HW2 Due date

Next Tuesday (October 14)

Lecture Objectives:

• Unsteady-state heat transfer - conduction

• Solve unsteady state heat transfer equation for a wall

Implicit methods - example

wioww TTTTT 2)(3

iwii TTTT )(2.0

woiw TTTT 3)23(

iiw TTT )12.0(

400 800 1200 1600 2000 24000

10

20

30

40

50

60

70

80

T[C

]

time

To Tw Ti

=0 To Tw Ti

=36 system of equation Tw Ti

=72 system of equation Tw Ti

After rearranging:

2 Equations with 2 unknowns!

Explicit methods - example

wioww TTTTT 2)(3

iwii TTTT )(2.0

3

)23( owi

w

TTTT

2.0

)12.0(

iw

i

TTT

=0 To Tw Ti

=360 To Tw Ti

=720 To Tw Ti

=36 sec

2 3 4 5 6 7 8 9 100

10

20

30

40

50

60

70

80

T [C

]

time

To Tw Ti

UNSTABILITY

There is NO system of equations!

Tim

e

Explicit method

Problems with stability !!!

Often requires very small time steps

Explicit methods - example

30

)230( owi

w

TTTT

2

)12(

iw

i

TTT

=0 To Tw Ti

=36 To Tw Ti

=72 To Tw Ti =36 sec

400 800 1200 1600 2000 24000

10

20

30

40

50

60

70

80

T[C

]

time

To Tw Ti

Stable solution obtainedby time step reduction

10 times smaller time step

Tim

e

Unsteady-state conduction - Wall

sourcep

qx

T

c

T

2

2

q

Ts

0

T

-L / 2 L /2

h

h

h

To

T

h omogenous wa ll

L = 0.2 mk = 0 . 5 W/ m Kc = 9 20 J/kgK

= 120 0 k g/mp

2

Nodes for numerical calculation

x

Discretization of a non-homogeneous wall structure

Fa

cad

e s

lab

Insu

latio

n

Gyp

sum

Section considered in the following discussion

Discretization in space

2

2

x

T

c

kT

pDiscretization in time

Internal node Finite volume method

2/

2/

2/

2/2

2I

I

I

I

XI

XI

XI

XI

pII dxdx

Tkdxd

Tc

2

2

x

Tk

Tcp

For node “I” - integration through the control volume

( x) I- 1 ( x)I

x I

I-1 I I+1q I -1 to I q I to I+1

Boundaries of control volume

Surface node

2/

2/

I

I

XI

XI

III TTxdxdT

1

111

2/2/

2/

2/

2/

2/2

2

I

III

I

iII

XIXI

XI

XI

XI

XI x

TTk

x

TTk

dx

dTk

dx

dTk

x

Tk

xx

Tk

II

I

I

I

I

Left side of equation for node “I”

Right side of equation for node “I”

dx

TTk

x

TTkdxd

x

Tk

I

III

I

IIIXI

XI

I

I

1

1112/

2/2

2

Mathematical approach(finite volume method)

- Discretization in Time

- Discretization in Space

Mathematical approach(finite volume method)

xx

dx

TTk

x

TTkdxd

x

Tk

I

III

I

III

XI

XI

I

I

1

111

2/

2/2

2

I

III

I

IIIIII x

TTk

x

TTkTTx 111

pc

Explicit method

and for uniform grid

Implicit method

I

III

I

IIIIII x

TTk

x

TTkTTx 111

pc

By Substituting left and right side terms of equation

2/

2/

2/

2/2

2I

I

I

I

XI

XI

XI

XI

pII dxdx

Tkdxd

Tc

we get the following equation for

Using

xdx

][1

111

I

III

I

III

x

TTk

x

TTk

Physical approach(finite volume method)( x) I- 1 ( x)I

x I

I-1 I I+1q I -1 to I q I to I+1

Boundaries of control volume

2

2

x

Tk

Tcp

Change of energy in x

Change of heat flux along x

xxT

kTcp

)(

qx

=

Change of energy in x

Sum of energy that goes in and out of control volume x

=

or

) (1

1 I toI I to1-I

qqx

Tcp

For finite volume x:

1 I toI I to1-I

qqT

xcp

( x) I- 1 ( x)I

x I

I-1 I I+1q I -1 to I q I to I+1

1 I toI I to1-I

qqT

xcp

1I toI I to1-I

qqT

xcp

)(/)(/ 111

IIIIII

p TTxkTTxkTT

xc II

)(/)(/ 11

IIIIII

p TTxkTTxkTT

xc

xx For uniform grid

Physical approach(finite volume method)

Internal node finite volume method

Explicit method

Implicit method

By defining time step on the right side we get

I

III

I

IIIII

IIpI

x

TTk

x

TTkTT

xc

111

I

II

I

IIIII

IIpI

x

TTk

x

TTkTT

xc

11

Internal node finite volume method

I

II

I

IIIII

IIpI

x

TTk

x

TTkTT

xc

11

Explicit method

Implicit method

I

III

I

IIIII

IIpI

x

TTk

x

TTkTT

xc

111

FTCTBTA III

11

),,( 11

IIII TTTfTRearranging:

Rearranging:

Implicit method(internal node)

2

112

1

I

III

I

IIIII

IpI

x

TTk

x

TTkTT

c

IIpI

II

II

I

IIpII

I

I Tc

Tx

kT

x

kcT

x

k

12212)()2()(

AI BI CIFI

Internal nodes

1 2 3 4 5

B1 C1

A2 B2 C2

A3 B3 C3

A4 B4 C4

A5 B5

x =

F1

F2

F3

F4

F5

T1

T2

T3

T4

T5

kI-1=kI+1=kI

Implicit method(surface nodes)

B1 C1

A2 B2 C2

A3 B3 C3

A4 B4 C4

A5 B5

x =

F1

F2

F3

F4

F5

Surface nodes

1 2 3 4 5external internal0 6

T O Air T I Air

F0

F6

B0 C0

A1

C5

B6A6

T1

T2

T3

T4

T5

T0

T6

Surface node: 0

x

For surface nodes: flux coming in = flux going out

2/

100_ x

TTkTTh AirOexternal

Surface node: 6

AirIernal TThx

TTk_6int

65

2/

Calculate B0 and C0

Calculate A6 and B6

Discretization of a non-homogeneous wall structure

Fa

cad

e s

lab

Insu

latio

n

Gyp

sum

Section considered in the following discussion