ME3407 Heat Transfer Coursework: Assignment Solution 2019/2020

This document presents a detailed solution to a mechanical engineering heat transfer assignment. The assignment covers the analysis of heat transfer in a composite wall, including calculations for thermal resistances, heat loss, and temperature distribution. It also involves the analysis of heat transfer from a CPU with attached rectangular fins, calculating heat transfer rates from fins and bare surfaces, and determining overall surface efficiency. The solution includes step-by-step calculations, formulas, and considerations for convection and conduction heat transfer. Furthermore, the assignment explores the application of these principles to practical engineering problems, providing a comprehensive understanding of heat transfer concepts and their application in real-world scenarios. The assignment is well-structured and provides a thorough understanding of the subject matter.

Mechanical Engineering Heat Transfer 1
MECHANICAL ENGINEERING HEAT TRANSFER
Name of student
Institution
Date

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Q1
i)
Beryllium oxide > liquid side >gas side

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ii)

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Q2
The heat transfer from 1 single fin is given by:
where
 h = heat transfer coefficient = 20 W/m2K (given)
 P = Perimeter of the base of the fin
 A(base) = area of the base
 k = thermal conductivity of the fin material = 220 W/mK
 = Temperature gradient


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 Lx = equivalent length = L + (thickness / 2) (for short fins with conducting ends)
Perimeter = ?
Rectangular fin = 2 x (L+B) = 2 x (40 + 2) = 84 mm = 0.084 m
A(base) = L x B = 40 x 2 = 80 mm2
= 90 - 20 = 70 C (given)
Thus, putting all the values in the above formula and calculating heat transfer FOR 1 FIN.
Q=1.5206 W
There are 8 such fins, thus 8 x 1.5206 W = 12.1648 W
This is the amount of heat that is conducted from the fins. Now, the heat sink also has bare surfaces
BETWEEN the fins. The heat sink conducts heat through these plane surfaces also.
Since the thickkness of 1 fin = 2 mm. And there are 8 such fins, thus a total of 16 mm x 40 mm = 640
mm2 is covered by the fins. The heat sink's total area is 40 mm x 40 mm = 1600 mm2. Thus, the
remaining 1600 mm2 - 640 mm2= 960 mm2 is left bare. Heat also conducts from this 960 mm2 area
through convection.
Q(convection) = h x Area x
Thus, Q = 20 x 960 e-6 x 70 = 1.344 W
Thus, a total of 12.1648 + 1.344 W = 13.5088 W is lost from the heat sink.
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To calculate overall surface efficiency, we need to assume that the entire surface is at the same
temperature.
Thus, we need to calculate the total surface area of this heat sink. For this, we will start with one
single fin (see the above attached 2 diagrams). The lentgh L = 40 mm, thickness = 2 mm and height
H= 12 mm.
For one fin, surface area = 2 x (L x H) + (L x T) + 2 x (H x T)
Thus, surface area = 2 x (40 x 12) + (80) + 2 x (24)= 1088 mm2.
There are 8 fins, thus 8 x 1040 = 8704 mm2
Also, we have already calculated the bare left area of the heat sink, that is 960 mm2.

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Thus, total area available for heat transfer is 9664 mm2.
Thus, heat transfer = h x Area x
Q = 20 x 9664 e-6 x 70
Q = 13.5296 W
Efficiency = 99.84%.

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Q3

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Q4

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