Truyen Khoi 3

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    07-Dec-2015

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Truyen Khoi 3

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  • *CHNG 3CHNG CT

  • *I. Khi qutChng l phng php dng tch cc hn hp cht lng cng nh cc hn hp kh lng thnh cc cu t ring bit da vo bay hi khc nhau ca cc cu t trong hn hp

  • *cc phng php chng Chng n gin Chng bng hi nc trc tip Chng chn khng Chng ct

  • *1. PHN LOI HN HP HAI CU T Cn c vo mc ho tan ta c th chia dung dch hai cu t thnh cc loi sau:Cht lng hon ton ho tan vo nhau theo bt c t l no.Cht lng ho tan mt phn vo nhau.Cht lng khng ho tan vo nhau.

  • *Xt hn hp gm hai cu t A v B. theo nh lut Rauolt ta c p sut ring phn ca cc cu t l:PA = PbhA.xA PB = PbhB.xB = PbhB.(1 xA) p sut tngP = PA + PB = PbhA.xA + PbhB.(1 xA)

  • *

    x5 x3 x1 x4 x2

  • *2. CHNG N GIN. 1 Nguyn tc v s chng n gin, th txy.

    XW XP XD

  • *Chng n gin c ng dng cho nhng trng hp sau:Khi nhit si ca hai cu t khc nhau xa:Khi khng i hi sn phm c tinh khit cao.Tch hn hp lng ra khi tp cht khng bay hi.Tch s b hn hp nhiu cu t.

  • *Tnh ton qu trnh chng n gin.Phng trnh cn bng vt cht cho ton b qu trnh l:

  • *3. CHNG CT 3.1. Nguyn tc lm vic ca thp. Hi i di ln qua cc l ca a, cht lng chy t trn xung di theo cc ng chy chuyn. Nng cc cu t thay i theo chiu cao ca thp, nhit si cng thay i tng ng vi s thay i nng .

  • *Trn mi a xy ra qu trnh chuyn khi gia pha lng v pha hi. Do mt phn cu t d bay hi chuyn t pha lng vo pha hi v mt phn t hn chuyn t pha hi vo pha lng.Qu trnh chng ct c thc hin trong thit b lai thp lm vc lin tc hay gin an.

  • *3.2. Chng ct lin tc 3.2.1 S h thng thit b chng ct lin tc.

  • *Phng trnh cn bng cho ton thp.F = W + D F.xF = W.xW + D.xD Trong : F, W, D - sut lng nhp liu, sn phm y v nh, kmol/hxF, xW, xD - phn mol ca cu t d bay hi trong nhp liu, sn phm y v nh.

  • *Phng trnh ng nng lm vic ca an ct. Gi thuyt:S mol ca pha hi i t di ln bng nhau trong tt c tit din ca thp.Hn hp u vo thp nhit si.Cht lng nhng trong thit b ngng c thnh phn bng thnh phn hi ra khi nh thp.un si y thp bng hi t gin tip.S mol cht lng khng i theo chiu cao ca an ct v chng.

  • *Phng trnh:

    Vi l ch s hi lu ca thp

    Gx- lng lng c hi lu, kmol/h

  • *Phng trnh ng nng lm vic ca an chng.

    Vi

    lng hn hp nhp liu so vi sn phm nh.

  • *Xc nh ch s hi lu RxCh s hi lu ti thiu

    Ch s hi lu: R = b.Rmin

  • *Xc nh s mm l thuytCc bc xc nh:Xc nh ch s hi luXc nh cc ng lm vic.Xc nh ng cn bng.V th ng lm vic v ng cn bng.V s bc thang thay i nng .S bc thang l s mm l thuyt.

  • *3. Cn bng nhit lng ca qu trnh chng ct

  • *Cn bng nhit lng cho thit b gia nhit nhp liu

  • *Phng trnh cn bng:Qnl = F.CpF.(tFv tFr) + Qm

  • *Cn bng nhit lng cho thit b ngng t

  • *Phng trnh

    Qng = D.(R+1).rD = G.C.(tr tv) + Qm

  • *Cn bng nhit lng ca thit b lm lnh

  • *Lm lnh sn phm nh

    QiiD = D.CpD.(tsD tD) = G1.C.(tr tv) + Qm

  • *Lm lnh sn phm y

  • *Phng trnhQiiW = W.CpW.(tsW tW) = G2.C.(tr tv) + Qm

  • *Cn bng nhit lng ton thp

  • *Phng trnhQF + QK + QL0 = QD + QW + Qm + Qng

    Hay:

    QK = QD + QW + Qm + Qng - QF - QL0

  • *V dThp chng ct lin tc di p sut thng dng sn xut 700kg/h axit acetic vi nng 80% mol. Nhp liu vo vi nng 35% mol axit acetic nhit si. Sn phm nh l nc cha 5% mol axit acetic. Thp c gia nhit bng hi bo ha kh.1.V s h thng chng ct, ng cn bng lng (x) hi(y) % mol

  • *2.Tnh lng sn phm nh v nhp liu (kmol/h) .3. Cho t s hon lu bng 1,5 Rmin. Xc nh s mm thc ca thp cho bit s mm thc tng ng vi mt bc thay i nng l 2.4.Tnh chiu cao ca thp bit khong cch gia cc a l 0,4m, b phn gia nhit di y thp l 1m, khong cch t a trn cng ln nh thp l 0,6m (B dy ca a l khng ng k)

  • *Thnh phn cn bng lng (x) hi(y) % mol nc acetic x051020304050 y09.216.730.342.553.062.6x60708090100y71.679.586.493.0100

  • *Tm tc W=500kg/hxw= 30%mol ncxF= 71%mol ncxD= 93%mol ncR= 1.5 Rmin

  • *1. V s h thng chng ct

  • *V ng cn bng lng hi

  • *2. Tnh lng sn phm nh v nhp liu-Tnh lu lng mol sn phm y:Khi lng trung bnh hn hp nhp liu:

    Lu lng mol sn phm y: W

  • *Phng trnh cn bng vt liuF = W + D F.xF = W.xW + D.xD Thay gi tr vo ta c:F = 11 + D Fx0.71 = 11x0.3 + Dx0.93Gii h phng trnh ta c:F = 31.5 kmol/hD = 20.5 kmol/h

  • *Xc nh ch s hon lu ti thiu

    xF= 0.71 => y*F = 0.80 mol/mol

  • *Ch s hon lu thc tCh s hon lu thc t =1.5 RminSuy ra:R = 1.5*1.44 = 2.16

  • *Xc nh s mm thc ca thpXc nh s bc thay i nng :- Phng trnh ng lm vic ca on ct:

  • *- Phng trnh ng lm vic ca on chng:Vi: L= 31.5/20.5=1.54Suy ra:

    Pt ng lm vic on chng:y = 1.17x - 0.05

  • *_on cty = 0. 68x + 0.3_on chngy = 1.17x - 0.05

  • *V th cc ng lm vic