56
R. S. Lebelo and O. D. Makinde
s
2 = −
1
r
∂s 1
∂r
− α(1 + εs 1 )
m e
[s 1 /(1+εs 1 )]
− αω(1 + εs 1 )
m e
[εs 1 /(1+εs 1 )]
+ ϕs 1 ,
(5.13)
s 2 (0) = 0, s 1 (1) = 0.
(5.14)
The solutions are illustrated in Figs. 5.7 and 5.8 showing the steady state situation
as t → ∞. It is noted that high values of the temperature are experienced at the
center of the cylinder and that the values decrease towards the surface thereof, due
to convective heat loss. Figure 5.7 demonstrates temperature distribution in a twodimensional domain, whereas Fig. 5.8 shows a three-dimensional one.
Fig. 5.7 Steady state 2-D
t → ∞
Fig. 5.8 Steady state 3-D
t → ∞
Précédent

- 64/290

Suivant