sch Q multitubularSchimbator de caldura condensator abur-apa multitubular cu mai multe treceriN=12Qu=320[kw]ps=2.6[bar]t'=88[C]t"=118.2[C]0.0070.00140.004341. Calculul de predimensionare4.22334a. din anexa 4 alegem in functie de ps =>r=2178[kJ/kg]ts=128.73[C]b. aleg iz= 0.993-0.995 =>iz=0.994c. calculam debitul de aburD=0.1478106472[kg/s]d. din tabel alegem cp2 in functie de t2m =>tm=103.1[C]cp2=4.22334[kJ/kgK]G=2.5089210175[kg/s]e. calculam tm =>tm=22.325114956t=40.73t=10.53f. alegem coeficentul globa de caldura k'=900-1200 [w/mK]k'=1050[w/mK]S=13.6510788573[m]g. aleg tipul de teava18 x 1.5dm=16.5[mm]de=18[mm]di=15[mm]h. aleg lungimea tevilor preliminara l'=2-4 [m]l'=2[m]n'tot=131.6749502227i. din anexa 5 in funtie de tm prin interpolare pe m =>m=956.037[kg/m]aleg w'=0.5-1.5 [m/s]w'=0.8[m/s]i. din ecuatia de contiuitate a scurgerii aflam n'trn'tr=18.5630742912n'tr=7.0933805552aleg 6 =>ntr=6ntr=21w=0.7071647349[m/s]ntot=126j. pentru a alege mantaua aleg:s=26[mm]k'=8[mm]avemk=9[mm]s=de+8-10[mm]k'=8-12[mm]calculam:Dim=398[mm]aleg 400cresc k0 cu 1mm scriu in desen ca facem comanda de o teava roluita de 400 mmDim=400[mm]aleg grosime manta intre 8-10[mm]m=9[mm]Dem=418[mm]2. Calculul DEFINITIVa. determinam :Pr *10 din anexa 5 in fct de tm aflam Pr, , 1001.750.2950.6827tm=103.11.69420.286630.6834131101.570.2680.685 [-][m/s][W/mK]lc=15[mm]Re= w*lc/Re=37007.5394185527[-]Re=0.9964088886Nu=133.7917335599[-] =Nu* /di =6095.6673338241[W/mK]b. determinam /: /= d/d+ol/olde la pag.15 aleg apa tratata de alimentare a cazanelor =>d/d=Rd=0.00018[mK/W]ol=0.015[m]/=0.00048[mK/W]ol=50[W/mK]c. determinam :din anexa 5 in functie de ts aflam: c, c, cccc*101200.6862943.10.244ts=128.730.6862935.85410.2282861300.6862934.80.226[W/mK][kg/m][m/s]=c*t^(-0.25)unde:c=23087.7410439016g=9.81[m/s]c=0.6862[W/mK]c=0.0006440509c=935.8541[kg/m]r=2178000[W*s/kg]c=0.0000002283[m/s]de=0.018[m]t se alege prin metoda injumatatirii intervalului:14.8696814883g(t)=c*c*t^0.75+t-tmtm=22.325114956g(t=0)= -22.325114956g(t=1.395313685)=-1.83975006531.3953196847g(t=2.790639369)=12.57102557492.7906393695g(t=5.581278739)=37.2509654565.581278739g(t=11.16255748)=79.645512860411.162557478g(t=22.32511)=152.7203616472t =2.7906393695=17863.0458702143[W/mK]d. calculam coficientul global de transfer de caldura k:k=1428.5052245378[W/mK]e. se modifica suprafata de transfer de caldura S:S=10.0340079644[m]f. se modifica lungimea tevilor l:l=1.5363[m]rotunjim doar "+"alegem l=1.6[m]3. Calculul izolatiei termicedeterminam grosimea necesara izolatiei pentru a avea randamentul propus=0.994a. aleg izolatia: vata minerala =>iz=20[mm]iz=0.05[W/mK]b. determinam em= f(Nue)= f(Pr, Gr) dar Gr depinde de => aleg te'=20-30 [C]te'=22[C]te=15[C]'em= 10.612te''=21.6843349603[C]c. verificam daca |te'-te''| 3C este bine daca nu alegem te'=te'' si recalculam em|te'-te''|=0.3156650397[C]km=2.0233374009Qm=529.7594126727[kW]Qp=794.6391190091[kW]Qcons=321.9315895372[kW]sc=0.99753165224. Calculul hidraulica. calculam pierderile de sarcina pentru agentul termic secundarRe=37007.5394185527=0.0222906586pierderile de sarcina liniare pp=3410.2700284509[N/m]l=9.6pierderile de sarcina locala pdimensionare racordaleg wr'=0,51,5 [m/s]wr'=0.75[m/s]dir=0.0770725877[m]aleg din anexa 12-13 o teava STAS cu didir => dirSTASdirSTAS=0.089[m]cu grosime peretelui de 0.006 [m]recalculam wr =>wr=0.4218347901[m/s]Tipbucatiw p=**w/2buc * p[-][-][-][kg/m][m/s][N/m][N/m]racord complet deschis20.5956.0370.421834790142.530403031485.0608060629colector de intrare sau iesire21956.0370.421834790185.0608060629170.1216121258intrare in spatiul din interiorul tevilor si iesirea din acesta61956.0370.7071647349239.04842949091434.2905769454intoarcere cu 180 dintr'o sectiune in alta, printr'o curba in preincalzitoare sectionale52956.0370.7071647349478.09685898182390.4842949089p4079.957290043pierderile de sarcina pentru agentul termic secundar sunt p=7490.2273184939[N/m]

