Tier II: Case Studies

2. Tier II: Case Studies Section 2: Heat Exchange Network Optimization by Thermal Pinch Analysis

3. Optimization Problems • There are many different types of optimization problems • It is important to recognize that an optimization problem exists even if it does not immediately or easily lend itself to one of the previously described analytical methods of optimization • Sometimes an alternative method that is more case specific must be used

4. Optimization Problems • A common example of one of these problems is the optimization of a heat exchange network • Without knowing what the maximum possible network integration is, and the minimum possible heating and cooling utilities required, it can be very difficult to design an optimized heat exchange network

5. Optimization of Utility Use in a Heat Exchange Network • Heating and cooling utilities consumption can be treated as an optimization problem • The goal is to minimize the amount of heating and cooling utilities being used by optimizing the heat exchange network • A different method will be used for this type of optimization than what was seen previously

6. Constraints • Total heating (QH) and total cooling (QC) used will still need to be minimized according to a set of constraints • These constraints are: • The target temperature of individual streams • The minimum approach temperature in a heat exchanger

7. Constraints • Objective function: Minimize QH + QC • Constraints: • T2i = ai , T1i = bi • t1i = ci , t2i = di • DTmin = k

8. t2 T1 T2 t1 Minimum approach temperature T2 T1 t2 oC t1 Minimum Approach Temperature T1 – hot out T2 – hot in t1 – cold in t2 – cold out

9. Minimum Approach Temperature • To get the outlet temperature of one stream closer to the inlet temperature of the other stream, exchanger area must be increased, increasing capital cost • Decreased exchanger area means decreased capital cost, but increased utilities cost to make up for lost heat exchange capacity

10. Using Minimum Approach Temperature to Tradeoff Capital vs. Operating Costs • This graph demonstrates the tradeoff between capital and operating costs – a decrease in one is met with an increase in the other

11. Minimum Approach Temperature • The optimum exchanger size exists where the total annualized cost is minimized • This typically will correspond to a minimum approach temperature, DTmin of about 10oC • This DTmin = 10oC is a rule of thumb – it can change depending on the fluid service and the type of heat exchanger employed

12. Minimum Approach Temperature Thermal Equilibrium T = t Practical Feasibility T = t + DTmin • This must be included in the coming analysis

13. Graphical Method – Thermal Pinch Analysis • To optimize a heat exchange network, an example of the graphical method to determine the thermal pinch point will first be examined • The same example will then be solved using the algebraic method for comparison

14. Stream Data • Using the stream supply and target temperatures, the enthalpy change of each stream must be calculated • Enthalpy change: • DH = FiCpi(T2i – T1i) = HHi = FiCpi(t2i – t1i) = HCi • FiCpi = flow rate x specific heat (kW/K)

15. Stream Data

16. Stream Data • Stream data is then plotted as a series of straight line segments in order of ascending temperature • Each consecutive segment begins at the enthalpy level where the previous segment finished • A “hot” stream is any that must be cooled, while a “cold” stream is any that must be heated, regardless of supply temperature

17. Hot Streams

18. Cold Streams

19. Composite Stream Curves • Next the composite curves of the hot and cold streams must be constructed • These composite curves represent the total amount of heat to be removed from the hot streams and the total amount of heat that must be added to the cold streams to reach the target stream temperatures

20. Hot Composite Stream Construction H3 H2 H1 T11 T21 T12 T13 T22 T23

21. Hot Composite Stream Construction Hot composite stream

22. Cold Composite Stream Construction C3 C2 C1 t13 t22 t21 t11 t12 t23

23. Cold Composite Stream Construction Cold composite stream

24. Optimizing the Heat Exchange Network • The cold composite stream must now be superimposed over the hot composite stream to perform the thermal pinch analysis • This will give the minimum amount of utilities required to reach the target states • Note how the temperature axis is shifted for the cold composite stream to account for the minimum approach temperature

25. 240 No Heat Integration Cold composite stream Total hot utility required QH,max = 65,000 kW Total cold utility required Hot composite stream QC,max = 67,000 kW QC + QH = 132,000 kW

26. No Heat Integration • With no heat integration, the amount of energy required to reach the target state is maximized • In this case the total amounts of energy required are: • Cooling utility, QC = 67,000 kW • Heating utility, QH = 65,000 kW • Total utilities = QC + QH = 132,000 kW • Clearly there is room for optimization

27. Partial Heat Integration • By moving the cold composite stream down a bit, a partially integrated heat exchange network is graphically represented • Some heat is transferred from hot streams to cold streams to approach the temperature targets

28. Total hot utility required QH = 50,000 kW Integrated heat exchange 15,000 kW Total cold utility required QC = 52,000 kW Partial Heat Integration Cold composite stream Hot composite stream QC + QH = 102,000 kW

