DOI : 10.17577/IJERTV5IS120128
- Open Access
- Total Downloads : 106
- Authors : Prof. Nirajkumar C. Mehta, Dr. Dipesh D. Shukla, Prof. Pragnesh D. Kandoliya
- Paper ID : IJERTV5IS120128
- Volume & Issue : Volume 05, Issue 12 (December 2016)
- DOI : http://dx.doi.org/10.17577/IJERTV5IS120128
- Published (First Online): 15-12-2016
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License:
This work is licensed under a Creative Commons Attribution 4.0 International License
Advanced Mathematical Modeling of Heat Transfer in Induction Furnace Wall of Zirconia
1Prof. Nirajkumar C. Mehta,
1Ph. D. Student,
Rai University, Ahmedabad, Gujarat, India
3Prof. Pragnesh D. Kandoliya
1, 3 Asst. Prof., Vadodara Institute of Engineering,
Vadodara, Gujarat, India
2Dr. Dipesh D. Shukla 2Director, Amity University, Jaipur, India
Abstract: Furnaces are useful for melting different materials for casting process. In this research paper, we had done advanced heat transfer analysis of induction furnace wall made of zirconia using explicit finite difference method. We have divided actual geometry of furnace refractory wall into
14 elements and 24 nodes. We have derived explicit finite difference equations for all 24 nodes. We have calculated temperature distribution and thermal stress distribution for all different nodes with respect to time. We have plotted graphs for maximum temperature v/s time and maximum stress v/s time. We found that results indicate the effect of thermal fatigue in the induction furnace wall for zirconia. The analysis is very helpful in understanding how thermal fatigue failure of refractory wall happens.
Key words: Advanced heat transfer analysis, Temperature distribution, Stress distribution, explicit finite difference method, Zirconia
-
INRODUCTION
Furnace is a term used to identify a closed space here heat is applied to a body in order to raise its temperature. The source of heat may be fuel or electricity. Commonly, metals and alloys and sometimes non-metals are heated in furnaces. The purpose of heating defines the temperature of heating and heating rate. Increase in temperature softens the metals. They become amenable to deformation. This softening occurs with or without a change in the metallic structure. Heating to lower temperatures of the metal softens it by relieving the internal stresses. On the other hand, metals heated to temperatures above the critical temperatures leads to changes in crystal structures and recrystallization like annealing. Further some metals and alloys are melted, ceramic products vitrified, coals coked, metals like zinc are vaporized and many other processes are performed in Furnaces. [13]
Induction furnaces are widely used in the iron industry for the casting of the different grades of cast iron products. Refractory wall of induction melting furnace is a key component which is used as insulation layer. [14] It is made of ramming mass like silica, zirconia, magnesia etc. The refractory wall is directly influenced by the thermal cycling of the high temperature molten iron in the furnace. Thermal fatigue failure is easy to happen for it because of the larger phase transformation thermal stresses and it has a
shorter life. This can cause serious production accidents. Therefore, the service life problem of the refractory wall has always been a focus of attention in the application of this to the industry. [15]
The research on the distribution rule of temperature and thermal stress field and on the fatigue life assessment method for the refractory wall will not only lay foundation for the study on the thermal fatigue of this kind of parts under thermal shock condition of low cycle and high phase transition stresses but also offers effective control for thermal fatigue failure.
Computational heat transfer, computational fluid dynamic analysis is done for induction melting furnace, refrigerator condenser, induction heating furnace using different numerical methods like finite volume method and finite element method by different researchers. [1-12]
Here, Explicit Finite Difference Method is used to find out temperature and thermal stress variation with respect to time.
-
DEVELOPMENT OF ADVANCED HEAT TRANSFER MODEL
We have divided Induction Furnace Wall into a Nodal Network as shown in Fig. 1. It is divided into 24 nodes. We have derived Explicit Finite Difference Equations for all nodes as per the boundary conditions applied to it. The furnace wall is having thermal conduction heat transfer between different nodes. It is having atmospheric heat convection ha applied from top side of the furnace wall which is open to atmosphere. It is having heat convection from molten metal from inside which is hi. It is having heat convection ho from cooling water which is circulating outside the furnace wall. [16]
To solve this advanced heat transfer problem of induction melting furnace wall which is made from Zirconia, the following initial and boundary conditions, material properties and basic assumptions are made:
-
Refractory Materials for induction furnace wall meets the basic assumptions in the science of mechanics.
-
Environmental Temperature is homogeneous at 27° C.
