CH EN 5253 – Process Design II Heat Integration February 22, 2019

Books Product and Process Design Principles: Synthesis, Analysis and Evaluation by Warren D. Seider , J. D Seader, Daniel R. Lewin, and Soemantri Widagdo Chapter 9 ( 3rd Edition) Chemical Engineering Design: Principles, Practice and Economics of Plant and Process Design by Gavin Towler, and Ray Sinnott Section 3.5 ( 2nd Edition)

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

inequality constraints Linear Programming objective function equality constraints inequality constraints w.r.t. design variables The ND design variables, d, are adjusted to minimize f{x} while satisfying the constraints

EXAMPLE LP – GRAPHICAL SOLUTION A refinery uses two crude oils, with yields as below. Products Volumetric Yields Max. Production Crude #1 (%) Crude #2 (%) (bbl/day) Gasoline 70 31 6,000 Kerosene 6 9 2,400 Fuel Oil 24 60 12,000 The profit on processing each crude is: Crude #1: $2/bbl Crude #2: $1.4/bbl What is the optimum daily processing rate for each grade? What is the optimum if 6,000 bbl/day of gasoline is needed?

EXAMPLE LP –SOLUTION (Cont’d) Step 1. Identify the variables. Let x1 and x2 be the daily production rates of Crude #1 and Crude #2. Step 2. Select objective function. We need to maximize profit: Step 3. Develop models for process and constraints. Only constraints on the three products are given: Step 4. Simplification of model and objective function. Equality constraints are used to reduce the number of independent variables (ND = NV – NE). Here NE = 0.

EXAMPLE LP –SOLUTION (Cont’d) Step 5. Compute optimum. Inequality constraints define feasible space. Feasible Space

EXAMPLE LP –SOLUTION (Cont’d) Step 5. Compute optimum. Constant J contours are positioned to find optimum. x1 = 0, x2 = 19,355 bbl/day J = 27,097 J = 20,000 J = 10,000

EXAMPLE LP –SOLUTION (Cont’d) A refinery uses two crude oils, with yields as below. Volumetric Yields Max. Production Crude #1 Crude #2 (bbl/day) Gasoline 70 31 6,000 Kerosene 6 9 2,400 Fuel Oil 24 60 12,000 The profit on processing each crude is: $2/bbl for Crude #1 and $1.4/bbl for Crude #2. What is the optimum daily processing rate for each grade? 19,355 bbl/d What is the optimum if 6,000 bbl/day of gasoline is needed? 0.7*x1+0.31*x2=6,000, equality constraint added 0.31*19,355=6,000

Application of Linear Programming in HEN

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 Constraints Qsteam=0, R3=-50

Linear Form All values are positive

Solution Region R2 R1 R4 R3 Feasible Space Q_CW Qsteam ,Qcw , R1 , R2 , R3 ,R4 0 Constraints

Solution Solution When Qsteam is minimum, Qcw is also minimum R1 =80 R2 =82.5 R3 =0 R4 =75 Qcw =60

Heat Integrated Distillation Trains

Effect of ΔTmin on total cost ( Summary) True pinch is approached Area of heat transfer   Utility  minimum ΔTmin   Area of heat transfer  0 Utility  maximum Trade-off between Capital cost and Utility cost

Distillation Columns Energy balance F HF+ Qreb= D HD+B HB+Qcond

Distillation Columns 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 Cold stream to Condenser Close boiling point species

Heuristic “Position a Distillation Column Between Composite Heating and Cooling Curves” Difficult to use both hot and cold streams Hot utility to reboiler Cold streams to condenser

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

Heat Integration for Direct Distillation Sequence

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)

Heat Pumps How do they work? Carnot Efficiency ηmax= 1-Tc/Th Endoreversible η =1-√(Tc/Th) Same as Air Conditioner Convert low temperature heat to high temperature heat. Must add work as heat can not go up hill.

Heat Pumps in Distillation

Heat Pumps/Heat Engines Heurisitcs When positioning heat engines, to reduce the cold utilities, place them entirely above or below the pinch When positioning heat pumps, to reduce the total utilities, place them across the pinch.

Heat Pump Location

Heat Engine Location Tp