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STEADY STATE HEAT CONDUCTION IN MULTI-DIMENSIONS
Now, to calculate the loading vector, we need to determine the relevant coefficients,
that is,
2πhT a l ij 147
= 120π (5.80)
6
Similarly,
2πql jk
= 1.8π (5.81)
6
Substituting the coefficients and other values into Equation 5.74, we obtain
3337.5
{f}= 2π 3879.5 (5.82)
− 18.5
5.7 Summary
In this chapter, an extension of the steady state heat conduction analysis to multi-dimensions
has been given. All commonly encountered approximations, namely, two-dimensional,
three-dimensional and axisymmetric, have been discussed. Most of the boundary conditions
have also been implemented and explained via examples. We trust the reader will appreciate
the difficulties associated with such multi-dimensional calculations and that the exercises
given in this chapter will prove useful for further understanding of multi-dimensional steady
state heat conduction.
5.8 Exercise
Exercise 5.8.1 A square plate of size 100 cm by 100 cm is subjected to an isothermal bound-
◦
ary condition of 500 C on the top and to a convection environment on all the remaining
2
◦
three sides of 100 C with a heat transfer coefficient of 10 W/m K. The thermal conduc-
2
tivity of the plate is 10 W/m K. Assume the thickness of the plate is 1 cm. Determine the
temperature distribution in the plate using (a) two triangles and (b) eight triangles. Cal-
culate the temperature and heat fluxes in the x and y directions at a location (x = 30 cm,
y = 30 cm).
3
Exercise 5.8.2 If in Exercise 5.8.1, there is a uniform heat generation of 2 W/cm , and
a line source of 5 W/cm at a location of (x = 30 cm and y = 30 cm) then, calculate the
new temperature distribution using (a) two triangles and (b) eight triangles. Calculate the
temperature at the location (x = 40 cm, y = 40 cm) and the heat fluxes in both the x and y
directions.
Exercise 5.8.3 Repeat Exercise 5.8.1 using (a) one rectangle (b) four rectangles.
Exercise 5.8.4 Repeat Exercise 5.8.2 using (a) one rectangle (b) four rectangles.