-
Ignore the influence of heat radiation.
-
Ignore the effect of gravity field.
-
The surface of induction melting furnace wall is clean.
Node 3:
-
The initial temperature of the induction melting
( )+
3
3
2
( ) + k
3
3
2 2
2
3+ k
2
6 3 =
furnace is set 27° C and it is agreement with the
+1
ambient temperature during solving the problem.
3 3
2 2
-
Heat convections are considered constant for this
+1= (( ( )+ ( ) + k 2 3+
analysis.
3
2 3 2
3
3
4
3 2
-
Scarp material input inside furnace is considered
k 6 3 )
2
) +
uniform for our analysis.
T[3][i+1] = (((ha*x*(To-T[3][i])*0.5) +( hi*y*(Th-
T[3][i])*0.5 ) +(0.5*k*y*(T[2][i]-
T[3][i])/x)+(0.5*k*x*(T[6][i]-T[3][i])/y))
*((4*t)/(r*c*x*y))) +T[3][i];
Node 4:
( ) + k 5 4+ k
1 4 + k 7 4=
2
4
4 4
4 4
+4
2
2
+4 = ((( ) + k 5 4+ k 1 4 +
4
4
2
2
4
4
k 7 4 )
2
) +
T[4][i+1] = ((( ho*y*(Ta-T[4][i])) +(k*y*(T[5][i]-
T[4][i])/x)+(0.5*k*x*(T[1][i]-T[4][i])/y) +
(0.5*k*x*(T[7][i]-T[4][i])/y)) *((2*t)/(r*c*x*y)))
+T[4][i];
Node 5:
k 4 5 + k 6 5 + k 2 5 + k 8 5 =
+1
Fig. 1. Nodal network for finite difference method
5 5
+1= ((k 4 5 + k 6 5 + k 2 5 +
Node 1:
5
k 8 5 )
) +
(
)+ (
) + k 2 1+ k 4 1
5
1 1
1 1
2 1
2 1
2
2
T[5][i+1] = (((k*y*(T[4][i]-T[5][i])/x)+(k*y*(T[6][i]-
=
+1
T[5][i])/x) + (k*x*(T[8][i]-T[5][i])/y))
2 2
*((t)/(r*c*x*y))) +T[5][i];
+1 = (( ( )+ ( ) + k 2 1+
1
4 1
4 1
k )
2
4
1 2
)+
1 2
Node 6:
2
1
(
) + k 2 6+ k 3 6 + k 9 6 =
T[1][i+1] = (((0.5*ha*x*(To-T[1][i])) +(ho*y*(Ta-
6
2
2
T[1][i])/2)+(0.5*k*y*(T[2][i]-
+1
T[1][i])/x)+(0.5*k*x*(T[4][i]-T[1][i])/y)
6 6
2
)*((4*t)/(r*c*x*y))) +T[1][i];
+1= (( ( ) + k 2 6+ k 3 6 +
6
6
2
2
Node 2:
6
6
k 9 6 )
2
) +
( )+ k 1 2 + k 3 2 + k 5 2 =
2
2 2
2 2
+1
2
2
T[6][i+1] = (((hi*y*(Th-T[6][i])) + (k*y*(T[2][i]-
T[6][i])/x)+(0.5*k*x*(T[3][i]-T[6][i])/y) +
2
(0.5*k*x*(T[9][i]-T[6][i])/y)) *((2*t)/(r*c*x*y)))
+1 = ((( )+ k 1 2 + k 3 2 +
+T[6][i];
2
5 2
5 2
k )
2
2 2
2
2
) +
2
T[2][i+1] = (((ha*x*(To-T[2][i]))+(0.5*k*y*(T[1][i]-
( ) + k 8 7+ k
4 7 + k 10 7=
T[2][i])/x)+(0.5*k*y*(T[3][i]-T[2][i])/x)
+(0.5*k*x*(T[5][i]-T[2][i])/y))*((2*t)/(r*c*x*y)))
+T[2][i];
2
7
7 7
7 7
+7
2
2
+ (k*x*(T[16][i]-T[11][i])/y)) *((t)/(r*c*x*y)))
+7 = (( ( ) + k 8
7+ k 4 7 +
7
1 7
1 7
k 0 )
2
7
) +
2
+T[11][i];
2
7
Node 12:
T[7][i+1] = (((ho*y*(Ta-T[7][i])) + (k*y*(T[8][i]-
T[7][i])/x)+(0.5*k*x*(T[4][i]-T[7][i])/y) +
( ) +
( ) + k 1 +
( ) + k 1 +
1 12
(0.5*k*x*(T[10][i]-T[7][i])/y)) *((2*t)/(r*c*x*y)))
2
12 2
12
3
+1
+T[7][i];
k 13
2
12 + k
2
9
12 + k 17 12 =
12 12
4
+1 = (( ( ) + ( ) +
Node 8:
12
2 12
2 12
k 11 12+ k
13 12 + k 9 12 +
k 7 8 + k 9 8 + k 5 8 + k 11 8=
