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Department of Energy and Process Engineering. TEP 4215 - Plan for Assignments with Guidance. Ass. Topic Supervised Deadline 1 Sequence of Distillation Columns 22.01 29.01 2 Minimum Energy Requirements and Pinch 29.01 05.02 3 Design of Heat Exchanger Networks (1) 05.02 12.02

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Presentation Transcript
slide4

Department of Energy and Process Engineering

TEP 4215 - Plan for Assignments with Guidance

Ass. Topic Supervised Deadline

1 Sequence of Distillation Columns 22.0129.01

2 Minimum Energy Requirements and Pinch 29.0105.02

3 Design of Heat Exchanger Networks (1) 05.0212.02

4 Optimization of Heat Exchanger Networks 12.0226.02

5 Retrofit Design of Heat Exchanger Networks 26.0205.03

6 Indirect Integration of Plants using Steam 05.0312.03

7 Integration of Distillation Columns 12.0309.04

8 Optimal Use of Heat Pump 09.0416.04

9 Area in Heat Exchanger Networks 16.0423.04

10 Heat Integration and Forbidden Matches 23.0430.04

11 Design of Heat Exchanger Networks (2) 30.04none

Guidance: ½ Scientific Assistant, 6 Student Assistants and the Lecturer

Truls Gundersen

14.01.13

slide5

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Reactor System (R)
    • No questions so far
      • Any Questions now?
    • Relevance for the Exam?
      • Lecturer to provide some wise words…..
      • Or: See the next Slide !!

Process, Energy and System

Question Session

T. Gundersen

Q - 01

slide6

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Reactor System (R) Q from 2010
    • What is expected by the Students on this topic
      • Q: “Chapter 6.1-6.3 is part of the list describing what is curriculum from the text book by Robin Smith. Are we expected to be able to reproduce (or derive) the formulas here, or should we be able to use them?”
      • A: These Sections are listed with importance “1”, meaning that this is Background material. No formulas need to be derived or used, these Sections are included (as “1”) to provide background for the discussion about effects of T and p in the lectures

Process, Energy and System

Question Session

T. Gundersen

Q - 02

slide7

FF

RF

P

RX

yR = XS

yP = XS(1+ω)

R

S

PX

BP

RR

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Reactor Separator Interface (R/S)
    • No questions so far
      • Any Questions now?
    • Relevance for the Exam?
      • Lecturer to provide some wise words…..

Process, Energy and System

Question Session

T. Gundersen

Q - 03

slide8

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Separation System (S)
    • No questions so far
      • Any Questions now?
    • Relevance for the Exam?
      • Lecturer to provide some wise words…..
      • Or: See next Slide !!

Process, Energy and System

Question Session

T. Gundersen

Q - 04

slide9

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Separation System (S) Q from 2009
    • Curriculum for Distillation Sequences
      • Q:“I would like to know what is relevant for the Exam in the topic Sequence of Distillation Columns. There is no (?) Exam Tasks where the entire process should be developed (such as in Assignment 1). Does this mean it is not very relevant?”
      • A:True, not given during 2005-2010  about time?
      • A:The topic is relevant, and one should know the Heuristic Rules, have a good Strategy for developing the sequence, and be able to handle Heat Integration.

Process, Energy and System

Question Session

T. Gundersen

Q - 05

slide10

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Separation System (S) Q from 2009
    • Hot and Cold Streams in Distillation Columns
      • Q:Why is the Reboiler identified as a Cold Stream, while the Condenser is identified as a Hot Stream, when the Reboiler has higher Temperature?
      • A: A mixture is boiling in the Reboiler (liquid to vapor) and condensing in the condenser (vapor to liquid), thus heat must be supplied to the Reboiler and removed from the Condenser
      • A: “Hot/Cold” refers to change in Thermodynamic State, not the absolute Temperature level of streams

Process, Energy and System

Question Session

T. Gundersen

Q - 06

slide11

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Separation System (S) Q from 2009
    • mCp Values in Distillation Columns & Utilities
      • Q:Why are mCp values of condensers and reboilers as well as utilities said to be “infinity”?
      • A: Consider the following (use Blackboard)
        • The slope of condensing/vaporizing streams in TQ diagrams
        • Enthalpy change for sensible vs. latent heat
      • A: Some Software Packages use ΔT=1°C for condensing and vaporizing streams (results in very large mCp values, but not “infinity”)

