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 bricks 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 200 mm high 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 the 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 the protrusions on the boundaries of the outer walls), then the 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 in terms of 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 with the help of 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 is the case with a 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. The 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.
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To protect yourself when working with household electrical appliances, you must first correctly calculate the cross section of the cable and wiring. Because if the cable is not chosen correctly, it can lead to a short circuit, which can cause a fire in the building, the consequences can be catastrophic.
This rule also applies to the choice of cable for electric motors.
Calculation of power by current and voltage
This calculation takes place upon the fact of power, it must be done even before the design of your home (house, apartment) begins.
- This value depends on the cable supply devices that are connected to the mains.
- According to the formula, you can calculate the current strength, for this you need to take the exact mains voltage and the load of the powered devices. Its value gives us an understanding of the cross-sectional area of veins.
If you know all the electrical appliances that should be powered from the network in the future, then you can easily make calculations for the power supply scheme. The same calculations can be performed for production purposes.
Single-phase network with a voltage of 220 volts
Current strength formula I (A - amperes):
I=P/U
Where P is the electrical full load (its designation must be indicated in the technical data sheet of this device), W - watt;
U - mains voltage, V (volts).
The table shows the standard loads of electrical appliances and the current they consume (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 |
| Vacuum cleaner | 680 - 1400 | 3,1 – 6,4 |
| Fan | 250 - 400 | 1,0 – 1,8 |
| TV | 125 - 180 | 0,6 – 0,8 |
| radio equipment | 70 - 100 | 0,3 – 0,5 |
| Lighting devices | 20 - 100 | 0,1 – 0,4 |
In the figure you can see a diagram of a house power supply device with a single-phase connection to a 220 volt network.
As shown in the figure, all consumers must be connected to the appropriate machines and a meter, then to a common machine that will withstand the total load of the house. The cable that will bring the current must withstand the load of all connected household appliances.
The table below shows hidden wiring in a single-phase scheme for connecting a dwelling for cable selection at a voltage of 220 volts.
| 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 shown in the table, the cross section of the cores also depends on the material from which it is made.
Three-phase network with a voltage of 380 V
In a three-phase power supply, the current strength is calculated using the following formula:
I = P / 1.73 U
P is the power consumption in watts;
U is the mains voltage in volts.
In a 380 V phase power supply, the formula is as follows:
I = P /657.4
If a three-phase 380 V network is connected to the house, then the connection diagram will look like this.
The table below shows the cross-sectional diagram of the cores in the supply cable at various loads at a three-phase voltage of 380 V for flush wiring.
| 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 |
For further calculation of power supply in load circuits characterized by a large reactive apparent power, which is typical for the use of power supply in industry:
- electric motors;
- induction furnaces;
- chokes of lighting devices;
- welding transformers.
This phenomenon must be taken into account in further calculations. In more powerful electrical appliances, the load goes much more, therefore, in the calculations, the power factor is taken as 0.8.
When calculating the load on household appliances, the power reserve should be taken as 5%. For the power grid, this percentage becomes 20%.
For durable and reliable operation of the electrical wiring, it is necessary to choose the right cable cross-section. To do this, you need to calculate the load in the power grid. When making calculations, it must be remembered that the calculation of the load of one electrical appliance and a group of electrical appliances differ somewhat.
Calculation of the current load for a single consumer
The choice of a circuit breaker and the calculation of the load for a single consumer in a 220 V residential network is quite simple. To do this, we recall the main law of electrical engineering - Ohm's law. After that, having set the power of the electrical appliance (indicated in the passport for the electrical appliance) and given the voltage (for household single-phase networks 220 V), we calculate the current consumed by the electrical appliance.
For example, a household electrical appliance has a supply voltage of 220 V and a nameplate power of 3 kW. We apply Ohm's law and get I nom \u003d P nom / U nom \u003d 3000 W / 220 V \u003d 13.6 A. Accordingly, to protect this consumer of electrical energy, it is necessary to install a circuit breaker with a rated current of 14 A. Since there are none, it is selected the nearest larger one, that is, with a rated current of 16 A.
