Question

We are interested in recovering acetone from 100 mol/s (initial temperature is 65 C) of a vapor mixture containing 66.9% (mole basis) acetone and rest nitrogen. For this purpose, the vapor mixture is cooled to 20 C where 95% of initial acetone leaves as liquid and the rest of acetone leaves as vapor along with all the nitrogen at 20 C. Calculate the required heat exchange rate.

At 65C, enthalpy of acetone is -214 kJ/mol and enthalpy of nitrogen is 1.2 kJ/mol. At 20 C, enthalpy of nitrogen is -0.15 kJ/mol. For acetone (liquid), enthalpy at 20 C is -250 kJ/mol.

Answer 1

Initial temp = 65 °C

Vapour mixture contains:

   66.9% acetone   , 33.1 % nitrogen

At 20°C

     95 % of acetone leaves as vapour = ( 95/100 )* 66.9    = 63.555%, N is same that is   33.1%

So acetone used

= 66.9- 63.555 = 3.345

At 65C, enthalpy of acetone is -214 kJ/mol and enthalpy of nitrogen is 1.2 kJ/mol. At 20 C, enthalpy of nitrogen is -0.15 kJ/mol. For acetone (liquid), enthalpy at 20 C is -250 kJ/mol.

So enthalpy exchange (according to above data,)

=66.9*214 + 33*1.2 -(63.555* *250 + 33.1*0.15) = 1537.395 kJ/mol