Glitserinni sovutish uchun turba ichida turba issiqlik almashinish qurilmasini xisoblang

Glitserinni sovutish uchun turba ichida turba issiqlik almashinish qurilmasini xisoblang

Berilgan;

• 
  Unumdorlik G=17.11tonna/soat
  
• 
  Glitserinning harorati;
  

• 
  t_bosh= 67⁰C
  
• 
  t_oxir=37 ⁰C
  

• 
  Suvning harorati
  

• 
  t_bosh=11⁰C
  
• 
  t_oxir=42 ⁰C
  

Yechish

Glitserinning o’rtacha harorati;

t_{1 o’rt}=0.5(67+37)=52 ⁰C

Glitserinning fizik-kimyoviy xassalari.

• 
  Zichlik: ρ₁=1109.9 kg/m³ (512 bet lVjad)
  
• 
  Qovushqoqlik: µ₁=2.4687∙10^-3 MPa/sek(516 bet IX jad)
  
• 
  Issiqlik sig’imi: C₁=2593.61 J/kg (562 bet Xl jad)
  
• 
  Issiqlik o’tkazuvchanlik: λ₁=0.3977 Vt/m∙K (561bet X jad) Suvning o’rtacha harorati;
  

t_{2 o’rt}=0.5(11+42)=26.5⁰C

Suvning fizik-kimyoviy xossalari.

• 
  Zichlik: ρ₂=996.03 kg/m³ (512 bet lVjad)
  
• 
  Qovushqoqlik: µ₂=0.8694∙10^-3 MPa/sek(516 bet lX jad)
  
• 
  Issiqlik sig’imi: C₂=4194 J/kg (562 bet Xl jad)
  
• 
  Issiqlik o’tkazuvchanlik: λ₂=0.6 Vt/m∙K (561 bet X jad)
  

Glitserin sarfini SI sistemasiga o’tkazamiz.

G₁==4.75 kg/s

Glitserin uchun issiqlik sarfini aniqlaymiz.

Q₁= G₁∙ C₁(t_1bosh- t_1oxir)= 4.75∙2593.61(67-37)=369589.4 BT

Suvning issiqlik balansi.

G₂==2.87kg/sek

Harorat tasirini tuzamiz.

67 37

42 11 =0.96>2

Δt_kat=25 Δt_kich=26

Δt_o’rt=⁰C

Qurilmaning isitish yuzasini taxminiy qiymatini topish.

F_max=²

Gliserin va suvning hajmiy sarfi

V₁===0.00427 m³/c (gliserin)

V₂===0.00288 m³/c (suv)

Issiqlik o’tkazish koeffitsienti.

Reynolds soni

𝜔₂₌ m/s

Diametrni turba ichida turba UAA ГОСТ9930-78 dan aparatni tayorlaymiz.

d₁=89х4=0.081 d₂=57х3.5=0.057

57х3.5 mm parallel ishchi turbalar sonini topamiz.

n^l=

nl=2 bo’lsa tezlikni va Reynolds kriteriyasini suvga nisbatan aniqlaymiz.

𝜔₂₌ m/s

Re₂=

𝜔₂₌ m/s

Re₂=

Gliserin uchun shunday qiymatlarni topamiz

𝜔₁₌ m/s

Re₂=

de=D₁-d₂=0.081-0.057=0.024 m

Nussel kriteriyasini aniqlaymiz

Nu=0.021∙Ee∙Re0.8∙Pr0.43∙(Pr/Prdev)0.25

Prandtli kriterisini aniqlaymiz

Pr=C1∙µ1/λ1=2593.61∙0.0024687/0.3977=16

Nu=0.021∙1∙8848*0.8∙16*0.43∙1*0.25=255.6

Glitserinda devorgacha issiqlik berish koeffitsienti.

⍺₁= Vt/m²∙K

Suvning issiqlik berish koeffitsienti. Prandtli kriteriyasi.

Pr=C2∙µ2/λ2=4194∙0.0008694/0.61=5.97

Nu=0.021∙1∙36569*0.8∙5.97*0.43∙1*0.25=394

Suvdan devorgacha issiqlik berish koeffitsienti

⍺₂= Vt/m²∙K

Devorning termik qarshligini aniqlash.