Calculation of current load for groups of consumers
Since the power supply of electricity consumers can be carried out not only individually, but also in groups, the issue of calculating the load of a group of consumers becomes relevant, since they will be connected to one circuit breaker.
To calculate a group of consumers, the demand coefficient K s is introduced. It determines the probability of simultaneous connection of all consumers of the group for a long time.
The value of K c = 1 corresponds to the simultaneous connection of all electrical appliances of the group. Naturally, the inclusion of all consumers of electricity in an apartment at the same time is extremely rare, I would say incredible. There are whole methods for calculating demand coefficients for enterprises, houses, entrances, workshops, and so on. The demand factor of an apartment will vary for different rooms, consumers, and will also largely depend on the lifestyle of residents.
Therefore, the calculation for a group of consumers will look somewhat more complicated, since this coefficient must be taken into account.
The table below shows the demand factors for electrical appliances in a small apartment:
The demand coefficient will be equal to the ratio of the reduced power to the total K from the apartment = 2843/8770 = 0.32.
We calculate the load current I nom \u003d 2843 W / 220 V \u003d 12.92 A. We select an automatic machine for 16A.
Using the above formulas, we calculated the operating current of the network. Now you need to select the cable section for each consumer or consumer groups.
PUE (rules for electrical installations) regulates the cable cross-section for various currents, voltages, powers. Below is a table from which, according to the estimated power of the network and current, the cable section for electrical installations with a voltage of 220 V and 380 V is selected:
The table shows only the cross sections of copper wires. This is due to the fact that aluminum wiring is not laid in modern residential buildings.
Also below is a table with the range of capacities of household electrical appliances for calculation in networks of residential premises (from the standards for determining the design loads of buildings, apartments, private houses, microdistricts).
Typical cable size selection
In accordance with the cable section, circuit breakers are used. Most often, the classic version of the wire section is used:
- For lighting circuits with a cross section of 1.5 mm 2;
- For circuits of sockets with a section of 2.5 mm 2;
- For electric stoves, air conditioners, water heaters - 4 mm 2;
A 10 mm 2 cable is used to enter the power supply into the apartment, although in most cases 6 mm 2 is enough. But a section of 10 mm 2 is chosen with a margin, so to speak, with the expectation of a larger number of electrical appliances. Also, a common RCD with a trip current of 300 mA is installed at the input - its purpose is fire, since the trip current is too high to protect a person or animal.
To protect people and animals, RCDs with a tripping current of 10 mA or 30 mA are used directly in potentially unsafe rooms, such as a kitchen, bath, and sometimes room outlet groups. The lighting network, as a rule, is not supplied with an RCD.
Theory calculation of electrical loads, the foundations of which were formed in the 1930s, aimed to determine a set of formulas that give an unambiguous solution for given electrical receivers and graphs (indicators) of electrical loads. In general, practice has shown the limitations of the "bottom-up" approach, based on the initial data for individual power consumers and their groups. This theory remains important when calculating the operating modes of a small number of power receivers with known data, when adding a limited number of graphs, when calculating for 2UR.
In the 1980s-1990s. the theory of calculating electrical loads increasingly adheres to non-formalized methods, in particular, the integrated method for calculating electrical loads, the elements of which are included in the "Guidelines for calculating the electrical loads of power supply systems" (RTM 36.18.32.0289). It is likely that working with information databases on electrical and technological indicators, cluster analysis and pattern recognition theory, construction of probabilistic and cenological distributions for expert and professional assessment can finally solve the problem of calculating electrical loads at all levels of the power supply system and at all stages of making a technical or investment decision .
Formalization of the calculation of electrical loads developed over the years in several directions and led to the following methods:
- empirical (method of demand coefficient, two-term empirical expressions, specific power consumption and specific load densities, technological schedule);
- ordered diagrams, transformed into calculation according to the calculated active power factor;
- actually statistical;
- probabilistic modeling of load curves.
