2&3-D, Steady State Conduction
Incropera chapter 3, Cengel chapter 3 p 163-184, Welty et al. p 233-244
This week:
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· What if conduction cannot be assumed one-dimensional?
· Common examples
· Fins – used to increase rate of heat transfer
· 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 problem?
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
Equation H-1
Efficiency of fin:
η = heat loss from fin
max. heat loss possible from fin
Equation H-2
Equation H-3
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 with fins
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
Efficiency of fin (Equation H-2 & Equation H-3):
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:
Integration from r1 to r2 yields (as shown last week):
This can be written in a different form:
Equation H-4
Where S is a shape factor.
In this example, S=
Units of S:
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.
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