2
2
1 12
1 12
+1
k 7 )
4
) +
8 8
3 12
T[12][i+1] = (( (0.5*hi*x*(Th-T[12][i]))+(
+1 = (( k 7 8 + k 9 8 + k 5 8 +
0.5*hi*y*(Th-T[12][i])) +(k*y*(T[11][i]-T[12][i])/x)+
8
(0.5*k*y*(T[13][i]-T[12][i])/x) + (0.5*k*x*(T[9][i]-
k 11 8 )
) +
T[12][i])/y) + (k*x*(T[17][i]-T[12][i])/y))
8
8
T[8][i+1] = (((k*y*(T[7][i]-T[8][i])/x)+( k*y*(T[9][i]-
T[8][i])/x ) +(k*x*(T[5][i]-T[8][i])/y) + (k*x*(T[11][i]-
T[8][i])/y)) *((t)/(r*c*x*y))) +T[8][i];
*((4*t)/(3*r*c*x*y))) +T[12][i];
Node 13:
( )+ k
12 13 + k 14 13 + k 18 13
Node 9:
=
2
13 2
13 13
13 13
+1
2
( 9) + k 8
9+ k 6 9 + k 12 9 =
+1 = (( (
)+ k 12 13 + k 14 13 +
+1
9
9
9
9
2
2
13
13
2
2
2
13
13
2
k 18 13)
) +
+1 = ((( ) + k 8 9+ k 6 9 +
T[13][i+1] = (( (hi*x*(Th-
9
1 9
1 9
k 2 )
2
9
) +
2
T[13][i]))+(0.5*k*y*(T[12][i]-T[13][i])/x)+
(0.5*k*y*(T[14][i]-T[13][i])/x)+ (k*x*(T[18][i]-
2 9
T[9][i+1] = (( (hi*y*(Th-T[9][i]))+(k*y*(T[8][i]-
T[9][i])/x)+ (0.5*k*x*(T[6][i]-T[9][i])/y) +
(0.5*k*x*(T[12][i]-T[9][i])/y)) *((2*t)/(r*c*x*y)))
T[13][i])/y)) *((2*t)/(r*c*x*y))) +T[13][i];
Node 14:
14
14
14
14
+T[9][i];
( )+
2
( ) + k
2 2
13
14+
+1
Node 10:
k 19
2
14 =
2
14 14
2
( 10) + k 11 10+ k 7 10 + k 15 10=
+1 = ((
( )+
( ) +
2
2
14 2
14 2 14
+10
4
10 10
2
k 13
2
14+ k
2
19
14)
) +
T[14][i+1] = (( (0.5*hi*x*(Th-
14
14
+10 = ((( ) + k 11 10+ k 7 10 +
10
1 10
1 10
k 5 )
2
10
) +
2
T[14][i]))+(0.5*hi*y*(Th-T[14][i]))
+(0.5*k*y*(T[13][i]-T[14][i])/x)+ (0.5*k*x*(T[19][i]-
2
10
T[14][i])/y)) *((4*t)/(r*c*x*y))) +T[14][i];
T[10][i+1] = (( (ho*y*(Ta-T[10][i]))+(k*y*(T[11][i]-
T[10][i])/x)+ (0.5*k*x*(T[7][i]-T[10][i])/y) +
(0.5*k*x*(T[15][i]-T[10][i])/y)) *((2*t)/(r*c*x*y)))
Node 15:
+T[10][i];
( ) + k 16 15+ k
10 15 +
15
+15
2
Node 11:
k 20
2
15=
2
15 15
k 10 11 + k 12 11 + k 8 11 + k 16 11=
+15 = (( ( ) + k 16 15+ k
10 15 +
+1
15
15
15
15
2
2
11 11
k 20
2
15)
) +
+1 = (( k 10 11 + k 12 11 + k 8 11 + T[15][i+1] = (( (ho*y*(Ta-T[15][i]))+(k*y*(T[16][i]-
11
T[15][i])/x)+ (0.5*k*x*(T[10][i]-T[15][i])/y)+(
k 16 11 )
) +
0.5*k*x*(T[20][i]-T[15][i])/y)) *((2*t)/(r*c*x*y)))
11
11
+T[15][i];
T[11][i+1] = (( (k*y*(T[10][i]-T[11][i])/x)+
(k*y*(T[12][i]-T[11][i])/x)+ (k*x*(T[8][i]-T[11][i])/y)
Node 16:
T[20][i+1] = (( (0.5*ho*x*(Ta-T[20][i]))+(
0.5*ho*y*(Ta-T[20][i]))+(0.5*k*y*(T[21][i]-
k 15 16 + k 17 16 + k 11 16 + k 21 16=
T[20][i])/x)+ (0.5*k*x*(T[15][i]-
16 16
16 16
+1
+1 = (( k
+ k
+ k
+
T[20][i])/y))*((4*t)/(r*c*x*y))) +T[20][i];
Node 21:
15 16
16
17 16
11 16
( )+ k
20 21 + k 22 21 + k 16 21
k 21 16)
) +
21 2
2
16
+1
T16][i+1] = (( (k*y*(T[15][i]-T[16][i])/x)+
=
21 21
2
(k*y*(T[17][i]-T[16][i])/x )+(k*x*(T[11][i]-