Demand factor method
The demand factor method is the simplest, most widespread, and the calculation of loads began with it. It consists in using the expression (2.20): according to the known (given) value Ru and the tabular values given in the reference literature (see examples in Table 2.1):
The value of Kc is assumed to be the same for power receivers of the same group (operating in the same mode), regardless of the number and power of individual receivers. The physical meaning is the fraction of the sum of the rated powers of electrical receivers, statistically reflecting the maximum practically expected and occurring mode of simultaneous operation and loading of some indefinite combination (implementation) of installed receivers.
The given reference data for Kc and Kp correspond to the maximum value, and not to the mathematical expectation. Summing up the maximum values, and not the averages, inevitably overestimates the load. If we consider any group of ES of the modern electrical economy (and not the 1930-1960s), then the conventionality of the concept of "homogeneous group" becomes obvious. Differences in the value of the coefficient - 1:10 (up to 1:100 and above) - are inevitable and are explained by the cenological properties of the electrical economy.
In table. 2.2 shows the LGS values characterizing the pumps as a group. When researching KQ4 further, for example only for raw water pumps, there may also be a spread of 1:10.
It is more correct to learn to evaluate Kc as a whole for the consumer (section, department, workshop). It is useful to perform an analysis of the calculated and actual values for all technology-related objects of the same level of the power supply system, similar to Table. 1.2 and 1.3. This will create a personal information bank and ensure the accuracy of calculations. The method of specific energy consumption is applicable for sections (installations) 2UR (second, third ... Level of the Energy System), departments of ZUR and workshops 4UR, where technological products are homogeneous and change little quantitatively (an increase in output usually reduces the unit consumption of electricity Aui).
Method "maximum power"
In real conditions, continuous operation of the consumer does not mean the constancy of the load at the point of its connection at a higher level of the power supply system. As a statistic value Lud, determined for some previously identified object by power consumption A and volume L /, there is some averaging over a known, often monthly or annual, interval. Therefore, the application of formula (2.30) gives not the maximum, but the average load. To select the ZUR transformers, one can take Рav = Рmax. In the general case, especially for 4UR (workshop), it is necessary to take into account Kmax as T to take the actual annual (daily) number of hours of production operation with the maximum use of active power.
Method of specific load densities
The method of specific load densities is close to the previous one. The specific power (load density) y is set and the area of the building of the structure or section, department, shop is determined (for example, for machine-building and metalworking shops y = 0.12 ... 0.25 kW/m2; for oxygen converter shops y = = 0.16 ... 0.32 kW/m2). A load exceeding 0.4 kW / m2 is possible for some areas, in particular, for those where there are single power receivers with a unit power of 1.0 ... 30.0 MW.
Process chart method
The technological schedule method is based on the schedule of the unit, line or group of machines. For example, the operation schedule of an arc steel-smelting furnace is specified: the melting time (27 ... 50 min), the oxidation time (20 ... 80 min), the number of melts, technological linkage with the operation of other steel-smelting units are indicated. The graph allows you to determine the total electricity consumption per melt, average per cycle (taking into account the time until the next melt), and the maximum load for calculating the supply network.
Ordered chart method
The method of ordered diagrams, which was applied in the directive in the 1960s - 1970s. for all levels of the power supply system and at all stages of design, in the 1980s-1990s. was transformed into the calculation of loads according to the calculated active power factor. If there is data on the number of power receivers, their power, modes of operation, it is recommended to use it for calculating the elements of the power supply system 2UR, ZUR (wire, cable, busbar, low-voltage equipment) supplying a power load with voltage up to 1 kV (simplified for the effective number of receivers of the entire workshop, i.e. for a network with a voltage of 6 - 10 kV 4UR). The difference between the method of ordered diagrams and calculation by the rated active power factor lies in the replacement of the maximum factor, always understood unambiguously as the ratio Pmax / Rav (2.16), by the rated active power factor Ap. The order of calculation for the node element is as follows:
A list (number) of power receivers is compiled, indicating their nominal PHOMi (installed) power;
The work shift with the highest electricity consumption is determined and the characteristic day is agreed (with technologists and the power system);
The features of the technological process that affect power consumption are described, power receivers with high load unevenness are distinguished (they are considered differently - according to the maximum effective load);
The following electrical receivers are excluded from the calculation (list): a) low power; b) reserve according to the conditions for calculating electrical loads; c) included sporadically;
Groups m of electrical receivers having the same type (mode) of operation are determined;
From these groups, subgroups are distinguished that have the same value of the individual utilization factor a:u/;
Electric receivers of the same operating mode are allocated and their average power is determined;
The average reactive load is calculated;
There is a group coefficient of utilization Kn of active power;
The effective number of power receivers in a group of n power receivers is calculated:
where the effective (reduced) number of power receivers is the number of power receivers of the same power that are homogeneous in terms of operation, which gives the same value of the calculated maximum P as a group of power receivers that are different in power and mode of operation.
