Page 131 - Fundamentals of The Finite Element Method for Heat and Fluid Flow

STEADY STATE HEAT CONDUCTION IN ONE DIMENSION
Assembly of the above equations results in

0.109 −0.103 0.0     0.148  123
T 1  
−0.103 0.187 −0.073 = 0.145 + 0.13 (4.98)
  T 2
0.0 −0.079 0.079    0.137 
T 3
On applying the relevant boundary conditions and solving the above system, we obtain
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T 1 = 100 C, T 2 = 85.39 C and T 3 = 83.52 C.
The heat dissipation can be calculated from the following relationship:
T i + T j
Q = 2 − T a (4.99)
e=1 hP e L e
2
Substituting the contribution from both elements results in a value of Q = 1.38 W.
4.5 Summary
In this chapter, examples of one-dimensional problems have been discussed in detail. In
most cases, analytical solutions were available as benchmarks for the finite element solu-
tions. There are many other application problems, which can be studied in one dimension.
However, the essential fundamentals of the finite element method for one-dimensional heat
conduction problems have been given, which may easily be extended to other forms of
one-dimensional heat conduction problems.

4.6 Exercise

Exercise 4.6.1 A composite wall with three different layers, as shown in Figure 4.2 gen-
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erates 0.25 GW/m of heat. Using the relevant data given in Example 4.2.1, determine
the temperature distribution across the wall using both linear and quadratic variations and
compare the results.

Exercise 4.6.2 An insulation system around a cylindrical pipe consists of two different lay-
ers. The first layer immediately on the outer surface of the pipe is made of glass wool and the
second one is constructed using plaster of Paris. The cylinder diameter is 10 cm and each
insulating layer is 1 cm thick. The thermal conductivity of the glass wool is 0.04 W/m C and
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that of the plaster is 0.06 W/m C. The cylinder carries hot oil at a temperature of 92 C,
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and the atmospheric temperature outside is 15 C. If the heat transfer coefficient from the
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outer surface of the insulation to the atmosphere is 15 W/m 2 ◦ C, calculate the temperature
at the interface between the two insulating materials and on the outer surface.
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Exercise 4.6.3 A solid cylinder of 10 cm diameter generates 0.3 GW/m of heat due to
nuclear reaction. If the outside temperature is 40 C and the heat transfer coefficient from the
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solid surface to the surrounding fluid is 30 W/m 2 ◦ C, calculate the temperature distribution
using quadratic elements. Assume a thermal conductivity of 15 W/m C.
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