Tier II: Case Studies

Process Optimization

Tier II: Case Studies

Section 2:

Heat Exchange Network Optimization by Thermal Pinch Analysis

Optimization Problems

• There are many different types of optimization problems

• It is important to recognize that an optimization problem exists even if it does not immediately or easily lend itself to one of the previously described analytical methods of optimization

• Sometimes an alternative method that is more case specific must be used

Optimization Problems

• A common example of one of these problems is the optimization of a heat exchange network

• Without knowing what the maximum possible network integration is, and the minimum possible heating and cooling utilities required, it can be very difficult to design an optimized heat exchange network

• Heating and cooling utilities consumption can be treated as an optimization problem

• The goal is to minimize the amount of heating and cooling utilities being used by optimizing the heat exchange network

• A different method will be used for this type of optimization than what was seen previously

Optimization of Utility Use in a Heat Exchange Network

Constraints

• Total heating (QH) and total cooling (QC) used will still need to be minimized according to a set of constraints

• These constraints are:– The target temperature of individual streams– The minimum approach temperature in a heat

exchanger

Constraints

• Objective function:Minimize QH + QC

• Constraints:• T2i = ai , T1i = bi

• t1i = ci , t2i = di

• Tmin = k

Minimum Approach Temperature

T1 T2

t2

t1

o C

T1 t2

T2

t1

Minimum approach temperature

T1 – hot out

T2 – hot in

t1 – cold in

t2 – cold out


• To get the outlet temperature of one stream closer to the inlet temperature of the other stream, exchanger area must be increased, increasing capital cost

• Decreased exchanger area means decreased capital cost, but increased utilities cost to make up for lost heat exchange capacity

Using Minimum Approach Temperature to Tradeoff Capital vs. Operating Costs

• This graph demonstrates the tradeoff between capital and operating costs – a decrease in one is met with an increase in the other

0

5

10

15

20

25

30

35

40

45

0.5 5.5 10.5 15.5 20.5

Tmin (oC)

An

nu

aliz

ed C

ost

($/

year

)

Annual operating cost

Annualized fixed cost

Total annualized cost

Minimum total annualized cost

Topt0


• The optimum exchanger size exists where the total annualized cost is minimized

• This typically will correspond to a minimum approach temperature, Tmin of about 10oC

• This Tmin = 10oC is a rule of thumb – it can change depending on the fluid service and the type of heat exchanger employed


Thermal Equilibrium

T = t

Practical Feasibility

T = t + Tmin

• This must be included in the coming analysis

Graphical Method – Thermal Pinch Analysis

• To optimize a heat exchange network, an example of the graphical method to determine the thermal pinch point will first be examined

• The same example will then be solved using the algebraic method for comparison

Stream Data

• Using the stream supply and target temperatures, the enthalpy change of each stream must be calculated

• Enthalpy change:• H = FiCpi(T2

i – T1i) = HHi

= FiCpi(t2i – t1

i) = HCi

• FiCpi = flow rate x specific heat (kW/K)

Stream Data

Hot Stream FiCpi Supply (oC) Target (oC) Enthalpy Change

(kW/oC) T2i T1

i HHi, (kW)

H1 400 340 260 32000

H2 350 400 360 14000

H3 300 450 380 21000

Cold Stream FiCpi Supply (oC) Target (oC) Enthalpy Change

(kW/oC) t1i t2

i HCi, (kW)

C1 250 240 290 12500

C2 300 300 400 30000

C3 450 350 400 22500

Stream Data

• Stream data is then plotted as a series of straight line segments in order of ascending temperature

• Each consecutive segment begins at the enthalpy level where the previous segment finished

• A “hot” stream is any that must be cooled, while a “cold” stream is any that must be heated, regardless of supply temperature

Hot Streams

Hot Streams

0

20000

40000

60000

80000

100000

220 270 320 370 420

T (oC)

H

(kW

)

H1

H2

H3

HH1

HH2

HH3

Cold Streams

Cold Streams

0

20000

40000

60000

80000

100000

220 270 320 370 420

t (oC)

H

(kW

)

HC1

HC2

HC3C3

C2

C1

Composite Stream Curves

• Next the composite curves of the hot and cold streams must be constructed

• These composite curves represent the total amount of heat to be removed from the hot streams and the total amount of heat that must be added to the cold streams to reach the target stream temperatures

Hot Composite Stream Construction

Hot Streams

0

20000

40000

60000

80000

100000

220 270 320 370 420

T (oC)

HH

(kW

)

