A pharmaceutical product is crystallized and washed with absolute ethanol. A 100

Question

A pharmaceutical product is crystallized and washed with absolute ethanol. A 100 kg batch of product containing 10% ethanol by weight is to be dried to 0.1% ethanol by passing 2 m3/min of nitrogen at 35 oC through the dryer. The dryer is operated at 35 oC and 0.1 MPa. Neither N2 nor the pharmaceutical are soluble in liquid ethanol (in other words, liquid ethanol behaves as if it were a pure component); the vapor pressure of the pharmaceutical is negligible. Estimate the rate (mols/min) that ethanol is removed from the crystals and the residence time of the crystals in the dryer:

Explanation / Answer

Vapor pressure of ethanol at 35 deg.c = 103 mm Hg=103/760 atm =0.1355

0,1Mpa= 0.9869 atm

Assuming the nitrogen gas leaving the dryer is saturated

Moles of nitrogen at 35 deg.c and 0.1Mpa can be calculated from ideal gas equation

n= PV/RT= 0.9869*2*1000/(0.0821*(35+273))= 78 moles/min

mass   = 78*29 gm = 2262 gm= 2.262 kg

mass of ethanol at inlet = 100*10/100 = 10 kg

Rest is dry product whose mass = 90

This correspond to 100-0.1 = 99.9% of final product.

Hence mass of final product = 90/0.999=90.1

Mass of ethanol to be removed = 100-90.1= 9.9 kg =9.9*1000 gm=9900 gm

Moles of ethanol removed =mass/molar mass = 9900/46= 215 moles

At the exit, partial pressure of ethanol vapour = vapour pressure = 103 mm Hg

Partial pressure of vapour/ partial pressure of nitrogen = moles of ethanol vapour/moles of dry nitrogen

0.1355/(0.9869-0.1355)= x/78

Hence x= 12 moles/min of ethanol vapour is carried along with nitrogen,

Residence time = moles of ethanol to be removed/ molar flow rate of ethanol =215/12 min=17.92 min

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