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Thermal Resistance Approach to
Analyze Temperature Distribution in
Hollow Cylinders Made of
Functionally Graded Material (FGM):
Under Dirichlet Boundary Condition
S M Shayak Ibna Faruqui
Abul Al Arabi &
M. S. Parvej
Bangladesh University of Engineering and Technology
Introduction to FGMs
• FGMs are superior composites with better performance.
• Continuous spatial distribution of 2 or more material in 1 ,2 or 3 directions.
• Scope of tailoring material property according to desire.
0
20
40
60
80
100
0 0.5 1
VolumeFraction
Thickness
Volume Fraction A
Volume Fraction B
Composite Vs. FGM
Composite layers Continuous material
gradient in FGMs
Historical Background
• Nature first introduced us with FGMS•
dense
porous
Organic bone Plants fiberHuman tooth
Historical Background
• First introduced in Japan during 1984-85.
• Challenge was finding a particular solution material which
could withstand a massive temperature gradient of 1000℃
with a 10mm thickness.
Problem with Traditional materials
and Composites
• In high temperature gradient condition they both fail.
Metals Buckle Composite Delaminates
ElectronicsAerospace
Chemical
Plants
Nuclear
Energy
Energy
Conversion
Optics
Biomaterials
Application of FGMs
Example of some Extreme Thermal Environment
Nuclear Reactor
Exhaust nozzle of rocketReentry Module
Types of FGMs
• Depends on material gradient which follows a
particular function.
1. pFGM: material distribution follows a power function
2. sFGM: material distribution follows sigmoid function
3. eFGM: material distribution follows an exponential
function
This Work: Motivation and Goals
• Introducing a new approach (TRA) to solve
temperature distribution in a Hollow FGM Cylinder.
• Bypassing problems caused by “Non-linearity” of
material gradient.
• Obtaining results analytically and applying FEM and
compare with TRA
𝑘 𝑜
𝑘𝑖
𝐾𝑓𝑔𝑚
𝑇𝑜
𝑇𝑖
This Work: Formulation of problemBoundaryCondition
A hollow cylinder is considered with an inner radius 𝑟𝑖 and outer radius 𝑟𝑜
where 𝑘𝑖 and 𝑘 𝑜 are the value of thermal conductivity at inner and outer
surfaces respectively. Thermal conductivity is assumed to vary from 𝑘𝑖 to 𝑘 𝑜
along radial direction following a Power Function where “n” is the
power law index.
𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖
𝑟 − 𝑟𝑖
𝑟𝑜 − 𝑟𝑖
𝑛
Power Function:
250
270
290
310
330
350
370
390
410
5 10 15 20 25
ThermalConductivity,K(W/mK)
Radial Distance, r(m)
n=0.1
n=0.5
n=1
n=2
n=4
n=10
n=0.25
This Work: Problem Statement
This Work: Governing Equation
Heat Conduction Equation in Cylindrical Co-ordinate:
This Work: Governing Equation
Heat Conduction Equation in Cylindrical Co-ordinate:
Due to symmetry & Steady State:
∂ 𝑇
∂φ
= 0 ,
∂ 𝑇
∂ 𝑧
=0 and
∂ 𝑇
∂ 𝑡
= 0
No Internal Heat
Generation
This Work: Governing Equation
The governing equation yields to:
𝑑
𝑑𝑟
𝐾𝑓𝑔𝑚 𝑟 . 𝑟
𝑑𝑇
𝑑𝑟
= 0
𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖
𝑟 − 𝑟𝑖
𝑟𝑜 − 𝑟𝑖
𝑛
• Thermal conductivity, K is not constant
• Follows power function of independent variable r
This Work: Governing Equation
The governing equation yields to:
𝑑
𝑑𝑟
𝐾𝑓𝑔𝑚 𝑟 . 𝑟
𝑑𝑇
𝑑𝑟
= 0
𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖
𝑟 − 𝑟𝑖
𝑟𝑜 − 𝑟𝑖
𝑛
• Thermal conductivity, K is not constant
• Follows power function of independent variable r
Power Law Index
𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖
𝑟 − 𝑟𝑖
𝑟𝑜 − 𝑟𝑖
𝒏
Function:
This Work: Problem Statement
250
270
290
310
330
350
370
390
410
5 15 25
ThermalConductivity,K
(W/mK)
Radial Distance, r(m)
n=0.1
n=0.5
n=1
n=2
n=4
n=10
n=0.25
This Work: Literature Review
1. S. Karampour et al. [2] solved the heat conduction equation for obtaining
temperature distribution analytically using Legendre polynomials and
Euler differential equations system.
