=∂∂ +1
Hence, Eq. (7.10) can be written in the algebraic form:
( ) tTT
xTTTk
li
li
li
li
li
Δ−
=Δ
+− +−+
1
211 2
(7.11)
( )li
li
li
li
li TTTTT 11
1 2 −++ +−+= λ (7.12)
where 2)( xtk ΔΔ=λ .
xi
Time differencing grid
Space differencing grid
xi−1 xi+1
tl
tl+1
FIGURE 7.6 Computational grid for the explicit method
Chapter 7 Partial Differential Equations / 10
Example 7.2
Use the explicit method to determine the temperature distribution for a slender rod having a length of 10 cm. At time t = 0, the temperature of the rod is 20°C and the boundary conditions are T(0) = 100°C and T(10 cm) = 50°C. Use the conduction coefficient k = 0.835 cm2/s, the time interval Δt = 0.5 s and the step size Δx = 2 cm. Solution
Calculate λ:
( )( )( ) 104375.0
25.0835.0
22 ==Δ
Δ=
xtkλ
At t = 0:
2004
03
02
01 ==== TTTT
Use Eq. (7.12). At t = 0.5 s:
[ ] 35.28100)20(2201044.02011 =+−+=T
[ ] 2020)20(2201044.02012 =+−+=T
[ ] 2020)20(2201044.02013 =+−+=T
[ ] 1313.2320)20(2501044.02014 =+−+=T
At t = 1 s:
[ ] 9569.34100)352.28(2201044.0352.2821 =+−+=T
[ ] 8715.20352.28)20(2201044.02022 =+−+=T
[ ] 3268.2020)20(2132.231044.02023 =+−+=T
[ ] 6089.2520)132.23(2501044.0132.2324 =+−+=T
The calculation can be continued to produce the result as in Fig. 7.7.
Chapter 7 Partial Differential Equations / 11
0
20
40
60
80
100
0 2 4 6 8 10 x (cm)
T (°C)
t = 0
t = 15 st = 10 s
t = 6 st = 3 s
t = 20 st → ∞
FIGURE 7.7 Result for Ex. 7.2
• For the explicit method, a more accurate result can be obtained if Δx and Δt approach zeros. However, stability of the results is only obtained if:
( )kxt
2
21
21
or Δ≤Δ
≤λ
If the stability condition is not fulfilled, the results are contaminated by oscillation.
• To prevent oscillation, the implicit method can be used, where the second order spatial derivative is approximated at time l+1:
( )2
11
111
2
2 2x
TTTxT l
il
il
i
Δ+−
=∂∂ +
−++
+
xi
Time differencing grid
Space differencing grid
xi−1 xi+1
tl
tl+1
FIGURE 7.8 Computational grid for the implicit method
Chapter 7 Partial Differential Equations / 12
Hence, Eq. (7.10) becomes
( ) tTT
xTTTk
li
li
li
li
li
Δ−
=Δ
+− ++−
+++
1
2
11
111 2
(7.13)
( ) li
li
li
li TTTT =−++− +
+++
−1
111
1 21 λλλ (7.14)
Eq. (7.14) leads to n simultaneous linear equations having n unknowns.
