To obtain a value of 15 kW how many amperes per 3 phases at a standard voltage of 380 volts and a power factor of 0.95, it is necessary to divide the power by the product of the root of three, voltage and cosine fi, which gives a result of approximately 24 Amps.
This base value is critical for the selection of an introductory circuit breaker, since it is the current strength that determines the thermal load on conductors and switching machines. An erroneous calculation can lead to false power outages or, worse, to overheating of the insulation and fire in the switchboard.
Understanding the physical nature of this process avoids common mistakes when users confuse kilowatts with amps, forgetting about the phase difference and reactive component of the load. In the three-phase network, the current is distributed evenly, so the calculation is not carried out on one phase, but taking into account the linear voltage.
Physical meaning and formula of calculation
To determine exactly how many amperes are 15 kW per 3 phases, you need to refer to Ohm's law for the AC circuit section. Unlike a single-phase household network, which uses a voltage of 220 volts, the three-phase system operates with a linear voltage of 380 volts (or 400 volts according to the new European standards). The formula for calculating the total power in a three-phase network is as follows: P = β3 Γ U Γ I Γ cosΟ. Here P is the power in Watts, U is the linear voltage, I is the required current strength, and cosΟ is the power factor.
Power factor cosΟ This is a key parameter that is often ignored in household calculations. For a purely active load, such as TENSIncandescent lamps or electric stoves, this coefficient is equal to one. However, in modern homes, a significant part of the load is made up of devices with a reactive component: refrigerators, washing machines, computers and, most importantly, electric motors of pumps or machines.
For such devices cosΟ It can range from 0.7 to 0.95. If you take it equal to 0.95, which is the standard for high-quality home appliances, the formula is transformed. Substituting the values, we get: 15000 W / (1.732 Γ 380 V Γ 0.95) β 23.98 A. That is why in technical tables for 15 kW the value of 25 Amps most often appears.
Selection of the nominal value of the automatic switch
Once you have calculated that 15 kW is how many amperes per 3 phases (about 24 A), you can not just take a 25 Amp machine and install it. Electrical networks require a margin of safety and accounting of initiating currents. The circuit breaker should protect the cable from overload, but not to disconnect the network during short-term consumption jumps, characteristic of the inclusion of powerful devices.
According to the rules of the device of electrical installations, the nominal current of the machine is selected from the standard series: 16, 20, 25, 32, 40 Amps and so on. Since the design current is 24 Amps, the 25A machine will be operating at its limit, which can lead to its heating and eventual operation. Moreover, there is the concept of "non-disconnected current", which is 1.13 of the nominal value, and "conditional disconnect current" - 1.45 of the nominal value for a certain time.
For a power of 15 kW, the optimal solution is often the installation of an introductory machine on the 25 Ampere with the characteristic "C" (for household networks) or "D" (if there are powerful engines). However, if the house is planned to work simultaneously all consumers at the power limit, it is safer to consider the option with a machine on the device. 32 Ampere., but only if the cable cross section allows such a current to pass without overheating.
Table of power and current correspondence
For the convenience of designing and checking existing networks, electricians use reference tables. They allow you to quickly determine how many amperes the wire or machine will hold at a given power. Below are the data for the 380V three-phase network, taking into account the power factor of 0.95.
| Power (kW) | Current (A) | Recommended automatic (A) | Min. copper section (mm2) |
|---|---|---|---|
| 10 kW | 16.2 A | 20 A | 2.5 mm2 |
| 15 kW | 24.3 A | 25-32 A | 4.0 mm2 |
| 20 kW | 32.4 A | 40 A | 6.0 mm2 |
| 30 kW | 48.6 A | 50-63 A | 10.0 mm2 |
It is important to understand that the table values are averaged. The actual capacity of the cable depends on the way it is laid (open, in the pipe, in the ground) and the ambient temperature. For example, a cable laid in a beam with other wires will warm more, and its load capacity will have to be reduced.
When choosing a section, always round in the big direction. If the calculation shows 3.8 mm2, you can not take a cable of 4 mm2, if it works in difficult conditions, it is better to take 6 mm2. This will ensure the durability of the wiring and no voltage loss in long areas.
Effect of phase distortion on calculations
In an ideal world, the load is distributed evenly over three phases. But in reality, especially in private homes, it is difficult to achieve a perfect balance. If you plug all the powerful consumers into one phase, there will be a skew and the current in this phase will exceed the estimated value for 15 kW of total power.
The skewed phase is dangerous because switch-off It may not work due to general overload, but due to excess current in one particular branch. The three-phase machine will shut down all three phases at once, even if the other two were underloaded. It is a difficult situation, especially in winter.
To minimize risks, it is necessary to correctly distribute devices into groups. For example, kitchen outlets can be divided into two or three groups and connected to different phases. Powerful single-phase consumers, such as air-conditioners or boilers, should also be "scattered" on L1, L2 and L3.