Process, Energy and System

Question Session

T. Gundersen

Q - 07

slide12

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Interface Separation System (S) and Heat Recovery System (H)
    • No questions so far
      • Any Questions now?
    • Relevance for the Exam?
      • Lecturer to provide some wise words…

Process, Energy and System

Question Session

T. Gundersen

Q - 08

slide13

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Targeting for Energy:
      • Q:“I understand that heat transfer across the Pinch temperature is unfavorable, but does such a process require more energy?”
      • A: Yes, the point is that the Pinch decomposes the process into two parts; one with heat deficit (above Pinch) and one with heat surplus (below Pinch). As a result, any heat transfer across the Pinch will make BOTH the deficit and surplus larger, and thus increase both external heating and cooling (i.e. energy)

Process, Energy and System

Question Session

T. Gundersen

Q - 09

slide14

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Targeting for Energy:
      • Q:“In the solution to Exam 2012, the “simplified heat cascade” is used. Stream C2 has a target temperature of 215C, while in the simplified cascade, the interval temperature is 245C. Of course, in kW, this is “accounted for”. However, I get the same result using all temperatures, and it is then more clear”
      • A: Yes, there are advantages (fewer intervals) and disadvantages (have to be careful with heat supply and rejection) with using the simplified heat cascade !!

Process, Energy and System

Question Session

T. Gundersen

Q - 10

slide15

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Targeting for Units:
      • Q:“In Assignment 7, how do we get Umin,MER = 9?”
      • A: See Grid Diagram below with Column A integrated above Pinch (i.e. Condenser H3)
      • Q: Why Utotal = 12
      • A: C3, C4 & H4
      • also need heat
      • exchangers !!
      • (with Utilities)

Process, Energy and System

Question Session

T. Gundersen

Q - 11

slide16

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Targeting for Units:
      • Q:“Euler’s Rule says U = N + L – S, where S stands for Subnetwork. What is a Subnetwork?”
      • A: A Subnetwork means that some streams and units are in heat balance, and that there are no connections with other streams/units. Notice that S = 1 means that the entire network is connected, while S = 2 means there are 2 independent Subnetworks.

Process, Energy and System

Question Session

T. Gundersen

Q - 12

slide17

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Targeting for Units:
      • Q:“Euler’s Rule says U = N + L – S; is N the number of heaters, coolers and streams?”
      • A: No!! N is the number of process streams and utility types !! (A very common misunderstanding)
      • Q:“My notes say that L = Uactual – Umin, and it is emphasized that Uactual ≠ Umin,MER. In the solution to Assignment 7, it is said that U = Umin,MER ??”
      • A: Distinguish Targets (before) from Design (after). Assign. 7 simply counts units. Remember Assign. 3

Process, Energy and System

Question Session

T. Gundersen

Q - 13

slide18

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Targeting for Units:
      • Q:“In the Exam for 2008 there is one heater and two coolers. One of the coolers operates across Pinch. The task is to find Umin,MER. How to count for the cooler operating across Pinch?
      • A: Yet another misunderstanding about the targeting formula for minimum number of units. The N-1 rule is applied at the targeting stage ahead of design, and should not be applied to a network. The background for the Exam question is to highlight that the network has 5 units while the minimum for MER is 7 units.

Process, Energy and System

Question Session

T. Gundersen

Q - 14

slide19

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Targeting for Units:

Process, Energy and System

Question Session

T. Gundersen

Q - 15

slide20

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Design of HENs:
      • Q:“In order to find the log mean temperature difference (LMTD) for heat exchangers one uses the inlet and outlet temperatures for the hot and cold side of the exchanger. What about utility exchangers (heaters and coolers) where we only have 2 temperatures?”
      • A:Utility exchangers have a hot and cold side in the same way as process/process exchangers. LMTD is calculated based on the supply and target temperatures of the streams and the utilities. See previous Slide !!

