Chapter 3. Economic Decision Making: Design of a Solvent Recovery System

Transcription

1 Chapter 3 Economic Decision Makg: Design of a Solvent Recover Sstem

2 Followg stream is currentl sent to flare: 10.3 mol/hr acetone 687 mol/hr air Is it beneficial to recover acetone? Economic Potential (EP) EP = Product value Raw material cost Acetone price = $ 0.27 /lb then: EP = (10.3 mol/hr)($ 0.27/lb)(58 lb/mol)(8150 hr/r) = $ /r

3 Workg hours: hr/r for contuous process 7500 hr/r for batch process solvent recover alternatives (vapor recover) 1 Condensation a) high pressure b) low temperature c) both 2 Absorption 3 Adsorption 4 Membrane 5 Reaction

4 J.R. Fair suggests that if the solvent concentration gas stream is less than 5% then the adsorption is the cheapest alternative. Harder tech. and less experienced process (adsorption) Which one? VS. Simpler tech. and more experienced process (absorption) We do not necessaril want to develop rigorous designs itiall because we want to build onl one of those processes (at best). Hence, we onl want to clude sufficient accurac for screeng the alternatives. For example we show a shortcut method for designg an absorption sstem for earl stage of design. (screeng)

5 Design 1: Design 2: solvent

6 Acetone is lost two places 1 - exit gas from absorber (with Air) 2 distillation bottom Trade-off : degree of recover of acetone and capital cost of the absorber because of higher number of tras. Rule of thumb: It is desirable to recover more than 99% of all valuable materials (normall 99.5%) We also have to fix the water flow rate to the gas absorber before we can calculate the material balances. For fixed recover of solvent, the greater the solvent flow rate, the fewer the tras that are required but the greater load on the distillation column.

7 Rule of thumb: For an isothermal, dilute absorber, choose L such that L mg =1.4 (ranges from 1.25 to 2.0) Material Balance For acetone water sstem at 77 F and 1 atm, (activit coefficient) γ = 6.7 and (partial pressure) P = 229/ 760 atm Henr s low for non-ideal solutions: m m T m = L =1.4 m G = 1.4 (2.02) (687) = 1943 mol/hr P i x i i = γ P o T i i i i o γ i Pi = slope of the equilibrium le = P = 2.02 x water flow to absorber

8 for 99.5% recover of acetone acetone loss = (10.3) = mol/hr acetone to distillation column = = mol/hr for 99.5% recover of acetone distillation column acetone at top = (10.25) = 10.2 mol/hr If the acetone composition specified to be 99% as product Top total flow = (10.2) 0.99 =10.3 water at top = = 0.1 mol/hr bottom products (10.25) = 0.05 mol/hr acetone = mol/hr water

9 For acetone-water no recclg process Acetone loss absorber overhead ($ 0.27 /lb)(58 lb/mol)( mol/hr)(8150 hr/r) = $ 6600/r Acetone loss Distillation bottom ($ 0.27 /lb)(58 lb/mol)( mol/hr)(8150 hr/r) $ 6600/r Pollution treatment cost [assume ($ 0.25 /lb BOD) and (1 lb acetone /lb BOD)] BOD = Biological Oxgen Demand ($ 0.25 /lb BOD)(1 lb BOD /lb acetone)(58 lb/mol)( mol/hr)(8150 hr/r) = $ 6100/r Sewer charge [assume $ 0.2 /1000 gal] ($ 0.2 /1000 gal)(1 gal /8.34 lb)(18 lb/mol)(1943 mol/hr)(8150 hr/r) = $ 6800/r Solvent (water) cost (assume $ 0.75 /1000 gal) ($ 0.75 /1000 gal)(1 gal /8.34 lb)(18 lb/mol)(1943 mol/hr)(8150 hr/r) = $ /r Now compare $ 6600 /r + $ 6600 /r + $ 6100 /r + $ 6800 /r +$ /r = $ /r With $ /r Thus we want to contue developg the design.

