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Pinch Technology and optimization of the use of utilities – part two

Pinch Technology and optimization of the use of utilities – part two. Maurizio Fermeglia maurizio.fermeglia@di3.units.it www.mose.units.it. Introduction to HEN Synthesis – Summary of part 1. Unit 1 . Introduction: Capital vs. Energy What is an optimal HEN design Setting Energy Targets

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Pinch Technology and optimization of the use of utilities – part two

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

  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. 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. 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 Uminand MER (minimize n. of HEx and MER target) • Design a HEN for “Threshold Problems”

  5. UNIT 4: Loops and Splits • The minimum number of units (Umin) in a network: • UMin = NStream + NUtil 1 (Hohman, 1971) • A HEN containing UHEX units (UHEX  Umin) has (UHEX  Umin) 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. Class Exercise 5 (Linnhoff and Flower, 1978) Example: Tmin = 10 oC. Step 1: Temperature Intervals (subtract Tmin from hot temperatures) Temperature intervals: 180oC  170oC 140oC 130oC 60oC 30oC

  7. Class Exercise 5 (Cont’d) Step 2: Interval heat balances For each interval, compute:Hi = (Ti  Ti+1)(CPHotCPCold )

  8. This defines: Cold pinch temp. = 140 oC QHmin = 60 kW QCmin = 160 kW Class Exercise 5 (Cont’d) Step 3: Form enthalpy cascade.

  9. MER Design above the pinch: MER Design below the pinch: Class Exercise 5 (Cont’d) UMin,MER = NStream + NUtil- 1 = 2 + 1 – 1 = 2  UMin,MER = 4 + 1 – 1 = 4MER design below pinch has 6 exchangers!i.e. There are two loops below pinch.

  10. Class Exercise 5 (Cont’d) Complete MER Design However, UMin = NStream + NUtil 1 = 4 + 2  1 = 5 The MER network has 8 units. This implies 3 independent “heat load loops”. We shall now identify and eliminate theseloops in order to design for UMin

  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 Tmin violation in the network:

  12. Class Exercise 5 (Cont’d) Identification and elimination of 1st loop (Cont’d): To restore Tmin, 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. 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 Tmin violation: 140 - x = 2(180 - 113.33 - Tmin)  x = 26.66 This is called “energy relaxation”

  14. Class Exercise 5 (Cont’d) Identification and elimination of 2nd loop: Since there is no Tmin violation, no adjustment of the loads of the exchangers is needed - we reduce the number of units by one with no energy penalty.

  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. Class Exercise 5 (Cont’d) Identification and elimination of 3rd loop: We use the heat path to restore Tmin:253.33 - x = 3(150 - Tmin- 60)  x = 13.33

  17. Class Exercise 5 (Cont’d) Therefore Umin Network is:

  18. Step 1: Perform MER Design for UHEX units. Try and ensure that design meets UMin,MER separately above and below the pinch. Loop Breaking - Summary Step 2: Compute the minimum number of units: UMin = NStream + NUtil 1 This identifies UHEX  Umin independent “heat loops”, which can be eliminated to reduce U. Step 3: For each loop, eliminate a unit. If this causes a Tmin violation, identify the “heat path” and perform “energy relaxation” by increasing the duties of the cooler and heater on the heat path. Loops improve energy recovery and heat load flexibility at the cost of added units (>Umin)

  19. Option 1. Stream Split Designs • Example.

  20. Option 3. Stream Splitting Stream Split Designs (Cont’d) • Option 2. Loops

  21. Loops: Improved energy recovery (normally) Heat load flexibility (normally) U > UMin (by definition) Stream Splitting: Maximum Energy recovery (always) Branch flowrate flexibility (normally) U = UMin (always) Loops vs. Stream Splits Stream splitting is a powerful technique for better energy recovery BUT - Don’t split unless necessary