How to check the skewed phases without devices
You can indirectly estimate the distortion in the brightness of the glow of incandescent lamps (if any) or by heating the wires in the shield. But the most reliable way is to use a multimeter or current mites to measure the current in each phase separately. The difference should not exceed 15-20%.
15 kW cable cross-section
Knowing that 15 kW for 3 phases consume about 25 Amps, you need to choose the cable of the appropriate cross section. Copper is the preferred material due to its high conductivity and flexibility. Aluminum is cheaper, but requires more cross-section and special measures on contact connections.
For current 25-32 Ampere minimum allowable cross-section of copper cable brand VWGng-LS or NYM It's 4 mm2. Such a cable in a single gasket can withstand current up to 35-40 Amps, which gives the necessary reserve. However, if the cable is laid in the ground or in high temperature conditions, it is better to increase the cross section to 6 mm2.
- π 2.5 mm2 - absolutely not enough for 15 kW, the cable will warm and melt.
- π 4.0 mm2 The minimum acceptable value for short-term work or ideal conditions.
- π 6.0 mm2 - the best choice for continuous use and stock for the future.
- π 10.0 mm2 - redundant for 15 kW, but justified at very long tracks (more than 80-100 meters) to reduce the voltage drop.
When calculating the length of the route, keep in mind that for every 100 meters of cable, the voltage drop should not exceed 5%. If the house is far from the pole, increasing the cable cross section is the only way to compensate for the loss and get a full 220/380 Volts at the entrance.
Useful tip: When buying a cable, always check the labeling on the insulation. Unscrupulous manufacturers can underestimate the cross section of the vein (the so-called "cable products according to TU", and not according to GOST). Use a bar to check the diameter of the vein before installation.
Errors in connection and operation
One common mistake is to use machines with a βBβ characteristic instead of a βCβ or βDβ. Automata type "B" are triggered by a short circuit current, which is 3-5 denominations, but sensitive to starting currents. For a network with engines (pumps, compressors) this will lead to constant knocking of traffic jams at the start of the equipment.
Another problem is poor contact at the connection points. If you use twists instead of terminals or loosely tightened screws in the machine, at the point of contact there will be a transition resistance. This will result in local heating, oxidation and eventual burnout of the contact, even if the total current in the network does not exceed 24 Amps.
β οΈ Warning: Never use aluminum and copper wires in the same twist. The aluminum-copper galvanic pair oxidizes rapidly, leading to heat and fire. Use only terminal pads or special transitional sleeves.
It is also worth remembering the temperature regime. Automatic switches have temperature compensation, but it works in a certain range. If the shield is installed in an unheated room or, conversely, under direct sunlight, the characteristics of the machine may shift.
βοΈ Checking readiness for connection of 15 kW
Differential protection and security
At 15 kW, the leakage currents can be more noticeable than in an apartment. To protect people from electric shock, protective shutdown devices must be installed (CCD) or differential automatics. The nominal leak current for the introductory RCD is usually 100 mA or 300 mA (fire protection), and for group lines is 30 mA.
It is important to choose the correct denomination of the RCD on the current. It must be equal to or greater than the denomination of the circuit breaker. If the machine costs 25 Amps, then the RCD should be at least 25 A, and better on a step higher - 40 A or 63 A. This will extend the life of the protection device.
The three-phase RCD controls the sum of currents in all three phases and the zero conductor. If the balance is disturbed (the current "leaks" through the person or damaged insulation), the device instantly breaks the chain. Ignoring this safety element at such facilities is unacceptable.
Main conclusion: For 15 kW for 3 phases, the design current is ~24A, but you need to choose an automatic and cable with a margin: automatic 25-32A, copper cable 4-6 mm2.
Frequently Asked Questions (FAQ)
Can I use a 16-amp machine for 15 kW?
No, the 16-Amp machine with a three-phase connection will withstand a power of about 10-11 kW. When you try to turn on a 15 kW load, it will constantly overheat and shut down after a while, since the current 24A is much higher than its nominal value.
What happens if you take a 2.5 mm2 cable for 15 kW?
The 2.5 mm2 cable is designed to handle a current of about 25-27 Amps only under ideal open gasket conditions. In reality, at a full load of 15 kW (24A), it will run at its limit, warm up strongly, the insulation will quickly age, which can lead to short circuit and fire.
How to distribute the load if 15 kW is the contract limit?
If you have a contract power of 15 kW, this means that the total current in the three phases should not exceed the calculated value. If the limit is exceeded, the power limiter will work (often built into the meter or a separate device), and the light will turn off. Watch for the uniformity of the loading phases, so that the machine does not knock out ahead of time.
Do I need a voltage stabilizer for 15 kW?
If the voltage in the mains is stable (380V +/- 10%), the stabilizer is not necessary. However, if there are strong jumps or "drawdowns" of voltage, a three-phase stabilizer with a capacity of at least 15-20 kVA (with a margin) will protect expensive equipment from failure.