CHE 165A L6 - HENs PDF [PDF]

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CHE 165 – Plant Design; D. Wagner



Heat Exchange Networks (HENs) Lecture 6 – Friday, September 27, 2019



1



2



Heat Integration Schematics



Simple Heat Exchange Network •



Look at a simple system: T1 t1



Cold Stream



t2



Hot Stream Steam t3



T2 T3 CW



•



How can we determine the optimal values for T2 and t2?



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Simple Heat Exchange Network •



We can plot temperature vs. duty: Qhot t3



Temperature



T1 ∆Tmin



t2



T2 T3 t1 Qcold



Qrec



Duty © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Simple Heat Exchange Network •



The maximum possible heat recovery is when the two curves “pinch” and ΔTmin = 0



TEMPERATURE



Qhot min ∆Tmin= 0



T1



t3 t2



T2 T3 t1 Qcold min



Qrec max



DUTY © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Simple Heat Exchange Network •



What happens as ∆Tmin approaches 0? – Hot utility (steam) consumption is the lowest. – Cold utility (cooling water) consumption is the lowest. – We still need three heat exchangers • • •



1 process-process exchanger. 1 process-hot utility exchanger. 1 process-cold utility exchanger.



•



What is ∆Tlm for the process-process exchanger?



•



How big is the process-process heat exchanger?



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Simple Heat Exchange Network •



We can see that changing ∆Tmin affects – Utility requirements. – Heat exchanger areas



•



How can we find an optimum ∆Tmin? – Design and cost the system for a range of ∆Tmin . • •



Determine capital costs. Determine operating costs.



– Combine capital and operating costs to determine an annualized cost – Plot annualized cost vs. ∆Tmin – Select the minimum © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Simple Heat Exchange Network ∆Tmin OPTIMIZATION 140



6



Cost (10 $/y)



120 100



∆Tmin opt



80 60 40 20 0 0



Utility Costs Total Cost



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



20



∆Tmin



40



60



Annualized Capital Cost



Chemical Engineering Design



Multi-stream Problems: Composite Curves •



How do we handle multiple streams that have temperature overlap?



•



Stream data must be combined in such a way as to represent the total energy sources and total energy demands in each temperature range.



•



The pinch method creates what is called a composite curve.



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves •



Consider a two stage reactor with reheat:



Feed



550°



Rctr #1



510°



A



550°



B



560°



Rctr #2



520° To Next Reactor



Streams A and B have overlapping duties between 520° and 550°.



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves •



Multistage reactor example - stream data Range 1 2



3



T in 510 520 520 520 550



T out 520 550 550 550 560



Streams A A B A+B B



M*Cp 1 1 1 2 1



Q 10 30 30 60 10



Plot T vs. Q for each temperature range.



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves •



Multistage reactor example - composite curve



TEMPERATURE (T)



570 560 550 540 530 520 510 500 0



20



40



60



80



100



Duty (Q)



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves •



There is an easy way to plot the composite curves: just add up the Q values over each range of T



TEMPERATURE (T)



570 560 550 540 530 520 510 500 0



20



40



60



80



100



Duty (Q)



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves for UOP Platforming Process 1000



Temperature (F)



800 600 400 200 0



0



50



100



150



200



250



Duty (MMBtu/h)



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves QH



1000



Temperature (F)



800



• We can set targets for hot and cold utilities using the composites, while paying attention to the process pinch



Pinch



600 400 200 QC 0



0



50



100



150



200



250



Duty (MMBtu/h)



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves QH



1000



Temperature (F)



800



• Since the duty scale is a difference in enthalpy, we can slide the composite curves horizontally, increasing or decreasing ∆Tmin



Pinch



600 400 200 QC 0



0



50



100



150



200



250



Duty (MMBtu/h)



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Composite Curves QH



1000



Temperature (F)



800



• If we decrease ΔTmin then our utility targets are reduced • What is the effect on capital cost though?



Pinch



600 400 200 QC 0



0



50



100



150



200



250



Duty (MMBtu/h)



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Optimization of ΔTmin



•



What happens as ∆Tmin is increased?



