You are designing a pin-style heatsink for forced cooling of an IC. A small fan blows RT air into the heatsink at about 2.3 ? ? . The aluminum pins are 2mm in diameter and have a constant surface temperature of 120°?. The heatsink is 100mm x 100mm and the pins have an inline arrangement with a transverse pitch of 4mm and a longitudinal pitch of 6mm. What is the total heat dissipation of this heatsink? The pins are 30mm high. Note RT=25 celsius ArrangementRange of RepCorrelation0-1000.25PrNu = 0.9Reg4P.0.36Prs100-1,0000.25P.Nu =0.52Reg5p,0.36PrsIn-Line1,000 - 2 x 1050.250.27Reg63Pr0.63 p0.36NuPrs2 x 105 - 2 x 1060.25PrNu = 0.033Reg*p.4Pr0-5000.251.04RE§4P0.36PreNu0.25100-1,000Pr0.71Reg5p,0.36P.NuStaggered0.25Pr0.21,000 - 2 x 105Nu = 0.35Reg6p,0.36Pre2 x 105 - 2 x 1060.20.25(ST0.031PrReg®p.0.36Nu =SLP.T's MaterialDensitySpecificLatent HeatLatent HeatThermalHeat (or)(kgof Fusion,ofConductivityWkgK,Vaporizationkg\mK)Air1.2257180.025Aluminum2712900396000200Ammonia68244001371000(Liquid)Cast Iron7300460247000408940390СopperЕроху206000400120010001.5G1018006700.3FiberglassGlass22007501Silicone107013000.2Steel785045024700020

GIVEN:v=2.3 m/sRe=V.Dν=2.3×2×10-31.56×10-5=294.8718

Answered: ou are designing a pin-style heatsink…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

You are designing a pin-style heatsink for forced cooling of an IC. A small fan blows RT air into the heatsink at about 2.3 ? ? . The aluminum pins are 2mm in diameter and have a constant surface temperature of 120°?. The heatsink is 100mm x 100mm and the pins have an inline arrangement with a transverse pitch of 4mm and a longitudinal pitch of 6mm. What is the total heat dissipation of this heatsink? The pins are 30mm high. Note RT=25 celsius

Arrangement
Range of Rep
Correlation
0-100
0.25
Pr
Nu = 0.9Reg4P.0.36
Prs
100-1,000
0.25
P.
Nu =
0.52Reg5p,0.36
Prs
In-Line
1,000 - 2 x 105
0.25
0.27Reg63
Pr
0.63 p0.36
Nu
Prs
2 x 105 - 2 x 106
0.25
Pr
Nu = 0.033Reg*p.4
Pr
0-500
0.25
1.04RE§4P0.36
Pre
Nu
0.25
100-1,000
Pr
0.71Reg5p,0.36
P.
Nu
Staggered
0.25
Pr
0.2
1,000 - 2 x 105
Nu = 0.35
Reg6p,0.36
Pre
2 x 105 - 2 x 106
0.2
0.25
(ST
0.031
Pr
Reg®p.0.36
Nu =
SL
P.
T's

Material
Density
Specific
Latent Heat
Latent Heat
Thermal
Heat (or)
(kg
of Fusion,
of
Conductivity
W
kgK,
Vaporization
kg
\mK)
Air
1.225
718
0.025
Aluminum
2712
900
396000
200
Ammonia
682
4400
1371000
(Liquid)
Cast Iron
7300
460
247000
40
8940
390
Сopper
Ероху
206000
400
1200
1000
1.5
G10
1800
670
0.3
Fiberglass
Glass
2200
750
1
Silicone
1070
1300
0.2
Steel
7850
450
247000
20

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