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Zona Radiatie2015

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Cuprins

Zona de radiatie (ZR)

1. Entalpia de reacție. Entalpia amestecului de reacție2. Calculul serpentinei in sectoarele zonei de radiatie

2.1. Calculul serpentinei în sectorul 1 din zona de radiație

2.2. Calculul serpentinei în sectorul 2 din zona de radiație

2.3. Calculul serpentinei în sectorul 3 din zona de radiație

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Parametrii de functionare:

-tiR= 6000 C-raport abur/ materie prima rs=0,6-tensiune termica Φ tR=250000 kJ/m2*h-conversie finala xf=0,84

MP (0,84 g )→ H 2 (0,005 g )+CH 4 (0,052 g )+C2 H 2 (0,004 g )+C2 H 4 (0,25 g )+C2 H 6 (0,041 g )+C3 H 6 (0,184 g )+C3 H8 (0,026 g )+C4 H 5 (0,077 g )+C4 H 8 ( 0,074 g )+C4 H 10 (0,047 g )+„comb ”(0,08g)

Compoziție, entalpie amestec de reacție funcție de conversia curentăNr Comp Conpozitie,

gfiMasa molara, Mig/mol

n fi=g fi

M i

mol/g

gi ( x )=g fi

x f

∗x

g/g MP alim

ni=gi ( x )M i

1 H2 0,005 2,016 2,480*10-3 0,0050,84

∗xc0,0050,84

∗x

2,0162 CH4 0,052 16,043 3,241*10-3 0,052

0,84∗xc

0,0520,84

∗x

16,0433 C2H2 0,004 26,038 0,153*10-3 0,004

0,84∗xc

0,0040,84

∗x

26,0384 C2H4 0,250 28,054 8,911*10-3 0,250

0,84∗xc

0,2500,84

∗x

28,0545 C2H6 0,041 30,070 1,363*10-3 0,041

0,84∗xc

0,0410,84

∗x

30,0706 C3H6 0,184 42,081 4,372*10-3 0,184

0,84∗xc

0,1840,84

∗x

42,0817 C3H8 0,026 44,097 0,589*10-3 0,026

0,84∗xc

0,0260,84

∗x

44,097

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8 C4H5 0,077 53,084 1,450*10-3 0,0770,84

∗xc0,0770,84

∗x

53,0849 C4H8 0,074 56,108 1,318*10-3 0,074

0,84∗xc

0,0740,84

∗x

56,10810 C4H10 0,047 58,123 0,808*10-3 0,047

0,84∗xc

0,0470,84

∗x

58,12311 MP

benzină102,481 102,481 1,561*10-3 1−xc 1−x

102,48112 Comb 250 250 0,320*10-3 0,080

0,84∗xc

0,0800,84

∗x

25013 H2O 60

(rs)18,015 33,305*10-3 60 (rs) ra

18,015Σ 1,60

(1+rs)59,871 1,60

(1+rs)0,020016∗x+0,0430635

Mm=∑i=1

13

y i∗M i=∑ ni(x )ntsp (x)

∗M i

Mms (x )= 1+rsn tsp(x )

= 1+0,60,020016∗x+0,0430635

= 1,60,020016∗x+0,0430635

1. Entalpia de reacție. Entalpia amestecului de reacție

Δ H fi0 (T )=Δ H f 25℃

0 +∫298

T

Cpi (T ) dT∗1

1000Cpi (T )=ai+bi∗T+ci∗T2

Kw Comb=11.8

d4 Comb20 =0.859⇒d15,6Comb

15,6 =0.863

Cp1 (T )=27,7047+3,38985∗10−3∗T

Δ H f 10 (T )=0+∫

298

T

(27,7047+3,38985∗10−3∗T ) dT∗11000

=1,69493∗10−6∗T2+0,0277047∗T−8,40652kJ /mol

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Cp2 (T )=22,3479+48,1275∗10−3∗T