– More heat exchangers are required (extra cost) – Log mean temperature differences are greater • •



Each heat exchanger is smaller The cost for each heat exchanger decreases (cost savings)



– More utilities are consumed • • • •



•



Cooling water demand increases Steam demand increases Utility costs increase Note: hot utility increase = cold utility increase



How do we decide on the appropriate ∆Tmin? – Same as the two-stream problem – Plot Total Annualized Cost vs. ∆Tmin for the process © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Optimization of ΔTmin ∆Tmin OPTIMIZATION 140



6



Cost (10 $/y)



120 100



∆Tmin opt



80 60 40 20 0 0



Utility Costs Total Cost



20



∆Tmin



40



60



Annualized Capital Cost



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Energy Costs •



Energy prices are often assumed to be well known •



•



In practice, energy prices are affected by: • • • •



•



See Ch6 & lecture on operating costs



Commodity nature of fuels Fuel mix Flaring of waste products (“fuel value” vs. disposal cost) Capital cost implications of fuel substitution



So the actual energy price varies with time and is seldom properly captured



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



THE CAPITAL – ENERGY TRADEOFF



Heat Exchanger Networks (HENs) design addresses the following problem:







Given:  







Design: 







NH hot streams, each with given heat-capacity flow rate, Ch supply temperature, Ths, and target temperature Tht NC cold streams, each with given heat-capacity flow rate, Cc supply temperature, Tcs, and target temperature Tct An optimal network of heat exchangers, connecting between the hot and cold streams and between the streams and cold/hot utilities.



What is optimal? 



Implies a trade-off between CAPITAL COSTS (Cost of equipment) and ENERGY COSTS (Cost of utilities).



21



STREAM REPRESENTATION IN Q-T DIAGRAMS T



Hot stream



T



∆H



∆H



TT



TS ∆H ∆T C = ∆T TT



Cold stream



H



∆T TS



TS = Stream supply temperature (oC) TT = Stream target temperature (oC) H = Stream enthalpy (MW)  ⋅ Cp (MW/ oC) C = m = Heat-capacity flow rate (MW/ oC) = Stream flow rate × specific heat capacity



H



22



HEX REPRESENTATION IN Q-T DIAGRAMS



60o



∆Tmin = 10o 50o



∆H



10o



Enthalpy



100



60o



∆Tmin = 20o 40o



60



(a)



70o 100o



80



40 0



Temperature



100o



Temperature



100



80o



(b)



80 60 40 0



∆H



Enthalpy



20o



23



Heuristics: Heat exchangers and furnaces 24



CHE 165 – Plant Design; D. Wagner



25.



26.



Unless required as part of the design of the separator or reactor, provide necessary heat exchange for heating or cooling process fluid streams, with or without utilities, in an external shell-and-tube heat exchanger using countercurrent flow. However, if a process stream requires heating above 750∘F, use a furnace unless the process fluid is subject to chemical decomposition. Near-optimal minimum temperature approaches in heat exchangers depend on the temperature level as follows: 10∘F or less for temperatures below ambient.  20∘F for temperatures at or above ambient up to 300∘F.  50∘F for high temperatures.  250 to 350∘F in a furnace for flue gas temperature above inlet process fluid temperature. 



Adapted from SSLW



HEX REPRESENTATION IN Q-T DIAGRAMS



60o



∆Tmin = 10o 50o



∆H



10o



Enthalpy



100



60o



∆Tmin = 20o 40o



60



(a)



70o 100o



80



40 0



Temperature



100o



Temperature



100



80o



(b)



