Composite Curve (Smith, R., 2005) Two-stream heat recovery problem

  

VIII

HEAT INTEGRATION

Dr. Eng. Yulius Deddy Hermawan Department of Chemical Engineering UPN “Veteran” Yogyakarta

  

Outline

  1. Heat Exchanger Network

  2. Reactor Heat Integration

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

VIII.1

HEAT EXCHANGER NETWORK

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

Composite Curve (Smith, R., 2005)

Two-stream heat recovery problem Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

T-H Diagram

160 150 140 130 120 110 100

  90

  80

  70

  60

  50

  40 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

D

H (MW)

  T ( o

  C)

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

A simple recovery problem with one hot stream and one cold stream Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

HE stream data for two hot streams and two cold streams

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

A simple flowsheet with two hot streams and two cold streams

  

The hot streams can be combined to obtain a composite curve

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

The cold streams can be combined to obtain a composite curve

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  

Plotting the hot and cold composite curves together allows the

targets for hot and cold utility to be obtained

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  Increasing DT min from 10 o

  C to 20 o

  C

increase the hot and cold utility targets o

  • 1
    • (
    • (

CP (MW.K

  80

  27.0 0.30 145 235 Reactor 2 product Hot 200 80 -30.0 0.25 195

  75 Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

The Problem Table Algorithm: Stream Population

Interval temp.

  245 250

235 240 230

195 200 190 200 185 180 190 180 190 145 140 150 140 150

  75

  70

  80

  30

  35

  0.20 25 185 Reactor 1 product Hot 250 40 -31.5 0.15 245

  40

  25

  1

  3

  

2

  4

  35 Reactor 2 feed Cold 140 230

  32.0

  Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

  T

  

Shifted temperatures

D T min

  = 10 o

  C Stream Type Supply temperature T

  S

  (

  o

  C) Target temperature T

  (

  C) Reactor 1 feed Cold 20 180

  o

  C) D H (MW)

  Heat capacity flowrate

  ) Shift temperature T

  S

  o

  C) Shift temperature T

  T

20 Stream Population

  • ∑CPH [MW.K
    • 1
    • 7.5

      2.0

      2.5

      2.0

      2.0 Deficit

      0.20

      10

      20

      25

      12.0

      2.0

      4.5

      2.0

      2.0 Deficit

      0.05

      40

      40

      30

      10.0 Q

      Cold Utility

      Cmin

      1. Dividing the problem at the pinch, and designing each part separately

      4. Maximising exchanger load

      (below)

      ≥ CP COLD

      (above) CP HOT

      ≤ CP COLD

      3. Immediately adjacent to the pinch, obeying the constraints: CP HOT

      2. Starting the design at the pinch and moving away

      3. Don’t use hot utilities below  Design is produced by:

      Cold Utility

      2. Don’t use cold utilities above

      1. Don’t transfer heat across the Pinch

      

    Simple Design for Maximum Energy Recovery

     To produce minimum utility load:

      4 Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      2

      3

      1

      Stream Population

      35

      70 -0.20 -14.0 Surplus -14.0 6.5 -14.0 14.0

      80

      C

      245 250

      Utility MW

      Utility MW Hot

      Deficit Hot

      [MW] Surplus/

      INTERNAL

      ] DH

      C) ∑CP

      10 -0.15 -1.5 Surplus -1.5 1.5 -1.5

      o

      (

      INTERNAL

      DT

      Interval temp.

      

    The Problem Table Algorithm: the problem table cascade

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      7.5 Q Hmin 235 240 230

      9

      80

      40

      70

      75

      0.0

      4.0

      4.0

      4.0 Deficit

      0.10

      145 140 150 140 150

      195 200 190 200

      4.0

      10 -0.10 -1.0 Surplus -1.0 -3.5 -1.0

      185 180 190 180 190

      3.0

      6.0

      6.0 Deficit 6.0 -4.5

      0.15

      40

      

    5. Supplying external heating only above the pinch, and

    external cooling only below the pinch

      

    Example problem stream data, showing pinch

    CP

      Pinch

    • 1

      [MW.K ]

      250 150 150

      40

      0.15

      2 200 150 150

      80

      0.25

      4 180 140 140

      20

      0.20

      1 230 140 140

      0.30

      3 Q

    7.5 Q

      10.0 Hmin Cmin

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    Heat Exchanger Above the Pinch

    Pinch

      250 150 203.3

      2

       Starting the design at the pinch and moving away

      200 150

      4

       Immediately adjacent to the pinch, obeying the constraints:

      180 140

      CP ≤ CP (above)

    HOT COLD

      7.0 MW CP ≥ CP (below)

    HOT COLD

      230 205 181.7 140

      H

      12.5 MW

      7.5 MW

      7.0 MW

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    Heat Exchanger Below the Pinch

    Pinch

      150 106.7

      40 C

       Starting the design at the pinch and moving away

      10 MW

      150

      80

       Immediately adjacent to the pinch, obeying the constraints:

      CP ≤ CP (above)

      140

      52.5

      20 HOT COLD

      1 CP ≥ CP (below)

    HOT COLD

      17.5 MW 6.5 MW

      140

      3 Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    A design that achieves the energy target:

    Grid Diagram

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      1.0

      T (

    o

      5 Cold 110 110

      4 Cold 80 130 2.5 0.050

      3 Cold 30 110 1.4 0.018

      

    90

    1.0 0.020

      2 Hot 140

      

    40

    3.0 0.025

      1 Hot 160

      )

      CP (MW.K

      Heat capacity flowrate

      D H (MW)

      C) Heat Duty

      T

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      C) Target temp.

      S ( o

      T

      Stream Type Supply temp.

      3. Develop grid diagram (heat exchanger network) for this case ! Stream data for Assignment 8.1.

      C !

      = 10 o

      

    2. From the composite curves, determine the target for hot and cold utility

    for DT min

      C !

      = 10 o

      1. Sketch the composite curves for DT min

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

    Assignment: 8.1.

      

    A design that achieves the energy target:

    Process Flow Diagram with Energy Integration Scheme

    • 1

      

    VIII.2

    REACTOR HEAT INTEGRATION

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    Characteristics of Reactor Heat Integration

      1. Adiabatic Operation: leads to an acceptable temperature rise for exothermic reactors or an acceptable decrease for endothermic reactors.

      2. Heat Carriers: If adiabatic operation produces an unacceptable rise or fall in temperature, then the option is to introduce a heat carrier. The operation is still adiabatic, but an inert material is introduced with the reactor feed as a heat carrier.

      3. Cold Shot: Injection of cold fresh feed for exothermic reactions or preheated feed for endothermic reactions to intermediate points in the

    reactor can be used to control the temperature in the reactor.

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    Reactor Heat Integration

    Effluent (Reactor Products) HEATER

      Feed REACTOR

      (UTILITIES) FEHE COOLER (UTILITIES)

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    A Complex Energy Integrated of HDA Process

      Gas recycle Purge

      Compressor Cooler

      H feed

    2 Furnace

      Separator Toluene

      FEHE-2 FEHE-3 FEHE-1

      PFR feed Quench

      Fuel

      e

      Methane

      cl cy

      CR

      re e en

      Product

      lu

      Column Benzene

      Recycle

      To

      Column Stabilizer

      R3 R2 R1 Column

      Diphenyl

    Source: Terrill, D. L. and Douglas, J. M. (1987), A T-H Method for Heat Exchanger Network Synthesis. Ind. Eng.

      Chem. Res. 26, 175-179

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      4841 lbmole/hr 1150 o

      F 521 psia

    4841 lbmole/hr

    146 o

      F 600.5 psia

      69 MMBtu/hr

    Feed of Reactor need to be heated

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    Effluent of Reactor need to be cooled

    4944.3 lbmole/hr 1150 o

      F 504 psia

      

    74.83 MMBtu/hr

    4944.3 lbmole/hr 113 o

      F 469.2 psia To next process

    65.7 MMBtu/hr

      

    Heat Duty of Heat Exchanger Processes

    (Comparation between process WITH and WITHOUT energy integration)

      65.46

      9.1 FEHE -

      

    74.83

      3.36 Condenser

      

    69

      Furnace

      

    With Energy

    Integration

    Without Energy Integration

      Heat Exchanger Heat Duty (MMBtu/hr)

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

      

    Process flow diagram of Feed-Effluent-Heat-Exchanger

    4944.3 lbmole/hr 1150 o

      9.1 MMBtu/hr

      F 469.2 psia

      F 564.7 psia 4944.3 lbmole/hr 113 o

      F 564.7 psia 4841 lbmole/hr 1150 o

      F 600.5 psia 4841 lbmole/hr 1106 o

      F 472.4 psia 4841 lbmole/hr 146 o

      3.36 MMBtu/hr 4944.3 lbmole/hr 222 o

      F 504 psia

      Dr. Eng. Y. D. Hermawan – ChemEng - UPNVY

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