PART 1 PINCH AND MINIMUM UTILITY USAGE

Transcription

1 PART 1 PINCH AND MINIMUM UTILITY USAGE

2 WHAT IS THE PINCH? The pinch point is a temperature. Typically, it divides the temperature range into two regions. Heating utility can be used only above the pinch and cooling utility only below it.

3 WHAT IS A PINCH DESIGN A heat exchanger network obtained using the pinch design method is a network where no heat is transferred from a hot stream whose temperature is above the pinch to a cold stream whose temperature is below the pinch.

4 IS PINCH TECHNOLGY CURRENT? YES and NO. It is a good first approach to most problems. Pinch technology is at the root of many other heat integration technologies. It is impossible to understand them without the basic concepts of pinch technology.

5 TEMPERATURE-ENTHALPY (T-H) DIAGRAMS Assume one heat exchanger. These are alternative representations T H,in T C,in Q T C,out T T H,in T C,out T H, out T H,in T H, out T H, out T C,in Slopes are the inverse of F*Cp. Recall that Q=F Cp T T C,out T C,in Q H Q

6 T-H DIAGRAMS Assume one heat exchanger and a heater T H,in T C,in Q Q H T C,out T T H,in T C,out T H, out T H,in T H, out T H, out T C,in T C,out H T C,in Q Q H H Q H Q

7 T-H DIAGRAMS Assume one heat exchanger and a cooler T H,in T C,in Q T C,out T T H,in T C,out Q C T H, out T H, out T H,in C T H, out T C,in T C,out Q C T C,in QC Q H Q

8 T-H DIAGRAMS Two hot-one cold stream T H1,in T C,in T H2,in T C,out T T H2,in Q 1 Q 2 T H1,in T C,out T H2,out T H2,out T H2,out T H1,in T H2,in T H1,out T H2, out T H1,out T C,in T C,out T C,in Q 1 Q 2 H Q 2 Q 1 Notice the vertical arrangement of heat transfer

9 Composite Curve T-H DIAGRAMS T Obtained by lumping all the heat from different streams that are at the same interval of temperature. T H Remark: By constructing the composite curve we loose information on the vertical arrangement of heat transfer between streams H

10 T-H DIAGRAMS Moving composite curves horizontally T T Cooling Heating Smallest T H Smallest T H T H1,in T H2,in T C,in T H1,in T H2,in Q 1 Q 2 T C,in Q 1 Q 2 Q H T C,out T C,out Q C T H2,out T H1,out T H2,out T H1,out

11 T-H DIAGRAMS T Cooling Moving the cold composite stream to the right Smallest T T C,in T H1,in T H2,in Q 1 Q 2 Q C T H2,out T H1,out Q H Heating H T C,out Increases heating and cooling BY THE SAME AMOUNT Increases the smallest T Decreases the area needed A=Q/(U* T ) Notice that for this simple example the smallest T takes place in the end of the cold stream

12 T-H DIAGRAMS T Cooling Heating In general, the smallest T can take place anywhere. H We call the temperature at which this takes place THE PINCH.

13 TEMPERATURE-ENTHALPY DIAGRAMS T Cooling Heating H From the energy point of view it is then convenient to move the cold stream to the left. However, the area may become too large. To limit the area, we introduce a minimum approach T min

14 GRAPHICAL PROCEDURE Fix T min Construct the hot and cold composite curve Draw the hot composite curve and leave it fixed Draw the cold composite curve in such a way that the smallest T= T min The temperature at which T= T min is the PINCH The non-overlap on the right is the Minimum Heating Utility and the non-overlap on the left is the Minimum Cooling Utility

15 HANDS ON EXERCISE H=27 MW H=-30 MW T=140 0 C T=230 0 C REACTOR 2 T=200 0 C T=80 0 C H=32 MW H=-31.5 MW T=20 0 C REACTOR 1 T=180 0 C T=250 0 C T=40 0 C Stream Type Supply T Target T H F*Cp ( o C) ( o C) (MW) (MW o C -1 ) Reactor 1 feed Cold Reactor 1 product Hot Reactor 2 feed Cold Reactor 1 product Hot T min =10 o C

16 Answer: Hot Streams FCp=0.15 FCp= FCp=0.4 FCp= FCp= H H

17 Answer: Cold Streams FCp=0.3 FCp=0.3 FCp=0.5 FCp=0.2 FCp= H H

18 Answer: Both Curves Together. Pinch T= T min Important observation: The pinch is at the beginning of a cold stream or at the beginning of a hot stream. 7.5 H

19 UTILITY COST vs. T min There is total overlap for some values of T min T COST Utility H TOTAL OVERLAP T T min H PARTIAL OVERLAP Note: There is a particular overlap that requires only cooling utility

20 PROBLEM TABLE Composite curves are inconvenient. Thus a method based on tables was developed. STEPS: 1. Divide the temperature range into intervals and shift the cold temperature scale 2. Make a heat balance in each interval 3. Cascade the heat surplus/deficit through the intervals. 4. Add heat so that no deficit is cascaded

21 PROBLEM TABLE We now explain each step in detail. Consider the example 1.1 Stream Type Supply T Target T H F*Cp ( o C) ( o C) (MW) (MW o C -1 ) Reactor 1 feed Cold Reactor 1 product Hot Reactor 2 feed Cold Reactor 2 product Hot T min =10 o C

22 PROBLEM TABLE Divide the temperature range into intervals and shift the cold temperature scale Hot streams Cold streams Hot streams Cold streams Now one can make heat balances in each interval. Heat transfer within each interval is feasible.