With the number of power receivers in a group of four or more, it is allowed to take pe equal to n (the actual number of power receivers), provided that the ratio of the rated power of the largest power receiver Pmutm to the rated power of the smaller power receiver Dom mm is less than three. When determining the value of p, it is allowed to exclude small power receivers, the total power of which does not exceed 5% of the rated power of the entire group;
According to the reference data and the heating time constant T0, the value of the calculated coefficient Kp is taken;
The calculated maximum load is determined:
Electrical loads individual nodes of the power supply system in networks with voltages above 1 kV (located at 4UR, 5UR) were recommended to be determined similarly with the inclusion of losses in.
The calculation results are summarized in a table. This completes the calculation of loads according to the calculated active power factor.
The calculated maximum load of a group of electrical receivers Рmax can be found in a simplified way:
where Рnom - group rated power (the sum of rated powers, with the exception of reserve ones according to the calculation of electrical loads); Рav.cm ~ average active power for the busiest shift.
The calculation according to formula (2.32) is cumbersome, difficult to understand and apply, and most importantly, it often gives a double (or more) error. The method overcomes non-Gaussian randomness, uncertainty and incompleteness of the initial information with the following assumptions: power receivers of the same name have the same coefficients, standby motors are excluded according to the conditions of electrical loads, the utilization factor is considered independent of the number of power receivers in the group, power receivers with an almost constant load schedule are distinguished, the smallest ones are excluded from the calculation power receivers. The method is not differentiated for different levels of the power supply system and for different stages of implementation (coordination) of the project. The calculated coefficient of the maximum active power Kmax is taken to tend to unity with an increase in the number of power receivers (in fact, this is not the case - statistics do not confirm this. For a department where there are 300 ... ,2… 1.4). The introduction of market relations, leading to automation, a variety of output, moves electrical receivers from group to group.
The statistical definition of Rav.cm for operating enterprises is complicated by the difficulty of choosing the busiest shift (transfer of the start of work for different categories of workers within a shift, four-shift work, etc.). Uncertainty appears in the measurements (superimposition on the administrative-territorial structure). Restrictions on the part of the power system lead to regimes when the maximum load Ptx occurs in one shift, while the consumption of electricity is greater in another shift. When determining Рр, it is necessary to abandon Рср.см, excluding intermediate calculations.
A detailed consideration of the shortcomings of the method is caused by the need to show that the calculation of electrical loads, based on classical ideas about the electrical circuit and load curves, theoretically cannot provide sufficient accuracy.
Statistical methods for calculating electrical loads are consistently defended by a number of specialists. The method takes into account that even for one group of mechanisms operating in a given production area, the coefficients and indicators vary widely. For example, the inclusion factor for non-automatic machine tools of the same type varies from 0.03 to 0.95, loading A3 - from 0.05 to 0.85.
The task of finding the maximum of the Рр function on a certain time interval is complicated by the fact that power receivers and consumers with different operating modes are fed from 2UR, ZUR, 4UR. The statistical method is based on measuring the loads of lines supplying characteristic groups of power receivers, without referring to the operating mode of individual power receivers and the numerical characteristics of individual graphs.
