Saturday 20 May C 180 C C 130 C C 60 C kw 50 C 30 C C 20 C

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1 Page 1 of 10 NORWEGIAN UNIVERSITY OF SCIENCE AND TECHNOLOGY (NTNU) - TRONDHEIM DEPARTMENT OF ENERGY AND PROCESS ENGINEERING PROPOSED SOLUTION EXAMINATION IN COURSE TEP 4215 PROCESS INTEGRATION Saturday 20 May 2006 ASSIGNMENT 1 (60%) The heat cascade for this problem is shown below. Since the Grand Composite Curve is needed to solve question (b), the full cascade is used based on both supply and target temperatures for the streams. The value for T min is specified in the text to be 20ºC. ST Q H 200 C 180 C H kw kw R C 130 C C kw 2800 kw kw 1750 kw H2 R 2 80 C 60 C kw kw C2 900 kw R 3 50 C 30 C 0 kw 300 kw Q C 20 C CW a) Since all temperature intervals have surplus of heat, there is no need for external heating. On this basis, the residuals can be calculated as follows: Q H = 0 R 1 = 500 R 2 = 2250 R 3 = 3600 Q C = 4300

2 Page 2 of 10 The need for external heating and cooling is then determined to be: Q H,min = 0 kw and Q C,min = 4300 kw The fact that there is no need for external heating means that this is a so-called threshold problem where only one type of utility is needed. The issue of a process Pinch becomes somewhat vague, even though one could define the Pinch to be in that end (temperature wise) of the process where there is no need for utilities, in this case the hot end of the process. Being a heat surplus process with only need for cooling brings around the idea of using this heat surplus for steam generation, and that is the background for the next question. b) The maximum amount of LP steam that can be produced can be found numerically from the heat cascade directly or read graphically from the Grand Composite Curve. First, the numerical way is shown: LP steam will enter the cascade as a cold stream at 120ºC. This means that surplus heat from the first interval (500 kw) and a fraction of the surplus of heat from the second interval can be used for steam production. Hot streams that can produce LP steam at 120ºC must have a temperature of at least 1ºC in order to satisfy the specified need for driving forces in the system. The heat surplus that is available in the second interval, at a temperature of 1ºC and higher, represents one seventh of the total heat surplus in that interval. This is easily found from the following calculation: (150 1) / (150 80) = 10 / 70 = 1/7 This means that 1750 / 7 = 250 kw of heat is available in the second interval for steam production. In total we have: Q LP = = 750 kw The same result can be found graphically from the Grand Composite Curve: T(ºC) LP CW Q(kW) In the Grand Composite Curve, modified temperatures are used to make it possible to represent both hot and cold process streams as well as utilities in the same diagram.

3 Page 3 of 10 Thus, hot process streams and utilities are drawn at temperatures that are ½ T min below their true temperatures, while cold process streams and utilities similarly are drawn at temperatures ½ T min above their true temperatures. In this case, LP steam acts as a cold utility (we are using hot process streams to vaporize boiler feed water), thus the modified temperature becomes: T LP = = 130ºC The other cold utility, cooling water, is heated from 10 to 15ºC, which in modified temperatures then is from 20 to 25ºC. The maximum amount of LP steam that can be generated can be approximately read from the Grand Composite Curve or calculated by linear interpolation between the two points 1ºC/500 kw and 70ºC/2250 kw when we know the modified temperature of LP steam to be 130ºC: Q LP = ((1 130)/(1 70)) ( ) = 750 kw Cooling water duty is reduced accordingly: Q CW = = 3550 kw Maximizing LP steam generation means that we have minimum driving forces in the point where the LP line touches the process Grand Composite Curve. As a result, a new Pinch point is introduced and is referred to as a Utility Pinch, since it is the result of decisions regarding the utility system (maximizing LP production to reduce CW consumption). This is important information for the next question. c) In order to design a heat exchanger network that achieves minimum external heating (Q H,min = 0 kw) and minimum external cooling (Q C,min = 4300 kw) as well maximum LP steam generation (Q LP = 750 kw and Q CW = 3550 kw), the Utility Pinch at 1ºC (for hot streams) and 120ºC (for cold streams) has to be respected. In fact, the design of the network should start at this Utility Pinch. Design of network below Pinch: We have two cold streams that should be heated to Pinch, and each of these require a Pinch exchanger that must satisfy the mcp-rule for hot and cold streams operating at and below the Pinch: mcp H,i mcp C,j Since the mcp values for the two hot streams ( and 30) are larger than both the cold streams (25 and 20), there are two feasible solutions: (H1 C1) and (H2 C2) or (H1 C2) and (H2 C1) The best distribution of driving forces (and thus smallest total area) is achieved when matching the hot stream with largest mcp with the cold stream with largest mcp. Thus, the first alternative (H1 C1) and (H2 C2) is selected for the design. The duties of these exchangers are determined by the tick-off rule:

4 Page 4 of 10 For the match between H1 and C1, we have: Q I = min (3000, ) = kw For the match between H2 and C2, we have: Q II = min (00, 2250) = 2250 kw Remaining cooling requirements for the hot streams must be covered by cooling water: Stream H1: Q Ca = 3000 = 1800 kw Stream H2: Q Cb = = 1750 kw Total cooling is then = 3550 kw (OK) Design of network above Pinch: We have two hot streams that should be cooled to Pinch, and each of these require a Pinch exchanger that must satisfy the mcp-rule for hot and cold streams operating at and above the Pinch: mcp C,j mcp H,i Here, only LP steam has large enough mcp (it is infinite, since LP steam is boiling at constant temperature) to handle H1 and H2. Two alternatives exist, one can split the LP stream and have two steam boilers (not very likely) or one could use H1 to produce 750 kw of LP steam, but then H2 has to be split. It turns out that after removing one unit, these designs become identical, thus we will not show both of these alternatives. When splitting the stream representing LP steam generation, the branch that is matching with H2 has to be small enough to allow H2 also to cover the 250 kw needed by cold stream C2. The following matches can thus be established above Pinch: Match between H2/C2: Q III = min (0, 250) = 250 kw Match between H2/LP: Q IV = min ((0-250), 750) = 150 kw Match between H1/LP: Q V = min (1800, ( )) = 600 kw Match between H1/C1: Q VI = min (( ), ) = kw The entire network is drawn below. The total number of units is 8, which corresponds to the minimum while achieving maximum LP steam generation: U min,mer = (N above 1) + (N below 1) = ( ) + ( ) = 8 The fewest number of units is: U min = (N total 1) = ( ) = 5 The number of loops in the network is thus: L = U U min = 8 5 = 3

5 Page 5 of 10 VI V C 160 C 100 C H1 C a C III IV II C C H2 C b C 60 C C1 130 C 30 C C C 120 C 600 LP I mcp C d) As indicated above, there are 3 heat load loops in the network, thus it should be possible in theory to reduce the number of units to simplify the network and (possibly) to reduce the total annual cost compared with the MER design. In normal Pinch type problems (this was a threshold problem until we introduced and maximized LP steam generation which resulted in a Utility Pinch), reduction in the number of units by breaking some of the heat load loops results in the need for more heat transfer area and/or need for more external heating and cooling. In this case, the adjustment on the energy side to reestablish temperature driving forces means reducing LP steam generation and increasing cooling water requirements. There are three independent heat load loops in the MER design: A: H1 (VI) C1 (I) H1 B: H2 (III) C2 (II) H2 C: H1 (V) LP (IV) H2 (Cb) CW (Ca) H1 The economically reasonable strategy for simplifying the network is to remove the smallest units first. In the MER design above, the smallest unit is the very small LP steam boiler represented as exchanger IV with a duty of 150 kw. By manipulating loop (C), this unit can be removed, and the stream split can then also be removed. The new duties of the heat exchangers in this heat load loop will then be: Q V Q IV Q Cb Q Ca = Q V = 750 kw = Q IV 150 = 0 kw = Q Cb = 1900 kw = Q Ca 150 = 1650 kw

6 Page 6 of 10 The corresponding network with updated duties and temperatures is shown below. 200 C H1 150 C H2 180 C VI 160 C 135 C 95 C C a C II 87.5 C C b C 60 C C1 III V 130 C 30 C 120 C C C 120 C LP 750 I mcp As expected, there are problems with the driving forces after removing one unit. Heat exchanger (I) has a T in the hot end that is only 15ºC, and the same T is observed in the cold end of heat exchanger (V). For heat exchanger V (the LP steam boiler), the cold temperature of 120ºC cannot be changed, thus we have to increase the temperature of hot stream H1 between heat exchangers (V) and (I) from 135ºC to 1ºC to satisfy the required T min of 20ºC. This temperature can be adjusted by using a path between LP (as one cold utility) and CW (another cold utility). In this case, the path is the shortest possible, since it only includes two units. Notice that while paths from a hot to a cold utility involves 3, 5, 7, etc. units, a path between two utilities of the same type (i.e. hot or cold) has 2, 4, 6, etc. units. In fact, these paths act like a loop and do not increase total utility consumption; they only change the distribution between utilities of the same type (hot or cold). In this case, the duty of the LP boiler (heat exchanger V) is reduced by y kw, while cooler Ca is increased by the same y kw. An equation to establish the value of y is to specify that the temperature of H1 after heat exchanger (V) should be 1ºC: 160 (750 y) / 30 = 1 y = 150 kw The new duties of the heat exchangers involved are then: Q V Q Ca = Q V 150 = 600 kw = Q Ca = 1800 kw Also notice that when heat exchanger (IV) was removed, the two heat exchangers (III) and (II) become close neighbors, and can thus be merged into one heat exchanger. Since heat exchanger (II) is the larger of the two, we will draw the network as if heat exchanger (III) has been removed from the network and increase the duty of heat