23 PROBLEM TABLE 2. Make a heat balance in each interval. (We now turn into a table format distorting the scale) F Cp=0.15 F Cp=0.25 F Cp=0.3 F Cp=0.2 T interval H interval Surplus/Deficit? Surplus Deficit Surplus Deficit Surplus Deficit Deficit Hot streams Cold streams

24 PROBLEM TABLE 3. Cascade the heat surplus through the intervals. That is, we transfer to the intervals below every surplus/deficit This interval has a surplus. It should transfer 1.5 to interval 2. This interval has a deficit. After using the 1.5 cascaded it transfers 4.5 to interval The largest deficit transferred is Thus, 7.5 MW of heat need to be added on top to prevent any deficit to be transferred to lower intervals

25 PROBLEM TABLE 4. Add heat so that no deficit is cascaded This is the minimum heating utility This is the position of the pinch This is the minimum cooling utility

26 IMPORTANT CONCLUSION DO NOT TRANSFER HEAT ACROSS THE PINCH THIS IS A GOLDEN RULE OF PINCH TECHNOLOGY. WE WILL SEE LATER HOW THIS IS RELAXED FOR DIFFERENT PURPOSES WHEN THIS HAPPENS IN BADLY INTEGRATED PLANTS THERE ARE HEAT EXCHANGERS WHERE SUCH TRANSFER ACROSS THE PINCH TAKES PLACE λ λ λ λ λ λ λ λ Heating utility is larger than the minimum Heat is transferred across the pinch Cooling utility is larger by the same amount

27 PROBLEM TABLE Heating utility of smaller temperature Heating utility at the largest temperature is now zero. These are the minimum values of heating utility needed at each temperature level.

28 MATHEMATICAL MODEL q 1 q 2 δ 0 δ 1 Let q i be the surplus or demand of heat in interval i. It is given by: H H C C q = F cp ( T T) F cp ( T T) i k Γ i H k k i 1 i s s i 1 s Γ The minimum heating utility is obtained by solving the following linear programming (LP) problem i C i q i S min = Minδ 0 q i+1 δ i δ i+1 st. δ = δ + q i=,... m i i 1 i 1 I δ 0 i q n δ n

29 PART 2 TOTAL AREA TARGETING

30 TOTAL AREA TARGETING In this part we will explore ways to predict the total area of a network without the need to explore specific designs. Because A=Q/(U* T ml ), one can calculate the area easily in the following situation. T T H1 T H2 T ml = ln T C2 (T H1 -T C2 )-(T H2 -T C1 ) (T H1 -T C2 ) (T H2 -T C1 ) T C1 Q H

31 TOTAL AREA TARGETING Since area=q/(u T ml ), the composite curve diagram provides one way of estimating the total area involved. Isolate all regions with a pair of straight line sections and calculate the area for each. T Heating utility, steam or furnace. Cooling water A 1 A 2 A 3 A 4 A 5 A 6 H The above scheme of heat transfer is called VERTICAL HEAT TRANSFER

32 EXERCISE Calculate the values of Q in each interval and estimate the corresponding area. Use U= MW m -2 o C T Furnace (300 o C) T= T Pinch min 140 Cooling Water H Q 1 Q 2 Q 3 Q 4 Q 5 Q 6 Obtain the total area estimate

33 EXERCISE 300 COMPOSITE CURVE Furnace (300 o C) I II III IV Pinch T, C 150 V VI 100 Cooling Water 50 0 T= T min Q, MW HOT COLD Units: Q= MW T= o C A= m 2 Interval Q T T H1 H2 T C1 T C2 I II III IV V VI

34 EXERCISE Interval Q T H1 T H2 T C1 T C2 T ml A I II III IV V VI Total Area Units: Q= MW T= o C U= MW m -2 o C, A= m2

35 TOTAL AREA TARGETING Drawbacks Fixed costs associated with the number of units are not considered. We will see later how the number of units can be calculated

36 QUESTIONS FOR DISCUSSION Is the total area predicted this way, realistic? That is, is it close enough to a value that one would obtain in a final design? Is the estimate, realistic or not, conservative? That is, is it larger than the one expected from a final design? How complex is a design built using the vertical transfer?

37 ANSWERS Is the total area predicted this way, realistic? That is, is it close enough to a value that one would obtain from a final design? YES, Within 10-15%

38 ANSWERS Is the estimate, realistic or not, conservative? That is, is it larger than the one expected from a final design? The area obtained is actually the minimum area needed to perform the heat transfer.

39 ANSWERS How complex is a design built using the vertical transfer? Very Complex. Take for example interval 4. There are four streams in this interval. Stream Type Supply T Target T H F*Cp (MW o C -1 ) ( o C) ( o C) (MW) Reactor 1 feed Cold Reactor 1 product Hot Reactor 2 feed Cold Reactor 1 product Hot This implies at least three heat exchangers, just in this interval.

40 HEAT EXCHANGER NETWORK FCp=0.16 FCp=0.09 FCp= R1 prod, FCp=0.15 R2 prod, FCp=0.25 R1 feed, FCp=0.2 R2 feed, FCp=0.3 I II FCp= III IV V VI

41 HEAT EXCHANGER NETWORK FCp=0.16 R1 prod, FCp=0.15 R2 prod, FCp= R1 feed, FCp=0.2 FCp=0.075 R2 feed, FCp=0.3 I II FCp= III IV V VI Called Spaghetti design TOTAL= 10 Exchangers

42 PREDICTING THE NUMBER OF UNITS We can anticipate very simply how many exchangers we should have!!! Consider the following warehouses, each containing some merchandise that needs to be delivered to the row of consumer centers. What is the minimum number of trucks needed? Warehouses Consumer Centers

43 ANSWER You need five trucks, possibly less in some other cases. Here is how you solve the problem specifically The general answer is N=S-1. When does one need less?

44 ANSWER When there is an exact balance between two streams or a subset of streams The general answer is N=S-P. P is the number of independent subsystems. (Two in this case)

45 GENERAL FORMULA FOR UNIT TARGETING N min = (S-P) above pinch + (S-P) below pinch If we do not consider two separate problems, above and below the pinch we can get misleading results.

46 SUPERTARGETING Economy of the system is dependent on T min COST Total Utility Capital Optimum T min

47 SPECIAL CASES There is total overlap for some values of T min T COST Total Utility TOTAL OVERLAP H T Capital T min PARTIAL OVERLAP Note: There is a particular overlap that requires only cooling utility H

48 PART 3 DESIGN OF MAXIMUM ENERGY RECOVERY NETWORKS

49 MER NETWORKS Networks featuring minimum utility usage are called MAXIMUM ENERGY RECOVERY (MER) Networks.

50 PINCH DESIGN METHOD RECALL THAT No heat is transferred through the pinch. This makes the region above the pinch a HEAT SINK region and the region below the pinch a HEAT SOURCE region.

51 Heat Sink Heat is obtained from the heating utility Minimum heating utility 9.0 T Pinch Heat Source H Minimum cooling utility Heat is released to cooling utility

52 CONCLUSION One can analyze the two systems separately, that is, Heat exchangers will not contain heat transfer across the pinch.