2. Celebi et al. [3] used Complimentary Function Method (CFM) to convert
the boundary value problem into an initial value problem and solved the
temperature problem.
3. M. Jabbari et al. [4], [5], [6], [7] & [8] obtained results for temperature
distribution directly by exact method and as he assumed the radial
distribution of material property to follow an exponential function, getting
exact solutions posed no complexity.
4. X.L. Peng et al. [9] also converted the boundary value problem associated
with the thermoelastic problem into a Fredholm integral equation and by
numerically solving the resulting equation, the distribution of the
temperature was obtained.
5. Shao Z.S [10] presented a study about mechanical and thermal stresses in
FGM hollow cylinder of finite length and Zimmerman et al [11] in
uniformly heated FGM cylinder.
This Work: Analytical Solution
For n=1 Integrating twice the solution yields,
It’s a 1st Order linear homogeneous ODE
𝑑
𝑑𝑟
𝐾𝑓𝑔𝑚 𝑟 . 𝑟
𝑑𝑇
𝑑𝑟
= 0
𝑇 𝑟 =
(𝑟𝑜 − 𝑟𝑖)𝐶1
𝑟𝑜 𝐾𝑜 − 𝑟𝑖 𝐾𝑖
𝑙𝑛
𝑟 𝐾𝑖 − 𝐾𝑜
𝐾𝑖 𝑟𝑜 − 𝑟 + 𝐾𝑜 𝑟 − 𝑟𝑖
+ 𝐶2
• 𝐶1 & 𝐶2 is integration constants.
This Work: Analytical Solution
For n=1 Integrating twice the solution yields,
It’s a 1st Order linear homogeneous ODE
𝑑
𝑑𝑟
𝐾𝑓𝑔𝑚 𝑟 . 𝑟
𝑑𝑇
𝑑𝑟
= 0
𝑇 𝑟 =
(𝑟𝑜 − 𝑟𝑖)𝐶1
𝑟𝑜 𝐾𝑜 − 𝑟𝑖 𝐾𝑖
𝑙𝑛
𝑟 𝐾𝑖 − 𝐾𝑜
𝐾𝑖 𝑟𝑜 − 𝑟 + 𝐾𝑜 𝑟 − 𝑟𝑖
+ 𝐶2
• It can be marked that the solution is valid if
𝐾 𝑜
𝐾 𝑖
≠
𝑟 𝑖
𝑟 𝑜
.
• 𝐶1 & 𝐶2 is integration constants.
Introducing: Thermal Resistance Approach
𝒌 𝒐
𝒌𝒊
𝑲 𝒇𝒈𝒎(r) 𝒌 𝟏
𝒌 𝟐
𝒌𝒋
𝒌 𝑵
Thermal Resistance Approach: Theoretical Background
L
𝑸 𝑸
𝑇1
𝑇2
𝑄=𝐾𝐴
𝑇1−𝑇2
𝐿
A
Thermal Resistance Approach: Theoretical Background
𝑸 𝑸
𝑇1
𝑇2
A
𝑄=
𝑇1−𝑇2
𝑅
R
𝑄=𝐾𝐴
𝑇1−𝑇2
𝐿
𝑅 = 𝐿
𝐾𝐴Thermal Resistance,
Thermal Resistance Approach: Methodology
𝑅𝑗=
𝑙𝑛
𝑟𝑗+1
𝑟𝑗
2𝜋𝐿𝐾𝑗
Thermal Resistance for any 𝑗 𝑡ℎ layer,
Total Thermal Resistance for the network, 𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑗 + ⋯ + 𝑅 𝑁
Temperature of any 𝑗 𝑡ℎ layer, 𝑇𝑗 =
1
𝑗
𝑅 𝑇𝑜 + 𝑗+1
𝑁
𝑅 𝑇𝑖
𝑅𝑡𝑜𝑡𝑎𝑙
Thermal Resistance Approach: Validation
• Obtained results were compared with [4] & [3]
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 1.05 1.1 1.15 1.2
DimensionlessTemperature
Dimensionless Radial distance
present Work
Jabbari
FCMCelebi (CFM)
Thermal Resistance Approach: Validation
Avg. Error < 0.000012%
Thermal Resistance Approach and FEM
0
200
400
600
800
1000
5 15 25
Temperature
distribution,T(ºC)
Radial distance, r(m)
FEM(ANSYS)
TRA(n=1)
Thermal Resistance Approach: Results and Discussion
250
270
290
310
330
350
370
390
400
450
500
550
600
650
700
750
800
850
900
0 10 20 30
Temperaturedistribution,T(K)
Radial distance,r (m)
n=0.5
n=0.25
n=0.1
250
270
290
310
330
350
370
390
400
450
500
550
600
650
700
750
800
850
900
5 15 25
Temperaturedistribution,T(k)
Radial distance, r (m)
n=1
n=4
n=10
Thermal Resistance Approach: Results and Discussion
250
270
290
310
330
350
370
390
400
450
500
550
600
650
700
750
800
850
900
0 10 20 30
Temperaturedistribution,T(K)
Radial distance,r (m)
n=0.5
n=0.25
n=0.1
250
270
290
310
330
350
370
390
400
450
500
550
600
650
700
750
800
850
900
5 15 25
Temperaturedistribution,T(k)
Radial distance, r (m)
n=1
n=4
n=10
-40
10
60
110
160
210
260
310
360
410
0
0.00001
0.00002
0.00003
0.00004
0.00005
0.00006
0.00007
3 13 23
