Navoiy davlat konchilik instituti kimyo metallurgiya fakulteti «metallurgiya» kafedrasi

S1 =	D1	=	850
= 4.25
	d1	200

Bunday maydalanish darajasiga 1 bosqichda erishish mumkin .

Maydalashdan so‘ng mahsulotning maksimal yirikligini aniqlaymiz:

D2 =	D1	=	850	= 200 mm
	S1		4.25

Ruda

May Maydalash

Sklad
maydalash sxemasi.


Maydalagichning chiqarish tuynugi kengligini aniqlaymiz: Z_I = 1,8

mm (36)

Tanlangan sxemaning qo‘llaniluvchi uskunaga mosligini tekshiramiz.

Tanlangan uskunalar 2-jadvalda keltirilgan talablarni qanoatlantirishi kerak.

Aniqlangan kirish chiqish tuynuklariga mos keluvchi maydalagich turi KSD 2200 gr [7] maydalagich markasi ekan.

Ko‘rsatkichlar	Maydalash jarayori
Ta’minlagichdagi eng katta bo‘lakning o‘lchami, mm CHiqarish tuynugining kengligi,mm Talab etiluvchi ishlab chiqarish unumi t/soat / m^3/soat	850 111 441 /155

2-jadval

Maydalash bosqichi	Maydalagich turi	Qabul qilish tuynugi kengligi, mm	Bo‘shatish tuynugining o‘zgarish kengligi, mm	Ishlab chiqarish unumdorligining o‘zgarishi, m^3/s	Loyixalashtirish uchun tanlangan ishlab chiqarish unumdorligi
					m^3/s	t/s
Birinchi	KKD1200/150	1200	130,150,180	560, 680, 800	680	1088
Ikkinchi	KSD 2200Gr	350	50 – 90	360-610	360	576
Uchinchi	KMD2200T1	130	5 -15	160-220	178	285

Maydalagich ishlab chiqarish unumini kuyidagi formula bo‘yicha aniqlaymiz :

Q_may. = Q_kat. · k_f · k_kr. · k_vl. · k_s , m³/s

Q_kat. – maydalagichning pasportidagi ishlab chiqarish unumi 360 m³/s;

ρ_n - rudaning sochma zichligi;

k_f, k_{kr ,} k_vl, k_s - rudaning qattiqlik, sochma zichlik , yiriklik va namlik bo‘yicha tuzatish koeffitsentlari ;

koeffitsentlarni miqdorini [5] dagi jadval bo‘yicha k_f-1.6 , k_kr-1.05 , k_vl-1, k_s-1.0 deb hisoblaymiz.

Berilgan loyiha bo‘yicha maydalagichning haqiqiy ishlab chiqarish unumini hisoblaymiz:

Q_may. = 360 · 1,00 · 1,05 · 1 · 1 = 378 m³/s

Natija bo‘yicha maydalagichlar sonini aniqlaymiz:

Talab qilingan unumdorlikni rudaning sochma zinchlikka bo‘lish orqali rudaning xajmiy unumdorligi aniqlanadi

V_c =661.76 /1.6 = 413.6

dona(37)
KSD 2200 markali maydalagichdan 1 dona olamiz.

2.1.3 YAnchish sxemasini hisoblash

Ruda

1

Maydalash

2 3

Yanchish

4

Tasniflash I

5 6

7

Cho’ktirish

8 9 18

Konsen. stol I Tasniflash II

13 14

11 12 Yanchish

Konsen. Stol II

Tasniflash III

23

22 15 16

17

Quyiltirish

19 20

Sianlash Texnik jarayonga

qaytariladi

21

Sorbsiyalash

24 25

To’yingan Eritma

qatron

Gravitatsiya sorbsiya usulida boyitishning texnologik sxemasi

Hisoblash tartibi:

Hisoblashlaqulay bo‘lishi uchun yanchish sxemasini xisoblashni gravitatsiya sxemasidan xoli amalga oshiramiz va umumiy sxemadagi raqamlarni o‘zgartirmaymiz.

Qabur qilingan dastlabki ma’lumotlarga asosan yanchish sxemasini hisoblaymiz.

YAnchish sexining ish unumdorligini aniqlaymiz.