80 60 40 0



∆H



Enthalpy



20o



25



HEX REPRESENTATION IN Q-T DIAGRAMS



60o



100o



∆Tmin = 10o



Temperature



100



80o



∆H



10o



Enthalpy



60o



∆Tmin = 10o



Temperature



100



80o



50o



60 40 0



50o



100o



80



80 60 40 0



∆H



Enthalpy



10o



26



HEX REPRESENTATION IN Q-T DIAGRAMS



60o



100o



∆Tmin = 10o



Temperature



100



80o



80 60



∆H



40 0



50o



10o



Enthalpy



80o H



67.5o 100o 50o



70o



C



60o



∆Tmin = 20o



Temperature



100 80



20o



60 QH



40 0



∆H - QH



Enthalpy



QC



27



DEFINITIONS



70o



Exchanger Duty. Data: Hot stream CH = 0.3 MW/ oC Cold stream CC = 0.4 MW/ oC



60o



100o 40o



∆Tmin = 20o



Check: T1 = 40 + (100 - 60)(0.3/0.4) = 70oC  Q = 0.4(70 - 40) = 0.3(100 - 60) = 12 MW Heat Transfer Area (A): A = Q/(U⋅∆Tlm) Data: Overall heat transfer coefficient, U=1.7 kW/m2 oC (Alternative formulation in terms of film coefficients) ∆Tlm = (30 - 20)/loge(30/20) = 24.66 So, A = Q/(U⋅∆Tlm) = 12000/(1.7×24.66) = 286.2 m2



28



EXAMPLE PROBLEM ∆ H = 160



o



60 C ∆ H = 100 o



80 C 180 oC C-1 R-1 100 oC o



130 C



120 oC



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC. Utilities. Steam@150 oC, CW@25oC Design a network of steam heaters, water coolers and exchangers for the process streams. Where possible, use exchangers in preference to utilities.



∆ H = 162 ∆ H = 180



40 oC



TS Stream (oC) H1 180 H2 130 C1 60 C2 30



30 oC



29



POSSIBLE SOLUTION TO EXAMPLE PROBLEM 60 oC



Summary of proposed design:



100



80 oC



180 oC



H



C-1



60



100 oC 130 oC



R-1



120 oC



Steam



CW



Units



60 kW



18 kW



4



Are 60 kW of Steam Necessary?



162



C 40 oC



18 30 oC



30



MER TARGETING WITH COMPOSITE CURVE Example:



TS Stream (oC) H1 180 130 H2 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC. (oC)



H2



∆T



H1



Tend (oC)



C2



Tstart (oC)



C1



Interval



CP



(kW/oC)



∆H



(kW)



31



MER TARGETING WITH COMPOSITE CURVE Example:



TS Stream (oC) H1 180 130 H2 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC.



Cold-1



120



100



20



H2



(oC)



H1



Tend (oC)



C2



Tstart (oC)



C1



∆T



Interval



CP



∆H



(kW/oC)



(kW)



1.8



36



32



MER TARGETING WITH COMPOSITE CURVE Example:



TS Stream (oC) H1 180 130 H2 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC.



Cold-1



120



100



Cold-2



100



60



H2



(oC)



H1



Tend (oC)



C2



Tstart (oC)



C1



∆T



Interval



CP



∆H



(kW/oC)



(kW)



20



1.8



36



40



5.8



232



33



MER TARGETING WITH COMPOSITE CURVE Example:



TS Stream (oC) H1 180 130 H2 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC.



Cold-1



120



100



Cold-2



100



Cold-3



60



H2



(oC)



H1



Tend (oC)



C2



Tstart (oC)



C1



∆T



Interval



CP



∆H



(kW/oC)



(kW)



20



1.8



36



60



40



5.8



232



30



30



1.8



54



34



MER TARGETING WITH COMPOSITE CURVE Example:



TS Stream (oC) H1 180 130 H2 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC.



Cold-1



120



100



Cold-2



100



Cold-3 Hot-1



H2



(oC)



H1



Tend (oC)



C2



Tstart (oC)



C1



∆T



Interval



CP



∆H



(kW/oC)



(kW)



20



1.8



36



60



40



5.8



232



60



30



30



1.8



54



180



130



50



1.0



50 35



MER TARGETING WITH COMPOSITE CURVE Example:



TS Stream (oC) H1 180 130 H2 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC.



Cold-1



120



100



Cold-2



100



Cold-3



H2



(oC)



H1



Tend (oC)



C2



Tstart (oC)



C1



∆T



Interval



CP



∆H



(kW/oC)



(kW)



20



1.8



36



60



40



5.8



232



60



30



30



1.8



54



Hot-1



180



130



50



1.0



50



Hot-2



130



80



50



3.0



150



36



MER TARGETING WITH COMPOSITE CURVE Example:



TS Stream (oC) H1 180 130 H2 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10 oC.



