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sergiy2304 [10]
3 years ago
15

Write a script (Program 2) to perform t he following matrix operations. Use output commands to clearly output each problem with

a problem description and result.
(a) Create the following three matrices: A = [ 9 3 5 /2 4 2/ 6 10 5 ] B = [ 1 4 4/ 0 9 4/ 2 6 1 ] C = [ 8 15 1/ 10 8 2/ 8 5 10 ]

(b) Calculate A + B and B + A to show that addition of matrices is commutative.

(c) Calculate A +( B + C ) and ( A + B )+ C to show that addition of matrices is associative.

(d) Calculate 8( A + C ) and 8A + 8C to show that multiplication by a scalar is distributive.

(e) Calculate A( B + C ) and AB + AC to show that matrix multiplication is distributive.

(f) Does AB = BA ?

(g) Find AT (transpose of A ).

(h) Calculate B-1 (inverse of B ).

(i) Calculate the product BB-1 .

(j) Create a 2x2 matrix consisting of the lower left elements of B .

(k) Create a row vector that is the vector sum of the first row of A and the third row of C .

(l) Calculate the dot product of the second row of B and the third column of C

Engineering
1 answer:
Simora [160]3 years ago
7 0

Find the attachments for complete solution

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A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a
OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS        A                B                    C

Benzene 1             4.72583     1660.652        -1.461

Toluene  2            4.07827     1343.943         -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5%    = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis     for all components

So for Benzene ; p × y1=x1 × p1s   ------let this be equation 1

for Toluene ; p × y2=x2 × p2s   ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

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