10 - 1 - Numerical PDE PDF
10 - 1 - Numerical PDE PDF
10 - 1 - Numerical PDE PDF
Chemical Engineering
AGUNG NUGROHO
UNIVERSITAS PERTAMINA
Numerical Solution
to Partial Differential
Equations (PDE)
What is a Partial Differential
Equation ?
Ordinary Differential Equations have only one independent variable
𝑑𝑦
3 + 5𝑦 2 = 3𝑒 −𝑥 , 𝑦(0) = 5
𝑑𝑥
Simplified
form 𝐴 𝑢𝑥𝑥 + 𝐵 𝑢𝑥𝑦 + 𝐶 𝑢𝑦𝑦 + 𝐷𝑢𝑥 + 𝐸𝑢𝑦 + 𝐹 𝑢 = 𝐺
𝑗 𝑗 𝑗 𝑗+1 𝑗
𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1 𝑇𝑖 − 𝑇𝑖
𝛼 =
Δ𝑥 2 Δ𝑡
𝛼 Δ𝑡
Solving for the temp at the time node 𝑛 + 1 gives choosing, 𝜆 = Δ𝑥 2
𝑗+1 𝛼 Δ𝑡 𝑗 𝑗 𝑗 𝑗
𝑇𝑖 = 𝑇 − 2𝑇 + 𝑇 + 𝑇
Δ𝑥 2 𝑖+1 𝑖 𝑖−1 𝑖
we can write the equation as,
𝑗+1 𝑗 𝑗 𝑗 𝑗
𝑇𝑖 = 𝜆 𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1 + 𝑇𝑖
This equation is called Forward Time Centered Space or FTCS
The Explicit Method (FTCS)
𝑗+1 𝑗 𝑗 𝑗 𝑗
𝑇𝑖 = 𝜆 𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1 + 𝑇𝑖
This equation can be solved explicitly because it can be written for each
internal location node of the rod for time node 𝑗 + 1 in terms of the
temperature at time node 𝑗.
This process continues until we reach the time at which we are interested in
finding the temperature.
Example 1: Explicit Method
Consider a steel rod that is subjected to a temperature of 100oC on the
left end and 25oC on the right end. If the rod is of length 0.05 m ,use the
explicit method to find the temperature distribution in the rod from 𝑡 = 0
and 𝑡 = 9 seconds. Use ∆𝑥 = 0.01 𝑚 , ∆𝑡 = 3 .
Given: 𝑘 = 54 𝑊 𝑚.𝐾 , 𝜌 = 7800 𝑘𝑔 𝑚3 , 𝑐 = 490 𝐽 𝑘𝑔.𝐾
The initial temperature of the rod is 20oC.
0 1 2 3 4 5
𝑇 = 100°𝐶 𝑇 = 25°𝐶
0.01𝑚
Example 1: Explicit Method
Recall, Number of time steps, 𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑘 =
𝛼= Δ𝑡
𝜌𝐶
9−0
= =3
therefore, 3
54 Boundary Conditions
𝛼 =
7800 × 490 𝑗
𝑇0 = 100°𝐶
= 1.4129 × 10−5 𝑚2 /𝑠 . 𝑗
for all 𝑗 = 0,1,2,3
𝑇5 = 25°𝐶
Then, All internal nodes are at 20°𝐶 for 𝑡 = 0 sec
Δ𝑡
𝜆=𝛼 This can be represented as,
Δ𝑥 2
3
= 1.4129 × 10−5 2
𝑇10 = 𝑇20 = 𝑇30 = 𝑇40 = 20°𝐶
0.01
= 0.4239
.