Radial Distance, r (m)
n=1
n=4
n=10
n=0.1
Distributionof𝐾_𝑓𝑔𝑚(W/mk)
Thermal Resistance Approach: Results and Discussion
Thermal Resistance Approach: Advantages
S-FGMP-FGM E-FGM
• Can be used for any type of material gradient.
• Gives you temperature profile for any kind of material
distribution when analytical solution is very complex.
• Helps to visualize the effect of material gradient
Thermal Resistance Approach: Future Scopes
• Determining material gradient for a desired stress profile
• Tailoring material gradient for desired temperature field
• Tailoring material gradient for desired stress field
Thermal Resistance Approach: Future Scopes
• Determining material gradient for a desired stress profile
• Tailoring material gradient for desired temperature field
• Tailoring material gradient for desired stress field
Thank You

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Thermal Resistance Approach to Analyze Temperature Distribution in Hollow Cylinders Made of Functionally Graded Material (FGM): Under Dirichlet Boundary Condition

  • 1. Thermal Resistance Approach to Analyze Temperature Distribution in Hollow Cylinders Made of Functionally Graded Material (FGM): Under Dirichlet Boundary Condition S M Shayak Ibna Faruqui Abul Al Arabi & M. S. Parvej Bangladesh University of Engineering and Technology
  • 2. Introduction to FGMs • FGMs are superior composites with better performance. • Continuous spatial distribution of 2 or more material in 1 ,2 or 3 directions. • Scope of tailoring material property according to desire. 0 20 40 60 80 100 0 0.5 1 VolumeFraction Thickness Volume Fraction A Volume Fraction B
  • 3. Composite Vs. FGM Composite layers Continuous material gradient in FGMs
  • 4. Historical Background • Nature first introduced us with FGMS• dense porous Organic bone Plants fiberHuman tooth
  • 5. Historical Background • First introduced in Japan during 1984-85. • Challenge was finding a particular solution material which could withstand a massive temperature gradient of 1000℃ with a 10mm thickness.
  • 6. Problem with Traditional materials and Composites • In high temperature gradient condition they both fail. Metals Buckle Composite Delaminates
  • 8. Example of some Extreme Thermal Environment Nuclear Reactor Exhaust nozzle of rocketReentry Module
  • 9. Types of FGMs • Depends on material gradient which follows a particular function. 1. pFGM: material distribution follows a power function 2. sFGM: material distribution follows sigmoid function 3. eFGM: material distribution follows an exponential function
  • 10. This Work: Motivation and Goals • Introducing a new approach (TRA) to solve temperature distribution in a Hollow FGM Cylinder. • Bypassing problems caused by “Non-linearity” of material gradient. • Obtaining results analytically and applying FEM and compare with TRA
  • 11. 𝑘 𝑜 𝑘𝑖 𝐾𝑓𝑔𝑚 𝑇𝑜 𝑇𝑖 This Work: Formulation of problemBoundaryCondition A hollow cylinder is considered with an inner radius 𝑟𝑖 and outer radius 𝑟𝑜 where 𝑘𝑖 and 𝑘 𝑜 are the value of thermal conductivity at inner and outer surfaces respectively. Thermal conductivity is assumed to vary from 𝑘𝑖 to 𝑘 𝑜 along radial direction following a Power Function where “n” is the power law index.
