Chapter12_B

# Chapter12_B - Chapter 12 - Section B - Non-Numerical...

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Chapter 12 - Section B - Non-Numerical Solutions 12.2 Equation (12.1) may be written: y i P = x i γ i P sat i . Summing for i = 1 , 2gives: P = x 1 γ 1 P sat 1 + x 2 γ 2 P sat 2 . Differentiate at constant T : dP dx 1 = P sat 1 µ x 1 d γ 1 1 + γ 1 + P sat 2 µ x 2 d γ 2 1 γ 2 Apply this equation to the limiting conditions: For x 1 = 0: x 2 = 1 γ 1 = γ 1 γ 2 = 1 d γ 2 1 = 0 For x 1 = 1: x 2 = 0 γ 1 = 1 γ 2 = γ 2 d γ 1 1 = 0 Then, µ 1 x 1 = 0 = P sat 1 γ 1 P sat 2 or µ 1 x 1 = 0 + P sat 2 = P sat 1 γ 1 µ 1 x 1 = 1 = P sat 1 P sat 2 γ 2 or µ 1 x 1 = 1 P sat 1 =− P sat 2 γ 2 Since both P sat i and γ i are always positive definite, it follows that: µ 1 x 1 = 0 ≥− P sat 2 and µ 1 x 1 = 1 P sat 1 12.4 By Eqs. (12.15), ln γ 1 = Ax 2 2 and ln γ 2 = 2 1 Therefore, ln γ 1 γ 2 = A ( x 2 2 x 2 1 ) = A ( x 2 x 1 ) = A ( 1 2 x 1 ) By Eq. (12.1), γ 1 γ 2 = y 1 x 2 P sat 2 y 2 x 1 P sat 1 = µ y 1 / x 1 y 2 / x 2 ¶µ P sat 2 P sat 1 = α 12 r Whence, ln 12 r ) = A ( 1 2 x 1 ) If an azeotrope exists, α 12 = 1a t0 x az 1 1. At this value of x 1 , ln r = A ( 1 2 x az 1 ) The quantity A ( 1 2 x 1 ) is linear in x 1 , and there are two possible relationships, depending on the sign of A . An azeotrope exhists whenever | A |≤| ln r | . NO azeotrope can exist when | A | < | ln r | . 12.5 Perhaps the easiest way to proceed here is to note that an extremum in ln γ 1 is accompanied by the opposite extremum in ln γ 2 . Thus the difference ln γ 1 ln γ 2 is also an extremum, and Eq. (12.8) becomes useful: ln γ 1 ln γ 2 = ln γ 1 γ 2 = d ( G E / RT 1 Thus, given an expression for G E / = g ( x 1 ) , we locate an extremum through: d 2 ( G E / ) 2 1 = d ln 1 2 ) 1 = 0 691

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For the van Laar equation , write Eq. (12.16), omitting the primes ( 0 ) : G E RT = A 12 A 21 x 1 x 2 A where A A 12 x 1 + A 21 x 2 Moreover, dA dx 1 = A 12 A 21 and d 2 A 2 1 = 0 Then, d ( G E / ) 1 = A 12 A 21 µ x 2 x 1 A x 1 x 2 A 2 1 d 2 ( G E / ) 2 1 = A 12 A 21 · 2 A x 2 x 1 A 2 1 x 1 x 2 A 2 d 2 A 2 1 1 µ 2 x 1 x 2 A 3 1 + x 2 x 1 A 2 ¶¸ = A 12 A 21 " 2 A 2 ( x 2 x 1 ) A 2 1 + 2 x 1 x 2 A 3 µ 1 2 # = 2 A 12 A 21 A 3 " A 2 ( x 2 x 1 ) A 1 + x 1 x 2 µ 1 2 # = 2 A 12 A 21 A 3 µ A + x 2 1 ¶µ x 1 1 A This equation has a zero value if either A 12 or A 21 is zero. However, this makes G E / everywhere zero, and no extremum is possible. If either quantity in parentheses is zero, substitution for A and / 1 reduces the expression to A 12 = 0or A 21 = 0, again making G E / everywhere zero. We conclude that no values of the parameters exist that provide for an extremum in ln 1 2 ) . The Margules equation is given by Eq. (12.9b), here written: G E = Ax 1 x 2 where A = A 21 x 1 + A 12 x 2 1 = A 21 A 12 d 2 A 2 1 = 0 Then, d ( G E / ) 1 = A ( x 2 x 1 ) + x 1 x 2 1 d 2 ( G E / ) 2 1 =− 2 A + ( x 2 x 1 ) 1 + ( x 2 x 1 ) 1 + x 1 x 2 d 2 A 2 1 2 A + 2 ( x 2 x 1 ) 1 = 2 · ( x 1 x 2 ) 1 A ¸ This equation has a zero value when the quantity in square brackets is zero. Then: ( x 2 x 1 ) 1 A = ( x 2 x 1 )( A 21 A 12 ) A 21 x 1 A 12 x 2 = A 21 x 2 + A 12 x 1 2 ( A 21 x 1 + A 12 x 2 ) = 0 Substituting x 2 = 1 x 1 and solving for x 1 yields: x 1 = A 21 2 A 12 3 ( A 21
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## This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter12_B - Chapter 12 - Section B - Non-Numerical...

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