PROPOSED SOLUTION - ASSIGNMENT 5

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Rudolf Bradford
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1 Norwegian University of Science and Technology Course: Process Integration Department of Energy and Process Engineering Number: TEP 5 Trondheim, 0..07, T. Gundersen Part: Heat Exchanger Networks PROPOSED SOLUTION - SSIGNMENT 5 Task : Modifications to an existing Heat Exchanger Network a) Energy consumption in the existing design (network) is: Q H = Q = ΔT = 0 (55 85) = 00 kw Q C = Q = ΔT = 0 (98 65) = 0 kw Minimum energy requirements (for a specified value of ΔT min = 0ºC) can be found by using the so-called Heat Cascade. Since this assignment does not require use of the Grand Composite Curve), a simplified heat cascade can be used where only the supply temperatures of the streams are used. This is related to the fact that only these temperatures are potential Pinch candidates, and that minimum energy requirements as well as the process Pinch point can be determined from this simplified heat cascade. ST H H 500 kw 650 kw 00 kw 50 kw Q H 75 C 55 C 000 kw kw 5 C R 05 C 975 kw kw 60 C C R 00 kw - 50 C C C Q C 0 C CW The heat cascade above gives us the following relations between the unknown variables: R = Q H 605, R = R + 775, Q C = R 50 We would like to minimize Q H and Q C while the residuals are non-negative (R i 0). n obvious solution is thus: R = 0 and R = 775 which gives the following results: T pinch = 5 C / 05 C, Q H,min = 605 kw, Q C,min = 55 kw The excess use of energy in the existing design is therefore: Page of 7

2 ΔQ H = Q H Q H,min = = 795 kw ΔQ C = Q C Q C,min = 0 55 = 795 kw b) The reason for excess use of energy is that heat is being transferred across the process Pinch point. This is illustrated by drawing the existing heat exchanger network in the so-called Stream Grid, where it is marked whether heat exchangers operate above, across or below the Pinch. Inlet and outlet temperatures on the hot and cold side of each exchanger relative to the hot and cold Pinch temperatures determine the drawing of each unit in the Stream Grid. 75 H H 0 85 C C [0] [0] [5] 05 Unit : QP H Unit : QP P Unit : QP P = C (T pinch,c T C,in ) = 0 (05 85) = 00 kw = 0 C (T C,out T pinch,c ) = 0 5 ( 05) = - 05 kw = H (T H,in T pinch,h ) 0 = 0 (75 5) 0 = 500 kw Notice that the small driving forces in the hot end of exchanger (ΔT = 5 = C, which is less than ΔT min = 0ºC) introduces heat transfer from the heat surplus area below Pinch to the heat deficit area above Pinch. s a result, this heat exchanger is acting as a heat pumping unit. s a result, QP P has a negative value (- 05 kw). Net heat transfer across the Pinch point (from above to below ) then becomes: QP = QP H + QP P + QP P = = 795 kw (q.e.d.) c) The minimum number of units for a new design (with and without strict Pinch decomposition) for this heat recovery problem can then be determined as follows: U min,mer = (N ) above + (N ) below = ( + + ) + ( + + ) = + = 7 U min = (N ) total = ( + + ) = 5 In comparison, the existing network has only units, which is one less than the estimate for U min. The reason is the existence of two sub-networks in the existing design: (H, C and ST) and (H, C and CW). Heat recovery in the existing design with only units is 080 kw in heat exchanger and 00 kw in heat exchanger, in total 80 kw. In comparison, there are 7 units in the Page of 7

3 MER design, which (according to our calculated minimum values) recover 75 kw. The extra 795 kw apparently requires more heat exchangers. This is an indication that our retrofit project not necessarily should aim for absolute minimum energy consumption, since new units will be quite expensive. d) Since heat exchanger has a heat pumping effect and therefore contributes in a negative and reducing way (-05 kw) to the net total heat transfer across the Pinch, there is no reason to study changes for this exchanger. This means that the problem can be simplified by cutting off those parts of the network (exchangers and parts of streams) that should not be changed in the retrofit project. This gives us the following network: 75 H 5 [0] H C [0] 05 We now see that H is the only heating resource above Pinch, and a retrofit project must therefore focus on utilizing this stream to heat up C above Pinch. The heating of C below Pinch can take place by heat exchanging with H which is an unutilized resource in this temperature region. In the existing network, this heat is lost entirely to cooling water. Idea no : Install a new heat exchanger () between H and C below Pinch, and use the following heat load path to reduce energy consumption: ST () C () H () CW The problem is, however, that the cold inlet temperature to heat exchanger cannot exceed 5 C in order to be able to cool stream H down to its target temperature of C. The savings in energy consumption would therefore be rather marginal: ΔQ H = ΔT = 0 (5 0) = 00 kw ccordingly, there is no reason to explore this project suggestion any further. Idea no. : Keep the idea of a new heat exchanger () between hot stream H and cold stream C, but at the same time install a new cooling water cooler (B) on hot stream H. In this way, the new heat exchanger () between H and C can be made large (in duty) without having to consider the target temperature of hot stream H. The resulting network is shown below, Page of 7