Process, Energy and System

Question Session

T. Gundersen

Q - 16

slide21

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Design of HENs:
      • Q:“In the Solution for Exam 2011(?), a match below Pinch is made between a hot stream with mCp=20 and a cold stream with mCp=40, while another cold stream with mCp=30 and a more favorable inlet temperature could have been used?”
      • A: The Pinch Design Method has 2 important elements; decompose the problem at the Pinch (i.e. design separate networks above and below Pinch) and start the design at the Pinch. Considering cold inlet temperatures means starting in the cold end! See next Slide !!

Process, Energy and System

Question Session

T. Gundersen

Q - 17

slide22

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Design of HENs:

Process, Energy and System

Question Session

T. Gundersen

Q - 18

slide23

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Design of HENs:
      • Q:“In the Solution for Exam 2010 (Task 1.c), cold stream C1 is split above Pinch. Why can we not do without this split?”
      • A:This is a crucial question and goes directly into the “heart” of the Pinch Design Method and the issues of Pinch Exchangers and mCp Rules. The next slide shows the split design and the split-free design. Unfortunately, this is a very common mistake in previous Exams !!

Process, Energy and System

Question Session

T. Gundersen

Q - 19

slide24

Ca

Cb

IV

I

III

II

H2

H

Network from the Solution to the Exam 2010

mCp

50

30

90

20

100°C

150°C

60°C

100°C

84°C

H1

1200

180°C

50°C

166.7°C

100°C

1500

134.4°C

90°C

α

140°C

C1

2500

β

Process, Energy and System

149.3°C

50°C

180°C

2000

110°C

90°C

C2

1400

400

800

90°C

An “MER” Design meeting Targets for Energy and Units

Question Session

T. Gundersen

Q - 20

slide25

Cb

Ca

IV

III

H2

H

Network from one of the Students

mCp

50

30

90

20

100°C

150°C

60°C

100°C

84°C

H1

II

2000

180°C

50°C

100°C

Process, Energy and System

II

700

90°C

117.8°C

140°C

C1

2500

2000

50°C

180°C

110°C

90°C

C2

1400

400

800

90°C

Violating mCp Rules for Pinch Exchangers

Question Session

T. Gundersen

Q - 21

slide26

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Design of HENs:
      • Q:“Could you say something about Utility Pinch, how it arise, and what kind of Effects does it have?”
      • A:Utility Pinch points arise only when we have more than one hot or more than one cold utility while at the same time maximizing the load (duty) of the cheaper utility to reduce total energy cost.
      • A: Utility Pinch will directly and strongly affect the design of the network (one more “Region” for each such Pinch), and will result in a more complex network (more units and stream splits) with increased total area.

Process, Energy and System

Question Session

T. Gundersen

Q - 22

slide27

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Advanced Design Tools for HENs:
      • Q:“How do we construct the Driving Force Plot (DFP) and what is it used for?”
      • A:Starting with the Composite Curves, the DFP is constructed by plotting corresponding values of cold composite temperatures and the temperature difference between the hot and the cold Composite Curves.
      • A: The DFP is used as a qualitative tool for good distribution of driving forces and thus minimizing total heat transfer area. More specifically it is used to find stream split ratios and to “tame” the tick-off rule.

Process, Energy and System

Question Session

T. Gundersen

Q - 23

slide28

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Advanced Design Tools for HENs:

Process, Energy and System

Nice, but not relevant for the Exam !!

Question Session

T. Gundersen

Q - 24

slide29

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Advanced Design Tools for HENs:
      • Q:“What is the Remaining Problem Analysis (RPA), and how/when do we use it?”
      • A:The RPA is used as a quantitative tool to provide feedback to the designer who makes one decision at a time. The effect of accepting a specific heater, cooler or process/process heat exchangers is measured by estimating the “final” values for Energy, Area and Units by combining data for the “accepted” unit and the target values for the Remaining Problem. See next Slide !!

Process, Energy and System

Question Session

T. Gundersen

Q - 25

slide30

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Advanced Design Tools for HENs:

Process, Energy and System

Nice, but not relevant for the Exam !!

Question Session

T. Gundersen

Q - 26

slide31

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Optimization of HENs:
      • Q:“In Assignment 7, can I disregard process and utility Pinch points during Optimization?”
      • A: Notice there is no Utility Pinch here, integrating the Condenser creates a new process Pinch !!
      • A: The observation is correct though. After the MER design, we “forget” about the Pinch point(s), but still focus on having feasible driving forces (ΔT ≥ ΔTmin). Energy consumption will increase when reducing units, and heat will flow across the Pinchpoint(s) !!

