PAR 1 PINCH AND MINIMUM UILIY USAGE
EMPERAURE-ENHALPY (-H) DIAGRAMS Assume one heat exchanger. hese are alternative representations H,in C,in Q C,out H,in C,out H, out H,in H, out H, out C,in Slopes are the inverse of F*Cp. (Recall that Q=F Cp Δ) C,out C,in Q Q
-H DIAGRAMS Assume one heat exchanger and a heater H,in C,in Q Q H C,out H,in C,out H,in H, out H, out H, out C,in C,out H C,in Q Q H Q H Q
-H DIAGRAMS Assume one heat exchanger and a cooler H,in C,in Q C,out H,in C,out Q C H, out H, out H,in C H, out C,in C,out Q C C,in Q C Q Q
-H DIAGRAMS wo hot-one cold stream H1,in H2,in C,in C,out H2,in Q 1 Q 2 H1,in C,out H2,out H2,out H2,out H1,in H2,in H1,out H2, out H1,out C,in C,out C,in Q 1 Q 2 Q 2 Q 1 Notice the vertical arrangement of heat transfer
Streams under phase change Phase change Slope change Single component Liquid Multicomponent We say this stream has variable Cp
Piece-wise linear representation
Composite Curves (-H DIAGRAMS) Obtained by lumping all the heat from different streams that are at the same interval of temperature. Remark: By constructing the composite curve we loose information on the vertical arrangement of heat transfer between streams
Composite Curves (-H DIAGRAMS) Moving composite curves horizontally Cooling Heating Smallest Δ Smallest Δ H1,in H2,in C,in H1,in H2,in Q 1 Q 2 C,in Q 1 Q 2 Q H C,out C,out Q C H2,out H1,out H2,out H1,out
Composite Curves (-H DIAGRAMS) Smallest Δ C,in Cooling Heating H1,in H2,in Q 1 Q 2 Q C H2,out H1,out Q H C,out Moving the cold composite stream to the right Increases heating and cooling BY EXACLY HE SAME AMOUN Increases the smallest Δ Decreases the area needed A=Q/(U* Δ ) Notice that for this simple example the smallest Δ takes place in the end of the cold stream
Composite Curves (-H DIAGRAMS) Cooling Heating In general, the smallest Δ can take place anywhere. We call the temperature at which this takes place HE PINCH.
Composite Curves (-H DIAGRAMS) Cooling Heating From the energy point of view it is then convenient to move the cold stream to the left. However, the area may become too large. o limit the area, we introduce a minimum approach Δ min Δ min is also known as HRA (Heat Recovery Approximation emperature)
GRAPHICAL PROCEDURE Fix Δ min (HRA) Draw the hot composite curve and leave it fixed Draw the cold composite curve in such a way that the smallest temperature difference is equal to Δ min he temperature at which Δ=Δ min is the PINCH he non-overlap on the right is the Minimum Heating Utility and the non-overlap on the left is the Minimum Cooling Utility
EXAMPLE =27 MW =-30 MW =140 0 C =230 0 C REACOR 2 =200 0 C =80 0 C =32 MW =-31.5 MW =20 0 C REACOR 1 =180 0 C =250 0 C =40 0 C Stream ype Supply arget F*Cp ( o C) ( o C) (MW) (MW o C -1 ) Reactor 1 feed Cold 20 180 32.0 0.2 Reactor 1 product Hot 250 40-31.5 0.15 Reactor 2 feed Cold 140 230 27.0 0.3 Reactor 2 product Hot 200 80-30.0 0.25 Δ min =10 o C
Hot Composite Curve 250 200 250 200 FCp=0.15 80 40 80 40 FCp=0.15 31.5 30 6 48 7.5
Cold Composite Curve 230 230 180 180 140 140 20 20 32 27 24 20 15
Pinch Diagram Pinch 250 230 200 180 140 Δ= Δ min he pinch is defined either as - he cold temperature (140 o ) - he corresponding hot temp (140 o +Δ min =150 o ) - he average (145 o ) 80 40 20 10 51.5 7.5 Observation: he pinch is at the beginning of a cold stream or at the beginning of a hot stream
UILIY COS vs. Δ min COS Utility OAL OVERLAP Δ min PARIAL OVERLAP Note: here is a particular overlap that requires only cooling utility
Special Overlap Cases Overlap leads only to cooling utility Different instances where the cold stream overlaps totally the hot stream. Case where only heating utility OAL OVERLAP We prefer this arrangement even if Δ>Δ min PARIAL OVERLAP
SUMMARY he pinch point is a temperature. ypically, it divides the temperature range into two regions. Heating utility can be used only above the pinch and cooling utility only below it.
