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Homework 4

Name : Nadya Amalia

Student ID : 20213042

Subject : Computational Physical Systems (FI5005)

Lecturer : Dr.rer.nat. Linus Ampang Pasasa

HEAT-CONDUCTION PROBLEMS

A thin rod insulated at all points except al its ends. Heat-conduction equation:

ߙ

ଶݑ

(,ݔ (ݐ

߲ݔ

2  −

 ߲ݑ

(,ݔ (ݐ

߲ݐ

= 0

length of the thin rod,   = 10 cm

final time, = 12 s

space step, ∆ݔ =  ℎ = 2 cm

time step, ݐ∆ =  = 0.1 s

boundary conditions: ,(0ݑ  ° ) = 100ݐ and , )ݑ 0) = 50 ° 

initial condition: ݑ(ݔ, ݅݊ݏ = (0 (ݔߨ)

thermal conductivity, ߙ = 0.835 cm2/s

using implicit finite-difference method and crank-nicolson method.

SOLUTION

Implicit Finite-Difference Method 

Heat-conduction equation:

ߙ

ଶ߲ݔ

2 =

߲ݐ

The heat-conduction equation requires approximation for the second derivative in space and

the first derivative in time. In implicit finite-difference method, the spatial derivative is

approximated at an advanced time level  + 1 by a centered finite-difference approximation

2

߲ݔ

2 ≅  +1

+1− 2

+1+ 

−1+1

)2ݔ∆)

Similarly, a forward finite-difference approximation is used to approximate the tie derivative

߲ݐ

≅ 

+1−

(ݐ∆)

Thus,

ߙ

ାଵାଵ  − 2 ାଵ  + ିଵାଵ)2ݔ∆)

  = ାଵ  − (ݐ∆)

which can be expressed as

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ܶߣ− ିଵାଵ  + (1 + (ߣ2 ାଵ  − ܶߣ ାଵାଵ  =  where

ߣ ݐ∆ߙ = ଶൗݔ∆For the thin rod at  =  ) = ):−ߣܶ ିଵଶ   (ߣ2 + 1) + ଶ  − ܶߣ ାଵଶ   =  ଵBoundary conditions: =  and + = Initial condition: 

() = Difference equation for the first interior node ( = 2), =  :ܶߣ− ଶିଵଶ   (ߣ2 + 1) + ଶଶ  − ܶߣ ଶାଵଶ   =  ଶଵ(1 + 2

ߣ

) ଶଶ  − ܶߣ ଷଶ ݅݊ݏ =  (ߨ2) ߣ100 +For the first interior node ( = 3), =

 :

ܶߣ− ଷିଵଶ   (ߣ2 + 1) + ଷଶ  − ܶߣ ଷାଵଶ   =  ଷଵܶߣ− 2

ଶ + (1 + 2ߣ) ଷଶ  − ܶߣ ସଶ ݅݊ݏ =  (ߨ4)For the first interior node ( = 4), =  :ܶߣ− ସିଵଶ   (ߣ2 + 1) + ସଶ  − ܶߣ ସାଵଶ   =  ସଵܶߣ− ଷଶ + (1 + 2ߣ) ସଶ  − ܶߣ ହଶ ݅݊ݏ =  (ߨ6)Similarly, for last interior node ( = 5) , = ૡ cm:ܶߣ−

ହିଵଶ   + (1 + 2

ߣ

)

ହଶ − ܶߣ

ହାଵଶ   =

 

ହଵ

ܶߣ− ସଶ + (1 + 2ߣ) ହଶ ݅݊ݏ =  We obtain the following systemߣ) + 50ߨ8) of the equations:

⎣

(ߣ2 + 1)   ߣ − 0 0

ߣ− (ߣ2 + 1)   ߣ − 0

0   ߣ − (ߣ2 + 1)   ߣ −

0 0   ߣ − (ߣ2 + 1)   ⎦ ⎣

22

32

42

52⎦

= ൦ݏ݅݊ ݅݊ݏߣ) + 100ߨ2) (ߨ4)݅݊ݏ

(ߨ6)

݅݊ݏ

(ߨ8) ߣ50 +

൪For the thin rod at  = . ) = ):

ܶߣ−

ିଵଷ   (ߣ2 + 1) +

ଷ − ܶߣ

ାଵଷ   = 

ଶ

Boundary conditions:  =  and + = Initial condition: 

() = Difference equation for the first interior node ( = 2), =  :ܶߣ− ଶିଵଷ   (ߣ2 + 1) + ଶଷ  − ܶߣ ଶାଵଷ   =  ଶଶ(ߣ2 + 1) ଶଷ  − ܶߣ ଷଷ  =  ଶଶ + ߣ(100)For the first interior node ( = 3), =  :ܶߣ− ଷିଵଷ   (ߣ2 + 1) + ଷଷ  − ܶߣ ଷାଵଷ   =  ଷଶ−

