I. Calculation of Motor Power and Wiring Diameter
First, the line current of a 100kW load needs to be calculated. For a three-phase balanced circuit, the formula for calculating the three-phase circuit power is: P = 1.732IUcosφ. From the three-phase circuit power formula, we can derive: Line current formula: I = P/1.732Ucosφ Where: P is the circuit power, U is the line voltage (380V for three phases), and cosφ is the power factor for inductive loads, generally taken as 0.8. Your 100kW load line current is: I = P/1.732Ucosφ = 100000/1.732 * 380 * 0.8 = 100000/526.53 = 190A. The current value also needs to be adjusted according to the nature and quantity of the load. If there are many large motors in the load, the starting current of the motor is very large, 4 to 7 times the operating current, so the starting current of the motor must also be considered. However, the starting current time is not very long, and generally only a coefficient of 1.3 to 1.7 is considered when selecting conductors. If we take a factor of 1.5, then the current is 285A. If there are many components in a 60KW load, and they are not used simultaneously, a factor of 0.5 to 0.8 can be used. Here, we take 0.8, and the current is 228A. We can then select wires, circuit breakers, contactors, thermal relays, and other equipment based on this current. Therefore, the step of calculating the current cannot be omitted.
Conductor selection: According to the current carrying capacity table of a certain wire manufacturer, select 50 square millimeter copper core rubber wire or 70 square millimeter copper core plastic wire.
Transformer Selection: Transformer selection involves many factors; here, we'll simply divide the total capacity by the power factor and round down. S=P/cosφ=100/0.8=125KVA. A transformer with a capacity greater than 125KVA will suffice. The current carrying capacity of a 50 square millimeter copper core cable depends on the laying method, ambient temperature, and cable structure. The long-term allowable current carrying capacity of a 50 square millimeter 10/35KV cross-linked polyethylene insulated cable is as follows: Air-laid long-term allowable current carrying capacity (10KV three-core cable): 231A; (35KV single-core cable): 260A. Direct burial long-term allowable current carrying capacity (soil thermal resistivity 100°C.cm/W): (10KV three-core cable): 217A; (35KV single-core cable): 213A.
II. Simplified Calculation Based on Power Distribution Cables
Given that the rated power of the motor is 22KW and the rated voltage is 380V, and the transformer is 400 meters away from the well site, what cross-sectional area of cable should be used? (Take the resistivity of copper P as 0.0175) (I) Calculate the rated current of the motor under rated power given the rated capacity. Solution: From P=S×COSφ, we get S=P/COSφ=22/0.8=27.5KVA. Here, P is the rated power and COSφ is the power factor, which is taken as 0.8 according to the motor nameplate. From S=I×U, we can calculate the rated current under rated power: I=S/U=27500/380=73A. From the calculation formula, we get the estimation formula: Multiply 2.5 by 9, and subtract 1 sequentially upwards. Multiply 35 by 3.5, and subtract 0.5 in pairs. Adjustments are made for changing conditions, and copper is upgraded by 10% for high temperature. For two, three, or four cables in a conduit, the full load current is reduced by 80%, 70%, or 60%.
Note: (1) The current carrying capacity (safe current) of various insulated wires (rubber and plastic insulated wires) in this section is not directly stated, but is expressed as "cross-section multiplied by a certain multiple", which can be obtained through mental calculation. The multiple decreases as the cross-section increases.
The saying "Multiply by nine for wires with cross-sections of 2.5mm² and below, then subtract one for each subsequent cross-section" refers to aluminum core insulated wires with cross-sections of 2.5mm² and below, whose current-carrying capacity is approximately nine times the cross-sectional area. For example, a 2.5mm² conductor has a current-carrying capacity of 2.5 × 9 = 22.5 (A). For conductors with cross-sectional areas of 4mm² and above, the current-carrying capacity is proportional to the cross-sectional area, decreasing by one for each subsequent cross-sectional area: 4 × 8, 6 × 7, 10 × 6, 16 × 5, 25 × 4. The saying "Thirty-five times three point five, pairs together minus point five" refers to the fact that the current-carrying capacity of a 35mm² conductor is 3.5 times its cross-sectional area, i.e., 35 × 3.5 = 122.5 (A). For conductors of 50mm² and above, the relationship between current-carrying capacity and cross-sectional area changes to pairing wire sizes together, with the multiple decreasing by 0.5 for each pair. That is, the current-carrying capacity of 50mm² and 70mm² conductors is 3 times their cross-sectional area; the current-carrying capacity of 95mm² and 120mm² conductors is 2.5 times their cross-sectional area, and so on.