Process, Energy and System

Question Session

T. Gundersen

Q - 27

slide32

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Optimization of HENs:
      • Q:“What is the definition of Loops and Paths?”
      • A: A Heat Load Loop is a “connection” between units and streams in such a way that the heat exchangers can change their duties (+/- x) without affecting the total enthalpy change of the streams involved.
      • A: A Heat Load Path is similar to a Loop in the sense that the enthalpy change of the streams remains the same. A Path is used to repair ΔT problems after breaking a Loop. Thus it involves Utility exchangers.

Process, Energy and System

Question Session

T. Gundersen

Q - 28

slide33

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Optimization of HENs:
      • Q:“Can a stream that is split into two branches be part of a Heat Load Loop or Path?”
      • A: Yes, as long as the stream enthalpy change remains the same, it does not matter whether the “+ x kW” unit is in serial or parallel arrangement with the “- x kW” unit.

Process, Energy and System

Question Session

T. Gundersen

Q - 29

slide34

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Optimization of HENs:
      • Q:“Is the objective of Optimization to break Loops by removing Units or is it to remove Units, and breaking Loops is a consequence of this?”
      • A: The objective of Optimization (or Evolution) is both to reduce Total Annual Cost and to simplify network complexity (number of units and stream splits). This can be done by removing small Units, since these have a considerable cost while not recovering much energy. Network complexity is also a consequence of the PDM with Pinch decomposition.

Process, Energy and System

Question Session

T. Gundersen

Q - 30

slide35

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Optimization of HENs:
      • Q:“Can you explain the procedure for removing units from the MER-(i) design in the solution for Exam 2007, in particular the arguments why cooler Ca of 100 kW can not be removed?”
      • A: The strategy is to try to remove the smallest units, and these can be both proc./proc. heat exchangers, heaters and coolers, but they have to be part of a loop. Quite often, however, some of these units can not be removed, since it turns out to be impossible to restore driving forces (ΔTmin). See next Slides !!

Process, Energy and System

Question Session

T. Gundersen

Q - 31

slide36

II

Pinch

α

IV

180°C

140°C

92ºC

90°C

140°C

III

Ca

H1

β

140°C

100 - x

I

150°C

120°C

140°C

H2

Cb

1200 + x

128.25ºC

γ

120°C

135°C

C1

600 - x

δ

153.0ºC

60°C

900 + x

160°C

120°C

H

C2

147.5ºC

500

1100

2400

mCp

50

60

100

40

Process, Energy and System

The smallest Unit Ca (100 kW) is part of a Loop:

H1 – (II) – C1 – (I) – H2 – (Cb) – CW – (Ca) – H1

Question Session

T. Gundersen

Q - 32

slide37

mCp

50

60

100

40

II

Pinch

α

IV

180°C

138°C

90°C

138°C

III

H1

β

138°C

I

150°C

120°C

141.67°C

H2

Cb

1300 + y

γ

120°C

135°C

C1

500 - y

δ

60°C

1000 + y

160°C

120°C

H

C2

147.5ºC

500 + y

1100

2400 - y

Process, Energy and System

When removing Cooler Ca (100 kW) by the Loop, Exchangers II, III and IV have ΔTmin Violations

 Use Path H – IV – II – I – Cb (short form)

Question Session

T. Gundersen

Q - 33

slide38

mCp

50

60

100

40

II

Pinch

α

IV

180°C

130°C

90°C

130°C

III

H1

β

130°C

I

150°C

120°C

141.67°C

H2

Cb

1700

γ

120°C

135°C

C1

100

δ

60°C

1400

160°C

110°C

H

C2

147.5ºC

900

1100

2000

Process, Energy and System

With y = 400 kW, driving forces are restored for heat exchangers III and IV, but the problems have increased for heat exchanger II. Thus, Cacan not be removed !!