Example 1: Explicit Method
Nodal temperatures when 𝑡 = 0 sec , 𝑗 = 0 : 𝑇00 = 100°𝐶
𝑇10 = 20°𝐶
𝑇20 = 20°𝐶
0 Interior nodes
𝑇3 = 20°𝐶
𝑇40 = 20°𝐶
𝑇50 = 25°𝐶
𝑗+1 𝑗 𝑗 𝑗 𝑗
𝑇𝑖 = 𝑇𝑖 + 𝜆 𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1
Example 1: Explicit Method
Nodal temperatures when 𝑡 = 0 sec , 𝑗 = 0 (Example Calculations)
setting 𝑖 = 0
𝑇01 = 100°𝐶 − Boundary Condition
𝑖=1 𝑖=2
𝑇11 = 𝑇10 + 𝜆 𝑇20 − 2𝑇10 + 𝑇00 𝑇21 = 𝑇20 + 𝜆 𝑇30 − 2𝑇20 + 𝑇10
= 20 + 0.4239 20 − 2(20) + 100 = 20 + 0.4239 20 − 2(20) + 20
= 20 + 0.4239 80 = 20 + 0.4239 0
= 20 + 33.912 = 20°𝐶
= 53.912°𝐶
Nodal temperatures when 𝑡 = 3sec , 𝑗 = 1:
𝑇01 = 100°𝐶 − Boundary Condition
𝑇11 = 53.912°𝐶
𝑇21 = 20°𝐶
Interior nodes
𝑇31 = 20°𝐶
𝑇41 = 22.120°𝐶
𝑇51 = 25°𝐶 − Boundary Condition
Example 1: Explicit Method
Nodal temperatures when 𝑡 = 3 sec , 𝑗 = 1 (Example Calculations)
setting 𝑖 = 0 𝑇 2 = 100°𝐶 − Boundary Condition
0
𝑖=1 𝑖=2
𝑇12 = 𝑇11 + 𝜆 𝑇21 − 2𝑇11 + 𝑇01 𝑇22 = 𝑇21 + 𝜆 𝑇31 − 2𝑇21 + 𝑇11
= 53.912 + 0.4239 20 − 2(53.912) + 100 = 20 + 0.4239 20 − 2(20) + 53.912
= 53.912 + 0.4239 12.176 = 20 + 0.4239 33.912
= 53.912 + 5.1614 = 20 + 14.375
= 59.073°𝐶 = 34.375°𝐶
𝑖=1 𝑖=2
𝑇13 = 𝑇12 + 𝜆 𝑇22 − 2𝑇12 + 𝑇02 𝑇23 = 𝑇22 + 𝜆 𝑇32 − 2𝑇22 + 𝑇12
= 59.073 + 0.4239 34.375 − 2(59.073) + 100 = 34.375 + 0.4239 20.899 − 2(34.375) + 59.073
= 59.073 + 0.4239 16.229 = 34.375 + 0.4239 11.222
= 59.073 + 6.8795 = 34.375 + 4.7570
= 65.953°𝐶 = 39.132°𝐶
Implicit
Method
𝛼 Δ𝑡 𝑗+1
𝑇𝑖+1 − 2𝑇𝑖
𝑗+1 𝑗+1
+ 𝑇𝑖−1 𝑇𝑖
𝑗+1
− 𝑇𝑖
𝑗
𝜆= 𝛼 =
Δ𝑥 2 Δ𝑥 2 Δ𝑡
𝑗+1 𝑗+1 𝑗+1 𝑗+1 𝑗
𝜆 𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1 = 𝑇𝑖 − 𝑇𝑖
𝑗+1 𝑗+1 𝑗+1 𝑗+1 𝑗
𝜆𝑇𝑖−1 − 2𝜆𝑇𝑖 + 𝜆𝑇𝑖+1 − 𝑇𝑖 = −𝑇𝑖
Rearranging yields 𝑗+1 𝑗+1 𝑗+1 𝑗
𝜆𝑇𝑖−1 − 2𝜆 + 1 𝑇𝑖 + 𝜆𝑇𝑖+1 = −𝑇𝑖
The rearranged equation can be written for every node during each time step. These
equations can then be solved as a simultaneous system of linear equations to find the
nodal temperatures at a particular time.
Example 2: Implicit Method
Consider a steel rod that is subjected to a temperature of 100oC on the left
end and 25oC on the right end. If the rod is of length 0.05 m, use the
implicit method to find the temperature distribution in the rod from 𝑡 = 0
and 𝑡 = 9 seconds. Use ∆𝑥 = 0.01 𝑚 , ∆𝑡 = 3 .
Given: 𝑘 = 54 𝑊 𝑚.𝐾 , 𝜌 = 7800 𝑘𝑔 𝑚3 , 𝑐 = 490 𝐽 𝑘𝑔.𝐾
The initial temperature of the rod is 20oC.
