Summary for TEP 4215 E&P/PI

R S H U Summary for TEP 4215 E&P/PI Reactor System (R) Endothermic vs. Exothermic Reactions Equilibrium vs. Kinetics Temperature Dependence of Equilibrium Constants and Reaction Rates (Arrhenius) Reactors play a key Role in the Thermal and the Mechanical Energy System of a Plant (or Site) Correct Integration of Reactors Sum 1

R S H U Summary for TEP 4215 E&P/PI Reactor / Separator Interface (R/S) Focus of the Discussion was based on the Definition and Use of the following Terms: à Degree of Conversion (Extent of Reaction) à Selectivity à Yield (Reactor and Process) à Recycle Rate Y P = X S (1 + W) F F R R R F R Y R = X S R X P X S P B P Sum 2

R S H U Summary for TEP 4215 E&P/PI Separation System (S) Economical Trade-offs in Distillation Columns à Operating Cost vs. Investment Cost à Number of Stages, Reflux and Pressure Combinatorial Issues & Heuristic Rules related to the Sequence of Distillation Columns Heat Integration Opportunities between Columns (Condenser / Reboiler) Briefly about Evaporators à Multi-effect, Forward/Backward Feed, BPR Sum 3

Sequence of Distillation Columns Process, Energy and System Problem Definition by Thompson and King, AIChE Jl, 1972: Given a mixture of N chemical components that is to be separated into N pure component products by using a selection of M separation methods Number of Sequences [ N ] 2( 1)! = = N!( N 1)! N Seq Number of Alternatives = N = N M Alt Seq ( N 1) Separation Systems Sum 4

A B C D A B C D A B C D A B C D B C D B C D A B A B C A B C C D B C C D B C A B Sequence of Columns 2 comps. è 1 sequence 3 comps. è 2 sequences 4 comps. è 2x(1+3 comps.) + 1x(2+2 comps.) è 2x2 + 1 = 5 sequences 5 comps. è 2x(1+4 comps.) + 2x(2+3 comps.) è 2x5 + 2x2 = 14 sequences 6 comps. è 2x(1+5 comps.) + 2x(2+4 comps.) + 1x(3+3 comps.) è 2x14 + 2x5 + 2x2 = 42 sequences Separation Systems Sum 5

Sequence of Distillation Columns Selected Heuristic Rules Process, Energy and System H1: Favor Separation of the most Volatile Component H2: Favor near-equimolar Separation H3: Favor Separation of the most Plentiful Component H4: Favor Simple Separations << H5: Delay Separation of Sharp Splits >> Heuristics cause Conflicts, some can be Quantified, others just cast a vote, their main use is to Eliminate Sequences!! Separation Systems Sum 6

Example Distillation Sequence Process, Energy and System Comp. Name Mole Frac. α=ki/kj CES A Propane 0.05 2.00 5.26 B i-butane 0.15 1.33 8.25 C n-butane 0.25 2.40 114.50 D i-pentane 0.20 1.25 13.46 E n-pentane 0.35 Nadgir & Liu, AIChE Journal, 1983: f = min (D/B, B/D) Δ = (α 1) 100 CES = f α Separation Systems Sum 7

R S H U Summary for TEP 4215 E&P/PI Heat Recovery System (H) Targets for best Performance à Minimum Energy from the Heat Cascade à Minimum Energy Cost with Multiple Utilities from the Grand Composite Curve à Fewest Number of Units from the (N 1) Rule à Minimum Area from Spaghetti Design ( Bath ) à Total Annual Cost vs. ΔT min 3-Way Trade-off (Area, Energy and Units) Sum 8

Heat Cascade as Algorithm/Procedure (1) Process, Energy and System (0) Given Set of Hot Stream Temperatures: TH, i = 1,n H, Set of Cold Stream Temperatures: TC, j = 1,n C, and Set of mcps, i = 1,n H, j = 1,n C (1) Calculate Shadow Temperatures from Hot Streams: THS, T H,S =T H ΔT min (2) Calculate Shadow Temperatures from Cold Streams: TCS, T C,S =T C +ΔT min (3) Obtain Total Set of Hot Stream Temperatures, THT, by merging and sorting TH and TCS è Notice that dim (THT) = n H + n C (4) Obtain Total Set of Cold Stream Temperatures, TCT, by merging and sorting TC and THS è Notice that dim (TCT) = n C + n H (5) Remove possible Duplicates in THT and TCT. The number of Temperature Intervals is then K = dim (THT) 1 (6) Temperature Intervals are now obtained by using one Temperature from THT and one from TCT starting at the highest Temperatures (7) Identify Heat Flows from all the Hot Streams to the respective Temperature Intervals based on mcp values and Interval Temperatures Sum 9

