TEP 4215 - Energy and Process The Course Content is primarily - - PDF document

Department of Energy and Process Engineering

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

 The Objective is to convey  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

Truls Gundersen 17.01.09

q j

 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 !!

Department of Energy and Process Engineering

 The Course Content is primarily

S b d St t f D i f i d

TEP 4215 - Energy and Process

 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 Utilit S t (h ti li d )

Truls Gundersen 17.01.09

 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

Department of Energy and Process Engineering

 The Curriculum for the Course is:

TEP 4215 - Energy and Process

 R. Smith: "Chemical Process Design and Integration", 2nd ed.,

John Wiley & Sons, January 2005.

 T. Gundersen: “Basic Concepts for Heat Recovery in Retrofit

Design of Continuous Processes”, Ch. 6 in “A Process Integration Primer”, IEA, 2000 (18 pages).

 Lectures and Assignments.  The Assignments are extremely Examination oriented (most of

th t ll i E i ti Q ti )

Truls Gundersen 17.01.09

them are actually previous Examination Questions).

 The Examination will test Understanding through Calculation

• Examples. This requires Insight and Training/Experience that can
• nly be established through working with the Assignments.
• Home Page: http://www.ivt.ntnu.no/ept/fag/tep4215/

Department of Energy and Process Engineering

Ass. Topic Supervised Deadline

TEP 4215 - Plan for Assignments with Guidance

1 Sequence of Distillation Columns (???) 27.01 03.02 2 Minimum Energy Requirements and Pinch 03.02 10.02 3 Design of Heat Exchanger Networks (1) 10.02 17.02 4 Optimization of Heat Exchanger Networks 17.02 03.03 5 Retrofit Design of Heat Exchanger Networks 03.03 10.03 6 Indirect Integration of Plants using Steam 10.03 17.03 7 Integration of Distillation Columns 17.03 24.03

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8 Optimal Use of Heat Pump 24.03 21.04 9 Area in Heat Exchanger Networks (???) 21.04 28.04 10 Heat Integration and Forbidden Matches (???) 28.04 05.05 11 Design of Heat Exchanger Networks (2) (???) 05.05 none

Guidance: One PhD Student, 4 Student Assistants and the Lecturer

Questions before Exam Process Integration TEP 4215

Proce

R S H U

• Reactor System (R)
• No questions so far

 Any Questions now?

• Relevance for the Exam?

 Lecturer to provide some wise words…..

ess, Energy and System

• T. Gundersen

Q - 01

Question Session Proce

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Questions before Exam Process Integration TEP 4215

ess, Energy and System

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

 Any Questions now?

• Relevance for the Exam?

 Lecturer to provide some wise words…..

• T. Gundersen

Q - 02

Question Session

Proce

R S H U

Questions before Exam Process Integration TEP 4215

ess, Energy and System

• Separation System (S)
• 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 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

• T. Gundersen

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 - 03

Question Session Proce

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Questions before Exam Process Integration TEP 4215

ess, Energy and System

• Separation System (S)
• 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 S S ft P k T 1C f

• T. Gundersen

 A: Some Software Packages use T=1C for condensing and vaporizing streams (results in very large mCp values, but not “infinity”)

Q - 04

Question Session

Proce

R S H U

Questions before Exam Process Integration TEP 4215

• Heat Recovery System (H)
• Stream Splitting (Assignment 4, Task 1):

 Q: A bit uncertain about how to find optimal split ratio, and how the economic parameters change based on this? (not sure I understood the last part)  A: Good starting value and parameters that affect the “optimal” split ratio (use the Blackboard?).

ess, Energy and System

• T. Gundersen

Remember the Cost Equation!  Also: See next Slide  And: Splits are often removed during Optimization

Q - 05

Question Session Proce

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Questions before Exam Process Integration TEP 4215

Q mCp

ess, Energy and System

H1 H2 C1

100°C 40°C 70°C 50°C

Q (kW) 1000 2700 2000

100°C

mCp (kW/°C) 20 90 40

Pinch

I II III I C

 

T T

500 kW 75°C

• T. Gundersen

Q - 06

Question Session

C2

50°C

1200

90°C

30

II III 2000 kW 700 kW 500 kW 66.67°C

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Network Optimization and Alternatives:

 Q: The proposed solutions to previous Exams present several alternatives for network optimization; how many solutions are expected during the Exam?  A: Solutions include many alternatives to make the grading easier and to show the 2nd Examiner (and later year students) that many alternatives exist.  A: With l 4 h th b t St t i t ti

ess, Energy and System

• T. Gundersen

 A: With only 4 hours, the best Strategy is to mention where alternatives exist and argue why one specific is chosen. Most important is the documentation of mastering the methodology (Loops and Paths).

