Why increasing the voltage can reduce line losses

It should be that under the condition that the transmission power p of the transmission line is constant, the higher the transmission voltage U, the lower the transmission current i on the line, so the line loss q=i rt is lower (i is the line current, r is the line resistance, and t is the time), so as to reduce the line loss.

The energy transferred is equal to the current multiplied by the voltage. i.e. w=i*v. When w is constant, i is inversely proportional to v.

As the voltage V increases, the smaller the current i will be. Whereas, the line loss is equal to the square times the resistance of the current. With the same wire, the smaller the current, the smaller the loss.

Reducing circuit consumption is to reduce the power consumed, and time accumulation is work.

Most people struggle with two formulas, two formulas:

1. P=U2 R On the surface, the line resistance does not change, the higher the voltage, the greater the damage, but note that at this time, i does not change. In fact, the formula p loss = u 2 r is based on the condition that i is constant, but here with the change of u, and the change of p leads to the change of i, so the formula p loss = u 2 r is not applicable to the study. But this equation still holds, but it cannot be used for qualitative analysis, only for calculations.

2. p=i 2*rBecause here r is constant, we have to reduce i, p=ui, so we increase u, then we decrease i.

It's a bit of a mess, I hope you understand.

Why does high voltage transmission reduce energy losses by increasing voltage? Is it possible to increase the current? Why?

Hello, dear, I am happy to answer for you, high voltage transmission can reduce energy loss, because the thermal effect of the current on the wire generates heat and consumes electrical energy. The amount of heat is proportional to the resistance and proportional to the square of the current. The power consumed by a single wire is p=i 2*r.

Therefore, it is necessary to find a way to reduce the resistance of the wire, especially to reduce the current on the wire, so as to reduce the power loss on the wire. In the transmission of a certain power, the higher the voltage, the smaller its current. When the resistance of the wire is constant, the transmission voltage can be increased, and the current in transmission can be reduced.

When the current is small, the loss of electrical energy is also reduced, so people use high voltage to transmit electric energy to reduce the loss of electrical energy in the process of electric energy transmission.

How to calculate high-voltage line losses? Hello dear, the concept and theory of line loss calculates the power loss that will occur on the road resistance when the load current passes through the line. Calculate the active power loss of a single line per mu p=i:

r。The operating parameters of the line are high. The rated voltage of the high-voltage line is relatively high, so that the electric field strength around the charged body is high.

The line is long and the geographical environment along the line is complex, and the high-voltage line often passes through the mountains and valleys, so there are many difficulties in transportation, and the maintenance workers are disturbed and hungry. I hope my answer is helpful to you, and finally I wish you good health and a happy mood again!

The reasons for the low losses in high-voltage transmission are lower currents, low resistivity, and technological advances.

1. Lower current.

Using a higher voltage (e.g. 500 kV) for a given electrical power and voltage will significantly reduce the required current and line resistance. By reducing the current, the resistance in wires and cables can be greatly reduced, and the losses can be reduced. This makes the transmission line run more efficiently.

2. Low resistivity.

In order to reduce the resistance of the wire, materials such as high-purity copper and aluminum with low resistivity are usually used, and thicker wires are used at the same time to reduce the resistance on the unit length of the conductor, thereby reducing the transmission loss. New technologies such as composite conductors and polished conductors have also been adopted for high-voltage current lines to reduce resistance.

3. Technological progress.

With the continuous development of technology, the design and construction of high-voltage transmission lines have also been greatly improved. Modern transmission lines use advanced wire materials, insulation materials and structural design, which greatly reduce power loss. For example, the conductor material of today's transmission lines has been greatly improved.

Improvements in high-voltage transmission

1. Improve the technical level and professionalism of high-voltage transmission equipment operation and maintenance personnel.

With the continuous development of high-voltage power transmission, high-voltage transmission equipment is also constantly refurbished and upgraded, how to truly achieve the maintenance and protection of high-voltage transmission equipment, the need for high-voltage transmission equipment operation and maintenance personnel to improve the corresponding technical level and professionalism, put an end to the behavior of unlicensed work, regular training for operation and maintenance personnel, after the training, but also uninterrupted to inform the corresponding ability test, improve the level of operation and maintenance team, and promote the development of high-voltage transmission equipment management.