"Adjustments for changing conditions, 10% discount for high temperatures, copper upgraded." This rule applies to aluminum core insulated wires laid openly in an environment with a temperature of 25℃. If the aluminum core insulated wire is laid openly in an area where the ambient temperature is consistently above 25℃, the current carrying capacity can be calculated using the above rule, then reduced by 10%. When using copper core insulated wire instead of aluminum, its current carrying capacity is slightly higher than that of aluminum wire of the same specification; the current carrying capacity can be calculated using the above rule by increasing the wire size by one notch. For example, the current carrying capacity of a 16mm² copper wire can be calculated based on a 25mm² aluminum wire: 16 × 5 = 80. Therefore, a 16mm² wire is suitable. Then, calculate the conductor cross-sectional area based on the allowable voltage drop. The minimum allowable operating voltage of the motor is 360V, and the secondary voltage of the transformer is 380V. The maximum allowable voltage drop under rated power is ΔU, which is 20V. The allowable resistance under rated power is R_wire = ΔU/I = 20/73 = 0.27Ω. From R_wire = ρL/S, the conductor cross-sectional area is calculated as: S = ρL/R_wire = 0.0175 × 400/0.27Ω = 24.62mm². Conclusion: A 25mm² copper cable should be selected.
III. How to select circuit breakers and thermal relays
How do we choose the appropriate cable cross-sectional area based on the current? We've chosen copper core cables. For example, if we're wiring an 18.5kW motor, its rated current is 37A. Based on experience, 1 square millimeter copper wire can handle 4-6A of current; we'll take the midpoint of 5A. Therefore, the cable cross-sectional area should be 37/5 = 6.4 square millimeters. Our standard cables are available in 6 square millimeter and 10 square millimeter sizes. To ensure reliability, we'll choose the 10 square millimeter cable. However, in practice, we might also choose 6 square millimeters. This depends on the power consumption of the load. If it's less than 60% of the rated power, this choice is acceptable. If the load is operating near the rated power, then the 10 square millimeter cable is the only option.
To determine the appropriate circuit breaker and thermal relay for a three-phase 220V motor, considering its rated voltage and current, the wiring should be optimized. A common three-phase 220V motor provides 3.5 amps per kilowatt. A common three-phase 380V motor provides 2 amps per kilowatt. A low-voltage 660V motor provides 1.2 amps per kilowatt. A high-voltage 3000V motor provides 1 amp for every four kilowatts. A high-voltage 6000V motor provides 1 amp for every eight kilowatts. For a three-phase motor, besides knowing its rated voltage, its rated power and rated current must also be known. For example, a 7.5kW, 4-pole three-phase asynchronous motor (commonly 2, 4, or 6 poles, with different pole numbers resulting in different rated currents) has a rated current of approximately 15A.
1. Circuit breaker: Generally, select one with a rated current of 1.5-2.5 times, commonly DZ47-6032A. 2. Wire: Select a wire with a suitable current carrying capacity based on the motor's rated current of 15A. If the motor starts frequently, select a relatively thicker wire, and vice versa. There are relevant calculation rules for current carrying capacity. Here we select 4 square millimeters. 3. AC contactor: Select a suitable size based on the motor power, 1.5-2.5 times. Generally, the model is listed in the selection manual. Here we select Chint CJX2--2510. Also, pay attention to the matching of auxiliary contacts, so that there are enough auxiliary contacts after purchase. 4. Thermal relay: The setting current can be adjusted. Generally, it is adjusted to 1-1.2 times the rated current of the motor. (1) Wiring multiple motors: Add up the total power of the motors and multiply by 2 to get their total current. (2) For lines within 50 meters, the conductor cross-section is: total current divided by 4. (Add a little margin if necessary) (3) For lines exceeding 50 meters, the conductor cross-section is: total current divided by 3. (Add a little margin if necessary) (4) The current density of large cables with a cross-section of 120 square millimeters or more should be lower.
IV. Summary
Circuit breaker selection: Multiply the motor's rated current by 2.5, and the setting current should be 1.5 times the motor's rated current. This ensures both frequent starting and sensitive short-circuit operation. The thermal relay's setting value should be 1.1 times the motor's rated current.
AC contactor: The AC contactor selected should be 2.5 times the motor current. This ensures long-term, frequent operation.
