Magnetically Driven Isentropic Compression of 45 Steel
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摘要: 在已有的众多准等熵加载技术中，磁驱动准等熵加载技术具有准等熵程度高、压力范围大、实验材料种类多、效费比高等特点。利用中物院流体物理研究所建成的磁驱动准等熵压缩和高速飞片实验装置CQ-1.5(最高加载压力为50 GPa), 成功开展了45钢的准等熵压缩实验，对装置的主要参数进行了介绍；利用激光干涉测试系统DPS获得了45钢飞片的自由面速度历史，通过反积分处理给出了材料准等熵压缩的p-V关系。通过分析实验数据，获得了45钢3种形式的等熵方程的参数。实验获得的最高等熵压力为47.5 GPa。
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关键词:
- 准等熵压缩 /
- 磁驱动 /
- 流场反演方法 /
- Lagrange分析 /
- 等熵方程
Abstract: The technique of magnetically driven isentropic compression has many advantages over previous experimental techniques of quasi-isentropic compression, such as the smoothly increasing pressure loading to a high magnitude without any initial shocks, the ability to investigate several samples under identical loading conditions in one shot, high benefit-cost ratio, etc.The main parameters of CQ-1.5 rebuilt by Institute of Fluid Physics, CAEP, are detailed.This apparatus aims at isentropic compression and driving high-speed flyers, its designed maximum pressure is 50 GPa.The samples of 45 steel are isentropically compressed by CQ-1.5 to a pressure as high as 47.5 GPa without any appearance of shock, a Doppler pins system (DPS) is used to record the velocity history of flyers.We use a backward integration method to treat the experimental data, and obtain thep-V relation of 45 steel under quasi-isentropic compression.We also ascertain the parameters of 3 forms of isentropic equation of 45 steel.The maximum isentropic pressure in this work is 47.5 GPa. -
图 1 小型磁驱动准等熵压缩和驱动飞片实验装置CQ-1.5结构示意图
Figure 1. Schematic of small magnetically driven quasi-isentropic compression and flyer- driving experimental device CQ-1.5
图 2 DPS记录的样品自由面速度历史
Figure 2. DPS records of free surface velocity of 45 steel samples
图 4 实验得到的准等熵压缩线与Hugoniot线比较
Figure 4. Comparison of the experimental quasi-isentropic compression curve with Hugoniot curve
表 1 实验得到的45钢准等熵压缩方程参数
Table 1. Parameters of quasi-isentropic compression equation of 45 steel determined by experiment
Isentropic
equationEquation form pmax/
(GPa)Parameters Birch $\begin{aligned} p(V)=& \frac{3}{2} B_{0}\left[\left(\frac{V_{0}}{V}\right)^{\frac{7}{3}}-\left(\frac{V_{0}}{V}\right)^{\frac{5}{3}}\right] \times \\ &\left\{1-\xi\left[\left(\frac{V_{0}}{V}\right)^{\frac{2}{3}}-1\right]\right\} \end{aligned}$ 41 ξ=0.5, B0=65.4 GPa Hugoniot $p=p_{0}+\frac{\rho_{0} c_{0}^{2}\left(1-\frac{V}{V_{0}}\right)}{\left[1-\lambda\left(1-\frac{V}{V_{0}}\right)\right]^{2}}$ 41 c0=3 104 m/s, λ=2.495 Polynomial $p=\sum\limits_{j=1}^{m} B_{j}\left(\frac{V_{0}}{V}-1\right)^{j}, m=1,2,3, \cdots$ 41 B1=37 GPa, B2=110 GPa, B3=5 100 GPa 
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