+1 = (((
)+ k
+ k +
T[16][i])/y)+( k*x*(T[21][i]-T[16][i])/y))
21
21
2
20 21
21
21
2
22 21
2
*((t)/(r*c*x*y))) +T[16][i];
k 16 21)
) +
Node 17:
T[21][i+1] = (( (0.5*ho*x*(Ta- T[21][i]))+(0.5*k*y*(T[20][i]-T[21][i])/x)+(
k 16 17 + k 18 17 + k 12 17 + k 22 17=
0.5*k*y*(T[22][i]-T[21][i])/x)+ (k*x*(T[16][i]-
17 17
17 17
+1
T[21][i])/y))*((2*t)/(r*c*x*y))) +T[21][i];
+1 = (( k 16 17 + k 18 17 + k 12 17 +
Node 22:
17
(
)+ k
21 22 + k 23 22 + k 17 22
k 22 17)
) +
22 2
+1
2
17
17
T[17][i+1] = (( (k*y*(T[16][i]-T[17][i])/x)+
(k*y*(T[18][i]-T[17][i])/x )+(k*x*(T[12][i]-
=
22 22
2
T[17][i])/y)+( k*x*(T[22][i]-T[17][i])/y))
22+1 = ((( 22)+ k
21 22 + k 23 22 +
*((t)/(r*c*x*y))) +T[17][i];
k 7 )
1 22
1 22
2
2
22
22
) +
2
Node 18:
T[22][i+1] = (( (ho*x*(Ta- T[22][i]))+(0.5*k*y*(T[21][i]-T[22][i])/x)+(
k 17 18 + k 19 18 + k 13 18 + k 23 18=
0.5*k*y*(T[23][i]-T[22][i])/x)+ (k*x*(T[17][i]-
18 18
18 18
+1
T[22][i])/y))*((2*t)/(r*c*x*y)))
+T[22][i];
+1 = ((k 17 18 + k 19 18 + k 13 18 +
18
23 18
23 18
k
)
) +
Node 23:
18
( )+ k
22 23 + k 24 23 + k 18 23
T[18][i+1] = (( (k*y*(T[17][i]-T[18][i])/x)+
(k*y*(T[19][i]-T[18][i])/x )+(k*x*(T[13][i]-
T[18][i])/y)+( k*x*(T[23][i]-T[18][i])/y))
=
2
23 2
23 23
23 23
+1
2
+1 = ((( )+ k
22 23 + k 24 23 +
*((t)/(r*c*x*y))) +T[18][i];
23
2
23 2
2
Node 19:
k 18 23)
) +
23
23
T[23][i+1] = (( (ho*x*(Ta-
( ) + k 18 19+ k 14 19 + k 24 19
=
2
19
19 19
19 19
+1
2
2
T[23][i]))+(0.5*k*y*(T[22][i]-T[23][i])/x)+(
0.5*k*y*(T[24][i]-T[23][i])/x)+ (k*x*(T[18][i]-
T[23][i])/y))*((2*t)/(r*c*x*y))) +T[23][i];
+1 = ((( ) + k 18 19+ k 14 19 +
19
2 19
2 19
k 4 )
2
19
) +
2
Node 24:
2 19
( )+
( ) + k
23 24+
T[19][i+1] = (( (hi*y*(Th-T[19][i]))+(k*y*(T[18][i]-
2
24
2 24
+1
2
T[19][i])/x)+ (0.5*k*x*(T[14][i]-T[19][i])/y)+(
k 19
2
24 =
2
24 24
2
0.5*k*x*(T[24][i]-T[19][i])/y)) *((2*t)/(r*c*x*y)))
+1 = (( ( )+ ( ) +
+T[19][i];
24
2 24
2 24
24
24
4
Node 20:
k 23
2
24+ k
2
19
24)
) +
( )+
( ) + k 1 +
( ) + k 1 +
2 20
T[24][i+1] = (( (0.5*ho*x*(Ta-T[24][i]))+(
0.5*hi*y*(Th-T[24][i])) +(0.5*k*y*(T[23][i]-
2
20
2 20
+20
2
T[24][i])/x)+ (0.5*k*x*(T[19][i]-
2
2
2
2
2
2
k 15 20 =
20 20
T[24][i])/y))*((4*t)/(r*c*x*y)))
+20 = (( ( )+ ( ) +
+T[24][i];
20
2 20
2 20
4
k 21 20+ k 15 20)
) +
2
2
20
-
-
PROGRAMMING & SOLUTION:
With the help of a computer program we can solve the matrix created by finite difference equations for 24 nodes. We can calculate temperature distribution and stress distribution with respect to time.
Table 1. Material Property and Boundary Conditions
Material Properties and Boundary Conditions for Zirconia
Unit
1
