STUDY OF FLOW RATES
IN CHEMICAL PLANTS
This chapter has 3 sections. Use of FRE to estimate flow rates in chemical plants forms the section one of this chapter. In section two fuzzy neural networks are used to estimate velocity of flow distribution in a pipe network. The final section estimates the three-stage counter current extraction unit again using fuzzy neural networks.
4.1 Use of FRE in Chemical Engineering
The use of fuzzy relational equations (FRE) for the first time has been used in the study of flow rates in chemical plants. They have only used the concept of linear algebraic equations to study this problem and have shown that use of linear equations does not always guarantee them with solutions. Thus we are not only justified in using fuzzy relational equation but we are happy to state by adaptation of FRE we are guaranteed of solutions to the problem. We have adapted the fuzzy relational equations to the problem of estimation of flow rates in a chemical plant, flow rates in a pipe network and use of FRE in a 3 stage counter current exaction unit [44].
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Experimental study of chemical plants is time consuming
expensive and need intensive labor, researchers and engineers prefer only theoretical approach, which is inexpensive and effective. Only linear equations have been used to study: (1). A typical chemical plant having several inter-linked units (2).
Flow distribution in a pipe network and (3). A three stage counter current extraction unit. Here, we tackle these problems in 2 stages. At the first stage we use FRE to obtain a solution.
This is done by the method of partitioning the matrix as rows. If no solution exists by this method we as the second stage adopt Fuzzy Neural Networks by giving weightages. We by varying the weights arrive at a solution which is very close to the predicted value or the difference between the estimated value and the predicted value is zero. Thus by using fuzzy approach we see that we are guaranteed of a solution which is close to the predicted value, unlike the linear algebraic equation in which we may get a solution and even granted we get a solution it may or may not concur with the predicted value.
To attain both solution and accuracy we tackle the problems using Fuzzy relational equations at the first stage and if no solution is possible by this method we adopt neural networks at the second stage and arrive at a solution.
Consider the binary relation P(X, Y), Q(Y, Z) and R(X, Z) which are defined on the sets X = {xi / i ∈ I} Y = {yi / j ∈ J}
and Z{zk / k ∈ K} where we assume that I = Nn, J = Nr and K =
Ns. Let the membership matrices of P, Q and R be denoted by P
= [pij], Q = [qik] and R = [rik] respectively, where pij = P(xi, yj), qik = Q(yj, zk) and rik = R(xi, zk) for i ∈ I (= Nn), j ∈ J (= Nm) and k ∈ K (= Ns). Entries in P, Q and R are taken from the interval
[0, 1]. The three matrices constrain each other by the equation P o Q = R
(1)
(where o denotes the max-min composition) known as the fuzzy relation equation (FRE) which represents the set of equation Max pijqjk = rik
(2)
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for all i ∈ Nn, k ∈ Ns. If after partitioning the matrix and solving the equation (1) yields maximum of qjk < rik for some qjk, then this set of equation has no solution. So at this stage to solve the equation 2, we use feed-forward neural networks of one layer with n-neurons with m inputs shown in Figure 4.1.1.
Inputs of the neuron are associated with real numbers Wij referred as weights. The linear activation function f is defined by
⎧0 if a < 0
⎫
⎪
⎪
f (a) = ⎨a if a ∈[0, 1]⎬
⎪
⎪
1 if a > 1
⎩
⎭
x1
w11
w21
wn1
x1
w12
w22
wn2
x1
w1n
w2n
wnn
ON
ON
1
ON2
m
Y1
Y2
Ym
Figure: 4.1.1
The output yi = f(max Wijxj), for all i ∈ Nn and j ∈ Nm.
Solution to (1) is obtained by varying the weights Wij so that the difference between the predicted value and the calculated value is zero.
FRE to estimate flow rates in a chemical plants
A typical chemical plant consists of several interlinked units.
These units act as nodes. The flowsheet is given in Figure 4.1.2.
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F2
D2
F5
F8
F
9
R
D1
M
F
F
3
1
F6
F4
D3
F7
Figure: 4.1.2
An experimental approach would involve measuring the
nine flow-rates to describe the state of the plant which would involve more money and labor.