Cazan ignitubularCazan ignitubular cu piese de intoarcere a gazelor de ardere in focarN=12[-]Qu=170[kW]tai=60[C]tae=80[C]Combustibilul gazos este caracterizat de compozitia volumetrica procentuala:

CH=92.6[%]CH=1.2[%]CH=0.6[%]CH=1.2[%]N=3.4[%]CO=1[%]verificare:100[%]1. Calculul arderii combustibilului gazoscalculul TEORETICa) volumul teoretic de aer necesar arderii stochiometrice, teoretice fara exces de aer, a unui metru cub normal de combustibil:V=9.53904[m/m]

b) componentele gazelor de ardere pentru arderea stochiometrica, teoretica fara exces de aer volumul de bioxid de carbon:VCO=1.0161[m/m] volumul de bioxid de sulf din gazele de ardere:VSO=0[m/m]

volumul gazelor triatomice CO si SO din gazele de ardere:VRO=1.0161[m/m] volumul de azot din gazele de ardere:VN=7.5698416[m/m] volumul de vapori de apa din gazele de ardere:VHO=2.12462464[m/m]

volumul teoretic de gaze de ardere fara exces de aer(=1):Vg=10.71056624[m/m]calculul REALALEGEM = coeficientul de exces de aer ca fiind raportul dintre aerul efectiv de ardere Va si cel teoretic necesar arderii unitatii de comustibil V0= 1.051.1= 1.05pentru focare cu ardere intensificata

calculul arderii cu exces de aer: volumul real de aer necesar arderii:V=10.015992[m/m] volumul real al gazelor de ardere:Vg=11.18751824[m/m]c) densitatea gazelor de ardere la stare normala:

=1.2469859138[kg/m]

presiunea partiala a gazelor triatomice din gazele de ardere:PRO=0.0908244329[bar]

presiunea partiala a vaporilor de apa:PHO=0.1899102727[bar]d) Puterea calorica inferioara a combustibilului gazos:H=36010.2[kJ/m]

2. Diagram I-tEntalpia gazelor de ardere rezultate din arderea unitatii de combustibil (kJ/) sau (kJ/kg)

tgVRO=1.0161VN=7.5698416VHO=2.12462464Vg=10.71056624V=9.53904= 1.05(-1)= 0.05iROiROxVROiNiNxVNiHOiHOxVHOIg=(ixV)iaeriaer x V=IV(-1) x IVI= Ig + (-1) x IV123456789101112000000000000100170172.737129.5980.2944872150.5319.756008321472.787495521301240.075262.003761534.79125552200357.5363.25575259.81966.64484768304.4646.7357404162976.636338096261.32492.551152124.62755763101.263895696400771.8784.22598526.53985.5216024626.21330.4399495686100.187531968531.65070.953664253.54768326353.73521516860012241243.70648046086.1526464968.52057.698963849387.55801024813.77761.916848388.09584249775.6538526480017041731.434410938273.836868813342834.2492697612839.52053856110710559.71728527.98586413367.50640256100022032238.4683139110529.649665617223658.6036300816426.72159568140913440.50736672.02536817098.74696368120027162759.7276169712846.021195221324529.6997324820135.44852768171916397.60976819.88048820955.32901568140032383290.1318200815200.241932825585434.7898291223925.16356192203319392.86832969.64341624894.80697792160037723832.7292231117493.903937630016375.9985446427702.63168224235322445.361121122.26805628824.89973824180043034372.2783264320007.091348834587346.9520051231726.32165392267525516.9321275.846633002.16825392200048434920.9723293422209.915254439248337.0270873635467.91464176300128626.659041431.33295236899.24759376