T11 T2

1 T12 T2

2T13 T2

3

H1

H2

H3

Hot Composite Stream Construction

Hot Streams

0

20000

40000

60000

80000

100000

220 270 320 370 420

T (oC)

H

(kW

)

Hot composite stream

Cold Composite Stream Construction

Cold Streams

0

20000

40000

60000

80000

100000

220 270 320 370 420

t (oC)

H

(kW

)

t11 t1

2

t23

t13 t2

2

t21

C1

C3

C2

Cold Composite Stream Construction

Cold Streams

0

20000

40000

60000

80000

100000

220 270 320 370 420

t (oC)

H

(kW

)

Cold composite stream

Optimizing the Heat Exchange Network

• The cold composite stream must now be superimposed over the hot composite stream to perform the thermal pinch analysis

• This will give the minimum amount of utilities required to reach the target states

• Note how the temperature axis is shifted for the cold composite stream to account for the minimum approach temperature

No Heat Integration

QC,max = 67,000 kW

QH,max = 65,000 kW

QC + QH = 132,000 kW

0

20000

40000

60000

80000

100000

220 270 320 370 420 T

H

(kW

)

210 260 310 360 410 t = T - Tmin240

Total hot utility required

Total cold utility required



No Heat Integration

• With no heat integration, the amount of energy required to reach the target state is maximized

• In this case the total amounts of energy required are:

• Cooling utility, QC = 67,000 kW• Heating utility, QH = 65,000 kW• Total utilities = QC + QH = 132,000 kW

• Clearly there is room for optimization

Partial Heat Integration

• By moving the cold composite stream down a bit, a partially integrated heat exchange network is graphically represented

• Some heat is transferred from hot streams to cold streams to approach the temperature targets


0

20000

40000

60000

80000

100000

220 270 320 370 420 T

H

(kW

)

210 260 310 360 410 t = T - Tmin

QC = 52,000 kW

QH = 50,000 kW

Integrated heat exchange 15,000 kW

Total hot utility required

Total cold utility required



QC + QH = 102,000 kW


• This heat exchange network is only partially optimized and already utility consumption is reduced by 30,000 kW

• The utilities required are:• Cooling utility, QC = 52,000 kW• Heating utility, QH = 50,000 kW• Total utilities = QC + QH = 102,000 kW

• Clearly further integration can provide significant energy savings

Optimized Heat Integration

• To determine the optimized heat exchange network, the thermal pinch point must be found

• This is accomplished by moving the cold composite stream down just until one point on the line meets a point on the hot composite line

• This point is the thermal pinch point

0

20000

40000

60000

80000

100000

220 270 320 370 420 T

H

(kW

)

210 260 310 360 410 t = T - Tmin


QH,min = 8,500 kW

QC,min = 10,500 kW

Integrated heat exchange = 56,500 kWHot composite

stream


Pinch point

QC + QH = 19,000 kW

240


• The heat exchange network is now fully optimized

• Total required utilities are minimized• Minimum cooling utility, QC,min = 10,500 kW• Minimum heating utility, QH,min = 8,500 kW• Minimum total utilities = QC + QH = 19,000 kW

• No heat is passed through the pinch point

Passing Heat through the Pinch Point

• To have an optimized heat exchange network, it is critical that no heat is passed through the thermal pinch point

• By passing an amount of heat, , through the pinch point, an energy penalty of 2 is added to the total utilities requirement

• It is very important to maximize integration in a heat exchange network

Passing Heat Through the Pinch Point

0

20000

40000

60000

80000

100000

220 270 320 370 420 T

H

(kW

)

210 260 310 360 410 t = T - Tmin

QH,min

QC,min

QH = QH,min +

QC = QC,min +

QH + QC = QH,min + QC,min + 2

Crossing the Pinch Point

• It would appear that extra energy can be saved by lowering the cold composite stream line further

• This does not work however because it creates a thermodynamically infeasible region

• For this to work, heat would have to flow from the cooled hot streams to the heated cold streams - from a cold source to a hot source

Crossing the Pinch Point

0

20000

40000

60000

80000

100000

220 270 320 370 420 T

H

(kW

)

210 260 310 360 410 t = T - Tmin

Pinch point

Infeasible region



Disregarding Tmin

• Another tempting error is to disregard the minimum approach temperature

• By disregarding a minimum approach temperature, the absolute minimum thermodynamically possible utility requirements are obtained

• Although this is thermodynamically possible, it is not practically feasible as it would require an infinitely large heat exchanger area

• This would obviously cost far more than the relatively small energy savings are worth

Disregarding Tmin

0

20000

40000

60000

80000

100000

220 270 320 370 420 T

H

(kW

)

QH,min thermo.