53 PINCH MATCHES Consider two streams above the pinch T H,in T p T C,out = T p - T min + Q/FCp C T H,in = T p +Q/FCp H But T H,in > T C,out + T min. Thus replacing one obtains Q/FCp H >Q/FCp C T C,out Q T p - T min FCp H <FCp C Golden rule for pinch matches above the pinch. T min T min Violation when FCpH >FCp C

54 PINCH MATCHES Consider two streams below the pinch T C,in = T p - T min -Q/FCp C T p T H,out T H,out = T p -Q/FCp H But T H,out > T C,in + T min. Thus replacing one obtains T p - T min Q T C,in FCp C <FCp H Golden rule for pinch matches below the pinch. T min T min Violation when FCp C >FCp H

55 CONCLUSION Since matches at the pinch need to satisfy these rules, one should start locating these matches first. Thus, our first design rule: START BY MAKING PINCH MATCHES

56 QUESTION Once a match has been selected how much heat should be exchanged?

57 ANSWER As much as possible! This means that one of the streams has its duty satisfied!! THIS IS CALLLED THE TICK-OFF RULE

58 HANDS ON EXERCISE H=27 MW H=-30 MW T=140 0 C T=230 0 C REACTOR 2 T=200 0 C T=80 0 C H=32 MW H=-31.5 MW T=20 0 C T=180 0 C REACTOR 1 T=250 0 C T=40 0 C Stream Type Supply T Target T H F*Cp ( o C) ( o C) (MW) (MW o C -1 ) Reactor 1 feed Cold Reactor 1 product Hot Reactor 2 feed Cold Reactor 1 product Hot T min =10 o C PINCH=150 o C

59 HANDS ON EXERCISE FCp=0.15 H C C 40 0 C FCp=0.25 H C 80 0 C FCp= C C 20 0 C C 1 FCp= C C 2 Stream Type Supply T Target T H F*Cp ( o C) ( o C) (MW) (MW o C -1 ) Reactor 1 feed Cold Reactor 1 product Hot Reactor 2 feed Cold Reactor 1 product Hot T min =10 o C PINCH=150 o C

60 ABOVE THE PINCH FCp=0.15 H C C FCp=0.25 H C FCp= C C 1 FCp= C C C 2 Which matches are possible?

61 ANSWER (above the pinch) FCp=0.15 H C C FCp=0.25 H C FCp=0.2 FCp= C C C C 1 C 2 The rule is that FCp H <FCp C. We therefore can only make the match H1-C1 and H2-C2.

62 ANSWER (above the pinch) FCp=0.15 H C C C FCp=0.25 H C FCp=0.2 FCp= C C C C C 1 C 2 The tick-off rule says that a maximum of 8 MW is exchanged in the match H1-C1 and as a result stream C1 reaches its target temperature. Similarly 12.5 MW are exchanged in the other match and the stream H2 reaches the pinch temperature.

63 BELOW THE PINCH FCp=0.15 H C 40 0 C FCp=0.25 H C FCp= C 20 0 C C 1 Which matches are possible? The rule is that FCp C <FCp H. Thus, we can only make the match H2-C1

64 ANSWER (below the pinch) FCp=0.15 H C 40 0 C FCp=0.25 H C FCp= C C 20 0 C C The tick-off rule says that a maximum of 17.5 MW is exchanged in the match H2-C1 and as a result stream H2 reaches its target temperature.

65 COMPLETE NETWORK AFTER PINCH MATCHES FCp=0.15 H C C C 40 0 C FCp=0.25 H C 80 0 C FCp=0.2 FCp= C C C C C C C C C Streams with unfulfilled targets are colored.

66 WHAT TO DO NEXT? FCp=0.15 H C C C 40 0 C FCp=0.25 H C 80 0 C FCp=0.2 FCp= C C C C C C C C C Away from the pinch, there is more flexibility to make matches, so the inequalities do not have to hold. The pinch design method leaves you now on your own!!!!! Therefore, use your judgment as of what matches to select!!

67 ANSWER FCp=0.15 H C C C 40 0 C FCp=0.25 H C 80 0 C FCp=0.2 FCp= C C C C C C C C C We first note that we will use heating above the pinch. Thus all hot streams need to reach their inlet temperature. We are then forced to look for a match for H1. Please locate it.

68 ANSWER The match is H1-C1. We finally put a heater on the cold stream FCp=0.15 H C C 40 0 C FCp=0.25 H C 80 0 C FCp=0.2 FCp= C C H C C C C C C

69 ANSWER Below the pinch we try to have the cold streams start at their inlet temperatures and we later locate coolers (one in this case). FCp=0.15 H C C C 40 0 C FCp=0.25 H C C FCp= C C 20 0 C C 1 FCp= C H C C

70 FCp=0.15 H 1 EXAMPLE C C 40 0 C C FCp=0.25 H C C FCp= C C 20 0 C C 1 FCp= C H C C N min = (S-P) above pinch + (S-P) below pinch = =(5-1) + (4-1) = 7 If we do not consider two separate problems Nmin= (6-1)= 5, which is wrong Note: A heat exchanger network with 5 exchangers exists, but it is impractical and costly. This is beyond the scope of this course.

71 UNEQUAL NUMBER OF STREAMS AT THE PINCH Indeed, if the number of hot streams is larger than the number of cold streams, then no pinch matches are possible. FCp=0.15 FCp=0.25 FCp=0.1 Target=170 O C Target=140 O C H 1 H 2 H 3 FCp=0.2 FCp= O C 100 O C 140 O C 130 O C O C 90 O C O C C 1 C 2 Assume the matches H 1 -C 1 and the matches H 2 -C 2 have been selected. Since H 3 needs to go to the pinch temperature, there is no cold stream left to match, even if there is portions of C 1 or C 2 that are left for matching. Such matching would be infeasible. What is then, the solution?