Heat Cascade as Algorithm/Procedure (2) Process, Energy and System (8) Identify Heat Flows from the respective Temperature Intervals to all the Cold Streams based on mcp values and Interval Temperatures (9) Calculate the Enthalpy (Heat) Balance (Surplus or Deficit) for each Temperature Interval (10) Cascade Heat from the first Interval to the second, and from the second to the third Interval. Continue to the end of the Cascade (11) If all Residuals (i.e. Heat from one Interval to the next) are non-negative (R k 0), then no External Heating is required, Q H,min = 0, and Minimum External Cooling is obtained as the Residual from the last Interval, i.e. Q C,min = R K (12) If at least one Residual is negative, then Minimum External Heating and Cooling are: Q H,min = min ( R k ), k = 1,K-1, Q C,min = R K + Q H,min (13) The Process Pinch is the Interval Temperature with the most negative Residual which has zero heat flow after adding Minimum External Heating to the Cascade Sum 10

Example: Stream Data from Assignment 3 Process, Energy and System Stream T s ( C) T t ( C) mcp (kw/ C) ΔH (kw) H1 170 60 3.0 330 H2 150 30 1.5 180 C1 20 135 2.0 230 C2 80 140 4.0 240 ΔT min = 10 C THT = 170, 150, 145, 90, 60, 30 TCT = 160, 140, 135, 80, 50, 20 K = 6 1 = 5 Sum 11

Example: Stream Data from Assignment 3 Process, Energy and System H1 170 C 160 C 60 kw + 60 15 kw 150 C R 1 =60 140 C + 2.5 20 kw 7.5 kw 145 C R 2 =62.5 135 C C2 165 kw 220 kw - 82.5 82.5 kw H2 90 C R 110 kw 3 = 20 80 C 45 kw 60 kw + 75 90 kw 60 C R 4 =55 50 C C1 45 kw - 15 30 C R 5 =40 20 C 60 kw Sum 12

350 300 250 200 T( C) A A 2 3 0 500 1000 vertical criss cross 1 4 WS-8 cont. Vertical Design: 2 3 and 1 4 Criss-Cross Design: 2 4 and 1 3 Q(kW) 500 500 = + = 0.00909 68.05 0.00909 68.05 1616.4 m 500 500 = + = 1000 + 250 = 1250 m 0.005 100 0.05 40 2 2 Explanation: Optimal Distribution of (U ΔT) - not only ΔT Investment Cost Sum 13

R S H U Summary for TEP 4215 E&P/PI Heat Recovery System (H) Design of Network using PDM à Decomposition at Pinch (Process and Utility Pinch) à Start the Design at the Pinch à Pinch Exchangers and Requirements Ø mcp Rules: mcp out mcp in Ø Population: n out n in à Focus on ΔT, not ΔH à Tick-off Rule à Check Design against Targets!! Sum 14

R S H U Summary for TEP 4215 E&P/PI Heat Recovery System (H) Optimization of Heat Exchanger Networks à Stream Splitting (start with: α/β = mcp 1 /mcp 2 ) à Heat Load Loops and Paths The HEN Design Process as a Flow Diagram Retrofit Design of Heat Exchanger Networks à Targeting for good value of HRAT à XP Analysis (QP = QP P + QP H + QP C ) à Shifting to reduce XP Heat Transfer à UA Analysis (existing and new) followed by Loops and Paths for maximum Reuse of existing Units Sum 15

Exam 2 June 2008 Retrofit (60%) mcp = 50 kw/ C C3 60ºC Process, Energy and System H2 180ºC mcp = 80 kw/ C C2 200ºC H1 I Ca Q I = 3500 kw Q Ca = 4300 kw 130ºC 100ºC C1 40ºC Q III = 1300 kw II III Cb 50ºC Q II = 3600 kw Q Cb = 300 kw mcp = 60 kw/ C 70ºC mcp = 40 kw/ C Q H = 3600 kw H 105ºC 190ºC mcp = 20 kw/ C ΔT min = 10 C Sum 16

Exam 2008 ST Q H 200 C 190 C H2 800 kw 800 1600 kw Process, Energy and System Q Q Q H,exist C,exist H,min = 3600 kw = 4600 kw = 1000 kw H1 2800 kw 4200 kw 2400 kw 1600 kw 1200 kw R 1 180 C 170 C 200 R 2 110 C 100 C + 1200 R 3 70 C 60 C + 800 Q C 50 C 40 C 2000 kw 5600 kw 800 kw 400 kw 100 kw C1 C2 1500 kw C3 Q C,min H = 2000 kw Δ Q =Δ Q = 2600 kw C CW Simplified Cascade with Supply Temperatures only Sum 17

Exam 2008 180 H1 200 H2 II 105 190 H 145 II 130 3600 I 3600 121.67 110 Cross-Pinch Analysis 110 Ca III I 4300 C2 1300 3500 III 77.5 Cb 300 50 70 40 C1 60 C3 mcp (kw/ C) [60] [40] [20] 100 [80] [50] 100 Q XP = 50 (100 60) + 60 (121.67 110) 20 (105 100) = 2000 + 700 100 = 2600 kw Sum 18

