STUDIO E PROGETTAZIONE DEL SISTEMA DI RAFFREDDAMENTO DEI MODULI IN SILICIO DEL TRACCIATORE DEL RIVELATORE CMS (COMPACT MUON SOLENOID)

Pr

=

20.991

Pr

µ

K

_f

⋅

Cp

:=

µ

1.32

poise

100

=

µ

:=

µ

( )

T

ρ

1.787

×

103

kg

m

3

=

ρ

:=

ρ

( )

T

K

_f

0.062

watt

m K

⋅

=

K

_f

:=

Kf T

( )

Cp

986.314

joule

kg K

⋅

=

Cp

:=

Cp T

( )

Calculated physical properties:

µ

( )

T

interp v4 T0

(

,

,

µ4

,

T

)

poise

100

⋅

:=

T

:=

−

18

Evaluation temperature:

Kf T

( )

interp v3 T0

(

,

,

K3

,

T

)

watt

m K

⋅

⋅

:=

v4

:=

lspline T0

(

,

µ4

)

µ4

_:=

Ppf

〈 〉

4

v3

:=

lspline T0 K3

(

,

)

K3

_:=

Ppf

〈 〉

3

ρ

( )

T

interp v2 T0

(

,

,

ρ2

,

T

)

kg

m

3

⋅

:=

v2

:=

lspline T0

(

,

ρ2

)

ρ2

_:=

Ppf

〈 〉

2

Cp T

( )

interp v1 T0

(

,

,

Cp1

,

T

)

joule

kg K

⋅

⋅

:=

v1

:=

lspline T0 Cp1

(

,

)

Cp1

_:=

Ppf

〈 〉

1

T0

_:=

Ppf

〈 〉

0

Ppf

20

−

15

−

10

−

5

−

0 5 10 15 20 25 30 35 983 991 998 1006 1014 1022 1030 1037 1045 1053 1061 1068 1792 1779 1766 1753 1740 1727 1714 1701 1688 1675 1662 1649 0.0622 .0617 0.0611 0.0606 0.0600 0.0595 0.0589 0.0584 0.0578 0.0573 0.0567 0.0562 1.373 1.243 1.130 1.032 0.946 0.870 0.803 0.743 0.690 0.643 0.600 0.562

























































:=

3M PF-5060 thermal properties

Data for per-fluoro-carbon C

6

F

14

, taken from the 3M calculation sheet

"3M Performance Fluids"

developed by Lew Tousignant 3M Company.

Evaluations of the pressure drops for TIB

cooling loop (single sided layers)

Q

_cl

=

89.82

watt

∆T

:=

1

⋅

K

Assumed thermal gradient.

b

:=

4.55

⋅

mm

h

:=

1.9

⋅

mm

External pipes dimension (real dimensions)

s

:=

0.35 10

⋅

−

3

m

Pipe thickness

R

₀

:=

12mm

Bending radii

D

4

⋅

(

b

−

2

⋅

s

)

⋅

(

h

−

2

⋅

s

)

b

−

2

s

(

)

+

(

h

−

2

s

)

[

]

⋅

2

:=

D

=

1.83

mm

Hydraulic Diameter

G

_CLmin

Q

cl

Cp

⋅

∆T

:=

G

_CLmin

=

5.464 kg

min

−

1

Q

_CLmin

G

CLmin

ρ

:=

Minimum flow needed for the cooling loop.

Q

_CLmin

=

3.058

liter min

⋅

−

1

G

_Rmin

G

CLmin

n

_r

:=

G

_Rmin

=

0.546

kg min

⋅

−

1

_{Minimum flow needed for a row.}

_Q

Rmin

Q

_CLmin

n

_r

:=

G

_Rmin

=

32.784

kg hr

⋅

−

1

Minimum flow needed for a row

.

QRmin

=

18.348

liter hr

⋅

−

1

Number of chips in module n

_c

:=

4

N° of Opto hybrid per module n

_HY

:=

1

Optohybrid power consumption P

_oh

:=

0.74watt

Percent of row power dissipated

in cables

P

%c

:=

50

n

:=

3

Number of modules per row.

n

_ir

:=

5

Number of internal cooling rows.

n

_er

:=

5

Number of external cooling rows.

n

_r

:=

n

_ir

+

n

_er

Number of rows per cooling loop.

Q

_m

(

n

_c

⋅

P

_c

+

P

_oh

⋅

n

_HY

)

100

+

P

%c

100

⋅

:=

Heat produced by a module:

Q

_m

=

2.994

watt

Q

_r

:=

Q

_m

⋅

n

Heat produced by a row:

Q

_r

=

8.982

watt

Re

2.733

×

103 8.448

×

103 8.448

×

103 7.987

×

103

























=

Re

_i

ρ

⋅

V

i

⋅

D

i

µ

:=

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

Reynold's number

V

1.103 1.2 1.2 0.536





















m sec

-1

=

V

G

_Rmin

ρ

⋅

A

1

G

_Rmin

⋅

n

_er

ρ

⋅

A

2

G

_Rmin

⋅

n

_ir

ρ

⋅

A

3

G

_CLmin

ρ

⋅

A

4

















































:=

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

Fluid velocity

A

4.62 21.237 21.237 95.033





















mm

2

=

D

1.83 5.2 5.2 11





















mm

=

L

1.5 0.8 0.8 0.4





















m

=

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

A

b

−

2

⋅

s

(

) h

⋅

(

−

2

⋅

s

)

π

⋅

(

6mm s

−

_d6

⋅

2

)

2 4

π

⋅

(

6

mm s

−

_d6

⋅

2

)

2 4

π

⋅

(

12mm s

−

_d12

⋅

2

)

2 4









































:=

D

D

6mm

−

2s

_d6

6

mm

−

2

s

_d6

12mm

−

2s

_d12

























:=

L

L

_r

L

_d6

L

_d6

L

_d12





















:=

D12 pipe thickness

s

_d12

:=

0.5mm

D6 pipe thickness

s

_d6

:=

0.4mm

Lenght of each D12 cooling pipes

L

_d12

:=

0.4m

Lenght of each D6 cooling pipes

L

_d6

:=

0.8m

Lenght of row cooling pipe

L

_r

:=

1500

⋅

mm

Evaluation of pressure losses

i

:=

1

..