72 ANSWER Split cold stream until the inequality is satisfied. FCp=0.15 FCp=0.25 FCp=0.1 H 1 H 2 H O C 100 O C 140 O C 130 O C Target=170 O C FCp= O C 90 O C C 1 Target=140 O C FCp=0.25 FCp= O C 110 O C C 2 Notice that different combinations of flowrates in the split satisfy the inequality. 3

73 INEQUALITY NOT SATISFIED Consider the following case: FCp=0.5 H O C 100 O C Target=170 O C FCp= O C C 1 Target=140 O C FCp=0.4 C 2

74 Split the hot stream ANSWER FCp=0.2 H O C 100 O C FCp=0.3 Target=170 O C FCp= O C 90 O C C 1 Target=140 O C FCp= O C 10 C 2 15

75 SOLVE THE FOLLOWING Below the Pinch : PROBLEM FCp=0.5 H O C Target=40 O C FCp=0.3 H 2 Target=20 O C FCp= O C 30 O C C 1

76 Below the Pinch : ANSWER FCp=0.5 H O C 40 O C Target=40 O C FCp=0.3 H 2 60 O C Target=20 O C FCp=0.5 FCp= O C O C C 1 12

77 COMPLETE PROCEDURE ABOVE THE PINCH Start Yes FCp H FCp C Yes S H S C at pinch? No No Split Cold Stream Place matches Split Hot Stream

78 COMPLETE PROCEDURE BELOW THE PINCH Start Yes FCp H FCp C at pinch? Yes S H S C No No Split Hot Stream Place matches Split Cold Stream

79 HANDS ON EXERCISE Type Supply T Target T F*Cp ( o C) ( o C) (MW o C -1 ) Hot Hot Cold Cold T min =50 o C Minimum Heating Utility= 9.2 MW Minimum Cooling Utility= 6.4 MW

80 ANSWER FCp= C FCp=0.045 H C C C FCp=0.04 H C C C 6 FCp= C C C 1 FCp= C C C C 2

81 TRANSSHIPMENT MODEL (Papoulias and Grossmann, 1983) q 1 δ 0 We will now expand the mathematical model we presented to calculate the minimum utility. q 2 δ 1 S min = Minδ 0 q i q i+1 δ i Where st. δ = δ + q i=,... m i i 1 i 1 I δ 0 i δ i+1 k Γ H q = F cp T ) i H i k H C C k ( Ti 1 Ti ) Fs cps ( Ti 1 C s Γ i i q δ n

82 TRANSSHIPMENT MODEL s i,k,1 r i,1 p k,1 δ i,1 s i,k,2 r i,2 p k,2 Assume now that we do the same cascade for each hot stream, while we do not cascade the cold streams at all. In addition we consider heat transfer from hot to cold streams in each interval. s i,k,j r i,j δ i,j s i,k,,j+1 r i,j+1 δ i,j+1 s i,k,n r i,n δ i,n p k,j p k,j+1 p k,n The material balances for hot streams are: δ δ i,0 i, j = 0 = δ i, j 1 + r s j = 1,... m i, j i, k, j The material balances for cold streams are: p = 1,... m k, j si, k, j j = i Where r i,j and p k,j are the heat content of hot stream I and cold stream k in interval j. k I I

83 TRANSSHIPMENT MODEL s i,k,1 r i,1 p k,1 δ i,1 s i,k,2 r i,2 p k,2 r i,j r i,j+1 s i,k,j δ i,j s i,k,,j+1 p k,j p k,j+1 Although we have a simpler model to solve it, in this new framework, the minimum utility problem becomes: Min s. t δ δ p i,0 i, j k, j = 0 = δ = δ U,0 i, j 1 i + r s i, j i, k, j k s i, k, j i, j = 1,... m i, j = 1,... m k, j = 1,... m I I I δ i,j+1 r i,n s i,k,n p k,n Note that the set of hot streams now includes process streams and the utility U. Cold streams include cooling water. δ i,n

84 TRANSSHIPMENT MODEL s i,k,1 r i,1 p k,1 δ i,1 s i,k,2 r i,2 p k,2 We would like to have a model that would tell us the s i,k,j such that the number of units is minimum. We now introduce a way of counting matches between streams. Let Y i,k be a binary variable (can only take the value 0 or 1). s i,k,j r i,j δ i,j s i,k,,j+1 r i,j+1 δ i,j+1 s i,k,n r i,n p k,j p k,j+1 p k,n Then we can force Y i,k to be one using the following inequality j s i, k, j ΓYi, k indicating therefore that heat has been transferred from stream i to stream k in at least one interval. 0 δ i,n

85 TRANSSHIPMENT MODEL s i,k,1 r i,1 p k,1 The complete model would be: δ s i,1 i,k,2 r i,2 p k,2 Min s. t i k Y i, k s i,k,j r i,j δ i,j s i,k,,j+1 r i,j+1 δ i,j+1 s i,k,n p k,j p k,j+1 δ δ δ U,0 p i,0 i, j k, j j = δ = 0 = δ = s i, j 1 i i, k, j * U,0 + r s i, j i, k, j ΓY i, k k 0 s i, k, j i, j = 1,... m i, j = 1,... m k, j = 1,... m i, k I I I r i,n δ i,n p k,n The model can only be solved above and below the pinch separately. Why???

86 TRANSSHIPMENT MODEL We are minimizing the number of matches. Different answers can be obtained if separate regions are not considered. These answers are not guaranteed to be realistic.

87 GAMS MODEL Min s. t δ δ δ U,0 i,0 i, j p k, j j = 0 = δ = s i, j 1 i i = δ i, k, j * U,0 + r s k i, j i, k, j ΓY i, k Y i, k k 0 s i, k, j i, j = 1,... m i, j = 1,... m k, j = 1,... m i, k I I I TABLE P(K,J) load of cold stream K1 in interval J J0 J1 J2 J3 C C W ; VARIABLES S(I,K,J) heat exchanged hot and cold streams D(I,J) heat of hot streams flowing between intervals Y(I,K) existence of match Z total number of matches ; POSITIVE VARIABLE S POSITIVE VARIABLE D BINARY VARIABLE Y ; GAMS MODEL SETS I hot streams above pinch / S, H1 / K cold streams above pinch / C1,C2,W/ J temperature intervals / J0*J3 / ; SCALAR GAMMA /10000/; TABLE R(I,J) load of hot stream I1 in interval K J0 J1 J2 J3 S H ; EQUATIONS MINMATCH objective function-number of matches HSBAL(I,J) heat balances of hot stream I in INTERVAL J CSBAL(K,J) heat balances of cold stream J1 in K HTINEQ1(I,K) heat transferred inequalities; MINMATCH.. Z =E= SUM((I,K), Y(I,K)); HSBAL(I,J)$(ORD(J) NE 0).. D(I,J)-D(I,J-1)+ SUM(K,S(I,K,J)) =E= R(I,J); CSBAL(K,J)$(ORD(J) NE 0).. SUM(I, S(I,K,J)) =E= P(K,J) ; HTINEQ1(I,K).. SUM(J, S(I,K,J))-GAMMA*Y(I,K) =L= 0 ; MODEL TSHIP /ALL/ ; SOLVE TSHIP USING MIP MINIMIZING Z; DISPLAY S.L, D.L, Y.L;

