1. Pinch Technology and optimization of the use of utilities – part two Maurizio Fermeglia maurizio.fermeglia@di3.units.it www.mose.units.it

2. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 2 Introduction to HEN Synthesis – Summary of part 1 Unit 1. Introduction: Capital vs. Energy What is an optimal HEN design Setting Energy Targets Unit 2. The Pinch and MER Design The Heat Recovery Pinch HEN Representation MER Design: (a) MER Target; (b) Hot- and cold-side designs Unit 3. The Problem Table for MER Targeting

3. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 3 Introduction to HEN: Part two Unit 4. Loops and Splits Minimum Number of Units by Loop Breaking Class Exercise 5 Stream Split Designs Class Exercise 6 Unit 5. Threshold Problems Class Exercise 7

4. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 4 Part Two: Objectives This Unit on HEN synthesis serves to expand on what was covered to more advanced topics. Instructional Objectives - You should be able to: Identify and eliminate “heat loops” in an MER design (lower the n. of HEx) Use stream splits to design for Umin and MER (minimize n. of HEx and MER target) Design a HEN for “Threshold Problems”

5. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 5 UNIT 4: Loops and Splits The minimum number of units (U min ) in a network: U Min = N Stream + N Util  1 (Hohman, 1971) A HEN containing U HEX units (U HEX  U min ) has (U HEX  U min ) independent “heat loops”. The HEN above has 2 “heat loops” Normally, when heat loops are “broken”, heat flows across the pinch - the number of heat exchangers is reduced, but the utility loads are increased.

6. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 6 Class Exercise 5 (Linnhoff and Flower, 1978)  T min = 10 o C. Example: Step 1: Temperature Intervals (subtract  T min from hot temperatures) Temperature intervals: 180 o C  170 o C  140 o C  130 o C  60 o C  30 o C

7. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 7 Class Exercise 5 (Cont’d) Step 2: Interval heat balances For each interval, compute:  H i = (T i  T i+1 )  (  CP Hot  CP Cold )

8. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 8 Class Exercise 5 (Cont’d) Step 3: Form enthalpy cascade. This defines: Cold pinch temp. = 140 o C Q Hmin = 60 kW Q Cmin = 160 kW

9. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 9 Class Exercise 5 (Cont’d) MER Design above the pinch: MER Design below the pinch: U Min,MER = N Stream + N Util - 1 = 2 + 1 – 1 = 2 U Min,MER = 4 + 1 – 1 = 4 MER design below pinch has 6 exchangers! i.e. There are two loops below pinch.

10. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 10 Class Exercise 5 (Cont’d) Complete MER Design However, U Min = N Stream + N Util  1 = 4 + 2  1 = 5 The MER network has 8 units. This implies 3 independent “heat load loops”. We shall now identify and eliminate these loops in order to design for U Min

11. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 11 Class Exercise 5 (Cont’d) Identification and elimination of 1st loop: To reduce the number of units, the 80 kW exchanger is merged with the 60 kW exchanger. This breaks the heat loop, but also creates a  T min violation in the network:

12. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 12 Class Exercise 5 (Cont’d) Identification and elimination of 1st loop (Cont’d): To restore  T min, the loads of the exchangers must be adjusted along a “heat path” by an unknown amount x. A “heat path” is a path through the network that connects heaters with coolers.

13. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 13 Class Exercise 5 (Cont’d) Identification and elimination of 1st loop (Cont’d): Performing a heat balance on H1 in the exchanger which exhibits the  T min violation: 140 - x = 2(180 - 113.33 -  T min )  x = 26.66 This is called “energy relaxation”

14. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 14 Class Exercise 5 (Cont’d) Identification and elimination of 2nd loop: Since there is no  T min violation, no adjustment of the loads of the exchangers is needed - we reduce the number of units by one with no energy penalty.

15. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 15 Class Exercise 5 (Cont’d) Identification and elimination of 3rd loop: Shifting the load of the smallest exchanger (93.33 kW) around the loop, the network is reduced to…

16. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 16 Class Exercise 5 (Cont’d) Identification and elimination of 3rd loop: We use the heat path to restore  T min: 253.33 - x = 3(150 -  T min - 60)  x = 13.33

17. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 17 Class Exercise 5 (Cont’d) Therefore U min Network is:

18. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 18 Step 3: For each loop, eliminate a unit. If this causes a  T min violation, identify the “heat path” and perform “energy relaxation” by increasing the duties of the cooler and heater on the heat path. Step 2: Compute the minimum number of units: U Min = N Stream + N Util  1 This identifies U HEX  U min independent “heat loops”, which can be eliminated to reduce U. Loop Breaking - Summary Step 1: Perform MER Design for U HEX units. Try and ensure that design meets U Min,MER separately above and below the pinch. Loops improve energy recovery and heat load flexibility at the cost of added units (>U min )

19. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 19 Stream Split Designs Example. Option 1.

20. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 20 Stream Split Designs (Cont’d) Option 2. Loops Option 3. Stream Splitting

21. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 21 Loops vs. Stream Splits Loops: Improved energy recovery (normally) Heat load flexibility (normally)  U > U Min (by definition) Stream Splitting: Maximum Energy recovery (always) Branch flowrate flexibility (normally) U = U Min (always) Stream splitting is a powerful technique for better energy recovery BUT - Don’t split unless necessary

22. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 22 Stream Splitting Rules 1. Above the pinch (at the pinch):  Cold utilities cannot be used (for MER). So, if N H > N C, MUST split COLD streams, since for feasibility N H  N C  Feasible matches must ensure CP H  CP C. If this is not possible for every match, split HOT streams as needed. If Hot steams are split, recheck  2. Below the pinch (at the pinch):  Hot utilities cannot be used (for MER). So, if N C > N H, MUST split HOT streams, since for feasibility N C  N H  Feasible matches must ensure CP C  CP H. If this is not possible for every match, split COLD streams as needed. If Cold steams are split, recheck 

23. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 23 Class Exercise 6 Design a hot-side HEN, given the stream data below: Solution: Since N H > N C, we must split C1. The split ratio is dictated by the rule: CP H  CP C (necessary condition) and by a desire to minimize the number of units (“tick off “streams) 500 200

24. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 24 Class Exercise 6 (Cont’d) x is determined by the following energy balances: x (T 1  90) = 500 (10  x )(T 2  90) = 200 subject to: 200  T 1   T min = 10 150  T 2   T min = 10 Best to make T 1 = T 2. Here, this is not possible. Why? We shall make T 2 = 140 (why?)

25. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 25 Class Exercise 6 (Cont’d) A possible solution is therefore: (10  x ) (140  90) = 200  x = 6 T 1 = 90 + 500/ x = 173.33 (satisfies constraint) Complete solution is: This is an MER design which also satisfies U Min (U Min = 3).

26. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 26 Practice Exercise 1 Design a hot-side network for MER and U Min given the stream data below.

27. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 27 Practice Exercise 2 Determine QHmin, QCmin and the pinch location. Design an MER network which satisfies energy targets Design a network for UMin by eliminating the heat loops in the network and performing energy relaxation. Data:  T min = 20 o F

28. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 28 UNIT 5: Threshold Problems Example - Consider the problem Networks with excess heat supply or heat demand may have MER targets with only one utility (i.e., either Q Hmin = 0 or Q Cmin = 0). Such designs are not separated at the pinch, and are called “Threshold Problems”

29. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 29 Threshold Problems (Cont’d) Assuming a value of  T min = 10 o C, the Problem Table gives the following result. Assuming a value of  T min = 105 o C:

30. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 30 Threshold Problems (Cont’d) Threshold problems do not have a pinch, and have non-zero utility duties only at one end.

31. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 31 Threshold Problems (Cont’d) However, increasing driving forces beyond the Threshold Value leads to additional utility requirements.

32. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 32 Threshold Design Guidelines 1. Establish the threshold  T min 2. Note the common practice values for  T min : 3. Compare the threshold  T min to  T min,Experience Classify as one of the following: Pinched - treat as normal pinched problem Threshold - must satisfy target temperatures at the “no utility end”

33. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 33 Class Exercise 7 The graph shows the effect of  T min on the required levels of Q Hmin and Q Cmin for a process consisting of 3 hot and 4 cold streams.

34. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 34 Class Exercise 7 (Cont’d) Design a network for U min and MER for the process. Hint: Identify two essential matches by satisfying target temperatures at the “no utility end”

35. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 35 Class Exercise 7 - Solution Note: U Min = N Streams + N Utilities  1 = 7

36. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 36 Advanced HEN Synthesis - Summary Unit 4. Loops and Splits Minimum Number of Units by Loop Breaking - Umin Stream Split Designs - Umin and MER Unit 5. Threshold Problems Problems with only hot or cold utilities (no pinch!)

