University of Rome – Tor Vergata Faculty of Engineering – Department of Industrial Engineering

“THERMODYNAMIC AND HEAT TRANSFER”

FIN DESIGN dr. G. Bovesecchi [email protected]

06-7259-7127 (7249)

Accademic Year 2012-2013 Tuesday, 11,June, 13

1

Fin Design Ex. 26 A square heat sink (l=5cm) is made of stainless steel (λ= 45 Wm-1K-1). The fin step is 10 mm and the thickness is 5mm. The core temperature should be lower than 60°C when the ambient temperature is 27°C. Design the fin if the profile is triangular and it dissipates 65W. The fin efficiency is given by the expression: Ω=

2 4 ( MH ) + 1 − 1 2

2h con M = λ ⋅t

Where: • t is the fin thickness; • h is the convective coefficient (8 Wm-2K-1). Assumption: • no contribution is given from the triangular surface; • the triangular diagonal (a) is equal to the height of the triangular (H). Tuesday, 11,June, 13

2

Fin Design

a

L

a=L

LH

5 cm

L 5 mm 5 cm

5 mm

l

!

Solution The first step is to calculate the number of fin in the heat sink. 3 Tuesday, 11,June, 13

Fin Design l 50 mm n= = =5 s 10 mm The heat flow is the sum of two terms, one due to each fin and on due to the empty space between the fin. Q = Q tot + Q tot tot

fin

e.s.

but the heat sink has 5 fin: Q tot = 5⋅ Q fin + Q e.s.

(

)

from this equation we can derive the heat flow of each fin, this value is what we need to design the fin.  Q Q fin = tot − Q e.s. 5 4 Tuesday, 11,June, 13

Fin Design the heat flow dissipated by the empty space is:

Q e.s. = h ⋅ Ae.s. ⋅ ΔTe.s.

(

)

Ae.s. = ( s − t ) ⋅ L = 10 ⋅10−3 m− 5⋅10−3 m ⋅5⋅10−2 m= 2.5⋅10−4 m2 Q e.s. = 8W ⋅ m−2 ⋅ K −1 ⋅ 2.5⋅10−4 m2 ⋅ ( 60 − 27 ) °C = 6.6 ⋅10−2 W now we can calculate the heat flow dissipated by each fin:  Q 65W tot   Q fin = − Q e.s. = − 6.6 ⋅10−2 W = 12.93W 5 5 we can write the fin heat flow as:

Q fin = Ω ⋅ h ⋅ A fin ⋅ ΔT fin Tuesday, 11,June, 13

5

Fin Design Ω=

Ω ⋅ A fin =

Q fin h ⋅ ΔT fin

2

2h M= λ ⋅t

4 ( MH ) + 1 − 1 2

12.93W 2 = = 0.049 m 8W ⋅ m−2 ⋅ K −1 ⋅ ( 60 − 27 ) °C

−2 −1 2h 2 ⋅8W ⋅ m ⋅ K −2 M2 = = = 71.1m λ t 45W ⋅ m−1 ⋅ K −1 ⋅5⋅10−3 m

Using the two assumptions the surface of one fin is:

(

A fin = 2 ⋅ ( L ⋅ H ) = 2 ⋅ H ⋅5⋅10−2 m Ω ⋅ A fin = Tuesday, 11,June, 13

2 ⋅ A fin 4 ( MH ) + 1 − 1 2

=

(

)

2 ⋅ 2 ⋅ H ⋅5⋅10−2 m 4M 2H 2 +1 −1

) = 0.049m

2 6

Fin Design Ω ⋅ A fin = =

(

4 ⋅ H ⋅5⋅10−2 m

) = 0.049m

2

4M H + 1 − 1 2

H ⋅0.2 m 4 ⋅71.1m2 H 2 + 1 − 1 H = 0.245m

(

2

=

=

H 2841.4m2 H 2 + 1 − 1

= 0.245m

)

2841.4m2 H 2 + 1 − 1

H + 0.245= 0.245m 2841.4m2 H 2 + 1 H 2 + 0.49 ⋅ H + 0.06 = 17.06H 2 + 0.06 16.06 ⋅ H 2 + 0.49 ⋅ H = 0 H =0 Tuesday, 11,June, 13

0.490 H= = 3.05cm 16.06

7

University of Rome – Tor Vergata Faculty of Engineering – Department of Industrial Engineering

“THERMODYNAMIC AND HEAT TRANSFER”

FIN EFFICIENCY dr. G. Bovesecchi [email protected]

06-7259-7127 (7249)

Accademic Year 2012-2013 Tuesday, 11,June, 13

1

Fin Efficiency Ex. 27 A 120x50mm2 heat sink (see figure) is made of stainless steel (λ= 45 W/m K). The fin thickness is 5mm and the step is 10mm. The core temperature is 75°C. The heat sink is cooled by an air mass flow rate at 21°C (h = 37 W/m2K). Calculate the heat flow if the fin profile is square (A), triangular (C) and curved (D) using the attached graphic. Assumption: To calculate the curved surface of the fin consider the profile as it is triangular.

