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40956
The real power in kilowatts (kW) is equal to the apparent power in kilovolt-amps (kVA), times the power factor PF. The equation to use, for converting kW from kVA is, kW = kVA × PF. Using this equation the power factor needs to be known. If it is not known then use .8 and this will give you results close enough. kW = 2 x .8 = 1.6 kW or 1600 watts.
If your power source is 120 V then 8000/120 = 66.7 Amps. If operated at 240 V then it is 33.3 Amps. In the first case you would need 3 AWG and in the second 8 AWG.
Depends on power factor, but it should be about 8 Amps.
10
The amount of BTUs needed to heat a room depends on various factors such as insulation, ceiling height, climate, and desired temperature. As a rough estimate, for a moderately insulated room with an 8-foot ceiling in a cold climate, you would need around 20-25 BTUs per square foot. So for a 600 sqft room, you would need approximately 12,000-15,000 BTUs to heat it effectively.
The typical size of a residential PV system is between 3 kW and 8 kW, but can vary based on energy needs, roof size, and budget. A 5 kW system is common for a medium-sized household.
In a DC circuit Power (P) in watts is equal to Voltage (E) times Current (I). 1000 Watt = 1 kW P = EI thus I= P/E For kW then I = (P x 1000)/E Example with Power of 2 kW and Voltage of 250 V DC: I = (2000/250) = 8 Amp
AnswerKiloWatt hours. Its a measure of energy use. It is the amount of power used over time.Energy is the ability to do work, while power is the rate at which work is done.Your water heater may have a power rating of 5 kW. If it runs for 8 hours in a day, you would use 40 kWh of energy doing the work of heating the water.
Efficiency = Output value / Input valueFor example, if a machine needs 10 KW to run and produces 8 KW, its power efficiency is 8/10 = 0.8 or 80%Efficiency is always between 0 and 1 (or 0 and 100 if expressed as a percentage.)
(4 in high school) + (4 in college) = At least 8.
5kw = 6.25 kva becoz kva = kw/ pf if we take pf is o.8