CHE150B PSET1 (Incomplete) - -D AB ∇ c A-D BA ∇(1-c A =...

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CHEMICAL ENGINEERING 150B PROBLEM SET #1 ARJAN SINGH PUNIANI Problem 1 (a) D AB = D BA . Proof. Assume D AB 6 = D BA . Start with (1) N Az = - cD AB dy A dz + y A ( N Az + N Bz ) Keeping in mind that (2) N Bz = - cD BA dy B dz + y B ( N Bz + N Az ) Since we are dealing with a binary mixture, we can extract N Bz by substituting 1 - y B into the N Az equation above. Thus: N Az = - cD AB d (1 - y B ) dz + (1 - y B )( N Az + N Bz ) Simplifying to N Az = - cD AB - dy B dz + ( N Az + N Bz ) - y B ( N Az + N Bz ) We can eliminate a pair of N Az s, leaving 0 = cD AB dy B dz + N Bz - y B ( N Az + N Bz ) (3) N Bz = - cD AB dy B dz + y B ( N Az + N Bz ) Equating (2) and (3), we find that: - cD BA dy B dz + y B ( N Bz + N Az ) = - cD AB dy B dz + y B ( N Az + N Bz ) Eliminating y B ( N Bz + N Az ) (by associativity), - cD BA dy B dz + cD AB dy B dz = 0 ⇒ - c dy B dz ( D BA - D AB ) = 0 Date : 03 Sept 2010. 1
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2 ARJAN SINGH PUNIANI Which means D BA = D AB The contradiction completes the proof. Note this result holds true for the general case, where the gradient operator, ~ , acts on c rather than d dz . (b) J A + J B = 0 . Proof. Rewrite above as (4) - D AB c A - D BA c B = 0 Since c A + c B = 1
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Unformatted text preview: -D AB ∇ c A-D BA ∇ (1-c A ) = 0 ⇒ -D AB ∇ c A + D BA ∇ c A = 0 From ( a ) above, D AB = D BA so-D 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 1-y A for y B yields-cD AB ~ ∇ y A + y A ( N A + N B )-cD BA ~ ∇ (1-y A ) + (1-y A )( N A + N B ) Cancellation ensues between-cD 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 ±...
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