Consider the problem of minimizing fuel costs in a boilerhouse. The boilerhouse contains two turbine generators, each of which can be simultaneously operated with two fuels: fuel oil and medium Btu gas (MBG); see Fig. 19.9. TheMBG is produced as a waste off-gas from another part of the plant, and i

Question

Consider the problem of minimizing fuel costs in a boilerhouse. The boilerhouse contains two turbine generators, each of which can be simultaneously operated with two fuels: fuel oil and medium Btu gas (MBG); see Fig. 19.9. TheMBG is produced as a waste off-gas from another part of the plant, and it must be flared if it cannot be used on site. The goal of the RTO scheme is to find the optimum flow rates of fuel oil and MBG and provide 50 MW of power at all times, so that steady-state operations can be maintained while minimizing costs. It is desirable to use as much of the MBG as possible (which has zero cost) while minimizing consumption of expensive fuel oil. The two turbine generators [latex](G_{1} , G_{2} )[/latex] have different operating characteristics; the efficiency of [latex]G_{1}[/latex] is higher than that of [latex]G_{2}[/latex] . Data collected on the fuel requirements for the two generators yield the following empirical relations:

[latex]P_{1} = 4.5x_{1} + 0.1 x^{2}_{1} + 4.0x_{2} + 0.06 x^{2}_{2}[/latex] (19-33)
[latex]P_{2} = 4.0x_{3} + 0.05 x^{2}_{3}+ 3.5x_{4} + 0.02 x^{2}_{4}[/latex] (19-34)

where

[latex]P_{1}[/latex] = power output (MW) from [latex]G_{1}[/latex]
[latex]P_{2}[/latex] = power output (MW) from [latex]G_{2}[/latex]
[latex]x_{1}[/latex] = fuel oil to [latex]G_{1}[/latex] (tons ∕ h)
[latex]x_{2}[/latex] = MBG to [latex]G_{1}[/latex] (fuel units ∕ h)
[latex]x_{3}[/latex] = fuel oil to [latex]G_{2}[/latex] (tons ∕ h)
[latex]x_{4}[/latex] = MBG to [latex]G_{2}[/latex] (fuel units ∕ h)

The total amount of MBG available is 5 fuel units/h. Each generator is also constrained by minimum and maximum power outputs: generator 1 output must lie between 18 and 30 MW, while generator 2 can operate between 14 and 25 MW.
Formulate the optimization problem by applying the methodology described in Section 19.2. Then solve for the optimum operating conditions ([latex]x_{1}, x_{2}, x_{3}, x_{4}, P_{1}, P_{2}[/latex] ) using the Excel Solver.

Accepted Answer

Step 1. Identify the variables. Use [latex]x_{1}[/latex] through [latex]x_{4}[/latex] as the four process variables. Variables [latex]P_{1}[/latex] and [latex]P_{2}[/latex] are dependent because of the equality constraints (see Steps 3 and 4).

Step 2. Select the objective function. The way to minimize the cost of operation is to minimize the amount of fuel oil consumed. This implies that we should use as much MBG as possible, because it has zero cost. The objective function can be stated in terms of variables defined above; that is, we wish to minimize

[latex]f = x_{1}+ x_{3}[/latex] (19-35)

Step 3. Specify process model and constraints. The constraints given in the problem statement are as follows:

(1) Power relation

[latex]P_{1} = 4.5x_{1} + 0.1 x^{2}_{1} + 4.0x_{2} + 0.06 x^{2}_{2}[/latex] (19-33)
[latex]P_{2} = 4.0x_{3} + 0.05 x^{2}_{3}+ 3.5x_{4} + 0.02 x^{2}_{4}[/latex] (19-34)

(2) Power range

[latex]18 ≤ P_{1} ≤ 30[/latex] (19-36)
[latex]14 ≤ P_{2} ≤ 25[/latex] (19-37)

(3) Total power [latex] 50 = P_{1} + P_{2}[/latex] (19-38)

(4) MBG supply [latex] 5 = x_{2} + x_{4} [/latex] (19-39)

Note that all variables defined above are nonnegative.

Step 4. Simplify the model and objective function. This step is not need because this is a fairly small NLP problem.

Step 5. Compute the optimum. Using Excel Solver, the optimum is [latex]f = 6.39[/latex] , [latex]x_{1} = 5.89[/latex] , and [latex]x_{3}= 0.50[/latex] , meaning that [latex] 5.89 tons/h[/latex] of fuel oil are delivered to generator [latex]G_{1}[/latex], while [latex]0.50 tons/h[/latex] are used in [latex]G_{2}[/latex]. [latex]G_{2}[/latex] utilizes all of the MBG ([latex]x_{4}[/latex]), while [latex]G_{1}[/latex] uses none ([latex]x_{2}= 0[/latex]).

Step 6. Perform a sensitivity analysis. Many operating strategies may be satisfactory, though not optimal, for the above problem. In fact, the Excel Solver will converge to a local minimum if a starting point of [latex]x_{1} = 2, x_{2} = x_{3} = 4, x_{4} = 1[/latex] is used ([latex]f = 6.54[/latex]). Parameters in the original constraint equations can change as plant operating conditions vary. For example, suppose the total power requirement is changed to [latex]55 MW[/latex]; as an exercise, determine whether any of the active constraints change for the increased power requirement.

Question 19.4: Consider the problem of minimizing fuel costs in a boilerhou...

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