29. Partial Heat Integration • This heat exchange network is only partially optimized and already utility consumption is reduced by 30,000 kW • The utilities required are: • Cooling utility, QC = 52,000 kW • Heating utility, QH = 50,000 kW • Total utilities = QC + QH = 102,000 kW • Clearly further integration can provide significant energy savings

30. Optimized Heat Integration • To determine the optimized heat exchange network, the thermal pinch point must be found • This is accomplished by moving the cold composite stream down just until one point on the line meets a point on the hot composite line • This point is the thermal pinch point

31. Optimized Heat Integration QH,min = 8,500 kW Cold composite stream Integrated heat exchange = 56,500 kW Pinch point Hot composite stream QC,min = 10,500 kW 240 QC + QH = 19,000 kW

32. Optimized Heat Integration • The heat exchange network is now fully optimized • Total required utilities are minimized • Minimum cooling utility, QC,min = 10,500 kW • Minimum heating utility, QH,min = 8,500 kW • Minimum total utilities = QC + QH = 19,000 kW • No heat is passed through the pinch point

33. Passing Heat through the Pinch Point • To have an optimized heat exchange network, it is critical that no heat is passed through the thermal pinch point • By passing an amount of heat, a, through the pinch point, an energy penalty of 2a is added to the total utilities requirement • It is very important to maximize integration in a heat exchange network

34. a QH = QH,min + a QH,min a a QC = QC,min + a QC,min Passing Heat Through the Pinch Point QH + QC = QH,min + QC,min +2a

35. Crossing the Pinch Point • It would appear that extra energy can be saved by lowering the cold composite stream line further • This does not work however because it creates a thermodynamically infeasible region • For this to work, heat would have to flow from the cooled hot streams to the heated cold streams - from a cold source to a hot source

36. Crossing the Pinch Point Cold composite stream Pinch point Hot composite stream Infeasible region

37. Disregarding DTmin • Another tempting error is to disregard the minimum approach temperature • By disregarding a minimum approach temperature, the absolute minimum thermodynamically possible utility requirements are obtained • Although this is thermodynamically possible, it is not practically feasible as it would require an infinitely large heat exchanger area • This would obviously cost far more than the relatively small energy savings are worth

38. Disregarding DTmin QH,min thermo. QC,min thermo. 240

39. Algebraic Method • This same problem will now be solved using the algebraic method • This will involve producing a temperature interval diagram, tables of exchangeable heat loads, and cascade diagrams

40. Stream Data From before:

41. Temperature Interval Diagram • The first step is to construct the temperature interval diagram • This diagram shows the starting and finishing temperatures of each stream • An interval begins at a stream’s starting or finishing temperature, and it ends where it encounters the next beginning or finishing temperature of a stream • Draw horizontal lines across the table at each arrow’s head and tail, with the intervals lying between these lines • Note how the cold stream temperature scale is staggered by 10 degrees

42. Temperature Interval Diagram

43. Table of Exchangeable Heat Loads • The next step is to construct tables of exchangeable heat loads for the hot and cold streams • These tables show the amount of energy that must be added or removed from a stream over a particular interval • These energy values are calculated as DHj,i = FCpjDTi, where DTi is the positive temperature difference across the interval, and j denotes the stream number

44. Table of Exchangeable Heat Loads • For the hot streams,

45. Table of Exchangeable Heat Loads • For the cold streams,

46. Cascade Diagrams • Using the information from the heat load tables, the cascade diagrams can now be constructed • These diagrams will be used to determine the pinch point and the minimum heating and cooling utilities required

47. First, the cascade diagram is drawn as it appears at right, with one box for each interval that appeared in the temperature interval diagram Cascade Diagram

48. Next, the total values from the exchangeable heat load tables are added to the cascade diagram Hot stream loads enter on the left, cold stream loads exit on the right Cascade Diagram

49. Now, by subtracting an interval’s cold load from the hot load, and adding the resulting value to the residual from the previous stage we get the residual value for the subsequent stage ri = HHi – HCi + ri-1 Cascade Diagram 0 12000 7500 5500 -2500 -8500 -5500 -1500 4500 9) 0 – 2500 + 4500 = 2000 7) 4000 – 0 – 5500 = -1500 8) 16000 – 10000 – 1500 = 4500 6) 12000 – 9000 – 8500 = -5500 4) 7000 – 15000 + 5500 = -2500 3) 13000 – 15000 + 7500 = 5500 2) 3000 – 7500 + 12000 = 7500 1) 12000 – 0 + 0 = 12000 5) 0 – 6000 -2500 = -8500 2000

50. The thermal pinch point occurs at the largest negative number The absolute value of this number is now added in at the top to cascade through Pinch Point Thermal Pinch Point