It depends upon at how much voltage level 400 kvar capacitor bank is used.
Cable sizing is based on amperage of the load. The rating of the capacitor bank and the voltage at which it operated need to be stated to give an answer.
we can use the Out Put Capacitor Ex Kvar
{| |- | capacitance of the capacitor is mentioned in KVAR. Formula : KVAR = KW*tan@ FOR tan@, First note the power factor & KW without connecting capacitor. The noted power factor is in cos@.Convert the cos@ value in tan@. for ex. If power factor is 0.6, KW = 200 cos@ = 0.6 cos-1 (0.6) = 53.1 tan (53.1) = 1.333 200*1.333 = 266.6 KVAR if you use 266 KVAR capacitor, Then the power factor improves to unity (1.000). |}
Normal power is the multiplication of current to combination of resistive and reactive or capacitive load. From the vector sum of Apparent power minus real power we can get reactive power(KVAr), which is basically lagging power due to reactive load. This will be the exact rating of capacitor bank. You can find it by cos $ of apparent power.
Use a voltmeter and ammeter to measure the supply voltage and load current drawn by the motor. The product of these two readings will give you the apparent power in volt amperes. Then use a wattmeter to determine the true power of the machine, in watts. The reactive power, in reactive volt amperes, will then be the vector difference between the apparent power and the true power. 4hp = 1 KVAR, so for 25 hp, 25/4= 6025 KVAR
Cable sizing is based on amperage of the load. The rating of the capacitor bank and the voltage at which it operated need to be stated to give an answer.
Generally the capacitor rating of a bank are decided on the load factor.ie higher the KVAR higher the capacity.KVAR is the reactive power in which load angle differs with the load variation.If we know load factor multiply it by the sine angle which gives us the capacity of the cpapcity of the load bank. Generally the capacitor rating of a bank are decided on the load factor.ie higher the KVAR higher the capacity.KVAR is the reactive power in which load angle differs with the load variation.If we know load factor multiply it by the sine angle which gives us the capacity of the cpapcity of the load bank. Generally the capacitor rating of a bank are decided on the load factor.ie higher the KVAR higher the capacity.KVAR is the reactive power in which load angle differs with the load variation.If we know load factor multiply it by the sine angle which gives us the capacity of the cpapcity of the load bank.
It depends on the power factor of the load, but for a load power factor of 0.7 on a 2000 kVA transformer the real power and reactive power are both 1400 kilo (watts and VAR). So a 1400 kVAR capacitor on the load would restore the power factor to 1, allowing 2000 kW to be drawn instead of only 1400 kW.
we can use the Out Put Capacitor Ex Kvar
{| |- | capacitance of the capacitor is mentioned in KVAR. Formula : KVAR = KW*tan@ FOR tan@, First note the power factor & KW without connecting capacitor. The noted power factor is in cos@.Convert the cos@ value in tan@. for ex. If power factor is 0.6, KW = 200 cos@ = 0.6 cos-1 (0.6) = 53.1 tan (53.1) = 1.333 200*1.333 = 266.6 KVAR if you use 266 KVAR capacitor, Then the power factor improves to unity (1.000). |}
You end up with a leading power factor. The Kvar meter will run backwards.
Normal power is the multiplication of current to combination of resistive and reactive or capacitive load. From the vector sum of Apparent power minus real power we can get reactive power(KVAr), which is basically lagging power due to reactive load. This will be the exact rating of capacitor bank. You can find it by cos $ of apparent power.
Use a voltmeter and ammeter to measure the supply voltage and load current drawn by the motor. The product of these two readings will give you the apparent power in volt amperes. Then use a wattmeter to determine the true power of the machine, in watts. The reactive power, in reactive volt amperes, will then be the vector difference between the apparent power and the true power. 4hp = 1 KVAR, so for 25 hp, 25/4= 6025 KVAR
If the power factor is 0.7 the reactive power equals the real power, so the capacitor should be 1.5 kVAr. That is a reasonable estimate for a small induction motor.
for power factor correction kVAR for lasers Joules if you make them with capacitors and know the voltage and frequency you can figure either from microfarads energy =joules=kg-meters=watt -seconds= (C * e^2)/2 var = X(c) * e
KVAR Kilovolt-Ampere Reactive KVAR Kilovolt-Ampere-Reactance {| ! Acronym ! Definition | Formular for calculation of kvar |}
kvar