PINCH ANALYSIS Part 1- Pinch and Minimum Utility Usage.pdf

Pinch analysis using the method of composite curves and problem table calculation

1. PART 1

PINCH AND MINIMUM
UTILITY USAGE

2. TEMPERATURE-ENTHALPY
(T-H) DIAGRAMS

° Assume one heat exchanger. These are
alternative representations

ou

ou Soros are the
on _ | inverse of F*Cp.

(Recall that Q=F Cp AT)

>

3. T-H DIAGRAMS

* Assume one heat exchanger and a heater

4. Te in

T-H DIAGRAMS

» Assume one heat exchanger and a cooler
Thin

Toon

A
x
+

>
AH

5. T-H DIAGRAMS

« Two hot-one cold stream

AH

QQ

Notice the vertical arrangement of heat transfer

6. Streams under phase change

Phase |

hi

change wg
Slope
change

Liquid
m,
>
AH AH

Single component Multicomponent

We say this stream
has “variable Cp”

7. Piece-wise linear representation

AH

8. Composite Curves
(T-H DIAGRAMS)

Obtained by lumping all the heat from different streams that are
at the same interval of temperature.

AH AH
Remark: By constructing the composite curve we loose information on
the vertical arrangement of heat transfer between streams

9. Composite Curves
(T-H DIAGRAMS)

+ Moving composite curves horizontally

Cooling

Smallest AT

Ti in je in
= Se
Te ‚out

| Tai al Typ, out

10. Composite Curves
(T-H DIAGRAMS)

Moving the cold composite
stream to the right

+ Increases heating and cooling BY
EXACTLY THE SAME
AMOUNT
Increases the smallest AT

+ Decreases the area needed
A=Q/(U* AT )

Notice that for this simple
example the smallest AT
takes place in the end of the
cold stream

11. Composite Curves
(T-H DIAGRAMS)

” » In general, the smallest AT

Code can take place anywhere.

» We call the temperature at
which this takes place THE
> PINCH.

12. Composite Curves
(T-H DIAGRAMS)

« From the energy point
of view it is then
convenient to move the
cold stream to the left.

Cooling

|

« However, the area may

become too large.

Heati EE
e « To limit the area, we

AH introduce a minimum
approach AT jp,

AT min is also known as HRAT (Heat Recovery Approximation Temperature)

13. GRAPHICAL PROCEDURE

Fix AT... (HRAT)
Draw the hot composite curve and leave it fixed

min

Draw the cold composite curve in such a way that
the smallest temperature difference is equal to AT in
The temperature at which AT=AT,,,,, is the PINCH
The non-overlap on the right is the Minimum
Heating Utility and the non-overlap on the left is the
Minimum Cooling Utility

14. EXAMPLE

AH=27 MW

AH=-30 MW
>( > ei
T=140 °C T=230 °C = T=80 °C
T=200 °C
_—

AH=32 MW

© À REACTOR 1
To N

AH=-31.5 MW
© T=40°C|
T=180°C T=250°C

Stream Supply T TargetT AH F*Cp

CC) CC) (MW) (MW °C")
Reactor | feed 20 180 32.0 0.2

Reactor | product 250 40 -31.5 0.15

Reactor 2 feed 140 230 27.0 0.3
Reactor 2 product 200 80 -30.0 0.25

15. 250
200

80

40 J

Hot Composite Curve

7 ECp=0.15

FCp-0.15

>
6 48 75 AH

16. Cold Composite Curve

17. Pinch Diagram

250
230
200
180
Pinch—>
140
The pinch is defined either as
- The cold temperature (140 °) 80
- The corresponding hot
temp (140 %+ATy;p=150 °) 40
- The average (145°) 20 =
+ >
—. ~—+ AH
10 51.5 75,

Observation: The pinch is at the beginning of a cold stream or at the
beginning of a hot stream

18. UTILITY COST vs. AT yin

T 4 W
COST a a
e y

Utility

> i
AT rin AH
PARTIAL OVERLAP

Note: There is a particular overlap that requires only cooling utility

19. Special Overlap Cases

+ Overlap leads only to cooling utility
Lo
Va

« Different instances where the cold stream overlaps totally the
hot stream. Case where only heating utility

T

AH

T T

Ur
y y

AH AH
N TOTAL / N PARTIAL

OVERLAP We prefer this arrangement OVERLAP
even if AT>AT min

20. SUMMARY

» The pinch point is a temperature.

+ Typically, it divides the temperature range
into two regions.

