CPE 633 Lecture W2.pdf

8/9/2019 CPE 633 Lecture W2.pdf

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SETTING PINCH DESIGNTARGETS

By Siti Shawalliah Idris, AMIChemE

SETTING PINCH DESIGN TARGETS



Setting the Energy Targets

Setting the Pinch Design Targets

COMPOSITECURVE

PROBLEMTABLE



Setting the Targets

Title of your presentation

StreamData - !Tmin

PinchAnalysis

EnergyTargets



Try This

Setting The Energy Targets for Pinch Design

! Do you think the amount heat required is reallynecessary?

STEAM COOLING

WATER

UNITS

1200 360 4



Enthalpy Balance


180o

100o

130o 40o

80o

120o 30o

60o

!H = 2000

!H = 3600

!H = -3240

!H = -3200

"#$

QH

1200

Qc 360

QH – QC =% (!H)= Q

recovery




! Heat recovery is constant

! It’s necessary/possible to reduce the amount of QH

or QC

QH

%!H

QH

Qc

%!H

Qc

QH – QC =% (!H) = constant



Enthalpy Balance


! Consider the lowest utilities requirement..

! Is this the best design?

No. Violation of !Tmin

QH

840



T-H Diagram


! Represent a stream data

! !" is constant

T

H

Ts

Tt

130o

40o

!H = 3600



Heat Recovery Problem


Mass Specific Heat Initial Final Heat

flowrate heat capacity (supply) (target) load

W (kg/s) capacity flowrate temperature temperature H (kW) CP (kJ/kgK) CP (kW/K) TS (°C) TT (°C)

Cold 0.25 4 1.0 20 200 180

stream

Hot 0.4 4.5 1.8 150 50 180

stream

Cooler

Heater200°

Product

Feed

Reactor

150°50°

20°H

C

! Consider a Two – Stream Heat Recovery Problem

Can we reduce energy consumption? Yes





!

Recover some heat from the hot stream" use it toheat the cold stream in a heat exchanger

!

Thus, less steam and water to satisfy the remaining

duties.

! Ideally – Recover all of 180 kW in the hot stream toheat the cold stream.

!

However, this is not possible because of temperaturelimitations.




Heat

Exchanger

E

Heater

Feed

Product

Cooler

Reactor

H

C

200°

150°50°

20°


! Ideally.. This is how it should look like on a PFD





! By the Second Law of Thermodynamics, we can’t use a hotstream at 150°C to heat a cold stream at 200°C!

As in the informal statement of the Second Law:

“you can’t boil a kettle on ice”

! So the question is, how much heat can we actually recover, howbig should the exchanger be, and what will be thetemperatures around it?





0

50

100

150

200

250

0 50 100 150 200 250

Heat load (kW)

T e m p e r a t u r e ( ° C )

Heating duty50 kW

Heat recovery130 kW

Cooling duty50 kW

Cold streamHot stream

!Tmin =0

! Plot T/H diagram

!

Is it possible to an approach temperature (!Tmin) = 0 for a heat exchanger design?

! !&min = 0" Infinitely large area of heat transfer





0

50

100

150

200

250

0 50 100 150 200 250 300

Heat load (kW)

Temperature difference∆T min 20°C

T e m p e r a t u r e ( ° C )

Cold stream

Hot stream

Heating duty70 kW

Heat recovery110 kW

Cooling duty70kW

! Streams can be shift horizontally" allow temp. difference

! !" is constant = 180kW





Stream Type Supply

Temp.

(oC)

Target

Temp.

(oC)

!H

(MW)

1

Cold

40

110

14

2 Hot 160 40 -12

Another example….

160oC

40oC

50oC40oC

95oC

110oC

T

H (MW)

QHmin = 3 MWQCmin = 1 MW

40o

QRec = 11MW

110o

Steam

CW

!Tmin = 10o

160o

20o

180o





! Shift cold stream horizontally – Increase !Tmin

T

H (MW)

QHmin = 4 MWQCmin = 2 MW

40o

QRec = 10MW

110o

Steam

CW

!Tmin = 20o

160o

20o

180o

160oC50oC 40oC

95oC

110oC

40oC





Two Basic Facts

! Correlation between !Tmin and QHmin and QCmin

! This means that if we choose a value of !T min, we have an energy

target for how much heating and cooling we should be using if wedesign our heat exchanger correctly.

! More in, More out

! the hot utility load is increased by any value ! , the cold utility isincreased by ! as well.

! As the stream heat loads are constant, this also means that the heat

exchanged falls by ! .

QHmin + ' " QCmin +'





! So far…

! We have seen how to apply single stream of hot

and cold.

! What if…multiple hot streams and cold streams?



COMPOSITE CURVE




Setting The Energy Target

StreamData - !Tmin

Compositecurves

EnergyTargets




Composite Curve


! The resulting T /H plot of multiple hot and coldstreams on a single curve known as a composite

curve.

