Heat and Power Integration CHEN 4460 – Process Synthesis, Simulation and Optimization Dr. Mario Richard Eden Department of Chemical Engineering Auburn.

Heat and Power Integration

CHEN 4460 – Process Synthesis, Simulation and Optimization

Dr. Mario Richard EdenDepartment of Chemical Engineering

Auburn University

Lecture No. 8 – Heat and Power Integration: Targeting

October 23, 2006

Contains Material Developed by Dr. Daniel R. Lewin, Technion, Israel

Lecture 8 – Objectives

Compute the pinch temperatures

Compute the Maximum Energy Recovery (MER) targets using graphical and/or algebraic methods

Given data on the hot and cold streams of a process, you should be able to:

Motivating Example

• What is wrong with this process from an energy viewpoint?

C1C2

H1

H2

300 K 550 K

520 K

330 K

380 K

300 K

320 K

380 K

AdiabaticReactor

Washing

Purification

Separation

ToRecovery

ToStorage

ToFinishing

Impurities

No integration of energy!!!!

Short Bibliography

• Early pioneers – Rudd @ Wisconsin (1968) – Hohmann @ USC (1971)

• Central figure – Linnhoff @ ICI/UMIST (1978)– Currently: President, Linnhoff-March

• Recommended text– Seider, Seader and Lewin (2004): Product and Process

Design Principles, 2 ed. Wiley and Sons, NY– Linnhoff et al. (1982): A User Guide on Process

Integration for the Efficient Use of Energy, I. Chem. E., London

• Most comprehensive review:– Gundersen, T. and Naess, L. (1988): The Synthesis of

Cost Optimal Heat Exchanger Networks: An Industrial Review of the State of the Art, Comp. Chem. Eng., 12(6), 503-530

Capital vs. Energy 1:3

• The design of Heat Exchanger Networks (HENs) deals with the following problem:

Given:

NH hot streams, with given heat capacity flowrate, each having to be cooled from supply temperature TH

S to targets TH

T

NC cold streams, with given heat capacity flowrate, each having to be heated from supply temperature TC

S to targets TC

T

Design:

An optimum network of heat exchangers, connecting between the hot and cold streams and between the streams and cold/hot utilities (furnace, hot-oil, steam, cooling water or refrigerant, depending on the required duty temperature)

Capital vs. Energy 2:3

• Optimality – Implies a trade-off between CAPITAL COSTS (cost of

equipment) and ENERGY COSTS (cost of utilities).

Network for minimal energy cost ?

Network for minimal equipment cost ?

H H H

C

C

C

CoolingWater

Steam

Tin

Tin

Tin

Tin Tin Tin

ToutTout Tout

Tout

Tout

Tout

CoolingWater

Steam

Tin

Tin

Tin

Tin Tin Tin

ToutTout Tout

Tout

Tout

Tout

Capital vs. Energy 3:3

• Numerical Example

Design A: (AREA) = 13.3

[ A = Q/UTlm ]

Design B: (AREA) = 20.4 [ A = Q/UTlm ]

CoolingWater (90-110oF)CoolingWater (90-110oF)

Steam (400oF)

300o300o

500500

150o

200o200o

150o 150o

200o200o

200o200o

100100

100100

100100

300o300o

300o300o

500500 500500

CP = 1.0CP = 1.0

CP = 1.0CP = 1.0

CP = 1.0CP = 1.0

CP = 1.0 CP = 1.0 CP = 1.0

100 100 100

300o300o

500500

150o

200o200o

150o 150o

200o200o

200o200o

300o300o

300o300o

500500 500500

CP = 1.0CP = 1.0

CP = 1.0CP = 1.0

CP = 1.0CP = 1.0

CP = 1.0 CP = 1.0 CP = 1.0

100

100

100

Some Definitions 1:3

T

HH

TS

TT

TTS = Supply temperature (oC)

TT = Target temperature (oC)

H = Stream enthalpy (MW) CP = Heat capacity flowrate (MW/

oC)= Flowrate x specific heat

capacity= m x Cp (MW/ oC)

Some Definitions 2:3

• Minimum Allowable Temperature Driving Force Tmin

• Which of the two counter-current heat exchangers illustrated below violates T 20°F (i.e. Tmin = 20°F) ?