Question Session

T. Gundersen

Q - 34

slide39

II

Pinch

α

IV

180°C

140°C

92ºC

90°C

140°C

III

Ca

H1

β

140°C

100 - x

I

150°C

120°C

140°C

H2

Cb

1200 + x + y

128.25ºC

γ

120°C

135°C

C1

600 – x - y

δ

153.0ºC

60°C

900 + x + y

160°C

120°C

H

C2

147.5ºC

500 + y

1100 - y

2400

mCp

50

60

100

40

Process, Energy and System

The 2nd smallest Unit H (500 kW) is not part of a Loop

The 3rd smallest Unit I is part of a Loop but has to be removed by a Path and thus y = x = 600 kW

Question Session

T. Gundersen

Q - 35

slide40

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Optimization of HENs:
      • Q:“When manipulating a Heat Load Path by adding or subtracting a certain duty, how do we know which temperatures to focus on to assure sufficient driving forces?”
      • A: Not easy to provide a simple answer here. Focus has to be on the heat exchanger(s) with driving force problems, and then to identify a heat load path that will improve these driving forces. Two situations: An easy one where one temperature is fixed and a more difficult one where both temperatures change.

Process, Energy and System

Question Session

T. Gundersen

Q - 36

slide41

Remember our 3 Stream Problem (WS-1)

mCp

(kW/°C)

1.5

5.0

4.0

III

I

300°C

100°C

200  186.7  195ºC

H1

C

130 + y

II

200°C

100°C

H2

Process, Energy and System

50°C

217.5°C

180  175°C

250°C

55°C

C1

H

130 + y

150  170 - y

20  0

500

The ”easy” case when one temperature (175°C) is constant

More complicated: Several Units with ΔT problems

and cases where both temperatures for a unit change

Question Session

T. Gundersen

Q - 37

slide42

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Forbidden Matches in HENs:
      • Q:“In Exam June 2010, Task 1.d a forbidden match is introduced for H2 and C1. Why is not Pinch and mCp rules relevant here, and (general) when to use Pinch?”
      • A: A forbidden match results in increased energy use and thus heat transfer across Pinch. Thus, Pinch decom-position and mCp rules cannot be used, and the match information comes from the extended heat cascade.
      • A: The Pinch Concept is only used for MER design (Grassroots), XP Analysis (Retrofit) and Correct Integr. of dist. columns, evaporators, heat pumps/engines.

Process, Energy and System

Question Session

T. Gundersen

Q - 38

slide43

H2-C1 is

Forbidden

Match

Process, Energy and System

Exam 2010, Task 1.d

See Calculations

on the Blackboard

Question Session

T. Gundersen

Q - 39

slide44

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Forbidden Matches in HENs:
      • Q:“In Exam June 2010, Task 1.d the forbidden match (H2-C1) results in extra steam and cooling water use of 800 kW each. How should this be referred to when the question asks about additional energy consumption?”
      • A: Please do not say “1600 kW”, since the two utility types are very different and with very different cost !!
      • Q: ”In the same task, it is stated (in the solution) that one advantage is larger driving forces in heat exchangers. How is this explained, and is it visible?”
      • A:See next Slide !!

Process, Energy and System

Question Session

T. Gundersen

Q - 40

slide45

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Forbidden Matches in HENs:

Process, Energy and System

Question Session

T. Gundersen

Q - 41

slide46

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Questions related to Forbidden Matches in HENs:
      • Q:“In the cases of heat recovery problems with Forbidden Matches, when can the simplified Heat Cascade based only on Supply Temperatures be used?”
      • A: Since we use the Heat Cascade to argue about how we can reach maximum heat recovery when there are forbidden matches, I strongly recommend the use of the complete heat cascade. This will make the reasoning easier and we do not have to worry about streams ending in the middle if temperature intervals !!

Process, Energy and System

Question Session

T. Gundersen

Q - 42

slide47

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Procedure for Retrofit Design:
      • Q:”A more clear “recipe” for how to propose changes to an existing network. What is important? Removal of small units? Reduce area? Reduce heating/cooling? Etc.”
      • A:The Objective of retrofit is to save energy, thus one should look for changes (new units, additional area, repiping, resequencing, etc., that will reduce external heating and cooling.
      • A: Unfortunately, retrofit requires creativity, and it is impossible to provide a water proof procedure.

