IS2020RKPSG2A处理器模块,DCS工控模块备件
..Q附录F计算散热量F F Sp：电源的计算电源的基本原理是高效。说明这一点的另一种方式是，电源以热量的形式为其提供的每一种功率分配瓦特功率。因此，您可以使用上面Sp中的方法计算特定电源服务的机架中所有设备的总功率需求，然后将该数字除以电源消耗值。您不能简单地使用电源的额定值进行计算，因为应用程序不需要电源的全部容量。如果你使用电源条上的+，你应该计算出所消耗的功率除以路值，再加上电源的总功率。由于每个机架都有自己的电源，因此每个机架都应单独计算。Sp：Discrees Discrees固态继电器的计算需要两次计算，一次用于已在Sp中完成的信号，另一次用于s。继电器s不需要此计算。由于这些s中的固态继电器开关装置将降低一定量的电压，因此可以计算其功耗。s es从单独的电源发出的功率，因此它不包括在所使用的图中，计算Sp中的电源损耗。calcula功耗：在规范中找到您特定的电压降值。获取每个设备所需的电流值，如继电器指示灯信号等，连接上的一个点，并估计其持续时间百分比。获取当前的路数，查看设备制造商的文档或电子产品目录。熟悉设备运作或将运作方式的人可以估算时间百分比。将电压降乘以电流值乘以估计的持续时间百分比，得出平均功耗。对上的所有s重复上述步骤。节省时间，如果当前的绘制和时间中有几个s相似，那么您可以进行dermine，这样您只需要进行一次计算。对机架中的所有差异重复这些计算。差异检查：本规范列出了以下差异C点：电压降电压。所有计算均使用该值。在本例中，两个点驱动正弦曲线，控制液压缸的前进和后退行程。soid制造商的数据表显示ea F安装和硬件八月Q F soid绘制。安培。在管路循环过程中，气缸每秒钟插入和缩回一次。它需要几秒钟的前进和几秒钟的后退。由于圆柱体需要相等的时间前进和后退，所以两个正弦曲线在相等的时间长度内都是打开的：每秒钟中有几秒钟是时间。因此，由于两个正弦曲线具有相等的电流消耗和时间，我们的计算可以应用于两个s。使用公式“平均功耗压降x电流消耗x百分比”表示为时间x的小数。然后将这个结果乘以，因为我们有两个相同的正弦x soid。在这个例子中，这一点上的其他点，歌剧指挥灯在歌剧演员的面板上。Ea指示灯需要。电流。七个指示灯准时亮着，七个正在等待。对于时间x的灯。x每盏灯然后将这个数值乘以x盏灯。第一盏灯的总功耗对于时间x处亮起的灯。x每盏灯然后将这个数值乘以x盏灯。其他灯光的总耗散量加上我们得到的各个计算值+。+对于总计算Sp：差异输入的输入计算差异输入需要两次计算，一次用于已在Sp中完成的信号，另一次用于输入。
Q Appendix F Calculating Heat Dissipation F F Sp : Calculation for Power Supplies A basic ru for power supplies is that they are efficient. Ather way of stating this is that the power supply dissipas watt of power in the form of heat for every of power it delivers the . Therefore you can calcula the tal power requirement for all of the s in the rack served by a particular power supply using the method in Sp above then divide that figure by arrive at the power supply dissipation lue. You cant simply use the rating of the power supply su as for this calculation because the application y t require the full capacity of the power supply. If you are using the + on the power supply’s strip you should calcula the power drawn divide the lue by and add it the tal for the power supply. Since ea rack has its own power supply ea rack should be calculad on an individual basis. Sp : Calculations for Discre s Discre solid sta s require two calculations one for the ’s siglvel s whi was already done in Sp and one for the s. This calculation is t required for the Relay s. Since the solid sta switing devices in these s will drop a measurab amount of voltage their power dissipation can be calculad. that the power dissipad by the s es from a separa power source so it is t included in the figure used calcula power supply dissipation in Sp . calcula power dissipation: In the Specifications find the lue for the Voltage Drop for your particular . Obtain the required current lue for ea device su as a relay pilot light soid etc. connecd an point on the and esti its percent of ontime. obtain the current lues eck the device nufacturer’s documentation or an ectronics catalog. The percent of ontime can be estid by someone familiar with how the equipment operas or will opera. Multiply the Voltage Drop times the current lue times the estid percent of ontime arrive at average power dissipation for that . Repeat for all s on the . save time you could dermine if several s were similar in current draw and ontime so that you would only have ke their calculation once. Repeat these calculations for all Discre s in the rack. Discre Examp: The Specifications lists the following for the Point Discre C : Voltage Drop Volts Use that lue for all of the calculations for this . In this examp two of the ’s points drive soids that control the adnce and retract travel of a hydraulic cylinder. The soid nufacturer’s data sheet shows that ea F Installation and Hardware August Q F soid draws . Amp. The cylinder adnces and retracts once every seconds that the ine is cycling. It takes seconds adnce and seconds retract. Since the cylinder takes equal time adnce and retract both soids are on for equal ngths of time: seconds out of every seconds whi is of the time. Therefore since both soids have equal current draws and ontimes our sing calculation can be applied both s. Use the formula Average Power Dissipation Voltage Drop x Current Draw in x Percent expressed as a decil of ontime x . x per soid Then multiply this result by since we have two identical soids x Soids . tal for the two soids Also in this examp the other points on this point opera pilot lights on an operar’s panel. Ea pilot light requires . of current. Seven of the pilot lights are on of the time and seven are on an estid . For the lights that are on of the time x . x per light Then multiply this lue by x lights . tal dissipation for the first lights For the lights that are on of the time x . x per light Then multiply this lue by x lights . tal dissipation for the other lights Adding up the individual calculations we get + . + for the ’s tal calculation Sp : Input Calculations for Discre Input s A Discre Input requires two calculations one for the ’s siglvel s whi was already done in Sp and one for the input s. that the power dissipad by the input s es from a separa power source so are t included in the figure used calcula power supply dissipation in Sp .
