Answer:
Explanation:
Let the products in the outlet streams be P(i.e. CH₃OH), Q(i.e. C₂H₅OH) and R(i.e. C₃H₇OH) respectively
Then the mass balance for CO, H and C₂H₅OH can be computed as follows:
For CO, The mass balance is;
100 - P - 3Q - 5R = 30 --- (1)
For H₂, The mass balance is;
100 - 2P - 3Q - 4R = 30 --- (2)
For C₂H₅OH, The mass balance is;
Q = 5 --- (3)
Replacing the value of equation(3) into equation (1) and (2); we have:
From equation (1):
100 - P - 3(5) - 5R = 30 --- (1)
100 - 15 - P - 5R = 30
85 - P - 5R = 30
85 - 30 = P + 5R
P + 5R = 55 ----- (4)
From equation (2):
100 - 2P - 3Q - 4R = 30 --- (2)
100 - 2P - 3(5) - 4R = 30
100 - 15 - 30 = 2P +4R
55 = 2P + 4R
2P + 4R = 55 ----- (5)
From equation (4) and (5); we have:
P + 5R = 55 ----- (4)
2P + 4R = 55 ----- (5)
From equation (4);
Let P = 55 - 5R
Then, replace P = 55 - 5R in equation (5)
2(55 - 5R) + 4R = 55
110 - 10R + 4R = 55
110 - 6R = 55
6R = 110 - 55
6R = 55
R = 55/6
R = 9.17
Substitute, the value of R in equation (5); we have:
2P + 4R = 55
2P + 4(9.17) = 55
2P = 55 - 4(9.17)
2P = 55 - 36.68
2P = 18.32
P = 18.32/2
P = 9.16
Therefore, the outlet stream rates are as follows:
Mass Feed Rate ( mol/hr) Molar ratio
CO 30 0.281
H₂ 30 0.281
CH₃OH 9.16 0.085
C₂H₅OH 5 0.047
C₃H₇OH 9.17 0.086
CO₂ 23.34 0.219
Total: 106.67