Qchas=	Qgod · Kn	,t/s
	343 · 24 · Kv

Bu erda: 343– bir yildagi ish kunlari soni

24 – haftadagi uzluksiz ish kunlari, 3-smena 8 soatdan (3x8=24 chasa)

K_v – dastgoxlardan foydalanish koeffitsienti

K_n – xom ashyolarning xususiyatlari har xilligini hisobga olish koeffitsienti,

Qabul qilamiz: K_v=0,9 K_n=1,0

Qchas=	12500 000· 1	=1687.18t/s
	343 · 24 · 0.9

O‘rta maydalangan ruda omboridagi ruda zaxirasi 2 sutkaga etadi:

V= 48 ∙ 1687.18 / 2,5 = 32393 m³
Dastlabki ma’lumotlarni qabul qilamiz

Hisoblash tartibi:

γ_n= Q_n : Q₁

Q₁=Q₂= 1687.18t t/s

Q₄=Q₃

γ₃ = γ₄

γ₂ = γ₁ = 100 %

γ₆ = 40 %;

γ₈ = 20 %;

γ₁₁ = 2 %;

γ₁₃ = 50 %;

γ₁₆ = 30 %;

γ₂₂= 0.05 %;

γ₄ = γ₂₊ γ₆ = 100 +40=140%

γ₅ = γ₄ – γ₆ =140-40=100%

γ₂₃ = γ₁₁– γ₂₂ =2-0.05=1,95%

γ₁₂ = γ₈ – γ₁₁ =20-2=18%

γ₁₀ = γ₁₂+ γ₂₂ =18+0,05=18,05

γ₇ = γ₂₃ +γ₁₂₊ γ₅ =1,95+18+100=119,95%

γ₉ = γ₇ – γ₈ =119,95-20=99,95

γ₁₈ = γ₉ + γ₁₆ =99,95+30=129,95%

γ₁₄ = γ₁₈ – γ₁₃ =129,95-50=79,95%

γ₁₅ = γ₁₄ – γ₁₆ =79,95-30=49,95%

γ₁₇ =γ₁₅ +γ₁₃ =49,95+50=99,95

γ₂₀=0

γ₁₇=γ₁₉ = γ₂₁

Q₄ = Q₁ *γ₄ /100 = 1687.18t *140/100 =2362.052 t/s;

γ₁₄ = 79,95%; γ₅ =100% γ₁₃=50%; γ₁₀ = 18,05 % ;

Q₅ = Q₁ * γ₅/100 =1687.18t *100/100=1687.18t /s;

γ_{11 =} 2 %

Q₆ = Q₁ *γ₆ /100 =1687.18t *40/100 =647.872 t/s

Q₇ = Q₁ *γ₇ /100 = 1687.18t *119,95/100 =728,55t/s

Q₈ = Q₁ *γ₈ /100 =1687.18t *20/100 =337.436 t/s

Q₉ = Q₁ *γ₉ /100 = 1687.18t *99,95/100 =1687.18t /s

Q₁₀=Q₁ * γ₁₀ /100=1687.18t *18,05/100=304.53 t/s

Q₁₁=Q₁ * γ₁₁ /100=1687.18t *2/100=33.74/s

Q₁₂=Q₁ * γ₁₂ /100=1687.18t *18/100=303.69 t/s

Q₁₃ = Q₁ *γ₁₃ /100 = 1687.18t *50/100 =843.59/s

Q₁₄ = Q₁ *γ₁₄ /100 = 1687.18t *79,95/100=1348.9 t/s

Q₁₅ = Q₁ *γ₁₅ /100 = 1687.18t *49,95/100 =842.74 t/s;