While studying this problem in practice researchers have has neglected density variations across each stream. The mass balance equations across each node at steady state can be written as
F3 – F2 = F1,
F2 – F4 = F5,
F4 – F7 = F6,
F2 + F8 = F5,
F8 = F9 – F6.
(3)
Here Fi represents the volumetric flow rate of the ith stream.
In equation (3) at least four variables have to be specified or determined experimentally.
The remaining five can then be estimated from the equation (3), which is generated by applying the principle of
conservation of mass to each unit. We assume F1, F5, F6 and F9
are experimentally measured, equation (3) reads with known values on the right-hand side as follows:
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⎡ 1
− 1 0 0 0⎤ ⎡F ⎤ ⎡ F
⎤
2
1
⎢
⎥ ⎢ ⎥ ⎢
⎥
0
1
1
−
0
0
F
F
3
5
⎢
⎥ ⎢ ⎥ ⎢
⎥
⎢ 0 0 1
1
− 0⎥ ⎢F ⎥ = ⎢ F
⎥ (4)
4
6
⎢
⎥ ⎢ ⎥ ⎢
⎥
1
0
0
0
1
F
⎢
⎥ ⎢ ⎥ ⎢ F
⎥
7
5
⎢⎣ 0 0 0 0 1⎥ ⎢ ⎥ ⎢
⎥
⎦ F
F − F
⎣ 8 ⎦ ⎣ 9
6 ⎦
P o Q = R
(5)
where, P, Q and R are explained. Using principle of
conservation of mass balanced equation we estimate the flow rates of the five liquid stream. We in this problem aim to minimize the errors between the measured and the predicted value. We do this by giving suitable membership grades p ∈
ij
[0,
1] and estimate the flow rates by using these pij’s in the equation 3. Now the equation 4 reads as follows:
⎡p
p
0
0
0 ⎤ ⎡F ⎤
⎡ F
⎤
11
12
2
1
⎢
⎥ ⎢ ⎥ ⎢
⎥
0
p
p
0
0
F
F
22
23
3
5
⎢
⎥ ⎢ ⎥ ⎢
⎥
⎢ 0
0
p
p
0 ⎥ ⎢F ⎥ = ⎢ F
⎥ (6)
33
34
4
6
⎢
⎥ ⎢ ⎥ ⎢
⎥
p
0
0
0
p
F
⎢
⎥ ⎢ ⎥ ⎢ F
41
45
⎥
7
5
⎢ 0
0
0
0
p ⎥ ⎢
⎥ ⎢
⎥
⎣
⎦
F
F − F
55
⎣ 8 ⎦ ⎣ 9
6 ⎦
where
P = (pij),
Q = (qik) = [F2 F3 F4 F7 F8]t and
R = (rik) = [F1 F5 F6 F5 F9 – F6]t.
We now apply the partitioning method of solution to
equation (6). The partitioning of P correspondingly partitions R, which is give by a set of give subsets as follows:
⎡F ⎤ ⎡ F
⎤
2
1
⎢ ⎥ ⎢
⎥
F
F
3
5
⎢ ⎥ ⎢
⎥
[p
⎢ ⎥ = ⎢
⎥
11 p12 0 0 0]
F
F
, …
4
9
⎢ ⎥ ⎢
⎥
F
⎢ ⎥ ⎢ F
⎥
7
5
⎢ ⎥ ⎢
⎥
F
F − F
⎣ 8 ⎦ ⎣ 9
6 ⎦
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⎡F ⎤ ⎡ F
⎤
2
1
⎢ ⎥ ⎢
⎥
F
F
3
5
⎢ ⎥ ⎢
⎥
[0 0 0 0 p
⎢ ⎥ = ⎢
⎥
55]
F
F
.
4
6
⎢ ⎥ ⎢
⎥
F
⎢ ⎥ ⎢ F
⎥
5
5
⎢ ⎥ ⎢
⎥
F
F − F
⎣ 8 ⎦ ⎣ 9
6 ⎦
Suppose the subsets satisfies the condition max qik < rik then it has no solution. If it does not satisfy, this condition, then it has a final solution. If we have no solution we proceed to the second stage of solving the problem using Fuzzy Neural Networks.
