產(chǎn)品細節(jié)介紹
1VCR007346操作面板
啟動時，涌入電流非常高，導致熱容量迅速增加。使用的熱容量
變量與啟動電機所需的熱容量進行比較。如果沒有足夠的熱量
369可用于啟動電機，在電機冷卻至熱容量水平才能成功啟動之前，它會阻止操作員啟動。假設(shè)電機需要40%的熱容量才能啟動。如果電機在停止前過載運行熱容量會有一定值；比如說80%。在這種情況下，369（啟用啟動禁止）將鎖定或防止操作員啟動電機，直到熱容量降低到60%，以便成功啟動電機可以實現(xiàn)啟動。該示例如第7-19頁的圖7-7:START INHIBIT FUNCTIONALITY（起動抑制功能示意圖）所示。本節(jié)說明了如何使用兩個CT感應三相電流。
使用兩個CT而不是三個CT檢測相電流的正確配置如下所示。兩個CT中的每一個充當電流源。來自“A”相CT的電流流入標記為“A”的繼電器上的插入式CT。從…起在這里，它與剛剛通過插入式CT的相位“C”上CT的電流相加
標記為“C”的繼電器。該“總和”電流流過標記為“B”的插入式CT，然后從該處分裂返回
其各自的來源（CT）。極性非常重要，因為相位“B”的值必須等于“A”+“C”表示所有矢量的總和等于零。請注意，只有一個接地連接。做兩個接地連接為電流創(chuàng)建了一條平行路徑。在三線電源上，此配置將始終工作，并且將正確檢測不平衡。如果發(fā)生單個相位，繼電器的插入CT將始終存在較大的不平衡。例如，如果相位“A”丟失，
相位“A”讀數(shù)為零，而相位“B”和“C”均讀數(shù)為相位“C”的幅值。另一方面，階段
“B”丟失，在電源處，“A”將與相位“C”相異180×，相位“B”的矢量相加將為零。
7.6.8未接地系統(tǒng)接地故障檢測
50:0.025接地故障輸入用于靈敏地檢測高電阻接地系統(tǒng)的故障。檢測1至10 A初級接地電流轉(zhuǎn)換為輸入0.5 mA至5 mA至50:0.025抽頭。理解力這允許以簡單的方式使用此輸入來檢測未接地系統(tǒng)上的接地故障。
下圖說明了如何使用星形開口三角形電壓互感器配置來檢測相位接地。
正常情況下，50:0.025輸入端和電阻器之間出現(xiàn)的三相凈電壓為接近于零。在故障條件下，假設(shè)VT的二次側(cè)電壓為69 V，則繼電器看到的凈電壓電阻器為3Vo或3×69 V=207 V。369需要從每個RTD帶回三根引線：熱、回路和補償。在某些情況下可能很貴。然而，可以減少引線的數(shù)量，以便第一個RTD需要三根引線每個連續(xù)RTD只有一個。接線配置見下圖。
Upon a start, the inrush current
is very high, causing the thermal capacity to rapidly increase. The Thermal Capacity Used variable is compared to the amount of the Thermal Capacity required to start the motor. If there is not enough thermal capacity available to start the motor, the 369 blocks the operator from starting until the motor has cooled to a level of thermal capacity to successfully start. Assume that a motor requires 40% Thermal Capacity to start. If the motor was running in overload prior to stopping, the thermal capacity would be some value; say 80%. Under such conditions the 369 (with Start Inhibit enabled) will lockout or prevent the operator from starting the motor until the thermal capacity has decreased to 60% so that a successful motor start can be achieved. This example is illustrated in Figure 7–7: ILLUSTRATION OF THE START INHIBIT FUNCTIONALITY on page 7–19.
This section illustrates how to use two CTs to sense three phase currents.
The proper configuration for using two CTs rather than three to detect phase current is shown below. Each of the two CTs acts as a current source. The current from the CT on phase ‘A’ flows into the interposing CT on the relay marked ‘A’. From there, the it sums with the current flowing from the CT on phase ‘C’ which has just passed through the interposing CT on the relay marked ‘C’. This ‘summed’ current flows through the interposing CT marked ‘B’ and splits from there to return to its respective source (CT). Polarity is very important since the value of phase ‘B’ must be the negative equivalent of 'A' + 'C' for the sum of all the vectors to equate to zero. Note that there is only one ground connection. Making two ground connections creates a parallel path for the current
On a three wire supply
this configuration will always work and unbalance will be detected properly. In the event of a single phase, there will always be a large unbalance present at the interposing CTs of the relay. If for example phase ‘A’ was lost, phase ‘A’ would read zero while phases ‘B’ and ‘C’ would both read the magnitude of phase ‘C’. If on the other hand, phase ‘B’ was lost, at the supply, ‘A’ would be 180× out of phase with phase ‘C’ and the vector addition would be zero at phase ‘B’. 7.6.8 GROUND FAULT DETECTION ON UNGROUNDED SYSTEMS The 50:0.025 ground fault input is designed for sensitive detection of faults on a high resistance grounded system. Detection of ground currents from 1 to 10 A primary translates to an input of 0.5 mA to 5 mA into the 50:0.025 tap. Understanding this allows the use of this input in a simple manner for the detection of ground faults on ungrounded systems. The following diagram illustrates how to use a wye-open delta voltage transformer configuration to detect phase grounding. Under normal conditions, the net voltage of the three phases that appears across the 50:0.025 input and the resistor is close to zero. Under a fault condition, assuming the secondaries of the VTs to be 69 V, the net voltage seen by the relay and the resistor is 3Vo or 3 × 69 V = 207 V.
The 369 requires three leads to be brought back from each RTD: Hot, Return, and Compensation. In certain situations this can be quite expensive. However, it is possible to reduce the number of leads so that three are required for the first RTD and only one for each successive RTD. Refer to the following diagram for wiring configuration.
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