Q₁₆ = Q₁ *γ₁₆/100 = 1687.18t *30/100= 506.154 t/s

Q₁₇ = Q₁ *γ₁₇ /100 = 1687.18t *99,95/100=1687.18t t/s

Q₁₈ = Q₁ *γ₁₈ /100 = 1687.18t *129,95/100=2192.49/s

Q₁₇= Q₁₉ = Q₂₁

Q₂₀=0

Q₂₂ = Q₁ *γ₂₂ /100 = 1687.18t *0,05/100=0.843 t/s

Q₂₃ = Q₁ *γ₂₃ /100 = 1687.18t *1,95/100=32.9/s

2.1.4. Gravitatsiya sxemasini hisoblash

 = ₁ = 2.5 g/t = 0,00025% - dastlabki rudadagi oltinning miqdori,

₁ = ₂ =100%

₁₆ = 30% - konsentrator konsentratining chiqish miqdori,

₁₆ = 25 % - konsentrator konsentratiga oltinning ajralish darajasi,

16 =	16 x 	=	25x 0,00025	= 0.000208%
	16		30

₂₀ = 0

₁₃ =60%

13 =	13 x 	=	60x 0,00025	= 0.0003%
	13		50

₈=60%

8 =	8 x 	=	60x 0,00025	= 0.00075%
	8		20

₁₁ =30%

11 =	11 x 	=	30x 0,00025	= 0.00375%
	11		2

₅ =100%

5 =	5 x 	=	100x 0,00025	= 0.00025%
	5		100

₂₂=25%

22=	22 x 	=	25x 0,00025	= 0.125%
	22		0,05

₂₃ = ₁₁ - ₂₂ =30-25=5%

23 =	23 x 	=	5 x 0,00025	= 0,000641%
	23		1,95

₁₂ = ₈ - ₁₁ =60-30=30%

12=	12 x 	=	30 x 0,00025	= 0,000416%
	12		18

₁₀ = ₁₂ + ₂₃ =30+5=35%

10 =	10 x 	=	35x 0,00025	= 0,0004847%
	10		18,05

₆ = ₄ - ₅ =140-100=40%

6 =	6 x 	=	40x 0,00025	= 0.00025%
	6		40

₇ = ₁₀ + ₅ =35+100=135%

7 =	7 x 	=	135x 0,00025	= 0.0002813%
	7		119,95

₉ = ₇ - ₈ =135-60=75%

9 =	9 x 	=	75x 0,00025	= 0.0001875%
	9		99,95

₁₈ = ₉ + ₁₆ =75+25=100%

18 =	18 x 	=	100x 0,00025	= 0.000192%
	18		129,95

₁₄ = ₁₈ - ₁₃ =100-60=40%

14=	14 x 	=	40x 0,00025	= 0.0000136%
	14		79,95

₁₅ = ₁₄ - ₁₆ =40-25=15%

15 =	15 x 	=	15x 0,00025	= 0.000075%
	15		49,95

₁₇ = ₁₅ +₁₃ =15+60=75%

17=	17 x 	=	75x 0,00025	= 0,000187%
	17		99,95

₁₉ =₁₇
₁₉ = ₁₇

₂₁ = ₁₉ = 75%

₂₀ =0

Boyitishning sifat – miqdor sxemasi balansini tuzish uchun xar bir maxsulot uchun quyidagi formuladan Q va P ni xicoblaymiz.

Qn =	Q1 ∙ n	; P = P1 x n,	gde P1 = Q1 x 1 =	Q1 ∙ 
	100			100

P1 =	1687.18∙ 0,00025	= 0,004217 t/c
	100

P ₂ = P ₁ ∙ ₂ / 100 = 4217 ∙ 100 / 100 =4217 gr/s

P ₃ = P ₁ ∙ ₃ / 100 = 4217 ∙ 140 / 100 =5903 gr/s

P ₄ = P ₁ ∙ ₄ / 100 =4217∙ 140 / 100 =5903 gr/s

P ₅ = P ₁ ∙ ₅ / 100 = 4217 ∙ 100 / 100 =4217gr/s

P ₆ = P ₁ ∙ ₆ / 100 = 4217 ∙ 40 / 100 =1686.8 gr/s

P ₇ = P ₁ ∙ ₇/ 100 = 4217 ∙ 135 / 100 =5692.95 gr/s

P ₈ = P ₁ ∙ ₈ / 100 = 4217 ∙ 60 / 100 =2530.2 gr/s

P ₉= P ₁ ∙ ₉/ 100 = 4217 ∙ 75/ 100 =3162.75 gr/s

P ₁₀ = P ₁ ∙ ₁₀ / 100 = 4217 ∙ 35 / 100 =1475.95 gr/s

P ₁₁ = P ₁ ∙ ₁₁ / 100 =4217∙ 30 / 100 =1265.1 gr/s

P ₁₂ = P ₁ ∙ ₁₂ / 100 = 4217 ∙ 30/ 100 =1265.1 gr/s

P ₁₃ = P ₁ ∙ ₁₃ / 100 = 4217 ∙ 60/ 100 =2530.2 gr/s

P ₁₄ = P ₁ ∙ ₁₄ / 100 = 4217 ∙ 40 / 100 =1686.8 gr/s

P ₁₅ = P ₁ ∙ ₁₅ : 100 = 4217 ∙ 129.95 / 100 = 5479.99 gr/s

P ₁₆ = P ₁ ∙ ₁₆ : 100 = 4217 ∙ 25 / 100 = 1054.25 gr/s

P ₁₇ = P ₁ ∙ ₁₇ : 100 = 4217 ∙ 75 / 100 = 3162.75 gr/s

P₁₈ = P₁ ∙ e₁₈ : 100 = 4217∙ 100 / 100 =4217 gr/s

P₁₉ = P₁ ∙ e₁₉ : 100 = 4217 ∙ 75 / 100 =3162.75 gr/s

P ₂₀ = 0

P ₂₁ = P ₁₉ =R₁₇

P ₂₁ = P ₁ ∙ e₂₁ : 100 = 4217 ∙75 /100 = 3162.75 gr/s

P ₂₂ = P ₁ ∙ e₂₂ : 100 = 4217∙ 25 / 100 =1054.25 gr/s

P ₂₃ = P ₁ ∙ e₂₃ : 100 = 4217 ∙ 5 / 100 = 210.85 gr/s

Olingan natijalarni sifat miqdor sxemasining jadvaliga kiritamiz. Tasniflash jarayonidagi qumning tarkibigi oltinning yig‘ilish xususiyatini xisobga olgan xolda qumning tarkibidagi Au miqdorini 1.5 gr / t deb qabul qilami

3-jadval