truls gundersen

Department of Energy and Process Engineering

Truls Gundersen 10.01.10

TEP 4215 - Energy Utilization and Process Integration in Industrial Plants, or for short: “Energy and Process”

The Objective is to convey Systems Thinking and Systematic Methods for:

Analysis and Design (and partly Operation) of Processes and Utility Systems, with focus on Efficient Use of Energy while considering Economy, Operation and (to some extent) Environment

Requirements to be able to join the Course None (meaning previous courses), but it is an advantage

to have some basic knowledge about the following: heat exchangers, distillation columns, evaporators turbines, heat pumps and “simple” thermodynamics

Fall 98: 100 students from 8 departments in 4 faculties !! From Spring 2009: Compulsory for the “PuP” Program


The Course Content is primarily System based Strategy for Design of integrated

Process Plants with corresponding Utility Systems Systematic Methods for Analysis and Design of

Reactor Systems (very limited and not in depth) Thermally driven Separation Systems, such as (primarily)

Distillation and (to a much less extent) Evaporation Heat Exchanger Networks and Correct Heat Integration Utility Systems (heating, cooling and power)

The Thermodynamically based Pinch Analysis Brief Introduction to the use of Optimization Environmental Issues related to Energy Usage New Design and Retrofit of Existing Plants

TEP 4215 - Energy and Process



The Curriculum for the Course is: R. Smith: “Chemical Process Design and Integration”, 2nd ed.,

John Wiley & Sons, January 2005. Alternative Text Book:I.C. Kemp: “Pinch Analysis and Process Integration”, Elsevier, Butterworth Heinemann, December 2006.

T. Gundersen: “Basic Concepts for Heat Recovery in RetrofitDesign of Continuous Processes”, Ch. 6 in “A Process Integration Primer”, IEA, 2000 (18 pages).

Lectures and Assignments. Assignments are Examination oriented (most are previous Ex Q’s) Examination will test Understanding through Calculation Examples.

This requires Training established by working with Assignments.

• Home Page: http://www.ivt.ntnu.no/ept/fag/tep4215/

TEP 4215 - Energy and Process



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 4 Optimization of Heat Exchanger Networks 12.02 26.02 5 Retrofit Design of Heat Exchanger Networks 26.02 05.03 6 Indirect Integration of Plants using Steam 05.03 12.03 7 Integration of Distillation Columns 12.03 09.04 8 Optimal Use of Heat Pump 09.04 16.04 9 Area in Heat Exchanger Networks 16.04 23.04 10 Heat Integration and Forbidden Matches 23.04 30.04 11 Design of Heat Exchanger Networks (2) 30.04 none

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

TEP 4215 - Plan for Assignments with Guidance


Questions before ExamProcess Integration

TEP 4215

T. Gundersen

• 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 !!

Q - 01

Process, Energy and System

Question Session

R S H U


TEP 4215

T. Gundersen

• 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

Q - 02


Question Session

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R S H U


TEP 4215

• Reactor Separator Interface (R/S) No questions so far


Lecturer to provide some wise words…..

Q - 03

Question Session

R SFF

RR

P

PX

RXRF

BP

yR = XS

yP = XS(1+ω)

T. Gundersen


R S H U


TEP 4215

• Separation System (S) No questions so far


Lecturer to provide some wise words….. Or: See next Slide !!

Q - 04

Question Session

T. Gundersen


R S H U


TEP 4215

• 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.

Q - 05

Question Session

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TEP 4215

• 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

Q - 06

Question Session

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TEP 4215

• 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”)

Q - 07

Question Session

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TEP 4215

• Interface Separation System (S) and Heat Recovery System (H) No questions so far


Lecturer to provide some wise words…

Q - 08

Question Session

T. Gundersen

• 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)



TEP 4215

Q - 09

Question Session

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T. Gundersen

• 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 !!



TEP 4215

Q - 10

Question Session

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• 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)



TEP 4215

Q - 11

Question Session

H1

H2

C1

C2

120°C

170°C

120°C 30°C

60°C

80°C

60°C200°C 120°C

100°C

140°C

120°C

H3

140°C 140°C

R S H U

T. Gundersen


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.