(xtypo_quote) The method uses two integral characteristics: the general load average PQp and the general standard deviation, where the dispersion DP is taken for the same averaging interval. (/xtypo_quote)
The maximum load is determined as follows:
The value of p is taken to be different. In probability theory, the three-sigma rule is often used: Pmax = Pavg ± Za, which, with a normal distribution, corresponds to a limiting probability of 0.9973. The probability of exceeding the load by 0.5% corresponds to р = 2.5; for p = 1.65, a 5% probability of error is provided.
The statistical method is a reliable method for studying the loads of an operating industrial enterprise, providing a relatively correct value of the maximum load Pi (miiX) declared by an industrial enterprise during the hours of the maximum in the power system. In this case, it is necessary to assume a Gaussian distribution of the operation of electrical receivers (consumers).
The method of probabilistic modeling of load graphs involves a direct study of the probabilistic nature of successive random changes in the total load of groups of power receivers over time and is based on the theory of random processes, with the help of which autocorrelation (formula (2.10)), cross-correlation functions and other parameters are obtained. Studies of the work schedules of electrical receivers of large unit capacity, the work schedules of workshops and enterprises determine the prospects for the method of controlling power consumption modes and leveling the schedules.
Calculator Weight-At Home-Online v.1.0
Calculation of the weight of the house, taking into account the snow and operational load on the floors (calculation of vertical loads on the foundation). The calculator is implemented on the basis of SP 20.13330.2011 Loads and impacts (current version of SNiP 2.01.07-85).
Calculation example
Aerated concrete house measuring 10x12m is one-story with a residential attic.
Input data
- Structural scheme of the building: five-wall (with one internal load-bearing wall along the long side of the house)
- House size: 10x12m
- Number of floors: 1 floor + attic
- Snow region of the Russian Federation (to determine the snow load): St. Petersburg - 3rd region
- Roof material: metal tile
- Roof pitch: 30⁰
- Structural scheme: scheme 1 (attic)
- Attic wall height: 1.2m
- Attic facade decoration: textured front brick 250x60x65
- Material of the outer walls of the attic: aerated concrete D500, 400mm
- Material of the inner walls of the attic: not involved (the ridge is supported by columns that are not included in the calculation due to their low weight)
- Operating load on floors: 195kg / m2 - residential attic
- Ground floor height: 3m
- Facade decoration of the 1st floor: textured front brick 250x60x65
- Material of external walls of the 1st floor: aerated concrete D500, 400mm
- Material of internal walls of the floor: aerated concrete D500, 300mm
- Plinth height: 0.4m
- Plinth material: solid brick (masonry in 2 bricks), 510mm
House dimensions
Exterior wall length: 2 * (10 + 12) = 44 m
Inner wall length: 12m
Total wall length: 44 + 12 = 56 m
The height of the house, taking into account the basement \u003d The height of the basement walls + The height of the walls of the 1st floor + The height of the attic walls + The height of the gables \u003d 0.4 + 3 + 1.2 + 2.9 \u003d 7.5 m
To find the height of the gables and the area of the roof, we use the formulas from trigonometry.
ABC is an isosceles triangle
AB=Sun - unknown
AC \u003d 10 m (in the calculator, the distance between the axes of the AG)
Angle BAC = Angle BCA = 30⁰
BC = AC * ½ * 1/ cos(30⁰) = 10 * 1/2 * 1/0.87 = 5.7 m
BD = BC * sin(30⁰) = 5.7 * 0.5 = 2.9 m (pediment height)
Area of triangle ABC (gable area) = ½ * BC * AC * sin(30⁰) = ½ * 5.7 * 10 * 0.5 = 14
Roof area \u003d 2 * BC * 12 (in the calculator the distance between the axes is 12) \u003d 2 * 5.7 * 12 \u003d 139 m2
The area of the outer walls = (Height of the basement + Height of the 1st floor + Height of the attic walls) * Length of the outer walls + Area of two gables = (0.4 + 3 + 1.2) * 44 + 2 * 14 = 230 m2
Area of internal walls = (Height of basement + Height of 1st floor) * Length of internal walls = (0.4 + 3) * 12 = 41m2 .