ܶߣ

2ଷ + (

ߣ2 + 1)

ଷଷ −

ܶߣ ସଷ  =  ଷଶFor the first interior node ( = 4), =  :

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ܶߣ− ସିଵଷ   (ߣ2 + 1) + ସଷ  − ܶߣ ସାଵଷ   =  ସଶܶߣ− ଷଷ + (1 + 2ߣ) ସଷ  − ܶߣ ହଷ  =  ସଶSimilarly, for last interior node ( = 5) , = ૡ cm:ܶߣ−

ହିଵଷ  + (1 + 2

ߣ

)

ହଷ − ܶߣ

ହାଵଷ  =

 

ହଶ−ߣܶ ସଷ + (1 + 2ߣ) ହଷ  =  ହଶ + 50ߣWe obtain the following system of the equations:

⎣

(1 + 2ߣ

)   ߣ − 0 0

ߣ− (ߣ2 + 1)   ߣ − 0

0   ߣ − (ߣ2 + 1)   ߣ −

0 0   ߣ − (ߣ2 + 1)   ⎦ ⎣

23

33

43

53⎦

=

⎣

22 ߣ100 +

32

42

52 + 50

ߣ ⎦

Simultaneous equations can be solved for the temperatures at  ݐ for = 3, 4, 5, … ,Crank-Nicolson Method 

The Crank-Nicolson methos provides an alternative implicit scheme in second-order accurate

in both space and time. To provide this accuracy, difference aproximation are developed at the

midpoint of the time increment. To do this, th etemporal first derivative can be approximated

at ݐ+1 2/

by

߲ݐ

≅ 

+1− 2

(ݐ∆)

The second derivatice in space can be determined at the mid point by averaging the difference

aproximation at the beginning (ݐ ) and at the end (ݐ +1) of the time increment ଶ߲ݔ

ଶ ≅  12 ቈ +1 − 2 +  )2ݔ∆)1−   + +1+1 − 2 +1 +  )2ݔ∆)1+1−   which can be expressed as

ܶߣ− ିଵାଵ (ߣ + 1)2 +  ାଵ  − ܶߣ ାଵାଵ  = ܶߣ−  ିଵ   + (ߣ + 1)2  − ܶߣ ାଵwhere

ߣ ݐ∆ߙ = ݔ∆

ଶൗNumber of space steps, =  ℎൗ   = 5,   = 1, 2, … ,Number of time steps, =  ൗ   = 120,   = 1, 2, … ,

ߣ

=ݐ∆ߙଶݔ∆

For the thin rod at  =  ) = ):ܶߣ− ିଵଶ   (ߣ + 1)2 + ଶ  − ܶߣ ାଵଶ   = ܶߣ−  ିଵଵ   (ߣ + 1)2 + ଵ  − ܶߣ ାଵଵ

Boundary conditions: =

 and + =

 Initial condition:  ) )Difference equation for the first interior node) =  = 2), =  :

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ܶߣ− ଶିଵଶ   (ߣ + 1)2 + ଶଶ  − ܶߣ ଶାଵଶ   = ܶߣ−  ଶିଵଵ   + 2(1 + (ߣ ଶଵ  − ܶߣ ଶାଵଵ(ߣ + 1)2 ଶଶ  − ܶߣ 3ଶ ݅݊ݏ(ߣ + 1)2 =  (ߨ2) − ݅݊ݏߣ (ߨ4) ߣ100 +For the first interior node ( = 3), =  :ܶߣ−

ଷିଵଶ  + 2(1 +

ߣ 

)

ଷଶ − ܶߣ

ଷାଵଶ  = ܶߣ− 

ଷିଵଵ  + 2(1 +

ߣ

)

ଷଵ − ܶߣ

ଷାଵଵ−ߣܶ ଶଶ + 2(1 + (ߣ ଷଶ  − ܶߣ ସଶ  = ݅݊ݏߣ−  + ) + 2(1ߨ2) ݅݊ݏ(ߣ (ߨ4) − ݅݊ݏߣ (ߨ6)For the first interior node ( = 4), =  :ܶߣ− ସିଵଶ   (ߣ + 1)2 + ସଶ  − ܶߣ ସାଵଶ   = ܶߣ−  ସିଵଵ   + 2(1 + (ߣ ସଵ  − ܶߣ ସାଵଵܶߣ− 3

