Fuse melt current calculation method - Database & Sql Blog Articles

Probe current voltage pin 420*4450 head diameter 5.0 over current current and voltage pin

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The fuse current includes two aspects, one is the rated current of the fuse tube, and the other is the rated current of the melt, which cannot be confused.

1. Melt current IRNIRN ≥Iq/K(A) for single motor start-up...... (l) IRN-melt rated current (A)Iq-motor starting current (A) K-factor or IRN = (1.5- 2.5) IH (A) ...... (2) wherein IH- rated motor current (A) or 380V (AC) IRN = 7PH (A) ...... (3) 22OV (AC) IRN = 12PH (A) ...... ( 4) where: PH-motor rated power (kW)

A plurality of distribution main motor 2 (provided several) Total fuse IRN = (1.5-2.5) IHmax + I (n-1) (A) ...... (5) where: IHmax maximum motor or a motor starting simultaneously Group rated current I( n-1) - the sum of the rated currents of the motors other than the largest or simultaneously started motor group or IRN = I(nl) + Iqmax / 2.5 (A) (6) : Iqmax—maximum motor starting current (A) or IRN = K1[Iqmax-I(nl)] (A)...(7) where: K1-load factor, generally 0.4, there is also information that K1 is 0.9 -l.

3. Electric lamp total fuse melt IRN = (l.3-1.5 × watt-hour meter rated current (A) ... (8) insurance installed on the meter outlet.

4. Electric lamp shunt insurance IRN = working current of all lights on the branch (A)......(9)5. Insurance with control transformer (control line) IRN≥PH + 0.1Pq/U2(A)......(10) Where: PH-control transformer rated capacity (VA) Pq - the starting capacity of the attracting coil of the largest appliance in the line or the sum of the starting capacity of several electrical attracting coils (VA) U2-control transformer secondary voltage (V) 6. No need to control the transformer's insurance (control circuit) IRN≥0.4[Iq +IH(n-1)](A)...(11) where: Iq-the largest electric appliance in the line (or the attracting coil that several appliances start at the same time) Start current (A)IH(n-1)-the sum of the rated currents of the remaining electrical attracting coils in the line (A) 7. The melt current of the load balancing control circuit IRN≥ IH(A)...(12) Medium: IH - sum of rated currents of all electrical appliance loads (A) 8. Selection of high voltage fuses Transformer capacity below 100kVA, IRN = (2-2.5) × transformer high side rated current (A)... (13 Transformer capacity is above 100kVA, IRN=(1.5-2)× transformer high-voltage side rated current (A)......(14)9. Transformer low-voltage side fuse transformer low-voltage side fuse is set according to low-voltage side rated current, also According to the actual load current selection, such as transformer rated current 910A, distribution aluminum row 80×8 (mm2), vertical, allowable current is 1320A, temperature correction coefficient is 0.88. The actual allowable current is 1320×0.88 = 1160A, which is larger than the rated current of the transformer and can be used.

When the above calculation finds that the fuse current value is different from the standard fuse, it can be selected according to the close standard as appropriate. For example, the S7-50/10-0.4 variable transformer high-voltage side current is 2.89A, calculated according to formula (13) (2-2.5) 2.89 = 5.78-8.23 A, high-voltage fuse standard: 2, 3, 5, 7.5, 10 15, 15, 20, 30, 40, 50, 75, 100, 150, 200, 300.

Therefore, 10A high voltage fuse can be used.