Q - 07

Question Session

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Pinch Design Method and Stream Splitting

(Exam 10 May 2003, Task 1.c):

 Q: I do not understand why C1 has to be split above Pinch? Both T, mCp and n rules are satisfied even if C1 is not split and matched with H1 & H2 directly.  A: This is the very heart of the Pinch Design Method and relates to the notion of Pinch

ess, Energy and System

• T. Gundersen
• Exchangers. Use Blackboard to explain.

 Also: See the next slides where the Stream Grid and the resulting design are shown.

Q - 08

Question Session

Proce

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Questions before Exam Process Integration TEP 4215

Q Q mCp

ess, Energy and System

H1 H2 C1

180°C 140°C 90°C 50°C 60°C 150°C

Q 2500 2400 4500 Q 2000 1500 —

100°C

mCp 50 30 90

• T. Gundersen

Q - 09

Question Session

C2

180°C 50°C

1800 800

90°C

20

Proce

R S H U

Questions before Exam Process Integration TEP 4215

150°C 100°C

mCp

ess, Energy and System

H1 H2 C1

180°C 140°C 90°C 50°C 60°C 150 C

50 30 90 20

Ca IV II I III 2500  

166.7°C 84°C

1200

134.4°C 100°C 149 3°C 100°C

Cb 1500

• T. Gundersen

Q - 10

Question Session

C2

180°C 50°C 90°C

20

H 400

110°C

1400 2000

90°C

800

149.3 C

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• More on the PDM and Stream Splitting (Exam

2 June 2007, Task 1.c):

 Q: In this case, two streams (H1 and C1) have to be split in two branches above Pinch. Why are the values

• f the branch flowrates ( and ) for H1 fixed when

the duties of the exchangers are given? Could we not have different temperatures for the two branches? After mixing the branches the hot Pinch temperature

ess, Energy and System

• T. Gundersen

After mixing the branches, the hot Pinch temperature is obtained! What about XP heat transfer?  A: It is the branch temperatures, not the mixing temperature that is important here, see next Slide!

Q - 11

Question Session Proce

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Questions before Exam Process Integration TEP 4215

mCp Pinch

II

mCp Pinch

II II

ess, Energy and System

mCp 50 60 100 H1 H2 C1

150°C 135°C 120°C 90°C 180°C 140°C 120°C

Pinch

140°C

Ca

   I IV

92ºC

III

600 128.25ºC 100 140°C 140°C

Cb

1200

mCp 50 60 100 H1 H2 C1

150°C 135°C 120°C 90°C 180°C 140°C 120°C

Pinch

140°C

Ca

   I IV

92ºC

III

600 128.25ºC 100 140°C 140°C

Cb

1200

• T. Gundersen

Q - 12

Question Session

40 C2

160°C 60°C 120°C

H 

147.5ºC 500 1100 900 153.0ºC 2400

40 C2

160°C 60°C 120°C

H 

147.5ºC 500 1100 900 900 153.0ºC 2400

• Heat Recovery System (H)

Proce

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Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Even more on the PDM and Stream Splitting

(Exam 2 June 2007, Task 1.c):

 Q: A good starting value for the split flowrates for stream C1 is (see previous Slide):

ess, Energy and System

2 H

mCp    

• T. Gundersen

 Why is this a good starting value (logic)?  A: This is an attempt to have near-optimal distribution of driving forces (see next Slide)

Q - 13

Question Session Proce

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Questions before Exam Process Integration TEP 4215

mCp

200°C

ess, Energy and System

H2 C1 C2

200°C 156.7°C 170°C 180°C

(kW/°C) 100 50 30

I II I II

 

T C

1000 kW 200 C

T C

• T. Gundersen

Q - 14

Question Session

190°C 1000 kW 1 2

62.5 , 37.5

C C

mCp mCp        

Proce

R S H U

Questions before Exam Process Integration TEP 4215

mCp

200°C

ess, Energy and System

H2 C1 C2

200°C 170°C 170°C 180°C

(kW/°C) 100 50 30

I II I II

 

T C

1000 kW 200 C

T C

• T. Gundersen

Q - 15

Question Session

190°C 600 kW 1 2

10 , 14

C I hot cold II C

mCp Q T T Q mCp      

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Retrofit Design and Cross Pinch Heat Transfer:

 Q: How should the formula for Cross Pinch Heat Transfer presented in the Lectures (Slide Heat 59) be used for various situations?

ess, Energy and System

, , , , XP H H in P H C C out P C

Q mCp T T mCp T T                = 0 

• T. Gundersen

 A: Unfortunately, this formula is not general and was developed to deal with the most (?) complicated case (see next Slide)

Q - 16

Question Session

• Heat Recovery System (H)

Proce

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Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Retrofit Design and Cross Pinch Heat Transfer:

ess, Energy and System

• T. Gundersen

Q - 17

Question Session

(a) (b) (c) (d) (e) (f) (g) The Equation only applies to Case (d), but can be expanded …?? Better: Use your “top floor”…!!

