Question

1.) A chemical species can take on three different forms: A, B and C. After some amount passes, the chemical can change between each form in a predictable way. If it is found in to remain as A. From form B, it has a 20% chance to become A, 30% chance to convert to C form A, it has a 10% chance to convert to B, 80% chance to convert to C and 10% chance and a 50% chance to remain as B. Finally, if it is in form C, it converts to A 60% of the time, to B 20% of the time and stays in the form of C 20% of the time, (a-2 points) If you start with 100 kg of pure form A and 50 kg of pure form B and allow the conversion above to take place, how many kg of form C are you likely to have? (b- 2 points) Create a matrix, M, that can be multiplied by any vector, v = kg of B kg of A] |kg of (c - 2 points) Find the Eigenvalues of M using Newton's Method to solve the roots of the characteristic equation. (d - 2 points) There is a composition of such that Mů = v. What will be the proportion of forms A, B and C at this composition? (Hint: Use one of the Eigenvectors to answer this) 2.) 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 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.

this is a chemical engineer question on process analysis.

Answer 1

1 chemical species Pg no 1 of 3 on three different take Can forms A,B,C. Given explanation in the question Transformed to mathematical Equations Can be A = 0.LA + 0.1 B t 0.80 to (1) (1) (16 found on form ^] B = 0.2 A +0.5 B + 0.3c +(2)(96 founden form B] 0.6A + 0.2B + 0.2 0–4 (3) 6 If found in form c] Given that 100 kg of pure form A and 50 kg of pure form B B = 50, -Y A = 100, Let us substitute Eq 0 and ① CY 100 11 0.1 Ctool + 01 (50) +0.8C 50 11 0.2 Cloul + 0.5(50) + 0.3c 50 - -10 - 20 + 0.50, 160 50+30=4 so 0%. 0.5c = iso the the above reaction will 160 kg Bc From the egne of 0 % § 6 matrix form, we can write in

Pg no 22 3 0:1 Y B 10 0:2 0.1 0.1 " 0.3 U To find the ralues Eigun Card I) :0 Scalar and called as Eigen value of -tlere A ir neighbour matriz o M. d is Hatria A y det CH-I ) ny characteristic &qn. 01d 0.8 02 0.5-d 0.3 11 o 0.6 0.2 02- se (0.1-d) 05 -1) (02-1) – (02.2003)| 01962102-d) - 0.380.6] 0.6 [Co2x01) - 10:5 --) (0-2)] + 0.8 by showing above re egn d -0.544, 0.347,

3 Pgno 1032 of ro such that Mine zu 0) Compost ition find the properties Z ABC at this point. let Find Eigen Vector at tal. (first guess =Y rol-1 0.1 0.8 A 0.2 0.5-1 0.3 B o 6. 0.2 02 с 13 -0.9 A toolB + 0.80 6.2 A -0.5B +0.36 let Asl 0-1 B +0.80 0.985 -0.5 B +0.36 0.2 For B =1, Cal above ean satisfies Eigen vector is (!!!,1) i The properties of A,B,C lilil at mu? MU=V