3. 2&3-D, STEADY STATE CONDUCTION
Incropera chapter 3, Cengel chapter 3 p 163-184, Welty et al. p 233- 244
THIS WEEK:
What if conduction cannot be assumed one-dimensional?
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Common examples
o Fins – used to increase rate of heat transfer
o 2D/3D problems with common geometries
Introduce concept of shape factors for conduction problems
Can YOU spot the difference between a 1D and a 2/3D
Learning objectives: By the end of week 3 you should be able to:
Define and determine fin efficiency, fin effectiveness, maximum possible heat loss from a fin, and heat loss in absence of a fin.
Identify and clearly justify whether a given problem should be analysed as 1-, 2- or 3-d conduction.
Explain the different strategies which can be used to solve 2 or 3-d conduction problems
Apply the shape factor method to solve heat transfer problems of common 2- and 3-d configurations, including resistances in series.
DEFINITION
A fin is a surface that extends from an object thereby extending the area of heat transfer to maximize heat flow by conduction and convection. Common fin geometries include:
Rectangular;
Annular (eg on pipes);
Plate fins; and
Triangular fins.
Fins are used where heat transfer is difficult (eg to gases), that is, where h is low. They are rarely worthwhile if h is large (eg in boiling liquids, or high fluid velocities).
Assumptions
1-d conduction, ignore fin thickness (it is thin);
Steady state heat transfer;
Neglect radiation;
Constant k value; and
Constant h value over the fin.
Typical boundary conditions from end
Convection from end
P = fin cross-section perimeter (m)
Ac = fin cross-section area (m2)
L = length of fin from base (m)
Tb = temperature at the base of the fin (K)
EFFECTIVENESS OF FIN:
∑ = heat loss from fin ˃˃1 heat loss w. no fin
EFFICIENCY OF FIN:
η = heat loss from fin
max. heat loss possible from fin
Equation H-3
Equation H-1
Equation H-2
Maximum heat loss is when resistance due to conduction along fin is negligible compared with the resistance due to convection.
Table 3-3(Çengel and Ghajar 2011)
Fig 3-43, 3-44 (Çengel and Ghajar 2011)
A hot surface at 100 oC is to be cooled by attaching 3 cm long pin fins, which are 0.25 cm in diameter, and made of aluminium (alloy 2024-T6). The centre of the fins are spaced 0.6 cm apart. The temperature of the surrounding air is 30 oC, and the heat transfer coefficient from the fins is 35 Wm-2K-1.
a) Determine the rate of heat transfer from a 1 m x 1 m section of the plate in absence of fins
b) Determine the maximum possible heat transfer of a single fin
c) Determine the efficiency of a single fin
d) Determine the heat loss from the fin
e) Determine the total rate of heat loss from all fins
f) Determine the total rate of heat loss from the finned surface.
Are the fins useful?
EQUATION H-1
qfin qfin
hAT T nofin cb
EFFICIENCY OF FIN (EQUATION H-2 & EQUATION H-3):
qfin qmax
Refers to Table 3-3 for Afin
qfin hAfinTb T
1.2 2- AND 3-D CONDUCTION
For example,
HOW TO DEAL WITH 2- AND 3-D CONDUCTION
1. Simplify: model as 1-d system 2. Solve analytically
3. Numerical solutions
4. Graphical solutions
5. Shape factors
SHAPE FACTORS
For example, from 1-d conduction, shape factor can be used to determine rate of heat loss through a pipe wall:
q kAdT dr
Integration from r1 to r2 yields (as shown last week):
Equation H-4
12 q 2kl T T
r , ln 2 r
This can be written in a different form:
Where S is a shape factor.
In this example, S=
Units of S:
A small cubicle furnace (50 x 50 x 50 cm3 on the inside) is constructed of fireclay brick (k = 1.04 W/(mK)) with a wall thickness of 10 cm. The inside of the furnace is maintained at 500C and the outside is maintained at 50C. Calculate the heat loss through the walls.
A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75C, and
the earth surface temperature is 5C. Assuming that the thermal conductivity of the earth is 0.8 W/(mK), calculate the heat loss by the pipe.
2, 3-D STEADY STATE CONDUCTION
2T 2T 2T
k x2 y2 z2 0
1. Simplify: model as 1-d system
2. Solve analytical solutions
3. Numerical solutions
4. Graphical solutions
5. Shape factors
From integrating area and path over a 2 or 3-d problem. 6.
EXAMPLE A: SPHERE BURIED BENEATH THE GROUND
A spherical container containing radioactive material, is buried beneath the ground. The container is 2m in diameter, with the centre placed 10 m below the surface. Radioactive decay of the contents generates heat at a rate of 900 W. The soil is 20oC. What is the temperature at the surface of the sphere?
Assumptions:
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