10 Energ balance Normall we assume Coolg water 90 F Coolg water 120 F 77 F and T=10 F means: cools the solvent to 100 F 77 F 90 F 120 F? 135 F Acetone B.P. 90 F 120 F 100 F 212 F Water B.P. 77 F 77 F 276 F, 25 Psia Steam 90 F 120 F 100 F

11 We still must specif the temperature of stream enterg the distillation column. Saturated-liquid feeds are the most common case. Thus, we guess a temperature of 200 F (10.3 mol of acetone boilg at 135 F and 1943 mol of water boilg at 212 F). If we do not heat the column feed, up to saturation temperature we have to pa more for reboiler of the column, therefore the total heat load of preheat and the reboiler will crease. Havg the temperature and flow rates of the streams, calculatg the energ balance is not ver difficult. (However we will give a procedure for HEN design later) flow rate heat capacit Q = F C i i i p ( T T )

12 Equipment Design Considerations Gas absorber Kremser equation: N + 1 = ln L mg 1 ln L mg mx mx + 1 We can now check the effect of the design variables: Effect of Column pressure: suppose we double the pressure of the absorber (P T ) γ P PT i i L = 1.4mG L = 1.4 G o From this equation b doublg P T we decrease L b a factor of 2, but we still use the same number of plates the absorber (because L/mG is still 1.4).

13 Trade offs: Lower values of L decreases the load on the distillation column. Therefore decreasg solvent flow to the absorber has a significant effect on distillation column but no effect on the number of tras required the absorber. However, the diameter of absorber will decrease. On the other hand, for doublg the pressure of the absorber we must use a compressor which is a ver expensive piece of equipment. Effect of solvent tpe: if we use MIBK (Methl Isobutl Ketone) as solvent stead of water, forms an essentiall ideal mixture with acetone with γ =1 (stead of γ = 6.7 for water-acetone sstem) which will decrease m b a factor of 6.7 for this sstem, so Cut the liquid rate (L =1.4 m G) and therefore still cost, but this solvent costs much much higher than water. Effect of operatg temperature: if we change the let water temperature to 40 C γ P i i (100 F) then γ = 7.8 and P = 421 mmhg. From L = 1.4 G we can see that L will crease so does the cost of distillation column. PT o

14 Shortcut calculation for sizg absorber Kremser equation: N + 1 = ln L mg 1 ln L mg mx mx + 1 For an expensive absorber which is worth to be considered at screeng stag it should have at least 10 to 20 plates. Therefore estimatg N+1 b N bares onl at most 10% error.

15 for pure solvent, x = 0 usg rule of thumb : L/mG = 1.4 and 99% recover ( 100 ) = = 41 at this stage of approximation : ln 41 ln 40 + = mg L ln mg L ln N mg L ln mg L ln mg L

16 from Talor's series if L/mG is between we can use ln (L/mG) L/mG 1 ln (L/mG) = = 0.4 N = ln log 0.4 = N = 6 log 0.4 = 6 log (0.4) + log = 2 + log N + 2 = 6 log This approximation for a recover of 99% gives 10tras stead of actual value of 10.1 and for recover of 99.9% gives 16 tras which is a ver good estimation.

17 Distillation column Almost the same approach has been chosen for distillation column and shortcut design of this equipment is presented appendix A.2 and A. 3 FUG Fenske Underwood Gilliland Fenske equation (for N m ) N m x D (1 x ln ln SF x (1 D ) xw = = ln α ln α Underwood equation (for R m ) R m W ) [ xd( q 1) + ZF( Rm + 1) ] [( R + 1)(1 Z ) + ( q 1)(1 x )] RmZ F + qxd α = ( 1 ZF) + q(1 xd) m F D Gilliland equation (for theoretical number of tras) N N m R R = N + 1 R + 1 m SF Separation Factor q heat required to vaporize 1 mole of feed divided b the heat of vaporization

18 Homework Due time: two weeks Chapter 3: 2, 3, 4, 5, 6, 7