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Question:

630KVA transformer can use the maximum number of power appliances

The total power of 590kw, this transformer can bear it? It is best to give a formula, do not copy the.

Answer:

No additional measures in the case, 630KVA transformer can use the maximum 420KW power appliances. 1, the transformer power factor is generally 0.8 (also 0.7), then, the total power transformer = 630 * 0.8 = 504KW. 2, according to the "power engineering design manual", the transformer capacity, for a smooth load of a single transformer, the load rate is generally about 85%, so the upper case, with electrical power = 504 * 85% = 428.4KW. 3, if it is a single high-power motor, starting current as large, even if the measures, but also a corresponding reduction in electricity consumption. 4, if the three-phase imbalance, but also a corresponding reduction in electricity consumption.
600KVA box can generally increase the number of households with electricity
General transformer power factor is 0.85 or so 630X0.85 = 535.5KW, you should be installed capacity of 590KW, if it is not necessarily open at the same time, there is no problem. On the contrary, you can also increase the capacitor to improve the power factor to run .300KW If not more than 150 meters, then available 3X120 +50 copper cable, the distance from the increase in the line by.
630KVA transformer generally according to the efficiency of 0.8, about 504KW load, the capacitance compensation will increase the power factor to 0.95, the load power also increased with about 590KW near the existing total power 590kw should be installed capacity, the actual load should Less than 90% can load. 630KVA Transformer rated current: I = P / 1.732 / U = 630 / 1.732 / 0.38 ≈ 957 (A) The total installed power of 590kw, load rate or full load rate of about 0.9, the capacitance compensation to the power factor increased to 0.95 when the average power factor of about 0.9, the load current: I = P × 0.9 / 1.732 / U / COSφ = 590 × 0.9 / 1.732 / 0.38 / 0.9 ≈ 896 (A) 300kw load current (when COSφ is 0.8): I = P / 1.732 / U / COSφ = 300 / 1.732 / 0.38 / 0.8 ≈ 570 (A) Large cross-section copper wire per square safe current carrying about 2.5 ~ 3A, need to wire cross-section: S = 570/3 = 190 (square)

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