88 GAMS MODEL SOLUTION ---- VARIABLE S.L J1 J2 J3 S.C H1.C H1.C FCp=0.045 H C C ---- VARIABLE D.L J0 J1 J2 FCp=0.04 H 2 S H FCp= C ---- VARIABLE Y.L C1 C2 FCp= C C S H EXECUTION TIME = SECONDS

89 PART 4 UTILITY PLACEMENT HEAT AND POWER INTEGRATION

90 UTILITY PLACEMENT We now introduce the GRAND COMPOSITE CURVE, which will be useful to analyze the placement of utilities Start at the pinch T Pinch H

91 GRAND COMPOSITE CURVE T Total heating utility Total cooling utility These are called pockets Process-to Process integration takes place here H

92 UTILITY PLACEMENT We now resort to a generic grand composite curve to show how utilities are placed. T MP Steam HP Steam LP Steam Cooling Water H

93 Transshipment model for multiple utilities I W n S m j m k I S m j m k j k i i j k I S m S m j n i j m j m I S m j n i j k i k j i j i j i I i W n W n n S m S m m m j k Q Q m j k Q s p m j i Q Q m j i Q s r m j i s t Q c Q c Min 1,..., 1,..., 1,..., 1,..., 1,..., 0.,,,,,,,,, 1,,,,,,, 1,,,0 = = = + = = + = = + = = = + δ δ δ δ δ Results HOT UTILITY 1 HOT UTILITY 2 COLD UTILITY Interval T( i ) Q( i ) δ( i ) HU 1 ( i ) HU 2 ( i ) CU 1 ( i ) I0 250 I I I HOT UTILITY 1 = HP STEAM HOT UTILITY 2 = MP STEAM

94 UTILITY PLACEMENT Hot Oil placement and extreme return temperatures T Hot oil from furnace Oil minimum return temperature Water maximum return temperature Cooling water Process H

95 UTILITY PLACEMENT Furnace Theoretical Flame Temperature T T stack T stack Air Fuel Process Stream Ambient Temperature Stack Loss Fuel heat value H

96 COMBINED HEAT AND POWER Integration of a Heat Engine below the Pinch. T T Q H,min Q HE Pinch Q HE -W W Note that in this case there is no gain. The heat engine can be arranged separately and the utility usage will not change. Q C,min +( Q HE W)

97 COMBINED HEAT AND POWER T Integration of a Heat Engine Across the Pinch. Q H,min -( Q HE W) Q HE W TOTAL HEN ENERGY INTAKE Q H,min -( Q HE W) (Smaller) Pinch Q HE -W SYSTEM TOTAL ENERGY INTAKE vs. (separate) Q H,min +W Q H,min +Q HE Q C,min

98 INTEGRATION WITH DISTILLATION Q H,min Placement below the pinch. T Pinch In this case there is a gain of (Q -Q cond reb ) in the cooling utility. Q reb Q cond Q C,min +(Q cond -Q reb )

99 HEAT PUMPS/REFRIGERATION CYCLES Q H,min -(Q HP + W) Q HP +W TOTAL HEN INTAKE Q H,,min (Q HP +W) (smaller) Pinch TOTAL SYSTEM INTAKE W Q H,,min -Q HP Q HP vs. Q H,,min +W (separate) Savings in cooling utility Q HP Q C,min -Q HP

100 COMBINED HEAT AND POWER Utility Placement T Q H,min ( Q HP +Q LP ) Q HE T Q HP Q HP W Q LP Pinch Q LP TOTAL HEN INTAKE Q H,min ( Q HP +Q LP ) (smaller) Q C,min Q C,min TOTAL SYSTEM INTAKE Q H,min +W vs. Q H,min +W+(Q HP + Q LP ) (separate) H

101 COMBINED HEAT AND POWER Gas Turbine Placement TOTAL ENERGY INTAKE T EX vs. Q H,min + W+ Q LOSS Q H,min + W+ Q LOSS +Q S T T Q H,min -Q S Q F W Pinch Q S Air T EX T 0 Q LOSS Q LOSS Q S Q F -W H Q C,min T 0

102 PART 5 DISTILLATION PLACEMENT

103 PLACEMENT OF DISTILLATION Placement across the pinch. Q H,min +Q reb T Q reb Pinch Q cond Note that in this case there is no gain. The distillation column can be arranged separately and the utility usage will not change. Q C,min +Q cond

104 PLACEMENT OF DISTILLATION Placement above the pinch. Q H,min +Q reb -Q cond T Q reb Q cond Pinch Note that in this case there is a possible gain in the heating utility. Q C,min

105 ADJUSTING PRESSURE FOR PROPER PLACEMENT Heating Utility T T Pinch Pinch P As pressure increases utility usage decreases.

106 CRUDE FRACTIONATION EXAMPLE We now show a complete analysis of an atmospheric crude fractionation unit. We start with the supply demand diagram CONDENSER water naphtha PA1 steam PA2 kerosene steam PA3 diesel water crude HEN sour w ater DESALTER steam steam gas oil HEN FURNACE residue

107 CRUDE FRACTIONATION EXAMPLE We now show how to determine the heat load of pumparounds. We start with a column with no pumparound (results from are from a rigorous simulation) 1.2 PINCH 1.0 M*Cp, MMW/C COND CRUDE 0.2 RES PRODUCTS TEMPERATURE, C

108 CRUDE FRACTIONATION EXAMPLE We move as much heat from the condenser to the first pumparound as possible. The limit to this will be when a plate dries up. If the gap worsens to much, steam is added PINCH M*Cp, MMW/C COND PA1 CRUDE RES PRODUCTS TEMPERATURE, C

109 CRUDE FRACTIONATION EXAMPLE We continue in this fashion until the total utility reaches a minimum PINCH 1.0 PA1 M*Cp, MMW/ C COND PA2 CRUDE 0.2 RES TEMPERATURE, C

110 CRUDE FRACTIONATION EXAMPLE Especially when moving heat from PA2 to PA3, steam usage increases so that the flash point of products is correct and the gap is within limits M*Cp, MMW/C COND PA1 PA1 PA2 CRUDE PA3 RES TEMPERATURE, C