  • 12. 𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖 𝑟 − 𝑟𝑖 𝑟𝑜 − 𝑟𝑖 𝑛 Power Function: 250 270 290 310 330 350 370 390 410 5 10 15 20 25 ThermalConductivity,K(W/mK) Radial Distance, r(m) n=0.1 n=0.5 n=1 n=2 n=4 n=10 n=0.25 This Work: Problem Statement
  • 13. This Work: Governing Equation Heat Conduction Equation in Cylindrical Co-ordinate:
  • 14. This Work: Governing Equation Heat Conduction Equation in Cylindrical Co-ordinate: Due to symmetry & Steady State: ∂ 𝑇 ∂φ = 0 , ∂ 𝑇 ∂ 𝑧 =0 and ∂ 𝑇 ∂ 𝑡 = 0 No Internal Heat Generation
  • 15. This Work: Governing Equation The governing equation yields to: 𝑑 𝑑𝑟 𝐾𝑓𝑔𝑚 𝑟 . 𝑟 𝑑𝑇 𝑑𝑟 = 0 𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖 𝑟 − 𝑟𝑖 𝑟𝑜 − 𝑟𝑖 𝑛 • Thermal conductivity, K is not constant • Follows power function of independent variable r
  • 16. This Work: Governing Equation The governing equation yields to: 𝑑 𝑑𝑟 𝐾𝑓𝑔𝑚 𝑟 . 𝑟 𝑑𝑇 𝑑𝑟 = 0 𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖 𝑟 − 𝑟𝑖 𝑟𝑜 − 𝑟𝑖 𝑛 • Thermal conductivity, K is not constant • Follows power function of independent variable r Power Law Index
  • 17. 𝐾𝑓𝑔𝑚 𝑟 = 𝑘𝑖 + 𝑘 𝑜 − 𝑘𝑖 𝑟 − 𝑟𝑖 𝑟𝑜 − 𝑟𝑖 𝒏 Function: This Work: Problem Statement 250 270 290 310 330 350 370 390 410 5 15 25 ThermalConductivity,K (W/mK) Radial Distance, r(m) n=0.1 n=0.5 n=1 n=2 n=4 n=10 n=0.25
  • 18. This Work: Literature Review 1. S. Karampour et al. [2] solved the heat conduction equation for obtaining temperature distribution analytically using Legendre polynomials and Euler differential equations system. 2. Celebi et al. [3] used Complimentary Function Method (CFM) to convert the boundary value problem into an initial value problem and solved the temperature problem. 3. M. Jabbari et al. [4], [5], [6], [7] & [8] obtained results for temperature distribution directly by exact method and as he assumed the radial distribution of material property to follow an exponential function, getting exact solutions posed no complexity. 4. X.L. Peng et al. [9] also converted the boundary value problem associated with the thermoelastic problem into a Fredholm integral equation and by numerically solving the resulting equation, the distribution of the temperature was obtained. 5. Shao Z.S [10] presented a study about mechanical and thermal stresses in FGM hollow cylinder of finite length and Zimmerman et al [11] in uniformly heated FGM cylinder.
  • 19. This Work: Analytical Solution For n=1 Integrating twice the solution yields, It’s a 1st Order linear homogeneous ODE 𝑑 𝑑𝑟 𝐾𝑓𝑔𝑚 𝑟 . 𝑟 𝑑𝑇 𝑑𝑟 = 0 𝑇 𝑟 = (𝑟𝑜 − 𝑟𝑖)𝐶1 𝑟𝑜 𝐾𝑜 − 𝑟𝑖 𝐾𝑖 𝑙𝑛 𝑟 𝐾𝑖 − 𝐾𝑜 𝐾𝑖 𝑟𝑜 − 𝑟 + 𝐾𝑜 𝑟 − 𝑟𝑖 + 𝐶2 • 𝐶1 & 𝐶2 is integration constants.
  • 20. This Work: Analytical Solution For n=1 Integrating twice the solution yields, It’s a 1st Order linear homogeneous ODE 𝑑 𝑑𝑟 𝐾𝑓𝑔𝑚 𝑟 . 𝑟 𝑑𝑇 𝑑𝑟 = 0 𝑇 𝑟 = (𝑟𝑜 − 𝑟𝑖)𝐶1 𝑟𝑜 𝐾𝑜 − 𝑟𝑖 𝐾𝑖 𝑙𝑛 𝑟 𝐾𝑖 − 𝐾𝑜 𝐾𝑖 𝑟𝑜 − 𝑟 + 𝐾𝑜 𝑟 − 𝑟𝑖 + 𝐶2 • It can be marked that the solution is valid if 𝐾 𝑜 𝐾 𝑖 ≠ 𝑟 𝑖 𝑟 𝑜 . • 𝐶1 & 𝐶2 is integration constants.
