This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: D AB ∇ c AD BA ∇ (1c A ) = 0 ⇒ D AB ∇ c A + D BA ∇ c A = 0 From ( a ) above, D AB = D BA soD AB ∇ c A + D AB ∇ c A vanishes identically. ± (c) N A + N B = c V. Proof.cD AB ~ ∇ y A + y A ( N A + N B )cD BA ~ ∇ y B + y B ( N A + N B ) Substituting 1y A for y B yieldscD AB ~ ∇ y A + y A ( N A + N B )cD BA ~ ∇ (1y A ) + (1y A )( N A + N B ) Cancellation ensues betweencD AB ~ ∇ y A and y A ( N A + N B ), leaving ( N A + N B ) Where N A = c A v A and N B = c B v B , or c A v A + c B v B Which can be compacted to n X i =1 c i v i And, by deﬁnition, means that c A v A + c B v B = N A + N B = c V ±...
View
Full Document
 Spring '10
 Bell
 Chemical Engineering, Mass Transfer, Substitute good, dz dz dz, ARJAN SINGH, dyB dyB dyB

Click to edit the document details