2. Establish and improve the assessment and evaluation methods for the comprehensive management of high-voltage transmission equipment.

On the basis of formulating documents, further education and refinement, further supervision, inspection and assessment of the comprehensive management of the use and maintenance of high-voltage transmission equipment in various departments, set the basic indicators of daily evaluation and year-end evaluation, and have clear standards for daily basic management, operation management, maintenance management, material management and professional equipment management, so as to supervise the relevant units, improve their comprehensive management capabilities of high-voltage transmission equipment, and promote the development of the operation and management of high-voltage transmission equipment.

The higher the transmission voltage, the smaller the loss.

It is understood in this way, when the transmission power is constant, p=ui, the higher the voltage, the smaller the current on the wire, and u does not refer to the transmission voltage, but the voltage drop on the guide line.

Wire loss power = i r, the wire resistance is unchanged, the less the wire loss power.

The P=UI formula is also true, the resistance of the wire is constant, the higher the voltage, the smaller the current on the wire, and the smaller the voltage on the wire, so the P=UI is also smaller.

1. Ohm's law is applicable only in pure resistance circuits, that is, in the same circuit, the current in the conductor is proportional to the voltage at both ends of the conductor, and inversely proportional to the resistance value of the conductor, i=u The power of the electrical appliance p=ui, that is, the product of the voltage at both ends of the electrical appliance and the current passing through the electrical appliance. 3. If the electrical appliance is resistive, p=U2 R, p=i 2R, can be derived according to Ohm's law. (Note:.)

i is the current in the resistor, u is the voltage at both ends of the resistor) 4, q refers to the heat loss of the resistance, because if the electrical appliance is a resistor, the work it does is to generate heat q=u 2 r, q=i In the transmission line, there is p=i (U is the voltage of the transmission line, that is, the voltage to the n line, and i is the current flowing through the transmission line). 6. The heat loss on the transmission line (that is, the heat generated on the line) q=i 2r, r refers to the resistance of the transmission line. Under the condition that the transmission power p is unchanged, increase the transmission voltage U, and the transmission current i decreases, thereby reducing the heat loss of the transmission line, if the transmission line is regarded as a conductor, Ohm's law i = u r, u should be the voltage at both ends of the line, that is, u at both ends = u at the beginning - at the end of you.

In this formula, the voltage is not the output high voltage, and the loss refers to the loss due to the resistance of the wire, so the voltage is the voltage distributed over the resistance of the wire. The power output from the power plant is constant before and after the transformer, p=ui, when the output is high-voltage, the current decreases q=i 2rt, and the power loss decreases. You have to misunderstand that the u of p=u 2 r is the high voltage of the output.

The voltage and current output of the generator are constant, that is to say, the total power of the circuit is certain, and when the power is certain when the power is certain by the p ui, the larger the voltage, the smaller the current. Then the smaller the current obtained by Q i RT, the smaller the loss q in the circuit, so that it is said to use high voltage transmission to reduce the loss.

What you say should have a premise, when. When it remains unchanged, there will be a later conclusion, and what you said is too one-sided.

Solution: I think you should understand it this way. There is resistance in the high-voltage transmission line, so there is a voltage drop (which would be equivalent to the loss of a portion of the voltage on the wire).

In essence, we are equivalent to the power consumed by the resistance of the wire, which can be thought of as a series circuit. The u in your formula above p=u2 r is not the voltage of the wire, but the total voltage delivered, and in fact the total voltage of the output u out=u-u line. The voltage of the wire U line = IR line while the current i of the wire is calculated based on the delivered power and the delivered voltage.

So the power consumption of the wire Q = I 2R wire.

You just need to know that in reality, increasing the voltage can actually reduce line losses. If you must know why, don't apply Ohm's law from your middle school. If you want to know the principle, go to the professional information, and if you want to know the calculation formula, it is estimated that the tutor at your university is also learning.

