A feed containing equimolar amounts of methanol and water is mixed with 10 moles of a 40 % mol aqueous methanol stream. The mixture enters a first separation unit that creates two streams. A top stream exits the separation unit that contains 70 % mol methanol and rest water. The bottom stream, which is 70 moles enters a second separation unit. A top stream exists the second separation unit as a 50 % methanol & 50 % water mixture. The other stream is unknown. If the fresh feed to the system is 100 moles (equimolar mixture) and the two top streams exiting the separation units have the same flow, draw the flow sheet and find the molar flow and the composition of the bottom streams existing the second separation unit?
To solve this problem, we can take a step-by-step approach and perform material balances for each separation unit.
Step 1: Material balance on the first separation unit:
Given:
- Fresh feed to the system = 100 moles, equimolar mixture of methanol and water.
- A 40% mol aqueous methanol stream = 10 moles.
Let's assume the molar flow rate of the top and bottom streams exiting the first separation unit is F1 (moles/time).
Material balance equation for methanol:
0.5 * 100 + 0.4 * 10 = 0.7 * F1
50 + 4 = 0.7F1
54 = 0.7F1
F1 = 54 / 0.7
F1 ≈ 77.14 moles/time
Hence, the molar flow rate of the top and bottom streams exiting the first separation unit is approximately 77.14 moles/time.
Step 2: Material balance on the second separation unit:
Given:
- Top stream exiting the second separation unit contains a 50% methanol and 50% water mixture.
- The molar flow rate of the top stream = F1 ≈ 77.14 moles/time
- Bottom stream entering the second separation unit = 70 moles.
Let's assume the molar flow rate of the bottom stream exiting the second separation unit is F2 (moles/time).
Material balance equation for methanol:
0.7 * 70 = 0.5 * F1 + 0.5 * F2
49 = 0.5 * 77.14 + 0.5 * F2
49 = 38.57 + 0.5 * F2
0.5 * F2 = 49 - 38.57
0.5 * F2 ≈ 10.43
F2 ≈ 10.43 / 0.5
F2 ≈ 20.86 moles/time
Hence, the molar flow rate of the bottom stream exiting the second separation unit is approximately 20.86 moles/time.
To find the composition of the bottom stream exiting the second separation unit, we need to determine the methanol and water compositions separately.
Composition of methanol in the bottom stream:
Amount of methanol = 0.5 * F2
Composition of methanol = (amount of methanol) / (total amount of mixture in the bottom stream)
Composition of methanol = (0.5 * F2) / (F2 + 70)
Composition of water in the bottom stream:
Amount of water = 0.5 * F2
Composition of water = (amount of water) / (total amount of mixture in the bottom stream)
Composition of water = (0.5 * F2) / (F2 + 70)
Plugging in the value of F2 ≈ 20.86 moles/time, we can calculate the compositions of methanol and water in the bottom stream exiting the second separation unit.