1 MW is 1000 kW therefore 10 MW is equal to 10,000 kW.
If a load takes 50 kW at a power factor of 0.5 lagging calculate the apparent power and reactive power Answer: Apparent power = Active power / Power Factor In this case, Active power = 50 kW and power factor = 0.5 So Apparent power = 50/0.5 = 100 KVA
Kw can be calculated from volts be using the formula p=(voltage)square/resistance*1000
KVA is the vector sum of real and reactive power; put differently, KVA at a specified power factor will tell you how many KW you have: KW = KVA * pf You must provide a power factor or power factor angle (if angle, replace pf with cos (pf) in above equation) or total reactive power to calculate.
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
Around 80 m tower height + 40m radius
p.f=kW/kV.A
You multiply the power by time, ie kW*h
condenser capacity(kw) = compressor cooling capacity (kw) + compressor input (kw) power condenser capacity(kw) = Pf + Pa
kw of 100kva=100*0.8 pf=80kw( if the power factor is 0.8)
Kw can be calculated from volts be using the formula p=(voltage)square/resistance*1000
If a load takes 50 kW at a power factor of 0.5 lagging calculate the apparent power and reactive power Answer: Apparent power = Active power / Power Factor In this case, Active power = 50 kW and power factor = 0.5 So Apparent power = 50/0.5 = 100 KVA
Power Factor = KVA/KW. This has no unit. Its value is always 1 or less.
The power of the engine of the Renault 10 1967 is 30.9 kW (41.4 hp)
KVA is the vector sum of real and reactive power; put differently, KVA at a specified power factor will tell you how many KW you have: KW = KVA * pf You must provide a power factor or power factor angle (if angle, replace pf with cos (pf) in above equation) or total reactive power to calculate.
For a single phase supply: you need to know the volts, kW and the power factor. If the power factor is unknown, assume 0.75, except for heating elements which have a power factor of 1. Multiply the kW number by 1000 then divide by the voltage. The answer should be divided by the power factor to find the current in amps.
If the town has natural gas for heating, then the power needed per household is between 5 kW and 10 kW. If only electric heat is available, then the combination of electric resistance (auxiliary - from 7 to 10 kW for small residential units) heating, combined with electric clothes dryer (from 2 to 5 kW), electric water heater (from 2.5 to 7.5 kW), and stove (5 kW to 7.5 kW) totaling a min of 15 kW to a max of 20 kW. That would put the max power requirements between 5,000 kW and 10,000 kW. As a matter of reality, a diversity factor of .6 to .7 would reduce the max power to 3,500 kW and 7,000 kW.
if power consumption is 130 KW. how can it convert into unit per hour.