0 1 2 3 4 5
𝑇 = 100°𝐶 𝑇 = 25°𝐶
0.01𝑚
Example 2: Implicit Method
Recall, Number of time steps, 𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑘 =
𝛼= Δ𝑡
𝜌𝐶
9−0
= =3
therefore, 3
54 Boundary Conditions
𝛼 =
7800 × 490 𝑗
𝑇0 = 100°𝐶
= 1.4129 × 10−5 𝑚2 /𝑠 . 𝑗
for all 𝑗 = 0,1,2,3
𝑇5 = 25°𝐶
Then, All internal nodes are at 20°𝐶 for 𝑡 = 0 sec
Δ𝑡
𝜆=𝛼 This can be represented as,
Δ𝑥 2
3
= 1.4129 × 10−5 2 𝑇10 = 𝑇20 = 𝑇30 = 𝑇40 = 20°𝐶
0.01
= 0.4239
.
Example 2: Implicit Method
Nodal temperatures when 𝑡 = 0 sec , 𝑗 = 0 : 𝑇00 = 100°𝐶
𝑇10 = 20°𝐶
𝑇20 = 20°𝐶
Interior nodes
𝑇30 = 20°𝐶
𝑇40 = 20°𝐶
𝑇50 = 25°𝐶
We can now calculate the temperature at each node explicitly using the
equation formulated earlier,
For the 1st time step we can write four such equations with four unknowns, expressing
them in matrix form yields
Example 2: Implicit Method
1.8478 −0.4239 0 0 𝑇11 62.390
1
−0.4239 1.8478 −0.4239 0 𝑇2 20
=
0 −0.4239 1.8478 −0.4239 𝑇31 20
0 0 −0.4239 1.8478 𝑇41 30.598
𝑇01 = 100°𝐶
𝑇11 = 39.451 Hence, the nodal temps 𝑇11 = 39.451°𝐶
𝑇21 = 24.792 at 𝑡 = 3 sec are 𝑇21 = 24.792°𝐶
𝑇31 = 21.438 𝑇31 = 21.438°𝐶
𝑇41 = 21.477 𝑇41 = 21.477°𝐶
𝑇51 = 25°𝐶
Example 2: Implicit Method
Nodal temperatures when 𝑡 = 3sec
from BC 𝑇02 = 100°𝐶 , and 𝑇52 = 25°𝐶
For the interior nodes setting 𝑗 = 1 and 𝑖 = 1,2,3,4 gives the following,
−𝜆𝑇02 + (1 + 2𝜆)𝑇12 − 𝜆𝑇22 = 𝑇11
𝑖=1 (−0.4239 × 100) + (1 + 2 × 0.4239)𝑇12 − 0.4239𝑇22 = 39.451
−42.39 + 1.8478𝑇12 − 0.4239𝑇22 = 39.451
𝟏. 𝟖𝟒𝟕𝟖𝑻𝟐𝟏 − 𝟎. 𝟒𝟐𝟑𝟗𝑻𝟐𝟐 = 𝟖𝟏. 𝟖𝟒𝟏
For the 2nd time step we can write four such equations with four unknowns, expressing
them in matrix form yields
Example 2: Implicit Method
1.8478 −0.4239 0 0 𝑇12 81.841
2
−0.4239 1.8478 −0.4239 0 𝑇2 24.792
=
0 −0.4239 1.8478 −0.4239 𝑇32 21.438
0 0 −0.4239 1.8478 𝑇42 32.075
𝑇02 = 100°𝐶
𝑇12 = 51.326 Hence, the nodal temps 𝑇12 = 51.326 °𝐶
𝑇22 = 30.669 at 𝑡 = 6 sec are 𝑇22 = 30.669 °𝐶
𝑇32 = 23.876 𝑇32 = 23.876 °𝐶
𝑇42 = 22.836 𝑇42 = 22.836 °𝐶
𝑇52 = 25°𝐶
Example 2: Implicit Method
Nodal temperatures when 𝑡 = 6sec
from BC 𝑇03 = 100°𝐶 , and 𝑇53 = 25°𝐶