4

ORIGIN

:=

1

f

0.3164

Re

0.25

:=

Blasius correlation for 2.320<Re<100.000 (smooth pipes)

- Row pipe

- External coolin rows

- Internal cooling rows

- Supply pipe D12mm

Friction factor

_f

0.044 0.033 0.033 0.033





















=

DISTRIBUTED pressure losses for each

segment of pipes

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

∆P

i

1₂

⋅

ρ

⋅

( )

V

i

2

L

_i

D

_i

⋅

⋅

f

_i

:=

_∆P

0.385 0.064 0.064 3.086

×

10

−

3

















atm

=

Concentrated pressure loss [Memento des pertes de charge

I.E Idel'cik pag. 196]

K

_lr

λ

δ

←

135

D

_H

←

D

1

Re

←

Re

1

β

Re

D

H

2

⋅

R

₀

⋅

←

λ1

20

Re

0.65

D

_H

2

⋅

R

₀













0.175

⋅

←

λ1

if

50

<

β

∧

β

<

600

λ2

10.4

Re

0.55

D

_H

2

⋅

R

₀













0.225

⋅

←

λ2

if

600

<

β

∧

β

<

1400

λ3

5

Re

0.45

D

_H

2

⋅

R

₀













0.275

⋅

←

λ3

if

1400

<

β

∧

β

<

5000

←

λ

−

f

1

(

)

⋅

0.0175

R

₀

D

1

⋅

⋅

δ

:=

K

₉₀

λ

δ

←

90

D

_H

←

D

1

Re

←

Re

1

β

Re

D

H

2

⋅

R

₀

⋅

←

λ1

20

Re

0.65

D

_H

2

⋅

R

₀













0.175

⋅

←

λ1

if

50

<

β

∧

β

<

600

λ2

10.4

Re

0.55

D

_H

2

⋅

R

₀













0.225

⋅

←

λ2

if

600

<

β

∧

β

<

1400

λ3

5

Re

0.45

D

_H

2

⋅

R

₀













0.275

⋅

←

λ3

if

1400

<

β

∧

β

<

5000

←

λ

−

f

1

(

)

⋅

0.0175

R

₀

D

1

⋅

⋅

δ

:=

!! Pressure losses in manifolds neglected !!

∆P

_tot

=

7.658

×

104

Pa

∆P

_tot

:=

∆P

1

+

(

∆P

2

+

∆P

3

+

∆P

4

)

⋅

2

+

∆P

Cr

+

∆P

C6e

+

∆P

C6i

+

∆P

C12

∆P

_C90

=

139.289

Pa

∆P

_C90

(

2

⋅

K

₉₀

)

1 2

⋅ ρ

⋅

⋅

( )

V

4 2

:=

∆P

_C12

=

51.391

Pa

∆P

_C12

(

K

_v

⋅

1

)

1 2

⋅ ρ

⋅

⋅

( )

V

4 2

:=

∆P

_C6i

=

1.046

×

103

Pa

∆P

_C6i

(

3

⋅

K

₉₀

)

1 2

⋅ ρ

⋅

⋅

( )

V

3 2

:=

∆P

_C6e

=

1.046

×

103

Pa

∆P

_C6e

(

3

⋅

K

₉₀

)

1 2

⋅ ρ

⋅

⋅

( )

V

2 2

:=

∆P

_Cr

=

8.686

×

103

Pa

∆P

_Cr

(

12

⋅

K

_lr

+

10

⋅

K

₉₀

+

2

K

_v

)

1 2

⋅ ρ

⋅

⋅

( )

V

1 2

:=

Section variation

K

_v

:=

.2

135° curve

K

_lr

=

0.407

90° curve

K

₉₀

=

0.271

Q

_Rmin

=

18.348

liter hr

⋅

−

1

Experimental pressure drops in circular pipe

∆P

_sround

( )

G

:=

.0002

⋅

G

2

+

.0075

⋅

G

−

0.015

Experimental pressure drops in rectangular pipe

∆P

_sflat

( )

G

:=

.0006

⋅

G

2

+

.0158

⋅

G

−

0.059 0 10 20 30 40 50 60 0 1 2 3 4

∆

P

_sflat

( )

G

∆

P

_sround

( )

G

G

∆P

_sflat

Q

_Rmin

hr

liter

⋅









=

0.433

Experimental pressure drop

∆P

1

+

∆P

Cr

=

0.471

atm

Theoretical pressure drop

Error (%):

∆P

1

+

∆P

Cr

(

)

∆P

_sflat

Q

_Rmin

hr

liter

⋅

















⋅

atm

−

∆P

_sflat

Q

_Rmin

hr

liter

⋅

















⋅

atm

100

⋅

=

8.731

Evaluation of the heat transfer coefficient inside the row

Gnielinsky's formula

valid for transient regime flow:

3.000<Re<5*10^5

h

_Gn

i

f

_i

8

Re

_i

−

1000

(

)

⋅

Pr

1 12.7

f

_i

8

⋅

⋅

(

Pr

0.66

−

1

)

+

⋅

K

f

D

_i

⋅

:=

h

_DB

i 0.023

( )

