LECTURE DAY 3 Timo Laukkanen.

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Presentation transcript:

LECTURE DAY 3 Timo Laukkanen

What was important in Lecture 2 Problem Table Algorithm (PTA) Heat Cascade Grand Composite Curve (GCC) Pinch violations Stream grid Maximum Energy Recovery (MER) Network Targeting for minimum number of Units

What is important in Lecture 3 A LP (linear programming) transshipment model for minimizing utility consumption A MILP (Mixed Integer Linear Programming) extended transhipment model for minimizing the number of units

Different optimization model types

Global and Local Optima

Convex, Concave and Non-Convex functions

Optimization with a mathematical model

Sequential Heat Exchanger Synthesis Problem Data (streams, cost) DT min value Optimization of utilities consumption LP-transshipment model Minimization of number of units MILP-transshipment model Optimization of network cost NLP-superstructure Result OK? NO, new DTmin value YES, STOP

Sequential Heat Exchanger Synthesis LP Model: Minimum Utility Consumption (Transshipment Model)

Temperature Intervals Tsupply (°C) Ttarget (°C) m cp (kW/K) H1 400 120 1.0 H2 250 2.0 C1 160 1.5 C2 100 1.3 ∆Tmin = 50°C Hot Utility: Steam at 450°C Cold Utility: Water at 20°C Find the starting temperatures of all streams (and utilities) Keep one column for hot temperatures and one for cold Add ∆Tmin to get hot, substract to get cold

Hot Utility: Steam at 450°C Cold Utility: Water at 20°C TI 1 TI 2 TI 3 Tsupply (°C) Ttarget (°C) m cp (kW/K) H1 400 120 1.0 H2 250 2.0 C1 160 1.5 C2 100 1.3 ∆Tmin = 50°C Hot Utility: Steam at 450°C Cold Utility: Water at 20°C 450 400 TI 1 350 TI 2 250 200 TI 3 210 160 TI 4 150 100 TI 5 70 20 H1 250 H2 C1 C2 120 120

H1 H2 C1 C2 ∆Tmin = 50°C Hot Utility: Steam at 450°C Tsupply (°C) Ttarget (°C) m cp (kW/K) H1 400 120 1.0 H2 250 2.0 C1 160 1.5 C2 100 1.3 ∆Tmin = 50°C Hot Utility: Steam at 450°C Cold Utility: Water at 20°C 450 400 H1 H2 C1 C2 TI 1 75 350 TI 2 150 225 65 250 200 TI 3 40 80 60 52 210 160 TI 4 120 78 100 TI 5 30 70 20 H1 250 H2 C1 C2 120 120

Qs TI 1 75 C1 225 R1 60 150 TI 2 H1 R2 65 40 80 TI 3 52 H2 C2 78 120 60 R3 60 TI 4 30 R4 TI 5 Qw

R1 + 75 = Qs R2 + 225 + 65 = R1 + 150 R3 + 60 + 52 = R2 + 40 + 80 R4 + 78 = R3 + 60 + 120 Qw = R4 + 30 + 60

R1 + 75 = Qs R2 + 225 + 65 = R1 + 150 R3 + 60 + 52 = R2 + 40 + 80 R4 + 78 = R3 + 60 + 120 Qw = R4 + 30 + 60 Min Z = Qs + Qw s.t. R1 – Qs = 75 R2 - R1 = -140 R3 - R2 = 8 R4 - R3 = 102 Qw - R4 = 90 Qs, Qw, R1, R2, R3, R4 ≥ 0

General Formulation index Sets Parameters Variables

Interval k hot process cold process hot utility cold utility

Sequential Heat Exchanger Synthesis Extended LP Transshipment Model

Qs TI 1 75 C1 225 R1 60 150 TI 2 H1 R2 65 40 80 TI 3 52 H2 C2 78 120 60 R3 60 TI 4 30 R4 TI 5 Qw

C1 Q112 Q113 H1 R12 Q122 Q123 C2 Q124 R13 R14

Energy Balances

Sequential Heat Exchanger Synthesis MILP Transshipment Model Minimum number of units

Simplifactions Hot utilities treated as hot streams Cold utilities treated as cold streams Divide the problem at the pinch Energy balance for temperature intervals:

If, then Either, or