图3-2 150万吨选煤厂工艺流程图第4章 工艺流程的计算4.1介质流程的计算选煤厂旋流器小时入料量为7275.227(/)Q t h =，要求分选比重31.40(/)p g cm δ=原煤水分 5.0%Qn W =加重剂中磁性物比重35.0(/)f g cm δ=4．1．1给料中煤泥水的计算:取煤泥比重31.5(/)c g cm δ= 100%cn r =9.96 4.007.0020.96%m r r r r =++++=浮原次＝给料中煤泥量：759.545(/)n m G Q r t h =⋅= 非磁性物的含量：59.545(/)c n G G t h == 原煤泥水量为：35275.22714.486(/)100100Qnn Qn W W Q m h W =⋅⨯=-入＝－5煤泥水的体积：59.54514.48654.1831.5nn n cG V W δ=+=+= 3(/)t m 煤泥水的密度：59.54514.4861.36654.183n n n n G W V ++∆===3(/)t m 煤泥水单位体积的固体含量：59.545g 1.09954.183n n n G V === 3(/)t m 4．1．2补加浓介质的性质的计算:设浓介质比重 2.0X ∆= 5.0f δ= 1.5c δ=浓介质中非磁性物的含量 5%cx r = 磁性物含量 95%fx r = 浓介质悬浮液的密度：5 1.54.47855% 1.595%f c X f cX c fX r r δδδδδ⋅⨯===⋅+⋅⨯+⨯ 3(/)t m补加介质中干介质质量1 2.010.28751 4.4781X X λδ∆--===-- 浓介质悬浮液的固体含量：0.2875 4.478 1.288X X g λδ=⋅=⨯=3(/)t m浓介质悬浮液的煤泥含量：31.2885%0.064(/)cX X cX g g r t m =⋅=⨯= 浓介质悬浮液的磁性物含量：31.2880.064 1.224(/)fX X cX g g g t m =-=-= 单位体积含水量：32.0 1.2880.712(/)X X X g t m ω=∆-=-=4．1．3确定工作介质的性质：要求分选比重：31.40(/)p g cm δ=取工作介质悬浮液的比重： 37 1.400.12 1.28(/)p t m δ∆=-∆=-= 查《旧手册》 Δ值在0.12～0.18之间 则工作介质中非磁性物含量最高极限值：77max 77()()100%()()n cn X x n cx n c n X X n n G r g V r r G g V ⋅⋅∆-∆+⋅⋅⋅∆-∆=⨯∆-∆+∆-∆59.545100%(2.0 1.28) 1.28854.1835%(1.28 1.366)100%59.545(2.0 1.28) 1.28854.183(1.28 1.366)115.47%⨯⨯-+⨯⨯⨯-=⨯⨯-+⨯⨯-= 取工作介质中非磁性物含量：7max 40%c c r r =< 则工作介质中磁性物含量：7710060%f c r r =-=7775 1.52.586540% 1.560%f c f c c f r r δδδδδ⋅⨯===⋅+⋅⨯+⨯ 3(/)t m介质中干介质的质量：7777(1)(1.281) 2.5860.4571 2.5861g δδ∆-⋅-⨯===--3(/)t m其中非磁性物含量：7770.45740%0.183c c g g r =⋅=⨯= 3(/)t m 磁性物含量：7770.4570.1830.274f c g g g =-=-= 3(/)t m 单位体积含水量：777 1.280.4570.823g ω=∆-=-= 3(/)t m4．1．4分选作业计算：确定循环介质量：选3GDMC1000/700 三产品重介旋流器，得：1.25275.2271.56220K Q n q ⋅⨯===7(台) 单台旋流器循环量为：V ＝6003(/)m h 总循环量：81200V = 3(/)m h 工作介质总量：78120054.1831254.183n V V V =+=+= 3(/)m h7770.4571254.183573.162G g V =⋅=⨯=(/)t h 777573.16240%229.265c c G G r =⋅=⨯=(/)t h 777573.162229.265343.897f c G G G =-=-=(/)t h7770.8231254.1831032.193W V ω=⋅=⨯=3(/)m h求循环介质其它参数：87573.16259.545513.617n G G G =-=-=(/)t h 87229.26559.545169.720c c n G G G =-=-=(/)t h 888513.617169.720343.897f c G G G =-=-=(/)t h871032.19314.4861017.707n W W W =-=-=3(/)m h 8888513.6171017.7971.2761200G W V ++∆===(/)Kg L 888169.720100%100%33.044%513.617c c G r G =⨯=⨯= 8810010033.04466.956%f c r r =-=-=旋流器一段分选作业计算：设一段旋流器溢流中的悬浮液密度比工作介质低0.1，底流比工作介质高0.4，即：970.1 