What is the normal loss of the distribution network into the home?

Mainly caused by line loss, each line element in the power grid will produce a certain amount of power after passing the current, and the work will inevitably lose energy, as well as the line, the original, etc. have a certain resistance, and there will also be a certain loss.

Loss of line elements:

Analysis of the above nouns:

Wire loss refers to the loss on the wire resistance when the current flows through the line wire, and the magnitude is related to the resistance of the wire and the amount of current flowing.

Corona loss refers to the loss of electrical energy caused by the ionization of the thin layer of air around the wire due to the fact that the surface field strength of the wire exceeds the breakdown strength of the surrounding air, and the magnitude is related to the surface electric field strength, surface condition, weather conditions, geographical conditions, etc.

When the rated voltage is applied to one winding of the transformer, and the rest of the windings are open, the active power absorbed by the transformer is called no-load loss. No-load loss refers to the hysteresis and eddy current of the material when the magnetic field lines pass through the material due to periodic changes in the transformer core lamination, and its magnitude is related to the operating voltage and tap voltage.

Stray losses are losses that occur on leads and housings, as well as other structural metal parts, and are load-dependent.

No-load loss refers to the hysteresis and eddy current of the material when the magnetic field lines pass through the material due to periodic changes in the transformer core lamination, and its magnitude is related to the operating voltage and tap voltage.

The capacitor has a certain active loss during operation, which is lost by the heating of the medium. When a reactor passes current, wire loss, hysteresis loss, and eddy current loss are generated. Theoretically, when the switch is in the conduction and disconnection state, the loss is zero, when the switch is in the actual switching state, there will be a period of time when voltage and current overlap exist, and at this time current and voltage will produce loss in the form of heat, light, noise, electromagnetic energy, etc.

There are two main types of oil, one is line loss, and the other is transformer loss. Line losses are generated by line voltage drops, and transformer losses are generated by transformers, mainly iron and copper losses.

Due to the resistance of various components (such as lines, transformers, etc.) that make up the power grid, in the power grid, the electric energy sent from the power plant has to go through multi-stage transformer voltage transformation and different voltage network transmission before it can reach the user for use. When the electrical energy passes through the resistance in the form of current, it will generate power loss and electric energy loss to heat the resistor; In addition, in the network of different voltage levels, in the process of electromagnetic energy conversion, in order to make the energy conversion form of electromagnetic induction continue to exist, it is necessary to provide an excitation electromotive force to the transformer core, and at the same time, the magnetic field will also produce eddy current and hysteresis loss in the core equipment, which will produce power loss and electric energy loss, so the active energy loss will inevitably occur when the electric energy is transmitted through the network. The power loss of the grid will not only cause the equipment in the grid to heat up, but also take up a part of the capacity of the power generation and power supply equipment because the energy consumed cannot be used.

Personally, there are two main parts of power grid loss:

Heat loss – the loss of current as it flows through a copper wire;

Electromagnetic loss - the charged conductor has an electromagnetic field in the surrounding space, which is dissipated in the space or material outside the electrical equipment in the form of excitation loss, iron loss, eddy current loss, radiation, etc.

The loss rate of wires is different from the different quota subheadings, such as copper core insulated wires and lighting circuits, and the loss rate is 16%, which should be specified according to the quota.

The fixed loss rate of the cable is 1%, the cable consumption is 1%, which is the waveform, bending, and radian coefficient, and the wire is taken as an example: 100 + reserved wire length) +.

The wire and cable products used in the power system mainly include overhead bare wires, busbars (busbars), power cables (plastic cables, oil-paper power cables (basically replaced by plastic power cables), rubber cables, overhead insulated cables, branch cables (replacing part of the busbar), electromagnetic wires and wires and cables for electrical equipment for power equipment.

Quality identification:

One has to see. See if there is a quality system certification; See whether the certificate is standardized; See whether there is the factory name, factory address, inspection seal, and production date; See if there are trademarks, specifications, voltages, etc. printed on the wires. It is also necessary to look at the cross-section of the copper core of the wire, the color of the high-grade copper is bright and the color is soft, otherwise it is a defective product.