ଶ + 2(1 + (ߣ ସଶ  − ܶߣ ହଶ  = ݅݊ݏߣ−  + ) + 2(1ߨ4) ݅݊ݏ(ߣ (ߨ6) − ݅݊ݏߣ (ߨ8)Similarly, for last interior node ( = 5) , = ૡ cm:ܶߣ− ହିଵଶ   (ߣ + 1)2 + ହଶ  − ܶߣ ହାଵଶ   = ܶߣ−  ହିଵଵ   + 2(1 + (ߣ ହଵ  − ܶߣ ହାଵଵܶߣ− ସଶ + 2(1 + (ߣ ହଶ  = ݅݊ݏߣ−  ) + 2(1ߨ6) ݅݊ݏ(ߣ + (ߨ8) ߣ50 +We obtain the following system of the equations:

⎣

(ߣ + 1)2   ߣ − 0 0

ߣ− (ߣ + 1)2   ߣ − 0

0   ߣ − (ߣ + 1)2   − ߣ

0 0   ߣ − ⎣ ⎦(ߣ + 1)2

22

32

42

52⎦

= ൦   ݅݊ݏ(ߣ + 1)2 (ߨ2) ݅݊ݏߣ −  ݅݊ݏߣ−ߣ) + 100ߨ4) ݅݊ݏ(ߣ + ) + 2(1ߨ2) ݅݊ݏߣ − (ߨ4) (ߨ6)݅݊ݏߣ− ݅݊ݏ(ߣ + ) + 2(1ߨ4) ݅݊ݏߣ − (ߨ6) (ߨ8)

݅݊ݏߣ− (ߨ6) ݅݊ݏ(ߣ + 1)2 + ߣ) + 50ߨ8)

൪For the thin rod at  = . ) = ):

ܶߣ− ିଵଷ   (ߣ + 1)2 + ଷ  − ܶߣ ାଵଷ   = ܶߣ−  ିଵଶ   (ߣ + 1)2 + ଶ  − ܶߣ ାଵଶBoundary conditions: 

=

 and

+

=

 Initial condition:  ) )Difference equation for the first interior node) =  = 2), =  :ܶߣ− ଶିଵଷ   (ߣ + 1)2 + ଶଷ  − ܶߣ ଶାଵଷ   = ܶߣ−  ଶିଵଶ   + 2(1 + (ߣ ଶଶ  − ܶߣ ଶାଵଶ(ߣ + 1)2 ଶଷ  − ܶߣ ଷଷ (ߣ + 1)2 =  ଶଶ  − ܶߣ ଷଶ + 100ߣFor the first interior node ( = 3), =  :ܶߣ− ଷିଵଷ   (ߣ + 1)2 + ଷଷ  − ܶߣ ଷାଵଷ   = ܶߣ−  ଷିଵଶ   + 2(1 + (ߣ ଷଶ  − ܶߣ ଷାଵଶܶߣ− ଶଷ + 2(1 + (ߣ ଷଷ  − ܶߣ ସଷ  = ܶߣ−  ଶଶ + 2(1 + ߣ) ଷଶ  − ܶߣ ସଶFor the first interior node ( = 4), =

 :

ܶߣ− ସିଵଷ   (ߣ + 1)2 + ସଷ  − ܶߣ ସାଵଷ   = ܶߣ−  ସିଵଶ   + 2(1 + (ߣ ସଶ  − ܶߣ ସାଵଶܶߣ− ଷଷ + 2(1 + (ߣ ସଷ  − ܶߣ ହଷ  = ܶߣ−  ଷଶ + 2(1 + ߣ) ସଶ  − ܶߣ ହଶSimilarly, for last interior node ( = 5) , = ૡ cm:ܶߣ− ହିଵଷ   (ߣ + 1)2 + ହଷ  − ܶߣ ହାଵଷ   = ܶߣ−  ହିଵଶ   + 2(1 + (ߣ ହଶ  − ܶߣ ହାଵଶܶߣ− 4

ଷ + 2(1 +ߣ

) ହଷ  = ܶߣ−  ସଶ + 2(1 + (ߣ ହଶ + 50ߣWe obtain the following system of the equations:

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⎣

(ߣ + 1)2   − ߣ 0 0

ߣ− (ߣ + 1)2   − ߣ 0

0   ߣ − (ߣ + 1)2   ߣ −

0 0   ߣ − ⎣ ⎦(ߣ + 1)2

23

33

43

53⎦

=

⎣

2(1 +ߣ 

) 22ܶߣ − 3

2 + 100ߣ

ܶߣ− 22 (ߣ + 1)2 + 3

2ܶߣ − 4

2

ܶߣ− 32 (ߣ + 1)2 + 4

2ܶߣ − 5

2

ܶߣ− 42 + 2(1 +

ߣ 

) 52 + 50

⎦ߣ

Simultaneous equations can be solved for the temperatures at  ݐ for = 3, 4, 5, … ,