Δ H f 20 (T )=−74,52+∫

298

T

(22,3479+48,1275∗10−3∗T ) dT∗11000

=0,0000240637∗T 2+0,0223479∗T−83,3166 kJ /mol

Cp3 (T )=27,9609+60,2362∗10−3∗T−19,46762∗10−6∗T2

Δ H f 30 (T )=228,2+∫

298

T

( 27,9609+60,2362∗10−3T−19,46762∗10−6∗T2 ) dT∗11000

=−6,48921∗10−9 T3+0,0000301181T 2+0,0279609∗T+217,365 kJ /mol

Cp4 (T )=10,8667+121,6808∗10−3∗T−37,56525∗10−6∗T2

Δ H f 40 (T )=52,51+∫

298

T

(10,8667+121,6808∗10−3 T−37,56525∗10−6 T2 ) dT∗11000

=−1,25217∗10−8 T 3+0,0000608404 T 2+0,0108667∗T +44,2002 kJ /mol

Cp5 (T )=10,2338+158,1230∗10−3∗T−45,56065∗10−6∗T 2

Δ H f 50 (T )=−83,82+∫

298

T

(10,2338+158,1230∗10−3 T−45,56065∗10−6 T2 ) dT∗11000

=−1,51869∗10−8 T3+0,0000790615T 2+0,0102338∗T−93,4888kJ /mol

Cp6 (T )=13,4925+190,9726∗10−3∗T−59,16355∗10−6∗T 2

Δ H f 60 (T )=19,71+∫

298

T

( 13,4925+190,9726∗10−3T−59,16355∗10−6T 2 ) dT∗11000

=−1,97212∗10−8T 3+0,0000954863 T2+0,0134925∗T+7,73156 kJ /mol

Cp7 (T )=10,8549+237,6317∗10−3∗T−72,91123∗10−6∗T2

Δ H f 70 (T )=−104,68+∫

298

T

(10,8549+237,6317∗10−3 T−72,91123∗10−6 T 2) dT∗11000

=−2,43037∗10−8T 3+0,000118816T 2+0,0108549∗T−117,823kJ /mol

Cp8 (T )=24,6162+185,6805∗10−3∗T−60,87526∗10−6∗T 2

Δ H f 80 (T )=186,11+∫

298

T

(24,6162+185,6805∗10−3 T−60,87526∗10−6 T2 ) dT∗11000

=−2,02918∗10−8 T3+0,0000928403T 2+0,0246162∗T +171,067 kJ /mol

Cp9 (T )=16,9533+258,8119∗10−3∗T−79,77959∗10−6∗T 2

Δ H f 90 (T )=−9,01+∫

298

T

(16,9533+258,8119∗10−3T−79,77959∗10−6 T 2) dT∗11000

=−2,65932∗10−8 T3+0,000129406T 2+0,0169533∗T−24,8501kJ /mol

Cp10 (T )=15,5790+312,0373∗10−3∗T−98,17201∗10−6∗T 2

Δ H f 100 (T )=−129,985+∫

298

T

(15,5790+312,0373∗10−3T−98,17201∗10−6T 2 ) dT∗11000

=−3,2724∗10−8∗T3+0,000156019∗T 2+0,015579∗T−147,617 kJ /mol

Δ H fMP0 (T )=Δ H f 11

0 (T )=−100+MMP

1000∗∫

298

T

C p ( t ) dT=−100+M MP

1000∫298

T

(−1,69578+0,29309 K w−0,45638d15,615,6 )dT +

M MP

1000∗10−3∫

298

T

(4,6894−1,18074 d15,615,6 ) (T−273 ) dT=−100+ 102,481

1000∫298

T

(−1,69578+0,29309∗11,791−0,45638∗0,745 )dT + 102,4811000

∗10−3∫298

T

(4,6894−1,18074∗0,745 ) (T−273 ) dT=0,000195213∗T 2+0,0389407∗T−128,94 kJ /mol

Δ H f Comb0 (T )=Δ H f 12

0 (T )=100+MComb

1000∫298

T

(−1,69578+0,29309 Kw Comb−0,45638 d15,6Comb15,6 )dT +

MComb

1000∗10−3∫

298

T

(4,6894−1,18074 d15,6Comb15,6 ) (T−273 ) dT=100+ 250

1000∫298

T

(−1,69578+0,29309∗11,1685−0,45638∗0,9189 )dT + 2501000

∗10−3∫298

T

(4,6894−1,18074∗0,9189 ) (T−273 ) dT=0,000450552∗T2+0,0435555∗T +47,0096 kJ /mol

Cp13 H 2 O (T )=34,4+0,62775∗10−3∗T+5,6079∗10−6∗T 2

Qabur=1

MH 2 O∫T1

T2

C p (T ) dT= 118,015∫T1

T2

(34,4+0,62775∗10−3∗T +5,6079∗10−6∗T2 ) dT=−1,03764∗10−7∗T 13−0,000017423∗T 1

2−1,90952∗T 1+1,03764∗10−7∗T23+0,000017423∗T 2

2+1,90952∗T 2 kJ /kg

Nr Comp Δ H fi0(T )

