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Formula for calculating power capacitors and description of product selection
View:86 Time:2018-06
Formula for calculating power capacitors and description of product selection


1. Compensation power (reactive power output): Q=3IU=2pi fCU2 (with n as rated value or nominal value, such as Q n, U n; without n as actual value, such as Q, U)

_For example, the parameters of BZMJ0.4-30-3 capacitor are as follows: Qn=30KVar Un=0.4KV In=43.3A f=50Hz Cn=596.8uF (the manufacturer produces capacitors according to this value, Cn generally remains unchanged)

2. When the grid voltage changes, the capacitor's actual reactive power output: Q=3IU=2pi fCnU2=(U/Un)2Qn (normally, 0.4KV capacitors are used on 400V lines)

_e.g. Un = 400V, U = 440V (i.e. 0.4KV capacitors are used on 440V lines)

Q=(440/400)2*Qn=1.21Qn(at this time, the capacitor is overloaded, the capacitor is heated seriously, and the life of the capacitor is shortened)

_e.g. Un = 450V, U = 400V (i.e. 0.45KV capacitors are used on 400V lines)

Q= (400/450) 2 *Qn = 0.79Qn (capacitor is used for reducing quota at this time, reactive power output is insufficient, user investment is not economical, but reliability is improved, capacitor life is prolonged. At present, all capacitor cabinets are automatically compensated by grouping. As long as the total capacitance is sufficient, increasing the rated voltage of capacitor does not affect the compensation effect of capacitor cabinet, the product life is about five years.

3. When the power grid has harmonics, the total current or the component of harmonic current increase.

_For example: I = 1.4In, U = Un Q = _3I U = _3 *1.4InUn = 1.4Qn (when the capacitor is seriously overloaded and the capacitor is quickly damaged and failed), so when users find harmonics in the power grid or use high-power loads (such as medium frequency furnace, large-scale converter, rectifier, etc.) or the protection device of capacitor upper level often operates. For example, thermal relay action, fuse fuse fuse, etc., if the current of the capacitor is more than 1.1 times of the rated current of the capacitor, it is suggested that the user use a capacitor with a higher rated voltage level, such as 0.525KV level: at this time, U=(400/525) Un = 0.76Un, Q=3IU=3 *1.4In *0.76=1.06Qn capacitor overload is not much, and it can be changed to a capacitor with a higher rated voltage level. Barely cope with the use. However, the influence of harmonics on the life of capacitors still exists, and its influence is quite complex, so it is inconvenient to discuss it here. The final solution is to remove the harmonic of power grid (add harmonic filter) (series tuned reactor), purify the power grid, and ensure the safe operation of capacitors and other electrical appliances.
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