TEP 4215

Q - 12

Question Session

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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



TEP 4215

Q - 13

Question Session

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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.



TEP 4215

Q - 14

Question Session

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TEP 4215

Q - 15

Question Session

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Pinch180°

C2210° 160°

C1210° 50°

H2220° 60°

H1270° 160°

160°

Ca

2

2

H

1

1

1000 kW

2500 kW

Cb

980 kW

1320 kW

2200 kW

160°

214.4°

120°

mCp(kW/°C)

18.0

22.0

20.0

50.0

T. Gundersen

• 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 !!



TEP 4215

Q - 16

Question Session

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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 !!



TEP 4215

Q - 17

Question Session

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TEP 4215

Q - 18

Question Session

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mCp

20

80

30

40

H1

H2

C1

C2

160°C

250°C

220°C 60°C

80°C

50°C300°C160°C

50°C

Pinch

150°C

150°C

2800 kW

H3000 kW

CII

I

III

IV

3000 kW

3400 kW 200 kW

2000 kW

85.63C

65 C

72.5 C

150 C

T. Gundersen


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 !!



TEP 4215

Q - 19

Question Session

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Q - 20

Question Session

H1

H2

C1

C2

180°C

140°C

180°C 50°C

90°C

50°C

60°C150°C 100°C

90°C

mCp

50

30

90

20

CaIV

II

I

III

H

2500

400

110°C

1400

α

β2000

166.7°C

90°C

800

84°C

1200

134.4°C

100°C

149.3°C

100°C

Cb

1500

Network from the Solution to the Exam 2010

An “MER” Design meeting Targets for Energy and Units

T. Gundersen


Q - 21

Question Session

Network from one of the Students

Violating mCp Rules for Pinch Exchangers

II

H1

H2

C1

C2

180°C

140°C

180°C 50°C

90°C

50°C

60°C150°C 100°C

90°C

mCp

50

30

90

20

Ca

IVIII

H

2500

400

110°C

1400

200090°C

800

84°C

2000100°C

100°C

Cb

700117.8°C

II

T. Gundersen


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.



TEP 4215

Q - 22

Question Session

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T. Gundersen

• 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.



TEP 4215

Q - 23

Question Session

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T. Gundersen




TEP 4215

Q - 24

Question Session

R S H U

Nice, but not relevant for the Exam !!

T. Gundersen


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 !!



TEP 4215

Q - 25

Question Session

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T. Gundersen




TEP 4215

Q - 26

Question Session

R S H U

Nice, but not relevant for the Exam !!

T. Gundersen

• 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 Pinch point(s) !!



TEP 4215

Q - 27

Question Session

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T. Gundersen


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.



TEP 4215

Q - 28

Question Session

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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.



TEP 4215

Q - 29

Question Session

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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.



TEP 4215

Q - 30

Question Session

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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 !!



TEP 4215

Q - 31

Question Session

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Q - 32

Question Session

H1

H2

C1

C2

150°C

135°C

160°C 60°C

120°C

90°C180°C140°C

120°C

Pinch

140°C

120°CH

Caα

β

γ

δ

I

147.5ºC

IV 92ºC

500

III

1100

II

900 + x

600 - x

153.0ºC

128.25ºC

2400

100 - x

140°C

140°C

Cb1200 + x

The smallest Unit Ca (100 kW) is part of a Loop:H1 – (II) – C1 – (I) – H2 – (Cb) – CW – (Ca) – H1

mCp

50

60

100

40

T. Gundersen


Q - 33

Question Session

mCp

50

60

100

40

H1

H2

C1

C2

150°C

135°C

160°C 60°C

120°C

90°C180°C138°C

120°C

Pinch

141.67°C

120°CH

α

β

γ

δ

I

147.5ºC

IV

500 + y

III

1100

II

1000 + y

500 - y

Tδ

Tγ

2400 - y

138°C

138°C

Cb1300 + y

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)

