FEM solution for example problem of stationary heat flow

FEM solution for example problem of stationary heat flow

Piotr Pluciński

e-mail: Piotr.Plucinski@pk.edu.pl

Jerzy Pamin

e-mail: Jerzy.Pamin@pk.edu.pl

Chair for Computational Engineering

Faculty of Civil Engineering, Cracow University of Technology URL: www.CCE.pk.edu.pl

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N i (x, y) = α 1i + α 2i x ^e + α 3i y ^e





1 x i y i

1 x j y j

1 x k y k







 α 1i

α 2i

α 3i



 =



 1 0 0





W =      

1 x i y i

1 x j y j

1 x k y k

     

= 2P ₄

y x

i k

j e

y N _i (x ^e , y ^e )

x 1 i

k j

α 1i = x j y k − x k y j

2P ₄

α 2i = y j − y k

2P ₄

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N i (x, y) = α 1i + α 2i x ^e + α 3i y ^e





1 x i y i

1 x j y j

1 x k y k







 α 1i

α 2i

α 3i



 =



 1 0 0





W =      

1 x i y i

1 x j y j

1 x k y k

     

= 2P ₄

y x

i k

j e

y N _i (x ^e , y ^e )

x 1 i

k j

α 1i = x j y k − x k y j

2P ₄

α 2i = y j − y k

2P ₄

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N i (x, y) = α 1i + α 2i x ^e + α 3i y ^e





1 x _i y _i 1 x j y j

1 x k y k







 α _1i α 2i

α 3i



 =



 1 0 0





W =      

1 x i y i

1 x j y j

1 x k y k

     

= 2P ₄

W α

_1i

=      

1 x i y i

0 x j y j

0 x k y k

     

= x j y k − x k y j

α 1i = W _α

_1i

W = x _j y _k − x k y _j 2P ₄

y x

i k

j e

y N _i (x ^e , y ^e )

x 1 i

k j

α 1i = x j y k − x k y j

2P ₄

α 2i = y j − y k

2P ₄

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N i (x, y) = α 1i + α 2i x ^e + α 3i y ^e





1 x _i y _i 1 x j y j

1 x k y k







 α _1i α 2i

α 3i



 =



 1 0 0





W =      

1 x i y i

1 x j y j

1 x k y k

     

= 2P ₄

W α

_1i

=      

1 x i y i

0 x j y j

0 x k y k

     

= x j y k − x k y j

α 1i = W _α

_1i

W = x _j y _k − x k y _j 2P ₄

y x

i k

j e

y N _i (x ^e , y ^e )

x 1 i

k j

α 1i = x j y k − x k y j

2P ₄

α 2i = y j − y k

2P ₄

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N i (x, y) = α 1i + α 2i x ^e + α 3i y ^e





1 x _i y _i 1 x j y j

1 x k y k







 α _1i α 2i

α 3i



 =



 1 0 0





W =      

1 x i y i

1 x j y j

1 x k y k

     

= 2P ₄

W α

_2i

=      

1 1 y i

1 0 y j

1 0 y k

     

= y j − y k

α 2i = W _α

_2i

W = y _j − y k

2P ₄

y x

i k

j e

y N _i (x ^e , y ^e )

x 1 i

k j α 1i = x j y k − x k y j

2P ₄

α 2i = y j − y k

2P ₄

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N i (x, y) = α 1i + α 2i x ^e + α 3i y ^e





1 x _i y _i 1 x j y j

1 x k y k







 α _1i α 2i

α 3i



 =



 1 0 0





W =      

1 x i y i

1 x j y j

1 x k y k

     

= 2P ₄

W α

_2i

=      

1 1 y i

1 0 y j

1 0 y k

     

= y j − y k

α 2i = W _α

_2i

W = y _j − y k

2P ₄

y x

i k

j e

y N _i (x ^e , y ^e )

x 1 i

k j α 1i = x j y k − x k y j

2P ₄

α 2i = y j − y k

2P ₄

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N _i (x, y) = α _1i + α _2i x ^e + α _3i y ^e





