石墨模具及摩擦系数对热压烧结的影响译自飞羽简报2003年07期

1、石墨磨具及摩擦系数对热压烧结的影响（ John Bring 译自 Dr. Fritsch 公司简报，原文附后）对烧结参数而言，重要的是它们是如何获得的。一般用离样品78mm 远的热电偶来测量决定烧结曲线的参数。样品是一个直径为 25mm 的压块。如果刀头要马上生产， 所需的烧结压力通过已知的公式计算求得。 使（刀头） 获得最高烧结密度的温度则作为烧结温度。 但热压烧结之后发现， 刀头的硬度和 密度并不是所采用的烧结参数所预期的那样。 烧结曲线的参数由每次在不同烧结 温度下烧结而成的刀头决定，如下面的例子：胎体： BH-1 热压参数： 350Kg/cm2保温时间： 3min图 1+2：压块和刀头

2、烧结体硬度和密度的比较曲线如上所示，相同的烧结工艺下获得性能不同的刀头和烧结体。问题在于，是 什么造成了性能的差异？如果我们看一下压块和刀头所设定的热压烧结条件时， 即可发现其间有很大的不同。一方面，显然的，是热压的零件数量不一样。就压 块的而言，我们讨论的是一个直径为 25mm，高为 3.54mm，热压面积为 4.909cm2 的块体的热压烧结；对刀头而言，则是 50个尺寸为 403.2 9 的刀头，总的热压 面积是 64cm2。第二个很大不同的是对于压块没有其它的外力作用在其上，而对 于刀头则有，即要有外力使石墨模框保持在一起。 外部的摩擦力影响了烧结过程。 为了补偿摩擦力的影响作用，可以

3、提高烧结温度或是考虑摩擦因素。我们来看图表，它清楚的显示即便是提高烧结温度，刀头也不能完全达到烧 结体那样的性能。 因此更有效的方法是考虑摩擦系数。 在这个过程中， 一个由压 头数量决定的数值要加到所计算的模框压力 (mould pressure)上，计算公式如下： P(bar)=NAP (sp)/A(cyl)+ 1.5N这里， P(bar)=热压过程所设定的模框压力 (mould pressure)N=每框刀头数A= 单个刀头的压制面积 (cm2)2P(sp)= 单位压力 (kg/cm2)A (cyl) =液压缸截面积 (cm2)再看如下考虑了摩擦系数后的刀头烧结数据表：当进行胎体的热压烧结

4、时，如果在烧结刀头时考虑了摩擦系数，那么采用与 压块同样的或稍高于压块的烧结温度（约 40），刀头就能获得与压块同样的硬 度和密度。Effect of the graphite mould/friction factorFor the sinter data we pass on, it is important to know how they were obtained. When determining sinter curves, the values are measured on samples where temperature is measured by a thermoco

5、uple approx. 7 8 mm from the sample. The sample is a compressed body with a diameter of 25 mm.If segments are now to be produced, the pressure required is calculated in accordance with the nown formula for sinter pressure. The temperature with the highest density as per the sinter curve is selected

6、as sinter temperature. After hot pressing, it is found that the hardness and density achieved do not correspond to the values which would be expected. If a sinter data curve is produced whose values are formed from segments each produced at a different temperature, the following pattern results, for

7、 example:bond: BH-1specific pressure = 350 kg/cm2 retention time = 3 minutesAs can be seen, the same values are not achieved with segments and compressed bodies. The question is, how does this difference arise? If we look at both sets of hot-pressing conditions (segments v compressed body), signific

8、ant differences can be found. On the one hand, of course, it is the number of parts being hot-pressed. In the case of the compressed body, we are talking about one part with a diameter of 25 mm, a height of 3.5 to 4 mm and an area of 4.909 cm2. In the case of the segments of the example, there are 5

9、0 segments with dimensions 40 x 3.2 x 9 (mm) and a (total) area of 64 cm2. The second significant difference is that with the compressed body, there are no additional forces acting from the outside, but there certainly are with segments. The principle means that an external force acts to hold the mo

10、uld together. Friction brakes the sintering process as a result of this force. To ircumvent/compensate for this effect, one can either increase temperature or work with a“ friction factorIf we look at the diagram, it is clear that even increasing the temperature will not entirely achieve the same va

11、lues as the compressed body. It is much more effective to work with additional force, the socalled friction factor. In this process, a value depending on the number of dies is added to the result of the mould pressure calculated according to the formula below: whereP(bar) = mould pressure to be set

12、at the hot press N = number of segments per mouldA = area of one segment (cm2)P(sp) = specific pressure (kg/cm2)A(cyl) = area of the hydraulic cylinder (cm2)If we now look at a sinter data sheet produced with segments again but with the friction factor taken into account, the following pattern results.When hot-pressing bonds, it can be said that if the friction factor is taken into accoun