• Heat Recovery System (H)

Proce

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Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Log Mean Temperature Difference (LMTD):

 Q: Definition, use and background?  A Th E ti i d f t t h t

ess, Energy and System

1 2 1 2

LMTD ln

LM

T T T T T        

• T. Gundersen

 A: The Equation is used for counter-current heat exchangers (can also be used for co-current units), and is a result from solving ordinary differential equations with constant parameters (see next Slides)

Q - 18

Question Session

Proce

T

TH TC

H H C C

dQ mCp dT mCp dT dQ    

Heat Exchangers have 3 basic Equations:

ess, Energy and System

Q Q dQ ( )

H C

dQ dA U T T   

and 1 1

H C H C H C

dQ dQ dT dT mCp mCp dT dT dQ          

• T. Gundersen

Q - 19

Question Session

1 1 ( ) ( )

H C H C H C H C H C

Q mCp mCp d T T U T T dA mCp mCp                 

Proce

T

TH TC

, , , ,

, ,

1 1 1

H in C out H out C in

T T A H C H C H C T T H in H out H

dT dT U dA T T mCp mCp T T mCp Q

 

             

 

, , , , , , , ,

ln

H in C out H in H out C out C in H out C in

T T T T T T U A T T Q Q Q A                    ess, Energy and System

Q Q dQ

, ,

1

H C out C in C

mCp Q T T mCp Q  

NB: Derivation requires constant parameters: U, mCpH, mCpC    

, , , , , , , ,

. ln

H in C out H out C in H in C out H out C in

T T T T U T T T T               

• T. Gundersen

Q - 20

Question Session

, pH, pC

LMTD

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Log Mean Temperature Difference (LMTD):

 Q: What happens when mCpH = mCpC ?  A Th i li d d i th E ti (

ess, Energy and System

1 2 1 2

LMTD ?? ln

LM

T T T T T          

• T. Gundersen

 A: There is linear dependency in the Equations (see Slide Q-15). This is no problem what so ever for nature, the driving forces are constant throughout the

• exchanger. For close cases, AMTD could be used.

Q - 21

Question Session

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Slope of Composite Curves: Assignment 2,

Task 2.b:

 Q: Why is the slope of the hot Composite Curve given by:

ess, Energy and System

1 arctan

H

mCp          



• T. Gundersen

 A: It should be tan not arctan, and the question gives the Lecturer an opportunity to correct the confusing proposed solution to this Assignment

Q - 22

Question Session

 

Proces

Task 2.b: When Tmin is increased from 10C to 20C, what happens to Utility Consumption and the Pinch point?

T

tg = (T) / Q

100ºC 110ºC

T

tg = (T) / Q

100ºC 110ºC

ss, Energy and System The notation (T) is confusing! The Pinch is fixed at 90C (cold streams) for a large range of Tmin , while hot Pinch temperature of course increases with Tmin. The Solution Text of this Assignment is, however, OK:

Q

90ºC

Q

90ºC

1 tan "slope" T T H mCp T mCp         

 

50 10 60 kW/ C

H

mCp    



• T. Gundersen

Q - 23

Question Session

 Confusing figure, but the answer is Correct !!

, min,2 , min,1

( ) ( ) 20 10 10

H H P C P C

T T T T T C           

 

,min

60 10 600 kW

H H

Q     

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Simplified versus Complete Heat Cascade

 Q: If we do not need to draw the Grand Composite Curve, can we then always exclude the Target Temperatures and only use Supply Temperatures?  A: Yes, if the rest of the Assignment/Task does not contain Questions that require use of the Grand Composite Curve, we can always use the simplified Heat Cascade

ess, Energy and System

• T. Gundersen

Heat Cascade.  Note: We may have to add intervals related to very high cold or very low hot target temperatures!! And: Streams may now end in the “middle” of intervals.

Q - 24

Question Session

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Simplified versus Complete Heat Cascade

 Q: Is it only Supply Temperatures that are potential Pinch Candidates?  A: Yes, and the same applies to the start (or supply temperature) of Stream Segments. This is a bit hard to prove, but is related to the fact that the start of a new stream or stream segment will result in an increased total mCp th s a “con e kink” ill be

ess, Energy and System

• T. Gundersen

increased total mCp, thus a “convex kink” will be

• bserved in the Composite Curve.

