Question

After the conversion above nearly reaches equilibrium, the mixture is pumped to a distillation column where form C is desired. The feed stream is 34% A, 24% B and 42% C by weight. The bottom exit stream is 10% A 20% B and 70% C, while the top exit stream only contains 20% C by weight and flows at 20 kg/hr.

(a – 2 points) Arrange this information into a system of linear equations and assume the process runs at steady-state (In – Out = 0).

(b – 2 points) Convert your system of equations into matrix form and use Gauss-Jordan Elimination to solve for the mass flow rates of each exit stream and the remaining composition of the top exit stream in terms of %A and %B by weight.

(c – 2 points) Show that you obtain the same result using Cramer’s Rule.

Answer 1

Given data : D=20 kg/hr Feet, feed, F Topi Bottom, in b) 10 A 34 24 20 С 70 42 20 D A Material balance egne DISTL Input out & generating disappear COLUMN - Accumulation proces is steady Accanmuktion -- No generation F No dissopperena F-W-20 koler an stle given that = Input = oretput DtW component balance wert component & Input -output = 0 0.42F - (0.2*20+0.7*W)= 0 -) 0.426 - atW- 4 42F-IW = 40

R2 R2 mest flow note of w=15.71 totho ib Gauss Jordan method :- 3) (0) 649 4.2 Converting into reduced row ethelon folui Converting into identity metrin je 20 1 40 4.2 - 7 R2 R2-42 R, -44 - 2.8 2.8 1 - 20 - O 15.71 1 R, R+R₂ ! 35.7 1 15.7 O - mass flow rate of F = 35.71 ks/

(c) verifying by Cramer's method :- Ах — В 1 20 40 4.2 20 40 -7 mass flow Rate of F - Taking determinents, 42 -7 - 100 - 2.8 - 35.71 = 35.71 kser ie F 4.2 40 mals flow (W) = rate of bottom 42 -7 - 44 - 15.71 0.34 W - 15.71 ks Component balance wort Aomponent A'! 35.71* Med= 15.7* 0.1+ 20+A → JA= 0.52852 Component balance art component Bl: 354*0.24 = 1571* 0-2 + 20KB B = 0.27142 weight percentager A 752065) in top prodent B - 27.14
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