111 CRUDE FRACTIONATION EXAMPLE Situation for a heavy crude 0.8 M*Cp, MMW/C COND SW PA1 CRUDE PA2 RES PA TEMPERATURE, C

112 PART 6 ENERGY RELAXED NETWORKS

113 ENERGY RELAXATION Energy relaxation is a name coined for the procedure of allowing the energy usage to increase in exchange for at least one of the following effects : a) a reduction in area b) a reduction in the number of heat exchangers c) a reduction in complexity (typically less splitting)

114 ENERGY RELAXATION IN THE PINCH DESIGN METHOD LOOP: A loop is a circuit through the network that starts at one exchanger and ends in the same exchanger PATH: A path is a circuit through the network that starts at a heater and ends at a cooler

115 ENERGY RELAXATION Illustration of a Loop FCp=0.010 H C C C 280 FCp=0.040 H C FCp= C H 20 0 C C 1 FCp= C C C 2 (*) Heat exchanger loads are in kw Tmin=13 o C

116 ENERGY RELAXATION Illustration of a Path FCp=0.010 H C C C FCp=0.040 H C FCp= C H 20 0 C C 1 FCp= C C C (*) Heat exchanger loads are in kw

117 ENERGY RELAXATION Procedure : Find a loop and move around heat from exchanger to exchanger until one exchanger is eliminated. H C C C 280 H C C H 20 0 C C X X C 40 0 C C X 520 +X If one wants to eliminate one exchanger: X=500. Note that X could have been negative, but we chose the smallest possible in absolute value.

118 ENERGY RELAXATION Result: Notice that the result is infeasible!!! 73 0 C H C C 45 0 C H C 65 0 C C C H 20 0 C C C 40 0 C C C This exchanger is in violation of the minimum approach

119 ENERGY RELAXATION We use a path to move heat around to restore feasibility 73 0 C H C C 45 0 C H C 65 0 C 280 +X 65 0 C C H 20 0 C C X C -X 40 0 C C X X C The value of X needed to restore feasibility is X=795

120 ENERGY RELAXATION Final Network H C C 45 0 C H C 65 0 C C C H 20 0 C C C 40 0 C C C

121 TEMPERATURE APPROACH ( T) RELAXATION We recall stating NO HEAT ACROSS THE PINCH. Being more specific, we should say, NO NET HEAT ACROSS THE PINCH. Thus we allow the following situations. Pinch as long as the NET heat transferred across the pinch is zero. However this implies allowing the temperature difference to be lower than T min

122 T RELAXATION We thus define two types of Minimum Temperature Approach. HRAT: (Heat Recovery Approach Temperature): This is the T min we use to calculate minimum utility. EMAT: (Exchanger Minimum Approach Temperature): This is the minimum approach we will allow in heat exchangers. When EMAT< HRAT networks can have 1. less splitting 2. less number of units 3. No significant increase in the total area.

123 T RELAXATION Consider the following problem Stream Type Supply T Target T F*Cp ( o C) ( o C) (MW o C -1 ) H1 Hot H2 Hot C1 Cold C2 Cold We now consider HRAT=20 o C and EMAT= 13 o C. The corresponding minimum utility are: T min Hot Utility Cold Utility Pinch 20 (HRAT) MW MW 132 O C 13 (EMAT) MW MW 112 O C

124 PSEUDO-PINCH METHOD We now consider that the difference (245 kw= 605 kw-360 kw) needs to go across the pinch of a design made using EMAT. Thus we first look at the solution of the pinch design method (PDM) for T min =13 o C FCp=0.010 H C C 45 0 C 280 FCp=0.040 H C FCp= C H 20 0 C C 1 FCp= C C C 2 (*) Heat exchanger loads are in kw

125 PSEUDO-PINCH METHOD To relax this network by 245 kw we extend the only heat exchanger above the pinch by this amount. We then proceed below the pinch as usual. FCp=0.010 FCp=0.040 H 1 H C C C C C 45 0 C C 215 FCp= C H C 20 0 C C 1 FCp= C C C 2 (*) Heat exchanger loads are in kw Note that the matching rules (FCp inequalities) can be somewhat relaxed.

126 PSEUDO-PINCH METHOD We know the solution of the pinch design method (PDM) for T min =20 o C FCp=0.010 FCp=0.040 H 1 H C C C 45 0 C C FCp=0.020 FCp= C C H H C C 40 0 C C 1 C 2 (*) Heat exchanger loads are in kw The PDM produces one additional split, and two heaters, while the PPDM features two coolers

127 DIFFICULTIES IN THE PPDM METHOD No clear indication what to do when there is many hot streams above the pinch. How to distribute the difference in heat? Even if the above is clarified it is not practical for more than a few streams.

128 T RELAXATION USING AUTOMATIC METHODS One can use the Transshipment model fixing the level of hot utility and creating the intervals using EMAT instead of HRAT. More sophisticated methods add area estimation to the model (This has been called the Vertical model) The number of units can also be controlled and sophisticated techniques have been used to explore all the flowsheets with the same number of matches. Finally, once a flowsheet is obtained a regular optimization can be conducted. This will be explored in the next Part.

129 PART 7 MATHEMATICAL PROGRAMMING APPROACHES

130 RECENT REVIEW PAPER (until 2000) Furman and Sahinidis. A critical review and annotated bibliography for Heat Exchanger Network Synthesis in the 20 th Century. Ind. Eng. Chem. Res., 41 pp , (2002).

131 PINCH DESIGN METHOD It is a DECOMPOSITION approach (3 steps) Perform Supertargeting and obtain the right HRAT, the pinch (or pinches) and the minimum utility usage Pick matches away from the pinch using the tick-off rule Evolve into higher energy consumption solutions by loop breaking and adjusting loads on paths.

132 SUPERSTRUCTURE APPROACH CONCEPT: A single optimization model, if solved globally, provides all the answers simultaneously. Superstructure of matches.