  • 21. Introducing: Thermal Resistance Approach 𝒌 𝒐 𝒌𝒊 𝑲 𝒇𝒈𝒎(r) 𝒌 𝟏 𝒌 𝟐 𝒌𝒋 𝒌 𝑵
  • 22. Thermal Resistance Approach: Theoretical Background L 𝑸 𝑸 𝑇1 𝑇2 𝑄=𝐾𝐴 𝑇1−𝑇2 𝐿 A
  • 23. Thermal Resistance Approach: Theoretical Background 𝑸 𝑸 𝑇1 𝑇2 A 𝑄= 𝑇1−𝑇2 𝑅 R 𝑄=𝐾𝐴 𝑇1−𝑇2 𝐿 𝑅 = 𝐿 𝐾𝐴Thermal Resistance,
  • 24. Thermal Resistance Approach: Methodology 𝑅𝑗= 𝑙𝑛 𝑟𝑗+1 𝑟𝑗 2𝜋𝐿𝐾𝑗 Thermal Resistance for any 𝑗 𝑡ℎ layer, Total Thermal Resistance for the network, 𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅1 + 𝑅2 + ⋯ + 𝑅𝑗 + ⋯ + 𝑅 𝑁 Temperature of any 𝑗 𝑡ℎ layer, 𝑇𝑗 = 1 𝑗 𝑅 𝑇𝑜 + 𝑗+1 𝑁 𝑅 𝑇𝑖 𝑅𝑡𝑜𝑡𝑎𝑙
  • 25. Thermal Resistance Approach: Validation • Obtained results were compared with [4] & [3] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 1.05 1.1 1.15 1.2 DimensionlessTemperature Dimensionless Radial distance present Work Jabbari FCMCelebi (CFM)
  • 26. Thermal Resistance Approach: Validation Avg. Error < 0.000012%
  • 27. Thermal Resistance Approach and FEM 0 200 400 600 800 1000 5 15 25 Temperature distribution,T(ºC) Radial distance, r(m) FEM(ANSYS) TRA(n=1)
  • 28. Thermal Resistance Approach: Results and Discussion 250 270 290 310 330 350 370 390 400 450 500 550 600 650 700 750 800 850 900 0 10 20 30 Temperaturedistribution,T(K) Radial distance,r (m) n=0.5 n=0.25 n=0.1 250 270 290 310 330 350 370 390 400 450 500 550 600 650 700 750 800 850 900 5 15 25 Temperaturedistribution,T(k) Radial distance, r (m) n=1 n=4 n=10
  • 29. Thermal Resistance Approach: Results and Discussion 250 270 290 310 330 350 370 390 400 450 500 550 600 650 700 750 800 850 900 0 10 20 30 Temperaturedistribution,T(K) Radial distance,r (m) n=0.5 n=0.25 n=0.1 250 270 290 310 330 350 370 390 400 450 500 550 600 650 700 750 800 850 900 5 15 25 Temperaturedistribution,T(k) Radial distance, r (m) n=1 n=4 n=10
  • 30. -40 10 60 110 160 210 260 310 360 410 0 0.00001 0.00002 0.00003 0.00004 0.00005 0.00006 0.00007 3 13 23 Radial Distance, r (m) n=1 n=4 n=10 n=0.1 Distributionof𝐾_𝑓𝑔𝑚(W/mk) Thermal Resistance Approach: Results and Discussion
  • 31. Thermal Resistance Approach: Advantages S-FGMP-FGM E-FGM • Can be used for any type of material gradient. • Gives you temperature profile for any kind of material distribution when analytical solution is very complex. • Helps to visualize the effect of material gradient
  • 32. Thermal Resistance Approach: Future Scopes • Determining material gradient for a desired stress profile • Tailoring material gradient for desired temperature field • Tailoring material gradient for desired stress field
  • 33. Thermal Resistance Approach: Future Scopes • Determining material gradient for a desired stress profile • Tailoring material gradient for desired temperature field • Tailoring material gradient for desired stress field