O’zbekiston respublikasi oliy va o’rta maхsus

AMALIY MASHG‘ULOT №2


MAVZU: ODDIY SUPERFOSFAT OLISH TEXNOLOGIK XISOBLARI.
Oddiy superfosfatning tarkibida 21.4 % P₂O₅ um . bor va 20.3 % P₂O₅ sing . Xomashyoning tarkibida 39.4 % P₂O₅ bor . Xomashyoning parchalanish darajasini va 1000 kg xomashyodan ugitning chikish darajasini aniklang .
Echilishi;
Fosfaritning parchalanish darajasi kuyidagi formula asosida aniklanadi
K = C_sing/C_um C_sing - P₂O_{5 sing}
C _um - P2O5 _um
K = (20.3 / 21.4 ) * 100 = 94.86 %

Ugitning chikish darajasi

A = C_f / C_s = P₂O_{5 fosf} / P O_5super = (39.4 / 21.4 )* 1000=1841,12 kg

Masala № 3

Apatit konsentratining tarkibida :

Ca₃(PO₄)₂ - 86 % SiO₂ - 1.11 %

Fe₂O₃ - 0,63 % MgO - 0.19 %
K₂O - 0.23 % NaO - 0.63 %
CaF₂ - 6.20 % Namligi - 0,35 %
Al₂O₃ - 0.90 % erimagan modda –3,96 %
100 kg apatit konsentratini parchalash uchun sulfat kislotaning (monogidratning) sarflanishini aniklang . Apatining parchalanish darajasi 95 %.
Parchalanish darajasi 95 % bulganda :

Ca₃(PO₄)₂ = 86,0 * 0,95 = 81,7

CaF₂ = 6,20 * 0,95 =5,89
Fe₂O₃ = 0.63 * 0.95 =0.60
Al₂O₃ = 0.90 * 0.95 =0.86
Na₂O =0.63 * 0.95 =0.60
K₂O = 0.23 * 0.95 = 0.22
MgO = 0.19 * 0.95 =0.18 kg reaksiyaga kirishmadi .

X_k – sulfat kislotaning sarflanishi

Y_b – cuvning sarflanishi

Ca ₃(PO₄) + 2 H₂SO₄ + H₂O =Ca(H₂ PO₄)₂*H₂O +2CaSO₄

X₁ = (81,7*196)/310 =51,66 Y₁ = (81.7*18)/ 310=4.74
196 = 2 * 98
SaF₂ + H₂SO₄ = CaSO₄ + 2HF

X₂ = (5.89*98)/78 = 7.40 Y₂ = 0

Fe₂O₃ +3 H₂SO₄ + 6H₂O = Fe₂(SO₄)₃ * 9H₂O

X₃ = (0.6*294)/160 = 1.10
Y₃ = (0.6*6.18)/160 = 0.40
Al₂O₃+ 3 H₂SO₄+15 H₂O =Al₂(SO₄)₃*18 H₂O

X₄ = (0.86*3*98)/102 = 2.48 Y₄ = 2.28

MgO + 3 H₂SO₄ = MgSO₄* H₂O

X₅ = 0,44
Y₅ = 0
K₂O+ H₂SO₄ = K₂SO₄+ H₂O

X₆ = 0.23

Y₆ = -0.042
Na₂O+ H₂SO₄ = Na₂SO₄+ H₂O

X₇ =0.95

Y₇ = - 0.17
4 HF + SiO₂ = SiF₄ + 2 H₂O
60 SiO₂ --------- 36 H₂O
1.11 SiO₂ --------- x H₂O

x = (1.11*36 )/ 60 = 0.66

100 kg apatitni parchalach uchun H₂SO₄ sarflanadi:

X_n = 51.66 + 7.40 + 1.1 + 2.48 + 0.44 + 0.23 + 0.95 = 64.26

H₂O

Y₆ = 4.47 + 0.4 + 2.23 + 0.66 – 0.042 – 0.17 = 7.548


AMALIY MASHG‘ULOT №3


MAVZU: EKSTRAKTSION FOSFOR KISLOTASI OLISH MODDIY BALANSI

Masala №1

Apatit konsentratidan EFK olish moddiy balansini tuzish :
(EFK ning tarkibida 32 % P₂O₅ ) H₂SO₄ konsentratsiyasi – 92 % . Apatit konsentratning tarkibida 39,4 % P₂O₅ , 47,42 % CaO , 6,22 % Ca F₂ bor . H₂SO₄ normasi stexiometrik xisobiga (CaO mikdoriga )nisbatan 100 %.
P₂O₅ utish darajasi 96 % . Filtratsiya jarayonida P₂O₅ yuvilish darajasi 98 % .Xomashyoning tarkibidagi ftorning 25 % gaz xolatda ajralib chikadi . kuykaning aylanish nisbati 5,8 : 1 . Suyuk fazaning kuyka fazasiga nisbati 3 : 1 . Apparatning unumdorligi 5880 kg/ch EFK .
EFK ishlab chikarish moddiy balansini tuzish
Ca₅F(PO₄)₃ + 5 H₂SO₄+ 5_n H₂O =35CaSO₄ + 5 CaSO_4* H₂O + 3H₃PO₄ + HF ↑

Ca₅F(PO₄)₃ + 3,5 H₂SO₄ =35CaSO₄ + 15 Ca(H₂SO₄) + HF

Masalaning echilishi :

Apatit konsentratining sarflanishi .