PROBLEM ABLE Composite curves are inconvenient. hus a method based on tables was developed. SEPS: 1. Divide the temperature range into intervals and shift the cold temperature scale 2. Make a heat balance in each interval 3. Cascade the heat surplus/deficit through the intervals. 4. Add heat so that no deficit is cascaded
PROBLEM ABLE We now explain each step in detail using our example Stream ype Supply arget F*Cp ( o C) ( o C) (MW) (MW o C -1 ) Reactor 1 feed Cold 20 180 32.0 0.2 Reactor 1 product Hot 250 40-31.5 0.15 Reactor 2 feed Cold 140 230 27.0 0.3 Reactor 2 product Hot 200 80-30.0 0.25 Δ min =10 o C
PROBLEM ABLE 1. Divide the temperature range into intervals and shift the cold temperature scale 250 230 200 180 140 250 240 200 190 150 80 80 40 20 40 30 Hot streams Cold streams Hot streams Cold streams Now one can make heat balances in each interval. Heat transfer within each interval is feasible.
PROBLEM ABLE 2. Make a heat balance in each interval. 250 240 F Cp=0.15 Δ interval interval Surplus/Deficit 10 1.5 Surplus 200 190 F Cp=0.25 40-6.0 Deficit 10 1.0 Surplus 150 F Cp=0.3 40-4.0 Deficit 70 14.0 Surplus 80 40 30 Hot streams F Cp=0.2 Cold streams 40 Deficit 10-2.0 Deficit
PROBLEM ABLE 250 240 200 190 150 80 3. Cascade the heat surplus through the intervals. hat is, we transfer to the intervals below every surplus/deficit. 1.5-6.0 1.0-4.0 14.0 his interval has a surplus. It should transfer 1.5 to interval 2. his interval has a deficit. After using the 1.5 cascaded it transfers 4.5 to interval 3. 1.5-6.0 1.0-4.0 14.0 1.5-4.5-3.5-7.5 6.5 he largest deficit transferred is -7.5. hus, 7.5 MW of heat need to be added on top to prevent any deficit to be transferred to lower intervals 40-2.0 4.5 30-2.0 2.5
PROBLEM ABLE 250 240 200 190 150 80 40 30 4. Add heat so that no deficit is cascaded. 7.5 250 1.5 1.5 240 1.5 9.0-6.0-6.0 200-4.5 3.0 1.0 1.0 190-3.5 4.0-4.0-4.0 150-7.5 0.0 14.0 14.0 80 6.5 14.0 40 4.5 12.0 30 2.5 10.0 his is the minimum heating utility his is the position of the pinch his is the minimum cooling utility
If the heating utility is increased beyond 7.5 MW the cooling utility will increase by the same amount 250 240 200 190 150 80 40 30 1.5-6.0 1.0-4.0 14.0 7.5 9.0 3.0 4.0 0.0 14.0 12.0 10.0 PROBLEM ABLE 1.5-6.0 1.0-4.0 14.0 7.5 + λ 9.0 + λ 3. 0 + λ 4. 0 + λ 0. 0 + λ 14. 0 + λ 12. 0 + λ 10. 0 + λ Heating utility is larger than the minimum Heat is transferred across the pinch Cooling utility is larger by the same amount
IMPORAN CONCLUSION DO NO RANSFER HEA ACROSS HE PINCH HIS IS A GOLDEN RULE OF PINCH ECHNOLOGY. WHEN HIS HAPPENS IN BADLY INEGRAED PLANS HERE ARE HEA EXCHANGERS WHERE SUCH RANSFER ACROSS HE PINCH AKES PLACE 7.5 + λ 1.5 9.0 + λ - 6.0 3. 0 + λ 1.0 4. 0 + λ -4.0 0. 0 + λ 14.0 14. 0 + λ 12. 0 + λ 10. 0 + λ Heating utility is larger than the minimum Heat is transferred across the pinch Cooling utility is larger by the same amount
7.5 1.5 9.0-6.0 3.0 1.0 4.0-4.0 0.0 14.0 14.0 12.0 10.0 Multiple Utilities 0.0 1.5 1.5 + 4.5-6.0 0.0 1.0 1.0 + 3.0-4.0 0.0 14.0 14.0 12.0 10.0 Heating utility at the largest temperature is now zero. hese are the minimum values of heating utility needed at each temperature level.