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电压降乘以电流值乘以估计的接通时间百分比,以达到该输出的平均功耗。90–30系列散热F GFK–0898F附录F–90–30型散热F-3模块上的所有输出重复。为了节省时间,您可以确定多个输出在电流消耗和接通时间上是否相似,这样您只需进行一次计算。对机架中的所有离散输出模块重复这些计算。离散输出模块示例:第7章的IC693MDL340部分列出了IC693MDL40 16点离散120VAC输出模块的以下内容:输出电压降:大1.5伏将该值用于此模块的所有计算。在此示例中,输出模块的两个输出点驱动控制液压缸前进和后退行程的电磁阀。电磁阀制造商的数据表显示,每个电磁阀的电流为1.0安培。机器循环时,油缸每60秒前进和缩回一次。前进需要6秒,后退需要6秒。由于油缸前进和缩回所需的时间相等,两个电磁阀打开的时间长度相等:每60秒中有6秒,占时间的10%。因此,由于两个螺线管具有相等的电流消耗和接通时间,我们的单一计算可以应用于两个输出。使用公式平均功耗=电压降x电流消耗(以安培为单位)x接通时间百分比(以十进制表示):1.5 x 1.0 x 0.10=0.15瓦每个螺线管然后将此结果乘以2,因为我们有两个相同的螺线管:0.15瓦x 2个螺线管=0.30瓦两个螺线管的总功率。在本示例中
Voltage Drop times the current value times the estimated percent of on–time to arrive at average power dissipation for that output. Series 90–30 Heat Dissipation F GFK–0898F Appendix F – Series 90–30 Heat Dissipation F-3Repeat for all outputs on the module. To save time, you could determine if several outputs were similar in current draw and on–time so that you would only have to make their calculation once.Repeat these calculations for all Discrete Output modules in the rack. Discrete Output Module Example: The IC693MDL340 section of Chapter 7 lists the following for the IC693MDL340 16–Point Discrete 120VAC Output Module: Output Voltage Drop: 1.5 Volts maximum Use that value for all of the calculations for this module. In this example, two of the Output module’s output points drive solenoids that control the advance and retract travel of a hydraulic cylinder. The solenoid manufacturer’s data sheet shows that each solenoid draws 1.0 Amp. The cylinder advances and retracts once every 60 seconds that the machine is cycling. It takes 6 seconds to advance and 6 seconds to retract. Since the cylinder takes equal time to advance and retract, both solenoids are on for equal lengths of time: 6 seconds out of every 60 seconds, which is 10% of the time. Therefore, since both solenoids have equal current draws and on–times, our single calculation can be applied to both outputs. Use the formula Average Power Dissipation = Voltage Drop x Current Draw (in Amps) x Percent (expressed as a decimal) of on–time: 1.5 x 1.0 x 0.10 = 0.15 Watts per solenoid Then multiply this result by 2 since we have two identical solenoids: 0.15 Watts x 2 Solenoids = 0.30 Watts total for the two solenoids Also in this example, the other 14 output points on
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●使用前:
在操作和维护,安装设备之前,请阅读并严格遵守安全相关的信息。
Read and strictly follow safety-related information before operatingmaintainingand installing equipment.
●使用者要求:
必须从事安装,组装或维护设备等相关工控电气机械以及其他相关技术的人员。
Must be engaged in the installation,assembly or maintenance of equipment and other related industriacontroeectrica machinery and other related technica personne.应用范围:
应用于品牌本系列工业以及生产生活自动化设备产品。
Used in the brand of this series of industriaand production automation eauipment products
适用于场景:系统集成,非标定制,产线升级改造,原设备电气替换等等。
Suitable for scenarios:system integration,non-standard customizationproduction line upgrading and transformation,electrical replacement of original equipment,etc.
●操作人员:
应该由本品牌专业培训操作员来安装、操作和维修设备。专业人员具有技术能力强,熟悉安全信息和安全实施:
熟悉设备的安装加工、操作和维护:熟悉处理所有的潜在危险情况。
Equipment should be installed,operated and maintained by trained operators from the brand
●危险情况
警告
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●除了产品特性以及功能,还需了解产品的使用寿命环境。
●充分学习产品使用手册,否则非专业人员可能会造成人身危险。
Study the product instruction manual thoroughlyotherwise it may cause personal danger to non-professionals.
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Applicable to cementwood,stone,rubber,petroleumchemical, power generation,
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