3. Bilantul general al cazanuluitcos=130[C]Igcos=2004.7330475728[kJ/m]taer=20[C]170000Iaer=248.01504[kJ/m]qcos=0.0484395326qch=0.001qext=0.005=0.9455604674B=0.004992684[m/s]

4. Bilant partial pe suprafete

It=36234.605592[kJ/m]tt=1965.890250426[C]tf=550[C]If=8920.174193272[kJ/m]Qf=135.6904636069[kW]Qc=34.3095363931Ic=2013.6794166963tc=130.5711155684[C]

t=0.031108372[%]I=0.026136145[%]5.Calculul termic al focaruluialeg un flux radiant in camera de ardere qr=30-40 [kW/m]qr=35[kW/m]suprafata de transfer de caldura prin radiatieSRC=3.4891833499[m]Lf=1.1949760305[m]Lf=1.5 x dcdc=0.796650687[m]volumul focaruluiVf=0.5956414956[m]suprafata peretilor camerei de ardereSper=3.9876381142[m]grosimea stratului radiant de gazes=0.5974880153[m]gradul de ecranare=0.875caracteristica radianta a gazelor de ardere in camera de arderekg=22.955004701unde:Tmg=1530.945125213[K]coeficientul de absorbtie al gazelor de ardere din focarag=0.9999988946caracteristica radianta a flacariikfl=1.9495122003coeficientul de luminozitate pentru flacara neluminoasa de combustibil gazos=0.2coeficientul de absorbtie al flacariiafl=0.6880181217coeficientul de absorbtie al flacarii si gazelor de arderea=0.9376027401caracteristica radianta a camerei de ardereafc=0.7835007366=0.8factorul de pozitie a flacariiM=0.44aM=0.54bM=0.2h/Hf=0.5temperatura de la capatul focaruluiT*f=15.0914649862[K]Tt=2238.890250426[K]c=0.0000000576[W/mK]tmg=1257.945125213[C]Img=22096.6967346591cp=1.5701165899df'=0.8090118623[m]tgm=146.0457324931[C]w'g=5.5[m/s]rotunjimdf=0.81[m]reclaculam vitezawg=5.0897488396[m/s]coficientul de transfer de calduralc=0.013349313[m]Re=2544.441574371[-]Nu=8.1587290421[-]Prg g*10 g *101000.6921.53.13tgm=146.04573249310.680790853526.70316777173.53520244592000.6732.84.01 [-][m/s][W/mK]A=2.6847383853B=6137170.20592985=0.048130168coeficientul de schimb de caldura prin radiaties=0.0120143817[m]Tgm=419.0457324931[K]tmGF=2120.2088588188Tp=363[K]kg=40.929139856kGF=5.445847232ag=0.3884376142R=4.5252054599[W/mK]QGF=35110.59462786521=5.5752054599[W/mK]*1035.233w=5.08974883964.260570206763.837[mK/W]If=-7063704.45821782I*f=2003101.263895696t*f=257.90853501384042.99314212054006353.735215168