Cold-1



120



100



Cold-2



100



Cold-3



H2



(oC)



H1



Tend (oC)



C2



Tstart (oC)



C1



∆T



Interval



CP



∆H



(kW/oC)



(kW)



20



1.8



36



60



40



5.8



232



60



30



30



1.8



54



Hot-1



180



130



50



1.0



50



Hot-2



130



80



50



3.0



150



Hot-3



80



40



40



2.0



80



37



180 160



Temperature (oC)



140 120



Int



Tstart (oC)



Tend (oC)



∆T (oC)



CP (kW/oC)



∆H (kW)



Cold-1



120



100



20



1.8



36



Cold-2



100



60



40



5.8



232



Cold-3



60



30



30



1.8



54



Hot-1



180



130



50



1.0



50



Hot-2



130



80



50



3.0



150



Hot-3



80



40



40



2.0



80



100 80 60 40 20 0



0



50



100



150



200



Enthalpy (kW)



250



300



350



38



180 160



Temperature (oC)



140 120 100 80 60 40 20 0



0



50



100



150



200



Enthalpy (kW)



250



300



350



39



30 kW 180 160



Temperature (oC)



140 120 100



“Pinch”



80 60 40 20 0



0



50



100



150



200



Enthalpy (kW)



250



300



350



40



100 kW



180 160



Temperature (oC)



140



∆Tmin = 30o



120 100 80 60 40 20 0



0



50



100



150



200



Enthalpy (kW)



250



300



350



41



MER Targeting (∆Tmin = 20oC): Hot pinch temperature = 80oC Cold pinch temperature = 60oC QH,min = 54 kW QC,min = 12 kW



54 kW



180 160



Temperature (oC)



140 120



∆Tmin = 20o



100 80 60 40 20 0



12 kW 0



50



100



150



200



Enthalpy (kW)



250



300



350



42



180



MER Targeting (∆Tmin = 10oC): Hot pinch temperature = 70oC Cold pinch temperature = 60oC QH,min = 48 kW QC,min = 6 kW



48 kW



160



Temperature (oC)



140 120



∆Tmin = 10o



100 80 60 40 20 0



6 kW 0



50



100



150



200



Enthalpy (kW)



250



300



350



43



MER TARGETING USING THE TI METHOD The temperature-interval method (Linnhoff and Flower, 1978), a systematic procedure for determining the minimum utility requirements over all possible HENs, consists of the following steps:  Adjusting the hot and cold stream temperatures to bring them to the same reference.  Determining the temperature intervals and carrying out enthalpy balances in each interval.  Computing the enthalpy cascade, the residual enthalpy flows, determining the location of the "pinch," and computing MER targets. 44



MER TARGETING WITH TI : EXAMPLE PROBLEM ∆ H = 160



o



60 C ∆ H = 100 o



80 C 180 oC C-1 R-1 100 oC 130 oC



120 oC



TS Stream (oC) H1 180 H2 130 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10oC. Utilities. Steam@150oC, CW@25oC



∆ H = 162



Compute the MER targets for this process using the TI Method.



∆ H = 180



40 oC



30 oC



45



MER TARGETING WITH TI : STEP 1  Adjusting the hot and cold stream temperatures to bring them to the same reference. o Need to adjust temperatures so that both hot and cold streams are on the same terms of reference for a given desired ∆Tmin. Arbitrarily, this is accomplished by subtracting ∆Tmin from all hot stream temperatures.