Exam 2008 Shifting Process, Energy and System 180 H1 200 190 130 H 3600-y IV IV 0 + y T H1 I I 3500 121.67 145 C2 Ca 4300-y 50 60 C3 mcp (kw/ C) [60] [80] [50] y can be found by ΔT min requirements y = 1500 kw Next: What about Investments?? Sum 19

Exam 2008 180 H1 200 H2 105 190 130 H 2100 IV 163.75 IV 1500 145 155 II II 3600 I I 3500 110 100 C2 96.67 2800 77.5 Cb Next: UA Analysis for maximum Reuse of existing Exchangers III III 1300 Ca 300 50 70 40 C1 60 C3 mcp (kw/ C) [60] [40] [20] [80] [50] Sum 20

R S H U Summary for TEP 4215 PI Separation/Heat Recovery Interface (S/H) Columns integrated above/below Pinch à Condenser above, Reboiler below Which Pinch Columns often create Pinch à Extended Grand Composite Curve (Andrecovich) à Distinguish Columns from Background Process Evaporators and Heat Integration à The Tool is again the Grand Composite Curve à Play with Pressure and the Number of Effects Sum 21

R S H U Summary for TEP 4215 PI Heat Recovery / Utility Interface (H/U) Correct Integration of Heat Pumps (open/closed) Correct Integration of Turbines (back pressure or extraction vs. condensing turbines) Co-production of Heat & Power (cogeneration) The quantitative Tool with Information about Load (heat duty) and Level (temperature) is: à The Grand Composite Curve à Modified Temperatures are important!! Sum 22

R S H U Summary for TEP 4215 PI Utility System (U) Not treated in much Detail in this Course Topics could (or should?) have been: à Design of Steam Systems (turbines, boilers, deaerators, etc.) à Design of fired Heaters (Furnaces) with optimal preheat of Combustion Air à Design of Refrigeration Cycles including Integration with the Process ( economizers ) à Etc., etc. Sum 23

R S H U Summary for TEP 4215 PI Other Topics Optimization: Only Demo with Examples from Heat Recovery using Math Programming à Forbidden Matches & Extended Cascade is relevant Operational Aspects (especially related to Flexibility and Controllability) The Importance of Topology (Structure) Extensions of the Pinch Principle à Heat Pinch, Mass Pinch, Water Pinch and Hydrogen Pinch (whenever an amount has a quality ) Sum 24

More on the Grand Composite Curve Process, Energy and System 50 Feed C1 C2 210 210 Reactor Compressor 130 270 H1 160 160 Condenser Distillation Column 220 H2 Reboiler Product 60 Heat Integration Introduction Extra 01

H1 360 kw 720 kw 720 kw 180 kw H2 880 kw ΔT min = 20 C ST 270ºC - - - - - - - 250ºC 230ºC - - - - - - - 210ºC 200 kw 220ºC - - - - - - - 200ºC 2000 kw 180ºC - - - - - - - 160ºC 440 kw 160ºC - - - - - - - 140ºC 1980 kw 220 kw + 720-520 - 1200 + 400 + 180 70ºC - - - - - - - - 50ºC + 220 60ºC - - - - - - - - 40ºC CW 500 kw 800 kw 400 kw 1800 kw C1 Grand Composite Curve is based on the Heat C2 Cascade The necessary data are modified Temperatures and the corresponding Heat Flows Heat Integration Targeting Extra 02

ST T 0 = 260 Q H,min = 1000 + 720 T 1 = 220 R 1 = 1720-520 T 2 = 210 R 2 = 1200-1200 T 3 = 170 R 3 = 0 + 400 T 4 = 150 R 4 = 400 + 180 T 5 = 60 R 5 = 580 T' ( C) 250 200 150 100 50 Grand Composite Curve (or Heat Surplus Diagram) LP MP HP Question: Is this another Pinch? + 220 T 6 = 50 Q C,min = 800 0 CW Q (kw) CW 0 500 1500 Heat Integration Targeting Extra 03

ST Grand Composite Curve T 0 = 260 Q H,min = 1000 + 720 T' ( C) Process, Energy and System T 1 = 220 R 1 = 1720-520 T 2 = 210 R 2 = 1200-1200 T 3 = 170 R 3 = 0 + 400 T 4 = 150 R 4 = 400 + 180 T 5 = 60 R 5 = 580 250 200 150 100 50 LP MP HP Answer: No + 220 T 6 = 50 Q C,min = 800 0 CW Q (kw) CW 0 500 1500 Heat Integration Targeting Extra 04

300 250 200 150 100 New CCs based on Heat Surplus and Deficit Part of Gr.CC and balanced by Hot and Cold Utilities (not representative for Area demand) T( C) UP LP PP MP UP HP Another way of showing it is not another Process Pinch 50 0 CW 0 500 1000 1500 2000 2500 Q(kW) Heat Integration Targeting Extra 05

300 250 200 150 100 50 T( C) True Balanced Composite Curves with Utilities (Notice difference in shape and scale) UP PP UP Q(kW) 0 0 1000 2000 3000 4000 5000 6000 7000 Heat Integration Targeting Extra 06