HDA Process. compressor. Reactor. Flash. heat. cool. Stabilizer. Recycle. Product

Transcription

1 HDA Process compressor heat Reactor cool Flash Recycle Product Stabilizer

2 Why Energy Integration? (ICI Experience) 20 case studies In every case, there was a reduction in energy In almost every case, the energy savings required less capital Up to 60% energy savings, up to 25% capital savings, up to 15% lower product price Why spend additional capital to waste energy?

3 Energy Integration What are the minimum cooling and heating requirements? What is the minimum number of heat exchangers required? What are the main trade-offs?

4 An Example For the two hot streams being cooled and the two cold streams being heated shown below, find the minimum heating and cooling requirements, as well as the minimum number of heat exchangers if min = 10 deg F. Stream 1 2 F 120 F 200 F 100 F FC p (Btu/F) F 90 F F 130 F

5 Net Energy Required (First Law Calculation) F 120 F 200 F 100 F F 90 F F 130 F FC p (Btu/F) Q aval (MBtu/hr) hus, 10 x10 3 Btu/hr would have to be supplied from a hot utility if there were no requirement of min = 10 deg F. Question: What are the minimum heating and cooling requirements? -10

6 Net Energy Required at emperature Intervals Define hot and cold temperature scales shifted by 10 deg F and show streams H (F) C (F)

7 Net Energy Required at emperature Intervals Draw in intervals by breaking at stream starting and ending points H (F) C (F)

8 Net Energy Required at emperature Intervals Compute net available Q for each interval H (F) C (F) Q aval (MBtu/hr) FC p Note: Sum of net heat available = -10 MBtu/hr Same as first law

9 Cascade Diagram We satisfy the net requirement for each interval from an external utility. H (F) C (F) H O H =50 C O L D U I L I Y H =-40 H =-80 H =40 H =20 U I L I Y BU...

10 Cascade Diagram We can always transfer excess heat from high temperature intervals to lower temperature intervals without violating min = 10 F H (F) C (F) H O U I L I Y 70 H = H = H =-80 H = H =20 C O L D U I L I Y

11 Minimum Heating and Cooling H (F) C (F) H O U I L I Y 70 H = H = H =-80 H = H =20 C O L D U I L I Y Minimum heating = 70 MBtu/hr Minimum cooling = 60 MBtu/hr Net = 10 MBtu/hr of heating Same as first law

12 Pinch emperature H (F) C (F) H O U I L I Y 70 H = H = H =-80 H = H =20 C O L D U I L I Y he design problem can be decomposed into two separate problems: 1. A high temperature region where only heating is requred 2. A low temperature region where only cooling is required

13 Relationship to First Law Minimum Heat in = 70 HO COLD }Net = -10 Minimum Heat out = 60 Heat in = 70 + Q E HO }Net = -10 Heat out = 60 + Q E If we put excess heat into the process, we must also remove this heat!

14 Excess Heating and Cooling H (F) C (F) H O U I L I Y Q E 70 H = H = Q E H =-80 Q E H =40 40+Q E H =20 60+Q E C O L D U I L I Y Excess steam or furnace capacity requires excess cooling water. Don t transfer heat across the pinch!

15 Multiple Utilities H (F) C (F) H O U I L I Y 70 H = H = H =-80 H = H =20 C O L D U I L I Y hot water cooling water We prefer to add heat at the lowest possible temperature, and to remove heat at the highest possible temperature! Multiple utilities correspond to multiple pinches.

16 -H Diagram Hot 230 Enthalpy, MBtu/hr Enthalpy, MBtu/hr

17 -H Diagram 230 Cold Enthalpy, MBtu/hr Enthalpy, MBtu/hr

18 -H Diagram 230 Cold Enthalpy, MBtu/hr Enthalpy, MBtu/hr

19 -H Diagram 230 Hot Cold Enthalpy, MBtu/hr min = 10 F Q H,min Q C,min Enthalpy, MBtu/hr

20 Limitation of the Procedure We need to know FC p values of all streams Inlet and outlet temperatures of all streams BU the design variables that fix the process flows must be determined from optimization which in turn depends on heatexchanger network Solution: Heat-exchanger network as function of the flows Some iterations are needed

21 An Exercise Given the following stream data below, find (assume min = 10 deg C) the minimum amount heat added the minimum amount of cooling the pinch temperature Stream C 60 C C 30 C 135 C 20 C 140 C 80 C FC p (kw/c)

22 Minimum Number of Exchangers From results of minimum heating and cooling estimates Sources Hot Utility 70 MBtu/hr Stream MBtu/hr Stream MBtu/hr Sinks Stream 3 MBtu/hr Stream MBtu/hr Cold Utility 60 MBtu/hr Heat loads always balance as result of first law analysis Number of exchangers (N E ) = Number of streams (N S ) + Number of Utilities (N U ) - 1 N E = = 5

23 Minimum Number of Exchangers he previous solution is not always correct. Consider Sources Hot Utility 230 MBtu/hr 230 Stream MBtu/hr 130 Stream MBtu/hr Sinks Stream 3 MBtu/hr Stream MBtu/hr Cold Utility MBtu/hr We see that if we have an exact matching between some streams, we need less exchanger N E = N S + N U - N P N P = number of independent subproblems

24 Minimum Number of Exchangers he previous solution is still not always correct. Consider Sources Hot Utility 70 Stream MBtu/hr Stream MBtu/hr Sinks Q E 70-Q 20-Q E E Q E Stream 3 MBtu/hr Stream MBtu/hr Cold Utility MBtu/hr We see that there are now six exchangers and there is now a loop (i.e., HU -> S3 -> S1 -> S4 -> HU) N E = N S + N U + N L - N P N L = number of loops