Process, Energy and System

Question Session

T. Gundersen

Q - 43

slide48

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • “Procedure” for Retrofit Design (1):
      • Identify causes for excess energy consumption  find all occurrences of Cross Pinch heat transfer (3 comps.)
      • Look for Shifting opportunities (changes in operating temperatures for heat exchangers) that will eliminate or reduce cross pinch heat transfer. Address the largest cross pinch violations first.
      • Sometimes Repiping is a good option; look for cases where the driving forces are not properly utilized.
      • Add New Units that will reduce Energy consumption

Process, Energy and System

Question Session

T. Gundersen

Q - 44

slide49

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • “Procedure” for Retrofit Design (2):
      • The above mentioned new units are introduced after shifting that has opened up for unused heating/cooling.
      • Calculate new Duties and Temperatures in the network.
      • Perform a UA-analysis to evaluate the re-use of existing units and the need for additional area in existing units.
      • Use heat load Loops and Paths in a way that maximizes the utilization of existing units, minimizes investments (area of new units and/or additional area to existing units) and (if possible) maximizes energy savings.

Process, Energy and System

Question Session

T. Gundersen

Q - 45

slide50

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Procedure for Retrofit Design:
      • Q:”I find it difficult to know where to start and also when to stop. In the solution to previous exams, several different projects are presented. The exam in 2009 is one example, and I noticed that Design B2 violates ΔTmin.”
      • A:For the start, see previous slides. For the end, please focus on demonstrating understanding and ability to use the tools and procedures. Repeating for several alternative cases does not give much more score, but these alternatives could be briefly mentioned !! After XP analysis, we “forget” about ΔTmin !!

Process, Energy and System

Question Session

T. Gundersen

Q - 46

slide51

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Procedure for Retrofit Design:
      • Q:”How to identify the placement of heat exchangers above, across and below Pinch in Retrofit cases, such as Assignment 5?”
      • A:The Grid Diagram is used, and the key guideline for this “placement” of heat exchangers is the actual values of hot inlet and outlet temperatures and cold inlet and outlet temperatures relative to the hot and cold Pinch temperatures. For details, see the next Slide (from the solution to Assignment 5).

Process, Energy and System

Question Session

T. Gundersen

Q - 47

slide52

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Procedure for Retrofit Design:

Process, Energy and System

Question Session

T. Gundersen

Q - 48

slide53

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Procedure for Retrofit Design:

Process, Energy and System

Question Session

T. Gundersen

Q - 49

slide54

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Procedure for Retrofit Design:

Process, Energy and System

Question Session

T. Gundersen

Q - 50

slide55

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • UA Analysis in Retrofit Design:
      • Q:”In the Solution to Exam 2009, a path is used to make UAnew = UAexist for a heater that is moved from C1 to C2. We have noticed that ΔT < ΔTmin in the cold end of exchanger 1. Is this an error in the solution, or?”
      • A:In retrofit design, the specification of ΔTmin or HRAT is only used to identify cross-Pinch heat transfer. In the design phase, focus is on minimizing energy con-sumption and maximizing utilization of existing units. Loops and Paths are used for this. See next Slide !!

Process, Energy and System

Question Session

T. Gundersen

Q - 51

slide56

Questions before Exam

Process Integration

TEP 4215

R

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H

U

  • Heat Recovery System (H)
    • UA Analysis in Retrofit Design:

Process, Energy and System

- y

- y

+ y

Question Session

T. Gundersen

Q - 52

slide57

Questions before Exam

Process Integration

TEP 4215

R

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H

U

  • Heat Recovery System (H)
    • Cross Pinch Heat Transfer and “Heat Pumping”:

Process, Energy and System

Exam 2008, Task 1.c

See Calculations

on the Blackboard

Question Session

T. Gundersen

Q - 53

slide58

H1

C

C2

Questions before Exam

Process Integration

TEP 4215

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H

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110°

mCp

(kW/°C)

300°

50°

105°

I

[20]

1100kW

Process, Energy and System

50°

180°

I

[30]

3900 kW

90°

QXP = 20(300-110) - 30(180-90) = 3800 – 2700 = 1100 kW

QXP = 30(90-50) - 20(110-105) = 1200 – 100 = 1100 kW

More details on the Blackboard !!

