已阅读5页,还剩1页未读, 继续免费阅读
版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领
文档简介
MOLD MATERIALS MAKING THE MOST OF HIGHPERFORMANCE MOLD MATERIALSUnderstanding high conductivity alloys and optimizing their use can help you build better molds. By Douglas Veitch, Director, Brush Wellman Injection molders and blow molders can benefit from high conductivity alloys by achieving faster cycle times and better part quality. There are certain properties of the mold material and polymer that enable these efficiencies to be realized. Once these characteristics are understood, mold builders can optimize their use of high-performance materials to provide a durable, fast-cycling mold for their customers.Cooling TimeMold Alloy Thermal PropertiesSome characteristics of mold materials enable us to better understand the thermal process that occurs while molding. Three important properties are:1. Thermal ConductivityHigher thermal conductivity equates to the transfer of more thermal energy per unit of time under steady state conditions.2. Thermal DiffusivityHigher thermal diffusivity means that thermal equilibrium will be reached faster when the temperature changes. A good thermal diffuser will react more quickly to environmental temperature changes.3. Thermal Effusivity (conductivity divided by the square root of the diffusivity)Higher thermal effusivity is a measure of the materials efficiency at instantly removing heat from an object at a higher temperature with which it suddenly makes contact (see Chart 1). Chart 1Mold MaterialThermal Conductivity, W/(cm-oK)Thermal Diffusivity,cm2/secThermal EffusivitySteel - P20 0.29 0.0811.02Copper Beryllium 1.25 0.415 1.94The following explains what all of this means when molding plastics.1. Heat mold up to operating temperature (via water channels). The higher diffusivity allows the copper mold alloy to reach equilibrium faster, so the molding operation can begin sooner. 2. Inject hot plastic melt into the mold and cool. Higher effusivity means the mold will begin to instantly and efficiently remove heat from the plastic. Then the high diffusivity translates to reaching steady state, uniform temperature quickly. Finally, once at equilibrium the conductivity determines how fast the thermal energy will be removed from the plastic until the part reaches the desired ejection temperature. 3. Maintain setpoint temperature (equal to water temperature) during mold-open, ejection and mold-close portions of the cycle. Again, the high diffusivity enables the mold to maintain equilibrium at setpoint during mold open, ejection and mold close. Since the air is a poor thermal medium, the contact between the water and copper is the overriding factor. Figure 1: IR temp distribution. Images courtesy of Brush Wellman Inc. Figure 1 shows pictures from a thermal FEA illustrating the uniform temperature of a copper beryllium mold compared to that of a mold made of P-20 steel. Polymer TypesThe two main polymer familiessemi-crystalline and amorphousboth benefit from higher conductivity mold materials. Semi-crystalline polymers have a densely packed, uniform molecular structure and include materials such as polyamide (nylon), polyethylene, polypropylene and polyacetal. These polymers become amorphous when melted during processing and will become semi-crystalline again when cooled. Amorphous polymers have a loose and random molecular structure, so that in some cases amorphous materials are transparent. Both types of polymers can benefit from improved heat transfer and reduced cooling time. The following are some differences that need to be realized to provide a better understanding of the application. Crystalline materials have a sharp melting point, and thus a latent heat energy that must be added when melting, and removed when cooling. The plastic needs to be solidified and cooled below the heat deflection temperature before ejection from the mold. The heat deflection temperature (HDT) is available on most resin datasheets. Just getting below the melting point is not enough. The part has to be cooled to the point where it is stiff enough to eject. Glass and mineral fillers increase the crystallization rate and the HDT so the part can be ejected at a higher temperature without deformation. Amorphous polymers do not have a melting point, but as the heat input is increased above the glass transition temperature (Tg), the viscosity of the polymer decreases until it begins to flow. Heat is added until the plastic can flow adequately to fill the mold. Then the heat has to be removed until the polymer is below the