HEAT EXCHANGER NETWORK sYnThESIS 1: TARGETING OBJECTIVES 1 Determine the minimum energy inputs required ∫ or a process 2 Determine the minimum number of heat exchangers required to transfer this energy 3 Develop a methodology for determining where these heat exchangers should be placed in the process 2/21/99 HENS 1: Targeting
2/21/99 HENS 1: Targeting 1 HEAT EXCHANGER NETWORK SYNTHESIS 1: TARGETING OBJECTIVES: 1) Determine the minimum energy inputs required for a process 2) Determine the minimum number of heat exchangers required to transfer this energy 3) Develop a methodology for determining where these heat exchangers should be placed in the process
SINGLE HEAT EXCHANGER- REVIEW W COLD T COLDin W HOT HOTin HOTout HOLDout Stream W(kg/h) Cp(kJ/kg-K Tin(K) Tout(K) Q(gJ/h Hot20,000 4.2 180112.5-567 Cod15,00 4.2 60150+567 2199 HENS 1: Targeting
2/21/99 HENS 1: Targeting 2 SINGLE HEAT EXCHANGER - REVIEW W HOT T HOTin T COLDout W COLD T COLDin T HOTout Stream W(kg/h) Cp (kJ/kg-K) Tin(K) Tout(K) Q(gJ/h) Hot 20,000 4.2 180 112.5 -5.67 Cold 15,000 4.2 60 150 +5.67
Heat exchanger Q-T diagram 200 180 2 160 14 120 s40 Q(gJ/h) 一Thot-Tcol 2199 HENS 1: Targeting
2/21/99 HENS 1: Targeting 3 Heat Exchanger Q-T Diagram 0 20 40 60 80 100 120 140 160 180 200 0 1 2 3 4 5 6 Q (gJ/h) Temperature (K) Thot Tcold
IMPORTANT POINTS 1 The First law says that the heat transferred from the hot stream must equal that transferred to the cold stream Therefore, of the six process variables (two flow rates and four temperatures), only five can be specified independently 2) The second law says that heat can only be transferred from a hotter fluid to a colder one. Therefore, the temperature of the cold fluid must be less than that of the hot fluid at all points along the leng th of the exchanger 2/21/99 HENS 1: Targeting
2/21/99 HENS 1: Targeting 4 IMPORTANT POINTS 1) The First Law says that the heat transferred from the hot stream must equal that transferred to the cold stream. Therefore, of the six process variables (two flow rates and four temperatures), only five can be specified independently. 2) The Second Law says that heat can only be transferred from a hotter fluid to a colder one. Therefore, the temperature of the cold fluid must be less than that of the hot fluid at all points along the length of the exchanger
IMPORTANT POINTS (Contd. 3) As a practical matter, for heat to be transferred at an acceptable rate, the temperature of the cold fluid must be less than that of the hot fluid at each point in the exchanger by a reasonable amount. We refer to the minimum acceptable temperature difference as the MINIMUM APPROACH TEMPERATURE or ATmir This is also known in the literature as the pinch TEMPERATURE. The methodology for synthesizing heat exchanger networks is sometimes called PINCH TECHNOLOGY 2199 HENS 1: Targeting 5
2/21/99 HENS 1: Targeting 5 IMPORTANT POINTS (Cont’d.) 3) As a practical matter, for heat to be transferred at an acceptable rate, the temperature of the cold fluid must be less than that of the hot fluid at each point in the exchanger by a reasonable amount. We refer to the minimum acceptable temperature difference as the MINIMUM APPROACH TEMPERATURE or DTmin. This is also known in the literature as the PINCH TEMPERATURE. The methodology for synthesizing heat exchanger networks is sometimes called PINCH TECHNOLOGY
