8500W with 4 square copper wire, 50A switch, why also trip?

8500W appliances with 4 square copper lines, 50A switches, itself is unreasonable, let’s take a look at there is no relationship with trip.

8500W with 4 square copper wire, 50A switch, why also trip?

Copper wire safety load flow.

1.5 square copper core wires allow long-term load currents to be: 8A — 15A. 2.5 Square copper cores allow long-term load currents to: 16A — 25A. The 4 square copper wire allows long-term load currents to be: 25A — 32A.

6 square copper cores allow long-term load currents to: 32A — 40A.

The 10-square copper core allows long-term load currents to be: 50A — 60A.

8500W with 4 square copper wire, 50A switch, why also trip?

220V power supply is 8500W, with a large wire and empty open.

1, i: Current P: Power u: Voltage COSφ: Power factor cosφ: 0.8-0.9, if you don’t know, you can calculate it.

2, the load is resistive: calculate the load rated current formula, i = p ÷ u = 8500w ÷ 220V = 39A.

3, 8500W resistance load rating is 39A, the wires are equipped with 6 square copper wires, 4 square copper wires are too small, easy to burn, air open to configure 40A, select 50A empty open, not safe.

4, although the configuration is unreasonable, but there is no relationship with the trip, the air open is 50A, the load current is 39A, 39A <50A, so air open does not trip.

5, the load is inductive: calculate the load rated current formula, i = p ÷ (u × cosφ) = 8500w ÷ (220V × 0.8) = 48A.

6, 8500W inductive load rated current is 48A, the wires are configured to configure 10 square copper wires, 4 square copper wires are severely overloaded, easy to be burn, air open to configure 50A, select 50A empty open is right.

7, although the configuration wire is unreasonable, but it is not related to the trip, the air open is 50A, the load current 48a, 48a <50a, so the air is not tripped.

Remarks: Empty open and wires are reasonable, the air is too large, too small, the wire is easy to overload, but the air will not trip, the air is too small, the electric wire is too large, the air is too easy to trip.

8500W with 4 square copper wire, 50A switch, why also trip?

380V power supply is 8500W, with a large wire and empty open.

1, √3: 1.732 i: Current P: Power U: Voltage COSφ: Power factor cosφ: 0.8-0.9, if you don’t know, you can take 0.8.

2, the load is resistive: calculate the load rated current formula, i = p ÷ (u × √3) = 8500w ÷ (380V × 1.732) = 13A.

3, 8500W resistance load rating is 13A, the wire should be configured 1.5 square copper wire, 4 square copper wire is too large, more waste, air open to configure 16A, select 50A empty open too large, not safe.

4, although the configuration is unreasonable, but there is no relationship with the trip, the air open is 50A, the load current is 13A, 13A <50A, so the air does not trip.

5, the load is inductive: calculate the load rated current formula, i = p ÷ (u × cosφ × ×3) = 8500w ÷ (380V × 1.732 × 0.8) = 16A.

6, 8500W inductive load rated current is 16A, the wires are equipped with 2.5 square copper wire, 4 square copper wires, but a bit large, empty open to configure 20A, select 50A empty open, not safe.

7, although the configuration is unreasonable, but it is also related to the trip, the air open is 50A, the load currents 16a, 16a <50a, so air open does not trip.

8500W with 4 square copper wire, 50A switch, why also trip?

Through our calculation, load 8500W, configure 4 square copper wire, 50A air open, is unreasonable, but there is no relationship with the trip, the trip should be wire short circuit, wire leakage or load leakage.