2 Tuesday, 11,June, 13

40 mm

Fin Efficiency

mm 0 12 5 mm

5 mm

!

50 mm

3 Tuesday, 11,June, 13

Fin Efficiency

4 Tuesday, 11,June, 13

Fin Efficiency Solution In this exercise we must calculate the heat flow dissipated by the heat sink in 3 different configuration of the heat sink (3 different profile). For all the profile we can write:

Q tot = Q tot + Q tot fin

Q step

e.s.

Q tot  = = Q fin + Q e.s. n

Q e.s. = h ⋅ Ae.s. ⋅ ΔTe.s. Q fin = Ω ⋅ h ⋅ A fin ⋅ ΔT fin 5 Tuesday, 11,June, 13

Fin Efficiency to compare the different profile an a-dimensional parameter is defined:

2h L' = L λt −2 −1 2 ⋅37W ⋅ m ⋅ K L′ = 40 ⋅10−3 ⋅ m ⋅ = 0.725 −1 −1 −3 45W ⋅ m ⋅ K ⋅5⋅10 m

this parameter let me to evaluate the efficiency of the 3 fin profile:

6 Tuesday, 11,June, 13

Fin Efficiency

0.84 0.82 0.74

7 Tuesday, 11,June, 13

Fin Efficiency to compare the different profile an a-dimensional parameter is defined:

2h L' = L λt −2 −1 2 ⋅37W ⋅ m ⋅ K L′ = 40 ⋅10−3 ⋅ m ⋅ = 0.725 −1 −1 −3 45W ⋅ m ⋅ K ⋅5⋅10 m

this parameter let me to evaluate the efficiency of the 3 fin profile:

η A = 0.74 ηC = 0.82 η D = 0.84 Tuesday, 11,June, 13

8

Fin Efficiency The number of fin the heat sink is:

l 50 mm n= = =5 s 10 mm the heat dissipated through the empty spaces between the fins is the same for all the profile, the surface is:

(

)

Ae.s. = ( s − t ) ⋅ L = 10 ⋅10−3 m− 5⋅10−3 m ⋅120 ⋅10−3 m= 6 ⋅10−4 m2 Now we must divide the problem for each fin profile: Profile A

(

)

A A fin = P ⋅ H = 2 ⋅ 5⋅10−3 m+ 120 ⋅10−3 m ⋅ 40 ⋅10−3 m= 0.01m2 9 Tuesday, 11,June, 13

Fin Efficiency Profile C 2

⎛t⎞ r = ⎜ ⎟ + H2 = ⎝ 2⎠

(

−3

) ( 2

−3

)

2

2.5⋅10 m + 40 ⋅10 m = 40.1⋅10−3 m

⎡ ⎛t⋅H⎞⎤ A = 2 ⋅ ⎢( r ⋅ L ) + ⎜ ⎥= ⎟ ⎝ 2 ⎠⎦ ⎣ C fin

−3 −3 ⎡ ⎛ 5⋅10 m ⋅ 40 ⋅10 m⎞ ⎤ C −3 −3 A fin = 2 ⋅ ⎢ 40.1⋅10 m ⋅120 ⋅10 m + ⎜ ⎥= ⎟ 2 ⎝ ⎠⎦ ⎣

(

)

ACfin = 9.7 ⋅10−3 m2 10 Tuesday, 11,June, 13

Fin Efficiency Profile D D ACfin = A fin

Now we can calculate the heat flow for each profile, the heat flow of the empty spaces is the same for each profile.

Q e.s. = h ⋅ Ae.s. ⋅ ΔTe.s. = 37W ⋅ m−2 ⋅ K −1 ⋅6 ⋅10−4 m2 ⋅ ( 75 − 21) K = 1.2W tot Q e.s. = 5⋅ Q e.s. = 5⋅1.2W = 6W

11 Tuesday, 11,June, 13

Fin Efficiency Now we can calculate the total heat flow for the 3 profiles: Profile A Q fin = η ⋅ h ⋅ A fin ⋅ ΔT fin = 0.74 ⋅37W ⋅ m−2 ⋅ K −1 ⋅0.01m2 ⋅ ( 75 − 21) K = 14.8W

 = 5⋅14.8W = 74W Q tot = 5⋅ Q fin fin  tot = 74W + 6W = 80W Q totA = Q tot + Q fin e.s. Profile C Q fin = 0.82 ⋅37W ⋅ m−2 ⋅ K −1 ⋅9.7 ⋅10−3 m2 ⋅ ( 75 − 21) K = 15.9W

 = 5⋅16.3W = 79.5W Q tot = 5⋅ Q fin fin  tot = 79.5W + 6W = 85.5W Q totA = Q tot + Q fin e.s. 12 Tuesday, 11,June, 13

Fin Efficiency Profile D Q fin = 0.84 ⋅37W ⋅ m−2 ⋅ K −1 ⋅9.7 ⋅10−3 m2 ⋅ ( 75 − 21) K = 16.3W

Q tot = 5⋅ Q fin = 5⋅16.3W = 81.5W fin  tot = 81.5W + 6W = 87.5W Q totA = Q tot + Q fin e.s.