133 SUPERSTRUCTURE APPROACH Possible flow sheets embedded (recycles/by-passes excluded)

134 SUPERSTRUCTURE APPROACH Possible flow sheets embedded (continued)

135 SUPERSTRUCTURE APPROACH Model Constraints Objective (no fixed costs)

136 SUPERSTRUCTURE APPROACH This is an MINLP formulation with which several researches have struggled. (MINLP methods could not be easily solved globally until recently (?). Therefore it needs some initial points. ALTOUGH A ONE-STEP CONCEPT IT BECAME IN REALITY AN ITERATIVE PROCESS

137 SUPERSTRUCTURE APPROACH Some Strategies to overcome the curse of non-convexity Hasemy-Ahmady et al., Many other methodologies attempted this goal (provide good initial points) like evolutionary algorithms, simulated annealing, etc.

138 STAGE SUPERSTRUCTURE APPROACH With isothermal mixing. Yee and Grossmann (1990, 1998) Note on the side: This is a remarkable coming back to origins. From Grossmann and Sargent (1978) In the 90 s the mathematical programming/ superstructure approach emerged as the dominant methodology

139 SUPERSTRUCTURE APPROACH From Grosmann and Sargent (1978)

140 SUPERSTRUCTURE APPROACH

141 LATEST MILP APPROACH (Barbaro and Bagajewicz, 2005) Counts heat exchangers units and shells Approximates the area required for each exchanger unit or shell Controls the total number of units Implicitly determines flow rates in splits Handles non-isothermal mixing Identifies bypasses in split situations when convenient Controls the temperature approximation (HRAT/EMAT or Tmin) when desired Allows multiple matches between two streams

142 LATEST MILP APPROACH (Barbaro and Bagajewicz, 2005) m Hot 7 8 stream i ˆ z H q, ijm z q im, jn ˆ z C q, ijn Cold stream j n Transportation Model Approach

143 LATEST MILP APPROACH (Barbaro and Bagajewicz, 2005) Heat Exchanger counting ˆ z, H z, H K ijm Yijm First interval m 1 z,h Yijm 0 z, H K ijm 0 z,h Kˆ ijm 0 Kˆ z, H ijm 2 Y z H Kˆ, ijm Y Kˆ z, H ijm ˆ z, H K ijm Y 0 z, H ijm z, H ijm z, H ijm Y Y z, H ijm+ 1 z, H ijm+ 1 Rest of intervals m

144 LATEST MILP APPROACH (Barbaro and Bagajewicz, 2005) Flowrate Consistency Cp im qˆ (T z,h ijm U m T L m ) Cp im 1 qˆ (T z,h ijm 1 U m 1 T L m 1 ) Cp im qˆ ( T z, H ijm U m T L m ) Cp im 1 qˆ ( T z, H ijm 1 U m 1 T L m 1 ) Heat exchanger spanning m Cp im qˆ ( T z, H ijm U m T L m ) = Cp im 1 qˆ ( T z, H ijm 1 U m 1 T L m 1 )

145 PART 8 INDUSTRIAL IMPORTANCE OF USING THE RIGHT MODEL Crude fractionation case study

146 CRUDE FRACTIONATION EXAMPLE CONDENSER water naphtha PA1 steam PA2 kerosene steam PA3 diesel water crude HEN sour water DESALTER steam steam gas oil HEN FURNACE residue

147 CRUDE FRACTIONATION EXAMPLE Light Crude Heavy Crude Temperature ( o C) Temperature (C) Enthalpy (MW) Enthalpy (MW) Cooling water is already included in the graphs. The light crude exhibits what is called a continuous pinch. The heavy crude is unpinched.

148 CRUDE FRACTIONATION Cooling water C2 FURNACE H8 H2 H3 H7 H6 H5 H1 DESALTER H9 H10 H4 MER network for the light crude.

149 CRUDE FRACTIONATION C2 Cooling water FURNACE H8 H7 H6 H2 DESALTER H5 H3 H4 H10 H1 C1 MER network for the heavy crude.

150 CRUDE FRACTIONATION NOT USED FOR HEAVY CRUDE NOT USED FOR LIGHT CRUDE FURNACE C2 Cooling water H8 H2 H3 H7 H6 H5 H1 DESALTER H4 H9 H10 C1 MER network efficient for both crudes.

151 CRUDE FRACTIONATION It is clear from the previous results that efficient MER networks addressing multiple crudes can be rather complex and impractical. Alternatives to the Pinch Design Method (PDM) are clearly needed. This does not mean that the PDM fails all the time. It is still capable of producing good results in many other cases.

152 CRUDE FRACTIONATION EXAMPLE We now illustrate the use of HRAT/EMAT procedures for the case of crude fractionation units. We return to our example of two crudes. The problem was solved using mathematical programming. The networks have maximum efficiency for both crudes. Only the vertical model and the control on the number of units was used. No variation in the matches was done (not really necessary in this case) and no further optimization was performed.

153 CRUDE FRACTIONATION NOT USED FOR HEAVY CRUDE FURNACE C2 Cooling water 18 units H3 H8 H6 H5 H7 H2 DESALTER NOT USED FOR HEAVY CRUDE H9 H1 H10 H4 C1 Solution for HRAT/EMAT = 20/10 o F

154 CRUDE FRACTIONATION NOT USED FOR HEAVY CRUDE FURNACE Cooling water H3 H8 H6 H7 18 units H5 H2 DESALTER H1 NOT USED FOR HEAVY CRUDE H9 H4 H10 C1 Solution for HRAT/EMAT = 40/30 o F

155 CRUDE FRACTIONATION NOT USED FOR HEAVY CRUDE Cooling water FURNACE H3 H8 H6 H7 H2 17 units H1 DESALTER H5 NOT USED FOR HEAVY CRUDE H9 H4 H10 C1 Solution for HRAT/EMAT = 80/60 o F

156 CRUDE FRACTIONATION Cost, MM$/yr Combined Multiperiod Multiperiod+ Multip+Des.Temp Network Model Desalt.Temp. +Higher HRAT HRAT/EMAT 20/20 20/10 20/10 40/30 Operational Fixed Total

157 CRUDE FRACTIONATION EXAMPLE We now illustrate the use of more sophisticated models that allow the control of splitting. These are essentially transshipment models that are able to control the level of splitting (something that regular transshipment models cannot do).

158 CRUDE FRACTIONATION Two branches unrestricted HRAT = 40 o F EMAT = 30 o F FURNACE H8 H7 H3 23 units H2 H6 H5 H4 5% More Energy Consumption DESALTER H10 H1 H9 C1

159 CRUDE FRACTIONATION restricted HRAT = 40 o F EMAT = 30 o F FURNACE H7 H3 H8 H6 21 units H2 H1 H5 H4 DESALTER 5% More Energy Consumption H10 H9 C1

160 CRUDE FRACTIONATION Cost, MM$/yr Multiperiod Two-branch Two-branch Model Unrestricted Restricted Operational Fixed Total

161 CRUDE FRACTIONATION The two-branch design Is efficient for all feedstocks proposed. Consumes only a few more millions Btu/hr Has many solutions of similar energy consumption. Complexity can be reduced at a relatively small energy increase and with some reduction of capital. All these models suggest that there is a lot of flexibility to perform an effective retrofit because there are these many options of similar cost to explore.

162 PART 9 RETROFIT AND TOTAL SITE INTEGRATION

163 RETROFIT The big question in trying to do a retrofit of a HEN is whether one really wants to achieve maximum efficiency. Original System Operating Costs ($/yr) HORIZON Capital Investment ($) Usually retrofits are too expensive and have a long payout.

164 RETROFIT It is desired to produce the largest reduction of cost with the smallest capital investment Present Condition Retrofits HORIZON Capital Investment ($) The question is how to identify the most profitable.

165 EXAMPLE Retrofits involve a) relocation, and b) addition of new units. NEW-1 DESALTER 9T-1 NEW-3 NEW-2 9T-2 9T-7 RELOC REB-T7 NEW-4 9T-9 9T-1 DESALTER RELOC NEW-5 9T-9 D101 9T-7 This particular study produced a) 700,000 annual savings with 1.2 years payoff, b) Additional 12% capacity (not counted in the $700,000 savings)

166 TYPES OF RETROFIT By inspection. Perform pinch design or pseudo pinch design and determine heat exchangers to add Systematic methods using tables and graphs exist They are outside the scope of this course. Mathematical programming approaches also exist but they have not passed the test of usability and friendliness

167 RECENT WORK ON RETROFIT Asante, N. D. K.; Zhu, X. X. An Automated and Interactive Approach for Heat Exchanger Network Retrofit. Chem.Eng. Res. Des. 75 (A), (1997). Briones, V.; Kokossis, A. C. Hypertargets: A Conceptual Programming Approach for the Optimisation of Industrial Heat Exchanger Networks II. Retrofit Design. Chem. Eng. Sci. 54, (1999). Barbaro, Bagajewicz, Vipanurat, Siemanond. MILP formulation for the Retrofit of Heat Exchanger Networks. Proceedings of Pres 05. (2005)

168 TOTAL SITE INTEGRATION USE OF GRAND COMPOSITE CURVES TO PLACE UTILITIES. Plant 1 Plant 2 H H Pockets are eliminated and curves are shifted

169 TOTAL SITE INTEGRATION Site sink and source profiles are constructed. LP Steam HP Steam Source Profile Sink Profile Overlap region. These are savings. Energy integration is performed by placing utilities. We will see how this method can be wrong. H

170 TOTAL SITE INTEGRATION Consider two plants. We would like to know under what conditions one can send heat from one plant to the other. PLANT 1 PLANT 2 COMBINED PLANT Above both pinches Pinch Point Plant 2 Between pinches Possible Location of Pinch Pinch Point Plant 1 Below both pinches

171 TOTAL SITE INTEGRATION T U 1,min -Q T U 2,min Effective integration takes place between pinches. Pinch 2 Pinch 1 Q T From where plant 2 is heat source to where plant 1 is heat sink. W 1,min W 2,min -Q T

172 TOTAL SITE INTEGRATION T U 1,min -Q T U 2,min + Q T Pinch 1 Q T Pinch 2 Integration outside the inter pinch region leads to no effective savings. W 1,min W 2,min

173 GRAND COMPOSITE CURVES PLANT 1 PLANT 2 Pinch = = = = = 6-1 Pinch 0 = = = = = = 2 4 = 3 + 1

174 GRAND COMPOSITE CURVES Plant 1 Temperature Plant 2 Maximum possible savings Enthalpy

175 ASSISTED HEAT TRANSFER PLANT 1 PLANT 2 If heat is not transferred above the two pinches only 13 of the maximum 17 can be saved. Pinch = = = = = Pinch 0 = = = = = = 3 12 = 7 + 5

176 ASSISTED HEAT TRANSFER Temperature Plant 1 Assisted Transfer Pocket Maximum possible savings Enthalpy Plant 2

177 ASSISTED HEAT TRANSFER 140 Typical solutions call for sealing the pocket preventing thus the savings. Temperature Plant 1 Maximum possible savings Plant Enthalpy Temperature Plant 1 Assisted Transfer Pocket Maximum possible savings Plant Enthalpy

178 HEAT EXCHANGER NETWORKS Heat Exchanger networks should be such that both plants can work at maximum efficiency when integrated and when they stand alone. We now show a result of a case study. An integration between a Crude unit (heat sink ) and an FCC unit (heat source). H5 H9 H4 H3 H7 H6 H2 H1 H8 C1 Above the Pinch Plant 1: Crude Unit Direct Integration (two stream circuits) Pinch location Original HEN Additional heat exchangers used during integration H10 H11 H15 H13 H14 H12 CW Below the Pinch Plant 2: FCC Unit C2 CW

179 HEAT EXCHANGER NETWORKS Heat Exchanger networks should be such that both plants can work at maximum efficiency when integrated and when they stand alone. We now show a result of a case study. An integration between a Crude unit (heat sink ) and an FCC unit (heat source). H5 H9 H4 H3 H7 H6 H2 H1 H8 Above the Pinch C1 CW Utility (MMBtu/hr) Heating Cooling No Integration Direct Integration Plant 1: Crude Unit H10 H11 H15 H13 H14 Direct Integration (two stream circuits) Savings 1,000,000 $/year C2 Pinch location H12 Original HEN CW Additional heat exchangers used during integration Below the Pinch Plant 2: FCC Unit

180 TOTAL SITE INTEGRATION Multiple plants can also be analyzed. We show below one example of such studies. Alternative solutions exists. Grand composite curves cannot be used anymore. 300 C PLANT 1 PLANT 2 PLANT 3 PLANT 4 (Test Case #2) (Trivedi) (C & F) (4sp1) C 200 C C C 40 C

181 FUTURE TRENDS Mathematical programming will be the dominant tool. Software companies are struggling to make the proper choice of existing methods for each case and most important numerically reliable. The best option for the time being is to intelligently interact with experts while using existing software.