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Lecture-8_Heat Exchanger Network_Part3.ppt
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- 2. Heat Exchanger Network ChemicalProcess Design and Simulation Dr. Imran Nazir Unar Department of Chemical Engineering MUET Jamshoro Heat Exchanger Network-Part-2 (Overview) Example Problem of HEN
- 3. Designing of HEN •Designing a Heat Exchanger Network • Problem Statement Design a Heat Exchanger Network that has four process streams with inlet temperatures of 30, 80, 150, 180°C and respective outlet temperatures of 135, 140, 30, 60°C. Note: In this problem we will use Temperature Interval Method for Heat Exchanger Networks
- 4. Designing of HEN •Solution: This problem has two parts A and B. • A: Determine MER (Minimum Energy Requirement) using Temp. Interval Method (TI). Step: 1: Create the table of all streams Streams Ts (°C) Tt (°C) C (kW/°C) Q (kW) H1 180 60 3 360 H2 150 30 1 120 C1 30 135 2 210 C2 80 140 5 300 Ts = Source Temp. , Tt = Target Temp., C: Heat capacity flowrate = m (mass flowrate) x c (heat capacity) = Energy/temp Assumption: Constant w.r.t Temp. Q=cmΔT =CΔT Hot Load 480 Cold Load 510
- 5. Designing of HEN •Solution: This problem has two parts A and B. • A: Determine MER (Minimum Energy Requirement) using Temp. Interval Method (TI). Step: 2: Designate Δtmin = 10°C (A good start) Streams Ts (°C) Tt (°C) Ts(°C) New Tt (°C) New C (kW/°C) H1 180 60 170 50 3 H2 150 30 140 20 1 C1 30 135 30 135 2 C2 80 140 80 140 5 - Subtract the ΔTmin from Ts and Tt of Hot streams These are the temperatures used for making intervals
- 6. Designing of HEN •Solution: This problem has two parts A and B. • A: Determine MER (Minimum Energy Requirement) using Temp. Interval Method (TI). Step: 3: Development of Temperature Interval Scale H1 H2 C1 C2 20 30 50 80 135 140 170 C=3 C=1 C=2 C=5
- 7. Designing of HEN •Solution: This problem has two parts A and B. • A: Determine MER Step: 4: Enthalpy Calculations in each interval 1 2 3 4 5 6 Qsteam Qcold 20 30 50 80 135 140 170 ΔH1=(3)(30°C) = 90 kW ΔH2=(-1)(5°C) = -5 kW ΔH3=(-3)(55°C) = -165 kW ΔH4=(2)(30°C) = 60 kW ΔH5=(-1)(20°C) = -20 kW ΔH6=(1)(10°C) = 10 kW Enthalpy b/w Each Interval Residual 90 kW (+) 85 kW (+) -80 kW (+) -20 kW (+) -40 kW (+) -30 kW Most minimum negative value
- 8. Designing of HEN •Solution: This problem has two parts A and B. • A: Determine MER Step: 5: Final Pass Calculations - MER 1 2 3 4 5 6 Qsteam Qcold 20 30 50 80 135 140 170 Residual 90 kW 85 kW -80 kW -20 kW -40 kW -30 kW 80 kW Reverse of most min. negative residual (+) 170 kW (+) 165 kW (+) 0 kW (+) (+) (+) 60 kW 40 kW 50 kW Hot Utility Cold Utility MER Targets Pinch Point
- 9. Heat Exchanger Network ChemicalProcess Design and Simulation Dr. Imran Nazir Unar Department of Chemical Engineering MUET Jamshoro Heat Exchanger Network-Part-3 Example Problem of HEN
- 10. Step: 1: Developmentof Stream Diagram across Pinch Point Designing of HEN • Solution: This problem has two parts A and B. • B: Designing of Heat Exchanger Network. Hot Stream Pinch Temperature: 90 °C Cold Stream Pinch Temperature: 80 °C Pinch H1 90°C 80°C 90°C 90°C 90°C 80°C 80°C 80°C C2 H2 C1 C (kW/°C) 3 1 2 5 180°C 60°C 150°C 30°C 135°C 30°C 140°C 80°C
- 11. Some Basics aboutHeat Exchangers Designing of HEN • Solution: This problem has two parts A and B. • B: Designing of Heat Exchanger Network. Q Thi Tco Tho Tci ΔT2 ΔT1 Ch Cc Ch and Cc are Specific Heat flowrates (Energy/Temp.) Q=Ch (Thi – Tho) Q=Cc (Tco – Tci) (Thi – Tho)= Q/Ch (Tco – Tci)= Q/Cc (ΔT2 – ΔT1)= Q (Cc – Ch) Cc Ch Taking difference Hot side of Pinch: ΔT1=ΔTmin ΔT2=ΔTmin+ Q(Cc-Ch)/(Cc Ch) Cc C ≥ h Cold side of Pinch: ΔT2=ΔTmin ΔT1=ΔTmin- Q(Cc-Ch)/(Cc Ch) Ch C ≥ i
- 12. Step: 2: Connectingthe Hot and Cold Streams Designing of HEN • Solution: This problem has two parts A and B. • B: Designing of Heat Exchanger Network. Pinch H1 90°C 80°C 90°C 90°C 90°C 80°C 80°C C2 H2 C1 C (kW/°C) 3 1 2 5 180°C 60°C 150°C 30°C 135°C 30°C 140°C Hot Side Cold Side Cc C ≥ h Ch C ≥ i A1 A2 B1 B2 C D E1 E2 F1 F2 G 134°C 270kW 110°C 60kW 50kW 30kW 35°C 80°C 10kW 50kW 90kW
- 13. Step: 3: Calculatingthe Heat Loads and Temp at various Nodes Designing of HEN • Solution: This problem has two parts A and B. • B: Designing of Heat Exchanger Network. At Node A1 Q=3(180-90)=270 kW At Node A2 Q=5(140-80)=300 kW At Node A2 270=5(X-80)=> X=134°C At Node B1 Q=1(150-90)=60 kW At Node B2 Q=2(135-80)=110 kW At Node A2 60=2(X-80)=> X=110°C At Node C Q=2(135-110)=50 kW At Node D Q=5(140-134)=30 kW At Node E1 Q=3(90-60)=90 kW At Node E2 Q=2(80-30)=100 kW At Node E2 90=2(80-X)=> X=35°C At Node F1 Q=1(90-30)=60 kW At Node F2 Q=2(35-30)=10 kW At Node F2 10=1(90-X)=> X=80°C At Node G Q=1(80-30)=50 kW
- 14. Step: 4: FinalDiagram of Heat Exchanger Network Designing of HEN • Solution: This problem has two parts A and B. • B: Designing of Heat Exchanger Network. H1 90°C 80°C 90°C 80°C C2 H2 C1 180°C 60°C 150°C 30°C 135°C 30°C 140°C 134°C 270kW 110°C 60kW 50kW 30kW 35°C 80°C 10kW 50kW 90kW Total Heating Load = 50 + 30 = 80 kW Total Cooling Load = 50 kW
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