ecm 11 soln - the same as in part A and so is the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
ECM 5 Winter, 2009 Problem Set 11 SOLUTIONS Due 2/23/09 1. The reaction A B takes place at constant temperature in a steady-state backmix reactor with no volume change on reaction. The reactor volume was 50 L, and the reactor was operated so that the concentration of A is the feed was 1.00 mol/L and the concentration of A in the product was 0.05 mol/L; there was inert solvent in both streams. The volumetric flow rate of the feed was 60 L/min. A . What fraction of A was converted in the reactor? (Call this the conversion of A.) The fraction of A converted is ( C Afeed - C Aproduct )/ C Afeed = (1.00 – 0.05)/1.00 = 0.95. B. Determine the volume of a steady-state backmix reactor operated at the same temperature to give the same conversion of A with the same feed when the feed rate is 3,500 L/min. The rate depends only on the temperature and the composition. The temperature is
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the same as in part A and so is the conversion; the composition of the reactor is the same everywhere and equal to that of the product stream (because it is a backmix reactor [and operated at steady state]). Therefore the rate is the same as in part A and can be calculated from the given information: r = ( C A0 C Aproduct )/( V / v ) [this is the design equation for the backmix reactor] So we calculate the rate r = (1.00 0.05)/(50/60) = 1.14 mol/(liter min) Now, use the given information to solve for V : 1.14 = 0.95/( V /3,500); V = 2,917 liter C. A steady-state backmix reactor with a volume of 10,000 L was used with the same feed and gave a conversion of 95%. What was the volumetric flow rate of the feed? Again, the rate is the same and we use the design equation to solve for v : r = 1.14 = 0.95/(10,000/ v ); v = 12,000 liter/min...
View Full Document

Ask a homework question - tutors are online