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當故障正確發(fā)生時，會出現(xiàn)壞的情況
斷路器自身的下游。從故障點看，電路為
由帶變壓器的網(wǎng)絡系列表示。短路
電流與CB1相同。
電機的額定電流等于385 A；要選擇的斷路器
是Tmax T5H 400。
CB3的選擇
對于CB3，當故障發(fā)生在正下游時，壞的情況也會出現(xiàn)
斷路器本身。
從故障點看，電路由兩個并聯(lián)支路表示：
電機和系列的網(wǎng)絡和變壓器。根據(jù)
在前面的規(guī)則中，短路功率是通過使用以下公式確定的
公式：
電機//（網(wǎng)絡+變壓器）
負載L的額定電流等于1443 A；要選擇的斷路器
是Tmax T7S1600或Emax X1B1600。
示例2
圖中所示的電路由電源、兩個變壓器組成
并聯(lián)和三個負載。
上游網(wǎng)絡：Ur1=20000 V
Sknet=500 MVA
變壓器1和2:Sr
=1600 kVA
英國
% = 6%
U1r/U2r=20000/400
負載L1:Sr
=1500 kVA；cosν=0.9；
負載L2:Sr
=1000 kVA；cosν=0.9；
負載L3:Sr
=50 kVA；cosν=0.9。

the worst condition arises when the fault occurs right downstream of the circuit-breaker itself. The circuit, seen from the fault point, is represented by the series of the network with the transformer. The short-circuit current is the same used for CB1. The rated current of the motor is equal to 385 A; the circuit-breaker to select is a Tmax T5H 400. Selection of CB3 For CB3 too, the worst condition arises when the fault occurs right downstream of the circuit-breaker itself. The circuit, seen from the fault point, is represented by two branches in parallel: the motor and the series of the network and transformer. According to the previous rules, the short-circuit power is determined by using the following formula: Motor // (Network + Transformer) The rated current of the load L is equal to 1443 A; the circuit-breaker to select is a Tmax T7S1600 or an Emax X1B1600. Example 2 The circuit shown in the diagram is constituted by the supply, two transformers in parallel and three loads. Upstream network: Ur1=20000 V Sknet = 500 MVA Transformers 1 and 2: Sr = 1600 kVA uk % = 6% U1r /U2r =20000/400 Load L1: Sr = 1500 kVA; cos? = 0.9; Load L2: Sr = 1000 kVA; cos? = 0.9; Load L3: Sr = 50 kVA; cos? = 0.9.