# How to measure the degree of high voltage electric energy? Active watt hour meter: 349.7, reactive watt hour meter: 311.0, rated dynamic stability current ratio: 185 times, rated short-circuit thermal current ratio: 75 times, and the progress of connecting transformer is 40 / 5 amps Q: can the power consumption of the user be calculated according to the above data? Yes, what is the power consumption?

## How to measure the degree of high voltage electric energy? Active watt hour meter: 349.7, reactive watt hour meter: 311.0, rated dynamic stability current ratio: 185 times, rated short-circuit thermal current ratio: 75 times, and the progress of connecting transformer is 40 / 5 amps Q: can the power consumption of the user be calculated according to the above data? Yes, what is the power consumption?

The voltage transformation ratio of 10kV transformer is 100, your current transformation ratio is 40 / 5 = 8, so the multiple = 100x8 = 800 times. You have not compared the electricity degree of last month, so you can only calculate the total electricity, active electricity = 349.7x800 = 279760 degrees, reactive electricity = 311x800 = 248800 degrees, the power factor is only about 0.75, and the power rate adjustment electricity charge is about 15%

### Why use high voltage for power transmission? The book says that because the power is certain, the voltage increases, the current decreases, and the heat loss of the conductor can be reduced But I don't understand: the resistance of the conductor is certain. The greater the voltage, the greater the conductor current! Isn't the heat loss increasing? When the power is constant, the voltage increases and the current decreases, the heat loss of the conductor can be reduced. I think I understand in this regard What I asked is that the resistance of the conductor is certain. The greater the voltage, the greater the current of the conductor! Isn't the heat loss increasing?

With the increase of voltage, the current loss of conductor in current transportation is small. P = u * I * cos φ, P - actual power of electric appliance, u - actual voltage, I - actual current, cos φ- Power factor (the parallel capacitor in the circuit can increase the power factor accordingly). It can be seen that the voltage is not directly related to whether the consumer saves power or not

### To find the chemical equation for preparing carbon dioxide in junior middle school, we should balance it and write an expression

CaCO3+2HCl=CaCl2+H2O+CO2↑
Calcium carbonate + hydrochloric acid = = = calcium chloride + water + carbon dioxide

### Write out the chemical experiment steps that often need to be tested, what happens before the reaction, what happens after the reaction, instruments and chemical formulas. The more, the better. The more, the better I don't have any examples. What I want is examples. There are few steps and chemistry books

There are two chemical experiments to be tested! 1、 Preparation of oxygen in the laboratory 1. Use H2O2 (hydrogen peroxide) 2h2o2 = MnO2 = 2H2O + O2 ↑ device: test tube, conduit, water tank and gas collector 2. KClO3 heating and decomposition 2kclo3 = △ = 2KCL + O2 device: alcohol lamp, test tube, conduit, water tank, gas collector and iron frame 3. KMnO4 (high

### Who has the equation of junior middle school chemistry experiment Equation obtained in the test

Combined reaction magnesium burns in air: 2mg + O2 ignites 2MgO iron burns in oxygen: 3Fe + 2O2 ignites Fe3O4 aluminum burns in air: 4Al + 3O2 ignites 2al2o3 hydrogen burns in air: 2h2 + O2 ignites 2H2O red phosphorus burns in air: 4P + 5o2 ignites 2p2o5

### Both diamond and graphite are simple substances of carbon. Under certain conditions, graphite can be converted into diamond and needs to absorb energy. It is known that the heat released when 12g graphite or diamond is completely burned is Q1 and Q2 in turn. The following statement is wrong () A. Q1＜Q2 B. Graphite is not as stable as diamond C. Graphite has lower energy than diamond D. Complete combustion produces as much carbon dioxide

A. Graphite can be converted into diamond under certain conditions and needs to absorb energy, indicating that the energy of graphite is lower than that of diamond. The heat released when 12g graphite is completely burned is less than that released when 12g diamond is completely burned, i.e. Q1 < Q2, so a is not selected;
B. The conversion of graphite to diamond needs to absorb energy, which shows that the energy of graphite is low and graphite is more stable than diamond, so B is selected;
C. The conversion of graphite into diamond needs to absorb energy, indicating that the energy of graphite is low, so C is not selected;
D. Diamond and graphite are isomers of each other. When diamond and graphite of equal mass are completely burned, the mass of carbon dioxide generated is the same, so D is not selected
Therefore, B