QC,min thermo.

240

Algebraic Method

• This same problem will now be solved using the algebraic method

• This will involve producing a temperature interval diagram, tables of exchangeable heat loads, and cascade diagrams

Stream Data

Hot Stream FiCpi Supply (oC) Target (oC)

(kW/oC) T2i T1

i

H1 400 340 260

H2 350 400 360

H3 300 450 380

Cold Stream FiCpi Supply (oC) Target (oC)

(kW/oC) t1i t2

i

C1 250 240 290

C2 300 300 400

C3 450 350 400

From before:

Temperature Interval Diagram

• The first step is to construct the temperature interval diagram

• This diagram shows the starting and finishing temperatures of each stream

• An interval begins at a stream’s starting or finishing temperature, and it ends where it encounters the next beginning or finishing temperature of a stream– Draw horizontal lines across the table at each arrow’s head

and tail, with the intervals lying between these lines

• Note how the cold stream temperature scale is staggered by 10 degrees

Temperature Interval Diagram

Interval Hot Streams Cold StreamsT t

330340

400 390

400410

380 370

450 440

300310

290300

350360

240250

260 250

1

3

2

4

5

6

7

8

9

FC

p =

250

C1

H3

FC

p =

300

H1

FC

p =

400

H2FC

p =

350

C2F

Cp

= 300

C3

FC

p =

450

Table of Exchangeable Heat Loads

• The next step is to construct tables of exchangeable heat loads for the hot and cold streams

• These tables show the amount of energy that must be added or removed from a stream over a particular interval

• These energy values are calculated asHj,i = FCpjTi, where Ti is the positive

temperature difference across the interval, and j denotes the stream number


• For the hot streams,

Interval H1,i H2,i H3,i Total, HHi

i kW kW kW kW1 - - 12000 120002 - - 3000 30003 - 7000 6000 130004 - 7000 - 70005 - - - 06 12000 - - 120007 4000 - - 40008 16000 - - 160009 - - - 0

Total cooling required (kW) 67000

Table of Exchangeable Loads - Hot Streams


• For the cold streams,

Interval C1,i C2,i C3,i Total, HCi

i kW kW kW kW1 - - - 02 - 3000 4500 75003 - 6000 9000 150004 - 6000 9000 150005 - 6000 - 60006 - 9000 - 90007 - - - 08 10000 - - 100009 2500 - - 2500

Total heating required (kW) 65000

Table of Exchangeable Loads - Cold Streams

Cascade Diagrams

• Using the information from the heat load tables, the cascade diagrams can now be constructed

• These diagrams will be used to determine the pinch point and the minimum heating and cooling utilities required

Cascade Diagram

1

2

3

4

5

6

7

8

9

• First, the cascade diagram is drawn as it appears at right, with one box for each interval that appeared in the temperature interval diagram

Cascade Diagram

• Next, the total values from the exchangeable heat load tables are added to the cascade diagram

• Hot stream loads enter on the left, cold stream loads exit on the right

1

2

3

4

5

6

7

8

9

3000

12000

13000

7000

0

12000

0

16000

4000

0

7500

15000

15000

6000

9000

0

10000

2500

Cascade Diagram

1

2

3

4

5

6

7

8

9

3000

12000

13000

7000

0

12000

0

16000

4000

0

7500

15000

15000

6000

9000

0

10000

2500

• Now, by subtracting an interval’s cold load from the hot load, and adding the resulting value to the residual from the previous stage we get the residual value for the subsequent stage

• ri = HHi – HCi + ri-1

0

12000

7500

-2500

-8500

-5500

-1500

4500

20001) 12000 – 0 + 0 = 120002) 3000 – 7500 + 12000 = 75003) 13000 – 15000 + 7500 = 55005) 0 – 6000 -2500 = -85006) 12000 – 9000 – 8500 = -55008) 16000 – 10000 – 1500 = 45007) 4000 – 0 – 5500 = -15009) 0 – 2500 + 4500 = 2000

5500

4) 7000 – 15000 + 5500 = -2500

Thermal Pinch Point

• The thermal pinch point occurs at the largest negative number

12000

1

2

3

4

5

6

7

8

9

3000

12000

13000

7000

0

12000

0

16000

4000

0

7500

15000

15000

6000

9000

0

10000

2500

7500

5500

-2500

-8500

-5500

-1500

4500

2000

0

Pinch Point

• The absolute value of this number is now added in at the top to cascade through