T. Gundersen


Q - 34

Question Session

mCp

50

60

100

40

H1

H2

C1

C2

150°C

135°C

160°C 60°C

120°C

90°C180°C130°C

120°C

Pinch

141.67°C

110°CH

α

β

γ

δ

I

147.5ºC

IV

900

III

1100

II

1400

100

Tδ

Tγ

2000

130°C

130°C

Cb1700

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

T. Gundersen


Q - 35

Question Session

H1

H2

C1

C2

150°C

135°C

160°C 60°C

120°C

90°C180°C140°C

120°C

Pinch

140°C

120°CH

Caα

β

γ

δ

I

147.5ºC

IV 92ºC

500 + y

III

1100 - y

II

900 + x + y

600 – x - y

153.0ºC

128.25ºC

2400

100 - x

140°C

140°C

Cb1200 + x + y

The 2nd smallest Unit H (500 kW) is not part of a LoopThe 3rd smallest Unit I is part of a Loop but has to be

removed by a Path and thus y = x = 600 kW

mCp

50

60

100

40

T. Gundersen


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.



TEP 4215

Q - 36

Question Session

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Q - 37

Question Session

mCp(kW/°C)

1.5

5.0

4.0

H1

H2

C1

200°C

250°C 50°C

100°C

100°C300°C

180 175°C

CI III

II

200 186.7 195ºC

150 170 - y 20 0500

130 + y

H130 + y

217.5°C55°C

The ”easy” case when one temperature (175°C) is constantMore complicated: Several Units with ΔT problems

and cases where both temperatures for a unit change

Remember our 3 Stream Problem (WS-1)

T. Gundersen

• 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.



TEP 4215

Q - 38

Question Session

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Q - 39

Question Session

100°C 90°C

+ 300

1500

+ 2400

+ 300

CW

ST

C1

C2

H1

H2

900 kW

190°C 180°C

180°C 170°C

150°C 140°C

60°C 50°C

50°C 40°C

1500 kW

2500 kW

2000 kW

1200 kW

600 kW

1000 kW

4500 kW

800 kW

200

QH

R1

R4

QC

R2

R3

300 kW

200 kW

100°C 90°C

+ 300

1500

+ 2400

+ 300

CW

ST

C1C1

C2C2

H1H1

H2H2

900 kW

190°C 180°C

180°C 170°C

150°C 140°C

60°C 50°C

50°C 40°C

1500 kW

2500 kW

2000 kW

1200 kW

600 kW

1000 kW

4500 kW

800 kW

200

QH

R1

R4

QC

R2

R3

300 kW

200 kW

Exam 2010, Task 1.dSee Calculations

on the Blackboard

H2-C1 isForbidden

Match

T. Gundersen


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 !!



TEP 4215

Q - 40

Question Session

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TEP 4215

Q - 41

Question Session

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H1

H2

C1

C2

180°C

140°C

180°C 50°C

90°C

50°C

60°C150°CmCp

50

30

90

20

CaI

II

Hb

2000

2400

170°C

200

2500

2000

100°C

100°C

Cb

1500

Ha117.8°C

H1

H2H2

C1

C2

180°C

140°C

180°C 50°C

90°C

50°C

60°C150°CmCp

50

30

90

20

CaCaII

II

HbHb

2000

2400

170°C

200

2500

2000

100°C

100°C

CbCb

1500

HaHa117.8°C

T. Gundersen


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 !!



TEP 4215

Q - 42

Question Session

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T. Gundersen

• 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.


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TEP 4215

Q - 43

Question Session

T. Gundersen

• 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


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TEP 4215

Q - 44

Question Session

T. Gundersen

• 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.


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TEP 4215

Q - 45

Question Session

T. Gundersen


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 !!


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TEP 4215

Q - 46

Question Session

T. Gundersen


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).


