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  1. Tier III: Optimization Design Problems Derek McCormack Section 1: Sample Problems

  2. Introduction Three sample problems have been given here to work on. The first is a heat exchange network optimization problem. The second is a transportation optimization problem to be solved with Lingo. The third problem deals with optimizing a heat exchanger’s minimum approach temperature.

  3. Question #1 Optimization of a Heat Exchange Network by Thermal Pinch Analysis

  4. Optimization of a Heat Exchange Network A plant has the following stream data:

  5. HEN Problem Using the stream data given and a DTmin of 10 K, do the following : • Determine the optimum heating and cooling utilities required by using the algebraic thermal pinch analysis method. Do you notice anything special with this example? • Now solve this problem using the graphical method, keeping in mind the results obtained above. • Create a possible heat exchange network for this situation based on the optimized conditions.

  6. HEN Solution Attempt to solve this problem before proceeding to the solution.

  7. Temperature Interval Diagram

  8. Table of Exchangeable Heat Loads

  9. Table of Exchangeable Heat Loads

  10. Cascade Diagram

  11. No Pinch Point? Notice that in this case the cascade diagram has no residuals that fall below zero. In this case, all of the heating needs of the cold streams are met by the hot streams, with an excess of heat left over. No heating utility is required, and the minimum cooling utility is 11,000 kW.

  12. Hot Composite Stream

  13. Cold Composite Stream

  14. Optimized

  15. No Pinch Point? Here we can see that we do not get a typical pinch point. The head of the cold composite stream cannot be moved below the tail of the hot composite stream. In this case, all of the heating requirements can be met by the hot streams, but 11,000 kW of cooling utility are still needed.

  16. Question #2 Optimization of Transportation Route Problem

  17. Transportation Problem Five chemical plants produce a chemical to be shipped and sold at three different selling stations. Each plant has a different production cost and shipping cost, while each warehouse that receives the product sells it for a different price. Warehouse 1 sells for 95 $/tonne, warehouse 2 sells for 90 $/tonne, and warehouse 3 sells for 93 $/tonne. The cost of production at each of the plants are as follows: plant 1 costs 42 $/tonne, plant 2 costs 45 $/tonne, plant 3 costs 43 $/tonne, plant 4 costs 46 $/tonne, and plant 5 costs 55 $/tonne. To ship from plant 1 costs 0.30 $/tonne•km, from plant 2 costs 0.35 $/tonne•km, from plant 3 costs 0.31 $/tonne•km, from plant 4 costs 0.34 $/tonne•km, and plant 5 costs 0.29 $/tonne•km.

  18. Transportation Problem The distances between plants and warehouses, in km, are as follows: Plant 1 has a production capacity of 1300 tonnes, plant 2 can make 1200 tonnes, plant 3 can make 1700 tonnes, plant 4 can make 1400 tonnes, and plant 5 can make 1600 tonnes. Furthermore, market research suggests that the amount sold at each warehouse is limited. Warehouse 1 can receive 2400 tonnes, warehouse 2 can receive 2000 tonnes, and warehouse 3 can receive 2500 tonnes.

  19. Transportation Problem What combination of shipments will maximize the profit that this company can earn, and what is that profit? Use Lingo to solve this. Attempt to solve this problem before proceeding to the solution.

  20. Transportation Problem Solution Before Lingo can be used, this problem must be broken down into components: Profit = Revenue – Expenses What is revenue? Revenue = S(selling price)*(quantity sold) = SP1(Sx1j) + SP2(Sx2j) + SP3(Sx3j) (i refers to a warehouse property, while j refers to a plant property)

  21. Transportation Problem Solution What are the expenses? The cost of production and the cost of shipping. Expenses = Production cost + Shipping cost The costs of shipping from each plant to each warehouse are given below.

  22. Transportation Problem Solution Production cost = S(cost per unit)*(quantity produced) =SCj*x1j + SCj*x2j + SCj*x3j Shipping cost = S(quantity shipped)*(shipping price) =Sx1j*S1j + Sx2j*S2j + Sx3j*S3j

  23. Transportation Problem Solution The objective function is now: maximize

  24. Transportation Problem Solution The constraints: Sx1j = 2400 Sx2j = 2000 Sx3j = 2500 Sxi1 <= 1300 Sxi2 <= 1200 Sxi3 <= 1700 Sxi4 <= 1400 Sxi5 <= 1600

  25. Lingo Solution

  26. Lingo Solution

  27. Question #3 Optimizing Minimum Approach Temperature

  28. DTmin Optimization A hot process stream coming out of a distillation tower has a specific heat flow rate, FCp, of 200 kW/K and must be cooled from 400 K to 300 K. Another process stream with an FCp of 150 kW/K must be heated from 330 K to 430 K before it enters a processing unit. A significant savings in utility costs can be realized by passing these streams through a heat exchanger.

  29. DTmin Optimization Heating utility is available at a cost of approximately 90 $/kW•year, while cooling utility is available at approximately 40 $/kW•year. Based on an expected useful life of 10 years, the heat exchanger is estimated to have an annualized fixed cost of about 600 $/year•m2. If the heat exchanger is expected to have a heat exchange coefficient of U = 1.2 kW/m2, investigate where the optimum minimum approach temperature lies. Hint: It is between DTmin = 5 K and 20 K.

  30. DTmin Optimization What is the optimum minimum approach temperature in this case? Use DTmin = 5 K, 10 K, and 20 K to develop your solution. Attempt to solve this problem before proceeding to the solution.

  31. Optimum DTmin Solution Using the algebraic method, the utility requirements and exchanged heat are calculated for each DTmin.

  32. Optimum DTmin Solution Next, for each case the inlet and outlet temperatures of the heat exchanger are calculated so that the log mean temperature difference can be calculated.

  33. Optimum DTmin Solution Then the area of each heat exchanger is calculated.

  34. Optimum DTmin Solution Finally, the annual utilities cost, heat exchanger cost, and total cost are calculated and plotted as a function of DTmin.

  35. Optimum DTmin Solution

  36. The End This is the end of the Process Optimization module.