Re

i

0.8

Pr

0.3

K

f

D

_i

:=

_{Fully turbulent flow pattern}

Re

1

=

2.733

×

103

h

_Gn

1 954.488

watt

m

2

⋅

K

=

h

_DB

1 1.091 10 3

×

watt

m

2

⋅

K

=

Pr

=

20.991

Pr

µ

K

_f

⋅

Cp

:=

µ

1.32

poise

100

=

µ

:=

µ

( )

T

ρ

1.787

×

103

kg

m

3

=

ρ

:=

ρ

( )

T

K

_f

0.062

watt

m K

⋅

=

K

_f

:=

Kf T

( )

Cp

986.314

joule

kg K

⋅

=

Cp

:=

Cp T

( )

Calculated physical properties:

µ

( )

T

interp v4 T0

(

,

,

µ4

,

T

)

poise

100

⋅

:=

T

:=

−

18

Evaluation temperature:

Kf T

( )

interp v3 T0

(

,

,

K3

,

T

)

watt

m K

⋅

⋅

:=

v4

:=

lspline T0

(

,

µ4

)

µ4

_:=

Ppf

〈 〉

4

v3

:=

lspline T0 K3

(

,

)

K3

_:=

Ppf

〈 〉

3

ρ

( )

T

interp v2 T0

(

,

,

ρ2

,

T

)

kg

m

3

⋅

:=

v2

:=

lspline T0

(

,

ρ2

)

ρ2

_:=

Ppf

〈 〉

2

Cp T

( )

interp v1 T0

(

,

,

Cp1

,

T

)

joule

kg K

⋅

⋅

:=

v1

:=

lspline T0 Cp1

(

,

)

Cp1

_:=

Ppf

〈 〉

1

T0

_:=

Ppf

〈 〉

0

Ppf

20

−

15

−

10

−

5

−

0 5 10 15 20 25 30 35 983 991 998 1006 1014 1022 1030 1037 1045 1053 1061 1068 1792 1779 1766 1753 1740 1727 1714 1701 1688 1675 1662 1649 0.0622 .0617 0.0611 0.0606 0.0600 0.0595 0.0589 0.0584 0.0578 0.0573 0.0567 0.0562 1.373 1.243 1.130 1.032 0.946 0.870 0.803 0.743 0.690 0.643 0.600 0.562

























































:=

3M PF-5060 thermal properties

Data for per-fluoro-carbon C

6

F

14

, taken from the 3M calculation sheet

"3M Performance Fluids"

developed by Lew Tousignant 3M Company.

Evaluations of the pressure drops for TIB

cooling loops (double sided layers)

Q

_cl

=

236.16

watt

∆T

:=

1

⋅

K

Assumed thermal gradient.

b

:=

4.55

⋅

mm

h

:=

1.9

⋅

mm

External pipes dimension (real dimensions)

s

:=

0.35 10

⋅

−

3

m

Pipe thickness

R

₀

:=

12mm

Bending radii

D

4

⋅

(

b

−

2

⋅

s

)

⋅

(

h

−

2

⋅

s

)

b

−

2s

(

)

+

(

h

−

2s

)

[

]

⋅

2

:=

D

=

1.83

mm

Hydraulic Diameter

G

_CLmin

Q

cl

Cp

⋅

∆T

:=

G

_CLmin

=

14.366 kg

min

−

1

Q

_CLmin

G

CLmin

ρ

:=

Minimum flow needed for the cooling loop.

Q

_CLmin

=

8.04

liter min

⋅

−

1

G

_Rmin

G

CLmin

n

_r

:=

G

_Rmin

=

1.437

kg min

⋅

−

1

_{Minimum flow needed for a row.}

_Q

Rmin

Q

_CLmin

n

_r

:=

G

_Rmin

=

86.197

kg hr

⋅

−

1

Minimum flow needed for a row

.

QRmin

=

48.241

liter hr

⋅

−

1

Chip power consumption

P

_c

:=

0.314watt

Number of chips in module n

c

:=

12

N° of Opto hybrid per module n

HY

:=

2

Optohybrid power consumption P

oh

:=

0.74watt

Percent of row power dissipated

in cables

P

%c

:=

50

n

:=

3

Number of modules per row.

n

_ir

:=

5

Number of internal cooling rows.

n

_er

:=

5

Number of external cooling rows.

n

_r

:=

n

_ir

+

n

_er

Number of rows per cooling loop.

Q

_m

(

n

_c

⋅

P

_c

+

P

_oh

⋅

n

_HY

)

100

+

P

%c

100

⋅

:=

Heat produced by a module:

Q

_m

=

7.872

watt

Q

_r

:=

Q

_m

⋅

n

Heat produced by a row:

Q

_r

=

23.616

watt

Re

7.185

×

103 2.221

×

104 2.221

×

104 2.1

×

104

























=

Re

_i

ρ

⋅

V

i

⋅

D

i

µ

:=

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

Reynold's number

V

2.901 3.155 3.155 1.41





















m sec

-1

=

V

G

_Rmin

ρ

⋅

A

1

G

_Rmin

⋅

n

_er

ρ

⋅

A

2

G

_Rmin

⋅

n

_ir

ρ

⋅

A

3

G

_CLmin

ρ

⋅

A

4

















































:=

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

Fluid velocity

A

4.62 21.237 21.237 95.033





















mm

2

=

D

1.83 5.2 5.2 11





















mm

=

L

1.5 0.8 0.8 0.4





















m

=

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

A

b

−

2

⋅

s

(

) h

⋅

(

−

2

⋅

s

)

π

⋅

(

6mm s

−

_d6

⋅

2

)

2 4

π

⋅

(

6mm s

−

_d6

⋅

2

)

2 4

π

⋅

(

12mm s

−

_d12

⋅

2

)

2 4









































:=

D

D

6

mm

−

2

s

_d6

6mm

−

2s

_d6

12mm

−

2s

_d12

























:=

L

L

_r

L

_d6

L

_d6

L

_d12





















:=

D12 pipe thickness

s

_d12

:=

0.5

mm

D6 pipe thickness

s

_d6

:=

0.4

mm

Lenght of each D12 cooling pipes

L

_d12

:=

0.4

m

Lenght of each D6 cooling pipes

L

_d6

:=

0.8

m

Lenght of row cooling pipe

L

_r

:=

1500

⋅

mm

Evaluation of pressure losses

i

:=

1

..

