Multiple Unit Material balance problem 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?

molar flow at the bottom streams = 30 moles

composition of the bottom streams exiting the second separation unit

= 0.20 ( methanol )

= 0.80 ( water )

Explanation:

The molar flow and the composition of the bottom streams is calculated as follows below

First we have to balance the overall material across the Mixer unit

= F1 + F2 = A

= 100 + 10 = 110 mole

next we calculate Methanol balance

= Xe1 * F1 + Xe2 * F2  = AXea

= (0.5 * 100) + ( 0.4 * 10 ) = 110 Xea

= 50 + 4 = 110Xea

Xea = 0.491   therefore: Xaw = 1 - Xea = 1 - 0.491 = 0.509

Next we calculate the material balance of separator 1

A = B + C

where: A = 110 moles , B = ? , C = 70 moles

Therefore B= A - C = ( 110 - 70 ) = 40 moles

From here we will find the value of Xec using the Methanol balance relationship

Xea * A = Xeb*B + Xec * C

where: A = 110 moles , B = 40 moles , C = 70 moles

           Xea = 0.491, Xeb = 0.7 ,

Input these values into the equation above : Xec = 0.372

note: at the exit top stream both the separators have the same flow rate

i.e : B = D = 40 mole

Material Balance over the separator 2 can hence be calculated as

C = D + E

E = c - d = 70 - 40

E = 30 moles ( mole flow rate at the 2nd separator unit )

calculate the value of XeE

methanol balance : Xec * c = Xed * D + XeE * E

hence : XeE = [ ( 70 * 0.372 ) -  ( 0.5 * 40 ) ] /  30

             XeE  =  0.20

attached below is the flow sheet of the problem