Process Process Modelli Modelling, ng, Simula Simulatio tion n and Contro Controll for Chemic Chemical al Engineering. Engineering. Worked orked problems. Chapter Chapter 2: Fundamenta undamentals. ls. 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 Chemica Chemicall Engine Engineers ers”” Second Second Edition Edition,, by William William L. Luyben. Luyben. At such, such, I can’t guarantee guarantee that the proposed proposed solutions solutions are free from errors. errors. Think about them as a starting point for developing or as a means of checking your own soluti solutions ons.. Any Any com commen ments ts or correc correctio tions ns will be apprec appreciat iated. ed. Contac Contactt me at
[email protected]
Problem 1 Write the component continuity equations describing the CSTR of Figure 1 with: 1. Simultaneous reactions reactions (first-order, isothermal). k1
A −→ B k2
A −→ C 2. Reversible (first-order, (first-order, isothermal). k1
A− − B k2
Figure 1: CSTR Solution
For both parts 1 and 2, the quantities entering, leaving, and accumulating in the system are analogous for every chemical species: •
Entering: F 0 C j 0
•
Leaving: F C j 1
•
Accumulating:
d (V C j ) dt
1. For simultaneous reactions, the generation terms are: •
A:
•
B: k 1 V C A
•
C: k 2 V C A
−
k1 V C A − k2 V C A
The expressions for the continuity equations are: •
A:
d (V C A ) dt
= F 0 C A0 − k1 V C A − k2 V C A − F C A
•
B:
d (V C B ) dt
= F 0 C B0 + k1 V C A − F C B
•
C:
d (V C C ) dt
= F 0 C C 0 + k2 V C A − F C C
2. For reversible reactions, the generation terms are: •
A:
•
B: k 1 V C A − k2 V C B
−
k1 V C A + k2 V C B
The expressions for the continuity equations are: •
A:
d (V C A ) dt
= F 0 C A0 − k1 V C A + k2 V C B − F C A
•
B:
d (V C B ) dt
= F 0 C B0 + k1 V C A − k2 V C B − F C B
Problem 2 Write the component continuity equations for a tubular reactor (Figure 2), with consecutive reactions ocurring: k1
k2
A −→ B −→ C
Figure 2: Tubular reactor Solution
The quantities entering, leaving, and accumulating are analogous for every chemical especies: •
Entering: AT vC j
−
AT Dj
∂C j ∂z
2
•
Leaving: AT vC j
•
Accumulating:
+
∂ ( AT vC j ) ∂z
∂ (AT dzC j ) ∂t
dz
∂C j − AT Dj − ∂z
∂ ∂z
∂C AT Dj ∂zj
dz
The generation terms for the consecutive reactions are: •
A:
•
B: A T dzk 1 C A − AT dzk 2 C B
•
C: A T dzk 2 C B
−
AT dzk 1 C A
The expressions for the continuity equations are, after dividing for AT dz : •
A:
∂ ∂ C = − ∂z ( vC A ) ∂t A
+
∂ ∂z
•
B:
∂ ∂ C = − ∂z ( vC B ) ∂t B
+
∂ ∂z
•
C:
∂ C ∂t C
∂ = − ∂z ( vC C ) +
∂ ∂z
A DA ∂C ∂z
B DB ∂C ∂z
C DC ∂C ∂z
+ + −
k1 C A k1 C A − k2 C B
k2 C B
Problem 3 Write the component continuity equations for a perfectly mixed batch reactor (no inflow or outflow) with first-order isothermal reactions: 1. Consecutive 2. Simultaneous 3. Reversible Solution
For a batch reactor, the continuity equations are analogous to the CSTR example, without the inflow and outflow terms. Asuming the rection volume is constant we have: 1.
2.
3.