4

ORIGIN

:=

1

ε

:=

0.001

⋅

mm

Roughness for each pipe

f

0.3164

Re

0.25

:=

Blasius correlation for 2.320<Re<100.000 (smooth pipes)

- Row pipe

- External coolin rows

- Internal cooling rows

- Supply pipe D12mm

Friction factor

_f

0.034 0.026 0.026 0.026





















=

DISTRIBUTED pressure losses for each

segment of pipes

- Row pipe

- External cooling rows

- Internal cooling rows

- Supply pipe D12mm

∆P

_i

1 2

⋅

ρ

( )

V

i

2

⋅

L

i

D

_i

⋅

⋅

f

_i

:=

_∆P

2.09 0.35 0.35 0.017





















atm

=

Concentrated pressure loss [Memento des pertes de charge

I.E Idel'cik pag. 196]

K

_lr

λ

δ

←

135

D

_H

←

D

1

Re

←

Re

1

β

Re

D

H

2

⋅

R

₀

⋅

←

λ1

20

Re

0.65

D

_H

2

⋅

R

₀













0.175

⋅

←

λ1

if

50

<

β

∧

β

<

600

λ2

10.4

Re

0.55

D

_H

2

⋅

R

₀













0.225

⋅

←

λ2

if

600

<

β

∧

β

<

1400

λ3

5

Re

0.45

D

_H

2

⋅

R

₀













0.275

⋅

←

λ3

if

1400

<

β

∧

β

<

5000

←

λ

−

f

1

(

)

⋅

0.0175

R

₀

D

1

⋅

⋅

δ

:=

K

₉₀

λ

δ

←

90

D

_H

←

D

1

Re

←

Re

1

β

Re

D

H

2

⋅

R

₀

⋅

←

λ1

20

Re

0.65

D

_H

2

⋅

R

₀













0.175

⋅

←

λ1

if

50

<

β

∧

β

<

600

λ2

10.4

Re

0.55

D

_H

2

⋅

R

₀













0.225

⋅

←

λ2

if

600

<

β

∧

β

<

1400

λ3

5

Re

0.45

D

_H

2

⋅

R

₀













0.275

⋅

←

λ3

if

1400

<

β

∧

β

<

5000

←

λ

−

f

1

(

)

⋅

0.0175

R

₀

D

1

⋅

⋅

δ

:=

!! Pressure losses in manifolds neglected !!

∆P

_tot

=

3.902

×

105

Pa

∆P

_tot

:=

∆P

1

+

(

∆P

2

+

∆P

3

+

∆P

4

)

⋅

2

+

∆P

Cr

+

∆P

C6e

+

∆P

C6i

+

∆P

C12

∆P

_C90

=

401.535

Pa

∆P

_C90

(

2

⋅

K

₉₀

)

1 2

⋅ ρ

⋅

( )

V

4 2

⋅

:=

∆P

_C12

=

355.268

Pa

∆P

_C12

(

K

_v

⋅

1

)

1 2

⋅ ρ

⋅

( )

V

4 2

⋅

:=

∆P

_C6i

=

3.015

×

103

Pa

∆P

_C6i

(

3

⋅

K

₉₀

)

1 2

⋅ ρ

⋅

( )

V

3 2

⋅

:=

∆P

_C6e

=

3.015

×

103

Pa

∆P

_C6e

(

3

⋅

K

₉₀

)

1 2

⋅ ρ

⋅

( )

V

2 2

⋅

:=

∆P

_Cr

=

2.679

×

104

Pa

∆P

_Cr

(

12

⋅

K

_lr

+

10

⋅

K

₉₀

+

2K

_v

)

1 2

⋅ ρ

⋅

( )

V

1 2

⋅

:=

Section variation

K

_v

:=

.2

135° curve

K

_lr

=

0.17

90° curve

K

₉₀

=

0.113

Comaparison with experimental data of chapter 4

Q

_Rmin

=

48.241

liter hr

⋅

−

1

Experimental pressure drops in circular pipe

∆P

_sround

( )

G

:=

.0002

⋅

G

2

+

.0075

⋅

G

−

0.015

Experimental pressure drops in rectangular pipe

∆P

_sflat

( )

G

:=

.0006

⋅

G

2

+

.0158

⋅

G

−

0.059 0 10 20 30 40 50 60 0 1 2 3 4

∆

P

_sflat

( )

G

∆

P

_sround

( )

G

G

∆P

_sflat

Q

_Rmin

hr

liter

⋅









=

2.1

Experimental pressure drop

∆P

1

+

∆P

Cr

=

2.354

atm

Theoretical pressure drop

Error (%):

∆P

1

+

∆P

Cr

(

)

∆P

_sflat

Q

_Rmin

hr

liter

⋅

















⋅

atm

−

∆P

sflat



_

Q

Rmin

⋅

_liter

hr



_









⋅

atm

100

Evaluation of the heat transfer coefficient inside the row

Gnielinsky's formula

valid for transient regime flow:

3.000<Re<5*10^5

h

_Gn

i

f

_i

8

Re

_i

−

1000

(

)

⋅

Pr

1 12.7

f

_i

8

⋅

⋅

(

Pr

0.66

−

1

)

+

⋅

K

f

D

_i

⋅

:=

h

_DB

i 0.023

( )

Re

i

0.8

_Pr

0.3

K

f

D

_i

:=

_{Fully turbulent flow pattern}

Re

1

=

7.185

×

103

h

_Gn

1 2.965

×

103

watt

m

2

_⋅

_K

=

h

_DB

1 2.364

×

103

watt

m

2

_⋅

_K

=

Spessore kapton

kout_pCF 2 watt m K⋅

⋅ :=

sglue1:=0.1 mm⋅ Spessore colla tra tubo e ledge

kair 0.023 watt m K⋅

⋅ :=

sglue2:=0.1mm

Spessore colla tra CF e HY

ahy:=300 mm⋅ 2 _{kout_phy} 0.34 watt

m K⋅

⋅ :=

Area di metà ibrido achips:=50mm2 Area sotto metà chips

kin_phy 60 watt m K⋅

⋅ :=

aledge :=200mm2 Area di contatto ledge-CF

gapair:=5 10⋅ −3⋅mm Distanza media tra il frame e il ledge

htc50lh 1700 watt m2⋅K

⋅ :=

d:=20 mm⋅ Distanza tra i baricentri dell'area scaldata e della superficie di contatto del ledge

********************************************************************************************** Calcoli

********************************************************************************************** Definiamo la resistenza termica come la differenza di temperatura divisa per la potenza termica scambiata, (1 watt nel caso considerato).

Dal fluido al tubo di alluminio (strato limite)

A :=(b+2 h⋅ _{) lledge}⋅ A =175.2 mm2 superficie di contatto tubo-ledge

Calcolo delle resistenze termiche, dal fluido ai

chips dell'ibrido.

Data una potenza assegnata (2 watt) che deve fluire dal fluido ai chips di un modulo "single sided" si determinano le cadute di temperatura nei vari strati che si incontrano.

Nota: tutte le considerazioni seguenti sono fatte per metà del

ledge di read-out

W:= 1watt potenza di riferimento

Calcoli esegiti con l'ibrido in FR4

Dati geometrici della struttura: Conducibilità termiche

spipe:= 0.35mm kglue1 0.2watt m K⋅ := sledge :=2.3mm kglue2:=kglue1 b:=3.6mm Larghezza tubo dopo schiacciatura

h:=1.85mm Altezza tubo dopo schiacciatura

kkapton 0.2watt m K⋅

:=

lledge:=24mm Lunghezza cava di alloggiamento tubo nel ledge

kAl 220 watt m K⋅

⋅ :=

scf :=.7mm Spessore fibra di carbonio

shy:=0.38mm Spessore ibrido _{kin_pCF} 220 watt m K⋅

⋅ :=

Assonometria dell'ibrido con rappresentata l'area di scambio con il ledge "read out"

Resistenza della fibra di carbonio

NOTA: i pedici "inp" ed "outp" indica le resistenze termiche rispettivamente nel piano della lamina (in plane), e quelle attraverso lo spessore (out of plane).

R5 1.087 K watt = R5 gapair kair aledge⋅ :=

Resistenza di contatto tra ledge e frame R4 0.06 K watt = R4 sledge Aledge kAl⋅ := Aledge = 175.2 mm2 Aledge:=min A aledge

(

,

)

Resistenza dell'alluminio del ledge R3 2.854 K watt = R3 sglue1 kglue1 A⋅ :=

Resistenza della colla tra tubo e ledge R2 9.081×10−3 K watt = R2 spipe A kAl⋅ :=

Spessore del tubo di alluminio watt A htc50lh⋅

Temperatura ibrido 25 − ⋅K 1 8 i R_i

∑

= 1 ⋅ watt + =−9.396K

Temperatura media frame e silicio 25 − ⋅K 1 6 i R_i

∑

= 1 ⋅ watt + =−17.633K 1 6 i R_i

∑

= 7.367 K watt = i R_i

∑

15.604 K watt = R 0 3.358 9.081×10−3 2.854 0.06 1.087 4.512 3.725

















































K watt = R R₈ 3.725 K watt = R₈ shy kout_phy ahy⋅ := Resistenza dell'ibrido i:=1 8.. ORIGIN:=1 R₇ 1.667 K watt = R₇ sglue2 kglue2 ahy⋅ :=

Resistenza della colla tra l'ibrido e il frame

R₆ 4.512 K watt = R₆ 1 Rinp_cf 1 Rinp_hy +













1 − Routp_cf + :=

Per valutare la resistenza temica nel piano si è considerato il parallelo tra la fibra di carbonio e l'ibrido, a causa dei piccoli spessori in gioco rispetto alla distanza "d" (come mostrato in appendice E, oltre 8 mm di distanza il calore risente soprattutto della conducibilità nel piano dell'ibrido).

Routp_cf 1.998 K watt = Routp_cf scf kout_pCF Achips⋅ := Achips = 175.2 mm2 Achips:=max achips Aledge

(

,

)

Rinp_hy 19.493 K watt = Rinp_hy d kin_phy 45⋅ mm shy⋅ := Rinp_cf 2.886 K watt = Rinp_cf d kin_pCF 45⋅ mm scf⋅ :=

ahy:=300 mm⋅ 2

Area di metà ibrido _kair 0.023 watt m K⋅

⋅ :=

achips:=50mm2 Area sotto metà chips

kout_phy 34 watt m K⋅

⋅ :=

aledge :=200mm2 Area di contatto ledge-CF

kin_phy 34 watt m K⋅

⋅ :=

gapair:=5 10⋅ −3⋅mm Distanza media tra il frame e il ledge d:=20 mm⋅ Distanza tra i baricentri dell'area scaldata

e della superficie di contatto del ledge

********************************************************************************************** Calcoli

********************************************************************************************** Definiamo la resistenza termica come il gradiente, in °K, necessari per far passare la potenza di 1watt, ovvero la nostra potenza di riferimento.

