Zona Radiatie2015

1 | P a g e

Parametrii de functionare:

-tiR= 600

0 C

-raport abur/ materie prima rs=0,6

-tensiune termica tR = 250000 kJ/m2*h

-conversie finala xf=0,86

MP(0,84g) H2(0,005g) + CH4(0,052g) + C2H2(0,004g) + C2H4(0,25g) + C2H6(0,041g)+ C3H6(0,184g) + C3H8(0,026g) + C4H5(0,077g) + C4H8(0,074g)+ C4H10(0,047g) + comb(0,08g)

C7,24H15,41

Compoziie, entalpie amestec de reacie funcie de conversia curent.

Nr

.

Comp Conpozitie

100*gfi

Masa

molara,

Mi

g/mol

nfi =gfiMi

mol/g

gi(x) = gfixf

x

g/g MP alim

ni =gi(x)

Mi

1 H2 1,20 2,016 2,480*10-3

0,005

0,84 xc

0,005

0,84

x

2,016

2 CH4 7,20 16,043 3,241*10-3

0,052

0,84 xc

0,052

0,84

x

16,043

3 C2H2 0,40 26,038 0,153*10-3

0,004

0,84 xc

0,004

0,84

x

26,038

4 C2H4 27,50 28,054 8,911*10-3

0,250

0,84 xc

0,250

0,84

x

28,054

5 C2H6 4,00 30,070 1,363*10-3

0,041

0,84 xc

0,041

0,84

x

30,070

6 C3H6 13,90 42,081 4,372*10-3

0,184

0,84 xc

0,184

0,84

x

42,081

7 C3H8 3,60 44,097 0,589*10-3

0,026

0,84 xc

0,026

0,84

x

44,097

8 C4H5 8,70 53,084 1,450*10-3

0,077

0,84 xc

0,077

0,84

x

53,084

9 C4H8 6,60 56,108 1,318*10-3

0,074

0,84 xc

0,074

0,84

x

56,108

10 C4H10 4,90 58,123 0,808*10-3

0,047

0,84 xc

0,047

0,84

x

58,123

11 MP

benzin 14

102,481 1,561*10-3

1 xc 1 x

102,481

12 Comb 8 250 0,320*10-3

0,080

0,84 xc

0,080

0,84

x

250

2 | P a g e

13 H2O 60

(rs)

18,015 33,305*10-3

60 (rs) ra

18,015

1,60 (1+rs)

59,871 1,60

(1+rs) 0,020016 x+ 0,0430635

Mm = yi Mi = ni(x)

ntsp(x) Mi

13

i=1

Mms(x) =1 + rs

ntsp(x)=

1 + 0,6

0,020016 x + 0,0430635=

1,6

0,020016 x + 0,0430635

1. Entalpia de reacie. Entalpia amestecului de reacie

Hfi0(T) = Hf25

0 + Cpi(T) dT

T

298

1

1000

Cpi(T) = ai + bi T + ci T2

Kw Comb = 11.8

d4 Comb20 = 0.859 d15,6 Comb

15,6 = 0.863

Cp1(T) = 27,7047 + 3,38985 103 T

Hf10 (T) = 0 + (27,7047 + 3,38985 103 T)

T

298

dT 1

1000

= 1,69493 106 T2 + 0,0277047 T 8,40652 kJ mol Cp2(T) = 22,3479 + 48,1275 10

3 T

Hf20 (T) = 74,52 + (22,3479 + 48,1275 103 T)

T

298

dT 1

1000

= 0,0000240637 T2 + 0,0223479 T 83,3166 kJ mol Cp3(T) = 27,9609 + 60,2362 10

3 T 19,46762 106 T2

Hf30 (T) = 228,2 + (27,9609 + 60,2362 103T 19,46762 106 T2)

T

298

dT 1

1000

= 6,48921 109T3 + 0,0000301181T2 + 0,0279609 T+ 217,365 kJ mol

3 | P a g e

Cp4(T) = 10,8667 + 121,6808 103 T 37,56525 106 T2

Hf40 (T) = 52,51 + (10,8667 + 121,6808 103T 37,56525 106T2)

T

298

dT 1

1000

= 1,25217 108T3 + 0,0000608404T2 + 0,0108667 T+ 44,2002 kJ mol

Cp5(T) = 10,2338 + 158,1230 103 T 45,56065 106 T2

Hf50 (T) = 83,82 + (10,2338 + 158,1230 103T 45,56065 106T2)

T

298

dT 1

1000

= 1,51869 108T3 + 0,0000790615T2 + 0,0102338 T 93,4888 kJ mol

Cp6(T) = 13,4925 + 190,9726 103 T 59,16355 106 T2

Hf60 (T) = 19,71 + (13,4925 + 190,9726 103T 59,16355 106T2)

T

298

dT 1

1000

= 1,97212 108T3 + 0,0000954863T2 + 0,0134925 T+ 7,73156 kJ mol

Cp7(T) = 10,8549 + 237,6317 103 T 72,91123 106 T2

Hf70 (T) = 104,68 + (10,8549 + 237,6317 103T 72,91123 106T2)

T

298

dT 1

1000

= 2,43037 108T3 + 0,000118816T2 + 0,0108549 T 117,823 kJ mol Cp8(T) = 24,6162 + 185,6805 10

3 T 60,87526 106 T2

Hf80 (T) = 186,11 + (24,6162 + 185,6805 103T 60,87526 106T2)

T

298

dT 1

1000

= 2,02918 108T3 + 0,0000928403T2 + 0,0246162 T+ 171,067 kJ mol

Cp9(T) = 16,9533 + 258,8119 103 T 79,77959 106 T2

Hf90 (T) = 9,01 + (16,9533 + 258,8119 103T 79,77959 106T2)

T

298

dT 1

1000

= 2,65932 108T3 + 0,000129406T2 + 0,0169533 T 24,8501 kJ mol Cp10(T) = 15,5790 + 312,0373 10

3 T 98,17201 106 T2

Hf100 (T) = 129,985 + (15,5790 + 312,0373 103T 98,17201 106T2)

T

298

dT 1

1000

= 3,2724 108 T3 + 0,000156019 T2 + 0,015579 T 147,617 kJ mol

4 | P a g e

HfMP0 (T) = Hf11

0 (T) = 100 +MMP1000

Cp(t)

T

298

dT

= 100

+MMP1000

(1,69578 + 0,29309Kw 0,45638d15,615,6) dT +

MMP1000

T

298

103 (4,6894 1,18074d15,615,6)(T 273) dT

T

298

= 100

+102,481

1000(1,69578 + 0,29309 11,791 0,45638 0,745)dT

T

298

+102,481

1000 103 (4,6894 1,18074 0,745)(T 273) dT

T

298

= 0,000195213 T2 + 0,0389407 T 128,94 kJ mol Hf Comb

0 (T) = Hf120 (T)

= 100 +MComb1000

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

T

298

+MComb1000

103 (4,6894 1,18074d15,6 Comb15,6 )(T 273) dT

T

298

= 100 +250

1000(1,69578 + 0,29309 11,1685 0,45638 0,9189) dT

T

298

+250

1000 103 (4,6894 1,18074 0,9189)(T 273) dT

T

298

= 0,000450552 T2 + 0,0435555 T + 47,0096 kJ mol Cp13 H2O(T) = 34,4 + 0,62775 10

3 T + 5,6079 106 T2

Qabur =1

MH2O Cp(T) dT

T2

T1

=1

18,015 (34,4 + 0,62775 103 T + 5,6079 106 T2) dT

T2

T1

= 1,03764 107 T13 0,000017423 T1

2 1,90952 T1 + 1,03764 107 T2

3 + 0,000017423 T22 + 1,90952 T2 kJ kg

5 | P a g e

Nr Comp Hfi0(T)

kJ mol 1 H2 1,69493 106 T2 + 2,77047 102 T 8,40652 2 CH4 2,40637 105 T2 + 2,23479 102 T 83,3166 3 C2H2 6,48921 109T3 + 3,01181 102 T2 + 2,79609 102 T