When analyzing the relationship between any two quantities, the third quantity should be a definite number, otherwise the result is meaningless. In p ui, the higher the u, the smaller the i, provided that p should be a fixed number.

For example, the relationship between p(50) u(25) x i(2) and p(200) u(40) x i(5) is still true, and two conclusions can be drawn from the comparison of the two formulas: the voltage increases, the current increases, and the voltage decreases.

There is no conclusion that the higher the voltage, the lower the current. This is certainly wrong, because p in these two equations is not the same, and the change in p will affect the change in the other two values, and the conclusion is certainly meaningless. It is concluded that "according to Ohm's law i=u r, the higher the obtained voltage."

According to the formula p=u 2 r, the higher the voltage, the greater the power, is not this a contradiction", it is precisely that p is not a constant. After p is a fixed number, the value of u is determined, and the value of i is determined accordingly, and the i and u in i=u r should also be i and u in this case, which is also meaningful.

That's right?

To put it simply, the higher the voltage at the same power, the smaller the current can flow. Then the line loss is smaller.

In detail, because the power p=u*i, u=r*i, p=i*i*r. Here the edge resistance r is certain. So the smaller the i, the smaller the p, the squared relationship. This refers to the power, voltage, current, and resistance on the transmission line.

Another advantage is that the wires can be thinner, and the cost is greatly reduced.

But if it's a superconductor at room temperature.

If it really happens, these don't apply anymore, r=0 p=0. Think about it. Why are those scientists so stupid... How many years have I not figured it out.

It should be that the higher the voltage, the lower the loss.

The power transmitted by the transmission line is certain, then according to the formula p=ui, the higher the voltage, the smaller the current, q=i 2rt, so at the same power (load), on the same line, the higher the voltage, the smaller the power consumption.

High-voltage transmission is a small current and requires a small carrier, so it can be transported over long distances and has low cost. Alternating current can be transferred through alternating magnetic induction, and current can be realized in transformer design. Power is voltage x current. 100vx1a = also = 100w.

Use high school p=ui

When the network power is constant (p is constant), the network heat loss power δp=r*i 2 is δp=r*(p u) 2

It can be seen that when the output power p is constant, the u increases and the heat loss power δp decreases.

The law of conservation of energy, electrical energy = uit, time cannot be controlled, so the current can be reduced if the voltage is high. According to Joule's law, the less current there is, the less heat loss the power line. Voltage rise and fall are achieved through transformers.

The so-called increased voltage here refers to the voltage of the transformer coil, not the voltage at both ends of the resistor in the simple point path that we learned.

First of all, the power delivered is certain; Increasing the voltage according to w=u*i will definitely decrease the current ; Damage = i*i*r The current is reduced, and the loss is definitely reduced; So increase the voltage and the current will be lowered! Remember that when the power is constant; Current and voltage are inversely proportional.

In the process of long-distance transmission, in order to reduce the power loss on the transmission line, according to the formula w=i 2rt, the practical way is to increase the transmission voltage to reduce the current on the transmission line.

The disadvantages of increasing the diameter of the transmission line to reduce the resistance of the transmission line are: (1) the cost of the transmission line increases(2) Due to the thickening of the conductor and the increase in weight, the requirements for transmission facilities are increased, such as the high strength requirements of the transmission tower, which also increases the cost. Wait a minute.

Therefore, the electricity used in our daily life, the electricity used in factories, and the electricity transmitted from power plants are all high-voltage transmission electricity.

I hope it helps you, and if you have any questions, you can ask them

I wish you progress in your studies and go to the next level!(*

To increase the conductive area, the method is to thicken the wire electrically, and open the tin wire, that is, to open the solder mask window on the existing line, if it is the power bureau, it will increase the power supply voltage, because the power p=ui, the voltage increases, the current remains unchanged, and the power will increase.

Two methods.

First, reducing the resistance of the transmission line usually thickens the transmission line, but the cost is greatly increased and it is not recommended.

The second is to increase the transmission voltage, p=ui, because the output power is constant, when the voltage increases, the current decreases and the power consumption of the transmission line is reduced.