When the FRE has no solution by the partition method, we solve these FRE using neural networks. This is done by giving weightages of zero elements as 0 and the modified FRE now reads as
⎡F ⎤ ⎡ F
⎤
2
1
⎢ ⎥ ⎢
⎥
F
F
3
5
⎢ ⎥ ⎢
⎥
P
⎢
o F ⎥ = ⎢ F
⎥ .
1
4
6
⎢ ⎥ ⎢
⎥
F
⎢ ⎥ ⎢ F
⎥
7
5
⎢ ⎥ ⎢
⎥
F
F − F
⎣ 8 ⎦ ⎣ 9
6 ⎦
The linear activation function f defined earlier gives the output yi = f (max Wij xj) (i ∈ Nn) we calculate max Wijxj as follows: 1. W11x1 = 0.02F2, W12x2 = 0F2, W13x3 = 0F2 W14x4 = 0.045F2, W15x5 = 0F2
y1 = f (maxj∈Nm Wijxj) = f (0.02F2, 0F2, 0.045F2, 0F2)
2. W21x1 = 0.04F3, W22x2 = 0.045F3, W23x3 = 0F3, W24x4 = 0.0F3, W15x5 = 0F3
y2 = f (maxj∈Nm Wijxj) = f (0.04F3, 0.045F3, 0F3, 0.0F3, 0F3) 3. W31x1 = 0.0F4, W32x2 = 0.085F4, W33x3 = 0.15F4, W34x4 =
0.0F4 W35x5 = 0F4
y3 = f (maxj∈Nm Wijxj) = f (0F4, 0.085F4, 0.15F4, 0F4, 0F4) 84
4. W41x1 = 0.0F7, W42x2 = 0F7, W43x3 = 0.2F7, W44x4 = 0.0F7, W45x5 = 0F7
y4 = f (maxj∈Nm Wijxj) = f (0F7, 0F7, 0.2F7, 0.0F7, 0F7) 5. W51x1 = 0.0F8, W52x2 = 0F8, W53x3 = 0F8, W54x4 = 0.45F8, W55x5 = 0.5F8
y5 = f (maxj∈Nm Wijxj) = f (0F8, 0F8, 0F8, 0.45F8, 0.5F8) shown in Figure 4.1.2. Suppose the error does not reach 0 we change the weights till the error reaches 0. We continue the process again and again until the error reaches to zero.
Thus to reach the value zero we may have to go on giving different weightages (finite number of time) till say sth stage Ps o Qs whose linear activation function f, makes the predicted value to be equal to the calculated value. Thus by this method, we are guaranteed of a solution which coincides with the predicted value.
4.2 Fuzzy neural networks to estimate velocity of flow distribution in a pipe network
In flow distribution in a pipe network of a chemical plant, we consider liquid entering into a pipe of length T and diameter D
at a fixed pressure Pi, The flow distributes itself into two pipes each of length T1(T2) and diameter D1(D2) given in Figure 4.2.1.
Pa, D1, V1
T
Pa, D2, V2
Figure: 4.2.1
The linear equation is based on Ohm’s law, the drop in
voltage V across a resistor R is given by the linear relation V =
85
iR (Ohm’s law). The hydrodynamic analogue to the mean
velocity v for laminar flow in a pipe is given by ∇p = v (32μT/D2). This is classical-Poiseulle equation. In flow distribution in a pipe network, neglecting pressure losses at the junction and assuming the flow is laminar in each pipe, the macroscopic momentum balance and the mass balance at the junction yields,
P
2
1 – Pa = (32μT/D2)v + (32μT1D1 )v1,
P
2
i – Pa = (32μT/D2)v + 32μT/D2 )v2,
D2v = D 2
2
1 v1 + v2D2 .
(1)
Hence Pa is the pressure at which the fluid leaves the system at the two outlets. The set of three equation in (1) can be solved and we estimate v, v1, v2 for a fixed (Pi – Pa). The system reads as
2
2
⎡32 T
μ / D 32 T
μ / D
0
⎤ ⎡v ⎤ ⎡p − p ⎤
1
1
i
a
⎢
⎥ ⎢ ⎥ ⎢
⎥
2
2
32 T
μ / D
0
32 T
μ / D
v
= p − p
⎢
.