+ 217,365 4 C2H4 1,25217 108 T3 + 6,08404 108 T2 + 1,08667 102

T + 44,2002 5 C2H6 1,51869 108 T3 + 7,90615 105 T2 + 1,02338 102

T 93,4888 6 C3H6 1,97212 108T3 + 9,54863 T2 + 1,34925 102 T

+ 7,73156 7 C3H8 2,43037 108 T3 + 1,18816 104 T2 + 1,08549 102

T 117,823 8 C4H5 2,02918 108T3 + 9,28403 105 T2 + 2,46162 102 T

+ 171,067 9 C4H8 2,65932 108 T3 + 1,29406 104 T2 + 1,69533 102

T 24,8501 10 C4H10 3,2724 108 T3 + 1,56019 104 T2 + 1,5579 102 T

147,617 11 MP

benzin 1,95213 104 T2 + 3,89407 102 T 128,94

12 Comb 4,50552 104 T2 + 4,35555 102 T + 47,0096 13 H2O

6 | P a g e

Nr Comp Hfi0 1000 ni(x)

x

kJ kg MP reacionat 1 H2 2,953 [1,69493 106 T2 + 2,77047 102 T 8,40652] 2 CH4 3,859 [2,40637 105 T2 + 2,23479 102 T 83,3166] 3 C2H2 0,183 [(6,48921) 109 T3 + 3,01181 102 T2 + 0,0279609 T

+ 217,365] 4 C2H4 10,689 [(1,25217) 108 T3 + 6,08404 105 T2 + 1,08667 102 T

+ 44,2002] 5 C2H6 1,623 [(1,51869) 108 T3 + 7,90615 105 T2 + 1,02338 102 T

93,4888] 6 C3H6 5,205 [(1,97212) 108 T3 + 9,54863 105 T2 + 1,34925 102 T

+ 7,73156] 7 C3H8 0,702 [(2,43037) 108 T3 + 1,18816 104 T2 + 1,08549 102 T

117,823] 8 C4H5 1,727 [(2,02918) 108 T3 + 9,28403 108 T2 + 2,46162 102 T

+ 171,067] 9 C4H8 1,570 [(2,65932) 108 T3 + 1,29406 104 T2 + 1,69533 102 T

24,8501] 10 C4H10 0,963 [(3,2724) 108 T3 + 1,56019 104 T2 + 1,5579 102 T

147,617] 11 MP

benzin 9,758 (1,95213 104 T2 + 3,89407 102 T 128,94)

12 Comb 0,831 (4,50552 104 T2 + 4,35555 102 T + 47,0096) 13 H2O

3,86691 107 T3 + 2,38042 104 T2 + 1,0363 101 T + 1358,52

HR sp(T) = 3,86691 107T3 + 0,000238042 T2 + 0,10363T + 1358,52

7 | P a g e

Nr Comp Hh spi(x, T) = 1000 ni(x) Hfi0

kJ kg MP alim sau kJ kg prod de reacie 1 H2 5 x

1,693 [1,69493 106 T2 + ,2,77047 102 T 8,40652]

2 CH4 52 x

13,476 [2,40637 105 T2 + 2,23479 102 T 83,3166]

3 C2H2 4 x

21,872 [6,48921 109 T3 + 3,01181 104 T2 + 2,79609 102 T

+ 217,365] 4 C2H4 250 x

23,565 [1,25217 108 T3 + 6,08404 105 T2 + 1,08667 102 T

+ 44,2002] 5 C2H6 41 x

25,259 [1,51869 108 T3 + 7,90615 105 T2 + 1,02338 102 T

93,4888] 6 C3H6 184 x

35,348 [1,97212 108 T3 + 9,54863 105 T2 + 1,34925 102 T

+ 7,73156] 7 C3H8 26 x

37,041 [2,43037 108 T3 + 1,18816 105 T2 + 1,08549 102 T

117,823] 8 C4H5 77 x

44,590 [2,02918 108 T3 + 9,28403 102 T2 + 2,46162 102 T

+ 171,067] 9 C4H8 74 x

47,131 [2,65932 108T3 + 1,29406 104 T2 + 1,69533 102 T

24,8501] 10 C4H10 47 x

48,823 [3,2724 108 T3 + 1,56019 104 T2 + 1,5579 102 T

147,617] 11 MP

benzin

1000 (1 x)