E_2dev=.2∙10^-4

K^l= Vt/m²

Issiqlik oqimini yuza zichligi.

q₁¹= K^l∙Δt_o’rt=331.3∙25=8282.5 ℃

Δt^l_dev1vaΔt^l_dev2 taxminiy axamiyatini aniqlash

Δt_l= q¹/⍺_l^l=8282.5/423.5=19.56 ℃

Δt^l_dev= q¹/E_1dev=8282.5∙0.00042=3.48 ℃

Δt^l₂= q¹∙ =8282.5/4216.5=1.96 ℃

Tekshirish;

Δt_o’rt=Δt_l+Δt^l_dev+ Δt^l₂=19.56+3.48 + 1.96 =25 ℃

Bu yerda;

Δt^l_dev1=t₁- Δt₁¹=48.5-19.56=28.94 ℃

Δt^l_dev2=t₂- Δt₂¹=23.5+1.96=25.46 ℃ Issiqlik berish koeffitsientiga o’zgartirish kiritish ( Pr/Pr_dev)^0.25

Prandtli kriteriyasini aniqlash. Δt^l_dev1=32.7 ℃

Pr_dev1=C_dev1∙µ_dev1/λ_dev1=2220,7∙0.000282/0.165=3.8 ℃

Prandtli kriteriyasini suv uchun aniqlash. Δt^l_dev2=30.35 ℃

Pr_dev2=C_dev2∙µ_dev2/λ_dev2=4399.5∙0.00074/0.616=5.29 ℃

Atseton uchun issiqlik berish koeffitsienti

⍺₁=⍺₁¹( Pr₁/Pr_dev1)^0.25=423.5(16/3.8)^0.25=606.65 Vt/m²∙K

Suv uchun issiqlik berish koeffitsienti

⍺₂=⍺₂¹( Pr₂/Pr_dev2)^0.25=4216.5(5.97/5.29)^0.25=4345.92 Vt/m²∙K

K^l= Vt/m²

q=k∙Δt_o’rt=331.3∙25=8282.5 ℃

Δt^l_dev1 =t₁-=48.5-℃

Δt^l_dev2 =t₂-=23.5-.6 ℃

Issiqlik berish yuzasini topamiz.

F_p^l =

UAA larni sirt maydonini 1 ta elementar uzunligi 6 m deb

F₁=d_o’rt∙L=3.14∙0.0535∙6=1.01 m²

Elementlar soni xar biri 2 sektordan iborat

N= ta

Umumiy elementlar soni

9∙2=18 ta

Gidravlik Hisob

ΔP= (λ+Σξ)

λ= ==0.0325

ξ₁ = 0.5 ξ₂=13.2 ξ₃=1

Σξ = ξ₁+ξ₂+ξ₃=14.7 L=6 m

ΔP = (0.02+14.7) =10721 Pa

Ksiloleritmasini isitishu chun IAQ ni hisoblang.

Ber;

G=10 t/soat,tbosh=12℃,tox=60℃,d=50x3mm,P=0.2MPa

Yechish;

1.Suv bug’ining kondensatlanishi t⁰si

T_kon=119.6 ℃ [LVII-549 bet 0.2 MPa da olamiz]

2.Haroratlar tasirini tuzamiz.

119.6 119.6

12 60<2

Δt_kat=107.6 Δt_kich=59.6

3.Xaroratlar tasirini o’rtacha farqini aniqlaymiz

Δt_o’rt =⁰C

Agarda>2 bo’lsa, u holda Δt_o’rt=

4.Ksilolning o’rtacha harorati

t₂ =t_kon-Δt_o’rt=119.6-83.6=36 ⁰C

5.Ksilolning sarfi berilgan G=10t/soat, uni SI da aniqlaymiz

G₂ =

6.Ksilolning hajmiy sarfi

V= m³/s

=850 ksilolning 36℃dagi zichligi [512bet IV-jad]

7.Ksilolni isitish uchun issiqlik sarfi

Q=G₂∙C(tox-tbosh)=2.8∙1802∙(60-12)=242189 Kj/soat

C-36℃ dagi solishtirma issiqlik sig’imi [562bet XI-rasm]

8.Isituvchi bug’ning sarfi {7% yo’qotilish bor}

G₁ ===0.12 kg/s

r=2208∙10³suvning solishtirma bug’ hosil bo’lish issiqligi [549bet LVII-jad]

