Calculation of email loads. Calculation of electrical loads of apartments and cottages. Calculation of short circuit currents. The value of K m at K and
To ensure safety during the operation of household electrical appliances, it is necessary to correctly calculate the cross section of the supply cable and wiring. Since an erroneously selected cross-section of the cable cores can lead to a fire in the wiring due to a short circuit. This threatens to cause a fire in the building. This also applies to the choice of cable for connecting electric motors.
Current calculation
The magnitude of the current is calculated by power and is necessary at the stage of designing (planning) a dwelling - an apartment, a house.
- The value of this quantity depends choice of power cable (wire), through which power consumption devices can be connected to the network.
- Knowing the voltage of the electrical network and the full load of electrical appliances, it is possible by the formula calculate the amount of current that needs to be passed through the conductor(wire, cable). According to its size, the cross-sectional area of \u200b\u200bthe veins is selected.
If the electrical consumers in the apartment or house are known, it is necessary to perform simple calculations in order to properly mount the power supply circuit.
Similar calculations are performed for production purposes: definitions required area cross-section of the cable cores when connecting industrial equipment(various industrial electric motors and mechanisms).
Single-phase network with a voltage of 220 V
The current strength I (in amperes, A) is calculated by the formula:
I=P/U,
where P is the electrical full load (must be specified in technical passport devices), W (watt);
U is the voltage of the electrical network, V (volts).
The table below shows load values of typical household electrical appliances and their current consumption (for a voltage of 220 V).
electrical appliance | Power consumption, W | Current strength, A |
Washing machine | 2000 – 2500 | 9,0 – 11,4 |
Jacuzzi | 2000 – 2500 | 9,0 – 11,4 |
Electric floor heating | 800 – 1400 | 3,6 – 6,4 |
Stationary electric stove | 4500 – 8500 | 20,5 – 38,6 |
microwave | 900 – 1300 | 4,1 – 5,9 |
Dishwasher | 2000 - 2500 | 9,0 – 11,4 |
Freezers, refrigerators | 140 - 300 | 0,6 – 1,4 |
Meat grinder with electric drive | 1100 - 1200 | 5,0 - 5,5 |
Electric kettle | 1850 – 2000 | 8,4 – 9,0 |
Electric coffee maker | 6z0 - 1200 | 3,0 – 5,5 |
Juicer | 240 - 360 | 1,1 – 1,6 |
Toaster | 640 - 1100 | 2,9 - 5,0 |
Mixer | 250 - 400 | 1,1 – 1,8 |
hair dryer | 400 - 1600 | 1,8 – 7,3 |
Iron | 900 - 1700 | 4,1 – 7,7 |
A vacuum cleaner | 680 - 1400 | 3,1 – 6,4 |
Fan | 250 - 400 | 1,0 – 1,8 |
Television | 125 - 180 | 0,6 – 0,8 |
radio equipment | 70 - 100 | 0,3 – 0,5 |
Lighting devices | 20 - 100 | 0,1 – 0,4 |
The figure shows scheme of the apartment power supply device with a single-phase connection to a 220 V network.
As can be seen from the figure, various consumers of electricity are connected through the appropriate machines to the electric meter and then to the general machine, which must be designed for the load of devices that the apartment will be equipped with. The wire that supplies power must also satisfy the load of energy consumers.
The following is table for hidden wiring with a single-phase apartment connection scheme for picking up a wire at a voltage of 220 V
Wire core cross section, mm 2 | Conductor core diameter, mm | Copper conductors | Aluminum conductors | ||
Current, A | Power, W | Current, A | power, kWt | ||
0,50 | 0,80 | 6 | 1300 | ||
0,75 | 0,98 | 10 | 2200 | ||
1,00 | 1,13 | 14 | 3100 | ||
1,50 | 1,38 | 15 | 3300 | 10 | 2200 |
2,00 | 1,60 | 19 | 4200 | 14 | 3100 |
2,50 | 1,78 | 21 | 4600 | 16 | 3500 |
4,00 | 2,26 | 27 | 5900 | 21 | 4600 |
6,00 | 2,76 | 34 | 7500 | 26 | 5700 |
10,00 | 3,57 | 50 | 11000 | 38 | 8400 |
16,00 | 4,51 | 80 | 17600 | 55 | 12100 |
25,00 | 5,64 | 100 | 22000 | 65 | 14300 |
As can be seen from the table, the cross section of the conductors depends, in addition to the load, on the material from which the wire is made.
Three-phase network with a voltage of 380 V
With a three-phase power supply, the current strength I (in amperes, A) is calculated by the formula:
I = P / 1.73 U,
where P is the power consumption, W;
U - network voltage, V,
since the voltage for a three-phase power supply is 380 V, the formula will take the form:
I = P /657.4.
In the case of a three-phase 380 V power supply to the house, the connection diagram will look like this.
The cross section of the conductors in the supply cable at various loads with a three-phase circuit with a voltage of 380 V for hidden wiring is presented in the table.
Wire core cross section, mm 2 | Conductor core diameter, mm | Copper conductors | Aluminum conductors | ||
Current, A | Power, W | Current, A | power, kWt | ||
0,50 | 0,80 | 6 | 2250 | ||
0,75 | 0,98 | 10 | 3800 | ||
1,00 | 1,13 | 14 | 5300 | ||
1,50 | 1,38 | 15 | 5700 | 10 | 3800 |
2,00 | 1,60 | 19 | 7200 | 14 | 5300 |
2,50 | 1,78 | 21 | 7900 | 16 | 6000 |
4,00 | 2,26 | 27 | 10000 | 21 | 7900 |
6,00 | 2,76 | 34 | 12000 | 26 | 9800 |
10,00 | 3,57 | 50 | 19000 | 38 | 14000 |
16,00 | 4,51 | 80 | 30000 | 55 | 20000 |
25,00 | 5,64 | 100 | 38000 | 65 | 24000 |
To calculate the current in the load supply circuits, characterized by a large reactive apparent power, which is typical for the use of power supply in industry:
- electric motors;
- chokes of lighting devices;
- welding transformers;
- induction furnaces.
When calculating, this phenomenon must be taken into account. In powerful devices and equipment, the share of reactive load is higher and therefore for such devices in the calculations the power factor is assumed to be 0.8.
Before building a house, it is important to correctly design its load-bearing structures. The calculation of the load on the foundation will ensure the reliability of the supports under the building. It is carried out before the selection of the foundation after determining the characteristics of the soil.
The most important document in determining the weight of house structures is the SP "Loads and Impacts". It is he who regulates what loads fall on the foundation and how to determine them. According to this document, loads can be divided into the following types:
- permanent;
- temporary.
Temporary, in turn, are divided into long-term and short-term. Constants include those that do not disappear during the operation of the house (the weight of walls, partitions, floors, roofs, foundations). Temporary long-term is the mass of furniture and equipment, short-term - snow and wind.
Permanent loads
- the dimensions of the elements of the house;
- the material from which they are made;
- load safety factors.
Construction type | Weight |
Walls | |
From ceramic and silicate solid brick 380 mm thick (1.5 bricks) | 684 kg/m2 |
The same thickness 510 mm (2 bricks) | 918 kg/m2 |
The same 640 mm thick (2.5 bricks) | 1152 kg/m2 |
The same thickness 770 mm (3 bricks) | 1386 kg/m2 |
Made of ceramic hollow bricks 380 mm thick | 532 kg/m2 |
The same 510 mm | 714 kg/m2 |
The same 640 mm | 896 kg/m2 |
The same 770 mm | 1078 kg/m2 |
Made of silicate hollow brick 380 mm thick | 608 kg/m2 |
The same 510 mm | 816 kg/m2 |
The same 640 mm | 1024 kg/m2 |
The same 770 mm | 1232 kg/m2 |
From a bar (pine) 200 mm thick | 104 kg/m2 |
Same thickness 300mm | 156 kg/m2 |
Frame with insulation 150 mm thick | 50 kg/m2 |
Partitions and interior walls | |
Made of ceramic and silicate bricks (solid) 120 mm thick | 216 kg/m2 |
Same thickness 250mm | 450 kg/m2 |
Made of ceramic hollow bricks 120 mm (250 mm) thick | 168 (350) kg/m2 |
From silicate brick hollow 120 mm thick (250 mm) | 192 (400) kg/m2 |
From drywall 80 mm without insulation | 28 kg/m2 |
From drywall 80 mm with insulation | 34 kg/m2 |
Overlappings | |
Reinforced concrete solid 220 mm thick with cement-sand screed 30 mm | 625 kg/m2 |
Reinforced concrete from hollow core slabs 220 mm with screed 30 mm | 430 kg/m2 |
Wooden on beams with a height of 200 mm with the condition of laying insulation with a density of not more than 100 kg / m 3 (at lower values, a margin of safety is provided, since independent calculations do not have high accuracy) with laying parquet, laminate, linoleum or carpet as a floor covering | 160 kg/m2 |
Roof | |
Coated with ceramic tiles | 120 kg/m2 |
From bituminous tiles | 70 kg/m2 |
From metal tiles | 60 kg/m2 |
- soil freezing depth;
- groundwater level;
- the presence of a basement.
When lying on the site of coarse and sandy soils (medium, large), you can not deepen the sole of the house by the amount of freezing. For clays, loams, sandy loams and other unstable bases, it is necessary to bookmark the depth of soil freezing in winter. It can be determined by the formula in the Joint Venture "Foundations and Foundations" or by maps in the SNiP "Construction Climatology" (this document has now been canceled, but in private construction it can be used for informational purposes).
When determining the location of the sole of the foundation of the house, it is important to control that it is located at a distance of at least 50 cm from the groundwater level. If the building has a basement, then the base mark is taken 30-50 cm below the floor mark of the room.
Having decided on the depth of freezing, you will need to choose the width of the foundation. For tape and columnar, it is taken depending on the thickness of the wall of the building and the load. For slabs, they are assigned so that the supporting part extends beyond the outer walls by 10 cm. For piles, the section is assigned by calculation, and the grillage is selected depending on the load and thickness of the walls. You can use the definition recommendations from the table below.
foundation type | Weight determination method |
Tape reinforced concrete | Multiply the width of the tape by its height and length. The resulting volume must be multiplied by the density of reinforced concrete - 2500 kg / m 3. Recommended: . |
Slab reinforced concrete | The width and length of the building are multiplied (20 cm are added to each size for protrusions on the boundaries of the outer walls), then multiplication is performed by the thickness and density of reinforced concrete. Recommended: . |
Columnar reinforced concrete | The cross-sectional area is multiplied by the height and density of reinforced concrete. The resulting value must be multiplied by the number of supports. In this case, the mass of the grillage is calculated. If the foundation elements have a widening, it must also be taken into account in the volume calculations. Recommended: . |
pile bored | The same as in the previous paragraph, but you need to take into account the mass of the grillage. If the grillage is made of reinforced concrete, then its volume is multiplied by 2500 kg / m 3, if from wood (pine), then by 520 kg / m 3. When manufacturing a grillage from rolled metal, you will need to familiarize yourself with the assortment or passport for products, which indicate the mass of one linear meter. Recommended: . |
Pile screw | For each pile, the manufacturer specifies the weight. It is necessary to multiply by the number of elements and add the mass of the grillage (see the previous paragraph). Recommended: . |
The calculation of the load on the foundation does not end there. For each structure in the mass, it is necessary to take into account the load safety factor. Its value for various materials is given in the joint venture "Loads and effects". For metal, it will be equal to 1.05, for wood - 1.1, for reinforced concrete and reinforced masonry structures of factory production - 1.2, for reinforced concrete, which is made directly at the construction site - 1.3.
Live loads
The easiest way to deal with the useful here. For residential buildings, it is 150 kg / m2 (determined based on the floor area). The reliability coefficient in this case will be equal to 1.2.
Snow depends on the construction area. To determine the snowy area, the Construction Climatology Joint Venture will be required. Further, by the number of the district, the magnitude of the load is found in the joint venture “Loads and impacts”. The reliability factor is 1.4. If the roof slope is more than 60 degrees, then the snow load is not taken into account.
Determining the value for the calculation
When calculating the foundation of a house, not its total mass will be required, but the load that falls on a certain area. The actions here depend on the type of building support structure.
foundation type | Actions in the calculation |
Tape | To calculate the strip foundation according to bearing capacity you need a load per linear meter, based on it the area of \u200b\u200bthe sole is calculated for the normal transfer of the mass of the house to the base, based on the bearing capacity of the soil (the exact value of the bearing capacity of the soil can only be found using geological surveys). The mass obtained in the collection of loads must be divided by the length of the tape. At the same time, the foundations for internal load-bearing walls are also taken into account. This is the easiest way. For a more detailed calculation, you will need to use the method of cargo areas. To do this, determine the area from which the load is transferred to a certain area. This is a time-consuming option, so when building a private house, you can use the first, simpler, method. |
slab | You will need to find the mass per square meter of the slab. The load found is divided by the area of the foundation. |
Column and pile | Usually, in private housing construction, the section of piles is predetermined and then their number is selected. To calculate the distance between the supports, taking into account the selected section and the bearing capacity of the soil, you need to find the load, as in the case of strip foundation. Divide the mass of the house by the length of the load-bearing walls under which the piles will be installed. If the step of the foundations turns out to be too large or small, then the cross section of the supports is changed and the calculation is performed again. |
Calculation Example
It is most convenient to collect loads on the foundation of a house in tabular form. The example is considered for the following initial data:
- the house is two-storey, the floor height is 3 m, the dimensions in the plan are 6 by 6 meters;
- foundation tape reinforced concrete monolithic 600 mm wide and 2000 mm high;
- solid brick walls 510 mm thick;
- monolithic reinforced concrete floors 220 mm thick with cement-sand screed 30 mm thick;
- hip roof (4 slopes, which means that the outer walls on all sides of the house will be the same height) covered with metal tiles with a slope of 45 degrees;
- one inner wall in the middle of the house made of bricks 250 mm thick;
- the total length of drywall partitions without insulation with a thickness of 80 mm is 10 meters.
- snow construction area ll, roof load 120 kg/m2.
Load definition | Reliability factor | Estimated value, tons |
Foundation 0.6 m * 2 m * (6 m * 4 + 6 m) \u003d 36 m 3 - foundation volume 36 m 3 * 2500 kg / m 3 \u003d 90000 kg \u003d 90 tons |
1,3 | 117 |
Exterior walls 6 m * 4 pcs \u003d 24 m - the length of the walls 24 m * 3 m \u003d 72 m 2 - area within one floor (72 m 2 * 2) * 918 kg / m 2 - 132192 kg \u003d 133 tons - the mass of the walls of two floors |
1,2 | 159,6 |
Internal walls 6 m * 2 pcs * 3 m = 36 m 2 wall area over two floors 36 m 2 * 450 kg / m 2 \u003d 16200 kg \u003d 16.2 tons - weight |
1,2 | 19,4 |
Overlappings 6 m * 6 m \u003d 36 m 2 - floor area 36 m 2 * 625 kg / m 2 \u003d 22500 kg \u003d 22.5 tons - weight of one floor 22.5 t * 3 \u003d 67.5 tons - the mass of the basement, interfloor and attic floors |
1,2 | 81 |
Partitions 10 m * 2.7 m (here, not the height of the floor is taken, but the height of the room) \u003d 27 m 2 - area 27 m 2 * 28 kg / m 2 \u003d 756 kg \u003d 0.76 t |
1,2 | 0,9 |
Roof (6 m * 6 m) / cos 45ᵒ (roof slope angle) \u003d (6 * 6) / 0.7 \u003d 51.5 m 2 - roof area 51.5 m 2 * 60 kg / m 2 \u003d 3090 kg - 3.1 tons - weight |
1,2 | 3,7 |
Payload 36m 2 * 150 kg / m 2 * 3 \u003d 16200 kg \u003d 16.2 tons (the floor area and their number are taken from previous calculations) |
1,2 | 19,4 |
snowy 51.5 m 2 * 120 kg / m 2 \u003d 6180 kg \u003d 6.18 tons (roof area taken from previous calculations) |
1,4 | 8,7 |
To understand the example, this table must be viewed in conjunction with the one in which the masses of structures are given.
