n b n Mn3C = n Mn /3 = 0,043 mol free Fe n Fe = n Fe 3n Fe3C = 16,13 mol, no free C b Mn3C = -Δ f G for Mn 3 C +3 b Mn + b C = 1907,9 kj/mol

Μέγεθος: px

Εμφάνιση ξεκινά από τη σελίδα:

Download "n b n Mn3C = n Mn /3 = 0,043 mol free Fe n Fe = n Fe 3n Fe3C = 16,13 mol, no free C b Mn3C = -Δ f G for Mn 3 C +3 b Mn + b C = 1907,9 kj/mol"

Τελαμών Ἁνανίας Κοτζιάς
5 χρόνια πριν
Προβολές:

1 PET Exercses 3+4 f 4 10 Feb + 13 Feb 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 Exercses 3+4 f 4 10 Feb + 13 Feb 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

2 PET Exercses 3+4 f 4 10 Feb + 13 Feb 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, ,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 Exercses 3+4 f 4 10 Feb + 13 Feb (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.

3 PET Exercses 3+4 f 4 10 Feb + 13 Feb 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 Exercses 3+4 f 4 10 Feb + 13 Feb 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

4 PET Exercses 3+4 f 4 10 Feb + 13 Feb 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 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, ,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.

5 PET 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 Exam 9 March 009: Questn 30. PET 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

6 PET 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, 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 Exercses 3+4 f 4 10 Feb + 13 Feb 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, =,4 W/m Furer law Q = -λ dt/ gves Q = 0, = 3 W/m X L X m dt D D dc

7 PET Exercses 3+4 f 4 10 Feb + 13 Feb / 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, ,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

Σχετικά έγγραφα

2 nd Law: m.s in + Q in /T in + S gen = ds/dt + m.s out + Q out /T out. S gen /ṁ = -(Q in /m)/t + Δs,

2 nd Law: m.s in + Q in /T in + S gen = ds/dt + m.s out + Q out /T out. S gen /ṁ = -(Q in /m)/t + Δs, P 444 5 xercises + f 4 6 Jan + Jan 5. kg ice at 6 K kg ater at 9 K. Heat Q at = is supplie y the surrunings. Specific heat ice : c i = 4 kj/kgk; ater c = 48 kj/kgk melting heat Δ m H = 4 kj/kg n La: m.s