9.Qurilmani isitish yuzasini maksimal taxminiy qiymati

F_tax=²

K=140 BT/(m²∙K) {172bet 4.8jad}

10.Oqimni harakat rejimini aniqlaymiz

𝜔₌ m/s

Re=

=0.51∙10^-3 - 36℃ dagi ksilolning dinamik qovushqoqligi

{516bet IX-jad}

11.Trubilent rejim uchun Nussel kriteriyasini aniqlaymiz

Nu=0.021∙E_e∙Re^0.8∙Pr^0.43∙(Pr/Pr_dev)^0.25∙₁

( Pr/Pr_dev)^0.25=1.05=1 doimiy son Ɛ_c=1

Pr=C∙µ/λ=1802∙0.00051/0.127=7.24

λ =0.127 BT/(m∙K) Ksilolning issiqlik o’tkazish koeffitsienti

{561bet 10-rasm}

12.Nu=0.021∙16133^0.8∙7.24^0.43∙1.05∙1=118

13.Issiqlik uzatish koeffitsientini topamiz

K^l=₁=bug’dan devorgacha issiq berish ₂=devordan eritmagacha issiq berish

Ɛ_dev =devorning termik qarshiligi

₁=2.02∙Ɛ₁ ∙Ɛ_r ∙Bt∙()^1/3∙l^1/3=2.02∙0.62∙0.6∙1070∙()^1/3∙3^1/3=13604Bt/(m²∙K)

Ɛ₁=0.62 ga teng {162bet 4.7-rasm}

Ɛ_r=0.6 bug’dagi havo miqdori{164bet 4.9-rasm}

Bt=1070-119.6℃dagi suv bug’ining kondensatsiya tenglamasiga bog’liq kattalik{162bet 4.6-rasm}

n=trubalar soni l=trubalar uzunligi

₂===340

14.Devorning termik qarshiligi

Ɛ_rdev ={131bet XXXI-jad}r₁-suv bug’I tomonidan devordan ifloslanishdagi issiqlik o’tkazish 5800 Bt/(m²∙K)

r₂-eritma tomonidan devordan ifloslanishdagi issiqlik o’tkazish 5800 Bt/(m²∙K)

δ=0.002m po’latning yemirilishi

=po’latning issiqlik o’tkazish koeffitsienti

Ɛ_rdev = =2578 Bt/(m²∙K)

15. K^l= =298.5 Bt/(m²∙K)

16.Issiqlik almashinish qurilmasini issiqlik almashinish yuzasini topamiz

F= m²

Standart bo’yicha 1 yo’lli IAQ ni tanlab olamiz

F=10m²

D=385mm

n=62 ta

l=2 m

Toluolni sovutish uchun garizantal IAQ ni hisoblang

Yechish;

Ber;

Toluol G=18 t/soat,tb1=105℃,tox1=30℃,

Suv tb2=15℃,tox2=45℃,

Yechish;

1.Toluolning o’rtacha xarorati

Δt_2o’rt=⁰C

Yuqorida chiqqan qiymat asosida biz toluolni shu haroratdagi xususiyatlarini aks ettiruvchi birliklarni tanlab olamiz. Bu birliklar jadval asosida olinadi.

• 
  Zichlik: ρ₁=820.5 kg/m³ (512 bet lVjad)
  
• 
  Qovushqoqlik: µ₁=0.35∙10^-3 MPa/sek(512 bet lVjad)
  
• 
  Issiqlik sig’imi: C₁=1780.7 J/kg (562 bet Xl jad)
  

Issiqlik o’tkazuvchanlik: λ₁=0.151Vt/m∙K (561 bet X ja)

2.Suvning o’rtacha xarorati

Δt_2o’rt=⁰C

Yuqorida chiqqan qiymat asosida biz suv shu haroratdagi xususiyatlarini aks ettiruvchi birliklarni tanlab olamiz. Bu birliklar jadval asosida olinadi.