Second, we have to try. You can take a wire head and bend it repeatedly by hand, where it feels soft, has good fatigue strength, plastic or rubber feels elastic and there is no crack on the wire insulator is the best product.

Question: Is 380 the same as 220?

It's the copper wire that's the loss.

Question: Is this loss paid by another company correctly?

You can ask him to pay.

The question is 3% of the power of the total table?

The right 100 degrees is 3 degrees more.

Question: Is his watch in his room coming up from my watch, is it 3% of my watch or 3% of his power?

The basic electrical loss is very small.

Did you know that wire loss is calculated in this way?

Do you know what the loss rate of pencils is?

1. No loss is added when calculating the engineering quantity. 100m2.(116) in the main material of the 5m lighting wire quota, so whether the unit price is considered according to (116) is considered. That is, the unit price is multiplied by the coefficient.

2. When calculating the quantity set price, only the reserved amount of the incoming cabinet is considered, and the loss is not considered, and the loss is only added when the material consumption plan is raised.

3. If the calculated engineering quantity is used for valuation, then the engineering quantity = the reserved quantity of the length of the ten panel cabinet measured on the drawing, and the engineering quantity of the main material will automatically change when the loss is set of quotas; If it is used as a material consumption plan, then, the engineering quantity: (the length measured on the drawing is the reserved length of the ten panel cabinet) (1 + loss rate), the loss can not be added when the pipe threading engineering quantity is calculated, and only the reserved amount into the box or connected equipment is added; However, cable laying needs to calculate the laying slack of the cable.

It is considered according to the total cross-section, and the filling rate of the power bridge does not exceed 40%, and the controlled one does not exceed 50%.

40° ambient temperature.

The cable single-row non-gap laying coefficient is taken.

The cable double-row clearance-free laying coefficient is taken.

This is just a simple calculation formula for the 380V system, just one of the methods for selecting the wire, a reference quantity, the estimation is more complicated, and it will be calculated.

The voltage loss percentage u=pl cs is generally taken as 4-5%. You can't take 7%. If 7% is lost on the wires, plus other losses, the voltage is not enough.

Four conditions must be taken into account in the selection of the conductor cross-section: economic current density, safe ampacity, voltage loss, and mechanical strength. The calculation is based on one, and the other three conditions are qualified at the same time.

1) According to the economic current density, the initial investment cost plus operation and maintenance cost is the lowest (the most reasonable range).

2) Select the wire section according to the heating condition (safe current carrying capacity) (1. Balance between heat generation and heat dissipation, and the temperature does not rise or fall; 2. The temperature is within the tolerance range of the insulating material; 3. The temperature will not reduce the service life).

3) Select the wire cross-section according to the voltage loss condition, and the voltage of each load point is within the qualified range.

4) Considering the mechanical strength, the wire can withstand the tensile force.

Economic current density.

The calculation is more complicated, and the national standard is generally directly adopted:

3000 and below 3000-500 5000 and above.

Bare copper wire mm a mm a mm

Bare aluminum wire a mm a mm a mm

Copper cable a mm a mm a mm

Aluminum cable a mm a mm a mm

The economic current density and voltage loss are related to the length of the wire, that is, the same cross-sectional area, the same current, the longer the line, the greater the power loss and voltage, such as 50kw 50 meters distance, 50mm copper wire, the voltage loss is 2V, and the power loss is; If it is 100 meters, the voltage loss is 4V, and the power loss is; If it is 400 meters, the voltage loss is 8V, and the power loss is;

Safe ampacities of the wire.

In order to ensure the continuous operation of the wire for a long time, the allowable current density is called the safe ampacity.

The general regulations are: copper wire selection 5 8a mm; Aluminum wire selection 3 5a mm.