Sheet2Schimbator de caldura condensator abur-apa multitubular cu mai multe treceriN=3Qu=230[kw]0.3461538462ps=1.7[bar]20.4640813732t'=97[C]t"=108.3[C]0.0070.00140.004341. Calculul de predimensionare4.22334a. din anexa 4 alegem in functie de ps =>r=2178[kJ/kg]ts=115.47[C]b. aleg iz= 0.993-0.995 =>iz=0.995c. calculam debitul de aburD=0.1061321299[kg/s]d. din tabel alegem cp2 in functie de t2m =>tm=102.65[C]cp2=4.22271[kJ/kgK]G=4.8201231676[kg/s]e. calculam tm =>tm=11.9419750264t=18.47t=7.17f. alegem coeficentul globa de caldura k'=900-1200 [w/mK]k'=1000[w/mK]S=19.2597957618[m]19.2629815745g. aleg tipul de teava16 x 1.5dm=14.5[mm]de=16[mm]di=13[mm]h. aleg lungimea tevilor preliminara l'=2-4 [m]l'=3[m]n'tot=140.9329516517i. din anexa 5 in funtie de tm prin interpolare pe m =>m=956.3655[kg/m]aleg w'=0.5-1.5 [m/s]w'=0.8[m/s]i. din ecuatia de contiuitate a scurgerii aflam n'trn'tr=47.4643635605n'tr=2.9692371514aleg 6 =>ntr=4ntr=40w=0.9492872712[m/s]ntot=160j. pentru a alege mantaua aleg:s=23[mm]k=10.5[mm]avemk=9[mm]s=de+8-10[mm]k=8-12[mm]calculam:Dim=361[mm]aleg 400cresc k0 cu 1mm scriu in desen ca facem comanda de o teava roluita de 400 mmDem=377[mm]2. Calculul DEFINITIVa. determinam :Pr *10 din anexa 5 in fct de tm aflam Pr, , 1001.750.2950.6827tm=102.651.70230.2878450.68330951101.570.2680.685 [-][m/s][W/mK]lc=13[mm]Re= w*lc/Re=42872.8465866069[-]Re=0.9972443562Nu=150.7896514313[-] =Nu* /di =7925.8462557479[W/mK]b. determinam /: /= d/d+ol/olde la pag.15 aleg apa tratata de alimentare a cazanelor =>d/d=Rd=0.00018[mK/W]ol=0.015[m]/=0.00048[mK/W]ol=50[W/mK]c. determinam :din anexa 5 in functie de ts aflam: c, c, cccc*101200.6862943.10.244ts=115.470.6862946.85990.2521541300.6862934.80.226[W/mK][kg/m][m/s]=c*t^(-0.25)unde:c=23261.7466645652g=9.81[m/s]c=0.6862[W/mK]c=0.0006061695c=946.8599[kg/m]r=2178000[W*s/kg]c=0.0000002522[m/s]de=0.016[m]t se alege prin metoda injumatatirii intervalului:14.1005611711g(t)=c*c*t^0.75+t-tmtm=11.9419750264g(t=0)= -11.9419750264g(t=1.395313685)=0.12719040110.7463734391g(t=1.492)=8.59336223921.4927468783g(t=2.985)=23.06921071792.9854937566g(t=5.97)=47.88959166375.9709875132g(t=11.94)=90.5823359067t =1.4927468783=21044.8472940002[W/mK]d. calculam coficientul global de transfer de caldura k:k=1529.7839863066[W/mK]e. se modifica suprafata de transfer de caldura S:S=12.5898793125[m]f. se modifica lungimea tevilor l:l=1.7274[m]rotunjim doar "+"alegem l=1.8[m]3. Calculul izolatiei termicedeterminam grosimea necesara izolatiei pentru a avea randamentul propus=0.995a. aleg izolatia: vata minerala =>iz=20[mm]iz=0.05[W/mK]b. determinam em= f(Nue)= f(Pr, Gr) dar Gr depinde de => aleg te'=20-30 [C]te'=22[C]te=15[C]'em= 10.612te''=19.1561165345[C]c. verificam daca |te'-te''| 3C este bine daca nu alegem te'=te'' si recalculam em|te'-te''|=2.8438834655[C]km=2.0233374009Qm=479.3614931424[kW]Qp=719.0422397137[kW]Qcons=231.1557788945[kW]sc=0.99688936074. Calculul hidraulica. calculam pierderile de sarcina pentru agentul termic secundarRe=42872.8465866069=0.0215613989pierderile de sarcina liniare pp=5145.8283932504[N/m]l=7.2pierderile de sarcina locala pdimensionare racordaleg wr'=0,51,5 [m/s]wr'=0.75[m/s]dir=0.09249996[m]aleg din anexa 12-13 o teava STAS cu didir => dirSTASdirSTAS=0.092[m]cu grosime peretelui de 0.005 [m]recalculam wr =>wr=0.7581736712[m/s]Tipbucatiw p=**w/2buc * p[-][-][-][kg/m][m/s][N/m][N/m]racord complet deschis20.5956.36550.7581736712137.436253313274.872506626colector de intrare sau iesire21956.36550.7581736712274.872506626549.7450132519intrare in spatiul din interiorul tevilor si iesirea din acesta41956.36550.9492872712430.91262701851723.6505080739intoarcere cu 180 dintr'o sectiune in alta, printr'o curba in preincalzitoare sectionale32956.36550.9492872712861.8252540372585.4757621109p5133.7437900628pierderile de sarcina pentru agentul termic secundar sunt p=10279.5721833132[N/m]