Tier III: Optimization Design Problems

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(No Transcript)
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Tier III Optimization Design Problems

• Derek McCormack
• Section 1
• Sample Problems

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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 exchangers minimum approach
  temperature.

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Question 1

• Optimization of a Heat Exchange Network by
  Thermal Pinch Analysis

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Optimization of a Heat Exchange Network

• A plant has the following stream data

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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.

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HEN Solution

• Attempt to solve this problem before proceeding
  to the solution.

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Temperature Interval Diagram
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Table of Exchangeable Heat Loads
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Table of Exchangeable Heat Loads
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Cascade Diagram
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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.

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Hot Composite Stream
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Cold Composite Stream
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Optimized
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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.

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Question 2

• Optimization of Transportation
• Route Problem

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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 /tonnekm, from plant 2 costs 0.35
  /tonnekm, from plant 3 costs 0.31 /tonnekm,
  from plant 4 costs 0.34 /tonnekm, and plant 5
  costs 0.29 /tonnekm.

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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.

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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.

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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)

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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.

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Transportation Problem Solution

• Production cost S(cost per unit)(quantity
  produced)
• SCjx1j SCjx2j SCjx3j
• Shipping cost S(quantity shipped)(shipping
  price)
• Sx1jS1j Sx2jS2j Sx3jS3j

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Transportation Problem Solution

• The objective function is now
• maximize

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Transportation Problem Solution

• The constraints
• Sx1j 2400
• Sx2j 2000
• Sx3j 2500
• Sxi1 lt 1300
• Sxi2 lt 1200
• Sxi3 lt 1700
• Sxi4 lt 1400
• Sxi5 lt 1600

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Lingo Solution
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Lingo Solution
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Question 3

• Optimizing Minimum Approach Temperature

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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.

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DTmin Optimization

• Heating utility is available at a cost of
  approximately 90 /kWyear, while cooling utility
  is available at approximately 40 /kWyear. Based
  on an expected useful life of 10 years, the heat
  exchanger is estimated to have an annualized
  fixed cost of about 600 /yearm2. 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.

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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.

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Optimum DTmin Solution

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

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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.

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Optimum DTmin Solution

• Then the area of each heat exchanger is
  calculated.

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Optimum DTmin Solution

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

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Optimum DTmin Solution
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The End

• This is the end of the Process Optimization
  module.