If the cable is used on the high voltage side of the 630kva transformer, how much should the cable m

Upstairs is too mother-in-law, 10kv if the primary current is YJV22-10 3*25.

If it is 6kV, the primary current is about 50A.

Choose YJV22-10 3*35 on it.

Summary. If the cable is used on the high-voltage side of the 630kva transformer, how much should the cable model be chosen, hello dear, it should be according to the typical design requirements of the local power supply department YJV-3*70mm2 copper core power cable. Of course, if the power supply department does not have typical design requirements, you can choose 630* per square millimeter, and you can choose 3*25mm2 copper core cable through calculation, and you can choose 35 or 50mm2 considering the increment and margin.

If the cable is used on the high-voltage side of the 630kva transformer, how much should the cable model be selected

If the high-voltage side of the 630kva transformer adopts the cable, how much should the cable model be chosen, hello dear, should smile and quietly according to the typical design requirements of the local power supply department YJV-3*70mm2 copper core power cable. Of course, if the power supply department does not have a typical design to meet the slag requirements, you can choose according to each square millimeter, 630*, you can choose 3 * 25mm2 copper core cable by calculating and dismantling Huai, and you can choose 35 or 50mm2 to consider the increment and margin. Hope it helps.

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16m is required.

Need to choose 10kvzr-yjv3*70 square millimeter cable.

Because the selection of the cable does not depend on the current alone, the current of the cable is not fixed, and it is necessary to multiply a coefficient with the difference of ambient temperature and laying method, as well as the verification of dynamic stability and thermal stability.

The short-circuit current is also calculated, so the line should be slightly larger.

and below are low-voltage cables.

It is a medium voltage cable.

35kV 110kV is a high-voltage cable.

110 220kV is UHV cable.

UHV cable is a kind of power cable that appears with the continuous development of cable technology, UHV cable is generally used as a hub belt in large-scale transmission systems, and belongs to a kind of high-voltage cable with high technical content, which is mainly used for long-distance power transmission.

A transformer is a stationary electrical appliance made using the principle of electromagnetic induction.

When the original coil of the transformer is connected to the AC power supply, the alternating magnetic flux is generated in the core, and the alternating magnetic flux is universally expressed.

The primary and secondary coils are the same, and are also simple harmonic functions, and the table is = msin t.

According to Faraday's law of electromagnetic induction, the induced electromotive force in the primary and secondary coils is e1=-n1d dt and e2=-n2d dt. where n1 and n2 are the turns of the primary and secondary coils.

As can be seen from the figure, u1=-e1, u2=e2 (the physical quantity of the original coil is represented by the lower corner mark 1, and the physical quantity of the secondary coil is represented by the lower corner mark 2), and its complex effective values are u1=-e1=jn1, u2=e2=-jn2, so that k=n1 n2, which is called the transformation ratio of the transformer.

From the above formula, u1 u2=-n1 n2=-k, that is, the ratio of the effective value of the voltage of the primary and secondary coils of the transformer is equal to the ratio of its turns and the bit difference between the voltages of the primary and secondary coils is .

And then it is obtained: u1 u2=n1 n2

In the case that the no-load current is negligible, there is i1 i2=-n2 n1, that is, the magnitude of the effective value of the primary and secondary coil currents is inversely proportional to its number of turns, and the phase difference.

Furthermore, we get: i1 i2=n2 n1.

The power of the primary and secondary coils of the ideal transformer is equal p1=p2. It shows that the ideal transformer itself has no power loss.

The actual transformer total present loss has an efficiency of =p2 p1. The efficiency of power transformers is very high, up to more than 90%.

10kV 380V transformer with a capacity of 630kVA. Please use a 225 square millimeter copper cable at the end of the low-voltage outlet.

If the transformer loss is not taken into account, the calculation should be conservative to ensure that the cable cross-sectional area is sufficient).

630kva=630kw

According to the formula for calculating the three-phase electric power:

p= 3uicosa (cosa is the power factor, which is generally taken.)

i=p÷（√3ucosa）=630000÷（

According to the current carrying capacity of the wire, calculated as 5a mm:

1126 5 = 225 (mm²).

10kV 380V transformer with a capacity of 630kVA. What size cable should I use for the low-voltage outlet?

The transformer to the electrical cabinet should use busbar, and the cable should not be selected. Choose 8*80 copper busbar (minimum 6*80). If you must use a cable, it is recommended

YJV3*25 for high-voltage side cables, 2 (3*240+1*120) for voltage-side cables, the thinnest general industrial cables are squared, but they are also square cables, and they are mostly used for signal transmission. Transformer: A transformer is a device that uses the principle of electromagnetic induction to change the alternating voltage, and the main components are the primary coil, the secondary coil and the iron core (magnetic core).

According to the common calculation method of power collapse engineering, the conductor cross-sectional area of high-voltage cable can be calculated by the following formula: conductor cross-sectional area (a) = k s) wherein, k is the cable resistivity, and the value of 90 minutes is generally taken; s is the load power (kW) in kilowatts. In this problem, suppose that the high-voltage cable uses a 1kV cable, and the cable resistivity k is taken, and substituted into the formula, there is:

a = 300) Therefore, high-voltage cables with a cross-sectional area of about 4 square millimeters should be selected. It should be noted that factors such as cable material and laying environment should also be considered in the specific selection to ensure safety and reliability.