Example 7.3
Repeat Ex. 7.2 using the implicit method. Solution
From Ex. 7.2, λ = 0.104375. From Eq. (7.14):
( )[ ]( )[ ]
M
44.301044.01044.021
201044.01044.021)100(1044.012
11
12
11
=−+
=−++−
TT
TT
Hence, the following system of linear equations can be formed:
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
−−−
22.252020
44.30
2088.11044.0001044.02088.11044.00
01044.02088.11044.0001044.02088.1
14
13
12
11
TTTT
[ ]T6152.22,2800.20,6256.20,9634.26=T
• The combination of the explicit and implicit approaches produces the semi-implicit method, and one of its kind is the Crank-Nicolson method:
xi
Time differencing grid
Space differencing grid
xi−1 xi+1
tl
tl+1
FIGURE 7.9 Computational grid for the Crank-Nicolson method
Chapter 7 Partial Differential Equations / 13
In this method, the finite difference term for the spatial derivative is
( ) ( ) ⎥⎦
⎤⎢⎣
⎡
Δ+−
+Δ
+−=
∂∂ +
−++
+−+2
11
111
211
2
2 2221
xTTT
xTTT
xT l
il
il
il
il
il
i (7.15)
Hence, Eq. (7.10) becomes
( ) ( ) li
li
li
li
li
li TTTTTT 11
11
111 1212 +−
++
++− +−+=−++− λλλλλλ (7.16)
Example 7.4
Repeat Ex. 7.2 using the Crank-Nicolson method. Solution
From Ex. 7.2, λ = 0.104375. From Eq. (7.16):
7866.5810437.020874.2)20(10437.020)10437.01(2)100(10437.0
10437.0)10437.01(2)100(10437.0
12
11
12
11
=−
+−+=−++−
TT
TT
By considering other nodes, the following system of linear equations can be formed:
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
−−−
3496.484040
7866.58
2087.21044.0001044.02087.21044.00
01044.02087.21044.0001044.02087.2
14
13
12
11
TTTT
[ ]T8424.22,1517.20,3652.20,5778.27=T
Chapter 7 Partial Differential Equations / 14
7.4 Hyperbolic Equations
• Hyperbolic PDEs are generally related to transient problems with convection, e.g. the 1-D wave equation.
• Consider a 1-D wave equation, which is a hyperbolic PDE:
0=∂∂
+∂∂
xua
tu
(7.17)
• One of the methods is the MacCormack’s technique, which is an explicit finite-difference technique and is second-order-accurate in both space and time. By using the Taylor series:
ttuuu t
itt
i Δ⎟⎠⎞
⎜⎝⎛
∂∂
+=Δ+
avg (7.18)
• This method consists of two steps: predictor and corrector. In predictor step, use forward difference in the right-hand side:
⎟⎟⎠
⎞⎜⎜⎝
⎛Δ−
−=⎟⎠⎞
⎜⎝⎛
∂∂ +
xuua
tu t
iti
t
i
1 (7.19)
Thus, from the Taylor series, the predicted value of u is:
ttuuu
t
i
ti
tti Δ⎟
⎠⎞
⎜⎝⎛
∂∂
+=Δ+ (7.20)
• In corrector step, by replacing the spatial derivatives with rearward differences:
⎟⎟⎠
⎞⎜⎜⎝
⎛Δ−
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂ Δ+
−Δ+Δ+
xuua
tu tt
itt
i
tt
i
1 (7.21)
The average of time derivative can be obtained using
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
=⎟⎠⎞
⎜⎝⎛
∂∂
Δ+ tt
i
t
i tu
tu
tu
21
avg (7.22)
Chapter 7 Partial Differential Equations / 15
• Hence the final, “corrected” value at time t+Δt is:
ttuuu t
itt
i Δ⎟⎠⎞
⎜⎝⎛
∂∂
+=Δ+
avg
• The accuracy of the solution for a hyperbolic PDE is dependent on truncation and round off errors, and the term representing it is called artificial viscosity ( )ν−Δ 12
1 xa .
• The effect of artificial viscosity leads to numerical dissipation, which is originated by the even-order derivatives in the truncated term, but it improves stability.
FIGURE 7.10 Numerical dissipation: (a) t = 0, (b) t > 0
• Another opposite effect is known as numerical dispersion, which is originated by the odd-order derivatives in the truncated term and causes ‘wiggles’.
FIGURE 7.11 Numerical dispersion: (a) t = 0, (b) t > 0
Chapter 7 Partial Differential Equations / 16
7.5 Finite Element Method ⎯ An Introduction
• The Finite Element Method (FEM) is a computer aided mathematical technique used to obtain an approximate numerical solution of a response of a physical system which is subjected to an external loading.