Question Session

T. Gundersen

Q - 54

slide59

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Retrofit Design – General Question:
      • Q:“In a retrofit project, do we normally add a new unit, or can it be sufficient to “shift” a heat exchanger away from its cross Pinch situation?”
      • A:It is extremely seldom that only repiping will improve heat recovery, but it happens …. When “shifting” a heat exchanger, the purpose is to reduce Cross Pinch heat transfer, and this will result in the release of heating resources above or release of cooling resources below Pinch. To take advantage of this, new units (one or more) are needed.

Process, Energy and System

Question Session

T. Gundersen

Q - 55

slide60

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • General Question for Grassroot and Retrofit:
      • Q:”When do we use HRAT?”
      • A:In the early age of Pinch Analysis, people working closely with industry (Oil Refineries) noticed that there was a large difference between the ΔTmin corresponding to the level of heat recovery and the smallest ΔT in the individual heat exchangers. As a result, they introduced HRAT (Heat Recovery Approach Temperature) and EMAT (Exchanger Minimum Approach Temperature). Today, HRAT is most often used in Retrofit as a target value indicating level of ambitions in energy savings.

Process, Energy and System

Question Session

T. Gundersen

Q - 56

slide61

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Heat Recovery System (H)
    • Correct Integration:
      • Q: “Assignment 8, at the end of the proposed solution, it is stated that without integration of the distillation column and use of heat pump, QH = 1200 kW and QC = 1300 kW. Can you explain this in more detail?”
      • Q:This is simply because the distillation column has to be accounted for, not only the “background process”. The external heating and cooling requirements of the process is (see next slide) 800 and 900 kW, and then the distillation column (if not integrated) requires 400 kW.

Process, Energy and System

Question Session

T. Gundersen

Q - 57

slide62

Questions before Exam

Process Integration

TEP 4215

R

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H

U

  • Heat Recovery System (H)
    • Correct Integration:

Process, Energy and System

Question Session

T. Gundersen

Q - 58

slide63

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Interface between Utility System and Heat Recovery System (H/U)
    • Correct Integration of Turbines:
      • Q: “Could you repeat the topic about turbines (back pressure and condensing), both about the types and how these should be integrated?”
      • Q:The general “class” of equipment is referred to as Heat Engines (opposite of Heat Pump). For the sake of simplicity, the discussion here focuses (as in the question) on Steam Turbines, where there are 3 types; Condensing, Back Pressure and Extracting (the last one is simply a combination of the two first). See next Slides !!

Process, Energy and System

Question Session

T. Gundersen

Q - 59

slide64

Questions before Exam

Process Integration

TEP 4215

R

S

H

U

  • Interface between Utility System and Heat Recovery System (H/U)
    • Correct Integration of Turbines:

HP

HP

HP

Process, Energy and System

CW

LP

LP

CW

Back Pressure

Condensing

Extracting

Question Session

T. Gundersen

Q - 60

slide65

Questions before Exam

Process Integration

TEP 4215

R

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H

U

  • Utility System (U)
    • No questions so far
      • Any Questions now?
    • Relevance for the Exam?
      • Lecturer to provide some wise words…..

Process, Energy and System

Question Session

T. Gundersen

Q - 61

slide66

Questions before Exam

Process Integration

TEP 4215

R

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H

U

  • Other Topics
    • Mathematical Programming
      • Q:“What should we know about this topic? As an example, should we be able to answer Task 3.d from 2008 (What kind of Math Programming models do we have for minimum energy cost w/wo forbidden matches and for minimum number of units)?”
      • A:This topic has been considerably reduced in the course (lectures & assignments) over the years, and is now a demonstration topic only. Forbidden matches is dealt with using logic in the Heat Cascade

Process, Energy and System

Question Session

T. Gundersen

Q - 62

slide67

Questions before Exam

Process Integration

TEP 4215

R

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H

U

  • Other Topics
    • Weights for the different Exam Tasks
      • Q:“Are all sub-tasks given the same weight when grading the Exam?”
      • A:No, these weights can actually change during the grading process, either to avoid negative effects (for the students) or after discussions between the internal and external examiners. Basically, these weights reflect a combination of work load and difficulty. Also notice that work load is not counting pages in the proposed exam solution !!

Process, Energy and System

Question Session

T. Gundersen

Q - 63