For the interior nodes setting 𝑗 = 2 and 𝑖 = 1,2,3,4 gives the following,
−𝜆𝑇03 + (1 + 2𝜆)𝑇13 − 𝜆𝑇23 = 𝑇12
𝑖=1 (−0.4239 × 100) + (1 + 2 × 0.4239)𝑇13 − (0.4239𝑇23 ) = 51.326
−42.39 + 1.8478𝑇13 − 0.4239𝑇23 = 51.326
𝟏. 𝟖𝟒𝟕𝟖𝑻𝟑𝟏 − 𝟎. 𝟒𝟐𝟑𝟗𝑻𝟑𝟐 = 𝟗𝟑. 𝟕𝟏𝟔
For the 3rd time step we can write four such equations with four unknowns, expressing
them in matrix form yields
Example 2: Implicit Method
1.8478 −0.4239 0 0 𝑇13 93.716
3
−0.4239 1.8478 −0.4239 0 𝑇2 30.669
=
0 −0.4239 1.8478 −0.4239 𝑇33 23.876
0 0 −0.4239 1.8478 𝑇43 33.434
𝑇03 = 100°𝐶
𝑇13 = 59.043 Hence, the nodal temps 𝑇13 = 59.043 °𝐶
𝑇23 = 36.292 at 𝑡 = 9 sec are 𝑇23 = 36.292 °𝐶
𝑇33 = 26.809 𝑇33 = 26.809 °𝐶
𝑇43 = 24.243 𝑇43 = 24.243 °𝐶
𝑇53 = 25°𝐶
Summary Example 2:
Implicit Method
To better visualize the temperature variation at different locations at different
times, the temperature distribution along the length of the rod at different times is
plotted below.
The Crank-Nicolson Method
WHY:
𝜕2 𝑇
Using the implicit method our approximation of was of 𝑂(Δ𝑥)2 accuracy,
𝜕𝑥 2
𝜕𝑇
while our approximation of was of 𝑂(Δ𝑡) accuracy.
𝜕𝑡
We can achieve similar orders of accuracy by approximating the second derivative, on the
left hand side of the heat equation, at the midpoint of the time step. Doing so yields
The first derivative, on the right hand side of the heat equation, is approximated
using the forward divided difference method at time level 𝑗 + 1,
𝑗+1 𝑗
𝜕𝑇 𝑇𝑖 − 𝑇𝑖
≈
𝜕𝑡 𝑖,𝑗
Δ𝑡
Crank-Nicolson Discretization
Consider PDE : 𝜕 2 𝑇 𝜕𝑇
𝛼 2=
𝜕𝑥 𝜕𝑡 𝑗+1
𝑗
Using forward in time and central in space, we get :
Δ𝑡
𝜆=𝛼
Δ𝑥 2
0 1 2 3 4 5
𝑇 = 100°𝐶 𝑇 = 25°𝐶
0.01𝑚
Example 3: Crank-Nicolson
Recall, Number of time steps, 𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙
𝑘 =
𝛼= Δ𝑡
𝜌𝐶
9−0
= =3
therefore, 3
54 Boundary Conditions
𝛼 =
7800 × 490 𝑗
𝑇0 = 100°𝐶
= 1.4129 × 10−5 𝑚2 /𝑠 . 𝑗
for all 𝑗 = 0,1,2,3
𝑇5 = 25°𝐶
Then, All internal nodes are at 20°𝐶 for 𝑡 = 0 sec
Δ𝑡
𝜆=𝛼 This can be represented as,
Δ𝑥 2
3
= 1.4129 × 10−5 2 𝑇10 = 𝑇20 = 𝑇30 = 𝑇40 = 20°𝐶
0.01
= 0.4239
.
