Flow rates : Known

Obtain : heat capacities (Cp) heat of vaporization/condensation

Estimate : vapor loads in the column(design)

Obtain heat loads of all streams which change temperature/phase

Q. Is it possible to combine/integrate/match them?

Energy integration

Heat Exchanger Network(HEN)

Energy BalanceLevel 5

Synthesis of Heat Exchanger Networks

Questions

1. Why is it important to make a particular study of HENs ?• Considerable reductions can be made in the energy requirements of chemical process plants by the application of recently developed techniques ( often of the order of 20 ~ 302 )• The reductions in energy requirements can be achieved with reduced capital costs

2. Is it possible to predict the minimum utility requirements ( e.g. cooling water, steam, fuel, etc. ) in a HEN 3. Is it possible to predict the minimum number of HENs ? ( Heat exchange matches required are a network if the maximum amount of energy is to be required )4. Given affirmative answers to questions 2 and 3, can optimal networks be designed ?5. How can the design procedures be extended to incorporate other terms of equipment such as gas turbines, distillation columns

minT : HRAT (Heat Recovery Approach Temperature

냉각장치 : = 1~ 2℃

싼 utility 필요시 : = 30 ℃

minT

minT

Pinch - sequentialMIP - concurrent

Minimum utility requirements

Simple example – one hot and one cold stream

FCp Tsupply Ttarget

Stream (kw/ ) ( ) ( )℃ ℃ ℃(1) Hot 1 300 90(2) Cold 3 100 180

F : mass flow rate of stream ( kg / s )Cp ; heat capacity of stream ( kJ / kg )℃FCp [ kw / ] ℃

℃10min T

300℃

QH steam

180℃100℃

90℃

Qc Cooling water

Temperature – Enthalpy Plot

T℃

)(kwH

180℃

300

200

10090

100 200210

300260

Hotstream

10 ℃

Coldstream

0

T℃

)(kwH

300

200

10090

20 210 2600

Qc

QEX

PINCH

QH

“ PINCH “THOT – TCOLD=

minT

For

kwQkwQT

kwQkwQT

cH

cH

30;60;20

20;50;10

min

min

℃

℃

For

minTIncreasing cold stream curve moving to the right

Q vs Plot minT

10 20 30 40

20

40

60

80QH ( or QC if QC ＞ QH )

QC ( or QH if QC＞ QH )

Q(kw)

minT ( )℃

QH and QC are dependent upon QH increase QC increase ( more in more out ) QH + X QC + X

minT

Multiple stream HENs

Construct composite curves for the combined hotAnd cold streams

Example – two streams

FCP Tsupply Ttarget -ΔHStreams C( kw / ) ( ) ( ) (kw) ℃ ℃ ℃

(1) 3 170 60 330(2) 1.5 150 30 180

510

-ΔH(kw)100 200 300 0 100 300

400

500200

0

30

60

100

150170

Individualstreams

Compositestream

(1)(2)

C=1.5(1)+(2)

C=4.5

C=3

C=1.5510

General Procedure

1. Composite hot stream curve on T~ ΔH plot2. Subtract ΔTmin everywhere and draw TH – ΔTmin curve3. Move the composite cold stream curve to right until curve for TH – ΔTmin and TC touch at “pinch “

T

ΔH

TH

TH – ΔTmin

Composite cold curve( move to right )

T

ΔHQC QEX

QH

Fig 8.1-8

Situations in which the composite hot and cold curvesare ΔTmin apart at some point are referred to as “pinched” Note that both heating and cooling utilities are required

On the “ hot side “ of the pinch only heating utilities are required ( the addition of a cooler simply means that additional heating is necessary )

On the “ cold side “ only cooling utilities are requiredIf more energy is added above the pinch than is necessary, then more energy has to be removed below the pinch. ( effectively increases ΔTmin ). This situation corresponds to transferring energy across the pinch

Note

In some cases unless ΔTmin is made unrealistically large only one utility is required ( heating or cooling )

For example

T

ΔHQEX QH

Heater only

critTmin

ΔTmin

QC

T

ΔHQEX

critTmin

TH

Cooler only

QC( or QH)

QH( or QC)

critTmin

QH

QC

ΔTmin

: value of ΔTmin at which second utility is required

Situations in which the composite hot and cold curves are always greater than ΔTmin apart are referred to as “ unpinched “.

Pinched Problem

• Do not use cooling utilities above the pinch• Do not use heating utilities below the pinch• Avoid transferring energy across the pinch

Consider a simple design problem

Stream Tsupply Ttarget C(FCP) ( ) ( ) (kw / )℃ ℃ ℃

(1) 20 135 2.0(2) 170 60 3.0 (3) 80 140 4.0(4) 150 30 1.5

ΔTmin = 10 ℃

First find the minimum heating and cooling requirements and the pinch temperature.Rather than construct composite curve it is easier to construct a table of energy excesses in terms of cold stream temperature.Subtract ΔTmin from hot stream temperature to give the following table.

Stream Tsupply Ttarget FCP

(1) (cold) 20 135 2.0(2) (hot) 160 50 3.0(3) (cold) 80 140 4.0(4) (hot) 140 20 1.5

Convenient representation :

2

4

1

3

FCP

3.0

1.5

2.0

4.0160 140 135 80 50 20

3.0 0.5 -1.5 2.5 -0.5

coldphotp FCFC ,,

Construct table

Stream temp energy surplus Interval or deficit

coldphotp FCFC ,,

160 – 140 3.0 60 (surplus)140 - 135 0.5 2.5(surplus)135 – 80 -1.5 -82.5(deficit)80 – 50 2.5 75(surplus)50 – 20 -0.5 -15

Energy surpluses may be cascaded down to lover temperatures as may be shown diagrammatically as follows :

60

2.5

-82.5

75

-15

160 - 140

140 – 135

135 - 80

80 - 50

50 – 20

60

62.5

-20

55

40

Hot utility“ cascade diagram “

Deficit of 20 kwhence this is not a feasible heatexchange process

∴ add 20kw from hot stility

Cold utilit

60

2.5

-82.5

75

-15

160 - 140

140 – 135

135 - 80

80 - 50

50 – 20

80

82.5

0

75

60

Hot utility

Zero energy cascaded at cold stream temp.80 (hot stream 90 )℃ ℃Hence pinch is at cold stream temp. of 80℃

Cold utility

20

Pinch at 90 ( hot streams )℃ 80 ( cold streams )℃QH ( total hot utilities ) : 20 kwQC ( total cold utilities ) : 60 kw

Note

If more than 20 kw is added through the hot utility then the additional energy is cascaded through the entire process to the cold utility

Problem representation ( Linnhoff )

C

H

Hot stream

Cold stream

Streamnumbers load

cooler

load

Heater Heat exchanger

FCP

FCP

FCP

FCP

Direction of increasing temp.

Advantage : easy to change configuration of network without re-routing streams.

For the problem under consideration start at pinch

2

4

1

3

pinchQH = 20 QC = 60kw

60

30

170

150

135

140

20

90

90

80

80

No energy should be transferred across the pinch, The design problem is effectively decomposed into two separate problems : one above the pinch and one below the pinch.

Consider a heat exchanger on stream 2 or the hot side of the pinch.

Two possibilities

4

2

1

3

FCP

3.0

1.5

2.0

4.0

90170

4

2

1

3

(1) (2)

T

ΔH ΔH

T

ΔTminΔTmin

FCPhot ＞ FCPcold FCPhot ≤FCPcold

infeasible feasible

A heat exchanger may be placed above the pinch connecting streams 2 and 3 with a load of 240 kw. Similarly a heat exchanger may be placed connectingstreams 4 and 1 with a load of 90 kw leaving 20 kw (QH) to be provided by a heater to stream 1

Cold End Design

4

2

1

FCP

3.0

1.5

2.0

90

80

90

60

30

20

4

2

1

ΔTmin

T T

FCPhot ≥FCPcold FCPhot ＜ FCPcold

infeasiblefeasible

A heat exchanger with a load of 90 kw may be placed connecting streams 2 and 1

Complete Design

2

4

1

3

H

C

170˚ 90˚ 60˚

150˚ 90˚ 70˚ 30˚

130˚20˚ 120˚ 80˚ 35˚ 60˚

20˚

90˚ 90˚ 90˚140˚

240˚

80˚

FCP

3.0

1.5

2.0

4.0

The above design achieves maximum possible energy recovery for ΔTmin = 10 ℃

1. Start at the pinch and make matches “ outwards “2. Immediately above pinch make matches that meet the requirement

3. Immediately below pinch make matches

that meet the requirement

FCPhot ≤FCPcold

FCPhot ≥FCPcold

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