Original stream table TS Stream (oC) H1 180 H2 130 C1 60 C2 30



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



Adjusted stream table TS Stream (oC) H1 170 H2 120 C1 60 C2 30



TT (oC) 70 30 100 120 46



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



1 2 3 4 5



47



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



1



170 – 120 = 50



2



120 – 100 = 20



3



100 – 70 = 30



4



70 – 60 = 10



5



60 – 30 = 30



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



TT (oC) 70 30 100 120



C2



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



48



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



1



170 – 120 = 50



2



120 – 100 = 20



3



100 – 70 = 30



4



70 – 60 = 10



5



60 – 30 = 30



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



49



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



1



170 – 120 = 50



2



120 – 100 = 20



3



100 – 70 = 30



4



70 – 60 = 10



5



60 – 30 = 30



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



50



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



1



170 – 120 = 50



2



120 – 100 = 20



3



100 – 70 = 30



4



70 – 60 = 10



5



60 – 30 = 30



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



51



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



1



170 – 120 = 50



2



120 – 100 = 20



3



100 – 70 = 30



4



70 – 60 = 10



5



60 – 30 = 30



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



52



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



1



170 – 120 = 50



2



120 – 100 = 20



3



100 – 70 = 30



4



70 – 60 = 10



5



60 – 30 = 30



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC) 1



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW) 50



53



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



1



170 – 120 = 50



1



50



2



120 – 100 = 20



1 + 2 –1.8 = 1.2



24



3



100 – 70 = 30



4



70 – 60 = 10



5



60 – 30 = 30



54



MER TARGETING WITH TI : STEP 2  Determining the temperature intervals and carrying out enthalpy balances in each interval. Interval



Ti-1 –Ti (oC)



H1 H2 C1



TS Stream (oC) H1 170 H2 120 C1 60 C2 30



C2



TT (oC) 70 30 100 120



∆H (kW) 100 180 160 162



ΣCh –ΣCc (kW/oC)



C (kW/oC) 1.0 2.0 4.0 1.8



∆H (kW)



1



170 – 120 = 50



1



50



2



120 – 100 = 20



1 + 2 –1.8 = 1.2



24



3



100 – 70 = 30



1 + 2 –1.8 –4 = –2.8



–84



4



70 – 60 = 10



2 –1.8 –4 = –3.8



–38



5



60 – 30 = 30



2 –1.8 = 0.2



6



55



MER TARGETING WITH TI : STEP 3 Energy Flows Between  Computing the enthalpy cascade, the residual enthalpy flows, determining the location of the "pinch," and computing MER targets.



Intervals (kW)



First Pass T0 = 170 oC



QH



R0 = 0



50 R1 = 50



T1 = 120 oC 24



R2 =74



T2 = 100 oC -84



R3 =-10



T3 = 70 oC -38



R4 =-48



T4 = 60 oC 6



T5 = 30 oC



QC



56 R5 =-42



MER TARGETING WITH TI : STEP 3 Energy Flows Between  Computing the enthalpy cascade, the residual enthalpy flows, determining the location of the "pinch," and computing MER targets.



Intervals (kW)



T0 = 170 oC



QH



First Pass



Final Pass



R0 = 0



48



R1 = 50



98



R2 =74



122



50 T1 = 120 oC 24 T2 = 100 oC -84 R3 =-10



T3 = 70 oC



MER Targeting: Cold pinch temp. = 60oC



QHmin



38



-38 R4 =-48



T4 = 60 oC



Hot pinch temp. = 70oC



No heat transfers across the pinch



0



6



57



QH,min = 48 kW QC,min = 6 kW



T5 = 30 oC



QC



R5 =-42



6



QCmin



PINCH



DECOMPOSITION



58



THE SIGNIFICANCE OF THE PINCH QHmin T



QHmin Heat sink Pinch



Pinch Heat source



QCmin



QCmin



The “pinch” separates the HEN problem into two parts:



Heat sink - above the pinch, where at least QHmin utility must be used.  Heat source - below the pinch, where at least QCmin utility must be used.  At MER no heat passes across the pinch.



H







59



HEN DESIGN RULES FOR MER  MER Targeting. Define pinch temperatures, QHmin and QCmin.  Divide problem at the pinch.  Design hot-end, starting at the pinch: Immediately above the pinch, pair up streams such that: CH ≤ CC. “Tick off” streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch.  Design cold-end, starting at the pinch: Immediately below the pinch, pair up streams such that: CH ≥ CC. “Tick off” streams to minimize costs. Add cooling utilities as needed (up to QCmin). Do not use hot utilities below the pinch.  Done!



60



EXAMPLE MER SOLUTION: PROBLEM STATEMENT ∆ H = 160



o



60 C ∆ H = 100 o



80 C 180 oC C-1 R-1 100 oC o



130 C



120 oC



TT (oC) 80 40 100 120



∆H (kW) 100 180 160 162



C (kW/oC) 1.0 2.0 4.0 1.8



∆Tmin = 10oC. Utilities. Steam@150oC, CW@25oC Design a network of steam heaters, water coolers and exchangers for the process streams. Where possible, use exchangers in preference to utilities.