! A hot composite curve a single plot of all the hot

streams and a cold composite curve of all the cold

streams in a particular problem.



Composite Curve


! Example



Composite Curve

Setting The Pinch Design Targets

! Stream data for the flowsheet

Stream

Type

Supply

Temp. Ts

(oC)

Target

Temp. TT

(oC)

!H (kW)

Heat Capacity

Flowrate CP

(kW oC-1)

Reactor 1 Feed

Cold

20

180

3200

20

Reactor 1

Product

Hot

250

40

-3150

15

Reactor 2 Feed Cold 140 230 2700 30

Reactor 2Product

Hot

200

80

-3000

25



Composite Curve


1•

Plot T/H diagram for Hot Streams and Cold Streams Separately

2•

Combine all the hot streams to get composite hot stream

3•

Combine all cold streams to get composite cold stream

4

•

Combine composite hot and cold streams together to give target for

utilities and recovery



Composite Curve

! T/H diagram for hot streams ! Composite Hot Stream


T (oC)

H (MW)

80o

3150

200o

250o

40o

3000

6150

T (oC)

H (MW)

80o

Q=CP!T=4800

200o

250o

40o

6150

750

600

Add heat available in each temperature interval to formcomposite curve

H intervals

(250-200)*15

(200-80)*(15+25)

(80-40)*(15)



Composite Curve

! T/H diagram for cold streams ! Composite Cold Stream


T (oC)

H (MW)

140o

3200

180o

230o

20o

2700

5900 5900

2000

Similar approach for cold streams to obtain cold compositecurve

T (oC)

H (MW)

140o

2400

180o

230o

20o

1500



Composite Curve


! Plot of hot and cold composite curves together gives the targetfor hot at cold utility

Shift the two curves so that will get !Tmin = 10oCT (oC)

H (MW)

50

QREC=5150

250

0

100

150

200

Pinch

!Tmin =10

Qhmin = 750

Qcmin = 1000



Composite Curve


! Shifting of the curves leads to behavior similar tothat shown by the two-stream problem.

! Though, the “kinked” nature of the composites

means that !T min can occur anywhere in the inter-change region and not just at one end.

For a given value of !T min, the utility quantitiespredicted are the minima required to solve the heat

recovery problem.



Composite Curve


! Although there are many streams in the problem..

In general !T min occurs at only one point of closest approach, which iscalled the pinch (Linnhoff et al. 1979).

! This means that it is possible to design a network which uses theminimum utility requirements, where only the heat exchangers at thepinch need to operate at !T values down to !T min.

!

It will be seen later that the pinch temperature is of great practicalimportance, not just in network design but in all energy-relatedaspects of process optimisation.



Composite Curve


! Let’s try increase !Tmin =20oCT (oC)

H50

QREC=4750

250

0

100

150

200

Pinch

!Tmin =20

Qhmin = 1150

Qcmin = 1400

T (oC)

H (MW)

50

QREC=5150

250

0

100

150

200

Pinch

!Tmin =10

Qhmin = 750

Qcmin = 1000

QC and QH increasecorrespondingly



Economic Trade-off


! The correct setting of !Tmin fixed by economictrade-offs



Summary


! Temperature-enthalpy diagrams can be used todetermine heat recovery potential

! Composite curves can be used to target for many hotstreams and many cold streams

! Energy targets set from material and energy balanceandΔTmin

!

Can be varied to different targets



Setting Energy Target – Composite Curve

WORKING EXAMPLE




Example 1


! Consider a process



Example 1

Streams No &

Type

Suppy

Temperature , Ts

(oC)

Target

Temperature, Tt

(oC)

Heat Capacity

Flowrate (kW/oC)

1 Hot 180 80 20

2 Hot 130 40 40

3 Cold 60 100 80

4 Cold 30 120 36

!Tmin =10 oC

Construct the composite curves

Read off energy targets for !Tmin =10 oC

35




PROBLEM TABLE ALGORITHM

© your company name. All rights

reserved.Setting The Pinch Design Targets



Recap

StreamData - !Tmin

Compositecurves

•

Complicated nature

EnergyTargets

“graph paper and scissors”approach (for sliding thegraphs relative to one

another) which would be

messy and imprecise.




Alternative Solution


The Problem Table Method




Divide into temperature intervals:Within a particular interval the two composite curves are not !Tmin apart from

each other

T (oC)

H (MW)

50

250

0

100

150

200

Pinch !tmin $( !Tmin

Within interval

Temperature Intervals




To allow for the maximum possible amount of heat exchange within each

temperature interval.

The only modification needed is to ensure that within any interval, hot streams

and cold streams are at least !T min apart.T (oC)

H (MW)

50

250

0

100

150

200

Shifted curves

!Tmin

Setting up the intervals in this way guarantees that full heat inter-change within anyinterval is possible.