• By using this technique, the computational domain which is theoretically a continuum, is being discretised in form of simple geometries.
• The mesh is the computational domain which is an assembly of discrete elemental blocks known as finite elements, and the vertices defining the elements are called nodes.
• Governing equation is employed at each element to form a set of algebraic equations ⎯ local system.
• Local equations assembled to form a global system which is the solved to yield a vector of variables.
Node & element Sample problem Mesh
y
x
z
x,y
node
element
3-D
2-D
1-D
FIGURE 7.10 Examples of elements and their applications
Chapter 7 Partial Differential Equations / 17
Geometry, material properties, mesh,BC, IC
Discretisation of structures into elements
Formation of local stiffness matrix [k]
Formation of global stiffness matrix [K]
Boundary conditions as
constraints to the model
Load conditionsto the model
Solving the system of Linear equation
[K]{a}= {F}
Calculaton of strains,stresses and forces
Boundary conditions
Loads
Graphical visualisation
print-out & files
FIGURE 7.11 Algorithm for a force analysis using FEM
• Consider a 1-D steady state heat conduction:
( ) 02
2
=+∂∂ xQ
xTk (7.17)
Eq. (7.17) needs appropriate boundary conditions such that:
)(00
∞=
=
−==
TThqTT
LLx
x (7.18)
Chapter 7 Partial Differential Equations / 18
T0
L
hT∞
xT0 TL
1 2 3 4
1 2 3
FIGURE 7.12 Boundary condition for a 1D steady state heat conduction
For the first element:
1 2 x1
x x2
1 2
ξ = −1 ξ = +1
ξ
FIGURE 7.13 The first element in the finite element method
The transformation of the coordinate system to a local system is given by an isoparametric coordinate ξ, i.e.
( ) 121
12
−−−
= xxxx
ξ (7.19)
Thus, in order to calculate the temperature at the middle section, a linear interpolation function or a linear shape function can be used:
21)(
21)(
2
1
ξξ
ξξ
+=
−=
N
N (7.20)
Hence, the temperature can be interpolated using the shape function as followed:
eTNTNT NT=+= 2211)(ξ (7.21)
Chapter 7 Partial Differential Equations / 19
Ni
ξ = -1 ξ = +1
N1 N2
T
ξ = -1 ξ = +1T1
T2
T = N1T1 + N2T2
FIGURE 7.14 Linear shape function
Differentiation of Eq. (7.21) gives
dxxx
d12
2−
=ξ
Using a chain rule:
[ ]e
e
xx
dxd
ddT
dxdT
BT
T
=
−−
=
=
1,11
12
ξξ
where
[ ]1,11
12
−−
=xx
B
• In energy form, the heat conduction problem can be represented by
( ) 02
2
=Ω⎭⎬⎫
⎩⎨⎧
+∂∂
∫ΩdxQ
xTkT
i
( )∫ ∫ ∞−+−⎟⎠⎞
⎜⎝⎛=∏
L
L
L
T TThdxxTQdxdxdTk
0
221
0
2
21 )( (7.22)
Discretisation of Eq. (7.22) gives
( )221
1
1
1
1
T21
22T
∞−−−+⎥⎦
⎤⎢⎣⎡−⎥⎦
⎤⎢⎣⎡=∏ ∑ ∫∑ ∫ TThdlQdlk
Le
eeee
e
eee TNTBBT ξξ (7.23)
Chapter 7 Partial Differential Equations / 20
Hence, the stiffness matrix for the element is
⎥⎦
⎤⎢⎣
⎡−
−== ∫− 11
112
1
1
T
e
eee
lkdlk ξBBk (7.24)
and, the heat rate vector is
⎭⎬⎫
⎩⎨⎧
== ∫− 11
221
1eeee lQdlQ ξNr (7.25)
The global stiffness matrix can be produced via an assembly of all stiffness matrices for all elements, i.e.