•
A:
dC A dt
= −k1 C A
•
B:
dC B dt
= k 1 C A − k2 C B
•
C:
dC C dt
= k 2 C B
•
A:
dC A dt
= −k1 C A − k2 C A
•
B:
dC B dt
= k 1 C A
•
C:
dC C dt
= k 2 C A
•
A:
dC A dt
= −k1 C A + k2 C B
•
B:
dC B dt
= k 1 C A − k2 C B
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Problem 4 Write the energy equation for the CSTR of Problem 1 in which consecutive first order reactions occur with exothermic heats of reaction λ 1 and λ 2 . Solution
Assuming that the entalphy can be represented as h = C pT on a molar basis, the energy balance can be written, neglecting mixing effects as ( λ is negative for an exothermic reaction): d ( V T (C A C p,A + C B C p,B )) = F 0 T 0 (C A0 C p,A + C B,0 C p,B ) dt − F T (C A C p,A + C B C p,B ) −
V (k1 C A λ1 + k2 C B λ2 )
Remember that entalphies are measured against a reference state, so we are not saying here that the entalphy of A and B are the same at 0 temperature, instead, we are saying that at 0 temperature, the difference of entalphy between species A and B with their respective reference states are the same (equal to 0).
Problem 5 Charlie Brown and Snoopy are sledding down a hill that is inclined θ degrees from horizontal. The total weight of Charlie, Snoopy, and the sleed is M. The sled is essentially frictionless but the air resistance of the sledders is proportional to the square of their velocity. Write the equations describing their position x, relative to the top of the hill (x=0). Charlie likes to ”belly flop”, so their initial velocity at the top of the hill is v0 . What would happen if Snoopy jumped off the sled halfway down the hill without changing the air resistance? Solution
First, the forces experienced by the ensemble of mass M must be determined. One is the component of the weight directed parallel to the hill F g = Mgsenθ, the other is the air resistance F r = kv 2 . Now from Newton’s second law: d2 x M 2 = F g + F r dt d2 x M 2 = Mgsenθ − k dt
2
dx dt
With the initial conditions x t=0 = 0, dx = v 0 dt t=0 Assuming that half way the sled already reached his ”terminal velocity”, after Snoopy jumps, Charlie Brown will decelerate, because F g is momentarily smaller that F r . In any case, the final velocity reached by Charlie Brown alone will be smaller than the velocity that would have been reached if Snoopy remained in the sled.
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Problem 6 An automatic bale tosser on the back of a farmer’s hay baler must throw a 60-pound bale of hay 20 feet back into a wagon. If the bale leaves the tosser with a velocity vr in a direction θ = 45 above the horizontal, what must vr be? If the tosser must accelerate the bale from a dead start to v r in 6 feet, how much force must be exerted? What value of θ would minimize the acceleration force?
Figure 3: Tosser Solution
Assuming that the y coordinate at the exit of the tosser and at the wagon are equal, the half time of flight is equal to the time required for reaching the maximum altitude, which can be calculated dividing the y-velocity at the exit of the tosser by the gravity acceleration: vr senθ 1 tflight = g 2
The distance L must be covered in t flight : L tflight vr =
= v r cosθ
Lg ft = 25 .4 s 2cosθsenθ
With l = 6[f t], we have from the dynamic equations for the acceleration step: l =
at2acc
2
vr = at acc
Which permit to determine the value of the acceleration ( a): a =
vr2 gL = 2l 4lcosθsenθ
5
From Newton’s second law, F tosser − Mgsenθ = gc Mg F tosser = gc
aM : gc
L senθ + 4lcosθsenθ
The minimum value corresponds to 142[ lbf ] (38.9”). The force required in the case of θ = 45 is 140 [lbf ].
Problem 7 A mixture of two inmiscible liquids is fed into a decanter. The heavier liquid settles to the bottom fo the tank. The lighter liquid β forms a layer on the top. The two interfaces are detected by floats and are controlled by manipulating the two flows F α and F β . F α = K α hα F β = K β (hα + hβ )
The controllers increase or decrease the flows as the levels rise or fall. The total feed rate is F 0 . The weight fraction of liquid in the feed is x α . The two densities ρα and ρβ are constant. Write the equations describing the dynamic behavior of this system.
Figure 4: Solution Assuming
that the flows F 0 , F β and F α are volumetric flows, first a volumetric fraction is calculated as: xα ρα 1 −
xα,v =
xα ρα 1 − (1 − xα )ρβ 1 −
−
6
The dynamic equations for the height of each phase are: dhα 1 = (F 0 xα,v dt Ad
−
K α hα )
dhβ 1 = (F 0 (1 − xα ) − K β (hα + hβ )) dt Ad
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