fabiana de fátima giacomini - PG-Mec Programa de Pós-Graduação ...

119
a = 0
(B.5)
w
a e
= 1 (B.6)
∆x
a
p
= a
e
2
+
∆x
(B.7)
b
p
2 xPC
( C e )
C
( e −1)
2
= φ
a
− ∆xP
(B.8)
∆x
• para os volumes de controle entre P = 2 a P = N −1:
a w
= 1 (B.9)
∆x
a e
= 1 (B.10)
∆x
a = a + a
(B.11)
p
w
e
b
p
= −
2 xPC
( C e )
C
( e −1)
∆x
P
(B.12)
• para o volume de controle
P = N :
a w
= 1 (B.13)
∆x
a = 0
(B.14)
e
a
p
= a
w
2
+
∆x
(B.15)
b
p
2 xPC
( C e )
C
( e −1)
2
= φ
b
− ∆xP
(B.16)
∆x