Internal Film Co-efficient hi
200
W/m2 K
2
External Film Co-efficient ho
40
W/m2 K
3
Atmosphere Film Co-efficient ha
10
W/m2 K
4
Density
5000
Kg/m3
5
Time Interval t
10
Seconds
6
Thermal Conductivity k
1.2
W/m K
7
Temperature outside Furnace Wall
303
Kelvin
8
Temperature inside Furnace Wall
1873
Kelvin
9
Temperature of Air
303
Kelvin
10
Specific Heat
780
J/kg K
11
Elasticity Constant
240000
N/ m2
12
Thermal Expansion Co- efficient
0.000000
86
m/ K
13
Ultimate Stress
600
MPa
-
RESULTS AND DISCUSSION:
We can see from the Fig. 2 that maximum temperature is increasing from atmospheric temperature 300 K and reaches to maximum temperature 1827 K in 45 minutes and then starts reducing and reaches to 911 K in next 15 minutes. It again starts increasing and reaches to maximum 1827 K after 105 minutes and again starts reducing. There are 10 similar temperature cycles in one day.
2000
1800
1600
1400
1200
1000
800
600
400
200
0
2000
1800
1600
1400
1200
1000
800
600
400
200
0
Temperature (Kelvin)
Temperature (Kelvin)
We can see from the Fig. 3 that maximum thermal stress is increasing from initial condition 0 MPa and reaches to maximum stress 377 MPa in 45 minutes and then starts reducing and reaches to 188 MPa in next 15 minutes. It again starts increasing and reaches to maximum stress 377 MPa after 105 minutes and again it starts reducing. There are 10 similar thermal stress cycles in one day.
Maximum Temperature v/s Time Zirconia
Maximum Temperature v/s Time Zirconia/p>
Time (Minutes)
Time (Minutes)
0
39
78
117
156
195
234
273
312
351
390
429
468
507
546
585
0
39
78
117
156
195
234
273
312
351
390
429
468
507
546
585
Fig. 2. Maximum Temperature v/s Time Graph for Zirconia
Maximum Stress v/s Time Zirconia
400 350
300
250
200
150
100
50
0
Time (Minutes)
Maximum Stress v/s Time Zirconia
400 350
300
250
200
150
100
50
0
Time (Minutes)
Stress (MPa)
Stress (MPa)
0
42
84
126
168
210
252
294
336
378
420
462
504
546
588
0
42
84
126
168
210
252
294
336
378
420
462
504
546
588
Fig. 3. Maximum Stress v/s Time Graph for Zirconia
-
CONCLUSION
We have found deviation of maximum temperature and maximum thermal stress with reference to time for induction furnace wall made up of alumina ramming mass. From the graph, we can conclude that induction furnace wall of zirconia is under the effect of low cycle thermal fatigue loading. The reason for its low life span is low cycle thermal fatigue load.
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-
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-
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