102,481 [1,95213 104 T2 + 3,89407 102 T 128,94]

12 Comb 80 x

210 [4,50552 104 T2 + 4,35555 102 T + 47,0096]

13 H2O

3,86691 107 T3 x + 2,3804 104 T2 x + 1,90487 103 T2

+ 1,0363 101 T x + 3,7998 101 T + 1358,52 x 1258,18

Entalpia amestecului la o anumit temperatur:

Hh sp(x, T) = 3,86691 107 T3 x + 0,00023804 T2 x + 0,00190487 T2

+ 0,10363 T x + 0,37998 T + 1358,52 x 1258,18 kJ kg MP alim sau kJ kg prod de reacie

8 | P a g e

2. Calculul serpentine in sectoarele zonei de radiatie

2.1. Calculul sperpentinei n sectorul 1 din zona de radiatie (ZR)

tiR = 600 = 873,15 K piR = 7,2 bar GMP = 1400 kg h rs = 0,60

Bilan termic pe sector: Hh iR = GMP Hh sp(x, TiR) Hh iR = GMP Hh sp(xiR, TiR)

= 1400 (3,86691 107 903,153 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: ti2R = 715,35 xc1 = 0,100306

xci2 = 0 + xc1 xci2 = 0,100306 Hhci2 = GMP Hh sp(xci2, Ti2R)

= 1400 (3,86691 107 988,53 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

Qab1 = GMP rs Cp H2O(T) dT

725+273,15

650+273,15

= 1400 0,60 (1,03764 107 T1

3 1,7423 105 T12 1,90952 T1 + 1,03764

107 T23 + 1,7423 107 T2

2 + 1,90952 T2) = 150553,08 kJ h

QR1 = GMP xc1 HR sp (650 + 715,35

2+ 273,15) = 213019,42 kJ h

Calculm lungimea sectorului pe baza ecuaiei de bilan termic:

Hh ci + Qf 1R = Hh ci2 + Qab1 + QR1

1,07391004 106 + Qf 1R = 1,738158 106 + 150553,08 + 213019,42 Qf 1R

= 1,02782 106 kJ h

9 | P a g e

Qf 1R = tR de L1R

L1R =Hh ci2 + Qab1 + QR1 Hh ci

de tR

=1,738158 106 + 150553,08 + 213019,42 1,07391004 106

0,08 318470= 10,273 m

nc 1R = 1 Lech 1R = L1R + nc 1R 75 di = 10,273 + 1 75 0,07 = 16,273 m

Mms(xciR) =1,6

0,020016 0 + 0,0430635= 37,154

Mms(xci2) =1,6

0,020016 0,106382 + 0,0430635= 35,403

Se calculeaza caderea de presiune :

p1R = R Lech 1R GMP

2 (1 + rs)2

2 di5 36002

[piR 100 Mms(xciR)

TiR+

(105 piR p1R) Mms(xci2)

1000 Ti2R]

[TiR

piR 100 Mms(xciR)+

1000 Ti2R(105 piR p1R) Mms(xci2)

]2

= 0,0366 8,314 16,273 15002 (1 + 0,63)2

36002 2 0,075

[4,913 100 37,154

903,15+

(105 4,913 p1R) 35,403

1000 988,5]

[903,15

4,913 100 37,154+

1000 988,5

(105 4,913 p1R) 35,403]

2

p1R = 30706 Pa = 0,307 bar

Se calculeaza densitatea:

iR

=piR Mms(0)

