交流接触器设计毕业设计完整修改版

l 3= l 1+ l i=55.5+29=84.5mm

总磁压降为:

∑Hl= H1l1+H2l2+H3l 3 (4-42)

=786×(36.5+29+84.5)

=95106

Im=∑H l /N=95106×10-3/2298=0.0414A (4)导磁体损耗规划电流IP

V1=axbx(l 1+a)=19×19×(55.5+19)=26894.5 V2=abl 1+ aebe(l i+ l 1+a)

=19×19×29+19×19×(29+55.5+19) =10469+37363.5 =47832.5

衔铁损耗计算公式:pC1=3.8 W/kg (4-44) PFe1= PC1γV1

=3.8×7800×2.69×10-5=0.797w 静铁心损耗:

PFe2= Pc2γV2=3.8×7800×4.78×10-5=1.42w 则: Ip=∑PFe /UN=(0.797+1.42)/220=10.08×10-3A (5)分磁环规划电流Id

Ed=4.44fNkФδmγ2 =4.44×50×1×4.3×10-4×0.8 =0.0764 v

Id=2Ed2 /(UN×rk) =2×0.07642/(220×7.355×10-4)

=0.072 A (6)计算线圈电流I

磁化电流:

Ir=Iδ+If+Im =0.031+0.0627+0.0414

=0.1351 损耗电流:

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4-43)

4-45)

4-46)

4-47)

4-48) ((((( Ia=Ip+Id=0.01+0.072=0.082A (4-49)

2 I=Ir?Ia=0.1351?0.0822=0.158A

22 Ψ1=arctg

Ia0.082= arctg=31.260 Ir0.1351 U2=E2+I2R2+2E(IR)sinΨ1

2202=E2+0.1582×94.772+2×(0.158×94.77) ×0.519E

得:E2+15.54E-48624=0 (4-50)

?15.54?15.542?4?(?48624) E==212.88V

2E

4.44fN?212.88 ==3.1×10-4

4.44?50?2298?1.35 故Фδm1=

??m???m14.3?10?4?3.1?10?4 ε===2.79%<3% ?4??m4.3?10 (要求计算前后两次磁通近似值之差与前次磁通近似值之比,应小于给定误差

值3%)故满足条件 线圈发热功率P

P=I2R1100℃=0.1582×94.77=2.366w (4-51)

线圈温升计算

线圈包围部分铁心损耗 PFe3

PC3=PC2=3.8w/kg

PFe3=a×b×kc×h×ρFePC3 (4-52)

=19×19×10-6×0.93×0.028×7800×3.8 =0.279w (2)线圈外表面散热面积S:

S=(2A1+2B1+2piΔ) ×h (4-53) =(2×21.4+2×25.4+2×3.14×12 ) ×28 =4.73×10-3m2 (3)线圈稳定温升计算:ηw

(长期工作制线圈的功率过载系数:kp2=1)

ηw=(P+PFe3)/(kp2×KT×S) (4-54)

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=(2.366+0.279)/(1×10.99×4.73×10-3

) =50.880C<700C 温升合格

衔铁在设计点的气隙磁导计算: (1)工作气隙磁导∧δ

(因δ0/a<0.2,/b<0.2,故用解析法计算) 一个磁极的磁导: ∧?δ=0abKc? 0-6-6 =

1.257?10?19?19?10?0.933?10-3 =1.401×10-7H (2)非工作气隙磁导∧f

(因δ0/a<0.2,/b<0.2,故用解析法计算)

∧f=μ0aebeKcf =1.257?10-6?19?19?10-6?0.930.2?10-3

=2.11×10-6H

(3)铁心对铁轭单位长度漏磁导λ: 将方形截面积等效为圆形半径r 由π×r2=ab

r=ab/?=19?19/3.14=10.72mm λ=

2??0ln(k?k2

?1) k=y2?2r2l2?2r236.52?2?10.2r2=7222r2=2?10.722=4.8 y为铁心与铁轭的中心距=55.5-19=36.5mm λ=

2?3.14?12.57?10?7ln(4.8?4.82 ?1) =78.94?10?72.25

=35.08×10-7H/m

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4-55)4-56)4-57)4-58)( ( ( (

(4)计算等效漏磁导:

11 ∧`ld=λLi=×35.08×10-7×29×10-3=0.339×10-7H (4-59)

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计算线圈感抗XL:

?N2(????`ld)??f XL= (4-60) ` = = =266.5

计算线圈电流I: I=

=

=0.66A

计算反电动势E:

E= = =176.23V 计算工作气隙磁通:

=

=2.564.14 主要判断

平均吸力判断:

????ld??f100?3.14?22982?(1.401?0.339)?21.1?10?14(1.401?0.339?21.1)?10?7

6087.8?10?722.84?10?7 Ω UR22 (4-61)

1100C?XL0.85?22094.772?266.52

U2?(IR1100C)2 (4-62) (0.85?220)2?(0.66?94.77)2 ФEδm=

4.44fN? (4-63)

176.234.44?50?2298?1.35

×10-4

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