The current carrying capacity is related to the cross-sectional area of the wire, the larger the cross-sectional area, the smaller the unit current-carrying capacity, such as the current carrying capacity of 1mm copper wire is 18A, the current carrying capacity of 10mm copper wire is 80A, the current carrying capacity of 100mm copper wire is 320A, etc., the voltage loss percentage or the simple calculation formula of the cross-sectional area of the wire.

u=pl/cs s=pl/c△u

u – the percentage of voltage loss.

p – the active power delivered (kw).

l - distance conveyed (m).

c - constant, aluminum wire single-phase, aluminum wire three-phase four-wire 46; Copper wire single-phase, copper wire three-phase four-wire 77; Aluminum alloy single-phase, aluminum alloy, three-phase 39,s - wire cross-section (mm).

Formulas and data** "Electrician Network Access Operation Permit Examination Low Voltage Theory Part", "Fundamentals of Electrical Engineering", etc.

The upstairs is too professional, I'll give a simple one: bare soft wire; insulated wires; power cables; Control cables. (Excerpted from Appendix 1 of the Installation Works Pricing Schedule, Volume II, Installation Works of Electrical Equipment).

Calculation of electrical energy loss.

1) Line loss power p(kw).

p=3（kip）2（re+ret+rei）×10-3

If the accuracy requirements are not high, the temperature additional resistor RET and the load current additional resistor REI can be ignored.

2) The line loses power W

3) Line loss rate.

4) Distribution transformers lose power pb

5) Distribution transformers lose power wb

6) Loss rate b

7) The combined loss ratio is +b.

Low-voltage line loss calculation method.

Low-voltage lines are characterized by intricate and varied lines, which are more complex than high-voltage distribution lines. There are single-phase power supply, 3 3-phase power supply, 3 4-phase power supply line, and more is a combination of these lines. Therefore, it is difficult to accurately calculate the losses of low-voltage networks, and approximate simplifications are generally used.

The data is not all that cannot be calculated, the necessary data:

1》Wire length: L

2》Wire Material: (Copper wire resistivity:,Aluminum wire resistivity:

3》Wire cross-section: S

4》Current through the wire: i

5》Voltage drop due to wire loss: u

Find the resistance of the wire (single wire) according to the wire material, length, cross-section, and passing current

r＝ρ×l/s)

Voltage drop based on wire resistance and passing current (single wire):

u ri to find the wire loss power according to the voltage drop of the wire (single wire).

p＝ui

Use a multimeter to measure the total resistance r of the three wires, calculate the current i = p 380 (p is power, the unit is watts), and the electricity lost (unit is degrees) = i square * r * 10000 degrees 1000p (here the unit of power p is kilowatts). If you don't understand, send me a message.

The power loss of the wire is calculated as follows:

1. Calculate line resistance: 20, aluminum resistivity: ; Copper, the total length of the single-phase 220V circuit is 2*L, the total length of the three-phase is 2*L, and the cross-sectional area of the wire is S, then, the line resistance R = resistivity * total length S.

2. Line current: the current value can be measured or calculated, single-phase 220V line current = equipment rated power 220, three-phase line current = equipment rated power rated voltage power factor.

3. Wire loss: Single-phase 220V circuit wire loss power = line current square * r, three-phase circuit wire loss power = line current square * 3 * power factor square * r.

4. When the three-phase is unbalanced, it is calculated according to the maximum current.

It's useless to look at this directly, first of all, you have to determine what the luminous power is, and also determine what the distance between the light source and the optical power is, test whether the pigtail and optical power ports are good, the loss per kilometer of the optical cable is considered normal, and calculate it yourself after determining the above points, such as 50 kilometers of line, the luminous power of the equipment is +13db, 50 times, equal to 25, 13 minus 25 = -12, the 50 km line, as long as it is not less than minus 12, it is normal.

That's a broad question! First of all, there are many kinds of optical cables, and the attenuation of different optical cables is different due to their different characteristics. Secondly, there are many kinds of light waves transmitted in optical cables (850nm, 1310nm, 1550nm, etc.), and different light waves attenuate differently in the same optical cable.

The attenuation is proportional to the distance, and the splice attenuation is caused by the splicing of the optical fiber. To sum up the above factors, the attenuation of several typical cases is given: 1310nm light wave in the optical fiber:

Light waves in optical fibers: