PROPOSED SOLUTION - ASSIGNMENT 2

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Gwendolyn Warren
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1 Norwegian University of Science and Technology Course: Process Integration Department of Energy and Process Engineering Number: TEP 4215 Trondheim, , T. Gundersen Part: Heat Integration PROPOSED SOLUTION - ASSIGNMENT 2 Task 1: Graphical Diagrams for Minimum Energy Consumption a) Temperature intervals for the cooling curve are: C : H2 is cooled Total mcp = 10 kw/ C Total ΔQ = 1000 kw C : H1 and H2 are cooled Total mcp = 60 kw/ C Total ΔQ = 7800 kw C : H2 is cooled Total mcp = 10 kw/ C Total ΔQ = 100 kw Temperature intervals for the heating curve are correspondingly: C : C2 is heated Total mcp = 30 kw/ C Total ΔQ = 1500 kw C : C1 and C2 are heated Total mcp = 70 kw/ C Total ΔQ = C : C2 is heated Total mcp = 30 kw/ C Total ΔQ = 1800 kw This is the basis for the following heating and cooling curves if we start by drawing the lowest temperature for each curve at the value of zero on the enthalpy axis: T ( C) Q (kw) Page 1 of 6

2 As indicated, the driving forces between the cooling curve and the heating curve are very small at 90 C on the heating curve. This has relevance for the next sub-task where we have to shift the heating curve towards the right until the vertical distance (and thus the driving forces) between the curves is at least 10 C. b) The heating and cooling curves (Composite Curves) after a sufficient shift until the specified value of ΔT min = 10 C is satisfied, look as follows: T ( C) Q (kw) By reading directly from the graphical diagram, the following approximate values for minimum heating and cooling are obtained: Q H,min = kw and Q C,min = kw c) It is the knee on the heating curve at 90 C that is causing the process Pinch point and thereby limits further heat recovery. It is common to refer to the Pinch point by specifying the corresponding temperatures on both the heating and the cooling curve. In this case, one would express the Pinch point temperatures as: T pinch = 100 C / 90 C Nevertheless, it is the knee at 90 C on the heating curve that is regarded as the true or real Pinch point. This is related to the next sub-task. d) It is important to be aware of the fact that the process Pinch point in the general case may (and will) change (move) with the specified value of ΔT min. This is important for design of heat exchanger networks as will become apparent later in this course. The important issue is that critical design decisions are made at the process Pinch, and the network structure will therefore change whenever the Pinch point changes. Page 2 of 6

3 i) For this particular problem, the Pinch point will remain at 90 C for increasing values of ΔT min, until we reach the situation where the slope change at 90 C on the heating curve is vertically under the slope change at 200 C on the cooling curve. Since the cooling curve is steeper (less total mcp) in the region above 200 C than the case is for the heating curve under 90 C, the Pinch point will jump over to the cooling curve when the value of ΔT min passes 110 C. This is of course an unrealistic and uneconomic high value since heat recovery is drastically reduced. For all practical purposes we can therefore say that for this particular problem, the process Pinch will not change. The so-called hot Pinch temperature will of course change with the value of ΔT min, and this can be expressed by: T pinch, hot = T pinch, cold + ΔT min Here, T pinch, cold is constant and equal to 90 C for all interesting values of ΔT min. ii) As indicated under (i), the Pinch point may change with the value of ΔT min. In the general case much smaller changes in ΔT min are needed to cause a change in the Pinch point, and this must be accounted for in the design phase. Task 2: Numerical Methods and the Heat Cascade ST 500 kw 500 kw 300 C Q H 290 C C R C kw H2 100 kw 500 kw 900 kw 200 C R C C R C H kw 100 C 1500 kw R 4 90 C C R 5 60 C 100 kw kw 2700 kw 900 kw C2 C1 60 C R 6 50 C C Q C 40 C CW Page 3 of 6