13 Tuesday, 11,June, 13

University of Rome – Tor Vergata Faculty of Engineering – Department of Industrial Engineering

“THERMODYNAMIC AND HEAT TRANSFER”

FIN EXTRA EXERCISE #1 dr. G. Bovesecchi [email protected]

06-7259-7127 (7249)

Accademic Year 2012-2013 Tuesday, 11,June, 13

1

Fin Efficiency Ex. 28 Calculate the core temperature of a stainless steel surface (100x250 mm2), cooled by an air mass flow rate (27°C and 4m/ s) that dissipate a thermal power of 45W. Then consider a stainless steel heat sink (λ=45 W/m K) (see figure) in the same condition and calculate the temperature at the fin base. Use the empirical correlation to evaluate the convective coefficient:

NuL1 = 0.102 ⋅ Re0.675 ⋅ Pr1/3 The thermo-physical properties of air at 27°C are: υ = 15.68⋅10−6 m2 ⋅ s −1 λ = 0.026W ⋅ m−1 ⋅ K −1 Pr = 0.708 Tuesday, 11,June, 13

2

Fin Efficiency

3 Tuesday, 11,June, 13

Fin Efficiency

4 Tuesday, 11,June, 13

Fin Efficiency Solution In this exercise we have to calculate the core temperature in two cases, one without fin and one with. The first step is to calculate the convective coefficient from the empirical Nusselt correlation: 0.675 ⎛ u ⋅ L1 ⎞ 0.675 1/3 1/3 NuL1 = 0.102 ⋅ Re ⋅ Pr = 0.102 ⋅ ⎜ ⋅ Pr ⎝ υ ⎟⎠

u ⋅ L1 4 m ⋅ s −1 ⋅100⋅10−3 m Re L1 = = = 25510 −6 2 −1 υ 15.68⋅10 m ⋅ s NuL1 = 0.102 ⋅ ( 25510 )

0.675

h ⋅ L1 NuL1 = λ Tuesday, 11,June, 13

⋅ ( 0.708) = 85.73 1/3

5

Fin Efficiency NuL1 ⋅ λ 85.73⋅0.026W ⋅ m−1 ⋅ K −1 −2 −1 h= = = 22.29W ⋅ m ⋅ K L1 100 ⋅10−3 m Without Fin

Q = h ⋅ Asur ⋅ ΔTsur ΔTsur

Q sur 45W = = = 80.8°C −2 −1 −3 −3 2 h ⋅ Asur 22.29W ⋅ m ⋅ K ⋅ 100 ⋅10 ⋅ 250 ⋅10 m

(

)

Tcore = Tamb + ΔTsur = 27°C + 80.8°C = 107.8°C

6 Tuesday, 11,June, 13

Fin Efficiency With Fin

2⋅h L = 10 ⋅10 − 3m−1 λ ⋅t

(

)

2 ⋅ 22.29W ⋅ m−2 ⋅ K −1 = 0.141 −1 −3 45W ⋅ m ⋅ K ⋅5⋅10 m

7 Tuesday, 11,June, 13

Fin Efficiency

8 Tuesday, 11,June, 13

Fin Efficiency With Fin

2⋅h L = 10 ⋅10 − 3m−1 λ ⋅t

(

)

2 ⋅ 22.29W ⋅ m−2 ⋅ K −1 = 0.141 −1 −3 45W ⋅ m ⋅ K ⋅5⋅10 m

η = 0.95

Q = Q fin + Q e.s. Q = n ⋅Ω ⋅ h ⋅ A fin ⋅ ΔT fin + n ⋅ h ⋅ Ae.s. ⋅ ΔTe.s. L1 100 mm n= = = 10 s 10 mm ⎛ t ⋅ h⎞ A fin = 2 ⋅ ⎜ + 2 ⋅ ( L2 ⋅ a ) ⎟ ⎝ 2 ⎠ Tuesday, 11,June, 13

9

Fin Efficiency 2

2

⎛ 5⋅10 m ⎞ ⎛t⎞ 2 −3 a= ⎜ ⎟ +h = ⎜ + 10 ⋅10 m ⎟ 2 ⎝ 2⎠ ⎝ ⎠ −3

(

)

(

( m ⋅ (100 ⋅10

)

2

= 10.3⋅10−3 m

)

A fin = 5⋅10−3 ⋅100 ⋅10−3 m2 + 2 ⋅ 250 ⋅10−3 ⋅10.3⋅10−3 m2 = 5.7 ⋅10−3 m2 Ae.s. = L2 ⋅ ( L1 − n ⋅t ) = 250 ⋅10−3

−3

)

− 10 ⋅5⋅10−3 m = 1.25⋅10−2 m2

Q ΔT = n ⋅Ω ⋅ h ⋅ A fin + n ⋅ h ⋅ Ae.s. ∆T =

−2

−1

(

45W −3

−3

22.29W ⋅ m ⋅ K ⋅ 10 ⋅0.95⋅5.7 ⋅10 m + 10 ⋅1.25⋅10 m 2

2

)

Tcore = Tamb + ΔT = 27°C + 30.3°C = 57.3°C Tuesday, 11,June, 13

= 30.3°C

10