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kJ /mol1 H2 1,69493∗10−6∗T2+2,77047∗10−2∗T−8,406522 CH4 2,40637¿10−5∗T 2+2,23479¿10−2∗T−83,31663 C2H2 −6,48921∗10−9 T3+3,01181¿10−2¿T 2+2,79609¿10−2∗T +217,3654 C2H4 −1,25217∗10−8 ¿T 3+6,08404¿10−8∗T2+1,08667¿10−2∗T+44,20025 C2H6 −1,51869∗10−8 ¿T3+7,90615¿10−5∗T 2+1,02338¿10−2∗T−93,48886 C3H6 −1,97212∗10−8∗¿ T3+9,54863¿T 2+1,34925¿10−2∗T +7,73156¿

7 C3H8 −2,43037∗10−8¿T 3+1,18816¿10−4∗T2+1,08549¿10−2∗T−117,8238 C4H5 −2,02918∗10−8∗¿T3+9,28403¿10−5∗T2+2,46162¿10−2∗T +171,067¿

9 C4H8 −2,65932∗10−8 ¿T3+1,29406¿10−4 ¿T 2+1,69533¿10−2∗T−24,850110 C4H10 −3,2724∗10−8∗T 3+1,56019¿10−4∗T2+1,5579¿10−2∗T−147,61711 MP

benzină1,95213¿10−4∗T 2+3,89407¿10−2∗T−128,94

12 Comb 4,50552¿10−4∗T 2+4,35555¿10−2∗T+47,009613 H2O

Nr Comp Δ H fi0∗1000∗ni ( x )

xkJ /kg MP reac ț ionat ă

1 H2 2,953∗[1,69493∗10−6∗T 2+2,77047 ¿10−2∗T−8,40652 ]2 CH4 3,859∗[2,40637¿10−5∗T2+2,23479¿10−2∗T−83,3166 ]3 C2H2 0,183∗[(−6,48921)∗10−9 ¿T3+3,01181¿10−2∗T 2+0,0279609∗T +217,365 ]4 C2H4 10,689∗[(−1,25217)∗10−8 ¿T3+6,08404 ¿10−5 ¿T2+1,08667¿10−2∗T+44,2002 ]5 C2H6 1,623∗[(−1,51869)∗10−8 ¿T3+7,90615 ¿10−5∗T 2+1,02338 ¿10−2∗T−93,4888]

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6 C3H6 5,205∗[(−1,97212)∗10−8 ¿T3+9,54863¿10−5¿T 2+1,34925¿10−2∗T+7,73156]7 C3H8 0,702∗[(−2,43037)∗10−8¿T 3+1,18816∗10−4 ¿T2+1,08549 ¿10−2∗T−117,823]8 C4H5 1,727∗[(−2,02918)∗10−8¿T 3+9,28403∗10−8 ¿T2+2,46162∗10−2∗T +171,067]9 C4H8 1,570∗[(−2,65932)∗10−8 ¿T3+1,29406∗10−4∗T 2+1,69533∗10−2∗T−24,8501]10 C4H10 0,963∗[(−3,2724)∗10−8∗T 3+1,56019∗10−4∗T 2+1,5579∗10−2∗T−147,617]11 MP

benzină−9,758∗(1,95213∗10−4∗T2+3,89407∗10−2∗T−128,94)