∆ H = 162 ∆ H = 180



40 oC



TS Stream (oC) H1 180 H2 130 C1 60 C2 30



30 oC



61



EXAMPLE MER SOLUTION – STEP 1  MER Targeting.



H1



H2



180oC



80oC



40oC



130oC



100oC



60oC



o



30oC



120 C



C (kW/ oC) 1.0



2.0



C1



4.0



C2



1.8



62



EXAMPLE MER SOLUTION – STEP 1  MER Targeting. For ∆Tmin = 10oC: Pinch temperatures 70oC and 60oC, QH,min = 48 kW and QC,min = 6 kW



H1



H2



180oC



80oC



40oC



130oC



100oC



60oC



o



30oC



120 C



QH,min = 48 kW



C (kW/ oC) 1.0



2.0



C1



4.0



C2



1.8



QC,min = 6 kW



63



EXAMPLE MER SOLUTION – STEP 2  Divide problem at the pinch.



H1



H2



180oC



130oC



100oC



o



120 C



QH,min = 48 kW



C (kW/ oC)



80oC



1.0



70oC



40oC



60oC



C1



4.0



C2



1.8



60oC



30oC



QC,min = 6 kW



2.0



64



On hot @ pinch: EXAMPLE MER SOLUTION – side STEP 3  Design hot-end, starting at the pinch: Immediately above Pairing rule is ≤ min CC. “Tick off” the pinch, pair CH ≤up CCstreams such Th,out –that: Tc,in C=H∆T streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch.



pinch



H1



H2



80oC



180oC



130oC



1



100oC



o



120 C



QH,min = 48 kW



C (kW/ oC)



1



1.0



70oC



40oC



60oC



C1



4.0



C2



1.8



60oC



30oC



QC,min = 6 kW



2.0



65



On hot @ pinch: EXAMPLE MER SOLUTION – side STEP 3  Design hot-end, starting at the pinch: Immediately above Pairing rule is ≤ min CC. “Tick off” the pinch, pair CH ≤up CCstreams such Th,out –that: Tc,in C=H∆T streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch.



pinch



H1



H2



80oC



180oC



130oC



100oC



C (kW/ oC)







1



1



1.0



70oC



40oC



60oC



C1



4.0



C2



1.8



2.0



120 o



120 C



QH,min = 48 kW



60oC



30oC



QC,min = 6 kW



66



On hot @ pinch: EXAMPLE MER SOLUTION – side STEP 3  Design hot-end, starting at the pinch: Immediately above Pairing rule is ≤ min CC. “Tick off” the pinch, pair CH ≤up CCstreams such Th,out –that: Tc,in C=H∆T streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch.



pinch



H1



H2



180oC







80oC



2







130oC



C (kW/ oC)



100oC



1



1



1.0



70oC



40oC



60oC



C1



4.0



C2



1.8



2.0



120 o



120 C



QH,min = 48 kW



2



100



60oC



30oC



QC,min = 6 kW



67



EXAMPLE MER SOLUTION – STEP 3  Design hot-end, starting at the pinch: Immediately above the pinch, pair up streams such that: CH ≤ CC. “Tick off” streams to minimize costs. Add heating utilities as needed (up to QHmin). Do not use cold utilities above the pinch. H1



H2







180oC



100oC



o



120 C



80oC



2







130oC







1



H



1



40



120







H



2



8



100







QH,min = 48 kW



C (kW/ oC) 1.0



70oC



40oC



60oC



C1



4.0



C2



1.8



60oC



30oC



QC,min = 6 kW



2.0



68



EXAMPLE MER SOLUTION – Sside TEP@4pinch: On cold  Design cold-end, starting at the pinch: Immediately below up streams such Pairing that: Crule the pinch,Tpair isC. “Tick off” H ≥ C h,in – Tc,out = ∆Tmin streams to minimize costs. Add cooling as needed CH ≥ utilities CC hot utilities below the pinch. (up to QCmin). Do not use pinch H1



