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process modeling and simulation chapter 2 solutions...

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Process Modelling, Simulation and Control for Chemical Engineering. Worked problems. Chapter 2: Fundamentals. This document contains my own solutions to the problems proposed at the end of each chapter of the book ”Process Modelling, Simulation and Control for Chemical Engineers” Second Edition, by William L. Luyben. As such, I can’t guarantee that the proposed solutions are free from errors. Think about them as a starting point for developing or as a means of checking your own solutions. Any comments or corrections will be appreciated. Contact me at [email protected]

Changes from previous version Now only the problem solutions (no problem statement) for odd problems are included, to limit reproduction of material subject to copyright.

Problem 1 F0 ρ0 CA0 CB0 V ρ CA CB

F ρ CA CB

Figure 1: CSTR. For both, simultaneous and reversible reactions, the quantities entering, leaving, and accumulating in the system are, for a generic component j, as follows: • Entering the system: F0 Cj0 • Leaving the system: F Cj • Accumulating in the system:

d (V Cj ) dt

1. For simultaneous reactions, the generation terms are: A: −(k1 + k2 )V CA B: k1 V CA C: k2 V CA The expressions for the continuity equations are: A:

d (V CA ) = F0 CA0 − F CA − (k1 + k2 )V CA dt

B:

d (V CB ) = F0 CB0 − F CB + k1 V CA dt

C:

d (V CC ) = F0 CC0 − F CC + k2 V CA dt

2. For reversible reactions, the generations terms are: A: −k1 V CA + k2 V CB B: k1 V CA − k2 V CB 1

The expressions for the continuity equations are: A:

d (V CA ) = F0 CA0 − F CA − k1 V CA + k2 V CB dt

B:

d (V CB ) = F0 CB0 − F CB + k1 V CA − k2 V CB dt

Problem 3 For a batch reactor, the continuity equations are analogous to the CSTR example, without the inflow and outflow terms. 1. Consecutive d (V CA ) = −k1 V CA dt d (V CB ) = k1 V CA − k2 V CB B: dt d (V CC ) = k2 V CB C: dt A:

2. Simultaneous d (V CA ) = −(k1 + k2 )V CA dt d (V CB ) = k1 V CA B: dt d (V CC ) = k2 V CA C: dt

A:

3. Reversible d (V CA ) = −k1 V CA + k2 V CB dt d (V CB ) = k1 V CA − k2 V CB B: dt

A:

Problem 5 M x θ g sin θ g Figure 2: Diagram of the forces that affect sled velocity. First, the forces experienced by the ensemble of mass M must be determined. The first is the component of the weight parallel to the hill Fg = M g sin θ, the second is the air resistance Fr = kv 2 . Now from Newton’s second law:

M

dx2 = Fg + Fr dt = M g sin θ − k



dx dt

2

 dx = v0 . dt (t=0) Assuming that halfway the sled already reached his “terminal velocity”, after Snoopy jumps, Charlie Brown will decelerate, because Fg is momentarily smaller than Fr . In any case, the final velocity reached by Charlie Brown alone will be smaller than the velocity that would have been reached is Snoopy remained in the sled.

With initial conditions x(t=0) = 0,



2

LC Fβ

hβ LC F0 hα

Fα Figure 3: Decanter.

Problem 7 Assuming that F0 , Fα and Fβ are volumetric flows, first a volume fraction is calculated as: xα,v =

xα ρβ ρα − xα (ρα − ρbeta )

The dynamic equations for the height of each phase are: 1 dhα = (F0 xα,v − Kα hα ) dt Ad 1 dhβ = (F0 (1 − xα,v ) − Kβ (hα + hβ )) dt Ad Where Ad is the transverse area of the decanter.

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