4 where it is also indicated (with X and Y) how the duties of the various units can be manipulated by using the following heat load loop and path: Path: Loop: ST () C () H (B) CW H () C () H (B) CW () H 75 H T B [0] 55 T 0 + X - Y H T T 0 - X 0 C [0] 00 - Y 00 - X +Y 0 + X The heat recovery in the new heat exchanger () between H and C is limited by the temperature T (cold outlet temperature) that must be at least ΔT min lower than the hot inlet temperature (98ºC). This gives us an equation to determine the duty of the new heat exchanger (X kw): 98 0 = 78 = T = 0 + X / 0 which gives: X = 0 (78 0) = 60 kw The heat recovery in the existing heat exchanger () between H and C is correspondingly limited by the fact that the hot outlet temperature (T ) must be at least ΔT min higher than the cold inlet temperature (T ) which has already been determined to be equal to 78ºC. This gives us another equation to determine the value of Y: T = 75 (00 X + Y) / 0 = = 98 By using for X = 60 this equation gives: Y = 0 (75 98) = 60 kw The resulting network where also stream C and heat exchanger () are included is shown below. New values for exchanger duties and stream temperatures are listed and form the basis for a U analysis that is used to investigate heat exchanger re-use and purchase. By installing two new heat exchangers ( and B), external heating is reduced from 00 kw to 770 kw, which is only 65 kw above the minimum value of 605 kw. Next, we perform a U analysis to get an indication about the need for additional heat transfer area, and to what extent we are utilizing the existing heat exchangers. Such a U analysis can also indicate whether it would be worthwhile to move exchangers to completely new tasks. The U analysis is using the heat transfer equation: Q = U ΔT LM U = Q / ΔT LM Page of 7

5 Below, index ex is for the existing situation, while new means U requirements in the new situation, both for existing and new heat exchangers. 75 H B H C C [0] [0] [5] Existing network: = 00 / 5. = 6.70 kw/ºc = 080 / 0.09 = 5.89 kw/ºc = 00 / 50.7 = 5.6 kw/ºc = 0 / =.7 kw/ºc Retrofit network: U new = 770 /.0 = 8.6 kw/ºc = 080 / 0.09 = 5.89 kw/ºc (unchanged) = 770 / 5.87 =.7 kw/ºc = 60 / 6.9 =. kw/ºc = 60 /.6 = 5.8 kw/ºc B = 50 / 8.5 = 0.96 kw/ºc From these calculations, we notice that area requirements in cooler () on stream H are very small while there is some extra capacity in heat exchanger () between H and C that is not utilized (U is 5.6 kw/ C in the existing design versus.7 kw/ C in the modified network). Idea no. : Page 5 of 7

6 Utilize the following heat load path to eliminate heat exchanger (), the cooler, in the modified network: ST () C () H () CW We can now change the duty of each unit in this path by Z kw in the following way: Q = 770 Z Q = 60 + Z Q = 60 Z Of course, we choose to have Z = 60 kw. This gives the following network with modified duties and temperatures indicated: 75 H 5 H B 50 [0] C [0] C [5] 080 We realize that the requirement for ΔT min is violated in the hot end of exchanger () and in the cold end of the existing exchanger (). For the latter, this does not matter at all as long as the exchanger can handle the new situation (sufficiently large U value), while it may indicate that the new exchanger () will be more costly than we would like. New U values for the modified network are now (here, we only include those that have been changed after the last heat load path manipulation): U new = 770 / 6.79 = 8.7 kw/ºc = 0 /.97 = 5.87 kw/ºc From this, we realize that we have gone beyond the capacity of exchanger (), since the required U value is 8.7 kw/ C is larger than the existing value of 5.6 kw/ºc. t the same time, the U requirement in the new exchanger () increases from 5.8 kw/ºc to 5.87 kw/ C, which indicates an increase of 7.5% in area requirement compared to an increase in duty of.8%. The duty of exchanger () cannot be changed (since this is the only unit on stream C), and therefore exchanger () becomes this large unless we still keep the cooler on stream H. When it comes to the U value for exchanger (), this can be adjusted to match the existing value by utilizing the following heat load path: Page 6 of 7

7 ST () C () H (B) CW Now, we can change the duty with W kw for the units in this path as follows: Q = 60 + W Q = 770 W Q B = 50 + W We can find the value of W that perfectly matches exchanger () in the new situation (which means a U value of 5.6 kw/ºc) through a trial and error procedure (iteration): W = 00 kw W = 0 kw W = 5 kw = 670 / 6.0 = 8.5 kw/ºc = 70 / 9.76 =.86 kw/ºc = 75 / 9.8 = 5.5 kw/ºc We also realize that the cooling water cooler () can be used as the new cooler (B) on stream H through re-piping. The U value is more than large enough, and unless there are significant differences between streams H and H with respect to volumetric flow, need for material of construction, etc., such a re-piping is feasible. The final network then becomes the following, with exchanger () as the only new investment and after the move of exchanger/cooler (): 75 H 00.5 [0] 5 H C [0] C [5] 080 The energy consumption (both steam and cooling water) has been reduced by 765 kw compared to the existing network. The costs are related to the purchase and installation of a new heat exchanger () with U = 5.87 kw/ºc, and the re-piping of the cooler () from stream H to stream H. The profitability of this project must of course be evaluated through detailed cost calculations. Page 7 of 7

PROPOSED SOLUTION - ASSIGNMENT 3

PROPOSED SOLUTION - ASSIGNMENT 3 Norwegian University of Science and Technology Course: Process Integration Department of Energy and Process Engineering Number: TEP 45 Trondheim, 0..07, T. Gundersen Part: Heat Exchanger Networks PROPOSED