• 
  Zichlik: ρ₂=995.7 kg/m³ (512 bet lVjad)
  
• 
  Qovushqoqlik: µ₂=0.82∙10^-3 MPa/sek(512 bet lVjad)
  
• 
  Issiqlik sig’imi: C₂=4184 J/kg (562 bet Xl jad)
  

Issiqlik o’tkazuvchanlik: λ₂=0.616 Vt/m∙K (561 bet X ja)

3.Issiqlik yuklamasini aniqlaymiz

Q=G∙C₁(tb1-tox1)=5∙1780.7∙(105-30)=667763 BT

G=18 t/soat=5kg/s

4.Suvning sarfini issiqlik balans tenglamasidan aniqlaymiz

G₂==5.3 kg/sek

5.Temperaturalar farqi

105 30

45 15 =4>2

Δt_kat=60 Δt_kich=15

Δt_o’rt=⁰C

6.Turbulen rejimdagi oqim uchun 25x2mm o’lchamli trubalar sonini aniqlaymiz

n=G/(0.785∙µ₁∙d∙Re)=5(0.785∙0.35∙10^-3∙0.021∙15000)=57

7.Quyidagi tasnifni g’ilofsimon IAQ ni tanlaymiz

D=325mm n=61 ta 1 yo’llik {215bet 4.2-jad}

8.Reynolds kriteriyasini aniqlaymiz

Re=

9.Toluol uchun Prankli kriteriyasi

P₁===4.13 u holda

⍺₁= ∙Re^0.8∙Pr^0.43=0.151∙14206^0.8∙4.13^0.43=583.2 Vt/m²∙K

10.Suv uchun t_2o’rt=30⁰C bo’lganda Pr=5.56 ga teng

(564 bet XIII ras)

11.Trubalar orasidagi suvning harakat tezligini aniqlaymiz

W_suv=G₂/[ρ₂∙0.785(D²-n∙d²)]=5.3/[995.7∙0.785∙(0.325²-61∙0.025²)]=0.1m/s

12.Reynolds kriteriyasini topish uchun d_эni topamiz

d_э=D²-nd²\D+nd=0.325²-61∙0.025²\0.325+61∙0.025=0.0365m

13.Reynolds kriteriyasi

Re₂=oraliq

Reynolds kriterisini 10000 deb qabul qilamiz. Nu esa

Nu=1.72∙(d_э∙Re)^0.6∙Pr^0.33=1.72∙(0.0365∙10000)^0.6∙5.56^0.33=104.4

15.Devorning sovutuvchi suvga issiqlik berish koeffitsienti

⍺₂=Nu∙\d_э=104.4∙0.616\0.0365=1762 Vt/m²∙K

16.Issiqlik uzatish koeffitsientini aniqlaymiz

K^l=.8 Vt/m²∙K

17.ƐΔt qiymatini aniqlash uchun P va R katalik topiladi

P=t_oxsuv-t_bsuv/t_btol-t_bsuv=45-15/105-15=0.33

R=t_btol-t_oxtol/t_oxsuv-t_bsuv=105-30/45-15=2.5

Bu qiymatlardan ƐΔt=0.49 bo’ladi (560bet 8-rasm)

18.Uning Δt_o’rt2=ƐΔt∙Δt_o’rt1=0.94∙32.5=30.55℃

19.Talab etilgan issiqlik almashuv qurilmasi almashinish yuzasini topamiz.

F=.9 m²

Standart bo’yicha (215bet 4.12-jad) dan

F=65 m² D=600 mm n=206 ta

l=4 m

Uzluksiz ishlovchi elaksimon tarelkali rektifikatsion kalonnani hisoblang. Aralashma benzol-toluol (50%-50%).Unumdorligi G=10t/soat atmosfera bosimi ostida ishlaydi, distilyatsiyaga benzolning kerakli miqdori X_D=96%,kub xolidagi toluolni kerakli miqdori,X_K=98%,Par bosimi P=3 kg/m²(0.3MPa)

I.Moddiy xisob

Ditilyatning massaviy sarfini G_D(kg/soat) orqali kub qoldiq massaviy sarfini C_w(kg/soat) orqali aniqlaymiz. Ular quyidagi tenglama asosida topiladi.