12 Comb 0,831∗(4,50552∗10−4∗T2+4,35555∗10−2∗T+47,0096)13 H2OΣ −3,86691∗10−7∗T 3+2,38042∗10−4∗T 2+1,0363∗10−1∗T+1358,52

Δ H Rsp (T )=−3,86691∗10−7T 3+0,000238042∗T2+0,10363T +1358,52

Nr Comp H hspi ( x ,T )=1000∗n i ( x )∗Δ H fi0

kJ /kg MP alim saukJ /kg∏ dereac ț ie1 H2 5∗x

1,693∗[1,69493∗10−6∗T2+,2,77047∗10−2∗T−8,40652]

2 CH4 52∗x13,476

∗[2,40637∗10−5∗T 2+2,23479∗10−2∗T−83,3166]

3 C2H2 4∗x21,872

∗[−6,48921∗10−9 ¿T3+3,01181∗10−4 ¿T 2+2,79609∗10−2∗T +217,365]

4 C2H4 250∗x23,565

∗[−1,25217∗10−8¿T 3+6,08404∗10−5∗T2+1,08667∗10−2∗T+44,2002]

5 C2H6 41∗x25,259

∗[−1,51869∗10−8∗T3+7,90615∗10−5 ¿T 2+1,02338∗10−2∗T−93,4888]

6 C3H6 184∗x35,348

∗[−1,97212∗10−8 ¿T3+9,54863∗10−5∗T 2+1,34925∗10−2∗T+7,73156]

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7 C3H8 26∗x37,041

∗[−2,43037∗10−8 ¿T 3+1,18816∗10−5∗T 2+1,08549∗10−2∗T−117,823 ]

8 C4H5 77∗x44,590

∗[−2,02918∗10−8 ¿T 3+9,28403∗10−2∗T2+2,46162∗10−2∗T +171,067]

9 C4H8 74∗x47,131

∗[−2,65932∗10−8T 3+1,29406¿10−4∗T 2+1,69533∗10−2∗T−24,8501]

10 C4H10 47∗x48,823

∗[−3,2724∗10−8∗T3+1,56019∗10−4∗T 2+1,5579∗10−2∗T−147,617]

11 MPbenzină

1000∗(1−x )102,481

∗[1,95213∗10−4∗T2+3,89407∗10−2∗T−128,94 ]

12 Comb 80∗x210

∗[4,50552∗10−4∗T 2+4,35555∗10−2∗T +47,0096 ]

13 H2OΣ −3,86691∗10−7∗T 3∗x+2,3804∗10−4∗T2∗x+1,90487∗10−3∗T 2+1,0363∗10−1∗T∗x+3,7998∗10−1∗T +1358,52∗x−1258,18

Entalpia amestecului la o anumită temperatură:H hsp ( x ,T )=−3,86691∗10−7∗T 3∗x+0,00023804∗T2∗x+0,00190487∗T 2+0,10363∗T∗x+0,37998∗T +1358,52∗x−1258,18kJ /kg MPalim saukJ /kg∏ de reacție

2. Calculul serpentinei in sectoarele zonei de radiatie2.1. Calculul serpentinei în sectorul 1 din zona de radiatie (ZR)

t iR=600℃=873,15 KpiR=7,2 ¿̄GMP=1400kg /hrs=0,60

Bilanț termic pe sector:H h iR=GMP∗H hsp(x ,T iR)H h iR=GMP∗H hsp ( x iR ,T iR )=1400∗(−3,86691∗10−7∗903,153∗0+0,00023804∗903,152∗0+0,00190487∗903,152+0,10363∗903,15∗0+0,37998∗903,15+1358,52∗0−1258,18 )=1,887195∗106 kJ /h

Presupunem: t i 2R=715,35℃Δ xc1=0,100306

xci 2=0+ Δ xc1⇒ xci 2=0,100306

Hhci 2=GMP∗H hsp ( xci 2 ,T i 2R )=1400∗(−3,86691∗10−7∗988,53∗0,100306+0,00023804∗988,52∗0,100306+0,00190487∗988,52+0,10363∗988,5∗0,100306+0,37998∗988,5+1358,52∗0,100306−1258,18 )=1,738158∗106 kJ /h

Qab 1=GMP∗rs∗ ∫650+273,15

725+273,15

C p H2 O (T ) dT=1400∗0,60∗(−1,03764∗10−7∗T13−1,7423∗10−5∗T 1

2−1,90952∗T 1+1,03764∗10−7∗T 23+1,7423∗10−7∗T 2

2+1,90952∗T 2)=150553,08kJ /h

QR1=GMP∗Δ xc1∗HR sp( 60 0+715,352

+273,15)=213019,42kJ /h

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Calculăm lungimea sectorului pe baza ecuației de bilanț termic:

H hci+Q f 1R=H hci2+Qab 1+QR1

1 ,887195∗106+Qf 1R=1,738158∗106+150553,08+213019,42⇒Qf 1R=1,02782∗106 kJ /h

Qf 1R=Φ tR∗π∗de∗L1R

L1R=H hci 2+Q ab1+QR1−H hci

π∗de∗Φ tR

=1,738158∗106+150553,08+213019,42−1 ,887195∗106

π∗0,08∗318470=10,273m

nc1R=1Lech1 R=L1R+nc 1R∗75∗di=10,273+1∗75∗0,07=16,273m

Mms ( xciR )= 1,60,020016∗0+0,0430635

=37,154

Mms ( xci 2 )= 1,60,020016∗0,1 00306+0,0430635

=35,403

Se calculeaza caderea de presiune :

Δ p1 R=λ∗R∗Lech1R∗GMP

2 (1+rs )2

π2∗di5∗36002 ∗[ piR∗100∗Mms ( xciR )

T iR

+(105∗p iR−Δ p1 R )∗Mms ( xci 2 )

1000∗T i 2R]∗[ T iR

piR∗100∗Mms ( xciR )+

1000∗T i 2R

(105∗p iR−Δ p1 R )∗Mms ( x c i 2) ]2

=0,0366∗8,314∗16,273∗15002∗(1+0,63 )2

36002∗π2∗0,075 ∗[ 7,2∗100∗37,154903,15

+(105∗7,2−Δ p1 R )∗35,403

1000∗988,5 ]∗[ 903,157,2∗100∗37,154

+ 1000∗988,5

(105∗7,2−Δ p1R )∗35,403 ]2

Δ p1 R=30706 Pa=0,307 ¿̄

Se calculeaza densitatea:

ρiR=piR∗Mms (0 )

R∗T iR

= 7,2∗37,1540,083∗903,15

=2,382kg /m3

ρi 2R=( p iR−Δ p1 R )∗Mms ( xci 2 )

R∗T i 2R

=(7,2−0,307 )∗35,403

0,083∗988,5=1,968kg /m3

ρm1 R=ρiR+ρi 2R

2=2,382+1,968

2=2,175kg /m3

Se calculeaza viteza de curgere :

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w iR=4∗GMP∗(1+rs )

3600∗ρ iR∗π∗di2 =

4∗1500∗(1+0,6)3600∗2,382∗π∗0,072 =55,679m /s

w i 2R=4∗GMP∗(1+rs )

3600∗ρi 2R∗π∗d i2 =

4∗1500∗(1+0,6)3600∗1,968∗π∗0,072 =67,392m /s

wm1R=wiR+w i2 R

2=55,679+67,392

2=61,535m /s

Se calculeaza timpul de stationare:

τ1=L1R

wm1R

=10,27361,535

=0,166 s

2.2. Calculul serpentinei în sectorul 2 din zona de radiație (ZR)

GMP=1400kg /hT i 2R=988,5K

H hci 2=1,738158∗106 kJ /hpi 2R=p iR−Δ p1 R=7,213−0,307=6,906 ¿̄xci 2=0+ Δ xc1=0+0,100306=0,100306

Presupunem: t i3 R=788℃Δ xc2=0,486238

xci 3=xci 2+ Δ xc2=0,100306+0,4862387=0,586544

Bilanț termic pe sector:i: H hci2=GMP∗H hsp(xci 2 , T i 2R)

Hhci2=GMP∗Hhsp ( xci 2 ,T i 2R )=1400∗(−3,86691∗10−7∗988,53∗0,100306+0,00023804∗988,52∗0,100306+0,00190487∗988,52+0,10363∗988,5∗0,100306+0,37998∗988,5+1358,52∗0,100306−1258,18 )=1,738158∗106 kJ /h

H hci 3=GMP∗H hsp ( xci 3 ,T i 3R )=1400∗(−3,86691∗10−7∗1061,153∗0,586544+0,00023804∗1061,152∗0,586544+0,00190487∗1061,152+0,10363∗1061,15∗0,586544+0,37998∗1061,15+1358,52∗0,586544−1258,18 )=2,798098∗106 kJ /h