  22. 1. Above the pinch (at the pinch): Cold utilities cannot be used (for MER). So, if NH > NC, MUST split COLD streams, since for feasibility NH NC Feasible matches must ensure CPH CPC. If this is not possible for every match, split HOT streams as needed. If Hot steams are split, recheck  Stream Splitting Rules 2. Below the pinch (at the pinch): • Hot utilities cannot be used (for MER). So, if NC > NH, MUST split HOT streams, since for feasibility NC NH • Feasible matches must ensure CPC CPH. If this is not possible for every match, split COLD streams as needed. If Cold steams are split, recheck 

  23.  500 200 Class Exercise 6 Design a hot-side HEN, given the stream data below: Solution:Since NH > NC, we must split C1. The split ratio is dictated by the rule: CPH CPC (necessary condition) and by a desire to minimize the number of units (“tick off “streams)

  24. Class Exercise 6 (Cont’d) x is determined by the following energy balances: x(T1  90) = 500 (10 x)(T2 90) = 200 subject to: 200 T1 Tmin = 10 150 T2 Tmin = 10 Best to make T1= T2 . Here, this is not possible. Why?We shall make T2 = 140 (why?)

  25. Complete solution is: Class Exercise 6 (Cont’d) A possible solution is therefore:(10 x) (140  90) = 200 x = 6 T1 = 90 + 500/x = 173.33 (satisfies constraint) This is an MER design which also satisfies UMin (UMin = 3).

  26. 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 QHmin = 0 or QCmin = 0). Such designs are not separated at the pinch, and are called “Threshold Problems”

  27. Assuming a value of Tmin= 10 oC, the Problem Table gives the following result. Threshold Problems (Cont’d) Assuming a value of Tmin= 105 oC:

  28. Threshold Problems (Cont’d) Threshold problems do not have a pinch, and have non-zero utility duties only at one end.

  29. Threshold Problems (Cont’d) However, increasing driving forces beyond the Threshold Valueleads to additional utility requirements.

  30. 1. Establish the threshold Tmin 2. Note the common practice values for Tmin: Pinched - treat as normal pinched problem Threshold - must satisfy target temperatures at the “no utility end” Threshold Design Guidelines 3. Compare the threshold Tmin to Tmin,Experience Classify as one of the following:

  31. Class Exercise 7 The graph shows the effect of Tmin on the required levels of QHmin and QCmin for a process consisting of 3 hot and 4 cold streams.

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

  33. 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

  34. Algorithm: initial condition and step 1 Minimum Temperature Approach = Smallest DT 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

  35. Algorithm: construct Temperature interval diagram (step 2) • Process streams represented by vertical lines • Axes are shifted by the minimum T approach

  36. Pinch temperature 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

  37. Pinch temperature 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

  38. 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

  39. 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

  40. 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

  41. Algorithm: Design the network above the pinch • Start from the design at the pinch • To make sure that DT min is not violated, match streams such that • Note that we consider ONLYstreams present at the pinch • Each exchanger is represented by two circles connected with a line, each circle represent a side

  42. 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

  43. Algorithm: Design the network below the pinch • Similar to previous one: start from the design at the pinch • To make sure that DT 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 DT min is violated (see figure)

  44. Algorithm: Design the network below the pinch • Split stream into substreams to meet the DT min criterion

  45. Algorithm: Design the network below the pinch • Move away from the pinch • Look at the remaining streams

  46. 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°

  47. 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 DTln 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) Exchanger DT ln U Q F factor Area. °C W/m2/°C kW m2 1 24.1 129.8 100 0.8 40.0 2 20.0 69.5 300 0.8 270.3 3 47.5 153.8 700 0.8 119.7 4 24.1 80.0 500 0.8 324.6 6 10.0 61.7 100 0.8 202.5 7 17.0 69.5 100 0.8 195.8 TOTAL 1063.0 Exchanger 5 requires an hot utility (steam): DT = 76.8 °C, U= 76.9 W/m2/°C, Area = 16.9 m2 Exchanger 8 requires cooling water: DT = 23.2 °C, U= 346 W/m2/°C, Area = 7.8 m2 TOTAL Area: 1087.7 m2

  48. 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

  49. Construction of Composite T-H diagram

  50. Construction of Composite T-H diagram

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