I. Calculation of Motor Power and Wiring Diameter
First, calculate the line current of 100KW load.
For a three-phase balanced circuit, the formula for calculating the power of the three-phase circuit is: P=1.732IUcosφ.
From the power formula of three-phase circuit, it can be deduced that:
Line current formula: I=P/1.732Ucosφ.
Where: p is the circuit power, u is the line voltage, three-phase is 380V, and cosφ is the inductive load power factor. Generally, the line current of your 100KW load is taken as 0.8:
I=P/1.732Ucosφ=100000/1.732*380*0.8=100000/526.53=190A
The current value should also be corrected 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, which is 4 to 7 times of the working current, so the starting current of the motor should be considered, but the time of starting current is not very long. Generally, only the coefficient of 1.3 to 1.7 is considered when selecting the conductor.
If we take 1.5, the current is 285A. If there are a large number of 60KW loads, and people don’t use them at the same time, we can take the utilization coefficient as 0.5 to 0.8, and here take 0.8, and the current will be 228A. You can choose wires, air switches, contactors, thermal relays and other equipment according to this current. So the step of calculating current cannot be omitted.
Wire selection:
According to the allowable current table of a wire manufacturer, choose 50 square copper-core rubber wire or 70 square copper-core plastic wire.
Transformer selection:
There are many conditions for transformer selection. Here, simply divide the total capacity by the power factor and round it up. S=P/cosφ=100/0.8=125KVA
Choose a transformer larger than 125KVA.
How much current a 50-square-meter copper-core cable can withstand also depends on the laying method, environmental temperature, cable structure type and other factors.
50 square 10/35KV XLPE insulated cable with long-term allowable current carrying capacity for air laying.
(10KV three-core cable) 231A(35KV single-core cable) 260A Long-term allowable current carrying capacity (soil thermal resistance coefficient 100 C.cm/W) (10kV three-core cable) 217A(35KV single-core cable) 213A.
Second, according to the simple calculation of power distribution cable
It is known that the rated power of the motor is 22KW and the rated voltage is 380V. The transformer is 400 meters away from the well site. How much cable is required?
(The resistivity ρ of copper is 0.0175) (1) Calculate the rated current of the motor at rated power with rated capacity.
Solution: from P=S×COSφ, S=P/COSφ=22/0.8=27.5KVA, where p is the rated power, and COSφ is the power factor, which is 0.8 according to the motor brand.
There is S=I×U to calculate the rated current at rated power. I=S/U=27500/380=73A is obtained by calculation formula.
Estimation formula:
Multiply it by nine at 2: 50, and subtract one number from it to go up.
Thirty-five times 3.5, and both groups subtract 5.5.
Conditions have changed and converted, and high-temperature 10% discount copper has been upgraded.
The number of pipe-piercing roots is two, three, four, eight, seven, six fold full load flow.
Description:
(1) The formula in this section does not directly indicate the current carrying capacity (safe current) of various insulated wires (rubber and plastic insulated wires), but is expressed by "multiplying the cross section by a certain multiple" and is obtained by mental calculation.
The multiple decreases with the increase of cross section.
"Multiply two point five times nine, and subtract one number from the top" refers to aluminum core insulated wires with various cross sections of 2.5mm’ and below, and their current carrying capacity is about 9 times of the number of cross sections. Such as 2.5mm’ conductor, the current carrying capacity is 2.5×9=22.5(A). The multiple relationship between the current carrying capacity and the number of cross-sections of conductors with a diameter of 4mm’ or more is that they are arranged upward along the line number, and the multiple is reduced by L one by one, that is, 4×8, 6×7, 10×6, 16×5 and 25×4.
"Thirty-five times three point five, both groups minus five points" means that the current carrying capacity of 35mm "conductor is 3.5 times the number of sections, that is, 35×3.5 = 122.5 (a). From 50mm’ and above, the multiple relationship between the current carrying capacity and the number of sections becomes a group of two wire numbers, and the multiple is reduced by 0.5 in turn. Namely 50, 70mm’
The current carrying capacity of the conductor is 3 times of the number of sections; The current carrying capacity of 95 and 120mm "conductor is 2.5 times of its cross-sectional area, and so on.
"If the conditions change, it will be converted, and the high-temperature copper will be upgraded at 10% discount".