 Note: This is the Basis for the Simplified Cascade

Q - 25

Question Session

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Another Question related to Pinch Candidates

 Q: How do we find the “Pinch causing Stream”?  A: The Pinch point is established by using the PTA, the Heat Cascade, or any other method. The Pinch point is then either a “modified” temperature (PTA)

• r the border between two intervals (Heat Cascade)

and given by two “true” temperatures.

ess, Energy and System

• T. Gundersen

The Pinch “generator” is then any hot or cold stream with a Supply Temperature equal to the hot or cold Pinch Temperature (can be more than one stream). There is a “group responsibility” here, though….. !!

Q - 26

Question Session

• Heat Recovery System (H)

Proce

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Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Counting the Number of Loops

 Q: What is the meaning of the notion “Independent Loops”, and what happens if a Dependent Loop is chosen?  A: The most important issue here is to avoid confusion related to the number of Loops and the number of Units that can be removed. It is totally arbitrary hat Loops e refer to as dependent or

ess, Energy and System

• T. Gundersen

arbitrary what Loops we refer to as dependent or independent. An Example was used to Illustrate this during the Lectures (see the next Slide)

Q - 27

Question Session Proce

Number of Units and Independent Loops

Pinch 180° 220° 60° H1 270° 160° Ca

3 2

360

ess, Energy and System

C2 210° 160° C1 210° 50° H2

4 4 H 1 1 3 2

Cb 620 880 2200 1000 1000 440

L = U U = 7 5 = 2

• T. Gundersen

Q - 28

Question Session

1 Loop (A) : 2 Loops (A, B) – Combine (A,B) (Total: 3) 3 Loops (A,B,C): Combine (A,B), (A,C) & (B,C) (Total: 6) L = Unetwork  Umin = 7  5 = 2

• Heat Recovery System (H)

Proce

R S H U

Questions before Exam Process Integration TEP 4215 Heat Recovery System (H)

• Network Optimization and Driving Forces

 Q: When establishing an MER design, the specified Tmin is strictly obeyed. In the Optimization phase, the Pinch decomposition is no longer considered. Does this mean that we no longer need to obey Tmin, only to have T  0?  A: When reducing the number of Units, we should have re-adjusted the trade-off between Energy and

ess, Energy and System

• T. Gundersen

j gy Area (and thus the value of Tmin). To make this a manageable task during Assignments and Exams, we ask for re-establishing Tmin , which is contrary to Retrofit, where UA analysis is more important.

Q - 29

Question Session Proce

R S H U

Questions before Exam Process Integration TEP 4215

• Interface between Utility System and Heat

Recovery System (H/U)

• Integration of Process, Distillation Column and

Heat Pump (Exam June 2007, Task 2)

 Q: Here, different values are used for Tmin (20C for Process and Distillation Column and 10C for the Heat Pump), what is then correct T’ for the Heat Pump?

ess, Energy and System

• T. Gundersen

p), p  A: Unfortunately, there is an error in the proposed Solution, see next Slide and Blackboard discussion about stream individual “contributions” to Tmin .

Q - 30

Question Session

Proce

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Questions before Exam Process Integration TEP 4215

C l C d

ess, Energy and System

Column Condenser: Heat Pump Condenser:

'

0.5 20 210°C

cond cond

T T    

' ' , ' , ,

137.5°C 0.5 20 10 137.5°C 0.5 10 132.5°C

process HP cond process HP cond HP cond

T T T T T          

80 100 120 140 160 180 200 220

T'(ºC)

Dist. Column Condenser Evaporator

80 100 120 140 160 180 200 220

T'(ºC)

Dist. Column Condenser Evaporator

• T. Gundersen

Q - 31

Question Session

20 40 60 500 1000 1500 2000

Q (kW)

20 40 60 500 1000 1500 2000

Q (kW)

• Utility System (U)

Proce

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Questions before Exam Process Integration TEP 4215 Utility System (U)

• Heat Pumps and Heat Engines

 Possible Confusion about the Carnot Equations?

ess, Energy and System

Assume no 1st Law Losses:

Wcycle = QH – QC

Assume no 2nd Law Losses:

• T. Gundersen

Q - 32

Question Session

(QC /QH) = (QC /QH)rev = TC /TH

Carnot Efficiency: max = C = Wcycle / QH = 1 – TC / TH

• Other Topics

Proce

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Questions before Exam Process Integration TEP 4215 p

• Weights for grading the Exam

 Q: Do the subtasks have the same weights, and if not, will these be indicated in the Exam text?  A: The subtasks may have very different weights, and these will not be provided, in fact these are subject to discussion during grading the Exam

• Detailed Topics (related to heat transfer Area)

 Q: Should we be able to derive the Bath Formula?

ess, Energy and System

• T. Gundersen

 Q: Should we be able to derive the Bath Formula?  Q: Will we be asked to count Degrees of Freedom?  A: No, but the physical background is important!

Q - 37

Question Session