Dal fluido al tubo di alluminio (strato limite)

A :=(b+2 h⋅ _{) lledge}⋅ A =175.2 mm2 superficie di contatto tubo-ledge

R1 1

A htc50lh⋅

:= R1 3.358 K

watt

=

Spessore del tubo di alluminio

R2 spipe A kAl⋅ := R2 9.081×10−3 K watt =

Resistenza della colla tra tubo e ledge R3 sglue1 kglue1 A⋅ := R3 2.854 K watt =

Calcoli esegiti con l'ibrido in allumina

si vedano i commenti e le figure riportate per il caso con ibrido in FR4

fatti sopra

Dati geometrici della struttura: Conducibilità termiche

spipe :=0.35mm kglue1 0.2watt m K⋅ := sledge :=2.3mm kglue2:=kglue1 b:=3.6mm Larghezza tubo dopo schiacciatura

h:=1.85mm Altezza tubo dopo schiacciatura

kkapton 0.2watt m K⋅

:=

lledge:=24mm Lunghezza cava di alloggiamento tubo nel ledge

kAl 220 watt m K⋅

⋅ :=

scf :=.7mm Spessore fibra di carbonio

shy:=0.38mm Spessore ibrido _{kin_pCF} 220 watt m K⋅

⋅ :=

skapton:=0.1mm Spessore kapton

kout_pCF 2 watt m K⋅

⋅ :=

sglue1:=0.1 mm⋅ Spessore colla tra tubo e ledge

htc50lh 1700 watt m2⋅K

⋅ :=

sglue2:=0.1mm

R8 0.037 K watt = R8 shy kout_phy ahy⋅ := Resistenza dell'ibrido R7 1.667 K watt = R7 sglue2 kglue2 ahy⋅ :=

Resistenza della colla tra l'ibrido e il frame

R6 4.66 K watt = R6 1 Rinp_cf 1 Rinp_hy +













1 − Routp_cf + := Routp_cf 1.998 K watt = Routp_cf scf kout_pCF Achips⋅ := Achips = 175.2 mm2 Achips:=max achips Aledge

(

,

)

Rinp_hy 34.4 K watt = Rinp_hy d kin_phy 45⋅ mm shy⋅ := Rinp_cf 2.886 K watt = Rinp_cf d kin_pCF 45⋅ mm scf⋅ :=

Resistenza della fibra di carbonio R5 1.087 K watt = R5 gapair kair aledge⋅ :=

Resistenza di contatto tra ledge e frame R4 0.06 K watt = R4 sledge Aledge kAl⋅ := Aledge = 175.2 mm2 Aledge:=min A aledge

(

,

)

R R 9.081×10−3 2.854 0.06 1.087 4.66 1.667 0.037













































K watt = i R_i

∑

13.731 K watt = 1 6 i R_i

∑

= 12.027 K watt = 25 − ⋅K 1 6 i R_i

∑

= 1 ⋅ watt + =−12.973K

Temperatura media frame e silicio

25 − ⋅K 1 8 i R_i

∑

= 1 ⋅ watt + =−11.269K Temperatura ibrido

Kp 0.05 watt m K⋅ ⋅ := rockwwool _Krw 0.047 watt m K⋅ ⋅ := aria _Ka 0.0273 watt m K⋅ ⋅ :=

Condizioni ambientali:

Temperatura dell'aria nella stanza: _Tamb:=290K Temperatura dentro la scatola: _Tbox :=281K Coefficienti di scambio termico convettivo:

esterno alla scatola: _hext 5 watt m2⋅K

⋅ :=

interno alla scatola: _hin 1 watt m2⋅K ⋅ := ∆T:=_{Tamb Tbox}− ∆T =9 K Aext :=a b⋅ ⋅2+a h⋅ ⋅2+b h⋅ ⋅2 Aud a _srw 1⋅2









− _sp 1⋅2 −









⋅





b−





srw1⋅2





−sp1⋅2





⋅2+





a−





srw1⋅2





−sp1⋅2





⋅





h−srw2−srw3−sp2−sp3





⋅2 := Alati b _srw 1⋅2









− _sp 1⋅2 −









⋅





h−srw2−srw3−sp2−sp3





⋅2 :=

Aint:=Aud Alati+

Calcolo della potenza entrante nella scatola

coibentata usata nei test di raffreddamento dei

moduli

Dati geometrici della scatola

:

ORIGIN:=1 Pianta: esterna a:=50cm b:=40cm Altezza: esterna h:=29cm Lati perimetro Spessori polistirolo: sp 5 2.5 7













⋅cm+2.5cm := Coperchio Fondo Spessori

rockwool: Lati perimetro srw 4 4 0.0001













⋅cm := Coperchio Fondo

Caratteristiche fisiche dei materiali:

Conducibilità termiche: polistirolo

Pp = 1.18 m Area media dello strato di polistirolo

del perimetro Ap1 Pp h srw2 − _srw 3 − sp2 sp3 + 2 −









⋅ :=

Area del coperchio e del fondo della rockwool _Ap

2:=





a−sp1−2 srw⋅ 1





⋅





b−sp1−2 srw⋅ 1





Ap₃:=Ap₂

HTC equivalente dello strato di polistirolo sui lati della scatola

perimetro coperchio fondo Hp Kp sp := _Hp 0.667 1 0.526













kg s3K = perimetro coperchio fondo Aext =0.922 m2 Ap 0.209 0.085 0.085













m 2 = _Arw 0.443 0.166 0.166













m 2 = Aint =0.184 m2 i:=1 3.. ki 1 Hp_i⋅Ap_i 1 Hrw_i⋅Arw_i + 1 hext Aext⋅ + 1 hin Aint⋅ + 1 Ka 1cm⋅Aint +

