2
2 ⎥
1
i
a
⎢ ⎥ ⎢
⎥
2
1
2
⎢ −D
D
D
⎥ ⎢v ⎥ ⎢ 0 ⎥
⎣
2
2
⎦ ⎣ 2 ⎦ ⎣
⎦
We transform this equation into a fuzzy relation equation. We use a similar procedure described earlier and obtain the result by fuzzy relation equation. We get max (0.2v, 0.025v, 0.03v), max (0.035v, 0v1, 0.04v1), max (0v2, 0.04v2, 0.045v2) by using neural networks for fuzzy relation equation described in [11]. Suppose the error does not reach to 0, we change the weights till the error reaches 0. We continue the process again and again till the error reaches zero.
4.3 Fuzzy neural networks to estimate three stage counter current extraction unit
Three-stage counter extraction unit is shown in Figure 4.3.1.
The components A present in phase E (extract) along with a nondiffusing substance as being mixture .
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Stage 1
Stage 2
Stage 3
X0
X3 (R)
Y1
Y4 (E)
Figure: 4.3.1
It is extracted into R by a nondiffusing solvent. The 3 extraction stage is given by the three equation.
EsY4 + RsXs = RsX3 + EsY3,
EsY3 + RsX1 = Es + RsX2,
EsY2 + RsX0 = EsY1 + RsX1
(1)
Yi(Xi) = moles of A, The flow of each stage is denoted by Es(Rs) and this constant does not vary between the different stages. The assumption of a linear equilibrium relationship for the
compositions leaving the ith stage equations
Yi = KXi
(2)
for i = 1, 2, 3 reads as
⎡ R
E
0
−E
0
0 ⎤
⎡X ⎤ ⎡R X ⎤
s
s
s
1
s
0
⎢
⎥ ⎢
⎥ ⎢
⎥
K
1
−
0
0
0
0
Y
0
1
⎢
⎥ ⎢
⎥ ⎢
⎥
⎢−R
0
R
E
0
−E ⎥ ⎢X ⎥ ⎢ 0 ⎥
s
s
s
s
2
⎢
⎥ ⎢
⎥ = ⎢
⎥
0
0
K
1
−
0
0
Y
0
⎢
⎥ ⎢ 2 ⎥ ⎢
⎥
⎢ 0
0
−R
0
R
E ⎥
⎢
⎥
X
⎢E Y ⎥
2
s
s
3
s
4
⎢
⎥ ⎢
⎥ ⎢
⎥
⎢⎣ 0
0
0
0
K
1
− ⎥⎦ ⎢Y
⎣
⎥⎦ ⎢⎣ 0 ⎥⎦
3
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where {X1, Y1, X2, Y2, X3, Y3} can be obtained for a given Es, Rs and K. Since use of linear algebraic equation does not result in the closeness of the measured and predicted value, we use neural networks for fuzzy relation equations to estimate the flow-rates of the stream, moles of the three-stage counter extraction unit and velocity of the flow distribution in a pipe network. As neural networks is a method to reduce the errors between the measured value and the predicted value. This allows varying degrees of set membership (weightages) based on a membership function defined over the range of value. The (weightages) membership function usually varies from 0 to 1.
We use the similar modified procedure described earlier and get result by fuzzy relation equation. We get max (0.2X1, 0.25X1, 0.3X1, 0X1, 0X1, 0X1), max (0.35Y1, 0.4Y1, 0Y1, 0Y1, 0Y1, 0Y1) max (0X2, 0X2, 0.45X2, 0.5X2, 0.55X2, 0X2), max (0.6Y2, 0Y2, 0.65Y5, 0.7Y2, 0Y2, 0Y2) max (0X3, 0X3, 0X3, 0X3, 0.75X3, 0.8X3), max (0Y3, 0Y3, 0.85Y3, 0Y3, 0.9Y3, 0.95Y3) by neural networks for fuzzy relation equation. We continue this process until the error reaches zero or very near to zero.
Thus we see that when we replace algebraic linear equations by fuzzy methods to the problems described we are not only
guaranteed of a solution, but our solution is very close to the predicted value.
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