Qab 2=GMP∗rs∗ ∫715+273,15

788+273,15

C p H2 O (T ) dT=1400∗0,60∗(−1,03764∗10−7∗T13−0,000017423∗T 1

2−1,90952∗T1+1,03764∗10−7∗T 23+0,000017423∗T 2

2+1,90952∗T 2 )=73431,04 kJ /h

QR2=GMP∗Δ xc2∗HR sp( 715+7882

+273,15)=1400∗0,430083∗(−3,86691∗10−7∗1016,153+0,000238042∗1016,152+0,10363∗1016,15+1358,52 )=841168,85 kJ /h

Calculăm lungimea sectorului pe baza ecuației de bilanț termic

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H hci2+Qf 2R=H hci3+Qab 2+QR2

L2R=H hci3+Q ab 2+QR2−H hci2

π∗de∗Φ tR

=2,798098∗106+73431,04+841168,85−1,738158∗106

π∗0,1∗318470=19,73m

nc2R=2Lech2 R=L2R+nc 2R∗75∗d i=19,73+2∗75∗0,07=31,73m

Mms ( xci 2 )= 1,60,020016∗0,10360+0,0430635

=35,403

Mms ( xci 3 )= 1,60,020016∗0,586544+0,0430635

=29,739

Se calculeaza caderea de presiune :

Δ p2R=λ∗R∗Lech2R∗GMP

2 (1+rs )2

π2∗di5∗36002 ∗[ pi 2R∗100∗Mms ( xci 2 )

T i2 R

+( pi 2R∗105−Δ p2R )∗Mms ( xci 3 )

T i 3R∗1000 ]∗[ T i 2 R

pi 2R∗100∗Mms ( xci 2 )+

T i 3R

( pi 2R∗105−Δ p2R )∗Mms ( x ci 3 ) ]2

=0,0366∗8,314∗31,73∗15002∗(1+0,6 )2

36002∗π 2∗0,075 ∗[ 6,906∗100∗35,403988,5

+(105∗6,906−Δ p2R )∗29,739

1000∗1061,15 ]∗[ 988,56,906∗100∗35,403

+ 1000∗1061,15

(105∗6,906−Δ p2R )∗29,739 ]2

Δ p2R=85208,2 Pa=0,852 ¿̄

Se calculeaza densitatea:

ρi 2R=( p iR−Δ p1 R )∗Mms ( xci 2 )

R∗T i 2R

=(7,2−0,307 )∗35,403

0,083∗988,5=1,968 kg /m3

ρi 3R=( p i 2R−Δ p2R )∗Mms ( xci 3 )

R∗T i 3R

=(6,906−0,852 )∗29,739

0,083∗1034,15=1,3006kg /m3

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ρm2 R=ρi 2R+ρi 3 R

2=1,968+1,3006

2=1,6343 kg /m3

Se calculeaza viteza de curgere:

w i 2R=4∗GMP∗(1+rs )

3600∗ρi 2R∗π∗d i2 =

4∗1500∗(1+0,63)3600∗1,968∗π∗0,072 =67,392m /s

w i 3R=4∗GMP∗(1+rs )

3600∗ρi 3R∗π∗di2=

4∗1500∗(1+0,63)3600∗1,3006∗π∗0,072 =101,975m /s

wm2R=wi 2R+wi 3R

2=67,392+101,975

2=84,6835m / s

Se calculeaza timpul de stationare:

τ 2=L2R

wm2R

= 19,7384,6835

=0,232 s

2.3. Calculul serpentinei în sectorul 3 din zona de radiație (ZR)

GMP=1400kg /hT i3 R=1061,15 K

H hci 3=2,798098∗106 kJ /hpi 3R=pi 2R−Δ p2R=6,906−0,852=6,054 ¿̄xci 3=0+ Δ xc1+ Δ xc2=0,586544