! This is done by using shifted temperatures.

Shifting Rule:

Cold stream : + "!T min

Hot stream : -"!

T min




! For example:

!T min = 10oC

Thus, "!T min = 5oC

Actual temperatures Shifted temperatures

Stream number and type CP (kW/K) TS

(°C) TT

(°C) SS

(°C) ST

(°C)

1. Cold 2 20° 135° 25° 140°

2. Hot 3 170° 60° 165° 55°

3. Cold 4 80° 140° 85° 145°

4. Hot 1.5 150° 30° 145° 25°



2

4

1

3

165°C

145°C

140°C

85°C

55°C

25°C

170°C

150°C

145°C135°C135°C

90°C

60°C60

°C50

°C

80°C80°C 90°C

145°C

150°C140°C

30°C20°C

1

2

3

4

5


! Shifted temperatureintervals

! Arranged Ts* and Tt*

from highest to lowest

! Boundary is where the

stream starts or end! Min 2 boundaries

Real temperature




In each shifted temperature interval, calculate energybalance from

!Hi = (% CPH-% CPC )i !Ti




! Stream Data

Surplus

Interval Si Si1 ∑ CPHOT ∑ CPCOLD or

number i (°C) (kW/°C) ∆Hi (kW) deficit

S 1 165°C1 20 3.0 60 Surplus

S 2 145°C

2 5 0.5 2.5 Surplus

S 3 140°C

3 55 1.5 82.5 Deficit

S 4

85°C 4 30 2.5 75 Surplus

S 5 55°C

5 30 0.5 15 Deficit

S 6 25°C




! It would therefore be possible to produce a feasiblenetwork design based on the assumption:

All “surplus” intervals rejected heat to cold utility,

All “deficit” intervals took heat from hot utility.

However, this would not be very sensible, because it wouldinvolve rejecting and accepting heat at inappropriatetemperatures.



T 145°C ∆T min /2For all hot streams

T 145°C ∆T min /2For all cold streams

60kW

60kW145°C

2

1

!

Instead of cool down surplus energy

by using cooling water (cold utility),

the surplus heat is cascaded to the

next temp. interval.

!

Assuming no hot utility is supplied

(!H =0) to the hottest interval 1;

can set up cascade.


Key feature of the temperatureintervals:

‘Any heat available in interval i is hotenough to supply any duty in

interval i+1



From hot utility

0 kW

60kW

62.5 kW

20kW

55kW

40kW

To cold utility

1

2

3

4

5

165°C

145°C

140°C

85°C

55°C

25°C

∆H

60 kW

∆H

2.5 kW

∆H

82.5 kW

∆H 75 kW

∆H 15 kW

(a) Infeasible


Cascade any surplus heat from high temperature to low temperatureFrom hot utility

0 kW

60kW

62.5 kW

20kW

55kW

40kW

To cold utility

1

2

3

4

5

165°C

145°C

140°C

85°C

55°C

25°C

∆H 60 kW

∆H

2.5 kW

∆H 82.5 kW

∆H 75 kW

∆H 15 kW

(a) Infeasible



Heat flows cannot be negative! Add heat to make them at least 0!


From hot utility

82.5 kW

0 kW

75kW

60kW

20kW

80kW

To cold utility

1

2

3

4

5

∆H

60kW

∆H 2.5 kW

∆H 82.5 kW

∆H 75kW

∆H 15kW

(b) Feasible




! Minimum Hot Utility Requirement = 20 kW

! Minimum Cold Utility Requirement = 60 kW

! Pinch temperature

Hot Stream Pinch Temp = 85o

C + 5o

C = 90o

CCold Stream Pinch Temp = 85oC – 5oC = 80oC

! Maximum Heat Recovery = 450kW! Hot streams loads = -510 kW( -ve indicates heat rejects)

! Max Heat Recovery = 510-60 = 450 KW!

Cold streams loads = 470kW

! Max. Heat Recovery = 470 – 20 =450 kW



Summary


! The problem table algorithm:

Step 1: Adjust for !Tmin

Step 2: Set up temperature intervals

Step 3: Calculate interval heat balance

Step 4: Cascade for positive heat flows

Then, QHmin, QCmin and pinch locationwithout drawing graphs



Setting The Energy Targets – Problem Table

Algorithm

WORKING EXAMPLE





! For example:

!T min = 10oC

Thus, "!T min = 5oC

Stream Type Supply Temp.Ts (

oC) Target

Temp. TT (oC)

Shifted SupplyTemp. Ts*

Shifter TargetTemp.

Tt*

1

Cold

20

180 T*s = 20+ 5 =

25185

2 Hot 250 40 = 250 – 5 = 245 35

3 Cold 140 230 145 235

4 Hot

200

80

195 75



Temperature Interval Heat Balance


! !



Thank You …

CPE 633 Lecture W2.pdf

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