1 − :=

Calcolo delle conducibilità termiche dei lati della scatola: Perimetro medio dello strato di rockwool

del perimetro Prw a _srw 1 −









⋅2+





b−srw1





⋅2 := _Prw = 1.64 m

Area media dello strato di rockwool

del perimetro Arw1 Prw h

srw 2 2 − srw3 2 −









⋅ :=

Area del coperchio e del fondo della rockwool _Arw

2:=





a−srw1





⋅





b−srw1





Arw₃:=Arw₂ HTC equivalente dello strato di rockwool

sui lati della scatola

perimetro coperchio fondo Hrw Krw srw := _Hrw 1.175 1.175 4.7×104

















watt m2⋅K = ***************************************** Perimetro medio dello strato di polistirolo

del perimetro Pp a _{2 srw} 1 ⋅ − _sp 1 −









⋅2+





b−2 srw⋅ 1−sp1





⋅2 :=

ki 0.06 0.041 0.033 1 Kwatt = perimetro coperchio fondo Q ∆T i ki

∑













⋅ :=

Q =1.203 watt

Potenza entrante nella scatola

i

ki

∑

0.134watt K

Tave Tw Tb

+

2

:=

proprietà dell'aria in funzione della temperatura ∆T _{Tw Tb}− →  := θ θ 180⋅π := Ppf 50 − 40 − 30 − 20 − 10 − 0 5 10 15 20 25 30 35 40 50 1.582 1.514 1.452 1.395 1.342 1.292 1.269 1.247 1.225 1.204 1.284 1.165 1.146 1.127 1.109 1.01 1.04 1.10 1.17 1.22 1.32 1.36 1.41 1.47 1.51 1.56 1.60 1.63 1.66 1.76 1.402 1.401 1.401 1.401 1.401 1.401 1.401 1.401 1.401 1.401 1.401 1.400 1.400 1.400 1.400

























































































:= Tave 273K− 24.95 − 22.95 − 20.45 − 17.95 − 14.95 − 12.45 −

































K = ∆T 0.1 3.9 8.9 13.9 19.9 24.9

































K = T0:=

(

Ppf〈 〉0 +273

)

⋅K ρ1:= Ppf〈 〉1 v1:=lspline T0

(

,ρ1

)

ρ( )T interp v1 T0

(

, ,ρ1,T

)

kg m3 ⋅ := ν2:=Ppf〈 〉2 v2:=lspline T0

(

,ν2

)

ν( )T interp v2 T0

(

, ,ν2,T

)

⋅10−5 m 2 s ⋅ := γ3:=Ppf〈 〉3 v3:=lspline T0

(

,γ3

)

γ( )T :=interp v3 T0

(

, ,γ3,T

)

Calcolo del coefficiente di scambio termico convettivo in

aria per le piastre

Stefano Linari 27/02/2003

Temperatura della parete scaldante °C

Tw 25 21 16 11 5 0

































− :=

Temperatura di bulk dell'aria °C Tb:=−24.9 Angolo della piastra rispetto alla verticale θ:=89 Lunghezza caratteristica del sistema l:=0.08m Spessore della cella convettiva δ:=0.02m

********************************************************************************************

Tw:=(Tw+273) K⋅ _Tb:=(Tb+273) K⋅ g 9.81m

s2

µ 1.619×10−4 1.626×10−4 1.632×10−4 1.632×10−4 1.631×10−4 1.631×10−4













































poise = µ:=

( )

ν ρ⋅→ Cp 1.005×103 1.005×103 1.005×103 1.005×103 1.005×103 1.005×103













































J kg K⋅ = Cp:=_{Cp Tave}

( )

ν 0.114 0.115 0.117 0.118 0.119 0.12

































stokes = ν:=ν

( )

_Tave Ka 0.022 0.023 0.023 0.023 0.023 0.023

































watt m K⋅ = Ka:=Ka Tave

( )

ρ 1.423 1.411 1.397 1.384 1.368 1.355

































kg m3 = ρ:=ρ

( )

_Tave β 4.031 10 3 − × 1 K = β 1 Tb := γ 1.401 1.401 1.401 1.401 1.401 1.401

































= γ:=γ

( )

_Tave Cp T( ) interp w2 T1( , ,Cp2,T) 1000J kg K⋅ ⋅ := w2:=lspline T1 Cp2( , ) Cp2:= Kaa〈 〉2 Ka T( ) interp w1 T1( , ,K1,T) watt m K⋅ ⋅ := w1:=lspline T1 K1( , ) K1:=Kaa〈 〉1 T1:=

(

Kaa〈 〉0 +273

)

⋅K Kaa 100 − 50 − 0 20 40 60 0.016 0.0204 0.0243 0.0257 0.0271 0.0285 1.009 1.005 1.005 1.005 1.005 1.009

































:=

proprietà dell'aria in funzione della temperatura

ORIGIN:=1 Tm:=Tave 273K− h_δ Nu_δ⋅_Ka

(

→

)

δ := Nu_δ 12.136 − 2.123 2.641 2.901 3.112 3.245

































= Nu_δ 1 1.44 1 1708 Ra_δ −













⋅ + Raδ 5830













1 3 1 −













+





















→  := hL NuL Ka ⋅

(

→

)

l := NuL 2.1 5.213 6.356 7.056 7.656 8.042

































=

Formula di Fujii -Imura NuL 0.56

(

_{GrL Pr}⋅

)