THE HEAT EXCHANGER NETWORK PROBLEM a chemical plant generally has more than one hot steam to be cooled and more than one cold stream to be heated. To better understand this problem let us consider the following simple example 2199 HENS 1: Targeting
2/21/99 HENS 1: Targeting 6 THE HEAT EXCHANGER NETWORK PROBLEM A chemical plant generally has more than one hot steam to be cooled and more than one cold stream to be heated. To better understand this problem let us consider the following simple example
1000g,30C9,000kg/h,30C~lkgh 1,500kgh,100C Heat to 200C REACTOR Heat to 100C ABSORBER STRIPPER Cool to 30C] 500 kg/h 10,00kg/h, 250C 16,000kg/h 1500CC130C 2199 HENS 1: Targeting
2/21/99 HENS 1: Targeting 7 1000 kg/h, 30 C Heat to 200 C Cool to 30 C 10,000 kg/h, 250 C 9,000 kg/h, 30 C 1,500 kg/h, 100 C 16,000 kg/h 15,000 kg/h ~1 kg/h 500 kg/h Cool to 30 C REACTOR Heat to 100 C ABSORBER STRIPPER
THE PROBLEM 1) There are two hot streams that need to be cooled, name Reactor efluent and stripper bottoms 2) There are two cold streams that need to be heated, namely, Reactor Feed and Stripper feed What is the best way to do this? 2199 HENS 1: Targeting
2/21/99 HENS 1: Targeting 8 THE PROBLEM 1) There are two hot streams that need to be cooled, namely, Reactor Effluent and Stripper Bottoms 2) There are two cold streams that need to be heated, namely, Reactor Feed and Stripper Feed What is the best way to do this?
FIRST LAW ANALYSIS PROCESS DATA Stream W(kg/h) Cp(kJ/ko-K) Tin(K) Tout(k) Q(g/h) Hot10,000 1.3 25030 2.860 Hot215,000 10030-4.410 Cold 110.000 30200+1.870 Col216,000 4,2 30100+4.704 Total Heating Available=2.860+ (4.410)=7.270 g/h Total Heating required=-1870+(4.704)=-6574 g/h Net heat available =0. gJ/h (Subject to the Second law) 2199 HENS 1: Targeting
2/21/99 HENS 1: Targeting 9 FIRST LAW ANALYSIS PROCESS DATA: Stream W(kg/h) Cp (kJ/kg-K) Tin(K) Tout(K) Q(gJ/h) Hot 1 10,000 1.3 250 30 -2.860 Hot 2 15,000 4.2 100 30 -4.410 Cold 1 10,000 1.1 30 200 +1.870 Cold 2 16,000 4.2 30 100 +4.704 Total Heating Available = 2.860 + (4.410) = 7.270 gJ/h Total Heating Required = -1.870 + (-4.704 ) = - 6.574 gJ/h Net Heat Available = 0.696 gJ/h (Subject to the Second Law)
THE SIMPLE (BUT WASTEFUL SOLUTION The simplest solution is 1 to cool each hot stream to its specified temperature using a cold utility such as cooling tower water and 2 to heat each cold stream to its specified temperature using a hot utility such as steam. Typical utility costs(see Turton et alare MP Steam(10 barg, 1840)....$3. 66/gJ Cw(15 C temperature rise). S0.24 gJ ooling cos;:(7.270(6800024)=Sl4,000yr Heating cost:(6.574)(8000(3.66=S192,000yr 2/21/99 HENS 1: Targeting
2/21/99 HENS 1: Targeting 10 THE SIMPLE (BUT WASTEFUL) SOLUTION The simplest solution is: 1) to cool each hot stream to its specified temperature using a cold utility such as cooling tower water and 2) to heat each cold stream to its specified temperature using a hot utility such as steam. Typical utility costs (see Turton et al) are: MP Steam (10 barg, 184C)….$3.66/gJ CW (15 C temperature rise) .. $0.24/gJ Cooling cost: (7.270)(8000)(0.24) = $14,000/yr Heating cost: (6.574)(8000)(3.66) = $192,000/yr
