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No. 60 kVA is 60 kW or less depending on the phase angle between voltage and current, or more specifically the power factor which is the cosine of the phase angle. Once you reduce it to power in kW, you have to multiply by the hours of running time to get kWh, which is a unit of energy, not power.
33A x 480V x 1.73 (square root of 3) = 27434 VA = 27.43 kVA(kW) Now, if this is correct, our facility runs 40 hours a week with all the lights and computers on and 128 hours a week with the lights off and transformer at idle.
The conversion of KVA to KW is governed by the equation KVA = KW/PF) For example, if the power factor is 0.6 120 KVA·0.6 = 72 Kilowatts Converting Watts to KVA (watts to kilovolt-amps) The conversion of W to KVA is governed by the equation KVA=W/(1000*PF) For example 1500W/(1000*0.83) = 1.8 kVA (assuming a power factor of 0.83) F
How to convert 1,000-kwh to KVA assuming the power source is at 380-volt, 3-phase, 50-hertz and power factor at 0.85.
I'm designing the electrical feed to a well. I need to size the transformer & cable feeding the main switchboard based on the KVA. I also need to run my voltage drop calcs based on the KW. I have a single line consisting of multiple pumps, a 100A panel for misc power needs (lighting, recaptacles, etc...) & a unit heater.
You probably mean you have a 500 kW generator. That is a power rating. The energy is that power times the hours run. But in an hour it will produce 500 kWh 1 kWh = 3.6 MJ, so 500 kWh = 1.8 GJ Diesel has an energy content (higher heating value) of 137380 BTU/gal according to to US DoE. That converts to 38.29 MJ/L.
Input kilovolt-amperes (kVA) (approximately): Minimum configuration: 0.40 kVA (two power supplies) Maximum configuration: 5.30 kVA (four power supplies) Btu output: Ship configuration — 1365 Btu/hr (400 watts) Full configuration — 18084 Btu/hr (5300 watts) If my appliance uses 5.30 KVA how much would it be in kwh?
If we assume output is actually 13 kVA and power factor is 1 (worst case), it is 13 kW x 320 g/kW·h = 4.16 kg/h As density of diesel fuel is around 0.85 kg/L, that is 4.9 L/h at max power. If at less than max output power, consumption will be somewhat less.
Depending on power factor, 500 kVA is a maximum of 500 kW, less if the power factor is less than unity. For kWh, you have to load in kilowatts and run time in hours.
You need to know the power factor or phase angle to convert kVA to kW (multiply by the power factor or by the cosine of the phase angle. Second, you need to know the hours run at that power level to get kWh, which is a unit of energy not a unit of power.