→  cos

( )

θ ⋅









1 4 ⋅ := RaL 1.133×104 4.302×105 9.511×105 1.444×106 2.002×106 2.437×106













































= RaL

(

GrL Pr⋅

)

→  := Ra_δ 177.102 6.721×103 1.486×104 2.256×104 3.128×104 3.807×104













































= Ra_δ

(

Gr_δ⋅Pr

)

→  := Gr_δ 244.247 9.293×103 2.065×104 3.16×104 4.429×104 5.433×104













































= Gr_δ g⋅ ∆β⋅ Tδ 3 ⋅ ν2 →  := GrL 1.563×104 5.948×105 1.322×106 2.022×106 2.835×106 3.477×106













































= GrL g⋅ ∆β⋅ Tl 3 ⋅ ν2 →  := Pr 0.723 0.72 0.714 0.706 0.701

































= Pr µ Ka⋅Cp













→  :=

hL 0.589 1.472 1.81 2.026 2.22 2.351

































watt m2⋅K

=

Piastra libera con inclinazione variabile

scaldante sotto

h_δ 13.612 − 2.397 3.008 3.332 3.61 3.794

































watt m2⋅K

=

Cella convettiva orizzontale

con piastra inferiore scaldante.

HL a_i g hL_i 0.56









4 ← hL i g 1 10 5 ⋅ kg 4 s12K4 ⋅ > g 1 10⋅ 11 kg 4 s12K4 ⋅ < ∧ if 0 otherwise ← i∈1 6.. for a :=

Il programma esegue il controllo dei valori calcolati in base ai parametri di validità delle correlazioni e in caso i risultati non siano soddisfacenti pone uguale a 0 il valore

corrispondente.

Il valore h finale che viene restituito è poi la media tra i valori ottenuti dalle due correlazioni.

Se per una data temperatura entrambi i valori calcolati sono uguali a 0 viene impostato un valore dato dalla sola

conduzione; la distanza di riferimento è stata presa 2cm. H_δ ai hδ i Raδi> 1700∧hδi> 0 Raδi 1 10 8 ⋅ < ∧ if 0 otherwise ← i∈1 6.. for a := h q_i 0.0273 0.02 kg s3K ⋅ _HL i= 0∧Hδi= 0 if HL_i H_δ i + 2 if HLi>0∧Hδi>0 HL_i HL_i>0 H_δ i= 0 ∧ if H_δ i if HLi= 0∧Hδi> 0 ← i∈1 6.. for q := 0.0273 0.02 = 1.365

HL 0 0 0 0 0





























= H_δ 2.397 3.008 3.332 3.61 3.794





























kg s3K = _h 2.397 3.008 3.332 3.61 3.794

































kg s3K = TmT =(−24.95 −22.95 −20.45 −17.95 −14.95 −12.45) K hT (1.365 2.39733 3.0077 3.33211 3.61004 3.79427) kg s3K = 24 22 20 18 16 14 0 2.5 5 7.5 10 12.5 15 17.5 20 22.5 25 h ∆T Tm

Andamento del coefficinete di scambio termico e della differenza di temperatura tra l'aria e la superficie di riferimento in funzione della loro temperatura media.

0.002 0.0067 0.0113 0.016 0.0207 0.0253 0.03 0

0.1 0.2 0.3

0.4 Conducibilità - distanza baricentri

metri W/°K hal x( ) hFR4 x( ) x hFR4 x( ) Kzz Az ⋅ s













1 − Kxx Ax⋅ x













1 − +









1 − := hal x( ) Kal Az ⋅ s













1 − Kal Ax⋅ x













1 − +









1 − :=

calcolo della conduttanza del FR4

calcolo della conduttanza dell'allumina

Area di passaggio del calore nello spessore dell'ibrido Ax= 2×10−5m2

Ax:=s 40⋅ mm

Spessore dell'ibrido s:=500 10⋅ −6m

Az:=3.85 10⋅ −4m2 Area di passaggio del calore in direzione _{ortogonale all'ibrido (impronta dei chips)}

Conducibilità termiche dell'FR4 nel piano ed ortogonalmente ad esso Kzz 0.34 watt m K⋅ ⋅ := Kxx 50watt m K⋅ :=

Conducibilità termica dell'allumina Kal 34watt

m K⋅

:=

proprietà fisiche e geometriche dell'ibrio elettronico:

Si calcola la resistenza termica che incontra il calore per attraversare l'ibrido da una faccia all'altra, in funzione della distanza sul piano dei due baricentri termici.

Si sommano separatamente i contributi lungo Z (ortogonale al piano) e lungo X (nel piano). L'ibrido è stato schematizzato a larghezza costante in direzione Y, pari a 45mm, e l'area di riferimento per il flusso lungo Z è quella dell'impronta dei chips.

Al di là dei valori numerici delle conducibilità, che possono essere molto approssimati per le ipotesi fatte sulle aree, riveste particolare importanza il rapporto R(x) che invece è insensibile a tali valori.

Calcolo della conducibilità termica dell'ibrido al

variare del materiale del substrato

hal x( )

Assonometria dell'ibrido con rappresentata l'area di scambio con il ledge "read out" 0.001 0.0092 0.0173 0.0255 0.0337 0.0418 0.05 0 0.5 1 1.5 R(distanza) metri R 0.00804m( ) = 1

CONCLUSIONE: per distanze superiore a 8 mm l'ibrido in FR4 ha una conducibilità

migliore in virtù degli effetti del rame della metallizzazione delle piste dei circuiti