The above formula depends on the condition that the aluminum core insulated wire is exposed at the ambient temperature of 25℃.
If the aluminum core insulated wire is exposed in an area where the ambient temperature is higher than 25℃ for a long time, the current carrying capacity of the wire can be calculated according to the above formula calculation method, and then it can be discounted by 10%;
When copper-core insulated wire is used instead of aluminum wire, its current carrying capacity is slightly larger than that of aluminum wire with the same specification. The current carrying capacity is calculated by the formula above, which is one wire number larger than that of aluminum wire.
For example, the current carrying capacity of 16mm’ copper wire can be calculated as 25mm2 aluminum wire and 16×5=80, so it is appropriate to choose 16 square meters.
Calculate the cross-sectional area of the conductor from the allowable voltage drop. The allowable minimum working voltage of the motor is 360V, the secondary voltage of the transformer is 380V, the allowable maximum voltage drop under rated power is △U is 20V, and the allowable resistance under rated power is
R line =△U/I=20/73=0.27Ω
From R line = ρ L/s, calculate the cross-sectional area of the conductor: S = ρ L/R line.
=0.0175×400/0.27Ω=24.62mm square
Conclusion: 25 square copper cable should be selected.
Third, how to choose circuit breakers and thermal relays
How to choose a cable with a large cross-sectional area according to the current? The cable we choose is a copper core cable.
For example, we need to wire an 18.5KW motor, and it can be calculated that its rated current is 37A. It is also based on experience that a square millimeter of copper wire can pass a current of 4~6A. If we take the median value of 5A, the cross-sectional area of the cable should be 37/5=6.4 square millimeters. Our standard cables are 6 square millimeters and 10 square millimeters in total. In order to ensure reliability, we choose 10 square millimeters. In fact, we may choose 6 square meters in the specific selection, which should be considered comprehensively. If the power consumed by the load is less than 60% of the rated power, we can choose this. If we basically want to work near the rated power, we can only choose 10 square meters of cable.
How to choose circuit breakers and thermal relays according to the power of the motor and considering the rated voltage and current wiring of the motor?
Three-phase two hundred and twenty-two motor, kilowatt and three point five amps.
Commonly used 308 motors, one kilowatt and two amperes.
Low-voltage 666 motor, kilowatt 1.2 amps.
High-voltage three thousand volt motor, four kilowatts and one ampere.
High voltage six thousand volt motor, eight kilowatts per ampere.
A three-phase motor, besides its rated voltage, must also know its rated power and rated current. For example, a three-phase asynchronous motor, 7.5KW, 4 poles (commonly used in 2, 4 and 6 stages, with different stages and different rated currents), has a rated circuit of about 15A.
1. Circuit breaker:
Generally, its rated current is 1.52.5 times, and DZ476032A is commonly used.
2. Wire:
According to the rated current 15A of the motor, choose the wire with appropriate current carrying capacity. If the motor starts frequently, choose a relatively thick wire, otherwise, it can be relatively thin. The current carrying capacity is determined by relevant calculation. Here, we choose 4 square.
3. AC contactor:
Just choose the appropriate size according to the motor power, 1.52.5 times. Generally, there are models in the selection manual. Here, we choose Chint CJX22510, and we must pay attention to the matching of auxiliary contacts. Don’t buy back auxiliary contacts when the time comes.
4. Thermal relay
Its setting current can be adjusted, generally to 11.2 times the rated current of the motor.
(1) Multiple motors with wires: add the total power of the motors and multiply it by 2, which is their total current.
(2) Within 50m of the line, the cross section of the conductor is: the total current divided by 4. (Put a little margin appropriately).
(3) Conductor section when the line length exceeds 50 meters: the total current is divided by 3. (Put a little bit more traffic).
(4) The current density of a large cable of more than 120 square meters is lower.
Fourth, summary:
Circuit breaker selection:
The rated current of the motor is multiplied by 2.5 times, and the setting current is 1.5 times that of the motor, which ensures frequent starting and sensitive short-circuit action.
The setting value of thermal relay is that the rated current of motor is 1.1 times.
Ac contactor:
The selection of AC contactor is 2.5 times that of motor flow.
This can ensure long-term and frequent work.