Example 3: Crank-Nicolson
Nodal temperatures when 𝑡 = 0sec , 𝑗 = 0 : 𝑇00 = 100°𝐶
𝑇10 = 20°𝐶
𝑇20 = 20°𝐶
Interior nodes
𝑇30 = 20°𝐶
𝑇40 = 20°𝐶
𝑇50 = 25°𝐶
We can now calculate the temperature at each node explicitly using the
equation formulated earlier,
For the 1st time step we can write four such equations with four unknowns,
expressing them in matrix form yields
Example 3: Crank-Nicolson
2.8478 −0.4239 0 0 𝑇11 116.30
1
−0.4239 2.8478 −0.4239 0 𝑇2 40.000
=
0 −0.4239 2.8478 −0.4239 𝑇31 40.000
0 0 −0.4239 2.8478 𝑇41 52.718
𝑇01 = 100°𝐶
𝑇11 = 44.372 Hence, the nodal temps 𝑇11 = 44.372°𝐶
𝑇21 = 23.746 at 𝑡 = 3sec are 𝑇21 = 23.746°𝐶
𝑇31 = 20.797 𝑇31 = 20.797°𝐶
𝑇41 = 21.607 𝑇41 = 21.607°𝐶
𝑇51 = 25°𝐶
Example 3: Crank-Nicolson
Nodal temperatures when 𝑖 = 0 , 𝑡 = 6sec, 𝑇02 = 100°𝐶 − BoundaryCondition
(Example Calculations)
𝑖=1 For the interior nodes setting 𝑗 = 1 and 𝑖 = 1,2,3,4 gives the following,
For the 2nd time step we can write four such equations with four unknowns,
expressing them in matrix form yields
Example 3: Crank-Nicolson
2
2.8478 −0.4239 0 0 𝑇1 145.971
2
−0.4239 2.8478 −0.4239 0 𝑇2 54.985
=
0 −0.4239 2.8478 −0.4239 𝑇32 43.187
0 0 −0.4239 2.8478 𝑇 2 54.908
4
𝑇02 = 100°𝐶
𝑇12 = 55.883 Hence, the nodal temps 𝑇12 = 55.883 °𝐶
𝑇22 = 31.075 at 𝑡 = 6 sec are 𝑇22 = 31.075 °𝐶
𝑇32 = 23.174 𝑇32 = 23.174 °𝐶
𝑇42 = 22.730 𝑇42 = 22.730 °𝐶
𝑇52 = 25°𝐶
Example 3: Crank-Nicolson
Nodal temperatures when 𝑖 = 0, 𝑡 = 9 sec, 𝑇03 = 100°𝐶 − BoundaryCondition
(Example Calculations)
𝑖=1 For the interior nodes setting 𝑗 = 2 and 𝑖 = 1,2,3,4 gives the following,
For the 3rd time step we can write four such equations with four unknowns,
expressing them in matrix form yields
Example 3: Crank-Nicolson
2.8478 −0.4239 0 0 𝑇13 162.34
3
−0.4239 2.8478 −0.4239 0 𝑇2 69.318
=
0 −0.4239 2.8478 −0.4239 𝑇33 49.509
0 0 −0.4239 2.8478 𝑇43 57.210
𝑇03 = 100°𝐶
𝑇13 = 62.604 Hence, the nodal temps 𝑇13 = 62.604 °𝐶
𝑇23 = 37.613 at 𝑡 = 9 sec are 𝑇23 = 36.613 °𝐶
𝑇33 = 26.562 𝑇33 = 26.562 °𝐶
𝑇43 = 24.042 𝑇43 = 24.042 °𝐶
𝑇53 = 25°𝐶
Summary Example 3:
Crank-Nicolson Method
To better visualize the temperature variation at different locations at different
times, the temperature distribution along the length of the rod at different times is
plotted below.
Internal Temperatures at 9 sec.
The table below allows you to compare the results from all three methods
discussed in previously with the analytical solution.
Crank-
Node Explicit Implicit Analytical
Nicolson
𝑇13 65.953 59.043 62.604 62.510
𝑇23 39.132 36.292 37.613 37.084
𝑇33 27.266 26.809 26.562 25.844
𝑇43 22.872 24.243 24.042 23.610
Exercise
1.
at all time, t
∆𝑥 = 0.2 𝑐𝑚, ∆𝑡 = 2 𝑠
𝑥 = 0.4 𝑐𝑚
𝑡 =2𝑠
Exercise
2.
at all time, t
, 𝑢𝑠𝑖𝑛𝑔 𝑚𝑎𝑡𝑙𝑎𝑏
Data: h = 25 W/m2
Exercise 𝑘 = 54 𝑊 𝑚.𝐾
𝜌 = 7800 𝑘𝑔 𝑚3
𝐽
𝑐 = 490 𝑘𝑔. 𝐾
Dengan IC 𝑇(𝑥, 0) = 𝑇𝑎
𝜕𝑇(𝐿, 𝑡)
BC 𝑇(0, 𝑡) = 𝑇𝑠 dan −𝑘 = ℎ(𝑇 − 𝑇𝑠 )
𝜕𝑥
Tentukan distribusi suhu pada berbagai posisi dan waktu sampai t=100s (implicit
dan Crank Nicolson) dengan mengambil nilai ∆𝑥 = 0.02 𝑚, ∆𝑡 = 10 𝑠
𝑗+1 𝑗+1 𝑗+1 𝑗
𝜆𝑇𝑖−1 − 2𝜆 + 1 𝑇𝑖 + 𝜆𝑇𝑖+1 = −𝑇𝑖
𝑗+1 𝑗+1 𝑗+1 𝑗
𝑖 =1 𝜆𝑇0 − 1 + 2𝜆 𝑇1 + 𝜆𝑇2 = −𝑇1
𝑗+1 𝑗+1 𝑗+1 𝑗
𝑖 =2 𝜆𝑇1 − 1 + 2𝜆 𝑇2 + 𝜆𝑇3 = −𝑇2
𝑗+1 𝑗+1 𝑗+1 𝑗
𝑖 =3 𝜆𝑇2 − 1 + 2𝜆 𝑇3 + 𝜆𝑇4 = −𝑇3
𝑗+1 𝑗+1 𝑗+1 𝑗
𝑖 =4 𝜆𝑇3 − 1 + 2𝜆 𝑇4 + 𝜆𝑇5 = −𝑇4
𝑗+1 𝑗+1 𝑗+1 𝑗
𝑖 =5 𝜆𝑇4 − 1 + 2𝜆 𝑇5 + 𝜆𝑇6 = −𝑇5
𝑗+1 𝑗+1
𝑇6 −𝑇4 𝑗+1 2ℎ ∆𝑥
At the BC x=L −𝑘 = ℎ(𝑇5 − 𝑇𝑠 ) 𝑇6𝑗+1 = 𝑇4𝑗+1 − 𝑗+1
(𝑇5 − 𝑇𝑠 )
2∆𝑥 𝑘
0 00 𝜆 2ℎ𝜆∆𝑥 𝑇4
0 0 2𝜆 −−1 1
++2𝜆2𝜆
+ 𝑇𝑘4
𝑗+1
𝑗+1
𝑗
−𝑇4 −−𝑇
𝜆𝑇𝑗
𝑗+1 2ℎ ∆𝑥
𝑇5 5 5− 𝜆 𝑇𝑠
𝑘
𝑗+1 𝑗 𝑗+1
−1.7064 0.3532 0 0 0 𝑇1 −𝑇1 − 𝜆𝑇0
𝑗+1 𝑗
0.3532 −1.7064 0.3532 0 0 𝑇2 −𝑇2
𝑗+1 𝑗
0 0 0.3532 −1.7064 0.3532 𝑇4 −𝑇4
𝑗+1 𝑗
0 0 0 0.7064 −1.7130 𝑇5 −𝑇5 − 1.3082
Dengan mengambil j=0, kita bisa hitung semua nilai T di j=1 (t=10s)
−1.7064 0.3532 0 0 0
1
𝑇1 −95.6436 𝑇11 = 62.9265
1
0.3532 −1.7064 0.3532 0 0 𝑇2 −25
𝑇21 = 33.2271
0 0.3532 −1.7064 0.3532 0 𝑇3
1
= −25
𝑇31 = 26.8195
1
0 0 0.3532 −1.7064 0.3532 𝑇4 −25
1 𝑇41 = 25.5630
0 0 0 0.7064 −1.7130 𝑇5 −26.3082
𝑇51 = 25.9004
Dengan mengambil j=1, kita bisa hitung semua nilai T di j=2 (t=20s)
Dengan mengambil j=2, kita bisa hitung semua nilai T di j=3 (t=30s)
𝑇13
−1.7064 0.3532 0 0 0
3
−157.9910 𝑇13 = 103.8526
0.3532 −1.7064 0.3532 0 0 𝑇 −43.8331
2
3 𝑇23 = 54.4331
0 0.3532 −1.7064 0.3532 0 𝑇3 = −30.3455
0 0 0.3532 −1.7064 0.3532 𝑇43 −26.8407 𝑇33 = 35.0233
0 0 0 0.7064 −1.7130 𝑇 3
5
−28.2612 𝑇43 = 28.8570
𝑇53 = 28.3990
End