R TiR=

4,913 37,154

0,083 903,15= 2,382 kg m3

i2R

=(piR p1R) Mms(xci2)

R Ti2R=

(4,913 0,307) 35,403

0,083 988,5= 1,968 kg m3

10 | P a g e

m1R

=

iR+

i2R

2=

2,382 + 1,968

2= 2,175 kg m3

Se calculeaza viteza de curgere :

wiR =4 GMP (1 + rs)

3600 iR

di2 =

4 1500 (1 + 0,63)

3600 2,382 0,072= 55,679 m s

wi2R =4 GMP (1 + rs)

3600 i2R

di2 =

4 1500 (1 + 0,63)

3600 1,968 0,072= 67,392 m s

wm1R =wiR + wi2R

2=

55,679 + 67,392

2= 61,535 m s

Se calculeaza timpul de stationare:

1 =L1R

wm1R=

10,273

61,535= 0,166 s

2.2. Calculul serpentine n sectorul 2 din zona de radiaie (ZR)

GMP = 1400 kg h Ti2R = 988,5 K Hhci2 = 1,738158 10

6 kJ h pi2R = piR p1R = 7,213 0,307 = 6,906 bar xci2 = 0 + xc1 = 0 + 0,100306 = 0,100306

Presupunem: ti3R = 788 xc2 = 0,486238 xci3 = xci2 + xc2 = 0,100306 + 0,4862387 = 0,586544

Bilan termic pe sector: i: Hh ci2 = GMP Hh sp(xci2, Ti2R) Hh ci2 = GMP Hh sp(xci2, Ti2R)

= 1400 (3,86691 107 988,53 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

Hhci3 = GMP Hh sp(xci3, Ti3R)

= 1400 (3,86691 107 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

11 | P a g e


761+273,15

725+273,15

= 1400 0,60 (1,03764 107 T1

3 0,000017423 T12 1,90952 T1 + 1,03764

107 T23 + 0,000017423 T2

2 + 1,90952 T2) = 73431,04 kJ h

QR2 = GMP xc2 HR sp (725 + 761

2+ 273,15)

= 1400 0,430083 (3,86691 107 1016,153 + 0,000238042 1016,152 + 0,10363 1016,15 + 1358,52) = 841168,85 kJ h

Calculm lungimea sectorului pe baza ecuaiei de bilan termic

Hh ci2 + Qf 2R = Hh ci3 + Qab2 + QR2

L2R =Hh ci3 + Qab2 + QR2 Hh ci2

de tR

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

0,1 318470= 19,73 m


Mms(xci2) =1,6

0,020016 0,10360 + 0,0430635= 35,403

Mms(xci3) =1,6

0,020016 0,586544 + 0,0430635= 29,739

12 | P a g e


p2R = R Lech 2R GMP

2 (1 + rs)2

2 di5 36002

[pi2R 100 Mms(xci2)

Ti2R+

(pi2R 105 p2R) Mms(xci3)

Ti3R 1000]

[Ti2R

pi2R 100 Mms(xci2)+

Ti3R(pi2R 105 p2R) Mms(xci3)

]2

= 0,0366 8,314 31,73 15002 (1 + 0,63)2

36002 2 0,075

[6,906 100 35,403

988,5+

(105 6,906 p2R) 29,739

1000 1061,15]

[988,5

6,906 100 35,403+

1000 1061,15

(105 6,906 p2R) 29,739]

2

p2R = 85208,2 Pa = 0,852 bar


i2R

=(piR p1R) Mms(xci2)

R Ti2R=

(4,913 0,307) 35,403

0,083 988,5= 1,968 kg m3

i3R

=(pi2R p2R) Mms(xci3)

R Ti3R=

(4,606 0,852) 29,739

0,083 1034,15= 1,3006 kg m3

m2R

=

i2R+

i3R

2=

1,968 + 1,3006

2= 1,6343 kg m3



3600 i2R

di2 =

4 1500 (1 + 0,63)

3600 1,968 0,072= 67,392 m s


3600 i3R

di2 =

4 1500 (1 + 0,63)