1 x i y i

1 x _j y _j 1 x _k y _k







 α 1i

α _2i α _3i



 =



 1 0 0





W =      

1 x _i y _i 1 x _j y _j 1 x _k y _k

     

= 2P ₄

W _α

_3i

=      

1 x i 1 1 x j 0 1 x k 0

     

= x _k − x _j

α 3i = W α

_3i

W = x k − x j

2P ₄

y x

i k

j e

y N i (x ^e , y ^e )

x 1 i

k j α 1i = x _j y _k − x k y _j

2P ₄

α _2i = y j − y k

2P 4

Computational Methods, 2020 J.Paminc

Determination of shape functions for 3-node element

Shape function N _i (x, y) = α _1i + α _2i x ^e + α _3i y ^e





1 x i y i

1 x _j y _j 1 x _k y _k







 α 1i

α _2i α _3i



 =



 1 0 0





W =      

1 x _i y _i 1 x _j y _j 1 x _k y _k

     

= 2P ₄

W _α

_3i

=      

1 x i 1 1 x j 0 1 x k 0

     

= x _k − x _j

α 3i = W α

_3i

W = x k − x j

2P ₄

y x

i k

j e

y N i (x ^e , y ^e )

x 1 i

k j α 1i = x _j y _k − x k y _j

2P ₄

α _2i = y j − y k

2P 4

Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

q _n = 0

q n = 0 q _n = 5 J/m ² s

T = 20 ◦ C

4 m

3 m

k = 0.9 J/ms ^◦ C f = 2 J/m ² s h = 1 m

X Y

1 i

2 j

3 k

i k j 4

Discretization

1 2

Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

q _n = 0

q n = 0 q _n = 5 J/m ² s

T = 20 ◦ C

4 m

3 m

k = 0.9 J/ms ^◦ C f = 2 J/m ² s h = 1 m

X Y

1 i

2 j

3 k

i k j 4

Discretization

1 2

Computational Methods, 2020 J.Paminc

Heat flow in 2D

Z

A

(∇w) ^T Dh∇T dA = − Z

Γ

_q

wh qdΓ − b Z

Γ

_T

whq n dΓ + Z

A

whf dA

+ boundary condition

T = b T on Γ T

T = NΘ, w = Nw = w ^T N ^T , ∇T = BΘ

∇w = w ^T B ^T , D = kI, h = const

Computational Methods, 2020 J.Paminc

Heat flow in 2D

Z

A

(∇w) ^T Dh∇T dA = − Z

Γ

whq n dΓ + Z

A

whf dA

+ boundary condition

T = b T on Γ T

T = NΘ, w = Nw = w ^T N ^T , ∇T = BΘ

∇w = w ^T B ^T , D = kI, h = const

Computational Methods, 2020 J.Paminc

Heat flow in 2D

Z

A

(∇w) ^T Dh∇T dA = − Z

Γ

whq n dΓ + Z

A

whf dA

+ boundary condition

T = b T on Γ T

T = NΘ, w = Nw = w ^T N ^T , ∇T = BΘ

∇w = w ^T B ^T , D = kI, h = const

Computational Methods, 2020 J.Paminc

Heat flow in 2D

Z

A

(∇w) ^T Dh∇T dA = − Z

Γ

whq n dΓ + Z

A

whf dA

+ boundary condition

T = b T on Γ T

T = NΘ, w = Nw = w ^T N ^T , ∇T = BΘ

∇w = w ^T B ^T , D = kI, h = const

w ^T Z

A

B ^T kBdA Θ = −w ^T Z

Γ

N ^T q n dΓ + w ^T Z

A

N ^T f dA, ∀w

Computational Methods, 2020 J.Paminc

Heat flow in 2D

Z

A

(∇w) ^T Dh∇T dA = − Z

Γ

whq n dΓ + Z

A

whf dA

+ boundary condition

T = b T on Γ T

T = NΘ, w = Nw = w ^T N ^T , ∇T = BΘ

∇w = w ^T B ^T , D = kI, h = const Z

A

B ^T kBdA Θ = − Z

Γ

N ^T q n dΓ + Z

A

N ^T f dA

Computational Methods, 2020 J.Paminc

Heat flow in 2D

Z

A

(∇w) ^T Dh∇T dA = − Z

Γ

whq n dΓ + Z

A

whf dA

+ boundary condition

T = b T on Γ T

T = NΘ, w = Nw = w ^T N ^T , ∇T = BΘ

∇w = w ^T B ^T , D = kI, h = const Z

A

B ^T kBdA Θ = − Z

Γ

N ^T q n dΓ + Z

A

N ^T f dA

K = Z

A

B

^T

kBdA, f = Z

A

N

^T

f dA, f

b

= − Z

Γ

N

^T

q

n

dΓ

Computational Methods, 2020 J.Paminc

Heat flow in 2D

Z

A

(∇w) ^T Dh∇T dA = − Z

Γ

whq n dΓ + Z

A

whf dA

+ boundary condition

T = b T on Γ T

T = NΘ, w = Nw = w ^T N ^T , ∇T = BΘ

∇w = w ^T B ^T , D = kI, h = const Z

A

B ^T kBdA Θ = − Z

Γ

N ^T q n dΓ + Z

A

N ^T f dA

K = Z

A

B

^T

kBdA, f = Z

A

N

^T

f dA, f

b

= − Z

Γ

N

^T

q

n

dΓ KΘ = f + f _b

Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

K matrix – element 1 N ¹ = 

1 − ¹ ₄ x ¹ ₄ x − ¹ ₃ y ¹ ₃ y 

B ¹ = ∇N =

 −0.250 0.250 0.000 0.000 −0.333 0.333



K ¹ = Z

A

¹

B ^T kBdA = A ¹ B ^T kB

=





0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600





K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

K matrix – element 1 N ¹ = 

1 − ¹ ₄ x ¹ ₄ x − ¹ ₃ y ¹ ₃ y 

B ¹ = ∇N =

 −0.250 0.250 0.000 0.000 −0.333 0.333



K ¹ = Z

A

¹

B ^T kBdA = A ¹ B ^T kB

=





0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600





K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

K matrix – element 1 N ¹ = 

1 − ¹ ₄ x ¹ ₄ x − ¹ ₃ y ¹ ₃ y 

B ¹ = ∇N =

 −0.250 0.250 0.000 0.000 −0.333 0.333



K ¹ = Z

A

¹

B ^T kBdA = A ¹ B ^T kB

=





0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600





K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

K matrix – element 2 N ² = 

1 − ¹ ₃ y ¹ ₄ x ¹ ₃ y − ¹ ₄ x 

B ² = ∇N =

 0.000 0.250 −0.250

−0.333 0.000 0.333



K ² = Z

A

²

B ^T kBdA = A ² B ^T kB

=





0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938





K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

K matrix – element 2 N ² = 

1 − ¹ ₃ y ¹ ₄ x ¹ ₃ y − ¹ ₄ x 

B ² = ∇N =

 0.000 0.250 −0.250

−0.333 0.000 0.333



K ² = Z

A

²

B ^T kBdA = A ² B ^T kB

=





0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938





K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

K matrix – element 2 N ² = 

1 − ¹ ₃ y ¹ ₄ x ¹ ₃ y − ¹ ₄ x 

B ² = ∇N =

 0.000 0.250 −0.250

−0.333 0.000 0.333



K ² = Z

A

²

B ^T kBdA = A ² B ^T kB

=





0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938





K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

1 1

i

3 j k 4

2

f vector – element 1 and 2 - A ¹ = A ²

f ^e = Z

A

^e

N ^T f dA = f 3 A ^e



 1 1 1



 =



 4 4 4





K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

f b vector – element 1 f _b ¹ = −

Z

Γ

¹_ij

(N ¹ ) ^T q _n dΓ − Z

Γ

¹_jk

(N ¹ ) ^T q _n dΓ

− Z

Γ

¹_ki

(N ¹ ) ^T q n dΓ

f _b ¹ =



 0 0 0





K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

f b vector – element 1 f _b ¹ = −

Z

Γ

¹_ij

(N ¹ ) ^T q _n b.c. = 0

dΓ − Z

Γ

¹_jk

(N ¹ ) ^T q _n b.c. = 0

dΓ

flow continuity along edge 1-3 q n 1

ki = −q n 2 ij

− Z

Γ

¹_ki

(N ¹ ) ^T q n dΓ

f _b ¹ =



 0 0 0





K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

f b vector – element 1 f _b ¹ = −

Z

Γ

¹_ij

(N ¹ ) ^T q _n b.c. = 0

dΓ − Z

