提升机选型计算

1、绞车提升能力计算已知： =25º L=960M f1=0.015 f2=0.2 n=7 每米钢丝绳mP=2.129 ，车皮重600kg, 煤重850kg, 矸石重1600kg（1350KG）已知：电动机型号JR127-6型，电机额定功率Ne 185KW,滚筒直径2m,二级传动系数y=0.85,过负荷系数1.9，提升机最大提升速度V=3.14*2（滚筒直径）*979(转速)÷（60*30传动比）=3.42m/s。一、绳端负荷：求 Qj（提6个煤车）Qj=n .g（Sin25º+f1COS25º）+L.mP .g (Sin25º+f2 COS25&

2、#186;)=6*9.8(850+600)(0.423+0.015*0.906)+960*2.129*9.8(0.423+0.2*0.906)=37190 + 12093=49283N 提4个矸石车时：Qj = n .g（Sin25º+f1COS25º）+L.mP .g (Sin25º+f2 COS25º) =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：Qj = n .g（Sin25º

3、+f1COS25º）+L.mP .g (Sin25º+f2 COS25º) =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 6.5 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

4、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 ÷49.71KN = 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：378.5KN ÷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P=185KW, Ie=350A

5、, Y=0.925 ，cos=0.925, =1.9, U2e=254V, I2e=462A, GD²=49kg/m²，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 =(Gd²)d.t²、Dg² = 49 *30²/2²=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 + G

6、j =5317.6 +200 +11025 +8200=24742.6变位质量M = G/ g =24742.6 / 9.81 =2522(kg.s²/m)三、提升运动学计算取a0 = 0.3m /s² ，a = 0.5m /s² ，v0 = 1m /s²，1、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

7、 /0.5 =4.84 (s)L1 =1/2 +( vmax + v0 ) =1/2*4.84*(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

8、 = 240.6 (s)一次提升循环时间Tq = t0 + t01 +t1 + t2 +t3 + 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&

9、#176;）+2522*0.3 =4174 + 1234 +756 =6164初加速终了时F0´ = F0 - Pk . L0 .sina =6164 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

10、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´ = F2 - Pk . L2 . sina = 5374 2.129 * 823 * sin25 =4633减速开始F3 = F2´ - m .a1 =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

11、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 =3836末等速终了F5´ = F5 - Pk . L5 . sina =3836 2.129 * 1.65 * sin25 = 3834五、 电动机容量校验F²x = 1/2(F0² + F01²)t0 + 1/2(F01.1

12、8; + F01.1.1²) t01+1/2(F1²+ F1.01²) t1 + 1/3( F2.01 + F2. F2.01 + F2.01²) + （F3² + F3.01²）.t3 +1/2( F4² + F4.01²).t4 + 1/2( F5 + F5.01²) =125342481 + 669317031 + 213393785 +6034433021 +54869695 +708150198 +48533845 = 7854040057Td = 1/2(t0 + t01 +t1 +t3 +t

13、4 +t5 )+t2 +/3 =1/2(3.3 +23 +4.84 +4.84 +33.35+3.3) +24.06+25/3 =285(s)Fdx = F²x / 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、井底阶段提升开始时：F0=kn(Q2+Qk)(sinx +f1cosx)+Lt.Pk(sina+f2cosa)+ m.ao =

14、1.1*4*(600+1600)(sin25+0.015*cos25°)+960*2.129（sin25° +0.2*cos25°）+2522*0.3 =4222 + 1234 +756 =6212初加速终了时F0´ = F0 - Pk . L0 .sina =6212 2.129 *1.65 * sin25°=6210低等速开始F01 = F0´ - m.ao = 6210 2522 * 0.3 =5453低等速终了时F01´ = F01 - Pk . L01 . sina =5453 2.129 * 23.3 . sin

15、25° =54322、井筒运行主加速开始时F1 = F01´ + m.a1 =5432 + 2522 * 0.5 =6693主加速终了F1´= F1 - Pk . L1 . sina =6693 -2.129 * 11 * sin25 =6683等速开始时F2 = F1´ - m .a1 = 6683 2522 * 0.5 =5422等速终了时F2´ = F2 - Pk . L2 . sina = 5422 2.129 * 823 * sin25 =4681减速开始F3 = F2´ - m .a1 =4681 2522 * 0.5 =

16、3420减速终了F3´ = F3- Pk . L3 . sina =3420 2.129 * 11 * sin25 =34103、在栈桥运行阶段末等速开始F4 = F3´ + m .a1 = 3410 + 1261=4671末等速终了F4´ = F4 - Pk . L4 . sina = 4671- 2.129 * 33.35 * sin25 =4641末减速开始F5 = F4´ - m .a0 =4641 2522 * 0.3 =3885末等速终了F5´ = F5 - Pk . L5 . sina =3885 2.129 * 1.65 * s

17、in25 = 3883七、 电动机容量校验F²x = 1/2(F0² + F01²)t0 + 1/2(F01.1² + F01.1.1²) t01+1/2(F1²+ F1.01²) t1 + 1/3( F2.01 + F2. F2.01 + F2.01²) + （F3² + F3.01²）.t3 +1/2( F4² + F4.01²).t4 + 1/2( F5 + F5.01²) =127302522 + 681281079 + 216490146 +6150556

18、205 +56445290 +722980009 +49782008 = 8004837259Td = 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 = F²x / Td =5300 kg等效功率N dx = F dx . Vm / 102 * y = 5300 *3.4 / （102 *0.85） =208KW电动机验算1 = Fmax /fe = 6645 / 1.9 * 102y .N / Vmax =0.79八、 提升动力学计算

19、（按5个矸石车）1、井底阶段提升开始时：F0=kn(Q2+Qk)(sinx +f1cosx)+Lt.Pk(sina+f2cosa)+ m.ao =1.1*5*(600+1600)(sin25+0.015*cos25°)+960*2.129（sin25° +0.2*cos25°）+2522*0.3 =5278 + 1234 +756 =7268初加速终了时F0´ = F0 - Pk . L0 .sina =7268 2.129 *1.65 * sin25°=7266低等速开始F01 = F0´ - m.ao = 7266 2522 *

20、0.3 =6509低等速终了时F01´ = F01 - Pk . L01 . sina =6509 2.129 * 23.3 . sin25° =64882、井筒运行主加速开始时F1 = F01´ + m.a1 =6488+ 2522 * 0.5 =7749主加速终了F1´= F1 - Pk . L1 . sina =7749-2.129 * 11 * sin25 =7740等速开始时F2 = F1´ - m .a1 = 7740 2522 * 0.5 =6479等速终了时F2´ = F2 - Pk . L2 . sina = 647

21、9 2.129 * 823 * sin25 =5738减速开始F3 = F2´ - m .a1 =5738 2522 * 0.5 =4477减速终了F3´ = F3- Pk . L3 . sina =4477 2.129 * 11 * sin25 =44673、在栈桥运行阶段末等速开始F4 = F3´ + m .a1 = 4467 + 1261=5728末等速终了F4´ = F4 - Pk . L4 . sina = 5728- 2.129 * 33.35 * sin25 =5700末减速开始F5 = F4´ - m .a0 =5700 252

22、2 * 0.3 =4943末等速终了F5´ = F5 - Pk . L5 . sina =4943 2.129 * 1.65 * sin25 = 4941九、 电动机容量校验F²x = 1/2(F0² + F01²)t0 + 1/2(F01.1² + F01.1.1²) t01+1/2(F1²+ F1.01²) t1 + 1/3( F2.01 + F2. F2.01 + F2.01²) + （F3² + F3.01²）.t3 +1/2( F4² + F4.01²).t4 + 1/2( F5 + F5.01²) =174270657 + 971304087 + 290290134 +8988702677 +96794235 +1088877233 +80597104 = 1.169*1010Td = 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 = F²x / Td =6404kg等效功率