Presupunem: t eR=869,742℃Δ xc3=xc f−xci 3=0,84−0,586544=0,253456

t eR=869,742℃⇒TeR=869.742+273,15=1142,892 K

Bilanț termic pe sector:H hci3=GMP∗H hsp(xci 3 , T i3 R)H hci3=GMP∗H hsp ( xci 3 , T i 3R )=1400∗(−3,86691∗10−7∗1061,153∗0,586544+0,00023804∗1061,152∗0,586544+0,00190487∗1061,152+0,10363∗1061,15∗0,586544+0,37998∗1061,15+1358,52∗0,586544−1258,18 )=2,798098∗106 kJ /h

HhceR=GMP∗Hh sp ( xcf , T eR)=1400∗(−3,86691∗10−7∗1142,8923∗0,84+0,00023804∗1101,7722∗0,84+0,00190487∗1142,8922+0,10363∗1142,892∗0,84+0,37998∗1142,892+1358,52∗0,84−1258,18)=3,777234∗106 kJ /h

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Qab 2=GMP∗rs∗ ∫788+273,15

869+273,15

Cp H 2O(T ) dT=1400∗0,63∗(−1,03764∗10−7∗T1

3−0,000017423∗T12−1,90952∗T1+1,03764∗10−7∗T 2

3+0,000017423∗T22+1,90952∗T2 )=140092,80 kJ /h

QR3=GMP∗Δ xc3∗HR sp(788+869,6222

+273,15)=1400∗0,303535∗(−3,86691∗10−7∗1101,8113+0,000238042∗1101,8112+0,10363∗1101,811+1358,52 )=578088,14kJ /h

Calculăm lungimea sectorului pe baza ecuației de bilanț termicH hci3+Q f 3R=H hceR+Q ab3+QR3

L2R=H hceR+Qab 3+QR3−H hci3

π∗de∗Φ tR

=3,777234∗106+140092,80+578088,14−2,798098∗106

π∗0,08∗250000=16,964m

nc3R=2Lech3 R=L3 R+nc 3R∗75∗d i=16,964+2∗75∗0,07=28,964m

Mms ( xci 3 )= 1,60,020016∗0,586544+0,0430635

=29,739

Mms ( xcf )= 1,60,020016∗0,25+0,0430635

=26,721

Se calculeaza caderea de presiune :

Δ p3 R=λ∗R∗Lech3R∗GMP

2 (1+rs)2

π2∗d i5∗36002 ∗[ pi 3R∗100∗Mms ( xci 3 )

T i 3R

+( p i3 R∗105−Δ p3 R )∗Mms ( xc f )

TeR∗1000 ]∗[ T i 3R

pi 3R∗100∗Mms ( xci 3 )+

T eR

( p i 3R∗105−Δ p3R )∗Mms ( xc f ) ]2

=0,0366∗8,314∗28,964∗15002∗(1+0,63 )2

36002∗π2∗0,075 ∗[ 6,054∗100∗29,7391061,15

+(105∗6,054−Δ p3 R )∗26,721

1000∗1101,772 ]∗[ 1061,156,054∗100∗29,739

+ 1000∗1101,772

(105∗6,054−Δ p3R )∗26,721 ]2

Δ p3 R=148746 Pa=1,487 ¿̄

Se calculeaza densitatea:

ρi 3R=( p i 2R−Δ p2R )∗Mms ( xci 3 )

R∗T i 3R

=(6,906−0,852 )∗29,739

0,083∗1061,15=1,3006 kg /m3

ρeR=( pi 3R−Δ p3 R )∗Mms ( xcf )

R∗T eR

=(6,054−1,487 )∗26,721

0,083∗1142,892=0,6624 kg/m3

ρm3 R=ρi 3R+ ρeR

2=1,3006+0,6624

2=0,9815kg /m3

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Se calculeaza viteza de curgere :

w i 3R=4∗GMP∗(1+rs )

3600∗ρi 3R∗π∗di2=

4∗1500∗(1+0,6)3600∗1,3006∗π∗0,072 =101,975m /s

w eR=4∗GMP∗(1+rs)

3600∗ρeR∗π∗d i2 =

4∗1500∗(1+0,6)3600∗0,6624∗π∗0,072 =200,225m /s

wm3R=wi 3R+weR

2=101,975+200,225

2=151,1m/ s

Se calculeaza timpul de stationare:

τ3=L3 R

wm3 R

=16,964151,1

=0,112 s

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