欢迎登录材料期刊网

材料期刊网

高级检索

运用计算流体动力学(CFD)的方法在 FLUENT 平台上建立雾化气体淬火的三维非稳态模型,对介质不同速度工况下淬火过程中试件温度场、流体速度场、流体温度场和流体压力场进行了流固热耦合数值模拟,并将模拟结果与边界条件实验结果进行了对比分析。结果表明,同等工况下,介质进口速度越大,流体的速度也越大;同截面上200 m/s 速度工况下试件上下表面两区域的压力差值比100 m/s 速度工况下大;试件放在淬火区淬火80 s 后,200 m/s 速度工况下试件的最高温度为299.5℃,100 m/s 速度工况下试件的最高温度为336℃,由此得出淬火介质进口速度越大,淬火试件的温度下降越快,试件冷却速度越快,冷却效果越好。

By means of computational fluid dynamics (CFD)method,a three-dimensional unsteady state model of atomized gas quenching was established in FLUENT platform.Fluid-solid heat coupled simulations on specimen temperature field,fluid velocity field,fluid temperature field and fluid pressure field of quenching process were carried out at different flow rates,and the simulation results were compared with the experimental results.The results showed that the velocity of fluid increased with the rising quenchant inlet velocity under the same working condition. In the same section,the pressure difference in the two areas on the top and bottom surface of specimen under the speed condition of 200 m/s was larger than that of 100 m/s.After quenched for 80 seconds,the highest temperature of spe-cimen under the speed condition of 200 m/s was 299.5 ℃,the highest temperature of specimen under the speed condi-tion of 100 m/s was 336 ℃.It was concluded that greater quenchant inlet velocity results in lower quenching specimen temperature,faster cooling rate,and better cooling effect.

参考文献

[1] Florent Chaffotte;Linda L(e)fevre;Didier Domergue;Aymeric Goldsteinas;Xavier Doussot;Qingfei Zhang.Optimising Gas Quenching Technology through Modelling of Heat Transfer[J].材料热处理学报,2004(05):758-763.
[2] 程赫明;谢建斌;李建云.9SiCr合金刃具钢在不同介质淬火后性能比较[J].航空材料学报,2004(4):14-17.
[3] Gunnar Heskestad.Scaling the interaction of water sprays and flames[J].Fire safety journal: An international journal devoted to research on fire safety science and engineering,20026(6):535-548.
[4] 潘健生;张伟民;田东;顾剑锋;胡明娟.热处理数学模型与计算机模拟[J].中国工程科学,2003(5):47-54.
上一张 下一张
上一张 下一张
计量
  • 下载量()
  • 访问量()
文章评分
  • 您的评分:
  • 1
    0%
  • 2
    0%
  • 3
    0%
  • 4
    0%
  • 5
    0%