Sol 05lug2016

Pagina 1

Dipartimento di Chimica, Materiali e Ingegneria Chimica “Giulio Natta” POLITECNICO di MILANO Solution of the exercise of the written test of July 05th, 2016 Two nitrogen streams contain, respectively, 35 mol% (stream a) and 4 mol% (stream b) of n-butane. They are treated using a heavy oil containing no n -butane within an absorption tray column, operated at 25 °C and at 15 bar, so that the content of n-butane is reduced to a molar fraction of 0.005. The ratio between the total molar flow rate of stream a and that of stream b is given by: 14.52.5 27 a b   , where 46( 1) 2MM     . The stream with the higher concentration of n-butane (i.e., stream a) is fed to the bottom of the column, whereas stream b is fed to the tray which is characterized by a concentration of n-butane in the gas phase equal to that in stream b. Determine:  the minimum (L/G);  the number of theoretical stages required to perform the desired separation, assuming L = const· Lmin, with 561.3 540const   , where 46( 1) ( 2)MM     ;  the feed tray for stream b. Data and Properties M4 = fourth digit of identification number M6 = sixth digit of identification number Antoine equation: 0ln( ) BPA TC  , with T in [K] and P° in [mmHg] Component MW [kg/kmol] A B C N2 14.007 13.2150 386.67 -18.67 n-C4H10 58.124 15.6782 2154.9 –34.42 oil 250 16.151 4294.55 -124 Solution To solve the exercise, the following values have been considered for the fourth and sixth dig it of the identification number: M4 = 0, M6 = 0. As a consequence: α = 1, β = 2  (a/b) = 2.00, const = 1.2. Conversion of mole fractions into mole ratios: Yo(a) = 0.538462, Yo(b) = 0.041667, XN+1 = 0, YN = 0.005025. The column is divided into two sections: 1) the first one comprises the top of the column and ends where stream b is…