Next, you need to add up all the obtained values. Total load for this example on the foundation, taking into account its own weight, is 409.7 tons. To find the load per linear meter of the tape, it is necessary to divide the obtained value by the length of the foundation (calculated in the first line of the table in brackets): 409.7 tons / 30 m = 13.66 t / m.p. This value is taken for calculation.
When finding mass at home, it is important to follow the steps carefully. It is best to devote enough time to this design stage. If you make a mistake in this part of the calculations, then you may have to redo the entire calculation of the bearing capacity, and this is an additional cost of time and effort. Upon completion of the collection of loads, it is recommended to double-check it to eliminate typos and inaccuracies.
The calculation of the load on the foundation is necessary for right choice its geometric dimensions and the area of the base of the foundation. Ultimately, the strength and durability of the entire building depends on the correct calculation of the foundation. The calculation comes down to determining the load per square meter of soil and comparing it with the allowable values.
To calculate, you need to know:
- The region in which the building is being built;
- Soil type and groundwater depth;
- The material from which the structural elements of the building will be made;
- The layout of the building, the number of storeys, the type of roof.
Based on the required data, the calculation of the foundation or its final verification is carried out after the design of the building.
Let's try to calculate the load on the foundation for one-story house, made of solid brick, solid masonry, with a wall thickness of 40 cm. The dimensions of the house are 10x8 meters. The ceiling of the basement is reinforced concrete slabs, the ceiling of the 1st floor is wooden on steel beams. The roof is gable, covered with metal tiles, with a slope of 25 degrees. Region - Moscow region, soil type - wet loams with a porosity coefficient of 0.5. The foundation is made of fine-grained concrete, the wall thickness of the foundation for calculation is equal to the thickness of the wall.
Determining the depth of the foundation
The depth of laying depends on the depth of freezing and the type of soil. The table shows the reference values of the depth of soil freezing in various regions.
Table 1 - Reference data on the depth of soil freezing
Reference table for determining the depth of the foundation by region
The depth of the foundation should generally be more depth freezing, but there are exceptions due to the type of soil, they are indicated in table 2.
Table 2 - Dependence of the depth of the foundation on the type of soil
The depth of the foundation is necessary for the subsequent calculation of the load on the soil and determining its size.
We determine the depth of soil freezing according to table 1. For Moscow, it is 140 cm. According to table 2, we find the type of soil - loam. The laying depth must not be less than the estimated freezing depth. Based on this, the depth of the foundation for the house is selected 1.4 meters.
Roof load calculation
The load of the roof is distributed between those sides of the foundation, on which the truss system rests through the walls. For an ordinary gable roof, these are usually two opposite sides of the foundation, for a four-pitched roof, all four sides. The distributed load of the roof is determined by the area of the projection of the roof, referred to the area of the loaded sides of the foundation, and multiplied by specific gravity material.
Table 3 - Specific gravity of different types of roofing
Reference table - Specific gravity of different types of roofing
- We determine the area of the projection of the roof. The dimensions of the house are 10x8 meters, the projection area of the gable roof is equal to the area of the house: 10 8 = 80 m 2.
- The length of the foundation is equal to the sum of its two long sides, since the gable roof rests on two long opposite sides. Therefore, the length of the loaded foundation is defined as 10 2 = 20 m.
- The area of the foundation loaded with a roof with a thickness of 0.4 m: 20 0.4 \u003d 8 m 2.
- The type of coating is metal tiles, the slope angle is 25 degrees, which means that the calculated load according to table 3 is 30 kg / m 2.
- The load of the roof on the foundation is 80/8 30 \u003d 300 kg / m 2.
Snow load calculation
The snow load is transferred to the foundation through the roof and walls, so the same sides of the foundation are loaded as in the calculation of the roof. The area of snow cover is calculated equal to the area of the roof. The resulting value is divided by the area of the loaded sides of the foundation and multiplied by the specific snow load determined from the map.
Table - calculation of snow load on the foundation
- The length of the slope for a roof with a slope of 25 degrees is (8/2) / cos25 ° = 4.4 m.
- The roof area is equal to the length of the ridge multiplied by the length of the slope (4.4 10) 2 \u003d 88 m 2.
- The snow load for the Moscow region on the map is 126 kg / m 2. We multiply it by the area of the roof and divide by the area of the loaded part of the foundation 88 126 / 8 = 1386 kg / m 2.
Floor load calculation
Ceilings, like the roof, usually rest on two opposite sides of the foundation, so the calculation is carried out taking into account the area of \u200b\u200bthese sides. The floor area is equal to the area of the building. To calculate the floor load, you need to take into account the number of floors and the basement floor, that is, the floor of the first floor.
The area of each overlap is multiplied by the specific gravity of the material from table 4 and divided by the area of the loaded part of the foundation.
Table 4 - Specific gravity of floors
- The floor area is equal to the area of \u200b\u200bthe house - 80 m 2. The house has two floors: one of reinforced concrete and one of wood on steel beams.
- Multiplying the area reinforced concrete floor per specific weight from table 4: 80 500=40000 kg.
- We multiply the area of \u200b\u200bthe wooden floor by the specific gravity from table 4: 80 200 \u003d 16000 kg.
- We summarize them and find the load on 1 m 2 of the loaded part of the foundation: (40000 + 16000) / 8 = 7000 kg / m 2.
Wall load calculation
The load of the walls is determined as the volume of the walls, multiplied by the specific gravity from table 5, the result is divided by the length of all sides of the foundation, multiplied by its thickness.
Table 5 - Specific gravity of wall materials
Table - Specific weight of walls
- The wall area is equal to the height of the building multiplied by the perimeter of the house: 3 (10 2 + 8 2) = 108 m 2.
- The volume of the walls is the area multiplied by the thickness, it is equal to 108 0.4 \u003d 43.2 m 3.
- We find the weight of the walls by multiplying the volume by the specific gravity of the material from table 5: 43.2 1800 \u003d 77760 kg.
- The area of all sides of the foundation is equal to the perimeter multiplied by the thickness: (10 2 + 8 2) 0.4 \u003d 14.4 m 2.
- The specific load of the walls on the foundation is 77760/14.4=5400 kg.
Preliminary calculation of the foundation load on the ground
The load of the foundation on the soil is calculated as the product of the volume of the foundation and the specific density of the material from which it is made, divided by 1 m 2 of its base area. The volume can be found as the product of the depth of the foundation and the thickness of the foundation. The thickness of the foundation is taken in the preliminary calculation equal to the thickness of the walls.
Table 6 - Specific density of foundation materials
Table - specific gravity of the soil material
- The foundation area is 14.4 m 2, the laying depth is 1.4 m. The volume of the foundation is 14.4 1.4 \u003d 20.2 m 3.
- The mass of the foundation made of fine-grained concrete is equal to: 20.2 1800 = 36360 kg.
- Ground load: 36360 / 14.4 = 2525 kg / m 2.
Calculation of the total load per 1 m 2 of soil
The results of previous calculations are summarized, and the maximum load on the foundation is calculated, which will be greater for those sides on which the roof rests.
The conditional design soil resistance R 0 is determined according to the tables of SNiP 2.02.01-83 "Foundations of buildings and structures".
- We sum up the weight of the roof, the snow load, the weight of the floors and walls, as well as the foundation on the ground: 300 + 1386 + 7000 + 5400 + 2525 \u003d 16 611 kg / m 2 \u003d 17 t / m 2.
- We determine the conditional design soil resistance according to the tables of SNiP 2.02.01-83. For wet loams with a porosity coefficient of 0.5, R 0 is 2.5 kg/cm 2 or 25 t/m 2 .
It can be seen from the calculation that the load on the ground is within the permissible range.
Design of electrical installations of apartments and cottages (Schneider Electric)
2.1. Calculation of electrical loads
At the initial design stage, when the exact data of power receivers are practically unknown, but it is necessary to obtain technical conditions for connecting electrical power, the question arises of how to calculate the amount of installed power of consumers and, on this basis, determine the estimated load at the entrance to an apartment or cottage. At the same time, the concept of the calculated electrical load Рр of a consumer or a network element means a power equal to the expected maximum load in 30 minutes.
In the Standards for determining the calculated electrical loads of buildings (apartments), cottages, microdistricts (quarters) of development and elements of the urban distribution network (changes and additions to the Instructions for the design of urban electrical networks- RD 34.20.185-94) specific design loads are given.