7 Page 7 of 10 exchanger (II) accordingly. There is of course no penalty by merging these units even though the action may be regarded as breaking heat load loop B. The simplified network is shown below. 200 C H1 VI V 160 C C I C a mcp C H2 II C C b C 60 C 120 C C1 130 C 30 C C C 120 C LP The smallest unit in the remaining network is the LP steam boiler (exchanger V). This unit is not part of a loop and can only be removed by the same path we used above to fix driving forces in the network. Removing this unit (means no LP steam generation at all) simply results in an increased duty of cooler Ca from 1800 to 20 kw. One could also consider merging the two units (VI) and (I). Since we need a match between H1 and C1 in the hot end of the network to reach the target temperature for stream C1, exchanger (VI) will remain in the network and have an increased duty of 20 kw, while unit (I) is removed. The temperature of hot stream H1 after the increased unit (VI) is going to be 120ºC, which means there is no scope for LP production. These two modifications then are identical, and the last and very simple network without LP steam generation is shown below. 200 C H1 150 C H2 180 C VI C II 60 C 130 C 30 C 2500 C a C C b 1900 C1 C2 mcp

8 Page 8 of 10 In summary, three different designs have been developed and their main features are indicated in the table below: Design Units Cooling (kw) LP steam (kw) Without economic data it is of course difficult to evaluate these alternative solutions. Having two steam boilers for this limited amount of steam generated is not expected to be worthwhile. Design 1 is expected to be the least attractive; design 2 seems attractive, while design 3 has the beauty of simplicity. ASSIGNMENT 2 (%) a) The third temperature interval with a heat deficit of 12 MW is obviously the most demanding of the intervals in the heat cascade. As a result, the following heat flows (heating, cooling and residuals) can be established: R 3 = 0 R 2 = 12 R 1 = 14 Q H = 8 R 3 = 0 R 4 = 6 Q C = 8 Minimum external heating and cooling is then: Q H,min = 8 MW and Q C,min = 8 MW b) The Process Grand Composite Curve can now be drawn based on the so-called modified temperatures in the heat cascade and the heat flow values established in question (a). The distillation column is drawn as a rectangle (box) in the same diagram. With a specified T min = 10ºC, the modified temperatures for the reboiler (cold stream) and condenser (hot stream) are then 155ºC and 115ºC respectively. T(ºC) Q(kW)

9 Page 9 of 10 c) The Grand Composite Curve above clearly indicates that the distillation column could and should be integrated above Pinch by transferring heat from the condenser to cold streams above Pinch. The savings will be 2 MW in heating (steam) and 2 MW in cooling (cooling water). The reboiler operates at a modified temperature of 155ºC, which means that the true reboiler temperature is 150ºC, and that the heating medium for running the reboiler has to be at a temperature of at least 160ºC in order to satisfy the need for minimum driving forces indicated in the assignment text. This exactly matches the LP steam that can be extracted from the backpressure turbine. d) After integrating the distillation column with the background process, the need for external heating in the process has been reduced from 8 MW to 6 MW, and the temperature region for this heat input is indicated by the Grand Composite Curve to be from about 100ºC to about 130ºC in modified temperatures. These temperatures (corresponding to Q = 2 MW and Q = 8 MW) can be established exactly by using linear interpolation between the two known points on the Grand Composite Curve at 90ºC / 0 kw and 150ºC / 12 kw. Nevertheless, these temperatures are definitely in a range where LP steam at 160ºC can be used for external heating. As indicated under question (c), the reboiler of the integrated distillation column can also be run by this LP steam. From a systems point of view, the turbine to be selected for combined heat and power is the backpressure turbine. The condensing turbine has a higher efficiency when it comes to power production, but a large amount of thermal energy (heat) is lost to cooling water. The backpressure turbine has a lower efficiency when it comes to power production, but the thermal energy leaving the turbine can be used for heating purpose in the process. This will then replace other steam that would have to be produced in a separate boiler. First, we establish the amount of power that will be produced by the backpressure turbine (with a specified efficiency of 2 = 0.21): The need for heating in the process and distillation column is known: Q 2 = 8 MW We also have two equations to establish heat input to the turbine (Q 1 ) and work (or power) produced (W): Q 1 = W + Q 2 and W = 2 Q 1 These equations can be solved with respect to W eliminating Q 1 : W = ( 2 / (1 2 ) ) Q 2 W = 2.13 MW The amount of HP steam to this turbine is then: Q 1 = = MW The condensing turbine must then provide the remaining need for power: W = = 2.37 MW The amount of HP steam to run this turbine is: Q 1 = W / 1 = 2.37 / 0.43 = 5.51 MW

10 Page 10 of 10 In summary, the consumption and production in the utility system are: The amount of power produced: W = = 4.5 MW (as specified) The amount of HP steam consumed: Q HP = = MW e) In total then MW of HP steam is converted into 4.5 MW of work (power) and 8 MW of heat (LP steam to the process and the column reboiler). The rest of the energy ( = 3.14 MW) is lost to cooling water in the condensing turbine. The overall efficiency of the arrangement is then: = ( ) / = = 79.9% Notice that this efficiency is considerably larger than the efficiencies of the two turbine types (43% for condensing and 21% for backpressure). The reason is that the system efficiency above gives credit to the fact that energy (heat) is reused in the backpressure turbine arrangement, since HP steam after producing power is reused in the form of LP steam rather than being rejected to cooling water as the case is in the condensing turbine. Trondheim, Truls Gundersen