H2







180oC



100oC



o



120 C



80oC



2







130oC







1



H



1



40



120







H



2



8



100







QH,min = 48 kW



C (kW/ oC) 1.0



70oC



40oC



3



60oC



60oC



3







54



30oC



2.0



C1



4.0



C2



1.8



QC,min = 6 kW



69



EXAMPLE MER SOLUTION – STEP 4  Design cold-end, starting at the pinch: Immediately below the pinch, pair up streams such that: CH ≥ CC. “Tick off” streams to minimize costs. Add cooling utilities as needed (up to QCmin). Do not use hot utilities below the pinch. H1



H2







180oC



80oC



2







130oC



C (kW/ oC)



1



1.0



70oC



3



C







40oC



2.0



6 100oC



o



120 C







H



1



40



120







H



2



8



100







QH,min = 48 kW



60oC



60oC



3







54



30oC



C1



4.0



C2



1.8







QC,min = 6 kW



70



PREVIOUS SOLUTION TO EXAMPLE PROBLEM 60 oC



Summary of previous design:



100



80 oC



180 oC



H



C-1



60



100 oC 130 oC



R-1



120 oC



Steam



CW



Units



60 kW



18 kW



4



Are 60 kW of Steam Necessary?



162



C 40 oC



18 30 oC



71



MER SOLUTION TO EXAMPLE PROBLEM 60 oC



Summary of MER design:



100



80 oC



180 oC



H



C-1



60



100 oC 130 oC



R-1



Steam



CW



Units



48 kW



6 kW



6



120 oC



162



C 40 oC



18 30 oC



72



MER SOLUTION TO EXAMPLE PROBLEM 60 oC 100



80 oC



180 oC



120



C-1 R-1



H



Summary of MER design: Steam



CW



Units



48 kW



6 kW



6



40



100 oC



120 oC



o



130 C



H



8



54



C 6



40 oC



73



30 oC



Capital Targets •



How do we get the capital cost without designing the heat exchange network (HEN)?



Temperature



Qhot



Qcold



Qrec Duty



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Thermodynamic Significance of the Pinch “pinch” Temperature



Qhot min



Qrec max Qcold min Duty



•



When the process is pinched it is decomposed into two sub problems



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Pinch Decomposition “pinch” Temperature



Qhot min



Qrec max Qcold min



Above the pinch we only put in utility heat and the process acts as a heat sink



Below the pinch we only reject heat to cold utility and the process acts as a heat source



Duty



•



When the process is pinched it is decomposed into two sub problems



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Pinch Decomposition •



What if we put in extra heat above the pinch? Qextra “pinch” Temperature



Qhot min



Qextra



Heat sink is now out of energy balance and we have to reject Qextra to a lower temperature



Qrec max Qcold min Duty © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Pinch Decomposition •



What if we put in extra heat above the pinch? Qextra “pinch” Temperature



Qhot min



Qextra



Qextra Qrec max Qcold min



Now the heat source is also out of energy balance and we have to reject Qextra to cold utility



Duty © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Pinch Decomposition The overall effect is that both hot and cold utility are increased by the amount of heat transferred across the pinch = Qextra Qextra “pinch” Temperature



Qhot min



Qextra



Qextra Qcold min



Qrec max Duty



So a simple rule for achieving energy targets is don’t transfer heat across the pinch! © 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Pinch Design Method •



Use composite curves to find the pinch temperature



•



Optimize ∆Tmin, making process changes



•



Decompose the network design problem using the pinch temperatures



•



Design the Heat Exchanger Network (HEN) for each subsystem separately: – Transfer no heat “across” the pinch – ∆T > ∆Tmin for all heat exchangers – Start at the pinch and work outward



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design



Pinch Analysis: Overall Conclusions • The pinch techniques can give useful insights into process energy efficiency • The main value lies in identifying good process modifications and choosing the right utilities • It is seldom (if ever) necessary to work through all of the details of the analysis – most benefit is gained with a high-level view • Different techniques are used for retrofit of existing networks and these can have more value, particularly in petrochemicals



© 2012 G.P. Towler / UOP. For educational use in conjunction with Towler & Sinnott Chemical Engineering Design only. Do not copy



Chemical Engineering Design