Your question shows the importance of using the correct symbols, because your symbols are incorrect and, therefore, your question is confusing.
So, are you asking how many milliwatts (mW) there are in a kilowatt (kW), or are you asking how many megawatts (MW) there are in a kilowatt(kW)?
Notice that the symbol for a watt is an upper-case W. The symbol for a milli is a lower-case m, and the symbol for a mega is an upper-case M.
If the former, then there are one-thousand milliwatts in a watt, and there are one-thousand watts in a kilowatt, so there must be one-million milliwatts in a kilowatt.
If the latter, then there are one-thousand kilowatts in a megawatt, so a megawatt must be one-thousandth of a megawatt in a kilowatt.
With average usage it would supply about 25 houses. It could be assumed that some houses would draw more power than others at any moment, so the average would be about 3 to 4 kW.
I assume this is a trick question, but my guess is 100.
The power output of a 100 kw power plant is 100 kw.
To answer this question the voltage has to be known.
The total kW load is only a guide to the solar panel capacity. Any energy generated by the solar panel will reduce the energy you draw from the grid, which leads directly to less CO2 output at the power plants. But the power plant is still needed for when there is a cloudy day. When it's cloudy the solar panel output is drastically reduced, although they still produce power, as the salesmen are keen to emphasize, but what they don't tell you is that the power output is reduced by about 95%. Also, in many countries, if your panels generate excess electricity you sell it back to the grid, so the capacity of the panels is not critical.
If they are at the same voltage, and the same kW, the only thing left that will influence the output amperage is the power factor the generator is running at.
Yes, you can use both, but watts is more useful; it provides the total amount of power the generator can output.
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
The mechanical load of a motor determines the necessary output power rating of an electric motor. As mechanical loads are defined in terms of watts (or, in North America, horse power), then motor's output must be rated in watts, too.The so-called 'power' rating of a transformer is determined by the rated voltage and the rated current of its secondary winding. The product of these two quantities is the transformer's rated 'apparent power', expressed in volt amperes.Incidentally, the symbol for "kilowatts" is "kW", not"KW's"!
100 W .1 Kw
You can measure the output power of the motor (mechanical power or shaft power) by a speed and torque sensor. The product of speed (angular velocity to be precise) and torque is power. Is this case the output power. The SI unit of power is W, kW is 1000 W. So the power/1000=power [kW].
Motors are rated according to their output power, never their input power. This is because it is the output power that determines the load the motor will be able to handle. The output power, of course, is lower than the input power due to the machine's losses.In North America, the output power is generally expressed in horsepower, whereas its input power is measured in watts. In the rest of the world, the horsepower is considered obsolete, so both the output power and the input power are measured in watts.As there are 746 W to the horsepower, your 75 kW output motor would be equivalent to approximately 100 horsepower.
3 db is double the power, so an input of 1 KW would yield an output of 2 KW for a 3 db gain.
100 kW is not an energy, it is a power. It is the same as 100 kJ/sec (100,000 joules/second. The 100 kW - or 100 kJ/second - can be converted to 100 kW of mechanical power - or less, depending on the type of energy, and how much is wasted.
power in equals power output plus power loss so input power for this question is 100 kw now use efficiency formula 100/80 times 100 to get 80% efficient
kw of 100kva=100*0.8 pf=80kw( if the power factor is 0.8)
The total kW load is only a guide to the solar panel capacity. Any energy generated by the solar panel will reduce the energy you draw from the grid, which leads directly to less CO2 output at the power plants. But the power plant is still needed for when there is a cloudy day. When it's cloudy the solar panel output is drastically reduced, although they still produce power, as the salesmen are keen to emphasize, but what they don't tell you is that the power output is reduced by about 95%. Also, in many countries, if your panels generate excess electricity you sell it back to the grid, so the capacity of the panels is not critical.
1 hp = 0.7457 kW Therefore by dividing 1 by 0.7457 we get 1 kW = 1.341 hp Just multiply either equation by the desired amount to find the equivalent power output. Either Imperial or metric
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.)
Theoretically 22 kW is equivalent to 29.5 HP but there are some power losses in the motor and the probable maximum output of mechanical power is 25 HP.
It's an a.c. motor with an output power of 5 horsepower or roughly 3.75 kW.