PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 1. deal gas: Δs = s - s 1 = c p ln(t /T 1 ) - R ln(p /p 1 ) (T << T crt; p << p crt ) ΔĖx = ṅ T Δs T T 1 (0 1 bar, h h 1 ) ṅ = 1 kg/s / 0,03 kg/ml = 31,5 ml/s Δs = - R ln(1/0) = 4,9 /(mlk) ΔĖx = ṅ T Δs = 31,5 93 4,9 = 8,1 kw. 1 kg l = 1000 g / 7 g/ml = 37,0 ml 37 ml l + 1½ 37 ml O ½ 37 ml l O 3. Tables curse materal sectn 1.8: b chem, l = 795,7 k/ml b chem, lo3 = Δ f G lo3 + 1½ b chem, + b chem b chem, lo3 = 15,1 k/ml Δex = b chem, l - ½ b chem, lo3 = 795,7 ½ 15,1 = 788, k/ml l = 9, M/kg l (= 8,1 kwh/kg l) Ths s als the mnmum exergy nput need fr l prductn frm pure l O 3. PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 3. Per kg steel: n Fe = 17,55 ml, b Fe = 374,3 k/ml n C = 0,50 ml, b C = 410,3 k/ml n Mn = 0,13 ml, b Mn = 487,7 k/ml n S = 0,09 ml, b S = 609,6 k/ml n P = 0,03 ml, b P = 861,4 k/ml n S = 0,11 ml, b S = 854,9 k/ml n Mn3C = n Mn /3 = 0,043 ml n Fe3C = n C n Mn3C = 0,457 ml free Fe n Fe = n Fe 3n Fe3C = 16,13 ml, n free C b Mn3C = -Δ f G fr Mn 3 C +3 b Mn + b C = 1907,9 k/ml b Fe3C = -Δ f G fr Fe 3 C +3 b Fe + b C = 1539,3 k/ml b n b chem, n 700k / ml (RZ calculates 6997,7 k/kml wth Σn = 16,86 ml) Exergy decrease as a result f mxng: n R T Σx lnx = -0,56 k/ml ths purty!) (can be neglected fr
PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 4. Fr the feed, wth enthalpy = 0 fr lqud at T = 0 C: cndenser F = (x c p + x B c pb )(T - 0 C) = heater = (0,45 189 + 0,55 75,5) 178 C = 40,4 k/ml Flash e-balance (lever rule): L G L1 53,0 40,4 40,4 4,8 1 G1 F 1 F 1, & L 1 + G 1 = 1 ml/s L 1 = 0,45 ml/s, G 1 = 0,55 ml/s Cndenser: 50/50 splt G = L = 0,55/ = 0,75 ml/s G + L = 34,3 k/ml Q c = (50,3 34,3) k/ml 0,55 ml/s = 10,3 kw PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 5. (Earler curse Transprt prcesser (TRP), G. Öhman, exam questn) p x, y T 101,3kPa, p 90,18K fr N. Rault' s Law : p x x p p tt tt x p p p 8 T 101,3kPa, 77,35K (1 x) p and y y, x p p tt 8 L, ( T 73K) 65 k / mlk L, L (77K 73K) 65 k / mlk 01k / kg 8 kg / kml ( T 77K) 30,5 k / mlk (90K 73K) 65 k / mlk 3 k / kg 3 kg / kml ( T 90K) 30,5 k / mlk G y (1 y) Prcedure: pck a Tbl,N < T < Tb,l x, y, and crrespndng L and G. Nte: n mxng heat s cnsdered here.
PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 6. Curse materal: sectn 3.6; nt equlbrum z = D/F ½, B/F = 1/z; x F = z Per mle feed, f q = 1 (blng lqud) : L/F = r D/F = z r abve the feed L/F = r D/F = z r +1 at/belw the feed V/F = = z (r + 1) n the whle clumn Mnmum heat nput needed: Q mn = V Δ vap = z (r mn + 1) Δ vap Fr ths mxture then W sep = W mn = -R T (zlnz + (1-z)ln(1-z)) = Q Rmn T (1/T tp 1/T bttm ) = (Q n, mn T R/ Δ vap ) (- ln α) (zlnz + (1-z)ln(1-z)) = (Q n, mn / Δ vap ) (- ln α) = z (r mn + 1) (- ln α) r mn = -1 + [ (zlnz + (1-z)ln(1-z))/( (- ln α) ) ] If x F becmes lwer, then mre lqud flw dwnwards needed fr strppng the lght cmpnent frm (mre) heavy cmpnent. r mn Nt necessarly symmetrc wth peak at z = ½ W mn Check ths als fr q=0 (saturated gas feed), t see the effect f preheat! z= x F D/F PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 7. a L11 X L1 X m D11 D1 and L X L X D D m. m L X L X m D D L X L X D D m m 1 μ d d (Bnary dffusn, -dc / + dc B / = 0, T T where X, X m, L L1 (Onsager) c B + c = cnstant, m = - mb, c wth chemcal ptental μ μ RTlny μ RTln gves n extra nfrmatn) c (μ chemcal ptental fr pure ).Ths gves : dt/ = 0 m = -D dc /, Fck s L11 μ dt L1 R dc Law, D = D, L11 X L1 X m and T c m = 0 -D dt/ = D dc / L μ dt L R dc m L X L X m dc / = -D /D dt/ T c Q Q = -(D 11 - D 1 D /D ) dt/ λ = -(D 11 - D 1 D /D ) b. m = 0 dc / = -D /D dt/ dc /dt = -D /D r dc / = -D /D dt/ dt/ 0 means that dc / 0, whch can be used t measure D r D 1. bulk
PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 8. Exam 9 March 009: Questn 301. a. h D = 4, k/ml Q c /D = 57,8 M/mn / 0,5 kml/mn = 115,6 k/ml ple pnt N at x = x D = 0,95; h N = 4, + 115,6 = 119,8 k/ml Q B /B = 45,4 M/mn / 0,77 kml/mn = 6,5 k/ml ple pnt M at x = x B = 0,05; at h = 6,5 k/ml belw saturated lqud lne fr x = x B, h M = 31 6,5 = -31,5 k/ml PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 (8. Exam 9 March 009: Questn 301.) b. F = B + D, mass balance, F = 17 ml/mn methd 1: energy balance: F h F + Q B = Q C + B h B + D h D h F = (Q C + B h B + D h D - Q B ) / F = (57,8-45,4 + 0,004 500 + 0,031 77)/17 M/mn/ml/mn = 30, k/ml dagram: x F ~ 0,415 methd : lever rule fr fndng pnt F n lne MN length rat FM/MN = 500/17 x F ~ 0,415
PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 (8. Exam 9 March 009: Questn 301.) (b. cntnues) pnt F s almst n te-lne fr T = 100 C: T ~ 99 C. lever rule fr te-lne at 99 C: lqud /ttal = length frm F t saturated gas /ttal = 71% gas/ttal = 9%,.e. q = 0,71 c. Draw lnes as n Fgure: V = gas frm tp tray L = lqud frm tp tray etc. 5 stages (4 + rebler) T C x - L k/ml y - G k/ml 68.7 1 1.98 1 41.87 69.35 0.977 13. 0.9963 4.03 70 0.9546 13.45 0.993 4.18 75 0.8017 15.19 0.9609 43.39 80 0.6684 16.94 0.937 44.7 85 0.5566 18.65 0.8788 46.14 90 0.4566 0.36 0.835 47.74 95 0.3705.05 0.757 49.53 100 0.917 3.75 0.6756 51.57 105 0. 