1、=4*(1600+600)*9.8*(0.423+0.015*0.906)+960*2.129*9.8*(0.423 + 0.2 * 0.906)=37617 + 12093=49710 N 提5个矸石车时:=5*(1600+600)*9.8*(0.423+0.015*0.906)+960*2.129*9.8*(0.423 + 0.2 * 0.906)=47022 + 12093=59115 N 钢丝绳安全系数校验:1、 提6个煤车时,查表得出6*7FC 24.5mm,公称强度1700Mpa钢丝绳破断拉力总和为378.5KN,所以钢丝绳安全系数:378.5KN 49.28KN = 7.68 6
2、.5 符合煤矿安全规程要求。2、 提4个矸石车时,查表得出6*7FC 24.5mm,公称强度1700Mpa钢丝绳破断拉力总和为378.5KN,所以钢丝绳安全系数:49.71KN = 7.6 6.5 符合煤矿安全规程要求。3、 提5个矸石车时,查表得出6*7FC 24.5mm,公称强度1700Mpa钢丝绳破断拉力总和为378.5KN, ,所以钢丝绳安全系数:59.115KN = 6.4 6.5 ,不符合煤矿安全规程要求。一、 电动机初选(按4个矸石车):Ns =Fc * Vmax / (1000 * Y) = 49710*3.5 /(1000 * 0.85) =204KW选JR127-6型电动机
3、P=185KW, Ie=350A , Y=0.925 ,cos=0.925, =1.9, U2e=254V, I2e=462A, GD=49kg/m,Nd =980r/min, 所以Vmax = D. Nd / 60t =3.14*2*980/60*30=3.42m/s二、提升电动机变位质量 1、电动机 Gd =(Gd)d.t、Dg = 49 *30/2=110252、天轮取Gt = 200KG3、提升机变位质量Gj = 8200KG4、钢丝绳变位质量Pk .Lk = 2.129*960 = 2043kg G = Qj + Gt +Gd + Gj =5317.6 +200 +11025 +82
4、00=24742.6变位质量M = G/ g =24742.6 / 9.81 =2522(kg.s/m)三、提升运动学计算取a0 = 0.3m /s ,a = 0.5m /s ,v0 = 1m /s,1、初加速段t0 = v0 / a0 =1/0.3 =3.3(s)L0 = 1/2 * v0 * t0 =1/2 *1*3.3=1.65t01 = LL1 - L0 = 25-1.65 =23.3t01 = L01 / v0 =23(S)2、主加速度t1 = vmax v0 / a1 = 3.42 1 /0.5 =4.84 (s)L1 =1/2 +( vmax + v0 ) =1/2*4.84*(
5、3.42+1) =11(m)3、减速段t3 = t1 = 4.84(s)L3 = L1 = 11m4、末减速阶段t5 = t0 = 3.3 (s)L5 = L0 = 1.65m5、矿车在栈桥低速阶段L4 = LB - L5 = 30 1.65 = 33.35t4 = L4 / V 0 = 33.3 5(s)6、矿车在井筒等速运行L2 = Lt (LL1 + L13 + L1 + L3) = 900 (25 + 30 + 11 +11)= 823t2= L2 / V max = 823 / 3.42 = 240.6 (s)一次提升循环时间Tq = t0 + t01 +t1 + t2 +t3 +
6、t4 + t5 + t =3.3 + 23 + 4.84 + 240.6 + 4.84 + 33.35 + 3.3 + 25= 338(s)=338/ 60 = 5.63 min四、 提升动力学计算(按6个煤车)1、井底阶段提升开始时:F0=kn(Q2+Qk)(sinx +f1cosx)+Lt.Pk(sina+f2cosa)+ m.ao =1.1*6*(600+850)(sin25+0.015*cos25)+960*2.129(sin25 +0.2*cos25)+2522*0.3 =4174 + 1234 +756 =6164初加速终了时F0 = F0 - Pk . L0 .sina =616
7、4 2.129 *1.65 * sin25=6162低等速开始F01 = F0 - m.ao = 6162 2522 * 0.3 =5405低等速终了时F01 = F01 - Pk . L01 . sina =5405 2.129 * 23.3 . sin25 =53842、井筒运行主加速开始时F1 = F01 + m.a1 =5384 + 2522 * 0.5 =6645主加速终了F1= F1 - Pk . L1 . sina =6645 -2.129 * 11 * sin25 =6635等速开始时F2 = F1 - m .a1 = 6635 2522 * 0.5 =5374等速终了时F2
8、= F2 - Pk . L2 . sina = 5374 2.129 * 823 * sin25 =4633减速开始F3 = F2 =4633 2522 * 0.5 =3372减速终了F3 = F3- Pk . L3 . sina =3372 2.129 * 11 * sin25 =33623、在栈桥运行阶段末等速开始F4 = F3 + m .a1 = 3362 + 1261=4623末等速终了F4 = F4 - Pk . L4 . sina = 4623- 2.129 * 33.35 * sin25 =4593末减速开始F5 = F4 - m .a0 =4593 2522 * 0.3 =38
9、36F5 = F5 - Pk . L5 . sina =3836 2.129 * 1.65 * sin25 = 3834五、 电动机容量校验Fx = 1/2(F0 + F01)t0 + 1/2(F01.1 + F01.1.1) t01+1/2(F1+ F1.01) t1 + 1/3( F2.01 + F2. F2.01 + F2.01) + (F3 + F3.01).t3 +1/2( F4 + F4.01).t4 + 1/2( F5 + F5.01 =125342481 + 669317031 + 213393785 +6034433021 +54869695 +708150198 +4853
10、3845 = 7854040057Td = 1/2(t0 + t01 +t1 +t3 +t4 +t5 )+t2 +/3 =1/2(3.3 +23 +4.84 +4.84 +33.35+3.3) +24.06+25/3 =285(s)Fdx = Fx / Td =5250 kg等效功率N dx = F dx . Vm / 102 * y = 5250 *3.4 / (102 *0.85) =205KW电动机验算1 = Fmax /fe = 6645 / (1.9 * 102y .N) / Vmax =0.79六、 提升动力学计算(按4个矸石车) =1.1*4*(600+1600)(sin25+0
11、.015*cos25 =4222 + 1234 +756 =6212=6212 2.129 *1.65 * sin25=6210 = 6210 2522 * 0.3 =5453 =5453 2.129 * 23.3 . sin25 =5432 =5432 + 2522 * 0.5 =6693 =6693 -2.129 * 11 * sin25 =6683 = 6683 2522 * 0.5 =5422 = 5422 2.129 * 823 * sin25 =4681 =4681 2522 * 0.5 =3420 =3420 2.129 * 11 * sin25 =3410 = 3410 + 1261=4671 = 4671- 2.129 * 33.35 * sin25 =4641 =4641 2522 * 0.3 =3885 =3885 2.129 * 1.65 * sin25 = 3883七、 电动机容量校验) t1 + 1