Total floor area = House length * House width * (Number of floors + 1) = 10 * 12 * (1 + 1) = 240 m2
Load calculation
Roof
Building city: St. Petersburg
According to the map of the snow regions of the Russian Federation, the city of St. Petersburg belongs to the 3rd district. The estimated snow load for this area is 180 kg/m2.
Roof snow load = Estimated snow load * Roof area * Coefficient (depending on roof slope) = 180 * 139 * 1 = 25,020 kg = 25 t
(coefficient depending on roof slope. At 60 degrees snow load is not taken into account. Up to 30 degrees coefficient = 1, from 31-59 degrees coefficient is calculated by interpolation)
Roof mass = Roof area * Roof material mass = 139 * 30 = 4,170 kg = 4 t
Total attic wall load = Roof snow load + Roof weight = 25 + 4 = 29 t
Important!The material loads are shown at the end of this example.
Attic (loft)
Mass of exterior walls = (Area of attic walls + Area of gable walls) * (Mass of exterior wall material + Mass of facade material) = (1.2 * 44 + 28) * (210 + 130) = 27,472 kg = 27 tons
Weight of interior walls = 0
Attic floor mass = Attic floor area * Floor material mass = 10 * 12 * 350 = 42,000 kg = 42 tons
Total load on the walls of the 1st floor = Total load on the walls of the attic + Mass of the outer walls of the attic + Mass of the attic floor + Operational load of the floor = 29 + 27 + 42 + 23 = 121 t
1st floor
Mass of exterior walls of the 1st floor = Area of exterior walls * (Mass of exterior wall material + Mass of facade material) = 3 * 44 * (210 + 130) = 44,880 kg = 45 tons
Mass of internal walls of the 1st floor = Area of internal walls * Mass of material of internal walls = 3 * 12 * 160 = 5 760 kg = 6 tons
Plinth floor mass = Floor area * Weight of floor material = 10 * 12 * 350 = 42,000 kg = 42 tons
Ceiling Service Load = Design Service Load * Floor Area = 195 * 120 = 23,400 kg = 23 t
Total load on the walls of the 1st floor = Total load on the walls of the 1st floor + Mass of the external walls of the 1st floor + Mass of the internal walls of the 1st floor + Weight of the basement floor + Operational load of the floor = 121 + 45 + 6 + 42 + 23 = 237 t
plinth
Plinth mass = Plinth area * Plinth material mass = 0.4 * (44 + 12) * 1330 = 29,792 kg = 30 t
Total load on the foundation \u003d Total load on the walls of the 1st floor + Plinth mass \u003d 237 + 30 \u003d 267 t
The weight of the house, taking into account loads
The total load on the foundation, taking into account the safety factor = 267 * 1.3 = 347 t
Linear weight of the house with a uniformly distributed load on the foundation = Total load on the foundation, taking into account the safety factor / Total length of the walls = 347 / 56 = 6.2 t / m.p. = 62 kN/m
When choosing to calculate loads on load-bearing walls (five-walls - 2 external load-bearing + 1 internal load-bearing), the following results were obtained:
Linear weight of external load-bearing walls (axes A and D in the calculator) = Area of the 1st external load-bearing wall of the plinth * Mass of the basement wall material + Area of the 1st external load-bearing wall * (Mass of the wall material + Mass of the facade material) + ¼ * Total load attic walls + ¼ * (Attic floor material weight + Attic floor service load) + ¼ * Total attic wall load + ¼ * (Attic floor material weight + Basement floor service load) = (0.4 * 12 * 1.33) + (3 + 1.2) * 12 * (0.210 + 0.130) + ¼ * 29 + ¼ * (42 + 23) + + ¼ * (42 + 23) = 6.4 + 17.2 + 7.25 + 16.25 + 16.25 = 63t = 5.2 t/m. P. = 52 kN