These Standards are based on an analysis of the power consumption modes of a promising set of electrical appliances and machines in an apartment (cottage). Data on the installed capacity of devices and machines were taken into account, the daily consumption of electricity, the possible operating time of each device and machine was determined.
In specific design loads, it is taken as a basis that the design load of an individual apartment (cottage) or a small number of apartments (cottages) is determined by occasional use devices, but of a significant installed capacity. Such appliances include, for example, hot water washing machines, hot tubs, hot water dishwashers, electric kettles, electric saunas, etc. Demand factors were determined for these appliances and then summed up their calculated loads with the loads of all other low power appliances, which were determined using the average value of the demand coefficient.
The developers of the Standards adopted as basic initial data:
1. Average apartment area (total), m2:
in standard buildings of mass development 70
in buildings with superior apartments
(elite) on individual projects 150
2. Area (total) of the cottage, m2 50 - 600
3. Average family, persons 3.1
4. Installed power, kW:
apartments with gas stoves 21,4
apartments with electric stoves in standard buildings 32.6
apartments with electric stoves in luxury buildings 39.6
cottages with gas stoves 35.7
cottages with gas stoves and electric saunas 48.7
cottages with electric stoves 47.9
cottages with electric stoves and electric saunas 59.9
In table. 2.1 shows the specific design load of electrical receivers of apartments in residential buildings, and in table. 2.2 - cottages.
In the "Temporary Instructions for Calculating the Electrical Loads of Residential Buildings" RM2696-01, it is recommended that the design load at the entrance to an apartment for houses of category I be determined by the formula:
where Pz is the declared power of electrical receivers, determined by summing up the rated powers of household and lighting appliances, as well as the socket network;
Table 2.1 Specific calculated electrical load of electrical receivers of apartments in residential buildings
Electricity consumers | Specific design electrical load, kW/apartment, with the number of apartments |
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Flats with stoves: | ||||||||||||||
On natural gas: | ||||||||||||||
On liquefied gas (including for group installations) and on solid fuel: | ||||||||||||||
Electrical power up to 8.5 kW | ||||||||||||||
Superior apartments with electric stoves up to 10.5 kW |
Table 2.2 Specific calculated electrical load of electrical receivers of cottages
Electricity consumers | Specific design electrical load, kW / cottage, with the number of cottages |
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Cottage with natural gas stoves | ||||||||||
Cottages with natural gas stoves and electric sauna up to 12 kW | ||||||||||
Cottages with electric stoves up to 10.5 kW | ||||||||||
Cottages with electric stoves up to 10.5 kW and electric sauna up to 12 kW |
Кс - demand coefficient, depending on the value of the declared capacity in the apartment.
In accordance with the "Temporary Instructions ..." at the pre-design stages, it is recommended to determine the design loads according to the estimated specific loads in accordance with Table 2.3, depending on the various levels of household electrification, and at the detailed design stage, the loads are specified according to the above formula.
In table. 2.3 when determining the specific loads, the following power of electrical receivers was taken, kW: lighting 2.8, socket network 2.8, electric stoves 9-10.5, washing machine 2.2, dishwasher 2.2, heated jacuzzi 2.5, shower heated cabin 3, storage water heater 2, instantaneous water heater 8-18, air conditioners 3, household appliances 4, heated floors 1.
Table 2.3 Approximate specific loads for houses of category I
Characteristics of apartments | Specific load, kW/apartment with the number of apartments |
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1 Houses with electric stoves up to 9 kW without saunas, instantaneous water heaters and air conditioners | 600 and more |
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2 Houses with electric stoves up to 10.5 kW: | ||||||||||||||
2.1 Without saunas and instantaneous water heaters | ||||||||||||||
water heaters up to 12 kW | ||||||||||||||
2.2 Without saunas, but with flow | ||||||||||||||
2.3 Without saunas, but with instantaneous water heaters up to 18 kW | ||||||||||||||
2.4 With saunas up to 12 kW, without instantaneous water heaters | ||||||||||||||
2.5 With saunas up to 6 kW and instantaneous water heaters up to 8 kW | ||||||||||||||
2.6 With saunas up to 12 kW and instantaneous water heaters up to 12 kW |
It needs to be explained that main goal The developers of these Standards and Instructions were to determine the average design loads, reduced to input into residential buildings or cottage settlements, based on the initial data taken as the base.
In SP31-110-2003, it is recommended that the design load for apartments with increased comfort be determined in accordance with the design assignment or in accordance with the declared capacity and demand and simultaneity factors.
Demand factors for a superior apartment:
Declared power, kW Up to 14 20 30 40 50 60 70 and more
Demand factor 0.8 0.65 0.6 0.55 0.5 0.48 0.45
Coefficients of simultaneity Ko for an apartment of increased comfort:
Number of apartments 1-5 6 9 12 15 18
Simultaneity factor. . . 1 0.51 0.38 0.32 0.29 0.26
Number of apartments 24 40 60 100 200 400 600 or more
Simultaneity factor. . . . 0.24 0.2 0.18 0.16 0.14 0.13 0.11
The calculated load of supply lines, inputs and on the RU-0.4 kV busbars of the transformer substation from electrical receivers of luxury apartments Рr.kv kW is determined by the formula:
where Rkv is the load of electrical receivers of superior apartments; n - number of apartments; Ko - the coefficient of simultaneity for apartments of superior comfort.
In SP31-106-2002, for single-apartment residential buildings, the design load, in cases where there are no restrictions, is also recommended to be determined on the instructions of the customer. However, when the possibilities of power supply are limited, the calculated load of electrical receivers should be taken at least:
5.5 kW - for houses without electric stoves;
8.8 kW - for houses with electric stoves.
If the total area of the house exceeds 60 m2, the design load must be increased by 1% for each additional 1 m2.
In real cases, the areas of luxury apartments and cottages differ significantly from the basic ones and do not have an upper limit on the level of household electrification.
Each individual apartment or cottage with outbuildings is its own microcosm, filled not with average, but with actual consumers of electricity, the rated power of which may differ significantly from those adopted in regulatory materials.
In principle, the specific design loads could not take into account the use by the customer of various, more and more advanced consumers with a long operating mode (more than 30 minutes), constantly appearing on the market for the comfort of housing and people's lives.
In table. 2.4 compiled according to the data normative documents, the results of the analysis of a large number of projects, the passport data of household electrical appliances, the recommended power values of individual electrical receivers and the calculated coefficients are given.
Determination of the calculated value Рр.р of the load of group and supply lines from electrical receivers connected to sockets is supposed to be carried out according to the recommendation given in SP31-110-2003 for hostels, according to the formula:
where Rud - specific power per outlet, with the number of outlets up to 100, taken 0.1, over 100 - 0.06 kW;
np - number of outlets;
Ko.r - coefficient of simultaneity for a network of sockets, determined depending on the number
Up to 10 sockets. . . .1.0
Over 10 to 20 outlets. . . .0.9
Over 20 to 50 outlets. . . .0.8
Over 50 to 100 outlets. . . .0.7
Over 100 to 200 outlets. . .0.6
Over 200 to 400 outlets. . .0.5
Over 400 to 600 outlets. . .0.4
Over 650 outlets. . . .0.35
The main calculated coefficients are: demand factor Kc, utilization factor Ki and power factor cosf.
The load demand coefficient is understood as the ratio of the calculated electrical load to the rated (installed) power of electrical receivers:
where Рр is the calculated electrical load, kW (30-min maximum); Ru - installed power of electrical receivers, kW.
