kvar can be calculated as follows the a product KVA andt the sine of the angle between the KVA and KW.
In the case of DC, and in AC when current and voltage are in phase, a volt-ampere is the same as a watt (and therefore, a kilo-volt-ampere is the same as a kW). In the case of AC, when current and voltage are NOT in phase, power = voltage x current x power factor; the power factor is the cosine of the angle between current and voltage, and it is always less than or equal to one. In such a case, a kVA would be less than a kW.
1 Megawatt = 1000 Kilowatts so 30 Mw = 30*1000 kw = 30,000 kw.
The correct symbol is 'kW', and it stands for 'kilowatt' which is the SI unit of power.
"Watt" is a rate of using energy."4 kW" means 4,000 watts."4 kW for 6 hours" means 4,000 watts for 6 hours.If you use energy at the rate of 4 kW for 6 hours, then altogether you use24 kilowatt-hours, or 24,000 watt-hours, or 86,400,000 joules.
To find KVAR, we can use the formula: KVAR^2 = KVA^2 - KW^2. Plug in the values given: KVAR^2 = 1750^2 - 1225^2. Calculating this gives KVAR = 1050.
The same way, as you convert Appels to Carrots ........... There is a formula: KVAr = KVA / KW or cos=KW/KVA > Yes, we are treating KW, KVA, & KVAr as the 3 sides in a 90 deg TRIANGLE ! KW= vertical katede KVAr = horizontal katede KVA = hypotenuse
kV is kilovolts, kW is kilowatts, kVA is kilovolt amps and kVAR is kilovolt-amps reactive. A common formula is kVA-squared = kW-squared + kVAR-squared.
kvar can be calculated as follows the a product KVA andt the sine of the angle between the KVA and KW.
kvar can be calculated as follows the a product KVA andt the sine of the angle between the KVA and KW.
{| |- | 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). |}
To calculate kVAR, first find the apparent power (S) using the formula: S = √(P^2 + Q^2), where P is the real power (in kW) and Q is the reactive power (in kVAR). Given P = 560 kW and kVA = 700, kVA = √(P^2 + Q^2) implies 700 = √(560^2 + Q^2). Solve for Q to find the reactive power kVAR.
When the power factor is leading, the capacitive kVAr is more than the Inductive kVAr and this still has to be supplied by the source. As kVA is the vector sum of kW and kVAr, still for the given kW, you have to produce more kVA. Alternately, for the given kVA, you can only convert partially into useful work. Secondary effects are voltage boost in the system, availability of stored energy to feed the fault in case of a fault, increase in the asymmetrical component of fault current, increasing thus the peak value of the fault current, etc.
For normal power factors (pf=80%), you have 0.8 kW for every kva. In general however, kW = pf x kVA. Where pf is the power factor, it is the cosine of the angular difference between the voltage and the current of a circuit in alternating current circuits.
Depending upon the connected load ( R, RL, RC or RLC) with a transformer, the power goes ou from a transformer may be of two types: 1. Active Power; measured in kW 2. Reactive Power; measured in kVAR If the rating will be in kW, then kVAR rating would not be accounted but if the rating is in kVA then it is possible for us to calculate the total active and reactive current as well as the powers, at a particular system voltage!
To convert KVA to KW, you need to know the power factor of the system. Assuming a power factor of 0.8 (common for many systems), the formula for converting KVA to KW is KW = KVA * power factor. With a KVA of 150 and a power factor of 0.8, the KW would be 150 * 0.8 = 120 KW.
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.