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TEP 4215

Q - 47

Question Session

T. Gundersen



R S H U


TEP 4215

Q - 48

Question Session

H1

H2 4

C11

175°

98°125°

155°

112°

85°

40°

20°

65°

45°

mCp(kW/°C)

[10]

[40]

[20]

[15]2 C2

2

3

3

125°

105°

T. Gundersen



R S H U


TEP 4215

Q - 49

Question Session

H1

H2 4

C11

175°

98°125°

155°

112°

85°

40°

20°

65°

45°

mCp(kW/°C)

[10]

[40]

[20]

[15]2 C2

2

3

3

125°

105°

T. Gundersen



R S H U


TEP 4215

Q - 50

Question Session

H1

H2 4

C11

175°

98°

155°

40°

20°

65°

45°

mCp(kW/°C)

[10]

[40]

[20]

3

3

A

A

BT1

T4T3

T2

1400 - Y 0 + X

1320 - X

1300 - X +Y

0 + X - Y

T. Gundersen

• 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 !!


R S H U


TEP 4215

Q - 51

Question Session

T. Gundersen

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


R S H U


TEP 4215

Q - 52

Question Session

H1

H2

C

C1II

300°

110°

140°

180° 50°

40°

80°

50°

mCp(kW/°C)

[20]

[90]

[50]

[30]I C2

II

I

H

94°

70°

1600 kW

400kW

2700 kW

2300 kW

III

III

2300 kW126.7°

185°

- y

- y

+ y

T. Gundersen

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


R S H U


TEP 4215

Q - 53

Question Session

H1

H2 Cb

C1

180°

77.5°200°

105°

130°

121.67°

60°

40°

70°

50°

mCp(kW/°C)

[60]

[40]

[20]

[50]I C3

II

I

III

110°

100°

190°100° [80]II C2

Ca

H

III

145°

110°

3500

36003600

1300

4300

300

H1H1

H2H2 Cb

C1C1

180°

77.5°200°

105°

130°

121.67°

60°

40°

70°

50°

mCp(kW/°C)

[60]

[40]

[20]

[50]II C3C3

IIII

II

IIIIII

110°

100°

190°100° [80]IIII C2C2

CaCa

HH

IIIIII

145°

110°

3500

36003600

1300

4300

300Exam 2008, Task 1.c

See Calculationson the Blackboard

T. Gundersen


R S H U


TEP 4215

Q - 54

Question Session

H1 C

300°

180° 50°

50°

mCp(kW/°C)

[20]

[30]I C2

I

110°

90°

105°

1100kW

3900 kW

QXP = 20(300-110) - 30(180-90) = 3800 – 2700 = 1100 kWQXP = 30(90-50) - 20(110-105) = 1200 – 100 = 1100 kW

More details on the Blackboard !!

T. Gundersen

• 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.


R S H UQuestions before Exam

Process IntegrationTEP 4215

Q - 55

Question Session

T. Gundersen

• 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.


R S H U


TEP 4215

Q - 56

Question Session

T. Gundersen

• 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.


R S H U


TEP 4215

Q - 57

Question Session

T. Gundersen

• Heat Recovery System (H) Correct Integration:


R S H U


TEP 4215

Q - 58

Question Session

0

20

40

60

80

100

120

140

160

180

0 200 400 600 800 1000 1200 1400

Q (kW)

T' (°C)

DistillationColumn

QR = 400 kW

QP = 400 kW

QC = 900 kW

0

20

40

60

80

100

120

140

160

180

0 200 400 600 800 1000 1200 1400

Q (kW)

T' (°C)

DistillationColumn

QR = 400 kW

QP = 400 kW

QC = 900 kW

T. Gundersen

• 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 !!



TEP 4215

Q - 59

Question Session

R S H U

T. Gundersen

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



TEP 4215

Q - 60

Question Session

HP

LP

HP

CW

HP

CWLP

Back Pressure Condensing Extracting

R S H U

T. Gundersen


Q - 61

Question Session

R S H U


TEP 4215

• Utility System (U) No questions so far


Lecturer to provide some wise words…..

T. Gundersen

• 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


R S H U


TEP 4215

Q - 62

Question Session

T. Gundersen

• 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 !!


R S H U


TEP 4215

Q - 63

Question Session

truls gundersen

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