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Consider two aluminum (k=180 W/mK) pin-fin heat sinks with 1 mm and 2 mm square pins both 25 mm long. Assume that the gap between the pins = side of the pin, ie the one mm pin-fin heat sink has 1 mm gap between its pins. Assuming the module base is 50 mm x 50 mm and the heat sink cannot overhang due to lack of space, and h = 40 W/m?K with jet impingement, calculate:   (a) Cooling surface area ratio (1 mm heat sink/ 2 mm heat sink)   (b) Heat transfer ratio (1 mm heat sink/2 mm heat sink) Try to fit the maximum number of pins on the 50 mm x 50 mm base

Consider two aluminum (k=180 W/mK) pin-fin heat sinks with 1 mm and 2 mm square pins both 25 mm long. Assume that the gap between the pins = side of the pin, ie the one mm pin-fin heat sink has 1 mm gap between its pins. Assuming the module base is 50 mm x 50 mm and the heat sink cannot overhang due to lack of space, and h = 40 W/m?K with jet impingement, calculate:   (a) Cooling surface area ratio (1 mm heat sink/ 2 mm heat sink)   (b) Heat transfer ratio (1 mm heat sink/2 mm heat sink) Try to fit the maximum number of pins on the 50 mm x 50 mm base

Elements Of Electromagnetics
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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Consider two aluminum (k=180 W/mK) pin-fin heat sinks with 1 mm and 2 mm square pins both 25
mm long. Assume that the gap between the pins = side of the pin, ie the one mm pin-fin heat sink has 1
mm gap between its pins. Assuming the module base is 50 mm x 50 mm and the heat sink cannot
overhang due to lack of space, and h = 40 W/m?K with jet impingement, calculate:
 
(a) Cooling surface area ratio (1 mm heat sink/ 2 mm heat sink)
 
(b) Heat transfer ratio (1 mm heat sink/2 mm heat sink)
Try to fit the maximum number of pins on the 50 mm x 50 mm base.
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