5880/(0,39*0,96) = 15565
P₂O₅ = 39,4 % = 0,394*15565 = 6125 kg
CaO = 47,42 % =0,4742 *15565 = 7381 kg
Ca F₂ = 6,22 % = 968 kg
Fe₂O₃ = 0.95 % =148 kg
Fe₂O₃ = 0.82 %=128kg
SiO₂ = 1.67 % =260kg
R = 3.07 = 477kg
H₂O = 0.50 % =78 kg
1) Ca₅F(PO₄)₃ + nH₃PO₄+aq=Ca(H₂PO₄)₂+ (n-7) H₂O + HF ↑ +aq

2) Ca₅F(PO₄)₃ +H₂SO₄+aq= H₃PO₄ + 5 CaSO₄*2H₂O +aq

Umumiy reaksiya

3) Ca₅F(PO₄)₃ +5H₂SO₄+ nH₃PO₄+aq=3H₃PO₄ + 5 CaSO₄*2H₂O + nH₃PO₄+ HF ↑ +aq

Bu reaksiya asosida moddiy balans tuzib bulmaydi , shuning uchun apatit konsentratning tarkibidagi moddalarning H₂SO₄ va H₃PO₄ kimyoviy reaksiyalarini yozamiz :
(4) P2O5 + 3H₂O = 2H₃PO₄
(5) CaO +H₂SO₄ = CaSO₄+2H₂O
(6) Fe₂O₃ +2 H₃PO₄ = Fe₃PO₄ + 3H₂O
(7) Se₂O₃+2 H₃PO₄ = 2GrPO₄+ 3H₂O
(8) 6 HF + SiO₂ = H₂SiF₆ + 2 H₂O
(9) 4 HF + SiO₂ = SiF₄ + 2 H₂O
(10) CaF₂+H₂SO₄= CaSO₄+2 HF

(8) va (9) reaksiyalarda (10) reaksiyada xosil bulgan HF katnashadi .

(4) reaksiya asosida :

(6125*0,98)/142 = x/196 x= (6125*0,98*196)/142=8285 kg/soat.

Suvning sarflanishi :
X/(3*18) =8285/196 x= (3*18*8285 )/196 = 2283 kg/soat

5. 
   reaksiya asosida :
   

a) H₂SO₄ sarflanishi :
7381*0,98 --------- x
56 --------- 98
x= (7381*0,98*98)/56 = 12658 kg/soat
xosil bulgan CaSO₄ mikdori :
12658 ---------- x
98 ---------- 136
x= (136*12658)/98 = 17566 kg/soat CaSO₄
yoki 22216 kg/soat CaSO₄*2H₂O
suv xosil buladi :
(12658*18)/98 = 2325 kg/soat

10. 
    tenglama asosida : H₂SO₄ sarflanishi :
    

968/78 = x/98 x = (968*98*0,98)/78 = 1192 kg/soat H₂SO₄
CaSO₄ xosil bulishi :
1192/98 = x/136 x=(1192*136)/98 = 1655 kg/soat CaSO₄
2093 kg/soat CaSO₄*2H₂O
(5) va (10) reaksiyalar asosida H₂SO₄ sarflanishi mikdori :
12658 +1192 = 13850 kg/soat ∑ H₂SO₄
22216 + 2093 = 19221kg/soat ∑ CaSO₄
24309 kg CaSO₄*2H₂O tarkibida :
2403 – 19221 = 5088 kg H₂O bor .
(10) reaksiya asosida ajralib chikkan HF mikdori :
(1102*2*20)/98 = 486 kg HF

5. 
   reaksiya asosida H₃PO₄sarflanishi mikdori :
   

(128*0,98*196)/160 = 154 kg/soat H₃PO₄
Fe₃PO₄-mikdori ;
(154*2*151)/(2*98) = 237 kg/soat
xosil bulgan H₂O mikdori :
x= (154*3*18)/(2*98) = 42 kg