Name electrical receivers | Rated or installed active power | Estimated coefficients | Note |
||
Demand X | Ki usage |
||||
Electrical lighting in living rooms | Lighting fixtures with incandescent lamps |
||||
Electric lighting of living rooms (bedrooms) | |||||
Electric lighting for classrooms, libraries, playrooms, etc. | |||||
Electric lighting for kitchens | |||||
Electric lighting of halls, corridors, etc. | |||||
Household socket network (TV and radio equipment, refrigerators, vacuum cleaners, irons, floor lamps, sconces, table lamps, etc.) | 100W/socket | 1 socket per 6 m2 of total area Ki \u003d 0.7 - with the number of outlets more than 50; Ki \u003d 0.8 - with the number of outlets from 20 to 50; Ki \u003d 0.9 - with the number of outlets from 10 to 20; Ki \u003d 1 - with the number of sockets up to 10 |
|||
electric stove | 10.5 kW/ppt | ||||
Washing machine | |||||
Dishwasher | |||||
Heated Jacuzzi | |||||
Shower cabin with heating | |||||
Storage water heaters | |||||
Flowing water heaters | |||||
Air conditioners | |||||
Electric fireplaces | |||||
Food processors, coffee makers, electric kettles, etc. (total) | 4-5 kW/apartment | ||||
Underfloor heating in living room, kitchen, hallway | |||||
Heated floor in the bathroom, sauna, nursery | |||||
Electric heating boilers | |||||
Electric heating devices | |||||
fan heaters | |||||
Electric heaters | |||||
Lawn mowers | |||||
Submersible pumps | |||||
Personal computers |
The active power utilization factor of one or a group of power receivers is understood as the ratio of the actual power consumption P to the rated power Rn:
Table 2.5 Source data for example
Premises | Area, m2 | Installed electrical appliances | Rated (installed) power, kW | Note |
Electric stove | Tab. 2.4 item 7 |
|||
Dishwasher | Tab. 2.4 p. 9 |
|||
Fridge | According to passport data |
|||
Food processor | Tab. 2.4 item 17 |
|||
electric lighting | Tab. 2.4 p. 4 |
|||
1 socket for a current of 16 A, 4 sockets for a current of 6 A | Tab. 2.4 p. 6 |
|||
Hall and corridors | electric lighting | Tab. 2.4 p. 5 |
||
6 sockets for a current of 6 A | Tab. 2.4 p. 6 |
|||
Tab. 2.4 item 11 |
||||
Electric shower | Tab. 2.4 item 12 |
|||
Warm floor (4 m2) | Tab. 2.4 item 19 |
|||
Fan | According to passport data |
|||
electric lighting | Tab. 2.4 p. 5 |
|||
4 sockets for current 6 A | Tab. 2.4 p. 6 |
|||
Electric shower | Tab. 2.4 item 12 |
|||
Warm floor (4 m2) | Tab. 2.4 item 19 |
|||
Fan | According to passport data |
|||
Washing machine | Tab. 2.4 p. 8 |
|||
electric lighting | Tab. 2.4 p. 5 |
|||
2 sockets for current 6 A | Tab. 2.4 p. 6 |
|||
Living room | Electric fireplace | Tab. 2.4 item 16 |
||
Air conditioner | Tab. 2.4 item 15 |
|||
home theater | According to passport data |
|||
electric lighting | Tab. 2.4 p. 1 |
|||
10 sockets for a current of 6 A | Tab. 2.4 p. 6 |
|||
Bedroom 1 | Warm floor (12 m2) | Tab. 2.4 item 18 |
||
Air conditioner | Tab. 2.4 item 15 |
|||
electric lighting | Tab. 2.4 p. 2 |
|||
4 sockets for current 6 A | Tab. 2.4 p. 6 |
|||
Bedroom 2 | Warm floor (10 m2) | Tab. 2.4 item 18 |
||
Air conditioner | Tab. 2.4 item 15 |
|||
electric lighting | Tab. 2.4 p. 2 |
|||
4 sockets for current 6 A | Tab. 2.4 p. 6 |
|||
Children's room | Warm floor (20 m2) | Tab. 2.4 item 18 |
||
Air conditioner | Tab. 2.4 item 15 |
|||
Personal Computer | Tab. 2.4 item 26 |
|||
electric lighting | Tab. 2.4 p. 3 |
|||
4 sockets for current 6 A | Tab. 2.4 p. 6 |
|||
Air conditioner | Tab. 2.4 item 15 |
|||
Personal Computer | Tab. 2.4 item 26 |
|||
electric lighting | Tab. 2.4 p. 3 |
|||
4 sockets for current 6 A | Tab. 2.4 p. 6 |
|||
In practical cases, for a number of consumers, such as power outlets and electric lighting, the utilization factor coincides with the simultaneity factor Ko for this group of consumers.
Initial data:
Apartment with a total area of 200 m2 in apartment building. The apartment has 5 rooms, a kitchen,
2 bathrooms, hall and corridors. In table. 2.5 shows the initial data on the installed household electrical equipment. All consumers, with the exception of the electric stove, are single-phase.
Calculation of loads.
Based on the data in Table. 2.5 we make the calculation table of the table. 2.6, which includes the estimated coefficients of demand and use, taken from Table. 2.4.
The power factors are taken from the data given in §1.3.
In table. 2.6 installed capacities of the same type of electrical receivers (for example, electric lighting, household socket network, fans, underfloor heating) are summed up..
Table 2.6 Calculation table for example No. 1
Name of groups of electrical consumers or individual electrical receivers | Installed (rated) power, kW | Estimated coefficients | Estimated power | Note |
|||
demandKs | Ki usage | power cosph/tgph | active | complete |
|||
electric lighting | Accepted everywhere incandescent lamps |
||||||
Household socket network | |||||||
Electric stove | |||||||
Dishwasher | |||||||
Fridge | |||||||
Food processor | |||||||
Air conditioners | |||||||
Washing machine | |||||||
Warm floor | |||||||
Electric shower | |||||||
Fans | |||||||
Electric fireplace | |||||||
home theater | |||||||
Personal computers | |||||||
Estimated active power (kW) of each group of power receivers is determined by the formula
Total power of each group of electrical receivers, kVA:
Considering that all loads, except for the electric stove, are single-phase, and the supply network is three-phase, neglecting the uneven load of the phases, at the input to the apartment we get the estimated current:
We choose for installation at the entrance to the apartment a three-phase, four-pole circuit breaker for a rated current of 63 A.
In table. Tables 2.7 and 2.8 show the recommended power ratings for power consumers in elite apartments, cottages and individual buildings on household plots. The recommended values are determined based on the analysis of a large number of projects completed in recent years.
In table. 2.7 and 2.8, installed power means the total power of consumers, the duration of which usually exceeds 1 hour. Occasional consumers are taken into account in the total power of the outlet network. The design power takes into account reduction factors for individual consumers and a general factor of 0.8, which takes into account the simultaneous operation of all consumers.
Total area of the elite apartment, m2 | Plate | Note |
||
established | estimated |
|||
Kitchen, living room, bedroom, nursery, bathroom, hall |
||||
Electrical | ||||
Kitchen, living room, 2 bedrooms, nursery, 2 bathrooms, hall |
||||
Electrical | ||||
Kitchen, living room, 2 bedrooms, 2 bathrooms, jacuzzi, nursery, library, hall |
||||
Electrical | ||||
Kitchen, living room, 2 bedrooms, 2 bathrooms, jacuzzi, nursery, library, winter garden, hall |
||||
Electrical |
The total area of the cottage or individual buildings on the site, m2 | Plate, heating | Note |
||
Installed | Estimated |
|||
Cottage 150 | Electric heating, water heaters, submersible pump, underfloor heating |
|||
Electrical | ||||
Cottage 250 | Electric boiler, water heaters, submersible pump, heated floors |
|||
Electrical | ||||
Cottage 300 | ||||
Electrical | ||||
Cottage 400 | ||||
Electrical | ||||
Cottage 500 | ||||
Electrical | ||||
Cottage 600 | ||||
Electrical | ||||
Guest house 100 | ||||
Electrical | ||||
wood burning | Electric heating, water heaters, underfloor heating |
|||
Electrical | ||||
Garage for two cars 40 | ||||
Greenhouse with electric heating | ||||
Electric lighting of the territory and artistic lighting | Plot area 0.2 ha |
2.2. Calculation of short circuit currents
Short circuit current (SC) calculations are performed for:
Selection and testing of electrical equipment for electrodynamic and thermal resistance;
Determining the settings and ensuring the selectivity of the protection operation at the inputs to the apartment or cottage.
This primarily applies to the choice of circuit breakers.
The main documents regulating the procedure for calculating short-circuit currents are:
GOST 28249-93 "Short circuits in electrical installations. Calculation methods in electrical installations of alternating current with voltage up to 1 kV;
Guidelines for the calculation of short-circuit currents and the choice of electrical equipment - RD 153-34.0-20.527-98 RAO UES of Russia, (2002).