25 Effect of Pinch Analysis he pinch analysis indicates that we only use heat above the pinch and cooling below the pinch, so that the design problem can be decomposed into two subproblems. H (F) C (F) Assume no loops & no exact matches N E = N S + N U - 1 = = 4 N E = N S + N U - 1 = = 3 N OAL = 7

26 Effect of Pinch Analysis Minimum Exchangers: N = 5 Pinch Analysis: N = 7 (minimum energy) A trade-off between captial cost (minimum exchangers) and utility costs (minimum energy) We can decrease the number of exchangers if we transfer some heat across the pinch, but the energy usage increases

27 Design of Minimum- Energy HEN s H (F) C (F) Q = 110 Q = Q = Pinch Q = 60 Q = Q = 160 Q = Stream FC p Above the pinch: Q tot = 70 MBtu/hr, heat added Above the pinch: Q tot = 60 MBtu/hr, heat removed

28 Design above the Pinch H (F) C (F) Q = 110 Q = Q = 360 Pinch 160 Q = Feasible Pinch Matches: (FC p ) HO < (FC p ) COLD We can match stream 1 with either 3 or 4, and we can only match stream 2 with 4.

29 Design above the Pinch H (F) C (F) Q= Pinch Q = ransfer the maximum amount of heat possible for each match to attemp to eliminate streams from the problem.

30 Remaining Heat Loads Pinch Q = = F H (F) C (F) Q = 360- = F ransfer the remaining heat from stream 1 to stream 4.

31 Remaining Heat Loads H (F) C (F) Q = 50 Pinch

32 Heat from Hot Utility H (F) C (F) Pinch H = F

33 Design above the Pinch (Summary) H (F) C (F) Pinch Q = Q=50 Q= H= Put in the matches at the pinch 2. Maximize the heat loads to eliminate streams 3. See what is left

34 Design above the Pinch (Alternatives) H (F) C (F) Q=50 H=70 Pinch Q = Q=

35 Design below the Pinch H (F) C (F) 160 Pinch Q = Q = 160 Q =

36 Design below the Pinch (One Alternative) H (F) C (F) Pinch C = 20 C = 40 = Q =

37 A Complete Design H (F) C (F) Q = Q=50 Q= H=70 C = 20 C = Q = otal Number of Exchangers: 7

38 Remarks Minimum approach temperature So far, we have assumed a value of 10 deg F rade-off: Larger minimum approach temperature, smaller heat-exchanger area but larger minimum heating and cooling Additional complexities he design problem is not always as simple as the example considered Stream splitting Alternatives

39 Reducing the Number of Exchangers he number of exchangers required for the overall process is always less than or equal to that for the minimum energy network he minimum energy network normally contains loops across the pinch hese loops can be broken by transferring heat across the pinch, but we will introduce at least one violation of the specified min min can be restored by shifting heat along a path, which increases energy consumption of the process

40 Loops A set of connections that starts from one stream and returns to the same stream. H (F) C (F)

41 Loops A set of connections that starts from one stream and returns to the same stream. H (F) C (F)

42 Loops A set of connections that starts from one stream and returns to the same stream. H (F) C (F)

43 Loops A set of connections that starts from one stream and returns to the same stream. H (F) C (F)

44 Breaking Loops Break the loop with exchanger with the small load Remove the smallest heat load from loop Breaking a loop across the pinch normally violates the 2nd law H (F) C (F) Q = Q=50 Q= H=70 C = 20 C = Q =

45 H (F) C (F) Q = H=70 Q=50+20 Q=-20 C = Q =

46 BU... H (F) C (F) Q = Q=70 Q= H=70 C = Q = Hot Stream: 200 F to 120 F Cold Stream: F to 130 F Impossible according to 2nd Law!

47 Restoring min Shift heat along a path, which is a connection between a cooler and a heater Q=70-Q E H=70+Q E Q=60+Q E Q= =120 Q = 120- Q E < 110 F C = 60+Q E hus, 1000(60+Q E ) = 3000(-110) Q E = 60 MBtu/hr

48 Reducing Number of Exchangers = = =120 =200 =145 Q=120 Q=10 = H=130 Q= = =168.4 =166.6 C = 120 =115 =100 Q = 60 =130 =110 =90 otal number of Exchangers: 6

49 Restoring min (An Alternative Path) Q=70 H=70+Q E =120 Q=60 Q=-Q E Q=120 < 110 F C = 60+Q E But this path can t restore min!

50 Breaking Another Loop Q=10 H=130 Q=120 Q= C = 120 Q=60

51 Pinch Heat Engines Q in +W Q in Heat Engine W Q in Q in Q E Q in Energy Cascade Energy Cascade Energy Cascade Energy Cascade Energy Cascade Q out Energy Cascade Q out Energy Cascade Q out Heat Engine Q out -W W Energy Cascade Q E -W Q out +Q E -W Heat Engine W Efficiency: 100% 100% Sandard Alone Place Heat Engine either above or below the pinch (not across)!

52 Heat Pumps Q in Q in - W Energy Cascade Pinch Energy Cascade Q E +W Q E Heat Pump W Energy Cascade Energy Cascade Q out Q out Q in Q in - (Q E +W) Energy Cascade Energy Cascade Q E - W Q E Heat Pump W Energy Cascade Energy Cascade Q E +W Q E Heat Pump W Q out + W Q out - Q E Place heat pump across the pinch!

53 Distillation Column as a heat engine Heat In (Q Reb ) Column Heat Out (Q Cond ) Column either above or below the pinch Q in Q in Energy Cascade Energy Cascade Column Energy Cascade Column Energy Cascade Q out Q out