A vapor mixture of n-butane (B) and n-hexane (H) contains 50.0 mole% butane at 120C and 1.0

Question:

A vapor mixture of n-butane (B) and n-hexane (H) contains 50.0 mole% butane at 120°C and 1.0 atm, A stream of this mixture flowing at a rate of 150.0 L/s is cooled and compressed, causing some but not all of the vapor to condense. (Treat this process as a single-unit operation.) Liquid and vapor product streams emerge from the process in equilibrium at T(°C) and 1100 mm Hg. The vapor product contains 60.0mole% butane.

(a) Draw and label a flowchart. Perform a degree-of-freedom analysis to show that you have enough information to determine the required final temperature (T), the composition of the liquid product (component mole fractions), and the molar flow rates of the liquid and vapor products from the given information and Antoine expressions for the vapor pressures p*B (T) and p*H (T). Just identify the equations—for example, mole balance on butane or Raoult’s law for hexane—but don’t write them yet.

(b) Write in order the equations that you would use to determine the quantities listed in part (a) and also the fractional condensation of hexane (mol H condensed/mol H fed). In each equation, circle the variable for which you would solve. Do no algebra or calculations.

(c) Complete the calculations either manually or with an equation-solving program.

(d) State three assumptions you made that could lead to errors in the calculated quantities.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question

Elementary Principles of Chemical Processes

ISBN: 978-0471720638

3rd Edition

Authors: Richard M. Felder, Ronald W. Rousseau

Question Posted: