CH EN 5253 – Process Design II Heat Integration, Part 2 February 16, 2018

Books Product and Process Design Principles: Synthesis, Analysis and Evaluation by J. D Seader, Warren D. Seider and Daniel R. Lewin Chapter 9 Chemical Engineering Design: Principles, Practice and Economics of Plant and Process Design by Gavin Towler, and Ray Sinnott Section 3.5

Minimum Utility Target Temperature Interval (TI) Method Graphical Method: Composite heating and cooling curves Linear Programming (LP) method

Linear Programming (LP) method Qsteam - R1 + 30 =0 ….. (LP.1) R1 - R2+ 2.5 = 0 ….. (LP.2) R2 - R3 - 82.5 = 0 ….. (LP.3) R3-R4+75=0 ….. (LP.4) R4-Qcw-15=0 ….. (LP.5) Qsteam ,Qcw , R1 , R2 , R3 ,R4 0 …..(LP.6) Qsteam=0, R3=-50

Linear Form All values are positive When Qsteam is minimum, Qcw is also minimum Solution 1 Qsteam =0 R1 =30 R2 =32.5 R3 =-50 R4 =25 Qcw =10 Solution 2 Qsteam =50 R1 =80 R2 =82.5 R3 =0 R4 =75 Qcw =60 Solution 3 Qsteam =100 R1 =130 R2 =132.5 R3 =50 R4 =125 Qcw =110

Comparison ( Minimum Utilities) Simple HEN HEN with Min. Utilities HE = 6 ( 3 Interior + 3 Auxiliary) HE = 7 ( 4 Interior + 3 Auxiliary) Saves CW 7.5e4 BTU/hr Steam 7.5e4 BTU/hr

Solution to reduce number of HE NHx,min= Ns + NU – NNW Break Heat Exchanger Loops Stream Splitting 4+2-2=4

Solution 1: Break Heat Exchanger Loops Simplest Change Eliminate HE 1 Transfer Heat duty to HE 2 Heat duty C1 stream + H @Heater C2 stream - H @ Heater

Solution 1: Break Heat Exchanger Loops Attack small Heat Exchangers First Cp=K an where 1>n  0.6 Case 1: small HE = 20 m2, large HE = 80 m2 Cp = K 200.6+K 800.6 = 19.9 K Case 2: Single HE= 100 m2 ( combined area) Cp= K 1000.6 =15.8 K

Solution 2: Stream Splitting Two streams created from one In principle, splitting of heat capacity mass flowrate, mCp Example: Design minimum number of HE ΔTmin =10 OC MER target for hot utility : 300 KW

No Splitting HEN NHx,min= Ns + NU – NNW =3+1-1=3 Simple Network without splitting Where is the problem?

Violation of 2nd law of thermodynamics Cold Stream, C1 Hot Stream, H2

Stream Splitting X (T1-90)=500 (10-x) (T2-90)=200 and 200-T1  ΔTmin i.e., T1  190 150-T2  ΔTmin i.e., T2  140 T2= 140 T1= 173.3 x= 4 T2= 135 T1= 180 x= 5.56 T2= 130 T1= 190 x= 5 Many Solutions

Hot Stream, H2 Cold Stream, C1

Stream Splitting: Solution OC OC Non-isothermal mixing == maximize lost work

Improved design: Minimization of Lost work Isothermal mixing Location of Heater moved for isothermal mixing Mixing split streams isothermally minimizes lost work

Optimization of HEN CP=K(Area)0.6

Effect of ΔTmin on total cost True pinch is approached Area of heat transfer  ∞ Utility  minimum ΔTmin  ∞ Area of heat transfer  0 Utility  maximum Tradeoff between Capital cost and Utility cost

ΔTmin > ΔTthres Threshold ΔTmin If ΔTmin is such that no pinch exists Either hot or cold utility to be used, not both ΔTthres = minimum ΔTmin below no pinch exists A chemical company wants to design a HEN with a pinch so that both hot/cold can be used ΔTmin > ΔTthres

Threshold ΔTmin ΔTmin=10C ΔTmin=105C Case 1 Case 2 S T(C) T(C) C Q(kW) Qsteam, min = 0 Kw Qcw, min = 46 Kw Case 2 Qsteam, min = 6 Kw Qcw, min = 52 Kw

Sensitivity of ΔTmin CP=K(Area)0.6 Area=Q/(UF ΔTmin) ΔT =UA/Q Utilities increase due to Lost work since it increases as ΔT increases Tradeoff between Capital cost and Utility cost

Optimization based on total cost Minimum Capital cost goes down when A is less. This is caused by delta T being larger for Q to remain the same.

Distillation Columns F HF+ Qreb= D HD+B HB+Qcond Q  Qreb  Qcond F HF-D HD-B HB 0 If Q is reduced Utility cost reduced number of trays / height of packing increased Tradeoff between operating cost and capital cost

Heuristic “Position a Distillation Column Between Composite Heating and Cooling Curves” When utility cost is high Adjust pressure level to position T-Q between hot and cold composite curves Hot streams to reboiler Condenser to cold streams Difficult to position When utility cost is high and separating close boiling points Hot utility to reboiler Condenser to cold streams Difficult to position

Multi-effect Distillation When position a Distillation Column Between Composite Heating and Cooling Curves not possible Feed split Two towers operating at different pressures

Adjust Pressure in C2 for ΔTmin Good when utility cost high Down side Purchase cost for two towers pump cost process complexity

Variation on two-effect distillation Feed Splitting (FS) Light Splitting/ forward heat-integration (LSF) Light Splitting/ reverse heat-integration (LSR)