1.280.1 1.18∆=∆-=-=(/)Kg L70.4 1.280.4 1.68∆=∆+=+=底(/)Kg L7979 1.28 1.181254.183250.8371.68 1.18V V ∆-∆-=⋅=⨯=∆-∆-底底3(/)m h97V 1254.183250.8371003.346V V =--=底＝3(/)m h设底流中磁性物含量比工作介质高10％：710%60%10%70%f f r r =+=+=底1001007030%c f r r =-=-=底底5 1.52.941530% 1.570%f c f f c c f r r δδδδδ⋅⨯===⋅+⋅⨯+⨯底底底3(/)t m(1)(1.681) 2.941 1.0301 2.9411g δδ∆-⋅-⨯===--底底底底3(/)t m1.03030%0.309c c g g r =⋅=⨯=底底底3(/)t m 1.0300.3090.721f c g g g =-=-=底底底 3(/)t m1.68 1.0300.650g ω=∆-=-=底底底 3(/)t m1.030250.837258.362G g V =⋅=⨯=底底底(/)t h 258.36230%77.509c c G G r =⋅=⨯=底底底(/)t h 258.36277.509180.853f c G G G =-=-=底底底(/)t h 0.6650250.837163.044W V ω=⋅=⨯=底底底3(/)m h 9573.162258.362314.800G G G =-=-=7底(/)t h 9229.26577.509151.756c G G G =-=-=c7c 底(/)t h 9314.800151.756163.044f G G G =-=-=9c9(/)t h1032.193163.044869.149W W W =-=-=97底3(/)m h999314.8000.3141003.346G g V ===3(/)t m 999151.7560.1511003.346c c G g V ===3(/)t m 90.3140.1510.163f g g g =-=-=9c93(/)t m999869.1490.8661003.346W V ω===3(/)t m 9990.3140.866 1.180g ω∆=+=+=3(/)t m 9∆与假定值相同，说明以上计算无误。

999151.756100%100%48.21%314.800c c G r G =⨯=⨯= 9910010048.2151.79%f c r r =-=-=旋流器二段分选作业计算：设二段旋流器溢流中的悬浮液密度比工作介质低0.1，底流比工作介质高0.4，即：100.1 1.680.1 1.58∆=∆-=-=底(/)Kg L0.4 1.680.4 2.08∆=∆+=+=11底(/)Kg L1010 1.68 1.58250.83750.1672.08 1.58V V ∆-∆-=⋅=⨯=∆-∆-底11底113(/)m h10V 250.83750.167200.670V V =--=11底＝3(/)m h设底流中磁性物含量比工作介质高10％：1110%70%10%80%f f r r =+=+=底11111001008020%c f r r =-=-=1111115 1.53.409520% 1.580%f c f c c f r r δδδδδ⋅⨯===⋅+⋅⨯+⨯3(/)t m(1)(2.081) 3.4091.5281 3.4091g δδ∆-⋅-⨯===--111111113(/)t m1111 1.52820%0.306c c g g r =⋅=⨯=113(/)t m 1111 1.5280.306 1.222f c g g g =-=-=11 3(/)t m2.08 1.5280.552g ω=∆-=-=111111 3(/)t m1.52850.16776.665G g V =⋅=⨯=111111(/)t h 111176.66520%15.333c c G G r =⋅=⨯=11(/)t h111176.66515.33361.332f c G G G =-=-=11(/)t h0.55250.16727.692W V ω=⋅=⨯=1111113(/)m h 10258.36276.665181.697G G G =-==11底－(/)t h 1077.50915.33362.176c G G G =-==c11c 底－(/)t h 10181.69762.176119.521f G G G =-=-=10c10(/)t h 163.04427.692135.352W W W =-=-=1011底3(/)m h101010181.6970.905200.670G g V ===3(/)t m 10101062.1760.310200.670c c G g V ===3(/)t m 100.9050.3100.595f g g g =-=-=10c103(/)t m101010135.3520.675200.670W V ω===3(/)t m 1010100.9050.675 1.58g ω∆=+=+=3(/)t m 10∆与假定值相同，说明以上计算无误。