3600 1,3006 0,072= 101,975 m s

wm2R =wi2R + wi3R

2=

67,392 + 101,975

2= 84,6835 m s

13 | P a g e


2 =L2R

wm2R=

19,73

84,6835= 0,232 s

2.3. Calculul serpentine nsectorul 3 din zona de radiaie (ZR)

GMP = 1400 kg h Ti3R = 1061,15 K Hhci3 = 2,798098 10

6 kJ h pi3R = pi2R p2R = 6,906 0,852 = 6,054 bar xci3 = 0 + xc1 + xc2 = 0,586544

Presupunem: teR = 869,742 xc3 = xcf xci3 = 0,84 0,586544 = 0,253456

teR = 869,742 TeR = 869.742 + 273,15 = 1142,892 K

Bilan termic pe sector: Hh ci3 = GMP Hh sp(xci3, Ti3R) Hh ci3 = GMP Hh sp(xci3, Ti3R)

= 1400 (3,86691 107 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, TeR)

= 1400 (3,86691 107 1101,7723 0,84 + 0,00023804 1101,7722 0,84+ 0,00190487 1101,7722 + 0,10363 1101,772 0,84 + 0,37998 1101,772 + 1358,52 0,84 1258,18) = 3,777234 106 kJ h


828,622+273,15

761+273,15

= 1400 0,63 (1,03764 107 T1

3 0,000017423 T12 1,90952 T1 + 1,03764

107 T23 + 0,000017423 T2

2 + 1,90952 T2) = 140092,80 kJ h

14 | P a g e

QR3 = GMP xc3 HR sp (761 + 869,742

2+ 273,15)

= 1400 0,303535 (3,86691 107 1067,9613 + 0,000238042 1067,9612 + 0,10363 1067,961 + 1358,52) = 578088,14 kJ h

Calculm lungimea sectorului pe baza ecuaiei de bilan termic

Hh ci3 + Qf 3R = Hh ceR + Qab3 + QR3

L2R =Hh ceR + Qab3 + QR3 Hh ci3

de tR

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

0,08 250000= 16,964 m


Mms(xci3) =1,6

0,020016 0,586544 + 0,0430635= 29,739

Mms(xcf) =1,6

0,020016 0,84 + 0,0430635= 26,721


p3R = R Lech 3R GMP

2 (1 + rs)2

2 di5 36002

[pi3R 100 Mms(xci3)

Ti3R+

(pi3R 105 p3R) Mms(xcf)

TeR 1000]

[Ti3R

pi3R 100 Mms(xci3)+

TeR(pi3R 105 p3R) Mms(xcf)

]2

= 0,0366 8,314 28,964 15002 (1 + 0,63)2

36002 2 0,075

[6,054 100 29,739

1061,15+

(105 6,054 p3R) 26,721

1000 1101,772]

[1061,15

6,054 100 29,739+

1000 1101,772

(105 6,054 p3R) 26,721]

2

p3R = 148746 Pa = 1,487 bar

15 | P a g e


i3R

=(pi2R p2R) Mms(xci3)

R Ti3R=

(4,606 0,852) 29,739

0,083 1061,15= 1,3006 kg m3

eR

=(pi3R p3R) Mms(xcf)

R TeR=

(3,754 1,487) 26,721

0,083 1101,772= 0,6624 kg m3

m3R

=

i3R+

eR

2=

1,3006 + 0,6624

2= 0,9815 kg m3



3600 i3R

di2 =

4 1500 (1 + 0,63)

3600 1,3006 0,072= 101,975 m s

weR =4 GMP (1 + rs)

3600 eR

di2 =

4 1500 (1 + 0,63)

3600 0,6624 0,072= 200,225 m s

wm3R =wi3R + weR

2=

101,975 + 200,225

2= 151,1 m s


3 =L3R

wm3R=

16,964

151,1= 0,112 s

16 | P a g e

Zona Radiatie2015

Documents

Transcript of Zona Radiatie2015