Various methods for calculating short-circuit currents are reflected in sufficient detail in the technical literature. In this paper, based on published materials, only the data that are necessary for calculating short-circuit currents in the implementation of power supply projects for elite housing, and, first of all, for power supply of estates and cottages, are given.
When calculating short-circuit currents in electrical installations up to 1 kV, it is necessary to take into account the active and inductive resistances of all elements of the short-circuited circuit, including power transformers, current transformers, reactors, current coils of circuit breakers and conductors. It is also necessary to take into account:
Change in the active resistance of conductors in a short-circuited circuit due to their heating during a short circuit;
The resistance of the electric arc at the location of the short circuit.
When compiling equivalent equivalent circuits, the parameters of the elements of the original design scheme should be reduced to the network voltage level at which the short circuit point is located.
When calculating short-circuit currents, it is allowed:
Simplify as much as possible the entire external network in relation to the place of short circuit, presenting it with an infinite power system with zero resistance;
Take the transformation ratios of transformers equal to the ratio of the average rated voltages of those voltage steps that connect the transformers. Values of average rated voltages: 10.5; 6.3; 0.4; 0.23 kV.
In electrical installations that are powered directly from the power grid, it is generally accepted that step-down transformers are connected to a source of constant voltage amplitude through the equivalent inductive resistance of the system. The value of this resistance (xc), reduced to the lowest voltage stage of the network, is calculated by the formula (mΩ)
where Uav.n.n is the average rated voltage of the network connected to the low voltage winding of the transformer, V;
Uav.n - average rated voltage of the network to which the higher voltage winding of the transformer is connected, V;
Ikv.n \u003d In0.v.n - effective value of the periodic component of the current in a three-phase short circuit at the terminals of the higher voltage winding of the transformer, kA;
Sk - conditional short-circuit power at the terminals of the higher voltage winding of the transformer, MV^A.
In the absence of the specified data, the equivalent inductive reactance of the system can be calculated using the formula (mΩ):
where Iot.nom is the rated breaking current of the circuit breaker installed on the high voltage side of the step-down transformer, kA.
In cases where a step-down transformer is connected to the power system network through a reactor, overhead or cable line (more than 1 km long), it is necessary to take into account not only inductive, but also active resistances of these elements.
Calculations of short-circuit currents in electrical installations with voltage up to 1 kV are recommended to be made in named units.
The active and inductive resistance of the step-down transformer (RT, XT) reduced to the low voltage stage of the network is calculated by the formulas, mOhm:
where St.nom is the rated power of the transformer, kV * A; Рк.з - losses of short circuit in the transformer, kW; Un.n.nom - rated voltage of the low voltage winding of the transformer, kV; Uк - transformer short circuit voltage, %.
In table. 2.9 shows the active and inductive resistances of transformers, reduced to a voltage of 0.4 kV.
Table 2.9 Resistance of step-down transformers with a secondary voltage of 0.4 kV
Rated power, | connections | Voltage short closures | Resistance, mOhm |
||||||||
direct sequence | zero sequence | single-phase short circuit current |
|||||||||
active | inductive | active | inductive | active | inductive | ||||||
where R0sh and X0sh - specific active and reactive resistance of the busbar, Ohm / m;
lш - busbar length, m.
The resistances of prefabricated bus ducts of the ShRA and ShMA types are given in Table 2.10.
Table 2.10 Resistance values of complete busbars
busbar | Rated current, A | Phase resistance, mOhm/m | Zero conductor resistance, mOhm/m |
||
active | inductive | active | inductive |
||
In the absence of data, the resistance of the busbar from the transformer to the circuit breaker can be taken approximately: Rsh = 0.5 mOhm, Khsh = 0.25 mOhm.
Active and inductive resistance of overhead lines (VL):
Active resistance (Ohm)
where p is the specific resistance of the wire material, for copper p \u003d 0.0178 Ohm * mm2 / m, for aluminum p \u003d 0.0294.
l - line length, m;
S - wire section, mm2.
Inductive reactance per phase (mΩ/m) is determined by the formula:
where a is the distance between the conductors, mm;
dpr - conductor diameter, mm.
Active and inductive resistances of cables with aluminum and copper conductors are given in Table. 2.11-2.14, overhead lines - in table. 2.15.
The inductive resistance of the phase-zero loop (mΩ / m) with phase and neutral conductors made of round wires of the same cross section and laid in parallel is determined by the formula:
The phase-zero loop resistances, excluding grounding devices, are given in Table. 2.16, the total resistances of the phase-zero loop of overhead lines and cables are given in Table. 2.17.
Active and inductive resistances of devices installed in networks with voltage up to 1 kV are given in Table. 2.18 and 2.19. The given resistance values of the circuit breakers include the resistances of the current coils of the releases and the transition resistances of the moving contacts.
Table 2.11 Active and inductive resistances of a cable with aluminum conductors in a non-conductive sheath
Cable section, | Resistance of three and four-core cable in a non-conductive sheath, mOhm / m |
|||
direct sequence | Zero sequence |
|||
It should be borne in mind that each machine is connected to the circuit in series through two detachable contacts. For an approximate accounting of the transition resistance of electrical contacts, the following is accepted: Rk = 0.1 mOhm - for contact connections of cables; Rk = 0.01 mOhm - for busbars; Rk - 1.0 mOhm - for switching devices.
Below are the transient active resistances of fixed contact connections, mOhm:
Table 2.12 Active and inductive resistances of a cable with aluminum conductors in an aluminum sheath
Cable section, | Resistance of a three- and four-core cable in an aluminum sheath, mOhm/m |
|||
direct sequence | Zero sequence |
|||
Table 2.13 Active and inductive resistances of a cable with aluminum conductors in a lead sheath
Cable section, | Resistance of three and four-core lead-sheathed cable, mOhm/m |
|||
direct sequence | Zero sequence |
|||
Table 2.14 Active and inductive resistances of a cable with copper conductors in a steel sheath
Cable section, | direct sequence | Zero sequence |
||
When calculating short circuit currents, the active and inductive resistances of the primary windings of all multi-turn measuring current transformers (Kt.a, Xta) that are present in the short circuit circuit are taken into account. The parameters of some multi-turn current transformers are given in Table. 2.19. The active and inductive resistance of single-turn transformers (for currents over 500 A) can be neglected when calculating short-circuit currents.
The active resistance of the arc is given in Table. 2.20.
Consider the principles for calculating the currents of a three-phase and single-phase short circuit. A three-phase short circuit refers to a short circuit between three phases in an electrical system. A single-phase short circuit means a short circuit to earth of power elements in a three-phase electrical system with a dead-earthed neutral, in which only one phase is connected to earth.
Calculation of three-phase short circuit currents is to define:
The initial effective value of the periodic component of the short circuit current;
Aperiodic component of the short circuit current at the initial and arbitrary time;
Surge current short circuit.
When the consumer is powered from the power system through a step-down transformer, the initial effective value of the periodic component of the short-circuit current (7k0) without taking into account the recharge from electric motors is calculated by the formula (kA)
where Uav.n.n is the average rated voltage of the network in which the short circuit occurred, V;
- total resistance of the short-circuit circuit, mOhm;
x1kz - total active and inductive resistance of the direct sequence of the short circuit circuit, equal, respectively
where xc is the equivalent inductive resistance of the system up to the step-down transformer, reduced to the low voltage stage, mOhm;
gt and xt - active and inductive resistance of the direct sequence of the step-down transformer, mOhm;
rr and хр - active and inductive resistances of reactors, mΩ (according to the manufacturer's data);
rtt and xtt - active and inductive resistances of the primary windings of the current transformer, Ohm; gAV and xAV - active and inductive resistances of automatic switches, mOhm, including the resistances of the current coils of the releases and the transitional resistances of the moving contacts;
rsh and xsh - active and inductive resistance of busbars, mOhm;
rk - total active resistance of various contacts, mOhm;
gkb, gvl, and hkb, hvl - active and inductive resistances of cable and overhead lines, mOhm; rD - active resistance of the arc at the place of short circuit, mOhm.
Table 2.15 Active and inductive resistance of wires of overhead lines and cables (for voltage up to 500 V)
Resistance, mOhm/m |
|||||
active | inductive |
||||
aluminum | open wires | with belt paper insulation | wires in pipes, cables with rubber and PVC insulation |
||
Table 2.16 Phase-zero loop resistance values excluding grounding devices
Section of a phase wire, mm2 | Active (numerator) and inductive (denominator) resistance of the loop, mOhm, at the cross section of the neutral wire, mm2 |
||||
Table 2.17 Loop impedances phase-zero overhead lines and cables, mOhm/m
Wire section, mm2 | cable or wire | Wires on rollers and insulators | Overhead line wires |
||||
reverse | aluminum | aluminum | aluminum |
||||
Table 2.18 Switching resistances of current coils of receptors and transitional resistances of movable contacts of circuit breakers and detachable contacts of knife switches
Rated current, A | Resistance of circuit breakers at 65 С, mΩ | Resistance of detachable contacts of knife switches, mOhm |
|
active | inductive |
||
Table 2.19 Resistance of primary windings of multi-turn current transformers
Current transformer transformation ratio | Resistance, mOhm, primary windings of multi-turn current transformers of accuracy class |
|||
Table 2.20 The value of the active resistance of the arc
The aperiodic component of the short circuit current is equal to the amplitude of the periodic component of the current at the initial moment of the short circuit, i.e.:
The aperiodic component of the short-circuit current at an arbitrary point in time is determined by the formula:
where t - time, s;
Ta is the decay time constant of the aperiodic component of the short-circuit current, s, equal to
where XE and RE are the resulting inductive and active resistances of the short circuit circuit, mOhm; yus - synchronous angular frequency network voltage, rad/s.
The surge current of a three-phase short circuit in electrical installations with one energy source (power system or autonomous source) is calculated by the formula:
where - impact coefficient, determined from the curves shown on
Rice. 2.1
Ta is the damping time constant
aperiodic component of short circuit current;
An example of calculating a three-phase short circuit
Determine the short circuit current at the input to the house (cottage).
The village is powered by a distribution point (DP) of the power system via a 10 kV overhead line through a 10 / 0.4 kV transformer with a capacity of 400 kV * A.
The power supply of the cottage is carried out by a 0.4 kV cable line 300 m long.
Cable with copper conductors with a cross section of 4x50 mm2 (Fig. 2.2).
Short-circuit power on RP-10 tires Sk.z=200 MV*A.
The design scheme and the equivalent scheme are shown in fig. 2.3.
Given that the length of the 10 kV line from the 10 kV switchgear system to the transformer substation is less than 1 km, then, in accordance with GOST 28249-93, the line may not be taken into account in the calculations of short-circuit currents.
Rice. 2.2
Rice. 2.3
Determination of equivalent circuit resistances
System resistance:
Transformer resistance 400 kVA (Table 2.9):
Contact resistance of electrical contacts (see GOST 28249-93 p. 2.5), Rk = 0.1 mOhm;
Resistance of circuit breakers (Table 2.18)
Current transformer resistance 300 / 5A 1 (see table. 2.19)
Resistance CL-0.4 kV, section 4x50, length 300 m (Table 2.14)
Short circuit resistance:
active:
reactive:
Short circuit impedance:
The initial value of the periodic component of the current of a three-phase short circuit:
Aperiodic component of the short circuit current at the initial moment of the short circuit:
where Ia0 is the largest initial value of the aperiodic component of the short circuit current.
The aperiodic component at an arbitrary time t is calculated by the formula:
where t - time, s
Ta is the decay time constant of the aperiodic component of the short circuit current;
in our case
the aperiodic component decays after about 0.002 s and can be ignored.
Surge short-circuit current:
where where. = 1 - along the curve in fig. 2.1 from the ratio
Calculation of single-phase short-circuit currents in networks up to 1 kV, it is performed to ensure reliable operation of protection at minimum short-circuit current values at the end of the protected line.
The calculated point of a single-phase short circuit is the electrically most distant point of the network section, protected by the switch.
In accordance with the requirements of the “Electrical Installation Rules” (PUE), in order to reliably disconnect a damaged section of the network, the smallest rated short-circuit current must exceed the rated current of the fuse-link or the rated current of the release of the circuit breaker protecting this section of the network, with a characteristic inversely dependent on current of at least 3 times.
If the circuit breaker has only an instantaneous release (cut-off), then the smallest rated short-circuit current must exceed the cut-off setting by at least 1.4 times.
Compared to the calculation of three-phase short circuit currents, the calculation of single-phase short circuit currents is more complicated, because in this case, in addition to taking into account the resistance in the direct short circuit circuit (in phase), it is also necessary to take into account the resistance in the zeroing circuit (in the reverse circuit). When steel pipes, cable duct frames and other building construction, many uncertainties appear in solving the problem of short-circuit resistance.
In addition, single-phase short circuits are asymmetric, which introduces additional difficulties into the calculation.
The calculation of single-phase short-circuit currents can be performed by the method of symmetrical components or by the resistance of the phase-zero loop.
The method of symmetrical components is proposed to simplify the calculations of asymmetric short circuits. The essence of this method is to replace the asymmetric system of currents of a three-phase network with a single-phase short circuit by three symmetrical systems: direct, reverse and zero sequence. Symmetric systems are quite simple for theoretical calculation. In the practical use of this method, difficulties often arise due to the lack of reference materials on zero-sequence resistances for the accepted version of the neutral circuit.
When calculating the currents of a single-phase short circuit from the resistance of the phase-zero loop, Ohm's law is used, but the same difficulties are encountered with the initial data.
Both methods should give the same result and could theoretically be deduced from one another. The accuracy of the calculation is determined only by the accuracy of the source data.
In GOST 28249-93, the calculation of single-phase short-circuit currents is based on the method of symmetrical components, which is discussed in more detail below.
The calculation of a single-phase short circuit by the method of symmetrical components is carried out according to the formula:
where I1 is the effective value of the periodic component of the current of a single-phase short circuit, kA;
Ul - average nominal (linear) mains voltage, V;
R1E - total active resistance of the short-circuit phase circuit (positive sequence resistance), mOhm;
R0E - total active resistance of the short circuit for the zero sequence current (resistance of the zero sequence), mOhm;
X1E - total inductive resistance of the short-circuit phase circuit (positive sequence resistance), mOhm;
X0E - total inductive resistance of the short circuit for the zero sequence current (resistance of the zero sequence), mOhm.
The negative sequence resistances are equal to the positive sequence resistances and are taken into account in the above formula by a factor of 2 before R1E and X1E.
The total active and total inductive resistance of the short-circuit phase circuit are determined by the formulas:
where r1Т and Х1Т are the direct sequence resistances of the step-down transformer, mΩ;
r1L and X1L - direct sequence resistance of the line (phase conductor), mOhm;
rТТ and ХТТ - resistance of primary windings of current transformers, mΩ;
rА and ХА - resistances of circuit breakers, mΩ;
rK - total active resistance of various contacts in the short circuit phase circuit, mOhm;
rD - active resistance of the electric arc at the place of short circuit, mOhm.
The total active and total inductive resistance of the short circuit for the zero sequence current are determined by the formulas:
where r0Т and Х0Т are zero-sequence resistances of the step-down transformer, mΩ; r0L and X0L - resistance of the zero sequence of the line (resistance of busbars, wires, cables, taking into account the grounding circuit), mOhm;
rТТ, ХТТ, rА, ХА, rК and rД - resistance of the short-circuit phase circuit, mΩ.
The zero sequence resistance of the line is equal to the resistance of the phase conductor plus three times the resistance of the neutral circuit:
where rN and XN are the equivalent resistances of the grounding (zero) circuit from the short circuit point to the transformer, taking into account all grounding elements (neutral wire, cable sheath, steel pipes, etc.), mOhm.
