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Why can shunt capacitor improve the power factor of inductive load? Is series capacitor OK? Why
The method of shunt capacitor can improve the power factor of inductive load, because part of the reactive current of inductive load is provided by capacitor, and the reactive current in the line, that is, the reactive current provided by power supply, is reduced, so the power factor is improved
In order to change the power factor of the circuit, capacitors are usually connected in parallel with inductive load. At this time, a current branch is added to ask whether the total current of the circuit increases or decreases,
In order to change the power factor of the circuit, the capacitor is often paralleled on the inductive load. At this time, a current branch is added. Ask whether the total current of the circuit increases or decreases, and whether the current and power of the inductive load change? Why?
To change the power factor of a circuit, capacitors are usually connected in parallel with inductive loads. At this time, a current branch is added, and the total current of the circuit is reduced, because the reactive current component that used to be provided by an external power supply is now provided by the capacitor instead of an external power supply, But the current was originally supplied by the power supply, but now it is supplied by the capacitor
In order to improve the power factor of the circuit, the capacitor is usually paralleled on the inductive load. At this time, a circuit branch is added, whether the total current increases or decreases, and whether the inductive component current and power change
Hello: 1. The capacitor in parallel in the circuit is pure capacitive current and does not consume active power. So the active power of the circuit will not change. 2. The current in the inductive load circuit includes reactive current. When the shunt capacitor is used for compensation, the inductive current of the circuit is provided by the capacitor, so it is in parallel
Parallel capacitors can improve power factor. Is the more parallel, the better? Why?
The shunt capacitor can cancel the inductance of the inductive load in the power supply circuit, so as to improve the power factor of the circuit. If there are too many capacitors, the circuit will be capacitive, and its power factor will drop again. This time, it is the increase of reactive power caused by too much capacitive load
Therefore, the shunt capacitance should be slightly more than the inductance in the loop to ensure that no matter how the inductance load changes, the capacitance and inductance will not be exactly equal, so as to avoid loop resonance
Calculation of motor power factor
There is inductance in the motor, so when the motor is connected to the AC circuit, the voltage at both ends of the motor is different from the current flowing through it. The phase of the current will lag one angle behind the phase of the voltage. At the same time, the product of the voltage and current in the circuit is not equal to the power
The output power of a generator is 1000kW, and the resistance of the transmission line used is 10 Ω. When the voltage of the generator connected to the transmission line is 50kV, the thermal power lost on the wire is calculated
According to P = UI, when the voltage of the generator connected to the transmission line is 5kV, the current on the conductor is I = Pu = 1000 × 10350 × 103a = 20a, and the thermal power loss on the conductor is p '= I2R = 202 × 10W = 4 × 103w. A: the thermal power loss on the conductor is 4 × 103w
A generator, the output power is 1000kW, the resistance of the transmission line is 10, when the transmission voltage of the generator is 50kV, calculate the current on the wire
The output power of a generator is 1000kW, and the resistance of the transmission line is 10. When the transmission voltage of the generator is 50kV, calculate the current on the line and the voltage lost on the transmission line?
When working at full load (output 5kV), the output current is I = 1000 / 50 = 20a, and the lost voltage is 20 * 10 = 200V
A generator with output power of 1000kW is used for high-voltage transmission with transmission voltage of 100kV. The total resistance of transmission line is 10 Ω. The calculation is as follows: (1) current on transmission line; (2) thermal power lost on transmission line
(1) From the solution of P output = UI, the transmission current I = P output u = 106105a = 10A & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;; (2) the thermal power lost by the conductor P thermal = I2R = 102 × 10W = 1000W A: (1) the current on the transmission conductor is 10A; (2) the thermal power lost on the transmission conductor is 1000W
The output power of a generator is 1000kW, and the resistance of the transmission line is 10 ohm. When the generator is connected to the transmission line, the voltage is 5kV and 50kV respectively
1. Calculate the current on the conductor,
2. Thermal power lost on the conductor
At 5kV: rated current on conductor I = P / 1.732ucos Φ = 1000 / 1.732/5/0.8 = 144A
Lost voltage U = IR = 144x10 = 1440v
Power loss P = 1.732ui = 1.732x144x1440 = 359147w ≈ 360kw
Lost thermal power = 360x860 = 309600 kcal
Equivalent to 309600 / (4102 / 0.48) = 36.6 cubic meters of natural gas
50kV: rated current on conductor I = P / 1.732ucos Φ = 1000 / 1.732/50/0.8 = 14.4a
Loss voltage U = IR = 14.4x10 = 144v
Power loss P = 1.732ui = 1.732x14.4x1440 = 35914.7w ≈ 36kw
Lost thermal power = 36x860 = 30960 kcal
Equivalent to 30960 / (4102 / 0.48) = 3.66 cubic meters of natural gas
1 kwh = 1 kwh = 860 kcal = 1000 W × 60 min × 60 s = 3600000 joules
The heat released by burning 1 cubic meter of natural gas is usually calculated according to the low calorific value (8500 kcal / m3). Then 4.77kw is equivalent to 4102 kcal and 0.48 cubic meter of natural gas
The output power of a generator is 1000kW, and the resistance of the transmission line used is 10 Ω. When the voltage of the generator connected to the transmission line is 50kV, the thermal power lost on the wire is calculated
According to P = UI, when the voltage of the generator connected to the transmission line is 5kV, the current on the conductor is I = Pu = 1000 × 10350 × 103a = 20a, and the thermal power loss on the conductor is p '= I2R = 202 × 10W = 4 × 103w. A: the thermal power loss on the conductor is 4 × 103w
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