100o 60o

50o

80o

100o 60o

40o

70o

A B

20o 10o

20o 30o

Clearly, exchanger A violates the Tmin

constraint

Some Definitions 3:3

100o 60o

40o

T1 = 70oOK

OK

Exchanger Duty (Q):

Data: Hot stream CP = 0.3 MW/ oC Cold stream CP = 0.4 MW/ oC

Heat Transfer Area (A):Data: Overall heat transfer coefficient, U=1.7 kW/m2 oC

(Alternative formulation in terms of film coefficients)

Check: T1 = 40 + (100 - 60)(0.3/0.4) = 70oC Q = 0.4(70 - 40) = 0.3(100 - 60) = 12 MW

Tlm = (30 - 20)/loge(30/20) = 24.66

So, A = Q/(UTlm) = 12000/(1.724.66) = 286.2 m2

Simple Example

Stream TS

(oC) TT

(oC) H

(kW) CP

(kW/oC)

H1 180 80 100 1.0 H2 130 40 180 2.0 C1 60 100 160 4.0 C2 30 120 162 1.8

Design a network of steam heaters, water coolers and exchangers for the process streams. Where possible, use exchangers in preference to utilities.

Utilities:

Steam @ 150 oC, CW @ 25oC

30° 120°

180° 80°

130° 40°

60° 100°

ΔH=162

ΔH=160

ΔH=100

ΔH=180

Simple Example - Targets

30° 120°

180° 80°

40°

60°

ΔH=162

ΔH=60ΔH=1813

0°

ΔH=100

100°

Units: 4Steam: 60 kWCooling water: 18 kW

Are these numbers optimal??

Temperature-Enthalpy Diagram

Correlation between Tmin, QHmin and QCmin

More in, More out! QHmin + x QCmin + x

Tmin = 10

Steam

CWC

H

110oC

100oC

T

HQCmin = 30 QHmin = 50

Steam

CW

Tmin = 20

Steam

CWC

H

120oC

100oC

T

HQCmin = 50 QHmin = 70

Steam

CW

The Composite Curve 1:2Temperature

Enthalpy

T1

T2

T3

T4

T5

CP

= A

CP = B

C P =

C

H Interval

(T1 - T2)*B

(T2 - T3)*(A+B+C)

(T3 - T4)*(A+C)

(T4 - T5)*A

Three (3) hot streams

The Composite Curve 2:2

Three (3) hot streams

Temperature

Enthalpy

T1

T2

T3

T4

T5

H Interval

H1

H2

H3

H4

CP = B

CP = A + B + C

CP = A + B

CP = A

H=150

H

180

130

CP = 3.0

80

40

H=50

H=80

CP

= 1

.0

C P = 2

.0

T

Simple Ex. – Hot Composite

30° 120°

180° 80°

130° 40°

60° 100°

ΔH=162

ΔH=160

ΔH=100

ΔH=180

H=150

T

H

180

130

CP

= 1

.0

C P = 2

.0

80

40

H=50

H=80Not to scale!

!

Not to scale!

!

H=232

T

H

120

100

CP = 5.8

60

30

H=36

H=54

CP

= 1

.8

CP

= 1

.8

Simple Ex. – Cold Composite

30° 120°

180° 80°

130° 40°

60° 100°

ΔH=162

ΔH=160

ΔH=100

ΔH=180

H=232

T

H

120

100

CP

= 1

.8

CP = 4.0

60

30

H=36

H=54Not to scale!

!

Not to scale!

!

Thermal Pinch Diagram

T

H

QC,min

QH,min

Tmin

Tpinch,hot

Tpinch,cold

Move cold composite

horizontally until the two curves are

exactly ΔTmin apart

Simple Ex. - Pinch Diagram

0

20

40

60

80

100

120

140

160

180

200

0 50 100 150 200 250 300 350

Enthalpy

Te

mp

era

ture

QCmin = 6 kW QHmin = 48 kW

TCpinch = 60

THpinch = 70

Maximum Energy Recovery (MER) Targets!

The Pinch

The “pinch” separates the HEN problem into two parts:

Heat sink - above the pinch, where at least QHmin utility must be used

Heat source - below the pinch, where at least QCmin utility must be used.