37. Introduction to HEN software Ref. Turton et al. Analysis, Synthesis and Design of Chemical Processes

38. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 38 The MUMNE algorithm The Minimum Utility steps: Choose a minimum approach temperature (parametric optimization) Construct a temperature interval diagram Construct a cascade diagram and determine the minimum utility requirements and the pinch temperatures Calculate the minimum number of heat exchangers above pinch Calculate the minimum number of heat exchangers below pinch Construct the heat exchanger network The object is to obtain an heat exchanger network …That exchange the minimum energy between the streams and the utilities …That uses the minimum number of equipment

39. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 39 Algorithm: initial condition and step 1 Minimum Temperature Approach = Smallest  T of two streams leaving or entering an heat exchanger = 10°C Hot Stream Data Mass Flow Cp Temp In Temp Out Stream Enthalpy Film Heat Transf. Coef kg/s kJ/kg/°C °C °C kW W/m2/°C 10.00.8000 300.0 150.0 1200. 400.0 2.500.8000 150.0 50.00 200.0 270.0 3.000 1.000 200.0 50.00 450.0 530.0 Cumulative Hot Stream Energy Available = 1850.0 kW Cold Stream Data Mass Flow Cp Temp In Temp Out Stream Enthalpy Film Heat Transf. Coef kg/s kJ/kg/°C °C °C kW W/m2/°C 6.250.8000 190.0 290.0 -500.0 100.0 10.00.8000 90.00 190.0 -800.0 250.0 4.000 1.000 40.00 190.0 -600.0 80.00 Cumulative Cold Stream Energy Available = -1900.0 kW

40. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 40 Algorithm: construct Temperature interval diagram (step 2) Process streams represented by vertical lines Axes are shifted by the minimum T approach

41. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 41 Algorithm: construct a cascade diagram (1) Shows the net amount of energy in each interval diagram Cascade  if there is an excess energy in a T interval we may “cascade” it down Energy cannot be transferred up (II law) Line is the point at which no more energy can cascade down We need to resort to utilities NOTE: not all problems have a pinch condition: the algorithm is still valid Pinch temperature

42. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 42 Algorithm: construct a cascade diagram (2) Additional heat is transferred to the C interval (yellow line) Energy is cascaded down through the pinch and rejected to the cold utility If heat is transferred across the pinch, the net result will be that more heat will have to be added from the hot utility and rejected to the cold utility To minimize the hot and cold utility requirements, energy should NOT be transferred across the pinch Pinch temperature

43. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 43 Algorithm: minimum n. of exchangers Above the pinch Draw boxes representing energy in the hot and cold process streams and utilities Transfer energy is indicated by lines (with the amount) For each line an heat exchanger is required The problem is split into two sub problems

44. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 44 Algorithm: minimum n. of exchangers Below the pinch Draw boxes representing energy in the hot and cold process streams and utilities Transfer energy is indicated by lines (with the amount ) For each line an heat exchanger is required The problem is not split into sub problems

45. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 45 Algorithm: minimum n. of exchangers General relationship For any sub problem With or without a pinch Above or below the pinch Min. No. of exchangers = No. of hot streams + No. of cold streams + No. of utilities – 1

46. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 46 Algorithm: Design the network above the pinch Start from the design at the pinch To make sure that  T min is not violated, match streams such that Note that we consider ONLY streams present at the pinch Each exchanger is represented by two circles connected with a line, each circle represent a side

47. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 47 Algorithm: Design the network above the pinch Move away from the pinch Look at the remaining streams Criterion used at pinch not necessarily holds away from the pinch The following constraints are not violated: The minimum approach T is used throughout the design The number of exchangers must be that calculated in step 3 Heat is added form the coolest possible source

48. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 48 Algorithm: Design the network below the pinch Similar to previous one: start from the design at the pinch To make sure that  T min is not violated, match streams such that Note that we consider ONLY streams present at the pinch Each exchanger is represented by two circles connected with a line, each circle represent a side What happens if the  T min is violated (see figure)

49. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 49 Algorithm: Design the network below the pinch Split stream into substreams to meet the  T min criterion

50. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 50 Algorithm: Design the network below the pinch Move away from the pinch Look at the remaining streams

51. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 51 Final result The final network of heat exchangers is the following It has the minimum n. of exchangers (8) Minimum utility requirement (Qh = 100 kW and Qc = 50 kW) Using a minimum approach T = 10°

52. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 52 Heat exchangers design: area and costs Up to now, emphasis on the topology of the network … to complete the design, it is necessary to Estimate the heat transfer area (A= Q / (U  T ln F) And the cost estimate If heat transfer coefficients are known (including fouling)… Transfer coefficients form literature (Seider – Tate, Donahue, …) … exchanger area can be calculated (for streams exchangers) ExchangerDT lnUQF factorArea. °CW/m 2 /°CkWm 2 124.1129.81000.840.0 220.069.53000.8270.3 347.5153.87000.8119.7 424.180.05000.8324.6 610.061.71000.8202.5 717.069.51000.8195.8 TOTAL1063.0 Exchanger 5 requires an hot utility (steam):  T = 76.8 °C, U= 76.9 W/m 2 /°C, Area = 16.9 m 2 Exchanger 8 requires cooling water:  T = 23.2 °C, U= 346 W/m 2 /°C, Area = 7.8 m 2 TOTAL Area: 1087.7 m 2

53. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 53 Effect of the minimum approach temperature Calculations must be repeated for different approach T (step 1) Problem: step 5 (matching streams and exchanging energy) cannot be programmed easily An approximate approach is necessary for investigating the effect of the approach temperature on the total cost Based on the Composite temperature enthalpy diagram Constructed by plotting enthalpy of all streams as a function of T

54. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 54 Construction of Composite T-H diagram Temperature Interval T °CEnthalpy of COLD streams in Temperature interval (kW) Cumulative Enthalpy of COLD streams (kW) D4000 C90(4)(90-40)= 200200 B140(8+4)(140-90)= 600800 A190(8+4)(190-140)= 6001400 290(5)(290-190)= 5001900 Temperature Interval T °CEnthalpy of HOT streams in Temperature interval (kW) Cumulative Enthalpy of HOT streams (kW) D5000 C100(2+3)(100-50)= 250250 B150(2+3)(150-100)= 250500 A200(8+3)(200-150)= 501050 300(8)(300-200)= 8001850

55. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 55 Construction of Composite T-H diagram

56. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 56 Composite T – H diagram; pinch and no pinch

57. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 57 Using the T-H diagram to estimate heat exchanger area The working equation is (A= Q / (U  T ln F) Consider a portion of the diagram (figure)

58. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 58 Results of the heat transfer area calculations

59. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 59 Finalize the design Consider the F factor By calculating the number of shells in a 1-2 geometry exchanger The effect of the ‘economy of scale’: the cost of two 1-2 S&T in series is greater than the equivalent 1-2 S&T with the same total area Calculate the cost of equipments Knowing the number of shells for each exchanger Using cost correlations Applying an economic criteria (such as Equivalent Annual Operating Cost – EAOC) Approximations All heat exchangers have the same area  over estimation of the costs No effect of material of construction (corrections available) No effect of operating pressure (corrections available) No multiple utilities (alternative methods) No streams with phase change (correction available)

60. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 60 Calculation of the costs for the network

61. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 61 Typical relationship for heat transfer area, utilities and EAOC for a HEN

62. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 62 Mass – exchange networks Exchangers to use energy more efficiently Temp Interval Diagram Cascade diagram – pinch Min n. of eq. above and below Composite T exchange diag. Final T exchange network Utility: cold and hot source of energy Hot and cold Temperature Separators to use mass more efficiently Composition Interval Diagram Cascade diagram – pinch Min n. of eq. above and below Composite mass exchange diag. Final mass exchange network Utility: separation - addition of solute (from source or to sink) Rich and lean Concentration

63. Aspentech Energy Analyzer

64. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 64 The streams NAMET in (°C) T out (°C) MCp (kW/°C) H14001508 H2200505 H32501006 NAMET in (°C) T out (°C) MCp (kW/°C) C419039010 C5401408 C6902405 Hot streams: Cold streams:

65. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 65 Temperature interals (  T min= 10)

66. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 66 The composite curve

67. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 67 TiTi T i -T i+1  MCp HOT -  MCp COLD QiQi Q cumulato (Q H =0) Q cumulato (Q H =350 ) 390150-2-300 50 24050-50-3500 1905014700350700 14050-3-100250600 9050-2-150100450 40 Cumulative Heat COLD PINCH TEMPERATURE 19 0 +  T min 20 0 HOT PINCH TEMPERATURE

68. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 68 Design rules Geenral: Never exchge heat cross pinch Above the pinch  MCp HOT  MCp COLD  Do not use coolers Belove the pinch  MCp COLD  MCp HOT  Do not use heaters

69. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 69 Manual solution (HENSAD) Above the Pinch

70. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 70 Manual solution (HENSAD) Belove the Pinch

71. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 71 ASPEN ENERGY ANALIZER PROCESS STREAM DATA

72. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 72 ASPEN ENERGY ANALIZER PROCESS UTILITIES DATA

73. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 73 ASPEN ENERGY ANALIZER PROCESS ECONOMIC DATA

74. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 74 ASPEN ENERGY ANALIZER PROCESS TARGET SUMMARY

75. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 75 ASPEN ENERGY ANALIZER

76. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 76

77. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 77 ASPEN ENERGY ANALIZER AUTOMATIC HEN DESIGN OPTIONS

78. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 78 AUTOMATIC HEN DESIGN NETWORKS ASPEN ENERGY ANALIZER

79. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 79 ASPEN ENERGY ANALIZER AUTOMATIC HEN DESIGN REPORT

80. Progettazione di processo e di prodotto Trieste, 10 July 2016 - slide 80 Tutorials Aspen Energy analyzer: case https://www.youtube.com/watch?v=gV_uBz6CNtA