∑←e
ekK (7.26)
Likewise, the global heat rate vector is an assembly by local heat rate vectors:
∑←e
erR (7.27)
To combine with the boundary condition 01 TT = , a penalty method can be used:
( )
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
∞+
+
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
+
hTR
RCTR
T
TT
hKKK
KKKKKCK
LLLLLL
L
L
MM
L
MMM
L
L
2
01
2
1
21
22212
11211
(7.28)
or, in a matrix form
K⋅T = R (7.29)
where the penalty parameter C can be estimated as:
410max ×= ijC K (7.30)
Chapter 7 Partial Differential Equations / 21
Example 7.5
Fig. 7.15 shows plate of three composites, where the temperature at the right-hand side is T0 = 20°C and the left-hand side is subjected to convection with T∞ = 800°C and h = 25 W/m2°C. Obtain a temperature distribution across the plate using the finite element method.
k1 k2 k3
k1 = 20 W/m°Ck2 = 30 W/m°Ck3 = 50 W/m°C
T0 = 20°C
0.3m 0.15m 0.15m
1 2 3 4
1 2 3T4 = 20°CT1
T∞ = 800°C
RAJAH 7.15 Plate of three composites used in Ex. 7.5
Solution
Divide the domain into three elements and construct local stiffness matrices for all elements:
⎥⎦
⎤⎢⎣
⎡−
−=
⎥⎦
⎤⎢⎣
⎡−
−=
⎥⎦
⎤⎢⎣
⎡−
−=
1111
15.050
1111
15.030
1111
3.020
3
2
1
k
k
k
Combine all local stiffness matrices to form a global stiffness matrix:
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
−−−
=
5500583003410011
7.66K
Chapter 7 Partial Differential Equations / 22
whereas the heat rate vector is
[ ]T0,0,0,80025×=R
The penalty parameter C can be estimated by 44 1087.6610max ××=×= ijC K
Hence, the finite element system can be solved as followed:
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
×
×
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−−−
−−−
44
3
2
1
10672,100080025
005,8050058300341001375.1
7.66
TTTT
[ ]T0.20,1.57,0.119,6.304=T
Chapter 7 Partial Differential Equations / 23
7.6 Boundary Element Method ⎯ An Introduction
• The Boundary Element Method (BEM) is a relatively new for PDE, where it consists only boundary elements, either line (2-D) or surface (3-D) elements.
• Consider a Laplace equation for a steady-state potential flow problems (ψ is the stream function):
02 =∇ ψ
• For a multi-dimensional case, this equation leads to an analytical solution:
r1ln
21π
=∗ψ (7.31)
FIGURE 7.15 Boundary elements
• Eq. (7.31) can be discretised to yield:
⎟⎠⎞⎜
⎝⎛⎟
⎠⎞
⎜⎝⎛
∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂+ ∫∑∫∑ ∗
=
∗
=j
N
j jj
N
jji ds
nds
nψψψψψ
1121 (7.32)
FIGURE 7.16 Example of an analysis using BEM
Chapter 7 Partial Differential Equations / 24
Exercises
1. A quarter of disc having a radius of 8 cm as shown below has a variation of temperature at the boundary aligned with the principal axes, while the temperature is fixed at 100°C at its outer radius of 8 cm. If the temperature distribution follows the Laplace equation:
02 =∇ T
20°CT1 T2 T3
T4 T5 T6
T7 T8
100°C
0°C 40°C 65°C 85°C 100°C
50°C
75°C
100°C
100°C
By using the grid as shown:
a. Obtain a system of algebraic equations using the finite difference method, b. Solve the system to obtain Ti .
2. By using the implicit technique, solve the following heat conduction problem:
),,(),( 2
2
txTx
txTt ∂
∂=
∂∂ α ,100 << x t > 0
using the following boundary conditions:
.0)0,(,100),10(,0),0( === xTtTtT
Use the constant 10=α , the time step 1.0=Δt , and a model of 10 grid including the grid at the boundary. Compar this solution with the solution obtained by using 3.0=Δt .