H

T

QCmin

QHmin

“PI NCH”

H

T

QCmin

QHmin

HeatSource Heat

Sink

Tmin

+x

x

+x

Significance of the Pinch

• Do not transfer heat across pinch

• Do not use cold utilities above the pinch

• Do not use hot utilities below the pinch

Algebraic Targeting Method

• Temperature scales– Hot stream temperatures (T)– Cold stream temperatures (t)

• Thermal equilibrium– Achieved when T = t

• Inclusion of temperature driving force ΔTmin

– T = t + ΔTmin

– Thus substracting ΔTmin from the hot temperatures will ensure thermal feasibility at all times

Algebraic Targeting Method

• Exchangeable load of the u’th hot stream passing through the z’th temperature interval:

• Exchangeable capacity of the v’th cold stream passing through the z’th temperature interval:

, 1( )Hu z u z zQ C T T

, 1 1 min min

, 1

( ) (( ) ( ))

( )

Cv z v z z v z z

Cv z v z z

Q C t t C T T T T

Q C T T

Algebraic Targeting Method

• Collective load of the hot streams passing through the z’th temperature interval is:

• Collective capacity of the cold streams streams passing through the z’th temperature interval is:

,H Hz u z

u

H Q

,C Cz v z

u

H Q

Algebraic Targeting Method

• Heat balance around each temperature interval:

1H C

z z z zr H H r

zHeat added by hot streams

Heat removed by cold streams

Residual heat from preceding interval

Residual heat to subsequent interval

HzH

1zr

CzH

zr

Algebraic Targeting Method

• The enthalpy cascade– r0 is zero (no hot streams exist above the first

interval)

– Feasibility is insured when all the rz's are nonnegative

– The most negative rz corresponds to the minimum heating utility requirement (QHmin) of the process

– By adding an amount (QHmin) to the top interval a revised enthalpy cascade is obtained

Algebraic Targeting Method

• The revised enthalpy cascade– On the revised cascade the location of rz=0

corresponds to the heat-exchange pinch point

– Overall energy balance for the network must be realized, thus the residual load leaving the last temperature interval is the minimum cooling utility requirement (QCmin) of the process

Algebraic Targeting Method

Stream TS

(oF)

TT

(oF)

H

(kBtu/h)

CP

(kBtu/h oF)

H1 260 160 3000 30

H2 250 130 1800 15

C1 120 235 2300 20

C2 180 240 2400 40

• Example– Two hot streams and two cold streams

– ΔTmin = 10°F

Step 1: Temperature intervals– Substract ΔTmin from hot temperatures

– 250°F 240°F 235°F 180°F 150°F 120°F

Stream TS

(oF)

TT

(oF)

H

(kBtu/h)

CP

(kBtu/h oF)

H1 250 150 3000 30

H2 240 120 1800 15

C1 120 235 2300 20

C2 180 240 2400 40

Stream TS

(oF)

TT

(oF)

H

(kBtu/h)

CP

(kBtu/h oF)

H1 260 160 3000 30

H2 250 130 1800 15

C1 120 235 2300 20

C2 180 240 2400 40

Algebraic Targeting Method

Step 2: Interval heat balances– For each interval calculate the enthalpy load Hi = (Ti Ti+1)(CPHot CPCold )

I nterval T i T i T i+1

CPHot

CPCold Hi

1 250 10 30 300 2 240 5 5 25 3 235 55 15 825 4 180 30 25 750 5 150 30 5 150 6 120

Stream CP

(kBtu/h oF)

H1 30

H2 15

C1 20

C2 40

Algebraic Targeting Method

Step 3: Enthalpy cascade

ΔH = 300

QH = 0T1 = 250°F

T2 = 240°F

ΔH = 25

T3 = 235°F

ΔH = -825

T4 = 180°F

ΔH = 750

T5 = 150°F

ΔH = -150

T6 = 120°F

Q1 = 300

Q2 = 325

Q3 = -500

Q4 = 250

QC = 100

ΔH = 300

QH = 500T1 = 250°F

T2 = 240°F

ΔH = 25

T3 = 235°F

ΔH = -825

T4 = 180°F

ΔH = 750

T5 = 150°F

ΔH = -150

T6 = 120°F

Q1 = 800

Q2 = 825

Q3 = 0

Q4 = 750

QC = 600

Most negative residual

TCpinch = 180°F

QHmin

QCmin

I nterval T i T i T i+1

CPHot

CPCold Hi

1 250 10 30 300 2 240 5 5 25 3 235 55 15 825 4 180 30 25 750 5 150 30 5 150 6 120

Summary – Heat Integration

Compute the pinch temperatures

Compute the Maximum Energy Recovery (MER) targets using graphical and/or algebraic methods

On completion of this part, given data on the hot and cold streams of a process, you should be able to:

Other Business

• Review of Midterm Exam– Tuesday October 24 during lab– Will meet in Ross Hall Auditorium– You will get your tests back to look at during solution

review – Test MUST be returned after presentation

• Next Lecture – October 30– Heat and Power Integration: Network Design (SSL pp. 316-

346)