5. 
   H₃PO₄ sarflanishi :
   

X = (148*196*0,98)/328 = 87 kg/soat
Xosil bulgan GrPO₄mikdori
X= (2*235*87)/196 =208 kg/soat
Xosil bulgan H₂O
X=(87*3*18)/196 = 24 kg/soat
Xosil bulgan HF 25 % gaz xolatda ajralib chikadi . 75 % - H₂SiF₆ xolatda suyuk fazada koladi .
25 % HF- ning –20 % SiF₄ xolda ajralib chikadi 5%-- HF
(48*0,2) = 97 kg/soat SiF₄
(486*0,05) = 24 kg/soat HF

∑ 97 +24 =121 kg/soat HF gaz xolda ajralib chikadi .

suyuk fazada 486 – 121 kg = 365 kg/soat xolda

5. 
   tenglamada 365 kg HF reaksiyaga kirishadi.
   

SiO₂- sarflanishi
X= (365*60)/120 = 182 kg/soat
H₂SiF₆ xosil bulishi :
X= (182*144)/60 = 438 kg/soat
H₂O xosil bulishi :
X=(182*2*18)/60 = 109 kg/soat
(9) tenglama asosida 97 kg HF reaksiyaga kirishadi
SiO₂- sarflanishi:
X=(97*60)/80 = 73 kg/soat
SiF₄ xosil bulishi :
(73*104)/60 = 126 kg/soat H₂SO₄
xosil bulishi :
(73*2*18)/60 = 44 kg/soat
(5-9) reaksiyalar asosida :
2325 + 42 + 24 + 109 + 44 = 2544 kg/soat xosil buladi .
(6-7) reaksiyalar asosida H₃PO₄ sarflanishi:
154 + 87 = 241 kg/soat
Kuykaning tarkibida 8285 –241 = 8044 kg/soat H₃PO₄ koladi
(5) va (10) reaksiyalar asosida
12658 + 1192 = 13850 kg/soat monogidrat sarflanadi .
13850/0,92 = 15054 kg/soat 92 % H₂SO₄
15054 – 13850 = 1204 kg/soat H₂O
Suyuk fazaning kattik fazaga nisbati 8 bulishi uchun 92 % H₂SO₄ , 56% gacha suyultiriladi
13850/0,56 = 24273 kg/soat
24273 – 13850 = 10882 kg/soat H₂O
92 % H₂SO₄ 56% H₂SO₄ gacha suyultirish uchun
10882 – 1204 = 9678 kg/soat H₂O sarflanadi
Parchalanmay kolgan apatitning tarkibi :
P2O5 = 6125 * 0,02 = 123
CaO = 7381* 0,02 = 148
Ca F₂ = 968* 0,02 = 19
Fe₂O₃ = 128* 0,02 =3
SiO₂ = 260-(182+73)=5
Gr₂O₃ = 148*0,02 = 3
Jami 301 kg
Apatit bilan berilgan erimaydigan moddalarning mikdori 477 kg/soat
477 + 301 = 778 kg/soat
H₂O balansi : H₂SO₄ bilan 10882 kg/soat beriladi , (5--9) reaksiyalar asosida 2544 H₂O xosil buladi , apatit bilan 78 kg H₂O beriladi , ∑ 13504 H₂Oberiladi .
H₃PO₄ olish uchun (4) reaksiya asosida 2283 kg H₂O sarflanadi kuykaning tarkibida
13504 – 2283 = 11222 kg/soat
5088 kg CaSO₄*2H₂O bulganda
11222 – 5088 = 6133 kg H₂O kuykaning suyuk fazasida
Moddiy balans jadvali
Kirish , kg/soat chikish , kg/soat
I.Apatit konsentratsiyasi I. Kuykaningkattikfazasi
P2O5 = 6125 CaSO₄*2H₂O = 24309
CaO = 7381 Fe₃PO₄ = 237
Ca F₂ = 968 Se₂O₃ = 208
Fe₂O₃ = 128 H₂SiF₆ = 438
SiO₂ = 260 R = 778
Se₂O₃ = 148
______________________________________________________________

Jami : 15565 Jami : 25970 kat. faza

II. H₂SO₄ (monogidrat) 13850 II. Kuykaning suyuk fazasi

H₂O 10882 H₃PO₄ = 8044
H₂O = 6133
________________________________ _______________________
Jami : 40297 40147
III . Gaz faza :
SiF₄ = 126
HF = 24
____________________________
40297
Masala № 2 :
100 kg apatit konsentratidan EFK olish jarayonida xosil bulgan pulpaning , suyuk fazaning va suyultirilgan eritmaning mikdorini aniklang. Pulpaning suyuk fazasining kattik fazasiga nisbati 2,5/1 (j:t) , gips soni1,6 , ajralib chikgan ftorning mikdori 5 kg . 100 kgapatit konsentratiga 117,9 kg H₂SO₄ sarflandi.