TALAT Lecture 2503: Calculation Methods for Fire Design
Insulating Materials Guide
1. Special topics on insulating materials
Winter-Summer 2012
Sujets Spéciaux sur les matériaux isolants en
Electrotechnique
Hiver-Eté 2012
(6MDI855)
HOMEWORK No.2
Presented by:
Koutoua Simon KASSI
MASTER IN ENGINEERING
Professor:
Issouf FOFANA, PhD, P. ENG in QUEBEC, Senior
Member IEEE
Chairholder of Isolime
Applied Sciences Dept.
2. 1 Koutoua Simon KASSI
Table des Matières
Liste des Figures 2
Problème 1: Claquage de matériau isolant solide (15 pts) 3
1. Expression du champ critique EC requis pour causer un claquage thermique de
l’isolation. 3
2. Montrons que le champ critique Ec est indépendant de Tc. 3
Problem 2: Accelerated Ageing Tests for HV Cables (185 pts) 4
1. Computation of A, B of cable A for each endurance tests, using data provided in
figures 1 to 3 of the topic. 4
1.1 Expressions of A and B available for each endurance tests 4
1.2 Values of A, B for tensile strength (TS) ‘figure1 of the topic’ 5
1.3 Values of A, B for Folding Endurance (FE) ‘figure2 of the topic’ 5
1.4 Values of A, B for Degree of Polymerisation (DP) ‘figure3 of the topic ’ 5
2. Computation of the half-life for the selected properties 5
3. Conclusion 7
Bibliographie 8
Annexe 9
3. 2 Koutoua Simon KASSI
Liste des Figures
Figure 1: Half-life for each test (TS, FE, DP) for thickness equal to 0.17 mm ..............................6
Figure 2: Half-life for each test (TS, FE, DP) for thickness equal to 0.13 mm ..............................7
Liste des tableaux
Tableau 1: Half-lives for thickness of 0.13mm ............................................................. 6
Tableau 2: Half-lives for thickness of 0.17mm ............................................................. 6
4. 3 Koutoua Simon KASSI
Problème 1: Claquage de matériau isolant solide (15 pts)
1. Expression du champ critique EC requis pour causer un claquage
thermique de l’isolation.
La quantité de chaleur(déduite des pertes par conduction) produite est
2
. ( ) (1.1)dcW E T
et celle dissipée est [1][2] [3]:
= Qté de chaleur absorbée (par le diélectrique) + Qté de chaleur évacuée(à l'extérieur)
= ( )
T
T v
W
dT
W C div kgradT
dt
(1.2)
Avec ( ) 0div kgradT
La relation (1.2) représente le bilan énergétique mis en jeu dans le diélectrique
Le champ minimal entraînant le claquage thermique de l’isolation est atteint lorsque
dc TW W , ce qui implique :
2
2 0 0 0
0
. ( )
- - -
. . . . .
( )
-
.
C
C
C
C V
kTV V C V C V C
C
C c c C ckT
V C
C
C c
dT
E T C
dt
C C T T C T T C T TT
E e
T t t A t AkT t
Ae
C T T
E
AkT t
(1.3)
2. Montrons que le champ critique Ec est indépendant de Tc.
A l’aide de la relation (1.3) et en supposant 0....CT T , on en déduit que :
(1.4)V
C
c
C
E
Akt
5. 4 Koutoua Simon KASSI
Problem 2: Accelerated Ageing Tests for HV Cables (185 pts)
1. Computation of A, B of cable A for each endurance tests, using data
provided in figures 1 to 3 of the topic.
1.1 Expressions of A and B available for each endurance tests
For each endurance test, we will use two types of curves, corresponding to
temperatures T1 and T2.
For the curve of T1, we have:
1 1
0
ln
ln ln .T T
P
k t k
P t
and
1
1
(2.1)
B
T
Tk Ae
For the curve of T2, we have:
2 2
0
ln
ln ln .T T
P
k t k
P t
and
2
2
(2.2)
B
T
Tk Ae
Relation (2.1) divided by (2.2) yields:
1
2
1 2
1 2
1 2
1 2
ln
(2.3)
and A=
T
T
B B
T T
T T
k
TT
k
B
T T
k e k e
(2.4)
Using the curves of 125°C and 105°C, we have: T1 = 125 + 273 = 398 K;
T2 = 105 + 273 = 378 K.
6. 5 Koutoua Simon KASSI
1.2 Values of A, B for tensile strength (TS) ‘figure1 of the topic’
1
ln 4
0.04
100
Tk
t
;
2
ln 1
0.001666
600
Tk
t
;
1
2
1 2 23908,966
24398
1 2
0.04ln 398.378.ln
0.001666
= = 23908,966 and A= 0.04 4,91310
398 378
T
T
k
TT
k
B e
T T
1.3 Values of A, B for Folding Endurance (FE) ‘figure2 of the topic’
1
ln 0.1
0.001428
70
Tk
t
;
2
ln 0.1
0.0003448
290
Tk
t
;
1
2
1 2 10689,54
8398
1 2
0.001428ln 398.378.ln
0.0003448
= = 10689,54 and A= 0.001428 6,59.10
398 378
T
T
k
TT
k
B e
T T
1.4 Values of A, B for Degree of Polymerisation (DP) ‘figure3 of the topic ’
1
ln 0.25
0.005
50
Tk
t
;
2
ln 0.25
0.0005319
470
Tk
t
;
1
2
1 2 16855,277
16398
1 2
0.005ln 398.378.ln
0.0005319
= = 16855,277 and A= 0.005 1,234.10
398 378
T
T
k
TT
k
B e
T T
2. Computation of the half-life for the selected properties
The half-life [4] of the cables depends on the coefficients A, B and the
temperatures (T1= 398K, T2= 378 K, T3 = 363 K). For each type of test we will
determine the half-life for each temperature. A Matlab code was developped to
calculate automatically the values of the half-lives. Results are summarized in table 1
and 2, and are plotted in the figures 1and 2. The matlab code is presented in annex 1.
7. 6 Koutoua Simon KASSI
Tableau 1: Half-lives for thickness of 0.13mm
Insulation
Thickness
(mm)
Half-life for TS
(days)
Half-life for FE
(days)
Half-life for DP
(days)
T1 T2 T3 T1 T2 T3 T1 T2 T3
0.13 159.6 727.4 2532 6.46 80.38 638.6 126.8 676.7 2682
Tableau 2: Half-lives for thickness of 0.17mm
Insulation
Thickness
(mm)
Half-life for TS
(days)
Half-life for FE
(days)
Half-life for DP
(days)
T1 T2 T3 T1 T2 T3 T1 T2 T3
0.17 128.3 531.6 1711 4.186 72.58 757.94 119.1 785.2 3703
Figure 1: Half-life for each test (TS, FE, DP) for thickness equal to 0.17 mm
360 365 370 375 380 385 390 395 400
0
500
1000
1500
2000
2500
3000
3500
4000
half-life(days)
Temperature(K)
Tensile Strenght(TS)
Folding Endurance(FE)
Degree of Polymerisation(DP)
8. 7 Koutoua Simon KASSI
Figure 2: Half-life for each test (TS, FE, DP) for thickness equal to 0.13 mm
3. Conclusion
According to results above (tables and curves), we can notice that the half-life
decrease with the increasing of the temperature. With the same paper (cable B) the
values of the half-life of test TS and DP are greater than those of the test FE. As
definition the half-life of an insulation paper is the time after which the insulation paper
loses the half of its properties. Taking account this definition we can conclude that:
- Folding Endurance (FE) test is adaptable for predict the condition of a very aged
paper.
- Tensile strength (TS) test is adaptable for predict the condition of a moderately aged
paper.
- Degree of Polymerisation (DP) test is adaptable for predict the condition of a slightly
aged paper.
360 365 370 375 380 385 390 395 400
0
500
1000
1500
2000
2500
3000
half-life(days)
Temperature(K)
Tensile Strength(TS)
Folding Endurance(FE)
Degree of Polymerisation(DP)
9. 8 Koutoua Simon KASSI
Bibliographie
[1] Issouf Fofana, Notes de Cours Sujets Spéciaux sur les Isolants Electriques 6MDI
855, Ete 2012., 1 vols. Chicoutimi: UQAC(Université du Québec à Chicoutimi).
[2] Kwan Chi Kao, Dielectric Phenomena in solids, vol. 1. Elsevier Academic Press
525 B Street, Suite 1900, San Diego, California 92101-4495, USA 84 Theobald’s
Road, London WC1X 8RR, UK, 2004.
[3] Issouf Fofana, Notes de Cours Ingénierie de la haute tension, 6MIG 930, 6th ed.,
vol. 1. Chicoutimi: UQAC(Université du Québec à Chicoutimi), 2012.
[4] “Notion de demi-vie d’un matériau,” http://fr.wikipedia.org/wiki/Demi-vie, juillet-
2012. .
10. 9 Koutoua Simon KASSI
Annexe
Annexe 1: Matlab Code of question 2 for problem 2
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%DEV2_SSISOLANTS_IFofana
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%PARTIE1: epaisseur 0.17mm
clear all
close all
A_TS=1.81*10^9 ;A_FE=3.12e22 ; A_DP=1.27e13 ;
B_TS=10692 ;B_FE=21460 ; B_DP=14187 ;
T=[125 105 90]+273%température en K
for i=1:3
kTS(i)=A_TS*exp(-B_TS/T(i))
demivie_TS(i)=1/(2*kTS(i))
kFE(i)=A_FE*exp(-B_FE/T(i))
demivie_FE(i)=1/(2*kFE(i))
kDP(i)=A_DP*exp(-B_DP/T(i))
demivie_DP(i)=1/(2*kDP(i))
end
plot(T(1:3),demivie_TS,'+r')
hold on
plot(T(1:3),demivie_FE,'*g')
hold on
plot(T(1:3),demivie_DP,'+k')
ylabel('half-life(days) ')
xlabel('Temperature(K)')
legend('First','Second','Third','Location','NorthEastOutside')
grid
%%
%%%%%%PARTIE2: epaisseur 0.13mm
clear all
close all
A_TS=8.84*10^9; A_FE=3.77e19 ; A_DP=2.19e11 ; % valeurs à modifier
B_TS=11410 ;B_FE=18959 ; B_DP=12596 ;
T=[125 105 90]+273%température en K
for i=1:3
kTS(i)=A_TS*exp(-B_TS/T(i))
demivie_TS(i)=1/(2*kTS(i))
kFE(i)=A_FE*exp(-B_FE/T(i))
demivie_FE(i)=1/(2*kFE(i))
kDP(i)=A_DP*exp(-B_DP/T(i))
demivie_DP(i)=1/(2*kDP(i))
end
plot(T(1:3),demivie_TS,'+r')
hold on
plot(T(1:3),demivie_FE,'*g')
11. 10 Koutoua Simon KASSI
hold on
plot(T(1:3),demivie_DP,'+k')
ylabel('half-life(days)')
xlabel('Temperature(K)')
legend('First','Second','Third','Location','NorthEastOutside')
grid