Tgin many cases before the part will be stiff enough to be ejected. In general, crystalline polymers contain more heat energy due to the latent heat. For example polycarbonatewhich is amorphoushas a heat capacity of 1.2 J/(g oK) while polypropylenewhich is semi-crystallinehas a heat capacity of 1.9 J/(g oK) or 58 percent higher. Molders will experience cycle time reductions and improved uniformity of cooling for both families of plastics when using high conductivity mold alloys.Some semi-crystalline materialssuch as nylonrequire relatively high mold temperatures to provide good surface finish and maximum crystallinity. High conductivity mold alloys can improve both characteristics, and reduce cycle time as an added bonus. This effect is achieved by simply running the mold at the desired temperaturefor example 180oF. The high conductivity alloy will be able to remove heat faster than steel, but at the recommended temperature, and the heat removal will be more uniform. The result is reduced cooling time and more uniform crystallinity in the molded part. When molding amorphous plastics, uniform cooling also is very important. For clear polymerslike polycarbonatethe part will have better clarity and toughness.Water CoolingWith steel tools, molders often run chillers to reduce cycle times and to compensate for the reduced heat transfer of the steel. The cold tool will often result in condensation on the mold surface that can adversely affect part quality. With high-conductivity tool alloys, the cooling water can be set at a higher temperature to prevent condensation, and yet achieve much faster cycles than steel tools. Also, the surface temperature of the mold will be very close to the water temperature setpoint. The long-term heat transfer performance of copper alloys is very good, because copper resists corrosion and bio-fouling in the cooling channels.EconomicsCycle time reduction always has been a key effort for molders. Increasingly, molders are attempting to improve cycles to offset higher resin, energy and transportation costs that they have not been able to pass through to their customers. Using copper mold alloys allows molders to improve their production rate, avoid capital investment and minimize quality issues. Higher conductivity molds provide more uniform cooling than steel tools, resulting in better dimensional control, decreased warpage and part strength improvements. Payback analysis for molds using high-performance alloys yields very desirable numbers due to the reduced cooling times.Figure 2: Copper beryllium insert stands up well to the glass-filled nylon used in chair bases.Applications1. Recently, in the case of a large polyethylene lid, the molder calculated the payback at 10 days using a copper beryllium insert in a steel tool. The cycle time was reduced from 75 seconds to 52 seconds, and allowed the molder to avoid purchasing an additional molding machine to keep up with demand. Capital avoidance is sometimes overlooked, but can be of tremendous benefit in the long term. 2. Another example is a chair base made of glass-reinforced nylon. The manufacturer was able to obtain a 20 percent cycle time reduction using copper beryllium for a core insert in the chair base mold (see Figure 2). Prior to using copper beryllium, the manufacturer was using strictly steel in its molds. After switching to molds using copper beryllium inserts they have witnessed a decrease in cycle from 122 seconds to 98 secondsallowing for faster production throughout. Also, the dimensional control of the hub diameter was improved. By using copp
温馨提示
- 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
- 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
- 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
- 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
- 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
- 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
- 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
最新文档
- 初中七年级劳动技术《合理支配零用钱之学会储蓄》教学设计
- 高一化学教学设计-食品中的有机化合物探究
- 核心素养导向的初中英语八年级情境化教学设计-以人教版八年级下册Unit 7 Section B (1a-1e)为例
- 高三地理教学设计:大气组成与垂直分层核心考点突破
- 高中体育高一滑冰教学设计冰雪运动技能培养
- 初中八年级地理上册晋教版4.3谋划母亲河生态保护第二课时教学设计
- 初中八年级语文三峡写景艺术与情感表达教学设计
- 九年级物理“电流和电路”教学设计
- 新人教版数学四年级上册第三单元 《多位数乘两位数》3.6 积的变化规律教学设计
- 2026及未来5年中国榉木实木地板数据监测研究报告
- 人教PEP四年级英语上册阅读理解专项30篇(含答案)
- 2026临汾市侯马市招聘乡(街道)消防协管员考试备考试题及答案详解
- 江西省人才发展集团有限公司2026年春季集中招聘专题【11人】建设笔试备考题库及答案解析
- 2026年高考上海卷英语含解析及答案(新课标卷)
- 广东省2026年普通高中学业水平合格性考试数学试题(含答案)
- 钢管脚手架用量计算表 形式2
- 资产评估公司人事管理制度
- 轴类零件加工工艺过程课件
- 食堂蔬菜等食材的采购协议
- 第六讲行为建筑学
- 《点到直线的距离公式》教案(公开课)
评论
0/150
提交评论