Γ

¹_jk

(N ¹ ) ^T q _n b.c. = 0

dΓ

flow continuity along edge 1-3 q n 1

ki = −q n 2 ij

− Z

Γ

¹_ki

(N ¹ ) ^T q n dΓ

f _b ¹ =



 0 0 0





K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

f b vector – element 2

f _b ² = − Z

Γ

²_ij

(N ² ) ^T q n dΓ

− Z

Γ

²_jk

(N ² ) ^T q _n dΓ − Z

Γ

²_ki

(N ² ) ^T q _n dΓ

K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

f b vector – element 2

f _b ² = − Z

Γ

²_ij

(N ² ) ^T q n dΓ

flow continuity along edge 1-3 q n 1

ki = −q n 2 ij

− Z

Γ

²_jk

(N ² ) ^T q n dΓ − Z

Γ

²_ki

(N ² ) ^T q n dΓ

K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

f b vector – element 2 f _b ² = −

Z

Γ

²_jk

(N ² ) ^T q n dΓ − Z

Γ

²_ki

(N ² ) ^T q n dΓ

− Z

Γ²_jk

(N

²

)

^T

q

n

dΓ = − Z

4

0

N

²

(x, y = 3)

^T

(−5)dx

=

" ₀ 10 10

#

K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

f b vector – element 2 f _b ² =



 0 10 10



 − Z

Γ

²_ki

(N ² ) ^T q _n dΓ

− Z

Γ² ki

(N

²

)

^T

q

n

dΓ = − Z

3

0

N

²

(x = 0, y)

T

q

n

dx

=

" _f

b1

0 f

_b4

#

K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



,

f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

Assembly

K ¹ =





0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600





K =









0.338 −0.338 0.000 0.000

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.600 0.000 0.000 0.000 0.000 0.000









K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



, f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

Assembly

K ² =





0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938





K =









0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938









K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



, f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

Assembly

f ¹ =



 4 4 4





f =







 4 4 4 0









K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



, f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

i

2 j k

1 1

i

3 j k 4

2

Assembly

f ¹ =



 4 4 4



 f ² =



 4 4 4





f =







 8 4 8 4









K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



, f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc

2D example of heat flow – 3-node elements

qn = 0 qn=0 qn = 5 J/m2 s

T=20◦C

k = 0.9 J/ms◦C f = 2 J/m2s h = 1 m

X Y

Discretization

1 i

2 j

3

k

i 1 j k 4

2

Assembly

f _b ¹ =



 0 0 0





f b =







 0 0 0 0









K

¹

=

 0.338 −0.338 0.000

−0.388 0.938 −0.600 0.000 −0.600 0.600



K

²

=

 _0.600 0.000 −0.600 0.000 0.338 −0.338

−0.600 −0.338 0.938



f

¹

= f

²

=

 ₄

4 4



f

_b¹

=

 ₀

0 0



, f

_b²

=

 _f

b 1

10 f

_{b 4}

+ 10



K =



0.938 −0.338 0.000 −0.600

−0.388 0.938 −0.600 0.000 0.000 −0.600 0.938 −0.338

−0.600 0.000 −0.338 0.938



f =



8 4 8 4



,

fb =



_fb1

0 10 fb4 + 10



Θ =



20 48.040 57.145 20



Computational Methods, 2020 J.Paminc