4 a) The process Heat Cascade is drawn by using all supply/target temperatures (see the figure above). The Heat Cascade has hot process streams and a set of hot temperatures on the left-hand side, while cold process streams and a corresponding set of cold temperatures are given on the right-hand side. The driving forces are taken into account since corresponding hot and cold temperatures have a difference of ΔT min. We will start by setting the external heating from steam (ST) equal to Q H = 0 kw. This gives the following residual values in the heat cascade: R 1 = 500 kw R 2 = kw R 3 = kw R 4 = R 5 = kw R 6 = kw Q C = kw It is not acceptable to have negative heat flows, and minimum external heating (and cooling) is achieved when we use an amount of steam that exactly makes all residuals non-negative. This will obviously be the case for Q H = 1100 kw, since this amount of extra heat is required to make the residual R 4 equal to zero. The residual values after this adjustment become: R 1 = 1600 kw R 2 = 600 kw R 3 = 900 kw R 4 = 0 R 5 = 900 kw R 6 = 700 kw Q C = 400 kw From this exercise we can conclude the following: Q H,min = 1100 kw, Q C,min = 400 kw, T pinch = 100 C / 90 C (R 4 = 0) b) When the requirements for driving forces are increased from ΔT min = 10 C to 20 C it is expected that the energy demand will increase. In the heat cascade above, intervals, heat flows and temperatures will change. There is no simple way to calculate the increase in energy demand, and a complete new heat cascade calculation must be made. The adjusted heat cascade is shown in the figure below. ST 400 kw 600 kw Q H 300 C 280 C R C 240 C kw H2 H1 900 kw 4500 kw 400 kw 2000 kw 100 kw R C 180 C R C 90 C R 4 70 C 50 C C 40 C Q C 3600 kw 2700 kw 1200 kw C2 C1 CW Page 4 of 6

5 We notice that the number of intervals in the cascade is reduced by 2. This is the result of coinciding temperatures (the supply temperature of 200ºC for hot stream H1 coincides with the interval border caused by the target temperature of 180ºC for cold stream C1, and the same applies to the temperatures 60ºC for H2 and 40ºC for C2). By studying the heat cascade one will quickly realize that heat residual R 3 is the critical one. The heat surplus of 400 kw in the highest (warmest) interval is not enough to cover the total heat deficit of the two next intervals. Minimum external heating and cooling can thus be calculated as: Q H,min = = 1700 kw Corresponding calculation for minimum external cooling gives: Q C,min = = 1000 kw The Pinch point is at the same place in the process and still caused by cold stream C1 which has a supply temperature of 90ºC and a relatively (in this case) large mcp value of 40 kw/ºc. The warm Pinch temperature is of course changing with the value of ΔT min. This gives the following: T pinch = 110 C / 90 C (R 4 = 0) The increase in energy consumption caused by larger demand on driving forces is then: ΔQ H = = 600 kw ΔQ C = = 600 kw Finally a prayer to all examination candidates: Do not make the statement that energy demand increases with 1200 kw, since we do not wish to add apples and pears (or bananas for that matter)!! A more elegant way to solve question (b) is to realize that the Pinch point does not change when ΔT min is increased from 10 to 20ºC. Cold stream C1 will still be the stream causing the Pinch, which means that cold Pinch temperature remains constant at 90ºC. The hot Pinch temperature will of course increase from 100ºC to 110ºC. T 110ºC 100ºC 90ºC tgα = Δ(ΔT) / ΔQ Q In the figure above, the following should be noticed: Page 5 of 6

6 ( T) Tmin,2 Tmin,1 Δ Δ =Δ Δ Δ Q=Δ Q = min,1 min,2 H,min Δ T = = 10 C Δ T = = 20 C Change in position for Cold Composite An alternative expression for the slope of the hot Composite Curve is: tgα = 1 / Σ H (mcp) The change in driving forces (ΔT) is of course equal to the change in minimum approach temperature (ΔT min ), and the change in x axis variable Q is equal to the change in minimum external heating (Q H,min ) and minimum external cooling (Q H,min ). This gives us the following equation: tgα = 1 / Σ H (mcp) = Δ(ΔT min ) / ΔQ H,min Here, ΔQ H,min is the only unknown, since we know Σ H (mcp) = = 60 kw/ºc and Δ(ΔT min ) = = 10ºC. The result is: ΔQ H,min = Σ H (mcp) Δ(ΔT min ) = = 600 kw New external utility requirements when ΔT min = 20ºC then becomes: Q H,min = = 1700 kw Q C,min = = 1000 kw As shown, these results are in agreement with the calculations based on a modified heat cascade. It should be point out, however, that this way to calculate the increase in utility consumption resulting from an increase in the specified driving forces is only valid when: a) There is no change in the Pinch point b) There is no change in the slope of the Composite Curve that is opposite of the Pinch generating stream. Page 6 of 6

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