Model Development for Optimization

• The methods described in the previous three lectures require an objective function as well as the constriants.

• This requires the development of a mathematical model for the process under consideration.

A mathematical model provides quantitative expressions that enable us to use optimization techniques to extract useful information about process design and operation. Two types of models may be developed:

1. Fundamental Models

2. Empirical Models \underline{Fundamental Models}:

• These models are developed from fundamental principles (e.g. mass and energy balances).

• Typically simplifying assumptions are made to keep the size of the model small.

• Model parameters are obtained from independent experiments and are physically meaningful quantities (e.g density, viscosity).

• The model expressions have predictive capabilities.

• The accuracy of the model predictions depend on how good the assumptions are.

• A lot of process experience is required to develop good models. The challenge is to convert a word problem into a mathematical description.

• Optimization methods used on the model give the optimal solution of the model. Whether this is the optimal solution of the process depends on how accurately the model depicts the process.

\underline{Example 1}: Consider a three stage compressor with intercooling. The following process data is given:

P_{1} = 100 kP a P_{4} = 1000 kP a T_{1} = 300 K

Assume that the compressor is used to compress air. What should be the values of P_{2} and P_{3} so that the total work is minimized?

\underline{Solution}:

• Assume that the flow is ideal compressible adiabatic flow.

• Since the gas is assumed to be ideal, k = \frac{C_{p}}{C_{v}} = 1.4 and does not change in the pressure range P_{1} to P_{4}.

• The equation describing the work done for ideal compressible adiabatic flow can be found in standard text-books.

• The work done for each stage is given as follows:

\begin{aligned}W_{1} & = & \frac{kRT_{1}}{k − 1} [(\frac{P_{2}}{P_{1}} )^{\frac{k−1}{k}} − 1] \\ W_{2} & = & \frac{kRT_{1}}{k − 1} [(\frac{P_{3}}{P_{2}} )^{\frac{k−1}{k}} − 1] \\ W_{3} & = & \frac{kRT_{1}}{k − 1} [(\frac{P_{4}}{P_{3}} )^{\frac{k−1}{k}} − 1] \end{aligned}  (1)

• The total work done W is given by:

\begin{aligned}W & = & W_{1} + W_{2} + W_{3} \\ & = & \frac{kRT_{1}}{k − 1} [(\frac{P_{2}}{P_{1}} )^{\frac{k−1}{k}} + (\frac{P_{3}}{P_{2}} )^{\frac{k−1}{k}} + (\frac{P_{4}}{P_{3}} )^{\frac{k−1}{k}} − 3] \end{aligned}

(2)

• The above equation gives us the objective function to minimize, as a function of P_{2} and P_{3}.

• Clearly P_{2} and P_{3} will take values between P_{1} = 100 kP a and P_{4} = 1000 kP a. We first try to find the optimal values by posing the problem as an unconstrained minimization problem. If the solution is not in the range 100 − 1000 kP a, we will add

inequality constraints.

• To minimize W w.r.t. P_{2} and P_{3}, the necessary conditions of optimality are:

\begin{aligned}\frac{∂W}{∂P_{2}} & = & 0 \\ \frac{∂W}{∂P_{3}} & = & 0 \end{aligned}  (3)

• This leads to the following algebraic equations that need to be solved simultaneously:

\begin{aligned}[(P_{1})^{\frac{1−k}{k}} (P_{2})^{\frac{1}{k}} − (P_{3})^{\frac{k−1}{k}} (P_{2})^{\frac{1−2k}{k}} ] & = & 0 \\ [(P_{2})^{\frac{1−k}{k}} (P_{3})^{\frac{1}{k}} − (P_{4})^{\frac{k−1}{k}} (P_{3})^{\frac{1−2k}{k}} ] & = & 0 \end{aligned} (4)

• Substitute

k = 1.4 P_{1} = 100 kP a P_{4} = 1000 kP a

• Solving the two equations simultaneously for P_{2} and P_{3}, we obtain:

\begin{aligned}P_{2}^{∗} & = & 215.44 kP a \\ P_{3}^{∗} & = & 464.17 kP a \end{aligned}  (5)

• The sufficient conditions of optimality can be checked to ensure that the solution obtained is indeed a minimum and not a maximum or a saddle-point.

\underline{Example 2}: Consider the following distillation problem: The objective is to minimize Q_{1} by selecting F_{1}, F_{2}, F_{3}, F_{4}, Q_{1}.

The feed conditions are as follows:

\begin{aligned}T otal feed & = & 100 & lbmol/hr liquid \\ h_{f} & = & 4000 & Btu/lbmol \\ x_{1} & = & 0.05 & (C_{3}H_{8}) \\ x_{2} & = & 0.15 & (i − C_{4}H_{10}) \\ x_{3} & = & 0.25 & (n − C_{4}H_{10}) \\ x_{4} & = & 0.20 & (i − C_{5}H_{12}) \\ x_{5} & = & 0.35 & (n − C_{5}H_{12}) \end{aligned}  (6)

The desired distillate product specifications are:

\begin{matrix}10 lbmol/h liquid \\ with x_{5} ≤ 0.07 \end{matrix}  (7)

The reflux ratio used is 1.5 times the minimum reflux. Equality Constraints

• Total material balances (one for each stage k,

k = 1, 2, 3, 4)

F_{k}^{L} + F_{k}^{V} + V_{k−1} + L_{k+1} = V_{k} + L_{k}  (8)

• Component material balance (one for each component i for each stage k)

x_{i,k}^{F}F_{k}^{L}+y_{i,k}^{F}F_{k}^{V} +y_{i,k−1}V_{k−1}+x_{i,k+1}L_{k+1} = y_{i,k}V_{k}+x_{i,k}L_{k} (9)

• Energy balance (one for each stage)

Q_{k} + h_{k}^{F} F_{k} + H_{k−1}V_{k−1} + h_{k+1}L_{k+1} = H_{k}V_{k} + h_{k}L_{k} (10)

• Equilibrium relations for liquid and vapor at each stage (one for each stage)

y_{i,k} = K_{i,k}x_{i,k}  (11)

• Relation between equilibrium constant and p, T, x, y (one for each stage)

K_{i,k} = K_{i}(p_{k}, T_{k}, x_{k}, y_{k})  (12)

• Relation between enthalpies and p, T, x, y (one for each stage)

\begin{aligned}h_{k} & = & h(p_{k}, T_{k}, x_{k}) \\ H_{k} & = & H(p_{k}, T_{k}, y_{k}) \end{aligned}  (13)

There are other implicit equalities that must be satisfied.

• Overall liquid mole fraction at each stage:

\begin{matrix}∑_{i=1}^{m}x_{i,k} = 1 \\ ∑_{i=1}^{m}y_{i,k} = 1 \end{matrix}  (14)

• Sum of all the feeds is given in the problem statement:

∑_{i=1}^{m}F_{k} = 100  (15)

Inequality Constraints

\begin{aligned}Q_{1} & ≥ & 0 \\ Q_{4} & ≤ & 0 \\ x_{i,k} & ≥ & 0 \\ y_{i;k} & ≥ & 0 \\ F_{k} & ≥ & 0 \\ x_{5,4} & ≤ & 0.07 (given in problem statement) \end{aligned} (16)

Results of Optimization

Variable Initial Guess Optimal Value
F_{1}  25  23.7
F_{2}  25  0.0
F_{3}  25  0.0
F_{4}  25  76.3
Q_{1}  5.0 × 10^{6}  3.38 × 10^{5}

The optimal solution suggests that it is possible to use some of the cold feed as reflux in the top stage without voiding the product composition specification. This outcome is not an obvious choice for the problem specification.