+ Heating utility can be used only above the
pinch and cooling utility only below it.

21. PROBLEM TABLE

Composite curves are inconvenient. Thus
a method based on tables was developed.

° STEPS:

1.

Divide the temperature range into intervals and
shift the cold temperature scale

Make a heat balance in each interval

Cascade the heat surplus/deficit through the
intervals.

Add heat so that no deficit is cascaded

22. PROBLEM TABLE

+ We now explain each step in detail using
our example

Stream Supply T TargetT AH F*Cp
(°C) CC) (MW) (MW °C!)
Reactor 1 feed 20 180 32.0 0.2

Reactor 1 product 250 40 -31.5 0.15
Reactor 2 feed 140 230 27.0 0.3
Reactor 2 product 200 80 -30.0 0.25

AT miy=10 °C

23. 250 —

230

200
180

140

80

40
20

Hot
streams

PROBLEM TABLE

Divide the temperature range into intervals and shift the
cold temperature scale

250
240

200
190

Cold Hot Cold
streams streams streams

Now one can make heat balances in each interval. Heat transfer within each interval is feasible.

24. 250
240

200
190

150

80

gs

2.

F Cp=0.15

PROBLEM TABLE

Make a heat balance in each interval.

F Cp=0.25

Hot
streams

FCp=0.3

F Cp=0.2

Cold
streams

AT,

interval
10
40
10
40
70
40
10

AH

interval
1.5

- 6.0
1.0

-4.0

14.0

-2.0

-2.0

Surplus/Deficit

Surplus
Deficit
Surplus
Deficit
Surplus
Deficit

Deficit

25. PROBLEM TABLE

3. Cascade the heat surplus through the intervals. That is,
we transfer to the intervals below every surplus/deficit.

250

1.5 Lis y 15
# + This intervalibas a $s The largest deficit
-60 surplus. It should 60 ae transferred is -7.5.
en transfer 1.5 to
2 . u.
x interval 2. 2 45 Thus, 7.5 MW of
1.0 10 A heat need to be
= + This interval has a 4 39 added on _ ;
40 deficit. After using 46 a me! e de
150 I the 1.5 cascaded it E e trans E to
y
transfers 4.5 to 75 ower intervals
14.0 interval 3. 14.0
80
* = Y 65
-20 20
40
* 4s
2.0 en
30
y

26. 250

240

200

190

150

80

40

45

-3.5

-7.5

6.5

45

PROBLEM TABLE

4. Add heat so that no deficit is cascaded.

250

240

200

150

80

40

7.5

| i

15 This is the

y 90 minimum heating
-60 utility

1 30

1.0 This is the

| 40 en ofthe
4.0 Pine

00
14.0 e
This is the

— #9 minimum cooling
2.0, utility

y 120
2.0

Ÿ 10.0

27. PROBLEM TABLE

If the heating utility is increased beyond 7.5 MW the cooling utility will
increase by the same amount

7.5 7.5 +h
250 = Bg
15 15 Heating utility is
240 | 90 y 90+z larger than the
-60 “0 minimum
200 l 30 2 3.0+%
1.0 10 Heat is
190 lso En) [4043 transferred
40 Pr a across the pinch
= 00 70.047
© T uo Tr ; Cooling utility is
A 14.0 +2. larger by the
-2.0 20 same amount
“ y 12.0 412.041
2.0 2.0
30

y 10.0 Y 18.083

28. IMPORTANT CONCLUSION

DO NOT TRANSFER
HEAT ACROSS THE
PINCH

THIS IS A GOLDEN RULE OF PINCH
TECHNOLOGY.

«WHEN THIS HAPPENS IN BADLY
INTEGRATED PLANTS THERE ARE
HEAT EXCHANGERS WHERE SUCH
TRANSFER ACROSS THE PINCH
TAKES PLACE

75 +h
+
15 Heating utility is
| sos larger than the
minimum
6.0
L 3.0+%
1.0 Heat is
| 404 transferred
ae across the pinch
Y 0.0+2
ine Cooling utility is
Y 14.042 larger by the same
0 amount
y 12.04%
-2.0

10.0+%

29. 9.0

3.0

4.0

0.0

14.0

12.0

10.0

Multiple Utilities

0.0
a E

0.0

Ls

Heating utility at
the largest
temperature is
now zero.

These are the

1.
En) mm 4 minimum values
y of heating utility

4.0
00
14.0
y 14.0
2.0
y 120
2.0
vy 100

needed at each
temperature
level.