1

Fall 2009 -- ChE 612 -- QUIZ #1

• Open notes, open book (Fogler ONLY) • Use exam booklets -- no loose papers -- label all booklets • Turn in question sheet and booklets – please identify all booklets! • Show all work; state all assumptions; partial credit given • Do both questions – each worth 45 points; 10 points free ☺ ------------------------------------------------------------------------------------------------------- PROBLEM #1 The gas-phase reaction 2A(g) B(g,l) + C(g) is to be run in a steady-state CSTR (pictured below). Note that the product B can condense to a liquid phase. It is planned to run the reactor at a conversion XA without condensation.

FAo

Useful Information: • Elementary kinetics (for the reaction as written above) • k = 2 liters/mole-sec • Pv,B = 100 Torr • P = 1000 Torr • T = 450 K • To = 300 K Questions: (Each worth 9 points)

I. What is the mole fraction of B in the gas phase when condensate is present? Also, construct a stoichiometric table.

II. What is the conversion XA* when condensation occurs?

III. What is the space-time τ corresponding to this XA* ?

It has been suggested that adding inert gas to the feed could improve conversion while still avoiding condensation.

IV. What flow rate of inert gas FI, expressed as θI, will correspond to an XA

* = 0.6? Assume the same P, T, and FAo as used above.

V. What is the space-time τ corresponding to this new XA* ? Is it practical?

2

PROBLEM #2

hot solventTa

CAo, vo,solvent at To

T V

membrane

reactor tube

B annulus

dQ

The plug flow reactor above is used to process the all-liquid phase endothermic reaction A B + C. The reaction rate follows zero order kinetics. The feed is a dilute mixture of A in an inert solvent. The walls of the reactor tube are a membrane that allows only species B to diffuse out into the annulus. Hot solvent flowing through the annulus heats the reactor tube contents by conduction through the membrane. The flow rate of hot solvent is high enough that we can assume CB, annulus ≈ 0. The diffusion rate of B is small enough compared to the reactor solvent flow that v ≈ vo. The heating rate is given by: d Ý Q dV =Ua Ta −T( ). The removal of B allows for separation during reaction. Questions: (Each worth 9 points)

I. Write complete reactor species balances for A and B in terms of concentrations.

II. Write the complete energy balance for the reactor. Derive T as a function of reactor volume V.

III. Solve the balance for A, and derive CA vs. V. IV. Solve the balance for B, and derive CB vs. V. V. Using stoichiometry only, derive CC vs. V.

Useful Data k = koT = 0.001T moles/liter-min where T (K) vo = 100 liters/min CAo = 1 moles/liter MWSolvent = 60 Ua = 450 cal/min-liter-K kc = 5/min cpSolvent = 30 cal/mole-K ρSolvent = 800 g/liter ∆HrA = 100000 cal/mole A To = 300 K Ta = 400 K