Revised Cascade Diagram

12000

1

2

3

4

5

6

7

8

9

3000

12000

13000

7000

0

12000

0

16000

4000

0

7500

15000

15000

6000

9000

0

10000

2500

7500

5500

-2500

-8500

-5500

-1500

4500

2000

8500

+ 8500

+ 8500

+ 8500

+ 8500

+ 8500

+ 8500

+ 8500

+ 8500

+ 8500

Revised Cascade Diagram

20500

1

2

3

4

5

6

7

8

9

3000

12000

13000

7000

0

12000

0

16000

4000

0

7500

15000

15000

6000

9000

0

10000

2500

16000

14000

6000

0

3000

7000

13000

10500

8500Qmin,heating =

Qmin,cooling =

• We now have the final revised cascade diagram

• It can be seen that by adding additional energy at the top, it will cascade through and also be present at the bottom

PinchPoint

QH + QC = QH,min + QC,min + 2


• The heat exchange network is now fully optimized

• Total required utilities are minimized• Minimum cooling utility, QC,min = 10,500 kW• Minimum heating utility, QH,min = 8,500 kW• Minimum total utilities = QC + QH = 19,000 kW

• As expected, these values are the same as obtained by using the graphing method

Design Considerations

• Some design rules to optimize utility consumption:– Do not pass heat through the pinch point– Do not use cooling utilities at temperatures

above the pinch point– Do not use heating utilities at temperatures

below the pinch point

Constructing the Heat Exchange Network

• Now that the pinch analysis has been performed, the heat exchange network can be constructed

• It is a good idea to perform the pinch analysis first because it sets the performance goal of an optimized heat exchange network

• There is no quick method of reliably determining the minimum number of heat exchangers, but the following method should help to construct the network


• With QC,min and QH,min known, construct a plot similar to the temperature interval diagram, except instead of arrows, use boxes that have a width representing FCp

• The area of these boxes corresponds to the heat exchanged by the stream

• Draw a horizontal line across at the pinch point – remember, no heat is to be passed across this point


Pinch Point

Hot Streams Cold StreamsT t

330340

400 390

400410

380 370

450 440

300310

290300

350360

240250

260 250

FCp = 250

C1

H3

FCp = 300

H2

FCp = 350

C2

FCp = 300

C3

FCp = 450H1

FCp = 400


• Now, add QC,min to the lowest point on the coldest hot stream and determine the resulting T1 and T2 for this exchange. Note that T1, T2, t1, and t2 now do not necessarily correspond to the same values as used earlier and are different for each exchanger

QC,min = FCp(T2 – T1)• Do the same with QH,min, adding it to the highest

point on the hottest cold stream

QH,min = FCp(t2 – t1)



330340

400 390

400410

380 370

450 440

300310

290300

350360

240250

260 250

FCp = 250

C1

H3

FCp = 300

H2

FCp = 350

C2

FCp = 300

C3

FCp = 450H1

FCp = 400QC,min = 10500 kW

QH,min = 8500 kW

286.25

381.1

Pinch Point


• Now, working out from the pinch point, match up streams, remembering not to transfer heat across the pinch point and keeping Tmin in mind

• For each matched stream, determine the temperatures that exist for the inlet and outlet of the heat exchanger

Qex = FCp(T2 – T1) = FCp(t2 – t1)

• Having the table of stream data including enthalpy change on hand may be helpful for determining the best way to match a stream

Matched Streams


330340

400 390

400410

380 370

450 440

300310

290300

350360

240250

260 250

FCp = 250

C1

H3

FCp = 300

H2

FCp = 350

C2

FCp = 300

C3

FCp = 450H1

FCp = 400QC,min = 10500 kW

QH,min = 8500 kW

286.25

381.1

317.5

Pinch Point

21000 kW

14000 kW

9000 kW

12500 kW

Heat Exchangers

• 4 heat exchangers, plus a heater and a cooler are needed to meet the optimum heat exchange requirements of this system

Heat Exchanger T2 (oC) T1 (

oC) t2 (oC) t1 (

oC) Duty (kW)

H3-C2 450 380 400 330 21000

H2-C3 400 360 381.1 350 14000

H1-C2 340 317.5 330 300 9000

H1-C1 317.5 286.25 290 240 12500

QH,min-C3 na na 400 381.1 8500

QC,min-H1 286.25 260 na na 10500

Heat Exchange Network

Conclusion

• There is no quick method that is guaranteed to give the minimum number of heat exchangers required every time

• However, by first performing a thermal pinch analysis to determine the maximum heat exchange possibilities, designing an optimum network configuration is made a lot easier

References

• Dr. El-Halwagi lecture notes

Documents

Tier II: Case Studies