The increase in 3 times the resistance of the zeroing circuit for the zero sequence current of the damaged phase is due to the fact that, in accordance with the method of symmetrical components, zero sequence currents of all three phases equal in value are closed through the zeroing circuit. In this way:
When determining the minimum values of single-phase short-circuit currents to test the protection sensitivity, it is recommended to take into account the increase in the active resistance of the conductors as a result of their heating by the short-circuit current. For this, the resistance of conductors with a cross section of up to 16 mm2 (inclusive) is recommended to be brought to a temperature of 1200C, the resistance of conductors with a cross section of 25-95 mm2 - to a temperature of 1450C, the resistance of conductors with a cross section of 120-140 mm2 - to a temperature of 950C. Such (approximate) values of the temperature of the conductors at the end of the short circuit were obtained as a result of calculations taking into account the real time-current characteristics of the protection devices and under the condition of the adiabatic process of heating the conductor cores. State standard GOST 2824+-89 is allowed to take the value for all sections temperature coefficient electrical resistance equal to 1.5, which corresponds to a temperature of 1450C. But conductors of large cross-sections practically do not heat up to such a temperature during a short circuit.
The temperature coefficient for converting the conductor resistance at 20°C to the resistance at the final temperature is calculated by the formula:
where Ocon. - temperature of the conductor core at the end of the short circuit, 0С.
Conductor resistance at final temperature
where r20 is the resistance of the conductor at a temperature of 20 0C.
An example of calculating the current of a single-phase short circuit.
For the scheme according to fig. 2.2 determine the current of a single-phase short circuit at the input to the cottage.
The calculation is carried out by the method of symmetrical components.
When the electrical installation is powered from the system through a step-down transformer, the initial value of the periodic component of the current of a single-phase short circuit is calculated by the formula (kA):
where r1E , x1E - active and inductive total resistances of the direct sequence relative to the short circuit point. In our case (see the calculation of a three-phase short circuit) - r1E \u003d 137.5 mOhm, X1E \u003d 45.4 mOhm;
r0E , XOE. - active and inductive total resistances of the zero sequence relative to the short circuit point.
These resistances are:
where r0T, X0T - active and inductive resistances of the zero sequence of the step-down transformer;
rТТ, XТТ - active and inductive resistances of the current transformer;
rkv, HKB - active and inductive resistances of automatic switches;
rK - contact resistance.
For the example in question:
According to the table 2.9 zero-sequence resistance of a 400 kVA transformer is: Х0Т = 149 mΩ, r0Т = 55.6 mΩ.
where r’0 and x’0 are active and inductive resistances of 1 m of a copper cable with a cross section of 4x50 mm2 (Table 2.14);
In this way:
When designing electrical wiring in a room, you need to start by calculating the current strength in the circuits. An error in this calculation can then be costly. An electrical outlet may melt if the current is too strong for it. If the current in the cable is greater than that calculated for a given material and cross-section of the core, the wiring will overheat, which can lead to melting of the wire, breakage or short circuit in the network with unpleasant consequences, among which the need for a complete replacement of electrical wiring is not the worst.
Knowing the strength of the current in the circuit is also necessary for the selection of circuit breakers, which should provide adequate protection against network overload. If the machine is standing with a large margin at face value, by the time it is triggered, the equipment may already be out of order. But if the rated current of the circuit breaker is less than the current that occurs in the network at peak loads, the machine will drive you crazy, constantly de-energizing the room when you turn on the iron or kettle.
The formula for calculating the power of an electric current
According to Ohm's law, current (I) is proportional to voltage (U) and inversely proportional to resistance (R), and power (P) is calculated as the product of voltage and current. Based on this, the current in the network section is calculated: I = P / U.
In real conditions, one more component is added to the formula and the formula for a single-phase network takes the form:
and for a three-phase network: I \u003d P / (1.73 * U * cos φ),
where U for a three-phase network is assumed to be 380 V, cos φ is the power factor, reflecting the ratio of the active and reactive components of the load resistance.
For modern power supplies, the reactive component is insignificant, the value of cos φ can be taken equal to 0.95. The exceptions are powerful transformers (for example, welding machines) and electric motors, they have a large inductive resistance. In networks where it is planned to connect such devices, the maximum current strength should be calculated using a cos φ factor of 0.8 or the current strength should be calculated using the standard method, and then a multiplying factor of 0.95 / 0.8 = 1.19 should be applied.
Substituting the effective voltage values of 220 V / 380 V and a power factor of 0.95, we obtain I \u003d P / 209 for a single-phase network and I \u003d P / 624 for a three-phase network, that is, in a three-phase network with the same load, the current is three times less. There is no paradox here, since three-phase wiring provides for three phase wires, and with a uniform load on each of the phases, it is divided into three. Since the voltage between each phase and working neutral wires is 220 V, the formula can also be rewritten in a different form, so it is clearer: I \u003d P / (3 * 220 * cos φ).
We select the rating of the circuit breaker
Applying the formula I \u003d P / 209, we get that with a load with a power of 1 kW, the current in a single-phase network will be 4.78 A. The voltage in our networks is not always exactly 220 V, so it will not be a big mistake to calculate the current strength with a small margin as 5 A for each kilowatt of load. It is immediately clear that it is not recommended to turn on an iron with a power of 1.5 kW in an extension cord marked “5 A”, since the current will be one and a half times higher than the passport value. And you can immediately “calibrate” the standard ratings of the machines and determine what load they are designed for:
- 6 A - 1.2 kW;
- 8 A - 1.6 kW;
- 10 A - 2 kW;
- 16 A - 3.2 kW;
- 20 A - 4 kW;
- 25 A - 5 kW;
- 32 A - 6.4 kW;
- 40 A - 8 kW;
- 50 A - 10 kW;
- 63 A - 12.6 kW;
- 80 A - 16 kW;
- 100 A - 20 kW.
Using the "5 amperes per kilowatt" technique, you can estimate the current strength that occurs in the network when connecting household devices. We are interested in peak loads on the network, so for the calculation you should use the maximum power consumption, and not the average. This information is contained in the product documentation. It is hardly worthwhile to calculate this indicator yourself, summing up the nameplate capacities of compressors, electric motors and heating elements included in the device, since there is also such an indicator as efficiency, which will have to be estimated speculatively with the risk of making a big mistake.
When designing electrical wiring in an apartment or a country house, the composition and passport data of the electrical equipment that will be connected are not always known for certain, but you can use the indicative data of electrical appliances common to our everyday life:
- electric sauna (12 kW) - 60 A;
- electric stove (10 kW) - 50 A;
- hob (8 kW) - 40 A;
- instantaneous electric water heater (6 kW) - 30 A;
- dishwasher (2.5 kW) - 12.5 A;
- washing machine (2.5 kW) - 12.5 A;
- jacuzzi (2.5 kW) - 12.5 A;
- air conditioning (2.4 kW) - 12 A;
- microwave oven (2.2 kW) - 11 A;
- storage electric water heater (2 kW) - 10 A;
- electric kettle (1.8 kW) - 9 A;
- iron (1.6 kW) - 8 A;
- solarium (1.5 kW) - 7.5 A;
- vacuum cleaner (1.4 kW) - 7 A;
- meat grinder (1.1 kW) - 5.5 A;
- toaster (1 kW) - 5 A;
- coffee maker (1 kW) - 5 A;
- hair dryer (1 kW) - 5 A;
- desktop computer (0.5 kW) - 2.5 A;
- refrigerator (0.4 kW) - 2 A.
The power consumption of lighting fixtures and consumer electronics is small, in general, the total power of lighting fixtures can be estimated at 1.5 kW and a 10 A machine per lighting group is enough. Consumer electronics are connected to the same outlets as irons, it is not advisable to reserve additional power for it.
If you sum up all these currents, the figure is impressive. In practice, the ability to connect the load is limited by the amount of allocated electrical power, for apartments with electric stove in modern houses it is 10-12 kW and there is an automatic machine with a nominal value of 50 A at the apartment input. And these 12 kW must be distributed, given that the most powerful consumers are concentrated in the kitchen and in the bathroom. The wiring will be less of a concern if it is broken down into enough groups, each with its own machine. For an electric stove (hob), a separate input is made with a 40 A automatic machine and a power outlet with a rated current of 40 A is installed, nothing else needs to be connected there. For washing machine and other bathroom equipment is made a separate group, with a machine gun of the appropriate denomination. This group is usually protected by an RCD with a rated current 15% greater than the rating of the circuit breaker. Separate groups are allocated for lighting and for wall sockets in each room.
It will take some time to calculate the powers and currents, but you can be sure that the work will not be in vain. Properly designed and well-installed electrical wiring is the key to the comfort and safety of your home.