1. 
   G_D+G_w=10000
   

G_D∙0.96+C_w∙0.02=10000∙0.5

0.02-98% kub qoldiqdan qolgan miqdor

0.5-50%

2. G_D=10000-G_w

G_D∙0.96+G_w∙0.02=10000∙0.5=(10000- G_w)∙0.96+ G_w∙0.02=5000

9600-0.96G_w+0.02G_w=5000
0.94G_w=4600

2. 
   G_w=4894(kg/soat)
   

3.G_D=10000- G_w=10000-4894=5106(kg/soat)

4.Keyingi hisoblar uchun distilyat va kub qoldig’ini mol ulushda ifodalaymiz

X_f===0.541

5.Distilyat

X_d===0.966

6.Kub qoldig’i

X_w===0.022

7.Oziqlanishdagi solishtirma mol sarfi

F===1.82

8.Flegmaning minimal soni

R_min===1.135 Bu yerda Yf=0.74-benzolning bug’dagi mol ulushi,oziqlanishda suyuqlik bilan muvozanati. {327bet 7.7-rasm}

9.Flegma sonini topamiz

R=1.3R_min+0.3=1.3∙1.135+0.3=1.78

10.Ishchi tizim tenglamasi

a)Kalonnaningyuqoriqismida;

Y===0.640x+0.347

b)Kalonnaning quyi qismida;

Y===1.3x-0.006

Kalonna diametrik va bug’ tezligini topish

11.Suyuqlikning o’rtacha konsentratsiyasi

a)Kalonna yuqori qismida

X^I_cp=(X_F+X_D)/2=(0.541+0.966)/2=0.753

b)kalonna quyi qismida

X^II_cp=(X_F+X_W)/2=(0.541+0.022)/2=0.281

12.Bug’ning o’rtacha konsentratsiyasi ishchi tizim tenglamasi ⑩dan topamiz.

a)Kalonna yuqori qismida

Y^I_cp=0.640x+0.347=0.64∙0.753+0.347=0.828

b)kalonna quyi qismida

Y^II_cp=1.3x-0.006=1.3∙0.281-0.006=0.359

13.Bug’ning o’rtacha xaroratini t-x,y (7.6-rasm) diagrammadan topamizY^I_cp= 0.828 Y^I_cp= 88℃

Y^II_cp=0.361t^II_cp= 103℃

14.Bug’ning o’rtacha mol massasi va zichligi

a)M^I_cp=0.829∙78+0.171∙92=80.3 kg/k mol

Bu yerda

0.829-------------- Y^I_cp

0.171----1---0.829

Ρ^I_cp=M^I_cp∙T₀/22.4∙t^I_cp=80.3∙273/22.4∙(273+88)=2.71 kg/m³

b)M^II_cp=0.361∙78+0.639∙92=87kg/k mol

Ρ^II_cp= M^II_cp∙T₀/22.4∙t^II_cp=87∙273/22.4∙(273+103)=2.81 kg/m³

15.Kalonnadagi bug’ning o’rtacha zichligi

Ρ_o’r.p.=(Ρ^I_cp+Ρ^II_cp)/2=(2.71+2.82)/2=2.77 kg/m³

16.Suyuq benzol va toluolning zichliklari yaqin. Kalonna yuqori qismida Y_D0.965 bo’lganda xarorat 82℃,bug’latish kubida esa X_W=0.023 bo’lganda 109 ℃ ga teng. Suyuq benzolning 82℃ da gizichligi ρ_b=813 kg/m³; suyuq toluolning 109 ℃ da gizichligi ρ_t=783 kg/m³ Kalonnadagi suyuqlik o’rtacha zichligini qabul qilamiz;

ρ_ж=(ρ_b+ ρ_t)/2=(813+783)/2=800kg/m³

17.Bug’ning kalonnadagi tezligi. Tarelkalar orasidagi masofa <<Калоние Апарати>> katalog-malumotnoma kitobidan olamiz.

H=300mm elaksimon tarelkalar uchun C=0.032mm grafik 7.2 dan topamiz. {323bet П.Р.}

𝜔=C=0,032=0,54m/s

18.Kalonnadan o’tayotgan hajmiy sarfni toppish uchun

T_o’rt=(88+103)/2=96 ℃

Xajmiy sarf

V=G_D(R+1)22.4T_cbP_o/M_DT_O3600P=5106(2.78)∙22.4∙369∙1.033/78.5∙273∙3600∙1=1.57 m³/s

M_D=X_D∙M_b+(1-X_D)∙M_t=0.966∙78+0.034∙92=78.478

P₀=1.033 atm bosimidagi absalyut bosim.

T_cb=273+96=369 ℃

P=1 partsial bosim

19.Kalonna diametric

D===1.92 m

20.<<Калоние Апарати>> kitobidan D=1900mm ni olamiz unda kalonadagi parning tezligi

W=V/0.785∙D²=1.57/0.785∙1.92²=0.6 m/c