用200KV,5mA X射线机180kV管电压透照Φ51×3.5mm的管子环焊缝,已知焊缝宽度b=10mm。满足Ug=3Ui。椭圆开口间距δ=5mm时的透照距离I1和透照偏移距离S(设Ui=0.0013V0.8,胶片离管子底面距离3mm)
正确答案:
已知条件,Ui=0.0013V0.8=0.0013*1800.8=0.08而Ug=3Ui=3*0.08=0.24mm
∴L1=dfL2/Ug=3(51+3)/0.24=675mm又S/(b+δ)=L1/L2
∴S=L1(b+δ)/L2=(675/54)*(10+5)=187.5mm
∴L1=dfL2/Ug=3(51+3)/0.24=675mm又S/(b+δ)=L1/L2
∴S=L1(b+δ)/L2=(675/54)*(10+5)=187.5mm
答案解析:有
![](/editor/images/201705/qrcode_for_gh_573e2a458573_258.jpg)
微信扫一扫手机做题