5.44 0.584 53.8 110 0.1591 7.13 0.4754 56.33 115 0.1035 8.8 0.355 59.14 10 0.0538 30.46 0.088 6.33 1.9 0.06 31.43 0.1096 64.46 15.8 0 3.39 0 66.78 9. Exam 9 March 009: Questn 30. PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 a. α = 1.5 = p C5 / p C8 at 70 C, 1 bar (see als nmgram curse materal #3 sldes 14,15 Clausus Clapeyrn: ln α -Δ vap /R (1/T bttm 1/T tp ) Δ vap - R ln(1,5) / (1/T 398 1/T 309 ) = 9,0 k/ml 30 k/ml b. deal, per mle feed: Q n, mn = ½ (r mn + 1) Δ vap = 15,96 k/ml real, per mle feed: Q n, real = ½ (r real + 1) Δ vap = 16,68 k/ml deal, per mle feed: W n, mn = Q n, mn T (1/T tp 1/T bttm ) = 15,96 93 (1/309-1/398) = 3,38 k/ml
PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 (9. Exam 9 March 009: Questn 30.) c. real, per mle feed: W n,real = Q n, real T (1/T tp 1/T bttm ) = 16,68 93 (1/303-1/408) = 4,15 k/ml lsses clumn: (Q n, real - Q n, mn ) T (1/T tp 1/T bttm ) = 0,153 k/ml lsses rebler: Q n, real T (1/T bttm 1/T rebler ) = 0,301 k/ml lsses cndenser: Q n, real T (1/T cndenser,ut 1/T tp ) = 0,313 k/ml 15,96 16,68 1 138 1 1178 n,mn clumn Effcency = 0, 816. Q n, real T Q ( (1/ Nt bad, reflux rat s lw. T rebler (1/ (1/ clumn (1/ cndenser d. pparently r = unchanged (may nt be s!); but nw Q n, mn = 1/3 (r mn + 1) Δ vap, etc. Ths gves new values Q n, mn = 10,6 k/ml feed Q n, real = 11,1 k/ml feed W n, mn =,7 k/ml feed W n,real =,77 k/ml feed Ex lss, cl = 0,10 k/ml feed Ex lss, cnd = 0,01 k/ml feed Ex lss, reb = 0,08 k/ml feed Effcency 0,815 unchanged ) PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 10. Exam 9 March 009: Questn 401. a. dt/ = 0 m = 10-6 kg/(m /s) = - D dc / D = Ð = 10-5 m /s and dc / = 0,1 (kg/m 3 )/m. dt/ = -100 K/m gves m = 0, L11 X L1 Xm D 11 D1 and m L 1 μ d d T T where X, Xm,L L1 (Onsager) c wth chem cal ptent al μ μ RTlny μ RTln cbulk (μ chemcal ptental fr pure ). Ths gves : L11 μ dt L1 R dc L11 X L1 Xm and T c L μ dt L R dc m L X L Xm T c -D dt/ = D dc / D = -D (dc /)/(dt/) = 10-5 m /s 0,1 / 100 = 10-8 kg/(m s K) Q = (-D 11 +D 1 D /D ) dt/ = -λ dt/ λ = D 11 - D 1 D /D, where D 1 D BT L 1 = L D 1 = -L 1 R/c avg, and - D L /T avg = L 1 /T avg L 1 = - D T avg D 1 = D T avg R/c avg, = 10-8 (330) 5/0,01 = 7, W m /kg λ = 0,03 W/(m K) = D 11 - D 1 D /D D 11 = λ - D 1 D /D = 0,03 0,07 = 0,003 W/(m K) b. Q = -(D 11 - D 1 D /D ) dt/ = -(0,003-7, 10-8 /10-5 ) 100 = 0,04 100 =,4 W/m Furer law Q = -λ dt/ gves Q = 0,03 100 = 3 W/m X L X m dt D D dc
PET 44304 015 Exercses 3+4 f 4 10 Feb + 13 Feb 015 11/ Exam 18 May 011 questn 109 a. Δh = c p (100-5)K =,68 k/ml Δs = c p ln(383/98) - R ln(100/1) = 8,0-38,3 k/(mlk) 100 bar, 5ºC: ex phys = RT 0 ln(p/p 0 ) = 11,4 k/ml 100 bar, 100ºC: ex phys = Δh - T 0 Δs =,68 + T 0 (38,3 8,0) = 11,7 k/ml ex chem (5 bar, 100ºC) = 843,0 k/ml ex chem (100 bar, 100ºC) = 843,4 k/ml b. exº chem MeO = -166,7 + 410,6 (fr C) + ½ 3,97 (fr O) + 36,09 (fr ) = 718, k/ml exº chem C 4 = 831,6 k/ml (data gven under a.) a reductn f 13,6 % MeO: C 4 : 718, k/ml / 3 kg/kml =,44 M/kg 831,6 k/ml / 16 kg/kml = 51,98 M/kg